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- 1. 5-1 Chapter 5 MASS AND ENERGY ANALYSIS OF CONTROL VOLUMESConservation of Mass5-1C Mass, energy, momentum, and electric charge are conserved, and volume and entropy are notconserved during a process.5-2C Mass flow rate is the amount of mass flowing through a cross-section per unit time whereas thevolume flow rate is the amount of volume flowing through a cross-section per unit time.5-3C The amount of mass or energy entering a control volume does not have to be equal to the amount ofmass or energy leaving during an unsteady-flow process.5-4C Flow through a control volume is steady when it involves no changes with time at any specifiedposition.5-5C No, a flow with the same volume flow rate at the inlet and the exit is not necessarily steady (unlessthe density is constant). To be steady, the mass flow rate through the device must remain constant.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 2. 5-25-6E A garden hose is used to fill a water bucket. The volume and mass flow rates of water, the fillingtime, and the discharge velocity are to be determined.Assumptions 1 Water is an incompressible substance. 2 Flow through the hose is steady. 3 There is nowaste of water by splashing.Properties We take the density of water to be 62.4 lbm/ft3 (Table A-3E).Analysis (a) The volume and mass flow rates of water are V& = AV = (πD 2 / 4)V = [π (1 / 12 ft) 2 / 4](8 ft/s) = 0.04363 ft 3 /s m = ρV& = (62.4 lbm/ft 3 )(0.04363 ft 3 /s) = 2.72 lbm/s &(b) The time it takes to fill a 20-gallon bucket is V 20 gal ⎛ 1 ft 3 ⎞ Δt = = ⎜ ⎟ = 61.3 s V& 0.04363 ft 3 /s ⎜ 7.4804 gal ⎟ ⎝ ⎠(c) The average discharge velocity of water at the nozzle exit is V& V& 0.04363 ft 3 /s Ve = = 2 = = 32 ft/s Ae πDe / 4 [π (0.5 / 12 ft) 2 / 4]Discussion Note that for a given flow rate, the average velocity is inversely proportional to the square ofthe velocity. Therefore, when the diameter is reduced by half, the velocity quadruples.5-7 Air is accelerated in a nozzle. The mass flow rate and the exit area of the nozzle are to be determined.Assumptions Flow through the nozzle issteady.Properties The density of air is given to be V1 = 40 m/s AIR2.21 kg/m3 at the inlet, and 0.762 kg/m3 at A1 = 90 cm2 V2 = 180 m/sthe exit.Analysis (a) The mass flow rate of air isdetermined from the inlet conditions to be m = ρ1 A1V1 = (2.21 kg/m3 )(0.009 m 2 )(40 m/s) = 0.796 kg/s &(b) There is only one inlet and one exit, and thus m1 = m2 = m . & & &Then the exit area of the nozzle is determined to be m& 0.796 kg/s m = ρ 2 A2V2 ⎯ & ⎯→ A2 = = = 0.0058 m 2 = 58 cm 2 ρ 2V2 (0.762 kg/ m 3 )(180 m/s)PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 3. 5-35-8E Steam flows in a pipe. The minimum diameter of the pipe for a given steam velocity is to bedetermined.Assumptions Flow through the pipe is steady.Properties The specific volume of steam at thegiven state is (Table A-6E) Steam, 200 psia P = 200 psia ⎫ 3 600°F, 59 ft/s ⎬ v 1 = 3.0586 ft /lbm T = 600°F ⎭Analysis The cross sectional area of the pipe is 1 mv (200 lbm/s)(3.0586 ft 3/lbm) & m= & AcV ⎯ ⎯→ A = = = 10.37 ft 2 v V 59 ft/sSolving for the pipe diameter gives πD 2 4A 4(10.37 ft 2 ) A= ⎯ ⎯→ D = = = 3.63 ft 4 π πTherefore, the diameter of the pipe must be at least 3.63 ft to ensure that the velocity does not exceed 59ft/s.5-9 A water pump increases water pressure. The diameters of the inlet and exit openings are given. Thevelocity of the water at the inlet and outlet are to be determined.Assumptions 1 Flow through the pump is steady. 2 The specific volume remains constant.Properties The inlet state of water is compressed liquid. We approximate it as a saturated liquid at thegiven temperature. Then, at 15°C and 40°C, we have (Table A-4) T = 15°C ⎫ 3 ⎬ v 1 = 0.001001 m /kg x=0 ⎭ 700 kPa T = 40°C ⎫ 3 ⎬ v 1 = 0.001008 m /kg Water x=0 ⎭ 70 kPaAnalysis The velocity of the water at the inlet is 15°C mv1 4mv1 4(0.5 kg/s)(0.001001 m3/kg) & & V1 = = = = 6.37 m/s A1 πD12 π (0.01 m)2Since the mass flow rate and the specific volume remains constant, the velocity at the pump exit is 2 2 A1 ⎛D ⎞ ⎛ 0.01 m ⎞ V2 = V1 = V1⎜ 1 ⎟ = (6.37 m/s)⎜ ⎜D ⎟ ⎟ = 2.83 m/s A2 ⎝ 2⎠ ⎝ 0.015 m ⎠Using the specific volume at 40°C, the water velocity at the inlet becomes mv 1 4mv 1 4(0.5 kg/s)(0.001008 m 3 /kg) & & V1 = = = = 6.42 m/s A1 πD12 π (0.01 m) 2which is a 0.8% increase in velocity.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 4. 5-45-10 Air is expanded and is accelerated as it is heated by a hair dryer of constant diameter. The percentincrease in the velocity of air as it flows through the drier is to be determined.Assumptions Flow through the nozzle is steady.Properties The density of air is given to be1.20 kg/m3 at the inlet, and 1.05 kg/m3 atthe exit. V2 V1Analysis There is only one inlet and oneexit, and thus m1 = m2 = m . Then, & & & m1 = m 2 & & ρ 1 AV1 = ρ 2 AV 2 V2 ρ 1.20 kg/m 3 = 1 = = 1.14 (or, and increase of 14%) V1 ρ 2 1.05 kg/m 3Therefore, the air velocity increases 14% as it flows through the hair drier.5-11 A rigid tank initially contains air at atmospheric conditions. The tank is connected to a supply line,and air is allowed to enter the tank until the density rises to a specified level. The mass of air that enteredthe tank is to be determined.Properties The density of air is given to be 1.18 kg/m3 at thebeginning, and 7.20 kg/m3 at the end.Analysis We take the tank as the system, which is a controlvolume since mass crosses the boundary. The mass balancefor this system can be expressed asMass balance: V1 = 1 m3 m in − m out = Δm system → m i = m 2 − m1 = ρ 2V − ρ 1V ρ1 =1.18 kg/m3Substituting, mi = ( ρ 2 − ρ 1 )V = [(7.20 - 1.18) kg/m 3 ](1 m 3 ) = 6.02 kgTherefore, 6.02 kg of mass entered the tank.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 5. 5-55-12 A smoking lounge that can accommodate 15 smokers is considered. The required minimum flow rateof air that needs to be supplied to the lounge and the diameter of the duct are to be determined.Assumptions Infiltration of air into the smoking lounge is negligible.Properties The minimum fresh air requirements for a smoking lounge is given to be 30 L/s per person.Analysis The required minimum flow rate of air that needs to be supplied to the lounge is determineddirectly from V&air = V&air per person ( No. of persons) = (30 L/s ⋅ person)(15 persons) = 450 L/s = 0.45 m 3 /s SmokingThe volume flow rate of fresh air can be expressed as Lounge V& = VA = V (πD 2 / 4) 15 smokersSolving for the diameter D and substituting, 4V& 4(0.45 m 3 /s ) D= = = 0.268 m πV π (8 m/s)Therefore, the diameter of the fresh air duct should be at least 26.8 cm if the velocity of air is not to exceed8 m/s.5-13 The minimum fresh air requirements of a residential building is specified to be 0.35 air changes perhour. The size of the fan that needs to be installed and the diameter of the duct are to be determined.Analysis The volume of the building and the required minimum volume flow rate of fresh air are V room = (2.7 m)(200 m 2 ) = 540 m3 V& = V room × ACH = (540 m3 )(0.35/h ) = 189 m3 / h = 189,000 L/h = 3150 L/minThe volume flow rate of fresh air can be expressed as V& = VA = V (πD 2 / 4)Solving for the diameter D and substituting, House 4V& 4(189 / 3600 m 3 /s ) 0.35 ACH 200 m2 D= = = 0.106 m πV π (6 m/s)Therefore, the diameter of the fresh air duct should be at least 10.6 cmif the velocity of air is not to exceed 6 m/s.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 6. 5-65-14 A cyclone separator is used to remove fine solid particles that are suspended in a gas stream. Themass flow rates at the two outlets and the amount of fly ash collected per year are to be determined.Assumptions Flow through the separator is steady.Analysis Since the ash particles cannot be converted into the gas and vice-versa, the mass flow rate of ashinto the control volume must equal that going out, and the mass flow rate of flue gas into the controlvolume must equal that going out. Hence, the mass flow rate of ash leaving is mash = yash min = (0.001)(10 kg/s) = 0.01 kg/s & &The mass flow rate of flue gas leaving the separator is then mflue gas = min − mash = 10 − 0.01 = 9.99 kg/s & & &The amount of fly ash collected per year is mash = mash Δt = (0.01 kg/s)(365 × 24 × 3600 s/year) = 315,400 kg/year &5-15 Air flows through an aircraft engine. The volume flow rate at the inlet and the mass flow rate at theexit are to be determined.Assumptions 1 Air is an ideal gas. 2 The flow is steady.Properties The gas constant of air is R = 0.287 kPa⋅m3/kg⋅K (Table A-1).Analysis The inlet volume flow rate is V&1 = A1V1 = (1 m 2 )(180 m/s) = 180 m 3 /sThe specific volume at the inlet is RT1 (0.287 kPa ⋅ m 3 /kg ⋅ K)(20 + 273 K) v1 = = = 0.8409 m 3 /kg P1 100 kPaSince the flow is steady, the mass flow rate remains constant during the flow. Then, V&1 180 m 3 /s m= & = = 214.1 kg/s v 1 0.8409 m 3 /kgPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 7. 5-75-16 A spherical hot-air balloon is considered. The time it takes to inflate the balloon is to be determined.Assumptions 1 Air is an ideal gas.Properties The gas constant of air is R = 0.287 kPa⋅m3/kg⋅K (Table A-1).Analysis The specific volume of air entering the balloon is RT (0.287 kPa ⋅ m 3 /kg ⋅ K)(35 + 273 K) v= = = 0.7366 m 3 /kg P 120 kPaThe mass flow rate at this entrance is AcV πD 2 V π (1 m)2 2 m/s m= & = = = 2.132 kg/s v 4 v 4 0.7366 m3/kgThe initial mass of the air in the balloon is Vi πD3 π (3 m)3 mi = = = = 19.19 kg v 6v 6(0.7366 m3/kg)Similarly, the final mass of air in the balloon is V f πD3 π (15 m)3 mf = = = = 2399 kg v 6v 6(0.7366 m3/kg)The time it takes to inflate the balloon is determined from m f − mi (2399 − 19.19) kg Δt = = = 1116 s = 18.6 min m& 2.132 kg/sPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 8. 5-85-17 Water flows through the tubes of a boiler. The velocity and volume flow rate of the water at the inletare to be determined.Assumptions Flow through the boiler is steady.Properties The specific volumes of water at the inlet and exit are (Tables A-6 and A-7) P1 = 7 MPa ⎫ 3 ⎬ v 1 = 0.001017 m /kg T1 = 65°C ⎭ 7 MPa Steam 6 MPa, 450°C P2 = 6 MPa ⎫ 3 65°C 80 m/s ⎬ v 2 = 0.05217 m /kg T2 = 450°C ⎭Analysis The cross-sectional area of the tube is πD 2 π (0.13 m) 2 Ac = = = 0.01327 m 2 4 4The mass flow rate through the tube is same at the inlet and exit. It may be determined from exit data to be AcV 2 (0.01327 m 2 )(80 m/s) m= & = = 20.35 kg/s v2 0.05217 m 3 /kgThe water velocity at the inlet is then mv1 (20.35 kg/s)(0.001017 m3/kg) & V1 = = = 1.560 m/s Ac 0.01327 m 2The volumetric flow rate at the inlet is V&1 = AcV1 = (0.01327 m 2 )(1.560 m/s) = 0.0207 m 3 /sPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 9. 5-95-18 Refrigerant-134a flows through a pipe. Heat is supplied to R-134a. The volume flow rates of air atthe inlet and exit, the mass flow rate, and the velocity at the exit are to be determined. Q R-134a 200 kPa 180 kPa 20°C 40°C 5 m/sProperties The specific volumes of R-134a at the inlet and exit are (Table A-13) P1 = 200 kPa ⎫ 3 P1 = 180 kPa ⎫ 3 ⎬v 1 = 0.1142 m /kg ⎬v 2 = 0.1374 m /kg T1 = 20°C ⎭ T1 = 40°C ⎭Analysis (a) (b) The volume flow rate at the inlet and the mass flow rate are πD 2 π (0.28 m) 2 V&1 = AcV1 = V1 = (5 m/s) = 0.3079 m3 /s 4 4 1 1 πD 2 1 π (0.28 m)2 m= & AcV1 = V1 = 3 (5 m/s) = 2.696 kg/s v1 v1 4 0.1142 m /kg 4(c) Noting that mass flow rate is constant, the volume flow rate and the velocity at the exit of the pipe aredetermined from V&2 = mv 2 = (2.696 kg/s)(0.1374 m3/kg) = 0.3705 m3 /s & V&2 0.3705 m3 / s V2 = = = 6.02 m/s Ac π (0.28 m) 2 4PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 10. 5-10Flow Work and Energy Transfer by Mass5-19C Energy can be transferred to or from a control volume as heat, various forms of work, and by mass.5-20C Flow energy or flow work is the energy needed to push a fluid into or out of a control volume.Fluids at rest do not possess any flow energy.5-21C Flowing fluids possess flow energy in addition to the forms of energy a fluid at rest possesses. Thetotal energy of a fluid at rest consists of internal, kinetic, and potential energies. The total energy of aflowing fluid consists of internal, kinetic, potential, and flow energies.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 11. 5-115-22E Steam is leaving a pressure cooker at a specified pressure. The velocity, flow rate, the total and flowenergies, and the rate of energy transfer by mass are to be determined.Assumptions 1 The flow is steady, and the initial start-up period is disregarded. 2 The kinetic and potentialenergies are negligible, and thus they are not considered. 3 Saturation conditions exist within the cooker atall times so that steam leaves the cooker as a saturated vapor at 30 psia.Properties The properties of saturated liquid water and water vapor at 30 psia are vf = 0.01700 ft3/lbm, vg =13.749 ft3/lbm, ug = 1087.8 Btu/lbm, and hg = 1164.1 Btu/lbm (Table A-5E).Analysis (a) Saturation conditions exist in a pressure cooker at all times after the steady operatingconditions are established. Therefore, the liquid has the properties of saturated liquid and the exiting steamhas the properties of saturated vapor at the operating pressure. The amount of liquid that has evaporated,the mass flow rate of the exiting steam, and the exit velocity are ΔVliquid 0.4 gal ⎛ 0.13368 ft 3 ⎞ m= = ⎜ ⎟ = 3.145 lbm vf 0.01700 ft 3/lbm ⎜ ⎝ 1 gal ⎟ ⎠ m 3.145 lbm H2O m= & = = 0.0699 lbm/min = 1.165 × 10- 3 lbm/s Sat. vapor Q Δt 45 min P = 30 psia m& mv g (1.165 × 10-3 lbm/s)(13.749 ft 3 /lbm) ⎛ 144 in 2 ⎞ & V = = = ⎜ ⎟ = 15.4 ft/s ρ g Ac Ac 0.15 in 2 ⎜ 1 ft 2 ⎟ ⎝ ⎠(b) Noting that h = u + Pv and that the kinetic and potential energies are disregarded, the flow and totalenergies of the exiting steam are e flow = Pv = h − u = 1164.1 − 1087.8 = 76.3 Btu/lbm θ = h + ke + pe ≅ h = 1164.1 Btu/lbmNote that the kinetic energy in this case is ke = V2/2 = (15.4 ft/s)2 = 237 ft2/s2 = 0.0095 Btu/lbm, which isvery small compared to enthalpy.(c) The rate at which energy is leaving the cooker by mass is simply the product of the mass flow rate andthe total energy of the exiting steam per unit mass, E mass = mθ = (1.165 × 10 −3 lbm/s)(1164.1 Btu/lbm) = 1.356 Btu/s & &Discussion The numerical value of the energy leaving the cooker with steam alone does not mean muchsince this value depends on the reference point selected for enthalpy (it could even be negative). Thesignificant quantity is the difference between the enthalpies of the exiting vapor and the liquid inside(which is hfg) since it relates directly to the amount of energy supplied to the cooker.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 12. 5-125-23 Air flows steadily in a pipe at a specified state. The diameter of the pipe, the rate of flow energy, andthe rate of energy transport by mass are to be determined. Also, the error involved in the determination ofenergy transport by mass is to be determined.Properties The properties of air are R= 0.287 kJ/kg.K and cp = 1.008 300 kPa Air 25 m/skJ/kg.K (at 350 K from Table A-2b) 77°C 18 kg/minAnalysis (a) The diameter is determinedas follows RT (0.287 kJ/kg.K)(77 + 273 K) v= = = 0.3349 m 3 /kg P (300 kPa) mv (18 / 60 kg/s)(0.3349 m 3 /kg) & A= = = 0.004018 m 2 V 25 m/s 4A 4(0.004018 m 2 ) D= = = 0.0715 m π π(b) The rate of flow energy is determined from & Wflow = mPv = (18 / 60 kg/s)(300 kPa)(0.3349 m3/kg) = 30.14 kW &(c) The rate of energy transport by mass is & ⎛ 1 ⎞ Emass = m(h + ke) = m⎜ c pT + V 2 ⎟ & & ⎝ 2 ⎠ ⎡ 1 ⎛ 1 kJ/kg ⎞⎤ = (18/60 kg/s) ⎢(1.008 kJ/kg.K)(77 + 273 K) + (25 m/s)2 ⎜ ⎟ 2 2 ⎥ ⎣ 2 ⎝ 1000 m /s ⎠⎦ = 105.94 kW(d) If we neglect kinetic energy in the calculation of energy transport by mass & E mass = mh = mc p T = (18/60 kg/s)(1.00 5 kJ/kg.K)(7 7 + 273 K) = 105.84 kW & &Therefore, the error involved if neglect the kinetic energy is only 0.09%.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 13. 5-135-24E A water pump increases water pressure. The flow work required by the pump is to be determined.Assumptions 1 Flow through the pump is steady. 2 The state of water at the pump inlet is saturated liquid.3 The specific volume remains constant.Properties The specific volume of saturated liquid water at 10 psia is v = v f@ 10 psia = 0.01659 ft 3 /lbm (Table A-5E) 50 psiaThen the flow work relation gives wflow = P2v 2 − P1v 1 = v ( P2 − P1 ) Water 10 psia ⎛ 1 Btu ⎞ = (0.01659 ft 3 /lbm)(50 − 10)psia ⎜ ⎟ ⎜ 5.404 psia ⋅ ft 3 ⎟ ⎝ ⎠ = 0.1228 Btu/lbm5-25 An air compressor compresses air. The flow work required by the compressor is to be determined.Assumptions 1 Flow through the compressor is steady. 2 Air 1 MPais an ideal gas. 300°CProperties Combining the flow work expression with the idealgas equation of state gives Compressor wflow = P2v 2 − P1v 1 = R (T2 − T1 ) 120 kPa = (0.287 kJ/kg ⋅ K)(300 − 20)K 20°C = 80.36 kJ/kgPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 14. 5-14Steady Flow Energy Balance: Nozzles and Diffusers5-26C A steady-flow system involves no changes with time anywhere within the system or at the systemboundaries5-27C No.5-28C It is mostly converted to internal energy as shown by a rise in the fluid temperature.5-29C The kinetic energy of a fluid increases at the expense of the internal energy as evidenced by adecrease in the fluid temperature.5-30C Heat transfer to the fluid as it flows through a nozzle is desirable since it will probably increase thekinetic energy of the fluid. Heat transfer from the fluid will decrease the exit velocity.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 15. 5-155-31 Air is accelerated in a nozzle from 30 m/s to 180 m/s. The mass flow rate, the exit temperature, andthe exit area of the nozzle are to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Air is an ideal gas withconstant specific heats. 3 Potential energy changes are negligible. 4 The device is adiabatic and thus heattransfer is negligible. 5 There are no work interactions.Properties The gas constant of air is 0.287kPa.m3/kg.K (Table A-1). The specific heat of air atthe anticipated average temperature of 450 K is cp = P1 = 300 kPa1.02 kJ/kg.°C (Table A-2). T1 = 200°C AIR P2 = 100 kPa V1 = 30 m/s V2 = 180 m/sAnalysis (a) There is only one inlet and one exit, and A1 = 80 cm2thus m1 = m2 = m . Using the ideal gas relation, the & & &specific volume and the mass flow rate of air aredetermined to be RT1 (0.287 kPa ⋅ m 3 /kg ⋅ K )(473 K ) v1 = = = 0.4525 m 3 /kg P1 300 kPa 1 1 m= & A1V1 = (0.008 m 2 )(30 m/s) = 0.5304 kg/s v1 0.4525 m3/kg(b) We take nozzle as the system, which is a control volume since mass crosses the boundary. The energybalance for this steady-flow system can be expressed in the rate form as & & Ein − Eout = ΔEsystem 0 (steady) & =0 1 24 4 3 1442443 Rate of net energy transfer Rate of change in internal, kinetic, by heat, work, and mass potential, etc. energies & & Ein = Eout & & & & m(h1 + V12 / 2) = m(h2 + V 22 /2) (since Q ≅ W ≅ Δpe ≅ 0) V 22 − V12 V 2 − V12 0 = h2 − h1 + ⎯→ 0 = c p , ave (T2 − T1 ) + 2 ⎯ 2 2 (180 m/s) 2 − (30 m/s) 2 ⎛ 1 kJ/kg ⎞Substituting, 0 = (1.02 kJ/kg ⋅ K )(T2 − 200 o C) + ⎜ ⎟ 2 ⎜ 1000 m 2 /s 2 ⎟ ⎝ ⎠It yields T2 = 184.6°C(c) The specific volume of air at the nozzle exit is RT2 (0.287 kPa ⋅ m 3 /kg ⋅ K )(184.6 + 273 K ) v2 = = = 1.313 m 3 /kg P2 100 kPa 1 1 m= & A2V 2 ⎯ ⎯→ 0.5304 kg/s = A2 (180 m/s ) → A2 = 0.00387 m2 = 38.7 cm2 v2 1.313 m 3 /kgPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 16. 5-165-32 EES Problem 5-31 is reconsidered. The effect of the inlet area on the mass flow rate, exit velocity,and the exit area as the inlet area varies from 50 cm2 to 150 cm2 is to be investigated, and the final resultsare to be plotted against the inlet area.Analysis The problem is solved using EES, and the solution is given below.Function HCal(WorkFluid$, Tx, Px)"Function to calculate the enthalpy of an ideal gas or real gas" If Air = WorkFluid$ then HCal:=ENTHALPY(Air,T=Tx) "Ideal gas equ." else HCal:=ENTHALPY(WorkFluid$,T=Tx, P=Px)"Real gas equ." endifend HCal"System: control volume for the nozzle""Property relation: Air is an ideal gas""Process: Steady state, steady flow, adiabatic, no work""Knowns - obtain from the input diagram"WorkFluid$ = AirT[1] = 200 [C]P[1] = 300 [kPa]Vel[1] = 30 [m/s]P[2] = 100 [kPa]Vel[2] = 180 [m/s]A[1]=80 [cm^2]Am[1]=A[1]*convert(cm^2,m^2)"Property Data - since the Enthalpy function has different parametersfor ideal gas and real fluids, a function was used to determine h."h[1]=HCal(WorkFluid$,T[1],P[1])h[2]=HCal(WorkFluid$,T[2],P[2])"The Volume function has the same form for an ideal gas as for a real fluid."v[1]=volume(workFluid$,T=T[1],p=P[1])v[2]=volume(WorkFluid$,T=T[2],p=P[2])"Conservation of mass: "m_dot[1]= m_dot[2]"Mass flow rate"m_dot[1]=Am[1]*Vel[1]/v[1]m_dot[2]= Am[2]*Vel[2]/v[2]"Conservation of Energy - SSSF energy balance"h[1]+Vel[1]^2/(2*1000) = h[2]+Vel[2]^2/(2*1000)"Definition"A_ratio=A[1]/A[2]A[2]=Am[2]*convert(m^2,cm^2)PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 17. 5-17 A1 [cm2] A2 [cm2] m1 T2 50 24.19 0.3314 184.6 60 29.02 0.3976 184.6 70 33.86 0.4639 184.6 80 38.7 0.5302 184.6 90 43.53 0.5964 184.6 100 48.37 0.6627 184.6 110 53.21 0.729 184.6 120 58.04 0.7952 184.6 130 62.88 0.8615 184.6 140 67.72 0.9278 184.6 150 72.56 0.9941 184.6 1 0.9 0.8 0.7 m [1] 0.6 0.5 0.4 0.3 50 70 90 110 130 150 A[1] [cm ^2] 80 70 A[2] [cm ^2] 60 50 40 30 20 50 70 90 110 130 150 A[1] [cm ^2]PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 18. 5-185-33E Air is accelerated in an adiabatic nozzle. The velocity at the exit is to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Air is an ideal gas withconstant specific heats. 3 Potential energy changes are negligible. 4 There are no work interactions. 5 Thenozzle is adiabatic.Properties The specific heat of air at the average temperature of (700+645)/2=672.5°F is cp = 0.253Btu/lbm⋅R (Table A-2Eb).Analysis There is only one inlet and one exit, and thus m1 = m2 = m . We take nozzle as the system, which & & &is a control volume since mass crosses the boundary. The energy balance for this steady-flow system canbe expressed in the rate form as & & E in − E out = ΔE system 0 (steady) & =0 1 24 4 3 1442443 4 Rate of net energy transfer Rate of change in internal, kinetic, by heat, work, and mass potential, etc. energies 300 psia & & E in = E out AIR 250 psia 700°F 645°F 80 ft/s m(h1 + V12 / 2) = m(h2 + V 22 /2) & & h1 + V12 / 2 = h2 + V 22 /2Solving for exit velocity, [ V 2 = V12 + 2(h1 − h2 ) ]0.5 [ = V12 + 2c p (T1 − T2 ) ] 0.5 0.5 ⎡ ⎛ 25,037 ft 2 /s 2 ⎞⎤ = ⎢(80 ft/s) 2 + 2(0.253 Btu/lbm ⋅ R)(700 − 645)R ⎜ ⎟⎥ ⎢ ⎜ 1 Btu/lbm ⎟⎥ ⎣ ⎝ ⎠⎦ = 838.6 ft/sPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 19. 5-195-34 Air is decelerated in an adiabatic diffuser. The velocity at the exit is to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Air is an ideal gas withconstant specific heats. 3 Potential energy changes are negligible. 4 There are no work interactions. 5 Thediffuser is adiabatic.Properties The specific heat of air at the average temperature of (20+90)/2=55°C =328 K is cp = 1.007kJ/kg⋅K (Table A-2b).Analysis There is only one inlet and one exit, and thus m1 = m2 = m . We take diffuser as the system, & & &which is a control volume since mass crosses the boundary. The energy balance for this steady-flow systemcan be expressed in the rate form as & & E in − E out = ΔE system 0 (steady) & =0 1 24 4 3 1442443 4 Rate of net energy transfer Rate of change in internal, kinetic, by heat, work, and mass potential, etc. energies & & 100 kPa E in = E out 20°C AIR 200 kPa 500 m/s 90°C m(h1 + V12 / 2) = m(h2 + V 22 /2) & & h1 + V12 / 2 = h2 + V 22 /2Solving for exit velocity, [ V 2 = V12 + 2(h1 − h2 ) ]0.5 [ = V12 + 2c p (T1 − T2 ) ] 0.5 0.5 ⎡ ⎛ 1000 m 2 /s 2 ⎞⎤ = ⎢(500 m/s) 2 + 2(1.007 kJ/kg ⋅ K)(20 − 90)K⎜ ⎟⎥ ⎢ ⎜ 1 kJ/kg ⎟⎥ ⎣ ⎝ ⎠⎦ = 330.2 m/sPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 20. 5-205-35 Steam is accelerated in a nozzle from a velocity of 80 m/s. The mass flow rate, the exit velocity, andthe exit area of the nozzle are to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Potential energy changesare negligible. 3 There are no work interactions.Properties From the steam tables (Table A-6) 120 kJ/s P1 = 5 MPa ⎫ v 1 = 0.057838 m 3 /kg ⎬ T1 = 400°C ⎭ h1 = 3196.7 kJ/kg Steam 1 2and P2 = 2 MPa ⎫ v 2 = 0.12551 m 3 /kg ⎬ T2 = 300°C ⎭ h 2 = 3024.2 kJ/kgAnalysis (a) There is only one inlet and one exit, and thus m1 = m2 = m . The mass flow rate of steam is & & & 1 1 m= & V1 A1 = (80 m/s)(50 × 10 − 4 m 2 ) = 6.92 kg/s v1 0.057838 m3/kg(b) We take nozzle as the system, which is a control volume since mass crosses the boundary. The energybalance for this steady-flow system can be expressed in the rate form as & & Ein − Eout = ΔEsystem 0 (steady) & =0 1 24 4 3 1442443 Rate of net energy transfer Rate of change in internal, kinetic, by heat, work, and mass potential, etc. energies & & Ein = Eout & & & 2 & m(h1 + V12 / 2) = Qout + m(h2 + V2 /2) (since W ≅ Δpe ≅ 0) ⎛ V 2 − V12 ⎞ − Qout = m⎜ h2 − h1 + 2 & & ⎟ ⎜ 2 ⎟ ⎝ ⎠Substituting, the exit velocity of the steam is determined to be ⎛ V 2 − (80 m/s)2 ⎛ 1 kJ/kg ⎞ ⎞ − 120 kJ/s = (6.916 kg/s )⎜ 3024.2 − 3196.7 + 2 ⎜ ⎟⎟ ⎜ 1000 m 2 /s 2 ⎟ ⎟ ⎜ 2 ⎝ ⎝ ⎠⎠It yields V2 = 562.7 m/s(c) The exit area of the nozzle is determined from m= & 1 V 2 A2 ⎯ ⎯→ A2 = = ( mv 2 (6.916 kg/s ) 0.12551 m 3 /kg & ) = 15.42 × 10 −4 m 2 v2 V2 562.7 m/sPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 21. 5-215-36 [Also solved by EES on enclosed CD] Steam is accelerated in a nozzle from a velocity of 40 m/s to300 m/s. The exit temperature and the ratio of the inlet-to-exit area of the nozzle are to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Potential energy changesare negligible. 3 There are no work interactions. 4 The device is adiabatic and thus heat transfer isnegligible.Properties From the steam tables (Table A-6), P1 = 3 MPa ⎫ v 1 = 0.09938 m 3 /kg ⎬ T1 = 400°C ⎭ h1 = 3231.7 kJ/kgAnalysis (a) There is only one inlet and one exit, and thus m1 = m2 = m . We take nozzle as the system, & & &which is a control volume since mass crosses the boundary. The energy balance for this steady-flow systemcan be expressed in the rate form as & & Ein − Eout = ΔEsystem 0 (steady) & =0 1 24 4 3 1442443Rate of net energy transferby heat, work, and mass Rate of change in internal, kinetic, P1 = 3 MPa Steam P2 = 2.5 MPa potential, etc. energies T1 = 400°C V2 = 300 m/s & & Ein = Eout V1 = 40 m/s & & 2 & & m(h1 + V12 / 2) = m(h2 + V2 /2) (since Q ≅ W ≅ Δpe ≅ 0) V 22 − V12 0 = h2 − h1 + 2or, V 22 − V12 (300 m/s) 2 − (40 m/s) 2 ⎛ 1 kJ/kg ⎞ h2 = h1 − = 3231.7 kJ/kg − ⎜ ⎟ = 3187.5 kJ/kg 2 2 ⎜ 1000 m 2 /s 2 ⎟ ⎝ ⎠Thus, P2 = 2.5 MPa ⎫ T2 = 376.6°C ⎬ h2 = 3187.5 kJ/kg ⎭ v 2 = 0.11533 m3/kg(b) The ratio of the inlet to exit area is determined from the conservation of mass relation, 1 1 A1 v 1 V 2 (0.09938 m 3 /kg )(300 m/s) A2V 2 = A1V1 ⎯ ⎯→ = = = 6.46 v2 v1 A2 v 2 V1 (0.11533 m 3 /kg )(40 m/s)PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 22. 5-225-37 Air is decelerated in a diffuser from 230 m/s to 30 m/s. The exit temperature of air and the exit area ofthe diffuser are to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Air is an ideal gas withvariable specific heats. 3 Potential energy changes are negligible. 4 The device is adiabatic and thus heattransfer is negligible. 5 There are no work interactions.Properties The gas constant of air is 0.287 kPa.m3/kg.K (Table A-1). The enthalpy of air at the inlettemperature of 400 K is h1 = 400.98 kJ/kg (Table A-17).Analysis (a) There is only one inlet and one exit, and thus m1 = m2 = m . We take diffuser as the system, & & &which is a control volume since mass crosses the boundary. The energy balance for this steady-flow systemcan be expressed in the rate form as & & Ein − Eout = ΔEsystem 0 (steady) & =0 1 24 4 3 1442443 Rate of net energy transfer Rate of change in internal, kinetic, by heat, work, and mass potential, etc. energies & & 1 AIR 2 Ein = Eout & & 2 & & m(h1 + V12 / 2) = m(h2 + V2 /2) (since Q ≅ W ≅ Δpe ≅ 0) V 22 − V12 , 0 = h2 − h1 + 2or, h2 = h1 − V 22 − V12 = 400.98 kJ/kg − (30 m/s)2 − (230 m/s)2 ⎛ 1 kJ/kg ⎜ ⎞ ⎟ = 426.98 kJ/kg 2 2 ⎜ 1000 m 2 /s 2 ⎟ ⎝ ⎠From Table A-17, T2 = 425.6 K(b) The specific volume of air at the diffuser exit is v2 = RT2 = ( 0.287 kPa ⋅ m 3 /kg ⋅ K (425.6 K ) ) = 1.221 m 3 /kg P2 (100 kPa )From conservation of mass, 1 mv 2 (6000 3600 kg/s)(1.221 m 3 /kg) & m= & A2V 2 ⎯ ⎯→ A2 = = = 0.0678 m 2 v2 V2 30 m/sPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 23. 5-235-38E Air is decelerated in a diffuser from 600 ft/s to a low velocity. The exit temperature and the exitvelocity of air are to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Air is an ideal gas withvariable specific heats. 3 Potential energy changes are negligible. 4 The device is adiabatic and thus heattransfer is negligible. 5 There are no work interactions.Properties The enthalpy of air at the inlet temperature of 20°F is h1 = 114.69 Btu/lbm (Table A-17E).Analysis (a) There is only one inlet and one exit, and thus m1 = m2 = m . We take diffuser as the system, & & &which is a control volume since mass crosses the boundary. The energy balance for this steady-flow systemcan be expressed in the rate form as & & Ein − Eout = ΔEsystem 0 (steady) & =0 1 24 4 3 1442443 Rate of net energy transfer Rate of change in internal, kinetic, by heat, work, and mass potential, etc. energies & & Ein = Eout 1 AIR 2 & & & & m(h1 + V12 / 2) = m(h2 + V 22 /2) (since Q ≅ W ≅ Δpe ≅ 0) V 22 − V12 , 0 = h2 − h1 + 2or, V22 − V12 0 − (600 ft/s )2 ⎛ 1 Btu/lbm ⎞ h2 = h1 − = 114.69 Btu/lbm − ⎜ ⎟ 2 2 ⎜ 25,037 ft 2 /s 2 ⎟ = 121.88 Btu/lbm ⎝ ⎠From Table A-17E, T2 = 510.0 R(b) The exit velocity of air is determined from the conservation of mass relation, 1 1 1 1 A2V 2 = A1V1 ⎯ ⎯→ A2V 2 = A1V1 v2 v1 RT2 / P2 RT1 / P1Thus, A1T2 P1 1 (510 R )(13 psia ) V2 = V1 = (600 ft/s) = 114.3 ft/s A2 T1 P2 5 (480 R )(14.5 psia )PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 24. 5-245-39 CO2 gas is accelerated in a nozzle to 450 m/s. The inlet velocity and the exit temperature are to bedetermined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 CO2 is an ideal gas withvariable specific heats. 3 Potential energy changes are negligible. 4 The device is adiabatic and thus heattransfer is negligible. 5 There are no work interactions.Properties The gas constant and molar mass of CO2 are 0.1889 kPa.m3/kg.K and 44 kg/kmol (Table A-1).The enthalpy of CO2 at 500°C is h1 = 30,797 kJ/kmol (Table A-20).Analysis (a) There is only one inlet and one exit, and thus m1 = m2 = m . Using the ideal gas relation, the & & &specific volume is determined to be v1 = RT1 = ( 0.1889 kPa ⋅ m 3 /kg ⋅ K (773 K ) ) = 0.146 m 3 /kg P1 1000 kPa 1 CO2 2Thus, m= & 1 A1V1 ⎯ ⎯→ V1 = mv1 (6000/3600 kg/s ) 0.146 m3 /kg & = ( = 60.8 m/s ) v1 A1 40 × 10− 4 m 2(b) We take nozzle as the system, which is a control volume since mass crosses the boundary. The energybalance for this steady-flow system can be expressed in the rate form as & & Ein − Eout = ΔEsystem 0 (steady) & =0 1 24 4 3 1442443 Rate of net energy transfer Rate of change in internal, kinetic, by heat, work, and mass potential, etc. energies & & Ein = Eout & & 2 & & m(h1 + V12 / 2) = m(h2 + V2 /2) (since Q ≅ W ≅ Δpe ≅ 0) V 22 − V12 0 = h2 − h1 + 2Substituting, V 22 − V12 h2 = h1 − M 2 = 30,797 kJ/kmol − (450 m/s)2 − (60.8 m/s)2 ⎛ ⎜ 1 kJ/kg ⎞ ⎟(44 kg/kmol) 2 ⎜ 1000 m 2 /s 2 ⎟ ⎝ ⎠ = 26,423 kJ/kmolThen the exit temperature of CO2 from Table A-20 is obtained to be T2 = 685.8 KPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 25. 5-255-40 R-134a is accelerated in a nozzle from a velocity of 20 m/s. The exit velocity of the refrigerant andthe ratio of the inlet-to-exit area of the nozzle are to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Potential energy changesare negligible. 3 There are no work interactions. 4 The device is adiabatic and thus heat transfer isnegligible.Properties From the refrigerant tables (Table A-13) P1 = 700 kPa ⎫ v 1 = 0.043358 m 3 /kg ⎬ T1 = 120°C ⎭ h1 = 358.90 kJ/kg 1 R-134a 2and P2 = 400 kPa ⎫ v 2 = 0.056796 m 3 /kg ⎬ T2 = 30°C ⎭ h2 = 275.07 kJ/kgAnalysis (a) There is only one inlet and one exit, and thus m1 = m2 = m . We take nozzle as the system, & & &which is a control volume since mass crosses the boundary. The energy balance for this steady-flow systemcan be expressed in the rate form as & & Ein − Eout = ΔEsystem 0 (steady) & =0 1 24 4 3 1442443 Rate of net energy transfer Rate of change in internal, kinetic, by heat, work, and mass potential, etc. energies & & Ein = Eout & & 2 & & m(h1 + V12 / 2) = m(h2 + V2 /2) (since Q ≅ W ≅ Δpe ≅ 0) V22 − V12 0 = h2 − h1 + 2Substituting, V 22 − (20 m/s )2 ⎛ 1 kJ/kg ⎞ 0 = (275.07 − 358.90)kJ/kg + ⎜ ⎟ 2 ⎜ 1000 m 2 /s 2 ⎟ ⎝ ⎠It yields V2 = 409.9 m/s(b) The ratio of the inlet to exit area is determined from the conservation of mass relation, 1 A2V 2 = 1 A1V1 ⎯ ⎯→ A1 v 1 V 2 = = ( 0.043358 m 3 /kg (409.9 m/s ) = 15.65 ) v2 v1 A2 v 2 V1 ( 0.056796 m 3 /kg (20 m/s ) )PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 26. 5-265-41 Nitrogen is decelerated in a diffuser from 200 m/s to a lower velocity. The exit velocity of nitrogenand the ratio of the inlet-to-exit area are to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Nitrogen is an ideal gaswith variable specific heats. 3 Potential energy changes are negligible. 4 The device is adiabatic and thusheat transfer is negligible. 5 There are no work interactions.Properties The molar mass of nitrogen is M = 28 kg/kmol (Table A-1). The enthalpies are (Table A-18) T1 = 7°C = 280 K → h1 = 8141 kJ/kmol T2 = 22°C = 295 K → h2 = 8580 kJ/kmolAnalysis (a) There is only one inlet and one exit, and thus m1 = m2 = m . We take diffuser as the system, & & &which is a control volume since mass crosses the boundary. The energy balance for this steady-flow systemcan be expressed in the rate form as & & Ein − Eout = ΔEsystem 0 (steady) & =0 1 24 4 3 1442443 Rate of net energy transfer Rate of change in internal, kinetic, by heat, work, and mass potential, etc. energies & & 1 N2 2 Ein = Eout & & & & m(h1 + V12 / 2) = m(h2 + V 22 /2) (since Q ≅ W ≅ Δpe ≅ 0) V 22 − V12 h2 − h1 V 22 − V12 , 0 = h2 − h1 + = + 2 M 2Substituting, 0= (8580 − 8141) kJ/kmol + V22 − (200 m/s )2 ⎛ ⎜ 1 kJ/kg ⎞ ⎟ 28 kg/kmol 2 ⎜ 1000 m 2 /s 2 ⎟ ⎝ ⎠It yields V2 = 93.0 m/s(b) The ratio of the inlet to exit area is determined from the conservation of mass relation, 1 1 A1 v 1 V 2 ⎛ RT1 / P1 ⎞ V2 A2V 2 = A1V1 ⎯ ⎯→ = =⎜ ⎟ v2 v1 A2 v 2 V1 ⎜ RT2 / P2 ⎝ ⎟V ⎠ 1or, A1 ⎛ T1 / P1 ⎞ V 2 (280 K/60 kPa )(93.0 m/s ) =⎜ ⎟ ⎟ V = (295 K/85 kPa )(200 m/s ) = 0.625 A2 ⎜ T2 / P2 ⎝ ⎠ 1PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 27. 5-275-42 EES Problem 5-41 is reconsidered. The effect of the inlet velocity on the exit velocity and the ratio ofthe inlet-to-exit area as the inlet velocity varies from 180 m/s to 260 m/s is to be investigated. The finalresults are to be plotted against the inlet velocity.Analysis The problem is solved using EES, and the solution is given below.Function HCal(WorkFluid$, Tx, Px)"Function to calculate the enthalpy of an ideal gas or real gas" If N2 = WorkFluid$ then HCal:=ENTHALPY(WorkFluid$,T=Tx) "Ideal gas equ." else HCal:=ENTHALPY(WorkFluid$,T=Tx, P=Px)"Real gas equ." endifend HCal"System: control volume for the nozzle""Property relation: Nitrogen is an ideal gas""Process: Steady state, steady flow, adiabatic, no work""Knowns"WorkFluid$ = N2T[1] = 7 [C]P[1] = 60 [kPa]{Vel[1] = 200 [m/s]}P[2] = 85 [kPa]T[2] = 22 [C]"Property Data - since the Enthalpy function has different parametersfor ideal gas and real fluids, a function was used to determine h."h[1]=HCal(WorkFluid$,T[1],P[1])h[2]=HCal(WorkFluid$,T[2],P[2])"The Volume function has the same form for an ideal gas as for a real fluid."v[1]=volume(workFluid$,T=T[1],p=P[1])v[2]=volume(WorkFluid$,T=T[2],p=P[2])"From the definition of mass flow rate, m_dot = A*Vel/v and conservation of mass the area ratioA_Ratio = A_1/A_2 is:"A_Ratio*Vel[1]/v[1] =Vel[2]/v[2]"Conservation of Energy - SSSF energy balance"h[1]+Vel[1]^2/(2*1000) = h[2]+Vel[2]^2/(2*1000)PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 28. 5-28 ARatio Vel1 [m/s] Vel2 [m/s] 0.2603 180 34.84 0.4961 190 70.1 0.6312 200 93.88 0.7276 210 113.6 0.8019 220 131.2 0.8615 230 147.4 0.9106 240 162.5 0.9518 250 177 0.9869 260 190.8 1 0.9 0.8 0.7 0.6 A Ratio 0.5 0.4 0.3 0.2 180 190 200 210 220 230 240 250 260 Vel[1] [m /s] 200 180 160 Vel[2] [m /s] 140 120 100 80 60 40 20 180 190 200 210 220 230 240 250 260 Vel[1] [m /s]PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 29. 5-295-43 R-134a is decelerated in a diffuser from a velocity of 120 m/s. The exit velocity of R-134a and themass flow rate of the R-134a are to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Potential energy changesare negligible. 3 There are no work interactions.Properties From the R-134a tables (Tables A-11 through A-13) P1 = 800 kPa ⎫ v 1 = 0.025621 m 3 /kg 2 kJ/s ⎬ sat.vapor ⎭ h1 = 267.29 kJ/kg 1 R-134a 2and P2 = 900 kPa ⎫ v 2 = 0.023375 m 3 /kg ⎬ T2 = 40°C ⎭ h2 = 274.17 kJ/kgAnalysis (a) There is only one inlet and one exit, and thus m1 = m2 = m . Then the exit velocity of R-134a & & &is determined from the steady-flow mass balance to be 1 1 v 2 A1 1 (0.023375 m 3 /kg) A2V 2 = A1V1 ⎯ ⎯→ V 2 = V1 = (120 m/s ) = 60.8 m/s v2 v1 v 1 A2 1.8 (0.025621 m 3 /kg)(b) We take diffuser as the system, which is a control volume since mass crosses the boundary. The energybalance for this steady-flow system can be expressed in the rate form as & & Ein − Eout = ΔEsystem 0 (steady) & =0 1 24 4 3 1442443 Rate of net energy transfer Rate of change in internal, kinetic, by heat, work, and mass potential, etc. energies & & Ein = Eout & & & 2 & Qin + m(h1 + V12 / 2) = m(h2 + V2 /2) (since W ≅ Δpe ≅ 0) ⎛ V 2 − V12 ⎞ Qin = m⎜ h2 − h1 + 2 & & ⎟ ⎜ 2 ⎟ ⎝ ⎠Substituting, the mass flow rate of the refrigerant is determined to be & ⎛ 2 kJ/s = m⎜ (274.17 − 267.29)kJ/kg + (60.8 m/s)2 − (120 m/s)2 ⎛ 1 kJ/kg ⎞ ⎞ ⎜ ⎟⎟ ⎜ 2 ⎜ 1000 m 2 /s 2 ⎟ ⎟ ⎝ ⎝ ⎠⎠It yields m = 1.308 kg/s &PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 30. 5-305-44 Heat is lost from the steam flowing in a nozzle. The velocity and the volume flow rate at the nozzleexit are to be determined.Assumptions 1 This is a steady-flow process since thereis no change with time. 2 Potential energy change isnegligible. 3 There are no work interactions.Analysis We take the steam as the system, which is a 400°C STEAM 300°Ccontrol volume since mass crosses the boundary. The 800 kPa 200 kPaenergy balance for this steady-flow system can be 10 m/sexpressed in the rate form as QEnergy balance: & & Ein − Eout = ΔEsystem 0 (steady) & =0 1 24 4 3 1442443 Rate of net energy transfer Rate of change in internal, kinetic, by heat, work, and mass potential, etc. energies & & Ein = Eout ⎛ V2 ⎞ ⎛ V2 ⎞ & m⎜ h1 + 1 ⎟ = m⎜ h2 + 2 ⎟ + Qout & & & since W ≅ Δpe ≅ 0) ⎜ 2 ⎟ ⎜ 2 ⎟ ⎝ ⎠ ⎝ ⎠ V12 V2 Q &or h1 + = h2 + 2 + out 2 2 m&The properties of steam at the inlet and exit are (Table A-6) P = 800 kPa ⎫v1 = 0.38429 m3/kg 1 ⎬ T1 = 400°C ⎭ h1 = 3267.7 kJ/kg P2 = 200 kPa ⎫v 2 = 1.31623 m3/kg ⎬ T1 = 300°C ⎭ h2 = 3072.1 kJ/kgThe mass flow rate of the steam is 1 1 m= & A1V1 = 3 (0.08 m 2 )(10 m/s) = 2.082 kg/s v1 0.38429 m /sSubstituting, (10 m/s) 2 ⎛ 1 kJ/kg ⎞ V 2 ⎛ 1 kJ/kg ⎞ 25 kJ/s 3267.7 kJ/kg + ⎜ 2 2 ⎟ = 3072.1 kJ/kg + 2 ⎜ ⎟+ 2 ⎝ 1000 m /s ⎠ 2 ⎝ 1000 m 2 /s 2 ⎠ 2.082 kg/s ⎯ V2 = 606 m/s ⎯→The volume flow rate at the exit of the nozzle is V&2 = mv 2 = (2.082 kg/s)(1.31623 m3/kg) = 2.74 m3 /s &PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 31. 5-31Turbines and Compressors5-45C Yes.5-46C The volume flow rate at the compressor inlet will be greater than that at the compressor exit.5-47C Yes. Because energy (in the form of shaft work) is being added to the air.5-48C No.5-49 Air is expanded in a turbine. The mass flow rate and outlet area are to be determined.Assumptions 1 Air is an ideal gas. 2 The flow is steady.Properties The gas constant of air is R = 0.287 kPa⋅m3/kg⋅K (Table A-1). 1 MPaAnalysis The specific volumes of air at the inlet and outlet are 600°C 30 m/s RT1 (0.287 kPa ⋅ m 3 /kg ⋅ K)(600 + 273 K) v1 = = = 0.2506 m 3 /kg P1 1000 kPa Turbine RT2 (0.287 kPa ⋅ m 3 /kg ⋅ K)(200 + 273 K) v2 = = = 1.3575 m 3 /kg P2 100 kPa 100 kPa 200°CThe mass flow rate is 10 m/s A1V1 (0.1 m 2 )(30 m/s) m= & = = 11.97 kg/s v1 0.2506 m 3 /kgThe outlet area is mv 2 (11.97 kg/s)(1.3575 m 3 /kg) & A2 = = = 1.605 m 2 V2 10 m/sPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 32. 5-325-50E Air is expanded in a gas turbine. The inlet and outlet mass flow rates are to be determined.Assumptions 1 Air is an ideal gas. 2 The flow is steady.Properties The gas constant of air is R = 0.3704 psia⋅ft3/lbm⋅R (Table A-1E).Analysis The specific volumes of air at the inlet and outlet are 150 psia RT (0.3704 psia ⋅ ft 3 /lbm ⋅ R)(700 + 460 R) 700°F v1 = 1 = = 2.864 ft 3 /lbm P1 150 psia 5 lbm/s RT2 (0.3704 psia ⋅ ft 3 /lbm ⋅ R)(100 + 460 R) Turbine v2 = = = 13.83 ft 3 /lbm P2 15 psiaThe volume flow rates at the inlet and exit are then 15 psia 100°F V&1 = mv 1 = (5 lbm/s)(2.864 ft 3 /lbm) = 14.32 ft 3 /s & V&2 = mv 2 = (5 lbm/s)(13.83 ft 3 /lbm) = 69.15 ft 3 /s &PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 33. 5-335-51 Air is compressed at a rate of 10 L/s by a compressor. The work required per unit mass and the powerrequired are to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potentialenergy changes are negligible. 3 Air is an ideal gas with constant specific heats.Properties The constant pressure specific heat of air at the average temperature of (20+300)/2=160°C=433K is cp = 1.018 kJ/kg·K (Table A-2b). The gas constant of air is R = 0.287 kPa⋅m3/kg⋅K (Table A-1).Analysis (a) There is only one inlet and one exit, and thus m1 = m2 = m . We take the compressor as the & & &system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as & & E in − E out = ΔE system 0 (steady) & =0 1 24 4 3 1442443 4 Rate of net energy transfer Rate of change in internal, kinetic, 1 MPa by heat, work, and mass potential, etc. energies 300°C & & E in = E out Compressor & Win + mh1 = mh2 (since Δke ≅ Δpe ≅ 0) & & & = m(h − h ) = mc (T − T ) Win & 2 1 & p 2 1 120 kPaThus, 20°C 10 L/s win = c p (T2 − T1 ) = (1.018 kJ/kg ⋅ K)(300 − 20)K = 285.0 kJ/kg(b) The specific volume of air at the inlet and the mass flow rate are RT1 (0.287 kPa ⋅ m 3 /kg ⋅ K)(20 + 273 K) v1 = = = 0.7008 m 3 /kg P1 120 kPa V&1 0.010 m 3 /s m= & = = 0.01427 kg/s v 1 0.7008 m 3 /kgThen the power input is determined from the energy balance equation to be & Win = mc p (T2 − T1 ) = (0.01427 kg/s)(1.01 8 kJ/kg ⋅ K)(300 − 20)K = 4.068 kW &PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 34. 5-345-52 Steam expands in a turbine. The change in kinetic energy, the power output, and the turbine inlet areaare to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Potential energy changesare negligible. 3 The device is adiabatic and thus heat transfer is negligible.Properties From the steam tables (Tables A-4 through 6) P1 = 10 MPa ⎫ v 1 = 0.029782 m 3 /kg P1 = 10 MPa ⎬ T1 = 450°C T1 = 450°C ⎭ h1 = 3242.4 kJ/kg V1 = 80 m/sandP2 = 10 kPa ⎫ ⎬ h2 = h f + x 2 h fg = 191.81 + 0.92 × 2392.1 = 2392.5 kJ/kgx 2 = 0.92 ⎭ · STEAM m = 12 kg/s ·Analysis (a) The change in kinetic energy is determined from W V 22 − V12 (50 m/s)2 − (80 m/s) 2 ⎛ 1 kJ/kg ⎜ ⎞ ⎟ = −1.95 kJ/kg Δke = = 2 2 ⎜ 1000 m 2 /s 2 ⎟ ⎝ ⎠ P2 = 10 kPa(b) There is only one inlet and one exit, and thus m1 = m2 = m . We take the & & & x2 = 0.92turbine as the system, which is a control volume since mass crosses the V2 = 50 m/sboundary. The energy balance for this steady-flow system can be expressedin the rate form as & & Ein − Eout = ΔEsystem 0 (steady) & =0 1 24 4 3 1442443 Rate of net energy transfer Rate of change in internal, kinetic, by heat, work, and mass potential, etc. energies & & Ein = Eout & & m(h1 + V12 / 2) = Wout + m(h2 + V 22 /2) (since Q ≅ Δpe ≅ 0) & & ⎛ V 2 − V12 ⎞ Wout = −m⎜ h2 − h1 + 2 & & ⎟ ⎜ 2 ⎟ ⎝ ⎠Then the power output of the turbine is determined by substitution to be & Wout = −(12 kg/s)(2392.5 − 3242.4 − 1.95)kJ/kg = 10.2 MW(c) The inlet area of the turbine is determined from the mass flow rate relation, 1 mv 1 (12 kg/s)(0.029782 m 3 /kg) & m= & A1V1 ⎯ ⎯→ A1 = = = 0.00447 m 2 v1 V1 80 m/sPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 35. 5-355-53 EES Problem 5-52 is reconsidered. The effect of the turbine exit pressure on the power output of theturbine as the exit pressure varies from 10 kPa to 200 kPa is to be investigated. The power output is to beplotted against the exit pressure.Analysis The problem is solved using EES, and the solution is given below."Knowns "T[1] = 450 [C]P[1] = 10000 [kPa]Vel[1] = 80 [m/s]P[2] = 10 [kPa]X_2=0.92Vel[2] = 50 [m/s]m_dot[1]=12 [kg/s]Fluid$=Steam_IAPWS 130 120"Property Data" 110h[1]=enthalpy(Fluid$,T=T[1],P=P[1])h[2]=enthalpy(Fluid$,P=P[2],x=x_2) 100 T[2] [C]T[2]=temperature(Fluid$,P=P[2],x=x_2) 90v[1]=volume(Fluid$,T=T[1],p=P[1]) 80v[2]=volume(Fluid$,P=P[2],x=x_2) 70"Conservation of mass: " 60m_dot[1]= m_dot[2] 50 40"Mass flow rate" 0 40 80 120 160 200m_dot[1]=A[1]*Vel[1]/v[1] P[2] [kPa]m_dot[2]= A[2]*Vel[2]/v[2]"Conservation of Energy - Steady Flow energy balance"m_dot[1]*(h[1]+Vel[1]^2/2*Convert(m^2/s^2, kJ/kg)) =m_dot[2]*(h[2]+Vel[2]^2/2*Convert(m^2/s^2, kJ/kg))+W_dot_turb*convert(MW,kJ/s)DELTAke=Vel[2]^2/2*Convert(m^2/s^2, kJ/kg)-Vel[1]^2/2*Convert(m^2/s^2, kJ/kg) 10.25 P2 Wturb T2 [kPa] [MW] [C] 10 10.22 45.81 9.9 31.11 9.66 69.93 Wturb [Mw] 52.22 9.377 82.4 9.55 73.33 9.183 91.16 94.44 9.033 98.02 115.6 8.912 103.7 9.2 136.7 8.809 108.6 157.8 8.719 112.9 178.9 8.641 116.7 8.85 200 8.57 120.2 8.5 0 40 80 120 160 200 P[2] [kPa]PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 36. 5-365-54 Steam expands in a turbine. The mass flow rate of steam for a power output of 5 MW is to bedetermined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potentialenergy changes are negligible. 3 The device is adiabatic and thus heat transfer is negligible.Properties From the steam tables (Tables A-4 through 6) P = 10 MPa ⎫ 1 1 ⎬ h1 = 3375.1 kJ/kg T1 = 500°C ⎭ P2 = 10 kPa ⎫ ⎬ h2 = h f + x2 h fg = 191.81 + 0.90 × 2392.1 = 2344.7 kJ/kg H2O x2 = 0.90 ⎭Analysis There is only one inlet and one exit, and thus m1 = m2 = m . We & & &take the turbine as the system, which is a control volume since mass crossesthe boundary. The energy balance for this steady-flow system can beexpressed in the rate form as 2 & & Ein − Eout = ΔEsystem 0 (steady) & =0 1 24 4 3 1442443 Rate of net energy transfer Rate of change in internal, kinetic, by heat, work, and mass potential, etc. energies & & Ein = Eout & & mh1 = Wout + mh2 (since Q ≅ Δke ≅ Δpe ≅ 0) & & & Wout = − m(h2 − h1 ) &Substituting, the required mass flow rate of the steam is determined to be 5000 kJ/s = − m(2344.7 − 3375.1) kJ/kg ⎯ m = 4.852 kg/s & ⎯→ &PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 37. 5-375-55E Steam expands in a turbine. The rate of heat loss from the steam for a power output of 4 MW is tobe determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potentialenergy changes are negligible.Properties From the steam tables (Tables A-4E through 6E) 1 P1 = 1000 psia ⎫ ⎬ h1 = 1448.6 Btu/lbm T1 = 900°F ⎭ P2 = 5 psia ⎫ ⎬ h2 = 1130.7 Btu/lbm H2O sat.vapor ⎭Analysis There is only one inlet and one exit, and thus m1 = m2 = m . We & & &take the turbine as the system, which is a control volume since mass crossesthe boundary. The energy balance for this steady-flow system can be 2expressed in the rate form as & & Ein − Eout = ΔEsystem 0 (steady) & =0 1 24 4 3 1442443 Rate of net energy transfer Rate of change in internal, kinetic, by heat, work, and mass potential, etc. energies & & Ein = Eout & & mh1 = Qout + Wout + mh2 (since Δke ≅ Δpe ≅ 0) & & & = − m (h − h ) − W Qout & 2 1 & outSubstituting, & ⎛ 1 Btu ⎞ Qout = −(45000/3600 lbm/s)(1130.7 − 1448.6) Btu/lbm − 4000 kJ/s⎜ ⎜ 1.055 kJ ⎟ = 182.0 Btu/s ⎟ ⎝ ⎠PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 38. 5-385-56 Steam expands in a turbine. The exit temperature of the steam for a power output of 2 MW is to bedetermined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potentialenergy changes are negligible. 3 The device is adiabatic and thus heat transfer is negligible.Properties From the steam tables (Tables A-4 through 6) P1 = 8 MPa ⎫ ⎬ h1 = 3399.5 kJ/kg T1 = 500°C ⎭Analysis There is only one inlet and one exit, and thus m1 = m2 = m . We take the turbine as the system, & & &which is a control volume since mass crosses the boundary. The energy balance for this steady-flow systemcan be expressed in the rate form as & & Ein − Eout = ΔEsystem 0 (steady) & =0 1 24 4 3 1442443 1 Rate of net energy transfer Rate of change in internal, kinetic, by heat, work, and mass potential, etc. energies & & Ein = Eout & & mh1 = Wout + mh2 & & (since Q ≅ Δke ≅ Δpe ≅ 0) H2O & Wout = m(h1 − h2 ) &Substituting, 2500 kJ/s = (3 kg/s )(3399.5 − h2 )kJ/kg 2 h2 = 2566.2 kJ/kgThen the exit temperature becomes P2 = 20 kPa ⎫ ⎬T2 = 60.1 °C h2 = 2566.2 kJ/kg ⎭PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 39. 5-395-57 Argon gas expands in a turbine. The exit temperature of the argon for a power output of 250 kW is tobe determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Potential energy changesare negligible. 3 The device is adiabatic and thus heat transfer is negligible. 4 Argon is an ideal gas withconstant specific heats.Properties The gas constant of Ar is R = 0.2081 kPa.m3/kg.K. The constant pressure specific heat of Ar iscp = 0.5203 kJ/kg·°C (Table A-2a)Analysis There is only one inlet and one exit, and thus m1 = m2 = m . The inlet specific volume of argon & & &and its mass flow rate are v1 = RT1 = ( 0.2081 kPa ⋅ m3/kg ⋅ K (723 K ) ) = 0.167 m3/kg A1 = 60 cm2 P1 900 kPa P1 = 900 kPaThus, T1 = 450°C V1 = 80 m/s m= & 1 v1 A1V1 = 1 0.167 m3/kg (0.006 m )(80 m/s) = 2.874 kg/s 2We take the turbine as the system, which is a control volume since ARGONmass crosses the boundary. The energy balance for this steady-flow 250 kWsystem can be expressed in the rate form as & & Ein − Eout = & ΔEsystem 0 (steady) =0 1 24 4 3 1442443 4 4 Rate of net energy transfer Rate of change in internal, kinetic, by heat, work, and mass potential, etc. energies P2 = 150 kPa & & Ein = Eout V2 = 150 m/s & & m(h1 + V12 / 2) = Wout + m(h2 + V22 /2) (since Q ≅ Δpe ≅ 0) & & ⎛ V 2 − V12 ⎞ Wout = − m⎜ h2 − h1 + 2 & & ⎟ ⎜ 2 ⎟ ⎝ ⎠Substituting, ⎡ (150 m/s) 2 − (80 m/s) 2 ⎛ 1 kJ/kg ⎞⎤ 250 kJ/s = −(2.874 kg/s ) ⎢(0.5203 kJ/kg⋅ o C)(T2 − 450 o C) + ⎜ ⎟⎥ ⎢ 2 ⎜ 1000 m 2 /s 2 ⎟⎥ ⎣ ⎝ ⎠⎦It yields T2 = 267.3°CPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 40. 5-405-58 Helium is compressed by a compressor. For a mass flow rate of 90 kg/min, the power input requiredis to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potentialenergy changes are negligible. 3 Helium is an ideal gas with constant specific heats.Properties The constant pressure specific heat of helium is cp = 5.1926 kJ/kg·K (Table A-2a).Analysis There is only one inlet and one exit, and thus m1 = m2 = m . & & &We take the compressor as the system, which is a control volume since P2 = 700 kPamass crosses the boundary. The energy balance for this steady-flow T2 = 430 Ksystem can be expressed in the rate form as & & Ein − Eout = ΔEsystem 0 (steady) & =0 1 24 4 3 1442443 · Rate of net energy transfer Rate of change in internal, kinetic, Q by heat, work, and mass potential, etc. energies He & & Ein = Eout 90 kg/min · W & & Win + mh1 = Qout + mh2 (since Δke ≅ Δpe ≅ 0) & & & & Win − Qout = m(h2 − h1 ) = mc p (T2 − T1 ) & & P1 = 120 kPaThus, T1 = 310 K Win = Qout + mc p (T2 − T1 ) & & & = (90/6 0 kg/s)(20 kJ/kg) + (90/60 kg/s)(5.1926 kJ/kg ⋅ K)(430 − 310)K = 965 kWPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 41. 5-415-59 CO2 is compressed by a compressor. The volume flow rate of CO2 at the compressor inlet and thepower input to the compressor are to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potentialenergy changes are negligible. 3 Helium is an ideal gas with variable specific heats. 4 The device isadiabatic and thus heat transfer is negligible.Properties The gas constant of CO2 is R = 0.1889 kPa.m3/kg.K, and its molar mass is M = 44 kg/kmol(Table A-1). The inlet and exit enthalpies of CO2 are (Table A-20) T1 = 300 K → h1 = 9,431 kJ / kmol 2 T2 = 450 K → h2 = 15,483 kJ / kmolAnalysis (a) There is only one inlet and one exit, and thus m1 = m2 = m . The inlet specific volume of air and its volume flow & & & CO2rate are v1 = RT1 = ( 0.1889 kPa ⋅ m 3 /kg ⋅ K (300 K ) ) = 0.5667 m 3 /kg P1 100 kPa 1 V& = mv 1 = (0.5 kg/s)(0.5667 m 3 /kg) = 0.283 m 3 /s &(b) We take the compressor as the system, which is a control volume since mass crosses the boundary. Theenergy balance for this steady-flow system can be expressed in the rate form as & & Ein − Eout = ΔEsystem 0 (steady) & =0 1 24 4 3 1442443 Rate of net energy transfer Rate of change in internal, kinetic, by heat, work, and mass potential, etc. energies & & Ein = Eout & & Win + mh1 = mh2 (since Q ≅ Δke ≅ Δpe ≅ 0) & & & Win = m(h2 − h1 ) = m(h2 − h1 ) / M & &Substituting & Win = (0.5 kg/s )(15,483 − 9,431 kJ/kmol) = 68.8 kW 44 kg/kmolPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 42. 5-425-60 Air is expanded in an adiabatic turbine. The mass flow rate of the air and the power produced are tobe determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 The turbine is well-insulated, and thus there is no heat transfer. 3 Air is an ideal gas with constant specific heats.Properties The constant pressure specific heat of air at the average temperature of(500+150)/2=325°C=598 K is cp = 1.051 kJ/kg·K (Table A-2b). The gas constant of air is R = 0.287kPa⋅m3/kg⋅K (Table A-1).Analysis (a) There is only one inlet and one exit, and thus m1 = m 2 = m . We take the turbine as the & & &system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as & & E in − E out = ΔE system 0 (steady) & =0 1 24 4 3 1442443 4 Rate of net energy transfer Rate of change in internal, kinetic, 1 MPa by heat, work, and mass potential, etc. energies 500°C & & E in = E out 40 m/s ⎛ ⎞ ⎛ 2 ⎞ V2 ⎟ = m⎜ h2 + V 2 ⎟ + Wout Turbine m⎜ h1 + 1 & & & ⎜ ⎝ 2 ⎟ ⎠ ⎜ ⎝ 2 ⎟ ⎠ ⎛ V 2 − V 22 ⎞ ⎛ 2 2 ⎞ 100 kPa Wout = m⎜ h1 − h2 + 1 & & ⎟ = m⎜ c p (T1 − T2 ) + V1 − V 2 & ⎟ 150°C ⎜ 2 ⎟ ⎜ 2 ⎟ ⎝ ⎠ ⎝ ⎠The specific volume of air at the inlet and the mass flow rate are RT1 (0.287 kPa ⋅ m 3 /kg ⋅ K)(500 + 273 K) v1 = = = 0.2219 m 3 /kg P1 1000 kPa A1V1 (0.2 m 2 )(40 m/s) m= & = = 36.06 kg/s v1 0.2219 m 3 /kgSimilarly at the outlet, RT2 (0.287 kPa ⋅ m 3 /kg ⋅ K)(150 + 273 K) v2 = = = 1.214 m 3 /kg P2 100 kPa mv 2 (36.06 kg/s)(1.214 m 3 /kg) & V2 = = = 43.78 m/s A2 1m2(b) Substituting into the energy balance equation gives ⎛ V 2 − V 22 ⎞ Wout = m⎜ c p (T1 − T2 ) + 1 & & ⎟ ⎜ 2 ⎟ ⎝ ⎠ ⎡ (40 m/s) 2 − (43.78 m/s) 2 ⎛ 1 kJ/kg ⎞⎤ = (36.06 kg/s) ⎢(1.051 kJ/kg ⋅ K)(500 − 150)K + ⎜ ⎟⎥ ⎢ ⎣ 2 ⎝ 1000 m 2 /s 2 ⎠⎥ ⎦ = 13,260 kWPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 43. 5-435-61 Air is compressed in an adiabatic compressor. The mass flow rate of the air and the power input are tobe determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 The compressor isadiabatic. 3 Air is an ideal gas with constant specific heats.Properties The constant pressure specific heat of air at the average temperature of (20+400)/2=210°C=483K is cp = 1.026 kJ/kg·K (Table A-2b). The gas constant of air is R = 0.287 kPa⋅m3/kg⋅K (Table A-1).Analysis (a) There is only one inlet and one exit, and thus m1 = m2 = m . We take the compressor as the & & &system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as & & E in − E out = ΔE system 0 (steady) & =0 1 24 4 3 1442443 4 1.8 MPa Rate of net energy transfer Rate of change in internal, kinetic, by heat, work, and mass potential, etc. energies 400°C & & E in = E out Compressor ⎛ ⎞ ⎛ 2 ⎞ V2 ⎟ + Win = m⎜ h2 + V 2 ⎟ m⎜ h1 + 1 & & & ⎜ ⎝ 2 ⎟ ⎠ ⎜ ⎝ 2 ⎟ ⎠ 100 kPa ⎛ V 2 − V12 ⎞ ⎛ 2 2 ⎞ 20°C Win = m⎜ h2 − h1 + 2 & & ⎟ = m⎜ c p (T2 − T1 ) + V 2 − V1 & ⎟ 30 m/s ⎜ 2 ⎟ ⎜ 2 ⎟ ⎝ ⎠ ⎝ ⎠The specific volume of air at the inlet and the mass flow rate are RT1 (0.287 kPa ⋅ m 3 /kg ⋅ K)(20 + 273 K) v1 = = = 0.8409 m 3 /kg P1 100 kPa A1V1 (0.15 m 2 )(30 m/s) m= & = = 5.351 kg/s v1 0.8409 m 3 /kgSimilarly at the outlet, RT2 (0.287 kPa ⋅ m 3 /kg ⋅ K)(400 + 273 K) v2 = = = 0.1073 m 3 /kg P2 1800 kPa mv 2 (5.351 kg/s)(0.1073 m 3 /kg) & V2 = = = 7.177 m/s A2 0.08 m 2(b) Substituting into the energy balance equation gives ⎛ V 2 − V12 ⎞ Win = m⎜ c p (T2 − T1 ) + 2 & & ⎟ ⎜ 2 ⎟ ⎝ ⎠ ⎡ (7.177 m/s) 2 − (30 m/s) 2 ⎛ 1 kJ/kg ⎞⎤ = (5.351 kg/s) ⎢(1.026 kJ/kg ⋅ K)(400 − 20)K + ⎜ ⎟⎥ ⎢ ⎣ 2 ⎝ 1000 m 2 /s 2 ⎠⎥ ⎦ = 2084 kWPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 44. 5-445-62E Air is expanded in an adiabatic turbine. The mass flow rate of the air and the power produced are tobe determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 The turbine is well-insulated, and thus there is no heat transfer. 3 Air is an ideal gas with constant specific heats.Properties The constant pressure specific heat of air at the average temperature of (800+250)/2=525°F is cp= 0.2485 Btu/lbm·R (Table A-2Eb). The gas constant of air is R = 0.3704 psia⋅ft3/lbm⋅R (Table A-1E).Analysis There is only one inlet and one exit, and thus m1 = m2 = m . We take the turbine as the system, & & &which is a control volume since mass crosses the boundary. The energy balance for this steady-flow systemcan be expressed in the rate form as & & E in − E out = ΔE system 0 (steady) & =0 1 24 4 3 1442443 4 Rate of net energy transfer by heat, work, and mass Rate of change in internal, kinetic, 500 psia potential, etc. energies 800°F & & E in = E out ⎛ ⎞ ⎛ 2 ⎞ Turbine V2 ⎟ = m⎜ h2 + V 2 ⎟ + Wout m⎜ h1 + 1 & & & ⎜ 2 ⎟ ⎜ 2 ⎟ ⎝ ⎠ ⎝ ⎠ 60 psia ⎛ V 2 − V 22 ⎞ ⎛ 2 2 ⎞ Wout = m⎜ h1 − h2 + 1 & & ⎟ = m⎜ c p (T1 − T2 ) + V1 − V 2 & ⎟ 250°F ⎜ 2 ⎟ ⎜ 2 ⎟ 50 ft3/s ⎝ ⎠ ⎝ ⎠The specific volume of air at the exit and the mass flow rate are RT2 (0.3704 psia ⋅ ft 3 /lbm ⋅ R)(250 + 460 R) v2 = = = 4.383 ft 3 /lbm P2 60 psia V&2 50 ft 3 /s m= & = = 11.41 kg/s v 2 4.383 ft 3 /lbm mv 2 (11.41 lbm/s)(4.383 ft 3 /lbm) & V2 = = = 41.68 ft/s A2 1.2 ft 2Similarly at the inlet, RT1 (0.3704 psia ⋅ ft 3 /lbm ⋅ R)(800 + 460 R) v1 = = = 0.9334 ft 3 /lbm P1 500 psia mv 1 (11.41 lbm/s)(0.9334 ft 3 /lbm) & V1 = = = 17.75 ft/s A1 0.6 ft 2Substituting into the energy balance equation gives ⎛ V 2 − V 22 ⎞ Wout = m⎜ c p (T1 − T2 ) + 1 & & ⎟ ⎜ 2 ⎟ ⎝ ⎠ ⎡ (17.75 ft/s) 2 − (41.68 m/s) 2 ⎛ 1 Btu/lbm ⎞⎤ = (11.41 lbm/s)⎢(0.2485 Btu/lbm ⋅ R)(800 − 250)R + ⎜ ⎟⎥ 2 ⎜ 25,037 ft 2 /s 2 ⎟ ⎢ ⎣ ⎝ ⎠⎥⎦ = 1559 kWPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 45. 5-455-63 Steam expands in a two-stage adiabatic turbine from a specified state to another state. Some steam isextracted at the end of the first stage. The power output of the turbine is to be determined.Assumptions 1 This is a steady-flow processsince there is no change with time. 2 Kinetic and 12.5 MPapotential energy changes are negligible. 3 The 550°Cturbine is adiabatic and thus heat transfer is 20 kg/snegligible.Properties From the steam tables (Table A-6) STEAM P1 = 12.5 MPa ⎫ ⎬ h1 = 3476.5 kJ/kg 20 kg/s T1 = 550°C ⎭ I II P2 = 1 MPa ⎫ ⎬ h2 = 2828.3 kJ/kg T2 = 200°C ⎭ 100 kPa P3 = 100 kPa ⎫ 100°C ⎬ h3 = 2675.8 kJ/kg 1 MPa T3 = 100°C ⎭ 200°CAnalysis The mass flow rate through the second stage is 1 kg/s m3 = m1 − m 2 = 20 − 1 = 19 kg/s & & &We take the entire turbine, including the connection part between the two stages, as the system, which is acontrol volume since mass crosses the boundary. Noting that one fluid stream enters the turbine and twofluid streams leave, the energy balance for this steady-flow system can be expressed in the rate form as & & E in − E out = ΔE system 0 (steady) & =0 1 24 4 3 1442443 4 4 Rate of net energy transfer Rate of change in internal, kinetic, by heat, work, and mass potential, etc. energies & & E in = E out & & & & m1 h1 = m 2 h2 + m3 h3 + Wout & Wout = m1 h1 − m 2 h2 − m3 h3 & & &Substituting, the power output of the turbine is & W out = (20 kg/s)(3476.5 kJ/kg) − (1 kg/s)(2828.3 kJ/kg) − (19 kg/s)(2675.8 kJ/kg) = 15,860 kWPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 46. 5-46Throttling Valves5-64C Yes.5-65C No. Because air is an ideal gas and h = h(T) for ideal gases. Thus if h remains constant, so does thetemperature.5-66C If it remains in the liquid phase, no. But if some of the liquid vaporizes during throttling, then yes.5-67C The temperature of a fluid can increase, decrease, or remain the same during a throttling process.Therefore, this claim is valid since no thermodynamic laws are violated.5-68 Refrigerant-134a is throttled by a capillary tube. The quality of the refrigerant at the exit is to bedetermined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potentialenergy changes are negligible. 3 Heat transfer to or from the fluid is negligible. 4 There are no workinteractions involved.Analysis There is only one inlet and one exit, and thus m1 = m2 = m . We take the throttling valve as the & & &system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as & & & 0 (steady) E in − E out = ΔE system =0 & & E in = E out 50°C Sat. liquid mh1 = mh2 & & h1 = h2 & & R-134asince Q ≅ W = Δke ≅ Δpe ≅ 0 .The inlet enthalpy of R-134a is, from the refrigerant tables (Table A-11), T1 = 50°C ⎫ -12°C ⎬ h1 = h f = 123.49 kJ/kg sat. liquid ⎭The exit quality is T2 = −12°C ⎫ h2 − h f 123.49 − 35.92 ⎬ x2 = = = 0.422 h2 = h1 ⎭ h fg 207.38PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 47. 5-475-69 Steam is throttled from a specified pressure to a specified state. The quality at the inlet is to bedetermined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potentialenergy changes are negligible. 3 Heat transfer to or from the fluid is negligible. 4 There are no workinteractions involved.Analysis There is only one inlet and one exit, and thus m1 = m2 = m . We take the throttling valve as the & & &system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as & & & 0 (steady) E in − E out = ΔE system =0 & & E in = E out Throttling valve mh1 = mh2 & & Steam 100 kPa h1 = h2 2 MPa 120°C & &since Q ≅ W = Δke ≅ Δpe ≅ 0 .The enthalpy of steam at the exit is (Table A-6), P2 = 100 kPa ⎫ ⎬ h2 = 2716.1 kJ/kg T2 = 120°C ⎭The quality of the steam at the inlet is (Table A-5) P2 = 2000 kPa ⎫ h2 − h f 2716.1 − 908.47 ⎬ x1 = = = 0.957 h1 = h2 = 2716.1 kJ/kg ⎭ h fg 1889.8PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 48. 5-485-70 [Also solved by EES on enclosed CD] Refrigerant-134a is throttled by a valve. The pressure andinternal energy after expansion are to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potentialenergy changes are negligible. 3 Heat transfer to or from the fluid is negligible. 4 There are no workinteractions involved.Properties The inlet enthalpy of R-134a is, from the refrigerant tables (Tables A-11 through 13), P1 = 0.8 MPa ⎫ ⎬ h1 ≅ h f @ 25o C = 86.41 kJ/kg T1 = 25°C ⎭Analysis There is only one inlet and one exit, and thus m1 = m2 = m . We take the throttling valve as the & & &system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as & & & 0 (steady) P1 = 0.8 MPa E in − E out = ΔE system =0 & & T1 = 25°C E in = E out mh1 = mh2 & & h1 = h2 R-134a & & since Q ≅ W = Δke ≅ Δpe ≅ 0 . Then, T2 = −20°C ⎫ h f = 25.49 kJ/kg, u f = 25.39 kJ/kg (h2 = h1 ) ⎬ h g = 238.41 kJ/kg u g = 218.84 kJ/kg ⎭ T2 = -20°CObviously hf < h2 <hg, thus the refrigerant exists as a saturated mixture at the exit state, and thus P2 = Psat @ -20°C = 132.82 kPaAlso, h2 − h f 86.41 − 25.49 x2 = = = 0.2861 h fg 212.91Thus, u 2 = u f + x 2 u fg = 25.39 + 0.2861 × 193.45 = 80.74 kJ/kgPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 49. 5-495-71 Steam is throttled by a well-insulated valve. The temperature drop of the steam after the expansion isto be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potentialenergy changes are negligible. 3 Heat transfer to or from the fluid is negligible. 4 There are no workinteractions involved.Properties The inlet enthalpy of steam is (Tables A-6), P1 = 8 MPa P1 = 8 MPa ⎫ ⎬ h1 = 3399.5 kJ/kg T1 = 500°C T1 = 500°C ⎭Analysis There is only one inlet and one exit, and thus m1 = m2 = m . We take & & &the throttling valve as the system, which is a control volume since mass crosses H2Othe boundary. The energy balance for this steady-flow system can be expressedin the rate form as & & & 0 (steady) E in − E out = ΔE system =0 P2 = 6 MPa & & E in = E out mh1 = mh2 & & h1 = h2 & &since Q ≅ W = Δke ≅ Δpe ≅ 0 . Then the exit temperature of steam becomes P2 = 6 MPa ⎫ T = 490.1 °C (h2 = h1 ) ⎬ 2 ⎭PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 50. 5-505-72 EES Problem 5-71 is reconsidered. The effect of the exit pressure of steam on the exit temperatureafter throttling as the exit pressure varies from 6 MPa to 1 MPa is to be investigated. The exit temperatureof steam is to be plotted against the exit pressure.Analysis The problem is solved using EES, and the solution is given below."Input information from Diagram Window"{WorkingFluid$=Steam_iapws "WorkingFluid: can be changed to ammonia or other fluids"P_in=8000 [kPa]T_in=500 [C]P_out=6000 [kPa]}$Warning off"Analysis"m_dot_in=m_dot_out "steady-state mass balance"m_dot_in=1 "mass flow rate is arbitrary"m_dot_in*h_in+Q_dot-W_dot-m_dot_out*h_out=0 "steady-state energy balance"Q_dot=0 "assume the throttle to operate adiabatically"W_dot=0 "throttles do not have any means of producing power"h_in=enthalpy(WorkingFluid$,T=T_in,P=P_in) "property table lookup"T_out=temperature(WorkingFluid$,P=P_out,h=h_out) "property table lookup"x_out=quality(WorkingFluid$,P=P_out,h=h_out) "x_out is the quality at the outlet"P[1]=P_in; P[2]=P_out; h[1]=h_in; h[2]=h_out "use arrays to place points on property plot" Pout Tout Throttle exit T vs exit P for steam [kPa] [C] 495 1000 463.1 2000 468.8 490 3000 474.3 4000 479.7 485 5000 484.9 6000 490.1 480 T out [°C] 475 470 465 460 1000 2000 3000 4000 5000 6000 P [kPa] outPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 51. 5-515-73E High-pressure air is throttled to atmospheric pressure. The temperature of air after the expansion isto be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potentialenergy changes are negligible. 3 Heat transfer to or from the fluid is negligible. 4 There are no workinteractions involved. 5 Air is an ideal gas.Analysis There is only one inlet and one exit, and thus m1 = m2 = m . We take the throttling valve as the & & &system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as & & & 0 (steady) E in − E out = ΔE system =0 P1 = 200 psia & & E in = E out T1 = 90°F mh1 = mh2 & & h1 = h2 & &since Q ≅ W = Δke ≅ Δpe ≅ 0 . AirFor an ideal gas, h = h(T). P2 = 14.7 psiaTherefore, T2 = T1 = 90°FPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 52. 5-525-74 Carbon dioxide flows through a throttling valve. The temperature change of CO2 is to be determinedif CO2 is assumed an ideal gas and a real gas.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potentialenergy changes are negligible. 3 Heat transfer to or from the fluid is negligible. 4 There are no workinteractions involved.Analysis There is only one inlet and one exit, and thus m1 = m2 = m . We take the throttling valve as the & & &system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as & & & 0 (steady) E in − E out = ΔE system =0 & & E in = E out CO2 mh1 = mh2 & & 5 MPa 100 kPa 100°C h1 = h2 & &since Q ≅ W = Δke ≅ Δpe ≅ 0 .(a) For an ideal gas, h = h(T), and therefore, T2 = T1 = 100°C ⎯ ⎯→ ΔT = T1 − T2 = 0°C(b) We obtain real gas properties of CO2 from EES software as follows P1 = 5 MPa ⎫ ⎬h1 = 34.77 kJ/kg T1 = 100°C ⎭ P2 = 100 kPa ⎫ ⎬T2 = 66.0°C h2 = h1 = 34.77 kJ/kg ⎭Note that EES uses a different reference state from the textbook for CO2 properties. The temperaturedifference in this case becomes ΔT = T1 − T2 = 100 − 66.0 = 34.0°CThat is, the temperature of CO2 decreases by 34°C in a throttling process if its real gas properties are used.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 53. 5-53Mixing Chambers and Heat Exchangers5-75C Yes, if the mixing chamber is losing heat to the surrounding medium.5-76C Under the conditions of no heat and work interactions between the mixing chamber and thesurrounding medium.5-77C Under the conditions of no heat and work interactions between the heat exchanger and thesurrounding medium.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 54. 5-545-78 A hot water stream is mixed with a cold water stream. For a specified mixture temperature, the massflow rate of cold water is to be determined.Assumptions 1 Steady operating conditions exist. 2 The mixing chamber is well-insulated so that heat lossto the surroundings is negligible. 3 Changes in the kinetic and potential energies of fluid streams arenegligible. 4 Fluid properties are constant. 5 There are no work interactions.Properties Noting that T < Tsat @ 250 kPa = 127.41°C, thewater in all three streams exists as a compressed liquid,which can be approximated as a saturated liquid at the T1 = 80°Cgiven temperature. Thus, · m1 = 0.5 kg/s h1 ≅ hf @ 80°C = 335.02 kJ/kg H2O (P = 250 kPa) h2 ≅ hf @ 20°C = 83.915 kJ/kg T3 = 42°C h3 ≅ hf @ 42°C = 175.90 kJ/kg T2 = 20°CAnalysis We take the mixing chamber as the system, · m2which is a control volume. The mass and energybalances for this steady-flow system can be expressedin the rate form asMass balance: 0 (steady) min − mout = Δmsystem & & & =0 ⎯ ⎯→ m1 + m2 = m3 & & &Energy balance: & & Ein − Eout = ΔEsystem 0 (steady) & =0 1 24 4 3 1442443 Rate of net energy transfer Rate of change in internal, kinetic, by heat, work, and mass potential, etc. energies & & Ein = Eout & & & & & m1h1 + m2 h2 = m3h3 (since Q = W = Δke ≅ Δpe ≅ 0)Combining the two relations and solving for m2 gives & m1h1 + m2h2 = (m1 + m2 )h3 & & & & h1 − h3 m2 = & & m1 h3 − h2Substituting, the mass flow rate of cold water stream is determined to be m2 = & (335.02 − 175.90 ) kJ/kg (0.5 kg/s ) = 0.865 kg/s (175.90 − 83.915) kJ/kgPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 55. 5-555-79E Liquid water is heated in a chamber by mixing it with saturated water vapor. If both streams enter atthe same rate, the temperature and quality (if saturated) of the exit stream is to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potentialenergy changes are negligible. 3 There are no work interactions. 4 The device is adiabatic and thus heattransfer is negligible.Properties From steam tables (Tables A-5E through A-6E), h1 ≅ hf @ 50°F = 18.07 Btu/lbm h2 = hg @ 50 psia = 1174.2 Btu/lbmAnalysis We take the mixing chamber as the system, which is a control volume since mass crosses theboundary. The mass and energy balances for this steady-flow system can be expressed in the rate form asMass balance: min − m out = Δmsystem & & & 0 (steady) =0 T1 = 50°F m in = m out & & H2O m1 + m 2 = m3 = 2m & & & & (P = 50 psia) m1 = m 2 = m & & & T3, x3 Sat. vaporEnergy balance: · · m2 = m 1 & & Ein − Eout = ΔEsystem 0 (steady) & =0 1 24 4 3 1442443 Rate of net energy transfer Rate of change in internal, kinetic, by heat, work, and mass potential, etc. energies & & Ein = Eout & & & & & m1h1 + m2 h2 = m3h3 (since Q = W = Δke ≅ Δpe ≅ 0)Combining the two gives mh1 + mh2 = 2mh3 or h3 = (h1 + h2 ) / 2 & & &Substituting, h3 = (18.07 + 1174.2)/2 = 596.16 Btu/lbmAt 50 psia, hf = 250.21 Btu/lbm and hg = 1174.2 Btu/lbm. Thus the exit stream is a saturated mixture sincehf < h3 < hg. Therefore, T3 = Tsat @ 50 psia = 280.99°Fand h3 − h f 596.16 − 250.21 x3 = = = 0.374 h fg 924.03PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 56. 5-565-80 Two streams of refrigerant-134a are mixed in a chamber. If the cold stream enters at twice the rate ofthe hot stream, the temperature and quality (if saturated) of the exit stream are to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potentialenergy changes are negligible. 3 There are no work interactions. 4 The device is adiabatic and thus heattransfer is negligible.Properties From R-134a tables (Tables A-11 through A-13), h1 ≅ hf @ 12°C = 68.18 kJ/kg h2 = h @ 1 MPa, 60°C = 293.38 kJ/kgAnalysis We take the mixing chamber as the system, which is a control volume since mass crosses theboundary. The mass and energy balances for this steady-flow system can be expressed in the rate form asMass balance: 0 (steady) m in − m out = Δm system & & & =0 m in = m out & & T1 = 12°C · · m1 = 2m2 m1 + m 2 = m 3 = 3m 2 since m1 = 2m 2 & & & & & & R-134aEnergy balance: (P = 1 MPa) T3, x3 & & Ein − Eout = ΔEsystem 0 (steady) & =0 1 24 4 3 1442443 T2 = 60°C Rate of net energy transfer Rate of change in internal, kinetic, by heat, work, and mass potential, etc. energies & & Ein = Eout & & & & & m1h1 + m2 h2 = m3h3 (since Q = W = Δke ≅ Δpe ≅ 0)Combining the two gives 2m 2 h1 + m 2 h2 = 3m 2 h3 or h3 = (2h1 + h2 ) / 3 & & &Substituting, h3 = (2×68.18 + 293.38)/3 = 143.25 kJ/kgAt 1 MPa, hf = 107.32 kJ/kg and hg = 270.99 kJ/kg. Thus the exit stream is a saturated mixture sincehf < h3 < hg. Therefore, T3 = Tsat @ 1 MPa = 39.37°Cand h3 − h f 143.25 − 107.32 x3 = = = 0.220 h fg 163.67PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 57. 5-575-81 EES Problem 5-80 is reconsidered. The effect of the mass flow rate of the cold stream of R-134a onthe temperature and the quality of the exit stream as the ratio of the mass flow rate of the cold stream tothat of the hot stream varies from 1 to 4 is to be investigated. The mixture temperature and quality are to beplotted against the cold-to-hot mass flow rate ratio.Analysis The problem is solved using EES, and the solution is given below."Input Data""m_frac = 2" "m_frac =m_dot_cold/m_dot_hot= m_dot_1/m_dot_2"T[1]=12 [C]P[1]=1000 [kPa]T[2]=60 [C]P[2]=1000 [kPa]m_dot_1=m_frac*m_dot_2P[3]=1000 [kPa]m_dot_1=1"Conservation of mass for the R134a: Sum of m_dot_in=m_dot_out"m_dot_1+ m_dot_2 =m_dot_3"Conservation of Energy for steady-flow: neglect changes in KE and PE""We assume no heat transfer and no work occur across the control surface."E_dot_in - E_dot_out = DELTAE_dot_cvDELTAE_dot_cv=0 "Steady-flow requirement"E_dot_in=m_dot_1*h[1] + m_dot_2*h[2] 0.5E_dot_out=m_dot_3*h[3] 0.4"Property data are given by:"h[1] =enthalpy(R134a,T=T[1],P=P[1])h[2] =enthalpy(R134a,T=T[2],P=P[2]) 0.3T[3] =temperature(R134a,P=P[3],h=h[3])x_3=QUALITY(R134a,h=h[3],P=P[3]) x3 0.2 0.1 mfrac T3 [C] x3 1 39.37 0.4491 1.333 39.37 0.3509 0 1 1.5 2 2.5 3 3.5 4 1.667 39.37 0.2772 2 39.37 0.2199 m frac 2.333 39.37 0.174 40 2.667 39.37 0.1365 3 39.37 0.1053 3.333 39.37 0.07881 38 3.667 39.37 0.05613 4 39.37 0.03649 T[3] [C] 36 34 32 30 1 1.5 2 2.5 3 3.5 4 m fracPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 58. 5-585-82 Refrigerant-134a is condensed in a water-cooled condenser. The mass flow rate of the cooling waterrequired is to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potentialenergy changes are negligible. 3 There are no work interactions. 4 Heat loss from the device to thesurroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the coldfluid.Properties The enthalpies of R-134a at the inlet and the exit states are (Tables A-11 through A-13) P3 = 700 kPa ⎫ ⎬ h3 = 308.33 kJ/kg T3 = 70°C ⎭ Water P4 = 700 kPa ⎫ ⎬ h4 = h f @ 700 kPa = 88.82 kJ/kg 1 sat. liquid ⎭ R-134aWater exists as compressed liquid at both 3states, and thus (Table A-4) h1 ≅ hf @ 15°C = 62.98 kJ/kg 4 2 h2 ≅ hf @ 25°C = 104.83 kJ/kgAnalysis We take the heat exchanger as the system, which is acontrol volume. The mass and energy balances for this steady-flow system can be expressed in the rate form asMass balance (for each fluid stream): 0 (steady) min − mout = Δmsystem & & & = 0 → min = mout → m1 = m2 = mw and m3 = m4 = mR & & & & & & & &Energy balance (for the heat exchanger): & & Ein − Eout = ΔEsystem 0 (steady) & =0 1 24 4 3 1442443 Rate of net energy transfer Rate of change in internal, kinetic, by heat, work, and mass potential, etc. energies & & Ein = Eout & & & & & & m1h1 + m3h3 = m2 h2 + m4 h4 (since Q = W = Δke ≅ Δpe ≅ 0)Combining the two, m w (h2 − h1 ) = m R (h3 − h4 ) & & h3 − h4Solving for mw : & mw = & mR & h2 − h1Substituting, (308.33 − 88.82)kJ/kg mw = & (8 kg/min) = 42.0 kg/min (104.83 − 62.98)kJ/kgPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 59. 5-595-83E [Also solved by EES on enclosed CD] Air is heated in a steam heating system. For specified flowrates, the volume flow rate of air at the inlet is to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potentialenergy changes are negligible. 3 There are no work interactions. 4 Heat loss from the device to thesurroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the coldfluid. 5 Air is an ideal gas with constant specific heats at room temperature.Properties The gas constant of air is 0.3704 psia.ft3/lbm.R (Table A-1E). The constant pressure specificheat of air is cp = 0.240 Btu/lbm·°F (Table A-2E). The enthalpies of steam at the inlet and the exit states are(Tables A-4E through A-6E) P3 = 30 psia ⎫ ⎬ h3 = 1237.9 Btu/lbm T3 = 400°F ⎭ P4 = 25 psia ⎫ ⎬ h4 ≅ h f @ 212o F = 180.21 Btu/lbm T4 = 212°F ⎭Analysis We take the entire heat exchanger as the system, which is a control volume. The mass and energybalances for this steady-flow system can be expressed in the rate form asMass balance ( for each fluid stream): AIR 0 (steady) m in − m out = Δm system & & & =0 1 m in = m out & & Steam m1 = m 2 = m a and m 3 = m 4 = m s & & & & & & 3Energy balance (for the entire heat exchanger): 4 & & Ein − Eout = ΔEsystem 0 (steady) & =0 2 1 24 4 3 1442443 Rate of net energy transfer Rate of change in internal, kinetic, by heat, work, and mass potential, etc. energies & & Ein = Eout & & & & & & m1h1 + m3h3 = m2 h2 + m4 h4 (since Q = W = Δke ≅ Δpe ≅ 0)Combining the two, m a (h2 − h1 ) = m s (h3 − h4 ) & &Solving for ma : & h3 − h4 h3 − h4 ma = & ms ≅ & & ms h2 − h1 c p (T2 − T1 )Substituting, (1237.9 − 180.21)Btu/lbm ma = & (15 lbm/min) = 1322 lbm/min = 22.04 lbm/s (0.240 Btu/lbm ⋅ °F)(130 − 80)°FAlso, RT1 (0.3704 psia ⋅ ft 3 /lbm ⋅ R )(540 R ) v1 = = = 13.61 ft 3 /lbm P1 14.7 psiaThen the volume flow rate of air at the inlet becomes V&1 = m av 1 = (22.04 lbm/s)(13.61 ft 3 /lbm) = 300 ft 3 /s &PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 60. 5-605-84 Steam is condensed by cooling water in the condenser of a power plant. If the temperature rise of thecooling water is not to exceed 10°C, the minimum mass flow rate of the cooling water required is to bedetermined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potentialenergy changes are negligible. 3 There are no work interactions. 4 Heat loss from the device to thesurroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the coldfluid. 5 Liquid water is an incompressible substance with constant specific heats at room temperature.Properties The cooling water exists as compressed liquid at both states, and its specific heat at roomtemperature is c = 4.18 kJ/kg·°C (Table A-3). The enthalpies of the steam at the inlet and the exit states are(Tables A-5 and A-6) P3 = 20 kPa ⎫ ⎬ h3 = h f + x3h fg = 251.42 + 0.95 × 2357.5 = 2491.1 kJ/kg x3 = 0.95 ⎭ P4 = 20 kPa ⎫ ⎬ h4 ≅ hf @ 20 kPa = 251.42 kJ/kg sat. liquid ⎭Analysis We take the heat exchanger as the system, which is a control volume. The mass and energybalances for this steady-flow system can be expressed in the rate form asMass balance (for each fluid stream): 0 (steady) m in − m out = Δm system & & & =0 Steam m in = m out & & 20 kPa m1 = m 2 = m w and & & & m3 = m 4 = m s & & &Energy balance (for the heat exchanger): & & Ein − Eout = ΔEsystem 0 (steady) & =0 1 24 4 3 1442443Rate of net energy transfer Rate of change in internal, kinetic,by heat, work, and mass potential, etc. energies & & Ein = Eout Water & & & & & & m1h1 + m3h3 = m2 h2 + m4 h4 (since Q = W = Δke ≅ Δpe ≅ 0)Combining the two, m w (h2 − h1 ) = m s (h3 − h4 ) & &Solving for mw : & h3 − h 4 h3 − h 4 mw = & ms ≅ & & ms h 2 − h1 c p (T2 − T1 )Substituting, (2491.1 − 251.42)kJ/kg mw = & (20,000/3600 kg/s) = 297.7 kg/s (4.18 kJ/kg⋅ o C)(10°C)PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 61. 5-615-85 Ethylene glycol is cooled by water in a heat exchanger. The rate of heat transfer in the heat exchangerand the mass flow rate of water are to be determined.Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat lossto the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to thecold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Fluid propertiesare constant.Properties The specific heats of waterand ethylene glycol are given to be 4.18 Cold Waterand 2.56 kJ/kg.°C, respectively. 20°CAnalysis (a) We take the ethylene glycoltubes as the system, which is a control Hot Glycolvolume. The energy balance for thissteady-flow system can be expressed in 80°C 40°Cthe rate form as 2 kg/s & & Ein − Eout = ΔEsystem 0 (steady) & =0 1 24 4 3 1442443Rate of net energy transfer Rate of change in internal, kinetic,by heat, work, and mass potential, etc. energies & & Ein = Eout & & mh1 = Qout + mh2 (since Δke ≅ Δpe ≅ 0) & & Qout = mc p (T1 − T2 ) &Then the rate of heat transfer becomes & Q = [ mc p (Tin − Tout )] glycol = ( 2 kg/s)(2.56 kJ/kg. °C)(80 °C − 40°C) = 204.8 kW &(b) The rate of heat transfer from glycol must be equal to the rate of heat transfer to the water. Then, & Q 204.8 kJ/s & Q = [mc p (Tout − Tin )]water ⎯ mwater = & ⎯→ & = = 1.4 kg/s c p (Tout − Tin ) (4.18 kJ/kg.°C)(55°C − 20°C)PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 62. 5-625-86 EES Problem 5-85 is reconsidered. The effect of the inlet temperature of cooling water on the massflow rate of water as the inlet temperature varies from 10°C to 40°C at constant exit temperature) is to beinvestigated. The mass flow rate of water is to be plotted against the inlet temperature.Analysis The problem is solved using EES, and the solution is given below."Input Data"{T_w[1]=20 [C]}T_w[2]=55 [C] "w: water"m_dot_eg=2 [kg/s] "eg: ethylene glycol"T_eg[1]=80 [C]T_eg[2]=40 [C]C_p_w=4.18 [kJ/kg-K]C_p_eg=2.56 [kJ/kg-K]"Conservation of mass for the water: m_dot_w_in=m_dot_w_out=m_dot_w""Conservation of mass for the ethylene glycol: m_dot_eg_in=m_dot_eg_out=m_dot_eg""Conservation of Energy for steady-flow: neglect changes in KE and PE in each mass steam""We assume no heat transfer and no work occur across the control surface."E_dot_in - E_dot_out = DELTAE_dot_cvDELTAE_dot_cv=0 "Steady-flow requirement"E_dot_in=m_dot_w*h_w[1] + m_dot_eg*h_eg[1]E_dot_out=m_dot_w*h_w[2] + m_dot_eg*h_eg[2]Q_exchanged =m_dot_eg*h_eg[1] - m_dot_eg*h_eg[2]"Property data are given by:"h_w[1] =C_p_w*T_w[1] "liquid approximation applied for water and ethylene glycol"h_w[2] =C_p_w*T_w[2]h_eg[1] =C_p_eg*T_eg[1]h_eg[2] =C_p_eg*T_eg[2] 3.5 mw [kg/s] Tw,1 [C] 1.089 10 1.225 15 3 1.4 20 mw [kg/s] 1.633 25 2.5 1.96 30 2.45 35 3.266 40 2 1.5 1 10 15 20 25 30 35 40 Tw[1] [C]PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 63. 5-635-87 Oil is to be cooled by water in a thin-walled heat exchanger. The rate of heat transfer in the heatexchanger and the exit temperature of water is to be determined.Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat lossto the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to thecold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Fluidproperties are constant.Properties The specific heats of water and oil aregiven to be 4.18 and 2.20 kJ/kg.°C, respectively. Hot oilAnalysis We take the oil tubes as the system, which 150°Cis a control volume. The energy balance for this 2 kg/ssteady-flow system can be expressed in the rate Coldform as water 22°C & & Ein − Eout = ΔEsystem 0 (steady) & =0 1.5 kg/s 1 24 4 3 1442443Rate of net energy transfer Rate of change in internal, kinetic,by heat, work, and mass potential, etc. energies 40°C & & Ein = Eout & & mh1 = Qout + mh2 (since Δke ≅ Δpe ≅ 0) & & Qout = mc p (T1 − T2 ) &Then the rate of heat transfer from the oil becomes & Q = [mc p (Tin − Tout )]oil = ( 2 kg/s)(2.2 kJ/kg. °C)(150 °C − 40°C) = 484 kW &Noting that the heat lost by the oil is gained by the water, the outlet temperature of the water is determinedfrom & Q 484 kJ/s & Q = [mc p (Tout − Tin )]water ⎯ Tout = Tin + & ⎯→ = 22°C + = 99.2°C & mwaterc p (1.5 kg/s)(4.18 kJ/kg.°C)PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 64. 5-645-88 Cold water is heated by hot water in a heat exchanger. The rate of heat transfer and the exittemperature of hot water are to be determined.Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat lossto the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to thecold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Fluidproperties are constant.Properties The specific heats of cold and hot water are given to be 4.18 and 4.19 kJ/kg.°C, respectively.Analysis We take the cold water tubes as the system, whichis a control volume. The energy balance for this steady- Coldflow system can be expressed in the rate form as Water & & Ein − Eout = ΔEsystem 0 (steady) & =0 15°C 1 24 4 3 1442443 Hot water 0 60 k /Rate of net energy transfer Rate of change in internal, kinetic,by heat, work, and mass potential, etc. energies & & 100°C Ein = Eout 3 kg/s & Qin + mh1 = mh2 (since Δke ≅ Δpe ≅ 0) & & & = mc (T − T ) Qin & p 2 1Then the rate of heat transfer to the cold water in this heat exchanger becomes & Q = [ mc p (Tout − Tin )]cold water = (0.60 kg/s)(4.18 kJ/kg. °C)(45°C − 15°C) = 75.24 kW &Noting that heat gain by the cold water is equal to the heat loss by the hot water, the outlet temperature ofthe hot water is determined to be Q& 75.24 kW & Q = [mc p (Tin − Tout )]hot water ⎯ Tout = Tin − & ⎯→ = 100°C − = 94.0°C & cp m (3 kg/s)(4.19 kJ/kg.°C)PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 65. 5-655-89 Air is preheated by hot exhaust gases in a cross-flow heat exchanger. The rate of heat transfer and theoutlet temperature of the air are to be determined.Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat lossto the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to thecold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Fluidproperties are constant.Properties The specific heats of air and combustion gases are given to be 1.005 and 1.10 kJ/kg.°C,respectively.Analysis We take the exhaust pipes as the system, which is a control volume. The energy balance for thissteady-flow system can be expressed in the rate form as & & Ein − Eout = ΔEsystem 0 (steady) & & & = 0 → Ein = Eout 1 24 4 3 1442443 Rate of net energy transfer Rate of change in internal, kinetic, by heat, work, and mass potential, etc. energies & & mh1 = Qout + mh2 (since Δke ≅ Δpe ≅ 0) & & Qout = mc p (T1 − T2 ) &Then the rate of heat transfer from the Airexhaust gases becomes 95 kPa & Q = [ mc p (Tin − Tout )] gas & 20°C 0.8 m3/s = (1.1 kg/s)(1.1 kJ/kg.°C)(180°C − 95°C) = 102.85 kWThe mass flow rate of air is Exhaust gases PV& (95 kPa)(0.8 m 3 /s) 1.1 kg/s, 95°C m= & = = 0.904 kg/s RT (0.287 kPa.m 3 /kg.K) × 293 KNoting that heat loss by the exhaust gases is equal to the heat gain by the air, the outlet temperature of theair becomes Q& 102.85 kW & & Q = mc p (Tc,out − Tc,in ) ⎯ Tc,out = Tc,in + ⎯→ = 20°C + = 133.2°C & mc p (0.904 kg/s)(1.005 kJ/kg.°C)PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 66. 5-665-90 An adiabatic open feedwater heater mixes steam with feedwater. The outlet mass flow rate and theoutlet velocity are to be determined.Assumptions Steady operating conditions exist.Properties From a mass balance Water m3 = m1 + m 2 = 0.2 + 10 = 10.2 kg/s & & & 100 kPaThe specific volume at the exit is (Table A-4) 50°C 10 kg/s P3 = 100 kPa ⎫ 2 3 100 kPa ⎬ v 3 ≅ v f @ 60°C = 0.001017 m /kg 3 T3 = 60°C ⎭ 60°C Steam 1The exit velocity is then 100 kPa 160°C m3v 3 4m3v 3 & & 0.2 kg/s V3 = = A3 πD 2 4(10.2 kg/s)(0.001017 m 3 /kg) = π (0.03 m) 2 = 14.68 m/sPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 67. 5-675-91E An adiabatic open feedwater heater mixes steam with feedwater. The outlet mass flow rate and theoutlet velocity are to be determined for two exit temperatures.Assumptions Steady operating conditions exist.Properties From a mass balance m 3 = m1 + m 2 = 0.1 + 2 = 2.1 lbm/s & & & Water 10 psiaThe specific volume at the exit is (Table A-4E) 100°F P3 = 10 psia ⎫ 2 lbm/s 3 2 ⎬ v 3 ≅ v f @ 120° F = 0.01620 ft /lbm 10 psia T3 = 120°F ⎭ 3 120°F Steam 1The exit velocity is then 10 psia m 3v 3 4m3v 3 & & 200°F V3 = = 0.1 lbm/s A3 πD 2 4(2.1 lbm/s)(0.01620 ft 3 /lbm) = = 0.1733 ft/s π (0.5 ft) 2When the temperature at the exit is 180°F, we have m3 = m1 + m 2 = 0.1 + 2 = 2.1 lbm/s & & & P3 = 10 psia ⎫ 3 ⎬ v 3 ≅ v f @ 180° F = 0.01651 ft /lbm T3 = 180°F ⎭ m 3v 3 4m 3v 3 4(2.1 lbm/s)(0.01651 ft 3 /lbm) & & V3 = = = = 0.1766 ft/s A3 πD 2 π (0.5 ft) 2The mass flow rate at the exit is same while the exit velocity slightly increases when the exit temperature is180°F instead of 120°F.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 68. 5-685-92 Water is heated by hot oil in a heat exchanger. The rate of heat transfer in the heat exchanger and theoutlet temperature of oil are to be determined.Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat lossto the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to thecold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Fluidproperties are constant.Properties The specific heats of water and oil are given to be 4.18 and 2.3 kJ/kg.°C, respectively.Analysis We take the cold water tubes as the system, whichis a control volume. The energy balance for this steady- Oilflow system can be expressed in the rate form as 170°C & & E in − E out = ΔE system 0 (steady) & =0 10 kg/s 1 24 4 3 1442443 4Rate of net energy transfer Rate of change in internal, kinetic, 70°Cby heat, work, and mass potential, etc. energies & & E in = E out Water & Qin + mh1 = mh2 (since Δke ≅ Δpe ≅ 0) & & 20°C & = mc (T − T ) 4.5 kg/s Qin & p 2 1Then the rate of heat transfer to the cold water in thisheat exchanger becomes & Q = [ mc p (Tout − Tin )]water = ( 4.5 kg/s)(4.18 kJ/kg. °C)(70 °C − 20°C) = 940.5 kW &Noting that heat gain by the water is equal to the heat loss by the oil, the outlet temperature of the hot wateris determined from Q& 940.5 kW & Q = [mc p (Tin − Tout )]oil ⎯ Tout = Tin − & ⎯→ = 170°C − = 129.1°C & mc p (10 kg/s)(2.3 kJ/kg.°C)PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 69. 5-695-93E Steam is condensed by cooling water in a condenser. The rate of heat transfer in the heat exchangerand the rate of condensation of steam are to be determined.Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat lossto the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to thecold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Fluid propertiesare constant.Properties The specific heat of water is 1.0 Btu/lbm.°F(Table A-3E). The enthalpy of vaporization of water at Steam85°F is 1045.2 Btu/lbm (Table A-4E). 85°FAnalysis We take the tube-side of the heat exchanger 73°Fwhere cold water is flowing as the system, which is acontrol volume. The energy balance for this steady-flowsystem can be expressed in the rate form as & & Ein − Eout = ΔEsystem 0 (steady) & =0 1 24 4 3 1442443 Rate of net energy transfer by heat, work, and mass Rate of change in internal, kinetic, 60°F potential, etc. energies & & Ein = Eout Water & Qin + mh1 = mh2 (since Δke ≅ Δpe ≅ 0) & & & Qin = mc p (T2 − T1 ) & 85°FThen the rate of heat transfer to the cold water in this heat exchanger becomes & Q = [ mc p (Tout − Tin )] water = (138 lbm/s)(1.0 Btu/lbm.°F)(73°F − 60°F) = 1794 Btu/s &Noting that heat gain by the water is equal to the heat loss by the condensing steam, the rate ofcondensation of the steam in the heat exchanger is determined from Q & 1794 Btu/s & Q = (mh fg )steam = ⎯ msteam = & ⎯→ & = = 1.72 lbm/s h fg 1045.2 Btu/lbmPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 70. 5-705-94 Two streams of cold and warm air are mixed in a chamber. If the ratio of hot to cold air is 1.6, themixture temperature and the rate of heat gain of the room are to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potentialenergy changes are negligible. 3 There are no work interactions. 4 The device is adiabatic and thus heattransfer is negligible.Properties The gas constant of air Coldis R = 0.287 kPa.m3/kg.K. The airenthalpies of air are obtained from 5°Cair table (Table A-17) as Room 24°C h1 = h @278 K = 278.13 kJ/kg Warm h2 = h @ 307 K = 307.23 kJ/kg air 34°C hroom = h @ 297 K = 297.18 kJ/kgAnalysis (a) We take the mixing chamber as the system, which is a control volume since mass crosses theboundary. The mass and energy balances for this steady-flow system can be expressed in the rate form asMass balance: 0 (steady) min − mout = Δmsystem & & & = 0 → min = mout → m1 + 1.6m1 = m3 = 2.6m1 since m2 = 1.6m1 & & & & & & & &Energy balance: & & Ein − Eout = ΔEsystem 0 (steady) & =0 1 24 4 3 1442443 Rate of net energy transfer Rate of change in internal, kinetic, by heat, work, and mass potential, etc. energies & & Ein = Eout m1h1 + m2 h2 = m3h3 & & & & & (since Q ≅ W ≅ Δke ≅ Δpe ≅ 0)Combining the two gives m1 h1 + 1.6m1 h2 = 2.6m1 h3 or h3 = (h1 + 1.6h2 ) / 2.6 & & &Substituting, h3 = (278.13 +1.6× 307.23)/2.6 = 296.04 kJ/kgFrom air table at this enthalpy, the mixture temperature is T3 = T @ h = 296.04 kJ/kg = 295.9 K = 22.9°C(b) The mass flow rates are determined as follows RT1 (0.287 kPa ⋅ m3/kg ⋅ K)(5 + 273 K) v1 = = = 0.7599 m3 / kg P 105 kPa V& 1.25 m3/s m1 = 1 = & = 1.645 kg/s v1 0.7599 m3/kg m3 = 2.6m1 = 2.6(1.645 kg/s) = 4.277 kg/s & &The rate of heat gain of the room is determined from & Qcool = m3 (hroom − h3 ) = (4.277 kg/s)(297.18 − 296.04) kJ/kg = 4.88 kW &PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 71. 5-715-95 A heat exchanger that is not insulated is used to produce steam from the heat given up by the exhaustgases of an internal combustion engine. The temperature of exhaust gases at the heat exchanger exit andthe rate of heat transfer to the water are to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potentialenergy changes are negligible. 3 There are no work interactions. 4 Exhaust gases are assumed to have airproperties with constant specific heats.Properties The constant pressure specific heat of the exhaust gases is taken to be cp = 1.045 kJ/kg·°C(Table A-2). The inlet and exit enthalpies of water are (Tables A-4 and A-5) Tw,in = 15°C ⎫ ⎬ hw,in = 62.98 kJ/kg x = 0 (sat. liq.)⎭ Exh. gas Q Pw,out = 2 MPa ⎫ 400°C ⎬ hw,out = 2798.3 kJ/kg x = 1 (sat. vap.) ⎭ HeatAnalysis We take the entire heat exchanger as the system, exchangerwhich is a control volume. The mass and energy balances for 2 MPa Waterthis steady-flow system can be expressed in the rate form as sat. vap. 15°CMass balance (for each fluid stream): 0 (steady) min − mout = Δmsystem & & & = 0⎯ ⎯→ min = mout & &Energy balance (for the entire heat exchanger): & & Ein − Eout = ΔEsystem 0 (steady) & =0 1 24 4 3 1442443 Rate of net energy transfer Rate of change in internal, kinetic, by heat, work, and mass potential, etc. energies & & Ein = Eout & & & & & & mexh hexh,in + mw hw,in = mexh hexh,out + mw hw,out + Qout (since W = Δke ≅ Δpe ≅ 0)or & & & & & mexh c pTexh, in + mw hw,in = mexh c pTexh, out + mw hw, out + QoutNoting that the mass flow rate of exhaust gases is 15 times that of the water, substituting gives 15m w (1.045 kJ/kg.°C)(400°C) + m w (62.98 kJ/kg) & & & (1) = 15m w (1.045 kJ/kg.°C)Texh,out + m w (2798.3 kJ/kg) + Qout & &The heat given up by the exhaust gases and heat picked up by the water are & Q exh = m exh c p (Texh,in − Texh,out ) = 15m w (1.045 kJ/kg. °C)( 400 − Texh,out )°C & & (2) & Q w = m w (hw,out − hw,in ) = m w (2798.3 − 62.98)kJ/kg & & (3)The heat loss is & & & Qout = f heat loss Qexh = 0.1Qexh (4)The solution may be obtained by a trial-error approach. Or, solving the above equations simultaneouslyusing EES software, we obtain & Texh,out = 206.1 °C, Qw = 97.26 kW, m w = 0.03556 kg/s, mexh = 0.5333 kg/s & &PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 72. 5-725-96 Two streams of water are mixed in an insulated chamber. The temperature of the exit stream is to bedetermined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potentialenergy changes are negligible. 3 There are no work interactions. 4 The device is adiabatic and thus heattransfer is negligible.Properties From the steam tables (Table A-4), h1 ≅ hf @ 90°C = 377.04 kJ/kg h2 ≅ hf @ 50°C = 209.34 kJ/kgAnalysis We take the mixing chamber as the system, which is a control volume since mass crosses theboundary. The mass and energy balances for this steady-flow system can be expressed in the rate form as 0 (steady) m in − m out = Δm system & & & =0Mass balance: m in = m out & & Water m1 + m 2 = m 3 & & & 90°C 30 kg/s 1Energy balance: 3 & & E in − E out = ΔE system 0 (steady) & =0 2 1 24 4 3 1442443 4 4 Water Rate of net energy transfer Rate of change in internal, kinetic, by heat, work, and mass potential, etc. energies 50°C & & 200 kg/s E in = E out & & & & & m1 h1 + m 2 h2 = m3 h3 (since Q = W = Δke ≅ Δpe ≅ 0)Solving for the exit enthalpy, m1 h1 + m 2 h2 (30)(377.04) + (200)(209.34) & & h3 = = = 231.21 kJ/kg m1 + m 2 & & 30 + 200The temperature corresponding to this enthalpy is T3 =55.2°C (Table A-4)PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 73. 5-735-97 A chilled-water heat-exchange unit is designed to cool air by water. The maximum water outlettemperature is to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potentialenergy changes are negligible. 3 There are no work interactions. 4 Heat loss from the device to thesurroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the coldfluid. 5 Air is an ideal gas with constant specific heats at room temperature.Properties The gas constant of air is 0.287 kPa.m3/kg.K (Table A-1). The constant pressure specific heat ofair at room temperature is cp = 1.005 kJ/kg·°C (Table A-2a). The specific heat of water is 4.18 kJ/kg⋅K(Table A-3).Analysis The water temperature at the heat exchanger exit will be maximum when all the heat released bythe air is picked up by the water. First, the inlet specific volume and the mass flow rate of air are RT1 (0.287 kPa ⋅ m 3 /kg ⋅ K )(303 K ) v1 = = = 0.8696 m 3 /kg P1 100 kPa V& 5 m 3 /s ma = 1 = & = 5.750 kg/s v 1 0.8696 m 3 /kgWe take the entire heat exchanger as the system, which is a control volume. The mass and energy balancesfor this steady-flow system can be expressed in the rate form asMass balance ( for each fluid stream): 0 (steady) min − mout = Δmsystem & & & = 0 → min = mout → m1 = m3 = m a and m 2 = m 4 = m w & & & & & & & &Energy balance (for the entire heat exchanger): & & E in − E out = ΔE system 0 (steady) & =0 1 24 4 3 1442443 4 Rate of net energy transfer Rate of change in internal, kinetic, by heat, work, and mass potential, etc. energies & & E in = E out & & & & & & m1 h1 + m 2 h2 = m3 h3 + m 4 h4 (since Q = W = Δke ≅ Δpe ≅ 0)Combining the two, m a (h1 − h3 ) = m w (h4 − h2 ) & & m a c p ,a (T1 − T3 ) = m w c p , w (T4 − T2 ) & &Solving for the exit temperature of water, m a c p , a (T1 − T3 ) & (5.750 kg/s)(1.005 kJ/kg ⋅ °C)(30 − 18)°C T 4 = T2 + = 8°C + = 16.3°C & m w c p,w (2 kg/s)(4.18 kJ/kg ⋅ °C)PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 74. 5-74Pipe and duct Flow5-98 Saturated liquid water is heated in a steam boiler at a specified rate. The rate of heat transfer in theboiler is to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potentialenergy changes are negligible. 3 There are no work interactions.Analysis We take the pipe in which the water is heated as the system, which is a control volume. Theenergy balance for this steady-flow system can be expressed in the rate form as & & E in − E out = ΔE system 0 (steady) & =0 1 24 4 3 1442443 4 .Rate of net energy transfer Rate of change in internal, kinetic,by heat, work, and mass potential, etc. energies Qin & & E in = E out 5 MPa, sat. liq. Water 5 MPa & & mh1 + Qin = mh2 & 10 kg/s 350°C & Qin = m(h2 − h1 ) &The enthalpies of water at the inlet and exit of the boiler are (Table A-5, A-6). P1 = 5 MPa ⎫ ⎬ h1 ≅ h f @ 5 MPa = 1154.5 kJ/kg x=0 ⎭ P2 = 5 MPa ⎫ ⎬ h2 = 3069.3 kJ/kg T2 = 350°C ⎭Substituting, & Qin = (10 kg/s)(3069.3 − 1154.5)kJ/kg = 19,150 kWPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 75. 5-755-99 Air at a specified rate is heated by an electrical heater. The current is to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potentialenergy changes are negligible. 3 The heat losses from the air is negligible.Properties The gas constant of air is 0.287 kPa.m3/kg.K (Table A-1). The constant pressure specific heat ofair at room temperature is cp = 1.005 kJ/kg·°C (Table A-2a).Analysis We take the pipe in which the air is heated as the system, which is a control volume. The energybalance for this steady-flow system can be expressed in the rate form as & & E in − E out = ΔE system 0 (steady) & =0 1 24 4 3 1442443 4 Rate of net energy transfer by heat, work, and mass Rate of change in internal, kinetic, . potential, etc. energies We,in & & E in = E out & & mh1 + We,in = mh2 & 100 kPa, 15°C Air 100 kPa & We,in = m(h2 − h1 ) & 0.3 m3/s 30°C VI = mc p (T2 − T1 ) &The inlet specific volume and the mass flow rate of air are RT1 (0.287 kPa ⋅ m 3 /kg ⋅ K )(288 K ) v1 = = = 0.8266 m 3 /kg P1 100 kPa V& 0.3 m 3 /s m= 1 = & = 0.3629 kg/s v 1 0.8266 m 3 /kgSubstituting into the energy balance equation and solving for the current gives mc p (T2 − T1 ) & (0.3629 kg/s)(1.005 kJ/kg ⋅ K)(30 − 15)K ⎛ 1000 VI ⎞ I= = ⎜ ⎟ = 49.7 Amperes V 110 V ⎝ 1 kJ/s ⎠PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 76. 5-765-100E The cooling fan of a computer draws air, which is heated in the computer by absorbing the heat ofPC circuits. The electrical power dissipated by the circuits is to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potentialenergy changes are negligible. 3 All the heat dissipated by the circuits are picked up by the air drawn bythe fan.Properties The gas constant of air is 0.3704 psia·ft3/lbm·R (Table A-1E). The constant pressure specificheat of air at room temperature is cp = 0.240 Btu/lbm·°F (Table A-2Ea).Analysis We take the pipe in which the air is heated as the system, which is a control volume. The energybalance for this steady-flow system can be expressed in the rate form as & & E in − E out = ΔE system 0 (steady) & =0 1 24 4 3 1442443 4 Rate of net energy transfer Rate of change in internal, kinetic, . by heat, work, and mass potential, etc. energies We,in & & E in = E out & mh1 + We,in = mh2 & & 14.7 psia, 70°F Air 14.7 psia & 0.5 ft3/s 80°F We,in = m(h2 − h1 ) & & We,in = mc p (T2 − T1 ) &The inlet specific volume and the mass flow rate of air are RT1 (0.3704 psia ⋅ ft 3 /lbm ⋅ R )(530 R ) v1 = = = 13.35 ft 3 /lbm P1 14.7 psia V& 0.5 ft 3 /s m= 1 = & = 0.03745 lbm/s v 1 13.35 ft 3 /lbmSubstituting, & ⎛ 1 kW ⎞ W e,out = (0.03745 lbm/s)(0.240 Btu/lbm ⋅ R)(80 − 70)Btu/lbm⎜ ⎟ = 0.0948 kW ⎝ 0.94782 Btu ⎠PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 77. 5-775-101 A desktop computer is to be cooled safely by a fan in hot environments and high elevations. The airflow rate of the fan and the diameter of the casing are to be determined.Assumptions 1 Steady operation under worst conditions is considered. 2 Air is an ideal gas with constantspecific heats. 3 Kinetic and potential energy changes are negligible.Properties The specific heat of air at the average temperature of Tavg = (45+60)/2 = 52.5°C = 325.5 K is cp= 1.0065 kJ/kg.°C. The gas constant for air is R = 0.287 kJ/kg.K (Table A-2).Analysis The fan selected must be able to meet the cooling requirements of the computer at worstconditions. Therefore, we assume air to enter the computer at 66.63 kPa and 45°C, and leave at 60°C. We take the air space in the computer as the system, which is a control volume. The energybalance for this steady-flow system can be expressed in the rate form as & & Ein − Eout = ΔEsystem 0 (steady) & =0 1 24 4 3 1442443 Rate of net energy transfer Rate of change in internal, kinetic, by heat, work, and mass potential, etc. energies & & Ein = Eout & Qin + mh1 = mh2 (since Δke ≅ Δpe ≅ 0) & & & Qin = mc p (T2 − T1 ) &Then the required mass flow rate of air to absorb heat at a rateof 60 W is determined to be & Q 60 W & & Q = mc p (Tout − Tin ) → m = & = c p (Tout − Tin ) (1006.5 J/kg.°C)(60 - 45)°C = 0.00397 kg/s = 0.238 kg/minThe density of air entering the fan at the exit and its volume flow rate are P 66.63 kPa ρ= = = 0.6972 kg/m 3 RT (0.287 kPa.m 3 /kg.K)(60 + 273)K m 0.238 kg/min & V& = = = 0.341 m 3 /min ρ 0.6972 kg/m 3For an average exit velocity of 110 m/min, the diameter of the casing of the fan is determined from πD 2 4V& (4)(0.341 m 3 /min) V& = AcV = V →D= = = 0.063 m = 6.3 cm 4 πV π (110 m/min)PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 78. 5-785-102 A desktop computer is to be cooled safely by a fan in hot environments and high elevations. The airflow rate of the fan and the diameter of the casing are to be determined.Assumptions 1 Steady operation under worst conditions is considered. 2 Air is an ideal gas with constantspecific heats. 3 Kinetic and potential energy changes are negligible.Properties The specific heat of air at the average temperature of Tave = (45+60)/2 = 52.5°C is cp = 1.0065kJ/kg.°C The gas constant for air is R = 0.287 kJ/kg.K (Table A-2).Analysis The fan selected must be able to meet the cooling requirements of the computer at worstconditions. Therefore, we assume air to enter the computer at 66.63 kPa and 45°C, and leave at 60°C. We take the air space in the computer as the system, which is a control volume. The energybalance for this steady-flow system can be expressed in the rate form as & & Ein − Eout = ΔEsystem 0 (steady) & =0 1 24 4 3 1442443 Rate of net energy transfer Rate of change in internal, kinetic, by heat, work, and mass potential, etc. energies & & Ein = Eout & Qin + mh1 = mh2 (since Δke ≅ Δpe ≅ 0) & & & Qin = mc p (T2 − T1 ) &Then the required mass flow rate of air to absorb heat at a rateof 100 W is determined to be & & Q = mc p (Tout − Tin ) & Q 100 W m= & = = 0.006624 kg/s = 0.397 kg/min c p (Tout − Tin ) (1006.5 J/kg.°C)(60 - 45)°CThe density of air entering the fan at the exit and its volume flow rate are P 66.63 kPa ρ= = = 0.6972 kg/m 3 RT (0.287 kPa.m 3 /kg.K)(60 + 273)K m 0.397 kg/min & V& = = = 0.57 m 3 /min ρ 0.6972 kg/m 3For an average exit velocity of 110 m/min, the diameter of the casing of the fan is determined from πD 2 4V& (4)(0.57 m 3 /min) V& = AcV = V⎯ ⎯→ D = = = 0.081 m = 8.1 cm 4 πV π (110 m/min)PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 79. 5-795-103E Electronic devices mounted on a cold plate are cooled by water. The amount of heat generated bythe electronic devices is to be determined.Assumptions 1 Steady operating conditions exist. 2 About 15 percent of the heat generated is dissipatedfrom the components to the surroundings by convection and radiation. 3 Kinetic and potential energychanges are negligible.Properties The properties of water at room temperature are ρ = 62.1 lbm/ft3 and cp = 1.00 Btu/lbm.°F(Table A-3E).Analysis We take the tubes of the cold plate to be the system, which is a control volume. The energybalance for this steady-flow system can be expressed in the rate form as & & Ein − Eout = ΔEsystem 0 (steady) & =0 1 24 4 3 1442443 Cold plate Water Rate of net energy transfer Rate of change in internal, kinetic, by heat, work, and mass potential, etc. energies inlet & & Ein = Eout 1 & Qin + mh1 = mh2 (since Δke ≅ Δpe ≅ 0) & & & = mc (T − T ) Qin & p 2 1 2Then mass flow rate of water and the rate of heatremoval by the water are determined to be πD 2 π (0.25 / 12 ft) 2 m = ρAV = ρ & V = (62.1 lbm/ft3 ) (60 ft/min) = 1.270 lbm/min = 76.2 lbm/h 4 4 & & Q = mc p (Tout − Tin ) = (76.2 lbm/h)(1.00 Btu/lbm.°F)(105 - 95)°F = 762 Btu/hwhich is 85 percent of the heat generated by the electronic devices. Then the total amount of heat generatedby the electronic devices becomes & 762 Btu/h = 896 Btu/h = 263 W Q= 0.85PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 80. 5-805-104 [Also solved by EES on enclosed CD] The components of an electronic device located in a horizontalduct of rectangular cross section are cooled by forced air. The heat transfer from the outer surfaces of theduct is to be determined.Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant specific heats at roomtemperature. 3 Kinetic and potential energy changes are negligibleProperties The gas constant of air is R = 0.287 kJ/kg.°C (Table A-1). The specific heat of air at roomtemperature is cp = 1.005 kJ/kg.°C (Table A-2).Analysis The density of air entering the duct and the mass flow rateare 40°C P 101.325 kPa ρ= = = 1.165 kg/m 3 RT (0.287 kPa.m 3 /kg.K)(30 + 273)K m = ρV& = (1.165 kg/m 3 )(0.6 m 3 / min) = 0.700 kg/min &We take the channel, excluding theelectronic components, to be the system, Air 180 Wwhich is a control volume. The energy 30°Cbalance for this steady-flow system can be 0.6 m3/minexpressed in the rate form as 25°C & & & 0 (steady) Ein − Eout = ΔEsystem =0 1 24 4 3 1442443Rate of net energy transfer Rate of change in internal, kinetic,by heat, work, and mass potential, etc. energies & & Ein = Eout & Qin + mh1 = mh2 (since Δke ≅ Δpe ≅ 0) & & & Qin = mc p (T2 − T1 ) &Then the rate of heat transfer to the air passing through the duct becomes & Qair = [ mc p (Tout − Tin )]air = (0.700 / 60 kg/s )(1.005 kJ/kg. °C)( 40 − 30)°C = 0.117 kW = 117 W &The rest of the 180 W heat generated must be dissipated through the outer surfaces of the duct by naturalconvection and radiation, & & & Qexternal = Qtotal − Qinternal = 180 − 117 = 63 WPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 81. 5-815-105 The components of an electronic device located in a horizontal duct of circular cross section iscooled by forced air. The heat transfer from the outer surfaces of the duct is to be determined.Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant specific heats at roomtemperature. 3 Kinetic and potential energy changes are negligibleProperties The gas constant of air is R = 0.287 kJ/kg.°C (Table A-1). The specific heat of air at roomtemperature is cp = 1.005 kJ/kg.°C (Table A-2).Analysis The density of air entering the duct and the mass flow rate are P 101.325 kPa ρ= = = 1.165 kg/m 3 RT (0.287 kPa.m 3 /kg.K)(30 + 273)K m = ρV& = (1.165 kg/m 3 )(0.6 m 3 / min) = 0.700 kg/min &We take the channel, excluding the electronic components, to be the system, which is a control volume.The energy balance for this steady-flow system can be expressed in the rate form as & & Ein − Eout = ΔEsystem 0 (steady) & =0 1 24 4 3 1442443Rate of net energy transfer Rate of change in internal, kinetic, 1 2by heat, work, and mass potential, etc. energies Air & & Ein = Eout & Qin + mh1 = mh2 (since Δke ≅ Δpe ≅ 0) & & 30°C 0.6 m3/s & = mc (T − T ) Qin & p 2 1Then the rate of heat transfer to the air passing through the duct becomes & Qair = [ mc p (Tout − Tin )]air = (0.700 / 60 kg/s )(1.005 kJ/kg. °C)( 40 − 30)°C = 0.117 kW = 117 W &The rest of the 180 W heat generated must be dissipated through the outer surfaces of the duct by naturalconvection and radiation, & & & Qexternal = Qtotal − Qinternal = 180 − 117 = 63 WPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 82. 5-825-106E Water is heated in a parabolic solar collector. The required length of parabolic collector is to bedetermined.Assumptions 1 Steady operating conditions exist. 2 Heat loss from the tube is negligible so that the entiresolar energy incident on the tube is transferred to the water. 3 Kinetic and potential energy changes arenegligibleProperties The specific heat of water at room temperature is cp = 1.00 Btu/lbm.°F (Table A-2E).Analysis We take the thin aluminum tube to be the system, which is a control volume. The energy balancefor this steady-flow system can be expressed in the rate form as & & Ein − Eout = ΔEsystem 0 (steady) & =0 1 24 4 3 1442443 Rate of net energy transfer 2 by heat, work, and mass Rate of change in internal, kinetic, potential, etc. energies 1 Water & & Ein = Eout & Qin + mh1 = mh2 (since Δke ≅ Δpe ≅ 0) & & 55°F 180°F & =m 4 lbm/s Qin & water c p (T2 − T1 )Then the total rate of heat transfer to the water flowing through the tube becomes & Q total = mc p (Te − Ti ) = ( 4 lbm/s )(1.00 Btu/lbm.°F)(180 − 55)°F = 500 Btu/s = 1,800,000 Btu/h &The length of the tube required is & Qtotal 1,800,000 Btu/h L= = = 4500 ft & Q 400 Btu/h.ftPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 83. 5-835-107 Air enters a hollow-core printed circuit board. The exit temperature of the air is to be determined.Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant specific heats at roomtemperature. 3 The local atmospheric pressure is 1 atm. 4 Kinetic and potential energy changes arenegligible.Properties The gas constant of air is R =0.287 kJ/kg.°C (Table A-1). The specificheat of air at room temperature is cp = 1 21.005 kJ/kg.°C (Table A-2). Air 32°CAnalysis The density of air entering the 0.8 L/sduct and the mass flow rate are P 101.325 kPa ρ= = = 1.16 kg/m 3 RT (0.287 kPa.m 3 /kg.K)(32 + 273)K m = ρV& = (1.16 kg/m 3 )(0.0008 m 3 / s) = 0.000928 kg/s &We take the hollow core to be the system, which is a control volume. The energy balance for this steady-flow system can be expressed in the rate form as & & Ein − Eout = ΔEsystem 0 (steady) & =0 1 24 4 3 1442443 Rate of net energy transfer Rate of change in internal, kinetic, by heat, work, and mass potential, etc. energies & & Ein = Eout & Qin + mh1 = mh2 (since Δke ≅ Δpe ≅ 0) & & & Qin = mc p (T2 − T1 ) &Then the exit temperature of air leaving the hollow core becomes Q& 20 J/s & Qin = mc p (T2 − T1 ) → T2 = T1 + in = 32 °C + & = 53.4°C & mc p (0.000928 kg/s)(1005 J/kg.°C)PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 84. 5-845-108 A computer is cooled by a fan blowing air through the case of the computer. The required flow rateof the air and the fraction of the temperature rise of air that is due to heat generated by the fan are to bedetermined.Assumptions 1 Steady flow conditions exist. 2 Air is an ideal gas with constant specific heats. 3 Thepressure of air is 1 atm. 4 Kinetic and potential energy changes are negligibleProperties The specific heat of air at room temperature is cp = 1.005 kJ/kg.°C (Table A-2).Analysis (a) We take the air space in the computer as the system, which is a control volume. The energybalance for this steady-flow system can be expressed in the rate form as & & Ein − Eout = ΔEsystem 0 (steady) & =0 1 24 4 3 1442443 Rate of net energy transfer Rate of change in internal, kinetic, by heat, work, and mass potential, etc. energies & & Ein = Eout & & Qin + Win + mh1 = mh2 (since Δke ≅ Δpe ≅ 0) & & & + W = mc (T − T ) Qin & & p 2 1 inNoting that the fan power is 25 W and the 8 PCBs transfer a total of80 W of heat to air, the mass flow rate of air is determined to be & Q + Win & (8 × 10) W + 25 W & & Qin + Win = mc p (Te − Ti ) → m = in & & = = 0.0104 kg/s c p (Te − Ti ) (1005 J/kg.°C)(10°C)(b) The fraction of temperature rise of air that is due to the heat generated by the fan and its motor can bedetermined from Q& 25 W & & Q = mc p ΔT → ΔT = = = 2.4°C mc p (0.0104 kg/s)(1005 J/kg.°C) & 2.4°C f = = 0.24 = 24% 10°CPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 85. 5-855-109 Hot water enters a pipe whose outer surface is exposed to cold air in a basement. The rate of heatloss from the water is to be determined.Assumptions 1 Steady flow conditions exist. 2 Water is an incompressible substance with constant specificheats. 3 The changes in kinetic and potential energies are negligible.Properties The properties of water at the average temperature of (90+88)/2 = 89°C are ρ = 965 kg/m 3 andcp = 4.21 kJ/kg.°C (Table A-3).Analysis The mass flow rate of water is π (0.04 m) 2 m = ρAcV = (965 kg/m 3 ) & (0.8 m/s) = 0.970 kg/s 4We take the section of the pipe in the basementto be the system, which is a control volume. The & Qenergy balance for this steady-flow system can 1 2be expressed in the rate form as Water & & Ein − Eout = ΔEsystem 0 (steady) & =0 1 24 4 3 1442443 90°C 88°CRate of net energy transfer Rate of change in internal, kinetic,by heat, work, and mass potential, etc. energies 0.8 m/s & & Ein = Eout & & mh1 = Qout + mh2 (since Δke ≅ Δpe ≅ 0) & & Qout = mc p (T1 − T2 ) &Then the rate of heat transfer from the hot water to the surrounding air becomes & Qout = mc p [Tin − Tout ]water = (0.970 kg/s )(4.21 kJ/kg. °C)(90 − 88)°C = 8.17 kW &PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 86. 5-865-110 EES Problem 5-109 is reconsidered. The effect of the inner pipe diameter on the rate of heat loss asthe pipe diameter varies from 1.5 cm to 7.5 cm is to be investigated. The rate of heat loss is to be plottedagainst the diameter.Analysis The problem is solved using EES, and the solution is given below."Knowns:"{D = 0.04 [m]}rho = 965 [kg/m^3]Vel = 0.8 [m/s]T_1 = 90 [C]T_2 = 88 [C]C_P = 4.21[kJ/kg-C]"Analysis:""The mass flow rate of water is:"Area = pi*D^2/4m_dot = rho*Area*Vel"We take the section of the pipe in the basement to be the system, which is a control volume. Theenergy balance for this steady-flow system can be expressed in the rate form as"E_dot_in - E_dot_out = DELTAE_dot_sysDELTAE_dot_sys = 0 "Steady-flow assumption"E_dot_in = m_dot*h_inE_dot_out =Q_dot_out+m_dot*h_outh_in = C_P * T_1h_out = C_P * T_2 30 D [m] Qout [kW] 0.015 1.149 25 0.025 3.191 0.035 6.254 0.045 10.34 20 0.055 15.44 Q out [kW ] 0.065 21.57 15 0.075 28.72 10 5 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 D [m ]PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 87. 5-875-111 A room is to be heated by an electric resistance heater placed in a duct in the room. The power ratingof the electric heater and the temperature rise of air as it passes through the heater are to be determined.Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant specific heats at roomtemperature. 3 Kinetic and potential energy changes are negligible. 4 The heating duct is adiabatic, andthus heat transfer through it is negligible. 5 No air leaks in and out of the room.Properties The gas constant of air is 0.287 kPa.m3/kg.K (Table A-1). The specific heats of air at roomtemperature are cp = 1.005 and cv = 0.718 kJ/kg·K (Table A-2).Analysis (a) The total mass of air in the room is V = 5 × 6 × 8 m 3 = 240 m 3 200 kJ/min PV (98 kPa )(240 m 3 ) m= 1 = = 284.6 kg 5×6×8 m3 RT1 (0.287 kPa ⋅ m 3 /kg ⋅ K )(288 K )We first take the entire room as our system, which is a closedsystem since no mass leaks in or out. The power rating of the Weelectric heater is determined by applying the conservation ofenergy relation to this constant volume closed system: 200 W Ein − Eout = ΔEsystem 1 24 4 3 123 4 4 Net energy transfer Change in internal, kinetic, by heat, work, and mass potential, etc. energies We,in + Wfan,in − Qout = ΔU (since ΔKE = ΔPE = 0) (& & & ) Δt We,in + Wfan,in − Qout = mcv ,avg (T2 − T1 )Solving for the electrical work input gives We,in = Qout − Wfan,in + mcv (T2 − T1 ) / Δt & & & = (200/60 kJ/s) − (0.2 kJ/s) + (284.6 kg)(0.718 kJ/kg⋅o C)(25 − 15)o C/(15 × 60 s) = 5.40 kW(b) We now take the heating duct as the system, which is a control volume since mass crosses theboundary. There is only one inlet and one exit, and thus m1 = m2 = m . The energy balance for this & & &adiabatic steady-flow system can be expressed in the rate form as & & Ein − Eout = ΔEsystem 0 (steady) & =0 1 24 4 3 1442443 Rate of net energy transfer Rate of change in internal, kinetic, by heat, work, and mass potential, etc. energies & & Ein = Eout & & & We,in + Wfan,in + mh1 = mh2 (since Q = Δke ≅ Δpe ≅ 0) & & & & We,in + Wfan,in = m(h2 − h1 ) = mc p (T2 − T1 ) & &Thus, & & We,in + Wfan,in (5.40 + 0.2) kJ/s ΔT = T2 − T1 = = = 6.7o C & mc p (50/60 kg/s )(1.005 kJ/kg ⋅ K )PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 88. 5-885-112 A house is heated by an electric resistance heater placed in a duct. The power rating of the electricheater is to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Air is an ideal gas withconstant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible.Properties The constant pressure specific heat of air at room temperature is cp = 1.005 kJ/kg·K (Table A-2)Analysis We take the heating duct as the system, which is a control volume since mass crosses theboundary. There is only one inlet and one exit, and thus m1 = m2 = m . The energy balance for this steady- & & &flow system can be expressed in the rate form as & & Ein − Eout = ΔEsystem 0 (steady) & =0 1 24 4 3 1442443 Rate of net energy transfer Rate of change in internal, kinetic, by heat, work, and mass potential, etc. energies 300 W & & Ein = Eout & We,in & & + Wfan,in + mh1 = Qout + mh2 (since Δke ≅ Δpe ≅ 0) & & We & & & & We,in + Wfan,in = Qout + m(h2 − h1 ) = Qout + mc p (T2 − T1 ) & & 300 WSubstituting, the power rating of the heating element is determined to be & & & We,in = Qout + mc p ΔT − Wfan,in = (0.3 kJ/s ) + (0.6 kg/s) (1.005 kJ/kg ⋅ °C)(7°C) − 0.3 kW = 4.22 kW &PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 89. 5-895-113 A hair dryer consumes 1200 W of electric power when running. The inlet volume flow rate and theexit velocity of air are to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Air is an ideal gas withconstant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible. 4 Thepower consumed by the fan and the heat losses are negligible.Properties The gas constant of air is 0.287 kPa.m3/kg.K (Table A-1). The constant pressure specific heat ofair at room temperature is cp = 1.005 kJ/kg·K (Table A-2)Analysis We take the hair dryer as the system, which is a control volume since mass crosses the boundary.There is only one inlet and one exit, and thus m1 = m2 = m . The energy balance for this steady-flow system & & &can be expressed in the rate form as & & Ein − Eout = ΔEsystem 0 (steady) & =0 1 24 4 3 1442443 T2 = 47°C P1 = 100 kPaRate of net energy transfer Rate of change in internal, kinetic,by heat, work, and mass potential, etc. energies A2 = 60 cm2 T1 = 22°C & & Ein = Eout & & We,in + mh1 = mh2 (since Qout ≅ Δke ≅ Δpe ≅ 0) & & & · We,in = m(h2 − h1 ) = mc p (T2 − T1 ) & & We = 1200 WSubstituting, the mass and volume flow rates of air are determined to be & We,in 1.2 kJ/s m= = (1.005 kJ/kg⋅ C)(47 − 22) C = 0.04776 kg/s & c p (T2 − T1 ) o o v1 = RT1 = (0.287 kPa ⋅ m /kg ⋅ K )(295 K ) = 0.8467 m /kg 3 3 P1 (100 kPa ) V&1 = mv 1 = (0.04776 kg/s )(0.8467 m 3 /kg ) = 0.0404 m 3 /s &(b) The exit velocity of air is determined from the mass balance m1 = m2 = m to be & & & RT2 (0.287 kPa ⋅ m 3 /kg ⋅ K )(320K ) v2 = = = 0.9184 m 3 /kg P2 (100 kPa) 1 mv 2 (0.04776 kg/s)(0.9184 m 3 /kg) & m= & A2V 2 ⎯ V 2 = ⎯→ = = 7.31 m/s v2 A2 60 × 10 − 4 m 2PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 90. 5-905-114 EES Problem 5-113 is reconsidered. The effect of the exit cross-sectional area of the hair drier onthe exit velocity as the exit area varies from 25 cm2 to 75 cm2 is to be investigated. The exit velocity is tobe plotted against the exit cross-sectional area.Analysis The problem is solved using EES, and the results are tabulated and plotted below."Knowns:"R=0.287 [kPa-m^3/kg-K]P= 100 [kPa]T_1 = 22 [C]T_2= 47 [C]{A_2 = 60 [cm^2]}A_1 = 53.35 [cm^2]W_dot_ele=1200 [W]"Analysis: We take the hair dryer as the system, which is a control volume since masscrosses the boundary. There is only one inlet and one exit. Thus, theenergy balance for this steady-flow system can be expressed in the rate formas:"E_dot_in = E_dot_outE_dot_in = W_dot_ele*convert(W,kW) + m_dot_1*(h_1+Vel_1^2/2*convert(m^2/s^2,kJ/kg))E_dot_out = m_dot_2*(h_2+Vel_2^2/2*convert(m^2/s^2,kJ/kg))h_2 = enthalpy(air, T = T_2)h_1= enthalpy(air, T = T_1)"The volume flow rates of air are determined to be:"V_dot_1 = m_dot_1*v_1P*v_1=R*(T_1+273)V_dot_2 = m_dot_2*v_2P*v_2=R*(T_2+273)m_dot_1 = m_dot_2Vel_1=V_dot_1/(A_1*convert(cm^2,m^2))"(b) The exit velocity of air is determined from the mass balance to be"Vel_2=V_dot_2/(A_2*convert(cm^2,m^2)) A2 [cm2] Vel2 [m/s] 18 25 16 Neglect KE Changes 16 30 13.75 Set A 1 = 53.35cm^2 35 12.03 14 Include KE Changes 40 10.68 45 9.583 Vel 2 [m /s] 12 50 8.688 55 7.941 10 60 7.31 65 6.77 8 70 6.303 6 75 5.896 4 20 30 40 50 60 70 80 A 2 [cm^2]PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 91. 5-915-115 The ducts of a heating system pass through an unheated area. The rate of heat loss from the air in theducts is to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Air is an ideal gas withconstant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible. 4 Thereare no work interactions involved.Properties The constant pressure specific heat of air at room temperature is cp = 1.005 kJ/kg·K (Table A-2)Analysis We take the heating duct as the system, which is a control volume since mass crosses theboundary. There is only one inlet and one exit, and thus m1 = m2 = m . The energy balance for this steady- & & &flow system can be expressed in the rate form as & & Ein − Eout = ΔEsystem 0 (steady) & =0 1 24 4 3 1442443 Rate of net energy transfer by heat, work, and mass Rate of change in internal, kinetic, potential, etc. energies 120 kg/min AIR & & Ein = Eout · & & & & mh1 = Qout + mh2 (since W ≅ Δke ≅ Δpe ≅ 0) Q & Qout = m(h1 − h2 ) = mc p (T1 − T2 ) & &Substituting, & Qout = (120 kg/min)(1.005 kJ/kg⋅o C)(4o C) = 482 kJ/minPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 92. 5-925-116 Steam pipes pass through an unheated area, and the temperature of steam drops as a result of heatlosses. The mass flow rate of steam and the rate of heat loss from are to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potentialenergy changes are negligible. 4 There are no work interactions involved.Properties From the steam tables (Table A-6), P = 1 MPa ⎫ v1 = 0.25799 m3/kg 1 ⎬ 1 MPa STEAM 800 kPa T1 = 300°C ⎭ h1 = 3051.6 kJ/kg 300°C 250°C P2 = 800 kPa ⎫ · ⎬ h2 = 2950.4 kJ/kg T2 = 250°C ⎭ QAnalysis (a) The mass flow rate of steam is determined directly from m= & 1 v1 A1V1 = 1 0.25799 m 3 /kg [π (0.06 m) ](2 m/s) = 0.0877 kg/s 2(b) We take the steam pipe as the system, which is a control volume since mass crosses the boundary.There is only one inlet and one exit, and thus m1 = m2 = m . The energy balance for this steady-flow system & & &can be expressed in the rate form as & & Ein − Eout = ΔEsystem 0 (steady) & =0 1 24 4 3 1442443 Rate of net energy transfer Rate of change in internal, kinetic, by heat, work, and mass potential, etc. energies & & Ein = Eout & & mh1 = Qout + mh2 & & (since W ≅ Δke ≅ Δpe ≅ 0) & = m( h − h ) Q & out 1 2Substituting, the rate of heat loss is determined to be & Qloss = (0.0877 kg/s)(3051.6 - 2950.4) kJ/kg = 8.87 kJ/sPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 93. 5-935-117 Steam flows through a non-constant cross-section pipe. The inlet and exit velocities of the steam areto be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Potential energy changeis negligible. 3 There are no work interactions. 4 The device is adiabatic and thus heat transfer isnegligible.Analysis We take the pipe as thesystem, which is a control volumesince mass crosses the boundary. D1 D2The mass and energy balances for 200 kPa 150 kPa Steamthis steady-flow system can be 200°C 150°Cexpressed in the rate form as V1 V2Mass balance: 0 (steady) min − mout = Δmsystem & & & =0 V1 V1 πD12 V1 πD2 V2 2 min = mout ⎯ & & ⎯→ A1 = A1 ⎯ ⎯→ = v1 v1 4 v1 4 v2Energy balance: & & Ein − Eout = ΔEsystem 0 (steady) & =0 1 24 4 3 1442443 Rate of net energy transfer Rate of change in internal, kinetic, by heat, work, and mass potential, etc. energies & & Ein = Eout V12 V2 & & h1 + = h2 + 2 (since Q ≅ W ≅ Δke ≅ Δpe ≅ 0) 2 2The properties of steam at the inlet and exit are (Table A-6) P1 = 200 kPa ⎫v 1 = 1.0805 m 3 /kg ⎬ T1 = 200°C ⎭ h1 = 2870.7 kJ/kg P2 = 150 kPa ⎫v 2 = 1.2855 m 3 /kg ⎬ T1 = 150°C ⎭ h2 = 2772.9 kJ/kgAssuming inlet diameter to be 1.8 m and the exit diameter to be 1.0 m, and substituting, π (1.8 m) 2 V1 π (1.0 m) 2 V2 3 = (1) 4 (1.0805 m /kg ) 4 (1.2855 m 3 /kg ) V12 ⎛ 1 kJ/kg ⎞ V 2 ⎛ 1 kJ/kg ⎞ 2870.7 kJ/kg + ⎜ ⎟ = 2772.9 kJ/kg + 2 ⎜ ⎟ (2) 2 ⎝ 1000 m 2 /s 2 ⎠ 2 ⎝ 1000 m 2 /s 2 ⎠There are two equations and two unknowns. Solving equations (1) and (2) simultaneously using anequation solver such as EES, the velocities are determined to be V1 = 118.8 m/s V2 = 458.0 m/sPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 94. 5-945-118E Saturated liquid water is heated in a steam boiler. The heat transfer per unit mass is to bedetermined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potentialenergy changes are negligible. 3 There are no work interactions.Analysis We take the pipe in which the water is heated as the system, which is a control volume. Theenergy balance for this steady-flow system can be expressed in the rate form as & & E in − E out = ΔE system 0 (steady) & =0 1 24 4 3 1442443 4 Rate of net energy transfer Rate of change in internal, kinetic, by heat, work, and mass potential, etc. energies qin & & E in = E out 500 psia, Water 500 psia & & mh1 + Qin = mh2 & sat. liq. 600°F & Qin = m(h2 − h1 ) & q in = h2 − h1The enthalpies of water at the inlet and exit of the boiler are (Table A-5E, A-6E). P1 = 500 psia ⎫ ⎬ h1 ≅ h f @ 500 psia = 449.51 Btu/lbm x=0 ⎭ P2 = 500 psia ⎫ ⎬ h2 = 1298.6 Btu/lbm T2 = 600°F ⎭Substituting, q in = 1298.6 − 449.51 = 849.1 Btu/lbmPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 95. 5-955-119 R-134a is condensed in a condenser. The heat transfer per unit mass is to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potentialenergy changes are negligible. 3 There are no work interactions.Analysis We take the pipe in which R-134a is condensed as the system, which is a control volume. Theenergy balance for this steady-flow system can be expressed in the rate form as & & E in − E out = ΔE system 0 (steady) & =0 1 24 4 3 1442443 4 Rate of net energy transfer Rate of change in internal, kinetic, by heat, work, and mass potential, etc. energies qout & & E in = E out & 1200 kPa R134a 1200 kPa mh1 = mh2 + Qout & & 80°C sat. liq. & Qout = m(h1 − h2 ) & q out = h1 − h2The enthalpies of R-134a at the inlet and exit of the condenser are (Table A-12, A-13). P1 = 1200 kPa ⎫ ⎬ h1 = 311.39 kJ/kg T1 = 80°C ⎭ P2 = 1200 kPa ⎫ ⎬ h2 ≅ h f @ 1200 kPa = 117.77 kJ/kg x=0 ⎭Substituting, q out = 311.39 − 117.77 = 193.6 kJ/kg5-120 Water is heated at constant pressure so that it changes a state from saturated liquid to saturatedvapor. The heat transfer per unit mass is to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potentialenergy changes are negligible. 3 There are no work interactions.Analysis We take the pipe in which water is heated as the system, which is a control volume. The energybalance for this steady-flow system can be expressed in the rate form as & & E in − E out = ΔE system 0 (steady) & =0 1 24 4 3 1442443 4 Rate of net energy transfer Rate of change in internal, kinetic, by heat, work, and mass potential, etc. energies qin & & E in = E out & & mh1 + Qin = mh2 & 800 kPa Water 800 kPa Sat. Liq. sat. vap. & Qin = m(h2 − h1 ) & q in = h2 − h1 = h fgwhere h fg @ 800 kPa = 2047.5 kJ/kg (Table A-5)Thus, q in = 2047.5 kJ/kgPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 96. 5-96Charging and Discharging Processes5-121 Helium flows from a supply line to an initially evacuated tank. The flow work of the helium in thesupply line and the final temperature of the helium in the tank are to be determined.Properties The properties of helium are R = 2.0769 kJ/kg.K, cp = 5.1926 kJ/kg.K, cv = 3.1156 kJ/kg.K(Table A-2a).Analysis The flow work is determined from its definition butwe first determine the specific volume Helium 200 kPa, 120°C RT (2.0769 kJ/kg.K)(120 + 273 K) v = line = = 4.0811 m 3 /kg P (200 kPa) wflow = Pv = (200 kPa)(4.0811 m 3 /kg) = 816.2 kJ/kg InitiallyNoting that the flow work in the supply line is converted to evacuatedsensible internal energy in the tank, the final heliumtemperature in the tank is determined as follows u tank = hline hline = c p Tline = (5.1926 kJ/kg.K)(120 + 273 K) = 2040.7 kJ/kg u - tank = cv T tank ⎯ ⎯→ 2040.7 kJ/kg = (3.1156 kJ/kg.K)T tank ⎯ T tank = 655.0 K ⎯→Alternative Solution: Noting the definition of specific heat ratio, the final temperature in the tank canalso be determined from Ttank = kTline = 1.667(120 + 273 K) = 655.1 Kwhich is practically the same result.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 97. 5-975-122 An evacuated bottle is surrounded by atmospheric air. A valve is opened, and air is allowed to fillthe bottle. The amount of heat transfer through the wall of the bottle when thermal and mechanicalequilibrium is established is to be determined.Assumptions 1 This is an unsteady process since the conditions within the device are changing during theprocess, but it can be analyzed as a uniform-flow process since the state of fluid at the inlet remainsconstant. 2 Air is an ideal gas with variable specific heats. 3 Kinetic and potential energies are negligible. 4There are no work interactions involved. 5 The direction of heat transfer is to the air in the bottle (will beverified).Properties The gas constant of air is 0.287 kPa.m3/kg.K (Table A-1).Analysis We take the bottle as the system, which is a control volume since mass crosses the boundary.Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h andinternal energy u, respectively, the mass and energy balances for this uniform-flow system can beexpressed asMass balance: 100 kPa min − mout = Δmsystem → mi = m2 (since mout = minitial = 0) 17°CEnergy balance: Ein − Eout = ΔEsystem 1 24 4 3 123 4 4 8L Net energy transfer Change in internal, kinetic, Evacuated by heat, work, and mass potential, etc. energies Qin + mi hi = m2u2 (since W ≅ Eout = Einitial = ke ≅ pe ≅ 0)Combining the two balances: Qin = m2 (u2 − hi )where P2V (100 kPa )(0.008 m 3 ) m2 = = = 0.0096 kg RT2 (0.287 kPa ⋅ m 3 /kg ⋅ K )(290 K ) Table A -17 hi = 290.16 kJ/kg Ti = T2 = 290 K ⎯⎯ ⎯ ⎯ → ⎯ u 2 = 206.91 kJ/kgSubstituting, Qin = (0.0096 kg)(206.91 - 290.16) kJ/kg = - 0.8 kJor Qout = 0.8 kJDiscussion The negative sign for heat transfer indicates that the assumed direction is wrong. Therefore, wereverse the direction.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 98. 5-985-123 An insulated rigid tank is evacuated. A valve is opened, and air is allowed to fill the tank untilmechanical equilibrium is established. The final temperature in the tank is to be determined.Assumptions 1 This is an unsteady process since the conditions within the device are changing during theprocess, but it can be analyzed as a uniform-flow process since the state of fluid at the inlet remainsconstant. 2 Air is an ideal gas with constant specific heats. 3 Kinetic and potential energies are negligible. 4There are no work interactions involved. 5 The device is adiabatic and thus heat transfer is negligible.Properties The specific heat ratio for air at room temperature is k = 1.4 (Table A-2).Analysis We take the tank as the system, which is a control volumesince mass crosses the boundary. Noting that the microscopic energies Airof flowing and nonflowing fluids are represented by enthalpy h andinternal energy u, respectively, the mass and energy balances for thisuniform-flow system can be expressed asMass balance: initially evacuated min − mout = Δmsystem → mi = m2 (since mout = minitial = 0)Energy balance: Ein − Eout = ΔEsystem 1 24 4 3 123 4 4 Net energy transfer Change in internal, kinetic, by heat, work, and mass potential, etc. energies mi hi = m2u2 (since Q ≅ W ≅ Eout = Einitial = ke ≅ pe ≅ 0)Combining the two balances: u 2 = hi → cv T2 = c p Ti → T2 = (c p / cv )Ti = kTiSubstituting, T2 = 1.4 × 290 K = 406 K = 133 o CPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 99. 5-995-124 A rigid tank initially contains air at atmospheric conditions. The tank is connected to a supply line,and air is allowed to enter the tank until mechanical equilibrium is established. The mass of air that enteredand the amount of heat transfer are to be determined.Assumptions 1 This is an unsteady process since the conditions within the device are changing during theprocess, but it can be analyzed as a uniform-flow process since the state of fluid at the inlet remainsconstant. 2 Air is an ideal gas with variable specific heats. 3 Kinetic and potential energies are negligible. 4There are no work interactions involved. 5 The direction of heat transfer is to the tank (will be verified).Properties The gas constant of air is 0.287 kPa.m3/kg.K (Table A-1). The properties of air are (Table A-17) Ti = 295 K ⎯ ⎯→ hi = 295.17 kJ/kg T1 = 295 K ⎯ ⎯→ u1 = 210.49 kJ/kg T2 = 350 K ⎯⎯→ u 2 = 250.02 kJ/kgAnalysis (a) We take the tank as the system, which is a control volume since mass crosses the boundary.Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h andinternal energy u, respectively, the mass and energy balances for this uniform-flow system can beexpressed asMass balance: min − mout = Δmsystem → mi = m2 − m1 Pi = 600 kPa Ti = 22°CEnergy balance: Ein − Eout = ΔEsystem · 1 24 4 3 123 4 4 3 Q Net energy transfer Change in internal, kinetic, V1 = 2 m by heat, work, and mass potential, etc. energies P1 = 100 kPa Qin + mi hi = m2u2 − m1u1 (since W ≅ ke ≅ pe ≅ 0) T1 = 22°CThe initial and the final masses in the tank are PV 1 (100 kPa )(2 m3 ) m1 = = = 2.362 kg RT1 (0.287 kPa ⋅ m3/kg ⋅ K )(295 K ) P2V (600 kPa )(2 m3 ) m2 = = = 11.946 kg RT2 (0.287 kPa ⋅ m3/kg ⋅ K )(350 K )Then from the mass balance, mi = m2 − m1 = 11.946 − 2.362 = 9.584 kg(b) The heat transfer during this process is determined from Qin = −mi hi + m2u2 − m1u1 = −(9.584 kg )(295.17 kJ/kg ) + (11.946 kg )(250.02 kJ/kg ) − (2.362 kg )(210.49 kJ/kg ) = −339 kJ → Qout = 339 kJDiscussion The negative sign for heat transfer indicates that the assumed direction is wrong. Therefore, wereversed the direction.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 100. 5-1005-125 A rigid tank initially contains saturated R-134a liquid-vapor mixture. The tank is connected to asupply line, and R-134a is allowed to enter the tank. The final temperature in the tank, the mass of R-134athat entered, and the heat transfer are to be determined.Assumptions 1 This is an unsteady process since the conditions within the device are changing during theprocess, but it can be analyzed as a uniform-flow process since the state of fluid at the inlet remainsconstant. 2 Kinetic and potential energies are negligible. 3 There are no work interactions involved. 4 Thedirection of heat transfer is to the tank (will be verified).Properties The properties of refrigerant are (Tables A-11 through A-13) T1 = 8°C ⎫ v1 = v f + x1v fg = 0.0007887 + 0.7 × (0.052762 − 0.0007887 ) = 0.03717 m3/kg ⎬ x1 = 0.7 ⎭ u1 = u f + x1u fg = 62.39 + 0.7 × 172.19 = 182.92 kJ/kg P2 = 800 kPa ⎫ v 2 = v g @800 kPa = 0.02562 m3/kg ⎬ sat. vapor ⎭ u2 = u g @800 kPa = 246.79 kJ/kg Pi = 1.0 MPa ⎫ R-134a 1 MPa ⎬ hi = 335.06 kJ/kg 100°C Ti = 100°C ⎭Analysis We take the tank as the system, which is a control volumesince mass crosses the boundary. Noting that the microscopic energies 0.2 m3of flowing and nonflowing fluids are represented by enthalpy h and R-134ainternal energy u, respectively, the mass and energy balances for thisuniform-flow system can be expressed asMass balance: min − mout = Δmsystem → mi = m2 − m1Energy balance: Ein − Eout = ΔEsystem 1 24 4 3 123 4 4 Net energy transfer Change in internal, kinetic, by heat, work, and mass potential, etc. energies Qin + mi hi = m2u2 − m1u1 (since W ≅ ke ≅ pe ≅ 0) (a) The tank contains saturated vapor at the final state at 800 kPa, and thus the final temperature is thesaturation temperature at this pressure, T2 = Tsat @ 800 kPa = 31.31ºC(b) The initial and the final masses in the tank are V 0.2 m3 m1 = = = 5.38 kg v1 0.03717 m3/kg V 0.2 m3 m2 = = = 7.81 kg v 2 0.02562 m3/kgThen from the mass balance m i = m 2 − m1 = 7.81 − 5.38 = 2.43 kg(c) The heat transfer during this process is determined from the energy balance to be Qin = − mi hi + m2u2 − m1u1 = −(2.43 kg )(335.06 kJ/kg ) + (7.81 kg )(246.79 kJ/kg )− (5.38 kg )(182.92 kJ/kg ) = 130 kJPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 101. 5-1015-126 A cylinder initially contains saturated liquid-vapor mixture of water. The cylinder is connected to asupply line, and the steam is allowed to enter the cylinder until all the liquid is vaporized. The finaltemperature in the cylinder and the mass of the steam that entered are to be determined.Assumptions 1 This is an unsteady process since the conditions within the device are changing during theprocess, but it can be analyzed as a uniform-flow process since the state of fluid at the inlet remainsconstant. 2 The expansion process is quasi-equilibrium. 3 Kinetic and potential energies are negligible. 3There are no work interactions involved other than boundary work. 4 The device is insulated and thus heattransfer is negligible.Properties The properties of steam are (Tables A-4 through A-6)P1 = 200 kPa ⎫ ⎬ h1 = h f + x1 h fgx1 = 0.6 ⎭ (P = 200 kPa) = 504.71 + 0.6 × 2201.6 = 1825.6 kJ/kg m1 = 10 kgP2 = 200 kPa ⎫ H2O Pi = 0.5 MPa ⎬ h2 = h g @ 200 kPa = 2706.3 kJ/kgsat. vapor ⎭ Ti = 350°CPi = 0.5 MPa ⎫ ⎬ hi = 3168.1 kJ/kgTi = 350°C ⎭Analysis (a) The cylinder contains saturated vapor at the final state at a pressure of 200 kPa, thus the finaltemperature in the cylinder must be T2 = Tsat @ 200 kPa = 120.2°C(b) We take the cylinder as the system, which is a control volume since mass crosses the boundary. Notingthat the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internalenergy u, respectively, the mass and energy balances for this uniform-flow system can be expressed asMass balance: min − mout = Δmsystem → mi = m2 − m1Energy balance: Ein − Eout = ΔEsystem 1 24 4 3 123 4 4 Net energy transfer Change in internal, kinetic, by heat, work, and mass potential, etc. energies mi hi = Wb,out + m2u2 − m1u1 (since Q ≅ ke ≅ pe ≅ 0)Combining the two relations gives 0 = Wb, out − (m2 − m1 )hi + m2u2 − m1u1or, 0 = −(m2 − m1 )hi + m2h2 − m1h1since the boundary work and ΔU combine into ΔH for constant pressure expansion and compressionprocesses. Solving for m2 and substituting, m2 = hi − h1 m1 = (3168.1 − 1825.6) kJ/kg (10 kg ) = 29.07 kg hi − h2 (3168.1 − 2706.3) kJ/kgThus, mi = m2 - m1 = 29.07 - 10 = 19.07 kgPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 102. 5-1025-127E A scuba divers air tank is to be filled with air from a compressed air line. The temperature andmass in the tank at the final state are to be determined.Assumptions 1 This is an unsteady process since the conditions within the device are changing during theprocess, but it can be analyzed as a uniform-flow process since the state of fluid at the inlet remainsconstant. 2 Air is an ideal gas with constant specific heats. 3 Kinetic and potential energies are negligible. 4There are no work interactions involved. 5 The tank is well-insulated, and thus there is no heat transfer.Properties The gas constant of air is 0.3704 psia·ft3/lbm·R (Table A-1E). The specific heats of air at roomtemperature are cp = 0.240 Btu/lbm·R and cv = 0.171 Btu/lbm·R (Table A-2Ea).Analysis We take the tank as the system, which is a control volume since mass crosses the boundary.Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h andinternal energy u, respectively, the mass and energy balances for this uniform-flow system can beexpressed asMass balance: min − mout = Δmsystem → mi = m 2 − m1 Air 120 psia, 100°FEnergy balance: E in − E out = ΔE system 1 24 4 3 1 24 4 3 Net energy transfer Change in internal, kinetic, by heat, work, and mass potential, etc. energies 20 psia mi hi = m 2 u 2 − m1u1 70°F m i c p Ti = m 2 cv T2 − m1cv T1 2 ft3Combining the two balances: (m 2 − m1 )c p Ti = m 2 cv T2 − m1cv T1The initial and final masses are given by P1V (20 psia )(2 ft 3 ) m1 = = = 0.2038 lbm RT1 (0.3704 psia ⋅ ft 3 /lbm ⋅ R )(70 + 460 R ) P2V (120 psia )(2 ft 3 ) 647.9 m2 = = = RT2 (0.3704 psia ⋅ ft 3 /lbm ⋅ R )T2 T2Substituting, ⎛ 647.9 ⎞ 647.9 ⎜ T − 0.2038 ⎟(0.24)(560) = T (0.171)T2 − (0.2038)(0.171)(530) ⎜ ⎟ ⎝ 2 ⎠ 2whose solution by trial-error or an equation solver such as EES is T2 = 727.4 R = 267.4 °FThe final mass is then 647.9 647.9 m2 = = = 0.890 lbm T2 727.4PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 103. 5-1035-128 R-134a from a tank is discharged to an air-conditioning line in an isothermal process. The finalquality of the R-134a in the tank and the total heat transfer are to be determined.Assumptions 1 This is an unsteady process since the conditions within the device are changing during theprocess, but it can be analyzed as a uniform-flow process since the state of fluid at the exit remainsconstant. 2 Kinetic and potential energies are negligible. 3 There are no work interactions involved.Analysis We take the tank as the system, which is a control volume since mass crosses the boundary.Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h andinternal energy u, respectively, the mass and energy balances for this uniform-flow system can beexpressed asMass balance: min − m out = Δmsystem A-C line − m e = m 2 − m1 m e = m1 − m 2Energy balance: E in − E out = ΔE system Liquid R-134a 1 24 4 3 1 24 4 3 5 kg Net energy transfer Change in internal, kinetic, by heat, work, and mass potential, etc. energies 24°C Qin − m e he = m 2 u 2 − m1u1 Qin = m 2 u 2 − m1u1 + m e heCombining the two balances: Qin = m 2 u 2 − m1u1 + (m1 − m 2 )heThe initial state properties of R-134a in the tank are 3 T1 = 24°C ⎫ v 1 = 0.0008261 m /kg ⎬ u1 = 84.44 kJ/kg (Table A-11) x=0 ⎭ h = 84.98 kJ/kg eNote that we assumed that the refrigerant leaving the tank is at saturated liquid state, and found the exitingenthalpy accordingly. The volume of the tank is V = m1v 1 = (5 kg)(0.0008261 m 3 /kg) = 0.004131 m 3The final specific volume in the container is V 0.004131 m 3 v2 = = = 0.01652 m 3 /kg m2 0.25 kgThe final state is now fixed. The properties at this state are (Table A-11) v −v f 0.01652 − 0.0008261 T2 = 24°C ⎫ x2 = 2 ⎪ = = 0.5061 3 ⎬ v fg 0.031834 − 0.0008261 ⎪ v 2 = 0.01652 m /kg ⎭ u 2 = u f + x 2 u fg = 84.44 kJ/kg + (0.5061)(158.65 kJ/kg) = 164.73 kJ/kgSubstituting into the energy balance equation, Qin = m 2 u 2 − m1u1 + ( m1 − m 2 ) he = (0.25 kg )(164.73 kJ/kg ) − (5 kg )(84.44 kJ/kg ) + (4.75 kg )(84.98 kJ/kg ) = 22.64 kJPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 104. 5-1045-129E Oxygen is supplied to a medical facility from 10 compressed oxygen tanks in an isothermalprocess. The mass of oxygen used and the total heat transfer to the tanks are to be determined.Assumptions 1 This is an unsteady process but it can be analyzed as a uniform-flow process. 2 Oxygen isan ideal gas with constant specific heats. 3 Kinetic and potential energies are negligible. 4 There are nowork interactions involved.Properties The gas constant of oxygen is 0.3353 psia·ft3/lbm·R (Table A-1E). The specific heats of oxygenat room temperature are cp = 0.219 Btu/lbm·R and cv = 0.157 Btu/lbm·R (Table A-2Ea).Analysis We take the tank as the system, which is a control volume since mass crosses the boundary.Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h andinternal energy u, respectively, the mass and energy balances for this uniform-flow system can beexpressed asMass balance: min − m out = Δmsystem − m e = m 2 − m1 m e = m1 − m 2Energy balance: Oxygen E in − E out = ΔE system 2000 psia 1 24 4 3 1 24 4 3 Net energy transfer by heat, work, and mass Change in internal, kinetic, 80°F, 30 ft3 potential, etc. energies Qin − m e he = m 2 u 2 − m1u1 Qin = m 2 u 2 − m1u1 + m e he Qin = m 2 cv T2 − m1cv T1 + m e c p TeCombining the two balances: Qin = m 2 cv T2 − m1cv T1 + (m1 − m 2 )c p TeThe initial and final masses, and the mass used are P1V (2000 psia )(30 ft 3 ) m1 = = = 331.4 lbm RT1 (0.3353 psia ⋅ ft 3 /lbm ⋅ R )(80 + 460 R ) P2V (100 psia )(30 ft 3 ) m2 = = = 16.57 lbm RT2 (0.3353 psia ⋅ ft 3 /lbm ⋅ R )(80 + 460 R ) m e = m1 − m 2 = 331.4 − 16.57 = 314.8 lbmSubstituting into the energy balance equation, Qin = m 2 cv T2 − m1 cv T1 + m e c p Te = (16.57)(0.157)(540) − (331.4)(0.157)(540) + (314.8)(0.219)(540) = 10,540 BtuPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 105. 5-1055-130 The air in an insulated, rigid compressed-air tank is released until the pressure in the tank reduces toa specified value. The final temperature of the air in the tank is to be determined.Assumptions 1 This is an unsteady process since the conditions within the device are changing during theprocess, but it can be analyzed as a uniform-flow process. 2 Air is an ideal gas with constant specific heats.3 Kinetic and potential energies are negligible. 4 There are no work interactions involved. 5 The tank iswell-insulated, and thus there is no heat transfer.Properties The gas constant of air is 0.287 kPa·m3/kg·K (Table A-1). The specific heats of air at roomtemperature are cp = 1.005 kJ/kg·K and cv = 0.718 kJ/kg·K (Table A-2a).Analysis We take the tank as the system, which is a control volume since mass crosses the boundary.Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h andinternal energy u, respectively, the mass and energy balances for this uniform-flow system can beexpressed asMass balance: min − m out = Δmsystem − m e = m 2 − m1 m e = m1 − m 2Energy balance: Air E in − E out = ΔE system 4000 kPa 1 24 4 3 1 24 4 3 Net energy transfer by heat, work, and mass Change in internal, kinetic, 20°C, 0.5 m3 potential, etc. energies − m e he = m 2 u 2 − m1u1 0 = m 2 u 2 − m1u1 + m e he 0 = m 2 cv T2 − m1cv T1 + m e c p TeCombining the two balances: 0 = m 2 cv T2 − m1cv T1 + (m1 − m 2 )c p TeThe initial and final masses are given by P1V (4000 kPa )(0.5 m 3 ) m1 = = = 23.78 kg RT1 (0.287 kPa ⋅ m 3 /kg ⋅ K )(20 + 273 K ) P2V (2000 kPa )(0.5 m 3 ) 3484 m2 = = = RT2 (0.287 kPa ⋅ m 3 /kg ⋅ K )T2 T2The temperature of air leaving the tank changes from the initial temperature in the tank to the finaltemperature during the discharging process. We assume that the temperature of the air leaving the tank isthe average of initial and final temperatures in the tank. Substituting into the energy balance equation gives 0 = m 2 cv T2 − m1cv T1 + (m1 − m 2 )c p Te 3484 ⎛ 3484 ⎞ ⎛ 293 + T2 ⎞ 0= (0.718)T2 − (23.78)(0.718)(293) + ⎜ 23.78 − ⎜ ⎟(1.005)⎜ ⎜ ⎟ ⎟ T2 ⎝ T2 ⎟⎠ ⎝ 2 ⎠whose solution by trial-error or by an equation solver such as EES is T2 = 241 K = −32°CPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 106. 5-1065-131E Steam is supplied from a line to a weighted piston-cylinder device. The final temperature (andquality if appropriate) of the steam in the piston cylinder and the total work produced as the device is filledare to be determined.Assumptions 1 This is an unsteady process since the conditions within the device are changing during theprocess, but it can be analyzed as a uniform-flow process since the state of fluid at the inlet remainsconstant. 2 Kinetic and potential energies are negligible. 3 The process is adiabatic.Analysis We take the tank as the system, which is a control volume since mass crosses the boundary.Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h andinternal energy u, respectively, and also noting that the initial mass in the system is zero, the mass andenergy balances for this uniform-flow system can be expressed asMass balance: m in − m out = Δm system mi = m 2Energy balance: E in − E out = ΔE system 1 24 4 3 1 24 4 3 Net energy transfer Change in internal, kinetic, by heat, work, and mass potential, etc. energies m i hi − Wb,out = m 2 u 2 Wb,out = mi hi − m 2 u 2Combining the two balances: Wb ,out = m 2 ( hi − u 2 )The boundary work is determined from Wb ,out = P (V 2 − V 1 ) = P (m 2v 2 − m1v 1 ) = Pm 2v 2Substituting, the energy balance equation simplifies into Pm 2v 2 = m 2 (hi − u 2 ) Pv 2 = h i − u 2The enthalpy of steam at the inlet is P1 = 300 psia ⎫ ⎬ hi = 1226.4 Btu/lbm (Table A - 6E) T1 = 450°F ⎭Substituting this value into the energy balance equation and using an iterative solution of this equationgives (or better yet using EES software) T2 = 425.1 °F u 2 = 1135.5 Btu/lbm v 2 = 2.4575 Btu/lbmThe final mass is V2 10 ft 3 m2 = = = 4.069 kg v 2 2.4575 ft 3 /lbmand the work produced is ⎛ 1 Btu ⎞ Wb,out = PV 2 = (200 psia)(10 ft 3 )⎜ ⎟ = 370.1 Btu ⎜ 5.404 psia ⋅ ft 3 ⎟ ⎝ ⎠PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 107. 5-1075-132E Oxygen is supplied from a line to a weighted piston-cylinder device. The final temperature of theoxygen in the piston cylinder and the total work produced as the device is filled are to be determined.Assumptions 1 This is an unsteady process since the conditions within the device are changing during theprocess, but it can be analyzed as a uniform-flow process since the state of fluid at the inlet remainsconstant. 2 Kinetic and potential energies are negligible. 3 The process is adiabatic. 4 Oxygen is an idealgas with constant specific heats.Analysis We take the tank as the system, which is a control volume since mass crosses the boundary.Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h andinternal energy u, respectively, and also noting that the initial mass in the system is zero, the mass andenergy balances for this uniform-flow system can be expressed asMass balance: m in − m out = Δm system mi = m 2Energy balance: E in − E out = ΔE system 1 24 4 3 1 24 4 3 Net energy transfer Change in internal, kinetic, by heat, work, and mass potential, etc. energies m i hi − Wb,out = m 2 u 2 Wb,out = mi hi − m 2 u 2Combining the two balances: Wb ,out = m 2 ( hi − u 2 )The boundary work is determined from Wb ,out = P (V 2 − V 1 ) = P (m 2v 2 − m1v 1 ) = Pm 2v 2Substituting, the energy balance equation simplifies into Pm 2v 2 = m 2 (hi − u 2 ) Pv 2 = hi − u 2 RT2 = c p Ti − c v T2Solving for the final temperature, cp cp RT2 = c p Ti − cv T2 ⎯ T2 = ⎯→ Ti = Ti = Ti = 450°F R + cv cpThe work produced is ⎛ 1 Btu ⎞ Wb,out = PV 2 = (200 psia)(10 ft 3 )⎜ ⎟ = 370.1 Btu ⎜ 5.404 psia ⋅ ft 3 ⎟ ⎝ ⎠PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 108. 5-1085-133 A rigid tank initially contains saturated R-134a vapor. The tank is connected to a supply line, and R-134a is allowed to enter the tank. The mass of the R-134a that entered and the heat transfer are to bedetermined.Assumptions 1 This is an unsteady process since the conditions within the device are changing during theprocess, but it can be analyzed as a uniform-flow process since the state of fluid at the inlet remainsconstant. 2 Kinetic and potential energies are negligible. 3 There are no work interactions involved. 4 Thedirection of heat transfer is to the tank (will be verified).Properties The properties of refrigerant are (Tables A-11 R-134a 1.2 MPathrough A-13) 36°C P = 1 MPa ⎫ v 1 = v g @1 MPa = 0.02031 m 3 /kg 1 ⎬ sat.vapor ⎭ u1 = u g @1 MPa = 250.68 kJ/kg P2 = 1.2 MPa ⎫ v 2 = v f @1.2 MPa = 0.0008934 m 3 /kg R-134a ⎬ sat. liquid ⎭ u2 = u f @1.2 MPa = 116.70 kJ/kg 0.12 m3 1 MPa Q Pi = 1.2 MPa ⎫ ⎬ hi = h f @ 36°C = 102.30 kJ/kg Sat. vapor Ti = 36°C ⎭Analysis We take the tank as the system, which is a control volume since mass crosses the boundary.Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h andinternal energy u, respectively, the mass and energy balances for this uniform-flow system can beexpressed asMass balance: min − mout = Δmsystem → mi = m2 − m1Energy balance: Ein − Eout = ΔEsystem 1 24 4 3 123 4 4 Net energy transfer Change in internal, kinetic, by heat, work, and mass potential, etc. energies Qin + mi hi = m2u2 − m1u1 (since W ≅ ke ≅ pe ≅ 0)(a) The initial and the final masses in the tank are V1 0.12 m 3 m1 = = = 5.91 kg v 1 0.02031 m 3 /kg V2 0.12 m 3 m2 = = = 134.31 kg v 2 0.0008934 m 3 /kgThen from the mass balance mi = m 2 − m1 = 134.31 − 5.91 = 128.4 kg(c) The heat transfer during this process is determined from the energy balance to be Qin = − mi hi + m2u2 − m1u1 = −(128.4 kg )(102.30 kJ/kg ) + (134.31 kg )(116.70 kJ/kg ) − (5.91 kg )(250.68 kJ/kg ) = 1057 kJPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 109. 5-1095-134 A rigid tank initially contains saturated liquid water. A valve at the bottom of the tank is opened, andhalf of the mass in liquid form is withdrawn from the tank. The temperature in the tank is maintainedconstant. The amount of heat transfer is to be determined.Assumptions 1 This is an unsteady process since the conditions within the device are changing during theprocess, but it can be analyzed as a uniform-flow process since the state of fluid leaving the device remainsconstant. 2 Kinetic and potential energies are negligible. 3 There are no work interactions involved. 4 Thedirection of heat transfer is to the tank (will be verified).Properties The properties of water are (Tables A-4 through A-6) 3 T1 = 200°C ⎫ v1 = v f @ 200 o C = 0.001157 m /kg ⎬ sat. liquid ⎭ u1 = u f @ 200o C = 850.46 kJ/kg Te = 200o C ⎫ ⎪ ⎬ he = h f @ 200 o C = 852.26 kJ/kg sat. liquid ⎪ ⎭Analysis We take the tank as the system, which is a control volume since mass crosses the boundary.Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h andinternal energy u, respectively, the mass and energy balances for this uniform-flow system can beexpressed asMass balance: min − mout = Δmsystem → me = m1 − m2 H2OEnergy balance: Sat. liquid T = 200°C Ein − Eout = ΔEsystem V = 0.3 m3 Q 1 24 4 3 123 4 4 Net energy transfer Change in internal, kinetic, by heat, work, and mass potential, etc. energies Qin = me he + m2u2 − m1u1 (since W ≅ ke ≅ pe ≅ 0)The initial and the final masses in the tank are V1 0.3 m 3 m1 = = = 259.4 kg v 1 0.001157 m 3 /kg m 2 = 1 m1 = 2 1 2 (259.4 kg ) = 129.7 kgThen from the mass balance, me = m1 − m 2 = 259.4 − 129.7 = 129.7 kgNow we determine the final internal energy, V 0.3 m 3 v2 = = = 0.002313 m 3 /kg m2 129.7 kg v 2 −v f 0.002313 − 0.001157 x2 = = = 0.009171 v fg 0.12721 − 0.001157 T2 = 200°C ⎫ ⎬ u 2 = u f + x 2 u fg = 850.46 + (0.009171)(1743.7 ) = 866.46 kJ/kg x 2 = 0.009171 ⎭Then the heat transfer during this process is determined from the energy balance by substitution to be Q = (129.7 kg )(852.26 kJ/kg ) + (129.7 kg )(866.46 kJ/kg ) − (259.4 kg )(850.46 kJ/kg ) = 2308 kJPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 110. 5-1105-135 A rigid tank initially contains saturated liquid-vapor mixture of refrigerant-134a. A valve at thebottom of the tank is opened, and liquid is withdrawn from the tank at constant pressure until no liquidremains inside. The amount of heat transfer is to be determined.Assumptions 1 This is an unsteady process since the conditions within the device are changing during theprocess, but it can be analyzed as a uniform-flow process since the state of fluid leaving the device remainsconstant. 2 Kinetic and potential energies are negligible. 3 There are no work interactions involved. 4 Thedirection of heat transfer is to the tank (will be verified).Properties The properties of R-134a are (Tables A-11 through A-13) R-134a P = 800 kPa → v f =0.0008458 m3/kg, v g = 0.025621 m3/kg 1 Sat. vapor P = 800 kPa u f =94.79 kJ/kg, u g = 246.79 kJ/kg V = 0.12 m3 Q 3 P2 = 800 kPa ⎫ v 2 = v g @800 kPa = 0.025621 m /kg ⎬ sat. vapor ⎭ u2 = u g @800 kPa = 246.79 kJ/kg Pe = 800 kPa ⎫ ⎬ he = h f @800 kPa = 95.47 kJ/kg sat. liquid ⎭Analysis We take the tank as the system, which is a control volume since mass crosses the boundary.Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h andinternal energy u, respectively, the mass and energy balances for this uniform-flow system can beexpressed asMass balance: min − mout = Δmsystem → me = m1 − m2Energy balance: Ein − Eout = ΔEsystem 1 24 4 3 123 4 4 Net energy transfer Change in internal, kinetic, by heat, work, and mass potential, etc. energies Qin = me he + m2u2 − m1u1 (since W ≅ ke ≅ pe ≅ 0)The initial mass, initial internal energy, and final mass in the tank are V f Vg 0.12 × 0.25 m3 0.12 × 0.75 m3 m1 = m f + mg = + = + = 35.47 + 3.513 = 38.98 kg v f v g 0.0008458 m3/kg 0.025621 m3/kg U1 = m1u1 = m f u f + mg u g = (35.47 )(94.79 ) + (3.513)(246.79 ) = 4229.2 kJ V 0.12 m3 m2 = = = 4.684 kg v 2 0.025621 m3/kgThen from the mass and energy balances, me = m1 − m 2 = 38.98 − 4.684 = 34.30 kg Qin = (34.30 kg )(95.47 kJ/kg ) + (4.684 kg )(246.79 kJ/kg ) − 4229 kJ = 201.2 kJPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 111. 5-1115-136E A rigid tank initially contains saturated liquid-vapor mixture of R-134a. A valve at the top of thetank is opened, and vapor is allowed to escape at constant pressure until all the liquid in the tankdisappears. The amount of heat transfer is to be determined.Assumptions 1 This is an unsteady process since the conditions within the device are changing during theprocess, but it can be analyzed as a uniform-flow process since the state of fluid leaving the device remainsconstant. 2 Kinetic and potential energies are negligible. 3 There are no work interactions involved.Properties The properties of R-134a are (Tables A-11E through A-13E) P = 100 psia → v f = 0.01332 ft 3/lbm, v g = 0.4776 ft 3/lbm 1 u f = 37.623 Btu/lbm, u g = 104.99 Btu/lbm R-134a P2 = 100 psia ⎫ v 2 = vg @100 psia = 0.4776 ft 3/lbm Sat. vapor ⎬ Q sat. vapor ⎭ u2 = u g @100 psia = 104.99 Btu/lbm P = 100 psia V = 4 ft3 Pe = 100 psia ⎫ ⎬ he = hg @100 psia = 113.83 Btu/lbm sat. vapor ⎭Analysis We take the tank as the system, which is a control volume since mass crosses the boundary.Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h andinternal energy u, respectively, the mass and energy balances for this uniform-flow system can beexpressed asMass balance: min − mout = Δmsystem → me = m1 − m2Energy balance: Ein − Eout = ΔEsystem 1 24 4 3 123 4 4 Net energy transfer Change in internal, kinetic, by heat, work, and mass potential, etc. energies Qin − me he = m2u2 − m1u1 (since W ≅ ke ≅ pe ≅ 0)The initial mass, initial internal energy, and final mass in the tank are Vf Vg 4 × 0.2 ft 3 4 × 0.8 ft 3 m1 = m f + m g = + = + = 60.04 + 6.70 = 66.74 lbm vf vg 0.01332 ft 3 /lbm 0.4776 ft 3 /lbm U 1 = m1u1 = m f u f + m g u g = (60.04 )(37.623) + (6.70 )(104.99 ) = 2962 Btu V 4 ft 3 m2 = = = 8.375 lbm v 2 0.4776 ft 3 /lbmThen from the mass and energy balances, me = m1 − m 2 = 66.74 − 8.375 = 58.37 lbm Qin = me he + m2u2 − m1u1 = (58.37 lbm )(113.83 Btu/lbm ) + (8.375 lbm )(104.99 Btu/lbm ) − 2962 Btu = 4561 BtuPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 112. 5-1125-137 A rigid tank initially contains superheated steam. A valve at the top of the tank is opened, and vaporis allowed to escape at constant pressure until the temperature rises to 500°C. The amount of heat transferis to be determined.Assumptions 1 This is an unsteady process since the conditions within the device are changing during theprocess, but it can be analyzed as a uniform-flow process by using constant average properties for thesteam leaving the tank. 2 Kinetic and potential energies are negligible. 3 There are no work interactionsinvolved. 4 The direction of heat transfer is to the tank (will be verified).Properties The properties of water are (Tables A-4 through A-6) P1 = 2 MPa ⎫ v 1 = 0.12551 m 3 /kg ⎬ T1 = 300°C ⎭ u1 = 2773.2 kJ/kg, h1 = 3024.2 kJ/kg STEAM Q P2 = 2 MPa ⎫ v 2 = 0.17568 m 3 /kg 2 MPa ⎬ T2 = 500°C ⎭ u 2 = 3116.9 kJ/kg, h2 = 3468.3 kJ/kgAnalysis We take the tank as the system, which is a control volume since mass crosses the boundary.Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h andinternal energy u, respectively, the mass and energy balances for this uniform-flow system can beexpressed asMass balance: min − mout = Δmsystem → me = m1 − m2Energy balance: Ein − Eout = ΔEsystem 1 24 4 3 123 4 4 Net energy transfer Change in internal, kinetic, by heat, work, and mass potential, etc. energies Qin − me he = m2u2 − m1u1 (since W ≅ ke ≅ pe ≅ 0)The state and thus the enthalpy of the steam leaving the tank is changing during this process. But forsimplicity, we assume constant properties for the exiting steam at the average values. Thus, h1 + h2 3024.2 + 3468.3 kJ/kg he ≅ = = 3246.2 kJ/kg 2 2The initial and the final masses in the tank are V1 0.2 m 3 m1 = = = 1.594 kg v 1 0.12551 m 3 /kg V2 0.2 m 3 m2 = = = 1.138kg v 2 0.17568 m 3 /kgThen from the mass and energy balance relations, me = m1 − m 2 = 1.594 − 1.138 = 0.456 kg Qin = me he + m2u2 − m1u1 = (0.456 kg )(3246.2 kJ/kg ) + (1.138 kg )(3116.9 kJ/kg ) − (1.594 kg )(2773.2 kJ/kg ) = 606.8 kJPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 113. 5-1135-138 Steam is supplied from a line to a piston-cylinder device equipped with a spring. The finaltemperature (and quality if appropriate) of the steam in the cylinder and the total work produced as thedevice is filled are to be determined.Assumptions 1 This is an unsteady process since the conditions within the device are changing during theprocess, but it can be analyzed as a uniform-flow process since the state of fluid at the inlet remainsconstant. 2 Kinetic and potential energies are negligible. 3 The process is adiabatic.Analysis We take the tank as the system, which is a control volume since mass crosses the boundary.Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h andinternal energy u, respectively, and also noting that the initial mass in the system is zero, the mass andenergy balances for this uniform-flow system can be expressed asMass balance: min − m out = Δmsystem ⎯ ⎯→ mi = m 2Energy balance: E in − E out = ΔE system 1 24 4 3 1 24 4 3 Net energy transfer Change in internal, kinetic, by heat, work, and mass potential, etc. energies m i hi − Wb ,out = m 2 u 2 Wb,out = mi hi − m 2 u 2Combining the two balances: Wb ,out = m 2 ( hi − u 2 )Because of the spring, the relation between the pressure and volume is a linear relation. According to thedata in the problem statement, 2700 P − 300 = V 5The final vapor volume is then 5 V2 = (1500 − 300) = 2.222 m 3 2700The work needed to compress the spring is V2 ⎛ 2700 ⎞ 2700 2 ∫ Wb,out = PdV = ∫⎜ 0 ⎝ 5 V + 300 ⎟dV = ⎠ 2×5 V 2 + 300V 2 = 270 × 2.222 2 + 300 × 2.222 = 2000 kJThe enthalpy of steam at the inlet is P1 = 1500 kPa ⎫ ⎬ hi = 2796.0 kJ/kg (Table A - 6) T1 = 200°C ⎭Substituting the information found into the energy balance equation gives V2 2.222 Wb,out = m 2 (hi − u 2 ) ⎯ Wb,out = ⎯→ ( hi − u 2 ) ⎯ ⎯→ 2000 = (2796.0 − u 2 ) v2 v2Using an iterative solution of this equation with steam tables gives (or better yet using EES software) T2 = 233.2°C u 2 = 2664.8 kJ/kg v 2 = 0.1458 m 3 /kgPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 114. 5-1145-139 Air is supplied from a line to a piston-cylinder device equipped with a spring. The final temperatureof the steam in the cylinder and the total work produced as the device is filled are to be determined.Assumptions 1 This is an unsteady process since the conditions within the device are changing during theprocess, but it can be analyzed as a uniform-flow process since the state of fluid at the inlet remainsconstant. 2 Kinetic and potential energies are negligible. 3 The process is adiabatic. 4 Air is an ideal gaswith constant specific heats.Properties The gas constant of air is 0.287 kPa·m3/kg·K (Table A-1). The specific heats of air at roomtemperature are cp = 1.005 kJ/kg·K and cv = 0.718 kJ/kg·K (Table A-2a).Analysis We take the tank as the system, which is a control volume since mass crosses the boundary.Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h andinternal energy u, respectively, and also noting that the initial mass in the system is zero, the mass andenergy balances for this uniform-flow system can be expressed asMass balance: min − m out = Δmsystem ⎯ ⎯→ mi = m 2Energy balance: E in − E out = ΔE system 1 24 4 3 1 24 4 3 Net energy transfer Change in internal, kinetic, by heat, work, and mass potential, etc. energies m i hi − Wb ,out = m 2 u 2 Wb,out = mi hi − m 2 u 2Combining the two balances: Wb ,out = m 2 ( hi − u 2 )Because of the spring, the relation between the pressure and volume is a linear relation. According to thedata in the problem statement, 2700 P − 300 = V 5The final air volume is then 5 V2 = ( 2000 − 300) = 3.148 m 3 2700The work needed to compress the spring is V2 ⎛ 2700 ⎞ 2700 2 ∫ Wb,out = PdV = ∫⎝ 0 ⎜ 5 V + 300 ⎟dV = ⎠ 2×5 V 2 + 300V 2 = 270 × 3.148 2 + 300 × 3.148 = 3620 kJSubstituting the information found into the energy balance equation gives Wb,out = m 2 (hi − u 2 ) P2V 2 Wb,out = (c p Ti − cv T2 ) RT2 2000 × 3.148 3620 = (1.005 × 523 − 0.718 × T2 ) (0.287)T2The final temperature is then T2 = 595.2 K = 322.2 °CPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 115. 5-1155-140 A hot-air balloon is considered. The final volume of the balloon and work produced by the air insidethe balloon as it expands the balloon skin are to be determined.Assumptions 1 This is an unsteady process since the conditions within the device are changing during theprocess, but it can be analyzed as a uniform-flow process. 2 Air is an ideal gas with constant specific heats.3 Kinetic and potential energies are negligible. 4 There is no heat transfer.Properties The gas constant of air is 0.287 kPa·m3/kg·K (Table A-1).Analysis The specific volume of the air at the entrance and exit, and in the balloon is RT (0.287 kPa ⋅ m 3 /kg ⋅ K )(35 + 273 K ) v= = = 0.8840 m 3 /kg P 100 kPaThe mass flow rate at the entrance is then Ai Vi (1 m 2 )(2 m/s) mi = & = = 2.262 kg/s v 0.8840 m 3 /kgwhile that at the outlet is AeVe (0.5 m 2 )(1 m/s) me = & = = 0.5656 kg/s v 0.8840 m 3 /kgApplying a mass balance to the balloon, min − m out = Δmsystem mi − m e = m 2 − m1 m 2 − m1 = (mi − me )Δt = [(2.262 − 0.5656) kg/s](2 × 60 s) = 203.6 kg & &The volume in the balloon then changes by the amount ΔV = (m 2 − m1 )v = (203.6 kg)(0.8840 m 3 /kg) = 180 m 3and the final volume of the balloon is V 2 = V1 + ΔV = 75 + 180 = 255 m 3In order to push back the boundary of the balloon against the surrounding atmosphere, the amount of workthat must be done is ⎛ 1 kJ ⎞ Wb,out = PΔV = (100 kPa)(180 m 3 )⎜ ⎟ = 18,000 kJ ⎝ 1 kPa ⋅ m 3 ⎠PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 116. 5-1165-141 An insulated rigid tank initially contains helium gas at high pressure. A valve is opened, and half ofthe mass of helium is allowed to escape. The final temperature and pressure in the tank are to bedetermined.Assumptions 1 This is an unsteady process since the conditions within the device are changing during theprocess, but it can be analyzed as a uniform-flow process by using constant average properties for thehelium leaving the tank. 2 Kinetic and potential energies are negligible. 3 There are no work interactionsinvolved. 4 The tank is insulated and thus heat transfer is negligible. 5 Helium is an ideal gas with constantspecific heats.Properties The specific heat ratio of helium is k =1.667 (Table A-2).Analysis We take the tank as the system, which is a control volume since mass crosses the boundary.Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h andinternal energy u, respectively, the mass and energy balances for this uniform-flow system can beexpressed asMass balance: min − mout = Δmsystem → me = m1 − m2 m 2 = 1 m1 (given) ⎯ 2 ⎯→ me = m 2 = 1 m1 2 HeEnergy balance: 0.08 m3 Ein − Eout = ΔEsystem 2 MPa 1 24 4 3 123 4 4 Net energy transfer Change in internal, kinetic, 80°C by heat, work, and mass potential, etc. energies − me he = m2u2 − m1u1 (since W ≅ Q ≅ ke ≅ pe ≅ 0)Note that the state and thus the enthalpy of helium leaving the tank is changing during this process. But forsimplicity, we assume constant properties for the exiting steam at the average values.Combining the mass and energy balances: 0 = 2 m1he + 2 m1u2 − m1u1 1 1 T1 + T2Dividing by m1/2 0 = he + u 2 − 2u1 or 0 = c p + cv T2 − 2cv T1 2Dividing by cv: 0 = k (T1 + T2 ) + 2T2 − 4T1 since k = c p / cvSolving for T2: T2 = (4 − k ) T = (4 − 1.667) (353 K ) = 225 K (2 + k ) 1 (2 + 1.667)The final pressure in the tank is P1V m RT m T 1 225 = 1 1 ⎯ ⎯→ P2 = 2 2 P1 = (2000 kPa ) = 637 kPa P2V m 2 RT2 m1T2 2 353PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 117. 5-1175-142E An insulated rigid tank equipped with an electric heater initially contains pressurized air. A valveis opened, and air is allowed to escape at constant temperature until the pressure inside drops to 30 psia.The amount of electrical work transferred is to be determined.Assumptions 1 This is an unsteady process since the conditions within the device are changing during theprocess, but it can be analyzed as a uniform-flow process since the exit temperature (and enthalpy) of airremains constant. 2 Kinetic and potential energies are negligible. 3 The tank is insulated and thus heattransfer is negligible. 4 Air is an ideal gas with variable specific heats.Properties The gas constant of air is R =0.3704 psia.ft3/lbm.R (Table A-1E). The properties of air are(Table A-17E) Ti = 580 R ⎯→ ⎯ hi = 138.66 Btu / lbm T1 = 580 R ⎯→ ⎯ u1 = 98.90 Btu / lbm T2 = 580 R ⎯→ ⎯ u2 = 98.90 Btu / lbmAnalysis We take the tank as the system, which is a control volume since mass crosses the boundary.Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h andinternal energy u, respectively, the mass and energy balances for this uniform-flow system can beexpressed asMass balance: min − mout = Δmsystem → me = m1 − m2Energy balance: AIR Ein − Eout = ΔEsystem 60 ft3 1 24 4 3 123 4 4 75 psia Net energy transfer Change in internal, kinetic, We by heat, work, and mass potential, etc. energies 120°F We,in − me he = m2u2 − m1u1 (since Q ≅ ke ≅ pe ≅ 0)The initial and the final masses of air in the tank are m1 = P1V = (75 psia ) 60 ft 3 ) ( = 20.95 lbm RT1 ( ) 0.3704 psia ⋅ ft 3 /lbm ⋅ R (580 R ) m2 = P2V = (30 psia )(60 ft ) 3 RT2 (0.3704 psia ⋅ ft /lbm ⋅ R )(580 R ) = 8.38 lbm 3Then from the mass and energy balances, me = m1 − m2 = 20.95 − 8.38 = 12.57 lbm We,in = me he + m2u2 − m1u1 = (12.57 lbm )(138.66 Btu/lbm ) + (8.38 lbm )(98.90 Btu/lbm ) − (20.95 lbm )(98.90 Btu/lbm ) = 500 BtuPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 118. 5-1185-143 A vertical cylinder initially contains air at room temperature. Now a valve is opened, and air isallowed to escape at constant pressure and temperature until the volume of the cylinder goes down by half.The amount air that left the cylinder and the amount of heat transfer are to be determined.Assumptions 1 This is an unsteady process since the conditions within the device are changing during theprocess, but it can be analyzed as a uniform-flow process since the exit temperature (and enthalpy) of airremains constant. 2 Kinetic and potential energies are negligible. 3 There are no work interactions. 4 Air isan ideal gas with constant specific heats. 5 The direction of heat transfer is to the cylinder (will beverified).Properties The gas constant of air is R = 0.287 kPa.m3/kg.K (Table A-1).Analysis (a) We take the cylinder as the system, which is a control volume since mass crosses theboundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented byenthalpy h and internal energy u, respectively, the mass and energy balances for this uniform-flow systemcan be expressed asMass balance: min − mout = Δmsystem → me = m1 − m2Energy balance: AIR Ein − Eout = ΔEsystem 300 kPa 1 24 4 3 123 4 4 Net energy transfer Change in internal, kinetic, 0.2 m3 by heat, work, and mass potential, etc. energies 20°C Qin + Wb,in − me he = m2u2 − m1u1 (since ke ≅ pe ≅ 0)The initial and the final masses of air in the cylinder are m1 = P1V 1 = (300 kPa ) 0.2 m 3 ( ) = 0.714 kg RT1 ( 0.287 kPa ⋅ m 3 /kg ⋅ K (293 K ) ) m2 = P2V 2 = (300 kPa ) 0.1 m 3 ( ) = 0.357 kg = 1 m1 RT2 ( 0.287 kPa ⋅ m 3 /kg ⋅ K (293 K ) ) 2Then from the mass balance, me = m1 − m2 = 0.714 − 0.357 = 0.357 kg(b) This is a constant pressure process, and thus the Wb and the ΔU terms can be combined into ΔH to yield Q = me he + m2 h2 − m1h1Noting that the temperature of the air remains constant during this process, we have hi = h1 = h2 = h.Also, me = m2 = 1 m1 . 2Thus, Q= (1 m1 + 1 m1 − m1 )h = 0 2 2PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 119. 5-1195-144 A balloon is initially filled with helium gas at atmospheric conditions. The tank is connected to asupply line, and helium is allowed to enter the balloon until the pressure rises from 100 to 150 kPa. Thefinal temperature in the balloon is to be determined.Assumptions 1 This is an unsteady process since the conditions within the device are changing during theprocess, but it can be analyzed as a uniform-flow process since the state of fluid at the inlet remainsconstant. 2 Helium is an ideal gas with constant specific heats. 3 The expansion process is quasi-equilibrium. 4 Kinetic and potential energies are negligible. 5 There are no work interactions involvedother than boundary work. 6 Heat transfer is negligible.Properties The gas constant of helium is R = 2.0769 kJ/kg·K (Table A-1). The specific heats of helium arecp = 5.1926 and cv = 3.1156 kJ/kg·K (Table A-2a).Analysis We take the cylinder as the system, which is a control volume since mass crosses the boundary.Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h andinternal energy u, respectively, the mass and energy balances for this uniform-flow system can beexpressed asMass balance: min − mout = Δmsystem → mi = m2 − m1Energy balance: Ein − Eout = ΔEsystem 1 24 4 3 123 4 4 Net energy transfer Change in internal, kinetic, by heat, work, and mass potential, etc. energies mi hi = Wb, out + m2u2 − m1u1 (since Q ≅ ke ≅ pe ≅ 0) m1 = PV1 1 = (100 kPa ) 65 m3 ( = 10.61 kg ) RT1 ( 2.0769 kPa ⋅ m 3/kg ⋅ K (295 K ) ) P V1 1 = P2 V 2 P ⎯ V 2 = 2 V1 = ⎯→ P 150 kPa 100 kPa 65 m3 = 97.5 m3 ( ) 1 m2 = P2V 2 = (150 kPa ) 97.5 m3 ( = 7041.74) RT2 ( 2.0769 kPa ⋅ m /kg ⋅ K (T2 K ) 3 T2 ) kgThen from the mass balance, 7041.74 mi = m2 − m1 = − 10.61 kg T2Noting that P varies linearly with V, the boundary work done during this process is (V 2 −V1 ) = (100 + 150)kPa (97.5 − 65)m 3 = 4062.5 kJ P1 + P2 Wb = 2 2Using specific heats, the energy balance relation reduces to Wb,out = mi c pTi − m2cv T2 + m1cv T1Substituting, ⎛ 7041.74 ⎞ 7041.74 4062.5 = ⎜ ⎜ T − 10.61⎟(5.1926 )(298) − ⎟ (3.1156)T2 + (10.61)(3.1156)(295) ⎝ 2 ⎠ T2It yields T2 = 333.6 KPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 120. 5-1205-145 An insulated piston-cylinder device with a linear spring is applying force to the piston. A valve atthe bottom of the cylinder is opened, and refrigerant is allowed to escape. The amount of refrigerant thatescapes and the final temperature of the refrigerant are to be determined.Assumptions 1 This is an unsteady process since the conditions within the device are changing during theprocess, but it can be analyzed as a uniform-flow process assuming that the state of fluid leaving the deviceremains constant. 2 Kinetic and potential energies are negligible.Properties The initial properties of R-134a are (Tables A-11 through A-13) 3 P1 = 1.2 MPa ⎫v 1 = 0.02423 m /kg ⎬u1 = 325.03 kJ/kg T1 = 120°C ⎭ h1 = 354.11 kJ/kgAnalysis We take the tank as the system, which is a control volume since mass crosses the boundary.Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h andinternal energy u, respectively, the mass and energy balances for this uniform-flow system can beexpressed asMass balance: min − mout = Δmsystem → me = m1 − m2Energy balance: Ein − Eout = ΔEsystem 1 24 4 3 123 4 4 Net energy transfer Change in internal, kinetic, by heat, work, and mass potential, etc. energies Wb,in − me he = m2u2 − m1u1 (since Q ≅ ke ≅ pe ≅ 0) R-134a 0.8 m3The initial mass and the relations for the final and exiting masses are 1.2 MPa V1 0.8 m3 120°C m1 = = = 33.02 kg v1 0.02423 m3/kg V 2 0.5 m3 m2 = = v2 v2 0.5 m3 me = m1 − m2 = 33.02 − v2Noting that the spring is linear, the boundary work can be determined from P + P2 (1200 + 600) kPa W b,in = 1 (V 1 −V 2 ) = (0.8 - 0.5) m 3 = 270 kJ 2 2Substituting the energy balance, ⎛ 0.5 m 3 ⎞ ⎛ 3⎞ 270 − ⎜ 33.02 − ⎟he = ⎜ 0.5 m ⎟u2 − (33.02 kg)(325.03 kJ/kg) (Eq. 1) ⎜ v2 ⎟ ⎜ v ⎟ ⎝ ⎠ ⎝ 2 ⎠where the enthalpy of exiting fluid is assumed to be the average of initial and final enthalpies of therefrigerant in the cylinder. That is, h + h2 (354.11 kJ/kg) + h2 he = 1 = 2 2Final state properties of the refrigerant (h2, u2, and v2) are all functions of final pressure (known) andtemperature (unknown). The solution may be obtained by a trial-error approach by trying different finalstate temperatures until Eq. (1) is satisfied. Or solving the above equations simultaneously using anequation solver with built-in thermodynamic functions such as EES, we obtain T2 = 96.8°C, me = 22.47 kg, h2 = 336.20 kJ/kg, u2 = 307.77 kJ/kg, v2 = 0.04739 m3/kg, m2 = 10.55 kgPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 121. 5-1215-146 Steam at a specified state is allowed to enter a piston-cylinder device in which steam undergoes aconstant pressure expansion process. The amount of mass that enters and the amount of heat transfer are tobe determined.Assumptions 1 This is an unsteady process since the conditions within the device are changing during theprocess, but it can be analyzed as a uniform-flow process since the state of fluid entering the deviceremains constant. 2 Kinetic and potential energies are negligible.Properties The properties of steam at various states are (Tables A-4 through A-6) V1 0.1 m 3 v1 = = =0.16667 m 3 /kg m1 0.6 kg P2 = P1 P1 = 800 kPa ⎫ ⎪ Q 3 ⎬u1 = 2004.4 kJ/kg Steam v 1 = 0.16667 m /kg ⎪ ⎭ 0.6 kg 0.1 m3 P2 = 800 kPa ⎫v 2 = 0.29321 m 3 /kg 800 kPa Steam ⎬ 5 MPa T2 = 250°C ⎭u 2 = 2715.9 kJ/kg 500°C Pi = 5 MPa ⎫ ⎬hi = 3434.7 kJ/kg Ti = 500°C ⎭Analysis (a) We take the tank as the system, which is a control volume since mass crosses the boundary.Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h andinternal energy u, respectively, the mass and energy balances for this uniform-flow system can beexpressed asMass balance: min − mout = Δmsystem → mi = m2 − m1Energy balance: Ein − Eout = ΔEsystem 1 24 4 3 123 4 4 Net energy transfer Change in internal, kinetic, by heat, work, and mass potential, etc. energies Qin − Wb,out + mi hi = m2u2 − m1u1 (since ke ≅ pe ≅ 0)Noting that the pressure remains constant, the boundary work is determined from W b,out = P (V 2 −V 1 ) = (800 kPa)(2 × 0.1 − 0.1)m 3 = 80 kJThe final mass and the mass that has entered are V2 0.2 m 3 m2 = = = 0.682 kg v 2 0.29321 m 3 /kg m i = m 2 − m1 = 0.682 − 0.6 = 0.082 kg(b) Finally, substituting into energy balance equation Qin − 80 kJ + (0.082 kg)(3434.7 kJ/kg) = (0.682 kg)(2715.9 kJ/kg) − (0.6 kg)(2004.4 kJ/kg) Qin = 447.9 kJPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 122. 5-122Review Problems5-147 A water tank open to the atmosphere is initially filled with water. The tank discharges to theatmosphere through a long pipe connected to a valve. The initial discharge velocity from the tank and thetime required to empty the tank are to be determined.Assumptions 1 The flow is incompressible. 2 The draining pipe is horizontal. 3 The tank is considered tobe empty when the water level drops to the center of the valve.Analysis (a) Substituting the known quantities, the discharge velocity can be expressed as 2 gz 2 gz V= = = 0.1212 gz 1.5 + fL / D 1.5 + 0.015(100 m)/(0.10 m)Then the initial discharge velocity becomes V1 = 0.1212 gz1 = 0.1212(9.81 m/s 2 )(2 m) = 1.54 m/swhere z is the water height relative to the center ofthe orifice at that time. z D0(b) The flow rate of water from the tank can be Dobtained by multiplying the discharge velocity bythe pipe cross-sectional area, πD 2 V& = ApipeV 2 = 0.1212 gz 4Then the amount of water that flows through the pipe during a differential time interval dt is πD 2 dV = V&dt = 0.1212 gz dt (1) 4which, from conservation of mass, must be equal to the decrease in the volume of water in the tank, 2 πD0 dV = Atank (−dz ) = − dz (2) 4where dz is the change in the water level in the tank during dt. (Note that dz is a negative quantity since thepositive direction of z is upwards. Therefore, we used –dz to get a positive quantity for the amount of waterdischarged). Setting Eqs. (1) and (2) equal to each other and rearranging, 2 2 2 πD 2 πD 0 D0 dz D0 −1 0.1212 gz dt = − dz → dt = − =− z 2 dz 4 4 D2 0.1212 gz D 2 0.1212 gThe last relation can be integrated easily since the variables are separated. Letting tf be the discharge timeand integrating it from t = 0 when z = z1 to t = tf when z = 0 (completely drained tank) gives 1 0 2 2 2 tf D0 0 D0 z2 2 D0 1 ∫t =0 dt = − D 2 0.1212 g ∫ z = z1 z −1 / 2 dz → t f = - D 2 0.1212 g 1 2 = D 2 0.1212 g z12 z1Simplifying and substituting the values given, the draining time is determined to be 2 2 D0 z1 2(10 m) 2 2m tf = = = 25,940 s = 7.21 h D2 0.1212 g (0.1 m) 2 0.1212(9.81 m/s 2 )Discussion The draining time can be shortened considerably by installing a pump in the pipe.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 123. 5-1235-148 The rate of accumulation of water in a pool and the rate of discharge are given. The rate supply ofwater to the pool is to be determined.Assumptions 1 Water is supplied and discharged steadily. 2 The rate of evaporation of water is negligible.3 No water is supplied or removed through other means.Analysis The conservation of mass principle applied to the pool requires that the rate of increase in theamount of water in the pool be equal to the difference between the rate of supply of water and the rate ofdischarge. That is, dm pool dm pool dV pool = mi − m e & & → mi = & + me & → V&i = + V&e dt dt dtsince the density of water is constant and thus the conservation ofmass is equivalent to conservation of volume. The rate of dischargeof water is V&e = AeVe = (πD 2 /4)Ve = [π (0.05 m) 2 /4](5 m/s) = 0.00982 m 3 /sThe rate of accumulation of water in the pool is equal to the cross-section of the pool times the rate at which the water level rises, dV pool = Across-sectionVlevel = (3 m × 4 m)(0.015 m/min) = 0.18 m 3 /min = 0.00300 m 3 /s dtSubstituting, the rate at which water is supplied to the pool is determined to be dV pool V&i = + V&e = 0.003 + 0.00982 = 0.01282 m 3 /s dtTherefore, water is supplied at a rate of 0.01282 m3/s = 12.82 L/s.5-149 A fluid is flowing in a circular pipe. A relation is to be obtained for the average fluid velocity intherms of V(r), R, and r.Analysis Choosing a circular ring of area dA = 2πrdr as our differential area, the mass flow rate through across-sectional area can be expressed as R dr & ∫ ∫ m = ρV (r )dA = ρV (r )2π r dr 0 r A RSolving for Vavg, 2 ∫ V (r )r dr R Vavg = R2 0PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 124. 5-1245-150 Air is accelerated in a nozzle. The density of air at the nozzle exit is to be determined.Assumptions Flow through the nozzle is steady.Properties The density of air is given to be 4.18 kg/m3 at the inlet.Analysis There is only one inlet and one exit, and thus m1 = m2 = m . Then, & & & m1 = m 2 & & 1 AIR 2 ρ 1 A1V1 = ρ 2 A2V 2 A V 120 m/s ρ 2 = 1 1 ρ1 = 2 (4.18 kg/m 3 ) = 2.64 kg/m 3 A2 V 2 380 m/sDiscussion Note that the density of air decreases considerably despite a decrease in the cross-sectional areaof the nozzle.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 125. 5-1255-151E A heat exchanger that uses hot air to heat cold water is considered. The total flow power and theflow works for both the air and water streams are to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potentialenergy changes are negligible. 3 There are no work interactions. 4 Heat loss from the device to thesurroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the coldfluid. 5 Air is an ideal gas with constant specific heats at room temperature.Properties The gas constant of air is 0.3704 psia.ft3/lbm.R =0.06855 Btu/lbm.R (Table A-1E). The specificvolumes of water at the inlet and exit are (Table A-4E) P3 = 20 psia ⎫ 3 AIR ⎬ v 3 ≅ v f @ 50° F = 0.01602 ft /lbm T3 = 50°F ⎭ 1 P4 = 17 psia ⎫ 3 Water ⎬ v 4 ≅ v f @ 90° F = 0.01610 ft /lbm T4 = 90°F ⎭ 3Analysis The specific volume of air at the inlet and the mass flow rate are 4 RT (0.3704 psia ⋅ ft 3 /lbm ⋅ R)(200 + 460 R) 2 v1 = 1 = = 12.22 ft 3 /lbm P1 20 psia V&1 (100 / 60) ft 3 /s m= & = = 0.1364 lbm/s v 1 12.22 ft 3 /lbmCombining the flow work expression with the ideal gas equation of state gives wflow = P2v 2 − P1v 1 = R(T2 − T1 ) = (0.06855 Btu/lbm ⋅ R)(100 − 200)R = −6.855 Btu/lbmThe flow work of water is wflow = P4v 4 − P3v 3 [ ⎛ = (17 psia)(0.01610 ft 3 /lbm) − (20 psia)(0.01602 ft 3 /lbm) ⎜] 1 Btu ⎜ 5.404 psia ⋅ ft 3 ⎞ ⎟ ⎟ ⎝ ⎠ = −0.00864 Btu/lbmThe net flow power for the heat exchanger is & Wflow = m air wflow + m air wflow & & = (0.1364 lbm/s)(-6.855 Btu/lbm) + (0.5 lbm/s)(-0.00864 Btu/lbm) ⎛ 1 hp ⎞ = −0.9393 Btu/s⎜ ⎟ = −1.329 hp ⎝ 0.7068 Btu/s ⎠PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 126. 5-1265-152 An air compressor consumes 4.5 kW of power to compress a specified rate of air. The flow workrequired by the compressor is to be compared to the power used to increase the pressure of the air.Assumptions 1 Flow through the compressor is steady. 2 Air is an ideal gas.Properties The gas constant of air is 0.287 kPa·m3/kg·K (Table A-1). 1 MPaAnalysis The specific volume of the air at the inlet is 300°C RT1 (0.287 kPa ⋅ m 3 /kg ⋅ K )(20 + 273 K ) v1 = = = 0.7008 m 3 /kg Compressor P1 120 kPaThe mass flow rate of the air is 120 kPa V 0.010 m 3 /s 20°C m= 1 = & = 0.01427 kg/s v 1 0.7008 m 3 /kgCombining the flow work expression with the ideal gas equation of state gives the flow work as wflow = P2v 2 − P1v 1 = R(T2 − T1 ) = (0.287 kJ/kg ⋅ K)(300 − 20)K = 80.36 kJ/kgThe flow power is & Wflow = mwflow = (0.01427 kg/s)(80.36 kJ/kg) = 1.147 kW &The remainder of compressor power input is used to increase the pressure of the air: & & & W = W total,in − Wflow = 4.5 − 1.147 = 3.353 kWPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 127. 5-1275-153 Steam expands in a turbine whose power production is 9000 kW. The rate of heat lost from theturbine is to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potentialenergy changes are negligible.Properties From the steam tables (Tables A-6 and A-4) P1 = 1.6 MPa ⎫ 1.6 MPa ⎬ h1 = 3146.0 kJ/kg 350°C Heat T1 = 350°C ⎭ 16 kg/s T2 = 30°C ⎫ ⎬ h 2 = 2555.6 kJ / kg x2 = 1 ⎭ TurbineAnalysis We take the turbine as the system, which is a control volumesince mass crosses the boundary. Noting that there is one inlet and one 30°Cexiti the energy balance for this steady-flow system can be expressed in sat. vap.the rate form as & & E in − E out = ΔE system 0 (steady) & =0 1 24 4 3 1442443 4 4 Rate of net energy transfer Rate of change in internal, kinetic, by heat, work, and mass potential, etc. energies & & E in = E out & & m1 h1 = m 2 h2 + Wout + Qout & & & & Qout = m(h1 − h2 ) − Wout &Substituting, & Qout = (16 kg/s)(3146.0 − 2555.6) kJ/kg − 9000 kW = 446.4 kWPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 128. 5-1285-154E Nitrogen gas flows through a long, constant-diameter adiabatic pipe. The velocities at the inlet andexit are to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Nitrogen is an ideal gaswith constant specific heats. 3 Potential energy changes are negligible. 4 There are no work interactions. 5The re is no heat transfer from the nitrogen.Properties The specific heat of nitrogen at the room temperature iss cp = 0.248 Btu/lbm⋅R (Table A-2Ea).Analysis There is only one inlet and one exit, and thus m1 = m2 = m . We take the pipe as the system, & & &which is a control volume since mass crosses the boundary. The energy balance for this steady-flow systemcan be expressed in the rate form as & & E in − E out = ΔE system 0 (steady) & =0 1 24 4 3 1442443 4 Rate of net energy transfer Rate of change in internal, kinetic, by heat, work, and mass potential, etc. energies & & E in = E out 100 psia N2 50 psia 120°F 70°F m(h1 + V12 / 2) & = m(h2 + V 22 /2) & h1 + V12 / 2 = h2 + V 22 /2 V12 − V 22 = c p (T2 − T1 ) 2Combining the mass balance and ideal gas equation of state yields m1 = m 2 & & A1V1 A2V 2 = v1 v2 A v v T P V 2 = 1 2 V1 = 2 V1 = 2 1 V1 A2 v 1 v1 T1 P2Substituting this expression for V2 into the energy balance equation gives 0.5 ⎡ ⎤ ⎡ ⎤ 0.5 ⎢ ⎥ ⎢ ⎥ ⎢ 2c p (T2 − T1 ) ⎥ 2(0.248)(70 − 120) ⎛ 25,037 ft 2 /s 2 ⎞⎥ V1 = ⎢ 2⎥ =⎢ ⎢ ⎜ ⎟ = 515 ft/s 2 ⎜ ⎟⎥ ⎢ ⎛ T2 P1 ⎞ ⎥ ⎢ 1− ⎜⎛ 530 100 ⎞ ⎝ 1 Btu/lbm ⎠⎥ ⎜ ⎢1 − ⎜ ⎟ ⎥ ⎟ ⎟ ⎢ ⎣ ⎝ 580 50 ⎠ ⎥ ⎦ ⎣ ⎝ T1 P2 ⎠ ⎦The velocity at the exit is T2 P1 530 100 V2 = V1 = 515 = 941 ft/s T1 P2 580 50PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 129. 5-1295-155 Water at a specified rate is heated by an electrical heater. The current is to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potentialenergy changes are negligible. 3 The heat losses from the water is negligible.Properties The specific heat and the density of water are taken to be cp = 4.18 kJ/kg·°C and ρ = 1 kg/L(Table A-3).Analysis We take the pipe in which water is heated as the system, which is a control volume. The energybalance for this steady-flow system can be expressed in the rate form as & & E in − E out = ΔE system 0 (steady) & =0 1 24 4 3 1442443 4 Rate of net energy transfer Rate of change in internal, kinetic, by heat, work, and mass potential, etc. energies . & & E in = E out We,in & & mh1 + We,in = mh2 & 15°C water 20°C & We,in = m(h2 − h1 ) & 0.1 L/s VI = mc p (T2 − T1 ) &The mass flow rate of the water is m = ρV& = (1 kg/L)(0.1 L/s) = 0.1 kg/s &Substituting into the energy balance equation and solving for the current gives mc p (T2 − T1 ) & (0.1 kg/s)(4.18 kJ/kg ⋅ K)(20 − 15)K ⎛ 1000 VI ⎞ I= = ⎜ ⎟ = 19.0 A V 110 V ⎝ 1 kJ/s ⎠PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 130. 5-1305-156 Steam flows in an insulated pipe. The mass flow rate of the steam and the speed of the steam at thepipe outlet are to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potentialenergy changes are negligible. 3 There are no work and heat interactions.Analysis We take the pipe in which steam flows as the system, which is a control volume. The energybalance for this steady-flow system can be expressed in the rate form as & & E in − E out = ΔE system 0 (steady) & =0 1 24 4 3 1442443 4Rate of net energy transfer Rate of change in internal, kinetic,by heat, work, and mass potential, etc. energies & & E in = E out 1400 kPa, 350°C Water 1000 kPa mh1 = mh2 & & D=0.15 m, 10 m/s D=0.1 m h1 = h2The properties of the steam at the inlet and exit are (Table A-6) P = 1400 kPa ⎫ v 1 = 0.20029 m 3 /kg ⎬ T = 350°C ⎭ h1 = 3150.1 kJ/kg P2 = 1000 kPa ⎫ 3 ⎬ v 2 = 0.28064 m /kg h2 = h1 = 3150.1 kJ/kg ⎭The mass flow rate is A1V1 πD12 V1 π (0.15 m)2 10 m/s m= & = = = 0.8823 kg/s v1 4 v1 4 0.20029 m3/kgThe outlet velocity will then be mv 2 4mv 2 4(0.8823 kg/s)(0.28064 m 3 /kg) & & V2 = = 2 = = 31.53 m/s A2 πD 2 π (0.10 m) 2PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 131. 5-1315-157 The mass flow rate of a compressed air line is divided into two equal streams by a T-fitting in theline. The velocity of the air at the outlets and the rate of change of flow energy (flow power) across the T-fitting are to be determined.Assumptions 1 Air is an ideal gas with constant specific heats. 2 The flow is steady. 3 Since the outlets areidentical, it is presumed that the flow divides evenly between the two.Properties The gas constant of air is R = 0.287 kPa⋅m3/kg⋅K (Table A-1). 1.4 MPaAnalysis The specific volumes of air at the inlet and outlets are 36°C RT1 (0.287 kPa ⋅ m 3 /kg ⋅ K)(40 + 273 K)v1 = = = 0.05614 m 3 /kg P1 1600 kPa 1.6 MPa 3 RT2 (0.287 kPa ⋅ m /kg ⋅ K)(36 + 273 K) 40°Cv2 =v3 = = = 0.06335 m 3 /kg 50 m/s P2 1400 kPaAssuming an even division of the inlet flow rate, the energybalance can be written as 1.4 MPa 36°C A1V1 A2V 2 A1 v 2 V1 0.06335 50 =2 ⎯ V 2 = V3 = ⎯→ = = 28.21 m/s v1 v2 A2 v 1 2 0.05614 2The mass flow rate at the inlet is A1V1 πD12 V1 π (0.025 m)2 50 m/s m1 = & = = = 0.4372 kg/s v1 4 v1 4 0.05614 m3/kgwhile that at the outlets is m1 0.4372 kg/s & m2 = m3 = & & = = 0.2186 kg/s 2 2Substituting the above results into the flow power expression produces & Wflow = 2m2 P2v 2 − m1Pv1 & & 1 = 2(0.2186 kg/s)(1400 kPa)(0.06335 m3/kg) − (0.4372 kg/s)(1600 kPa)(0.05614 m3/kg) = −0.496 kWPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 132. 5-1325-158 Air flows through a non-constant cross-section pipe. The inlet and exit velocities of the air are to bedetermined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Potential energy changeis negligible. 3 There are no work interactions. 4 The device is adiabatic and thus heat transfer isnegligible. 5 Air is an ideal gas with constant specific heats.Properties The gas constant of air is R = 0.287 kPa.m3/kg.K. Also, cp = 1.005 kJ/kg.K for air at roomtemperature (Table A-2)Analysis We take the pipe as the system,which is a control volume since mass q D1 D2crosses the boundary. The mass and energy 200 kPa 150 kPabalances for this steady-flow system can be Air 50°C 40°Cexpressed in the rate form as V1 V2Mass balance: 0 (steady)min − mout = Δmsystem& & & =0 P πD12 P πD22 P P 2 min = mout ⎯ ρ1 A1V1 = ρ 2 A2V2 ⎯ & & ⎯→ ⎯→ 1 V1 = 2 ⎯→ 1 D12V1 = 2 D2V2 V2 ⎯ RT1 4 RT2 4 T1 T2Energy balance: & & Ein − Eout = ΔEsystem 0 (steady) & =0 & since W ≅ Δpe ≅ 0) 1 24 4 3 1442443 Rate of net energy transfer Rate of change in internal, kinetic, by heat, work, and mass potential, etc. energies 2 2 & & V V Ein = Eout ⎯ h1 + 1 = h2 + 2 + qout ⎯→ 2 2 V12 V2or c p T1 + = c p T2 + 2 + q out 2 2Assuming inlet diameter to be 1.8 m and the exit diameter to be 1.0 m, and substituting given values intomass and energy balance equations ⎛ 200 kPa ⎞ 2 ⎛ 150 kPa ⎞ 2 ⎜ ⎟(1.8 m) V1 = ⎜ ⎟(1.0 m) V 2 (1) ⎝ 323 K ⎠ ⎝ 313 K ⎠ V12 ⎛ 1 kJ/kg ⎞ V 2 ⎛ 1 kJ/kg ⎞(1.005 kJ/kg.K)(323 K) + ⎜ ⎟ = (1.005 kJ/kg.K)(313 K) + 2 ⎜ ⎟ + 3.3 kJ/kg (2) 2 ⎝ 1000 m 2 /s 2 ⎠ 2 ⎝ 1000 m 2 /s 2 ⎠There are two equations and two unknowns. Solving equations (1) and (2) simultaneously using anequation solver such as EES, the velocities are determined to be V1 = 28.6 m/s V2 = 120 m/sPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 133. 5-1335-159 A water tank is heated by electricity. The water withdrawn from the tank is mixed with cold water ina chamber. The mass flow rate of hot water withdrawn from the tank and the average temperature of mixedwater are to be determined.Assumptions 1 The process in the mixing chamber is a steady-flow process since there is no change withtime. 2 Kinetic and potential energy changes are negligible. 3 There are no work interactions. 4 The deviceis adiabatic and thus heat transfer is negligible.Properties The specific heat and density of water aretaken to be cp = 4.18 kJ/kg.K, ρ = 100 kg/m3 (TableA-3). 20°C Tank T1 = 80°CAnalysis We take the mixing chamber as the system, T2 = 60°Cwhich is a control volume since mass crosses theboundary. The energy balance for this steady-flow & msystem can be expressed in the rate form as Tmix MixingEnergy balance: chamber & & Ein − Eout = ΔEsystem 0 (steady) & =0 1 24 4 3 1442443 20°C Rate of net energy transfer Rate of change in internal, kinetic, 0.06 kg/s by heat, work, and mass potential, etc. energies & & Ein = Eout m1h1 + m2 h2 = m3h3 & & & & & (since Q ≅ W ≅ Δke ≅ Δpe ≅ 0)or m hot c p Ttank,ave + mcold c p Tcold = (m hot + mcold )c p Tmixture & & & & (1)Similarly, an energy balance may be written on the water tank as [We,in + m hot c p (Tcold − Ttank,ave )]Δt = m tank c p (Ttank,2 − Ttank,1 ) & & (2) Ttank,1 + Ttank,2 80 + 60where Ttank,ave = = = 70°C 2 2and m tank = ρV = (1000 kg/m 3 )(0.060 m 3 ) = 60 kgSubstituting into Eq. (2), [1.6 kJ/s + mhot (4.18 kJ/kg.°C)(20 − 70)°C](8 × 60 s) = (60 kg)(4.18 kJ/kg.°C)(60 − 80)°C & ⎯ m hot = 0.0577 kg/s ⎯→ &Substituting into Eq. (1), (0.0577 kg/s)(4.18 kJ/kg.°C)(70°C) + (0.06 kg/s)(4.18 kJ/kg.°C)(20°C) = [(0.0577 + 0.06)kg/s](4.18 kJ/kg.°C)Tmixture ⎯ Tmixture = 44.5°C ⎯→PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 134. 5-1345-160 Water is boiled at a specified temperature by hot gases flowing through a stainless steel pipesubmerged in water. The rate of evaporation of is to be determined.Assumptions 1 Steady operating conditions exist. 2 Heat losses fromthe outer surfaces of the boiler are negligible.Properties The enthalpy of vaporization of water at 150°C ishfg = 2113.8 kJ/kg (Table A-4). Analysis The rate of heat transfer to water is Watergiven to be 74 kJ/s. Noting that the enthalpy of 150°Cvaporization represents the amount of energyneeded to vaporize a unit mass of a liquid at a Heaterspecified temperature, the rate of evaporation of Hot gaseswater is determined to be & Qboiling 74 kJ/s mevaporation = & = = 0.0350 kg/s h fg 2113.8 kJ/kg5-161 Cold water enters a steam generator at 20°C, and leaves as saturated vapor at Tsat = 150°C. Thefraction of heat used to preheat the liquid water from 20°C to saturation temperature of 150°C is to bedetermined.Assumptions 1 Steady operating conditions exist. 2 Heat losses from the steam generator are negligible. 3The specific heat of water is constant at the average temperature.Properties The heat of vaporization of water at 150°C is hfg = 2113.8 kJ/kg (Table A-4), and the specificheat of liquid water is c = 4.18 kJ/kg.°C (Table A-3).Analysis The heat of vaporization of water represents theamount of heat needed to vaporize a unit mass of liquid at aspecified temperature. Using the average specific heat, the Steamamount of heat transfer needed to preheat a unit mass of waterfrom 20°C to 150°C isq preheating = cΔT Water = ( 4.18 kJ/kg ⋅ °C)(150 − 20)°C = 543.4 kJ/kg 150°Cand Heater Cold water q total = q boiling + q preheating 20°C = 2113.8 + 543.4 = 2657.2 kJ/kgTherefore, the fraction of heat used to preheat the water is qpreheating 543.4 Fraction to preheat = = = 0.205 (or 20.5%) qtotal 2657.2PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 135. 5-1355-162 Cold water enters a steam generator at 20°C and is boiled, and leaves as saturated vapor at boilerpressure. The boiler pressure at which the amount of heat needed to preheat the water to saturationtemperature that is equal to the heat of vaporization is to be determined.Assumptions Heat losses from the steam generator are negligible.Properties The enthalpy of liquid water at 20°C is 83.91 kJ/kg. Other properties needed to solve thisproblem are the heat of vaporization hfg and the enthalpy of saturated liquid at the specified temperatures,and they can be obtained from Table A-4.Analysis The heat of vaporization of water represents the amount of heat needed to vaporize a unit mass ofliquid at a specified temperature, and Δh represents the amount of heat needed to preheat a unit mass ofwater from 20°C to the saturation temperature. Therefore, q preheating = q boiling (h f@Tsat − h f@20°C ) = h fg @ Tsat h f@Tsat − 83.91 kJ/kg = h fg @ Tsat → h f@Tsat − h fg @ Tsat = 83.91 kJ/kgThe solution of this problem requires choosing a boilingtemperature, reading hf and hfg at that temperature, andsubstituting the values into the relation above to see if it is Steamsatisfied. By trial and error, (Table A-4)At 310°C: h f @Tsat − h fg@Tsat = 1402.0 – 1325.9 = 76.1 kJ/kg WaterAt 315°C: h f @Tsat − h fg@Tsat = 1431.6 – 1283.4 = 148.2 kJ/kgThe temperature that satisfies this condition isdetermined from the two values above by Heater Cold waterinterpolation to be 310.6°C. The saturation 20°Cpressure corresponding to this temperature is9.94 MPa.5-163 Water is boiled at Tsat = 100°C by an electric heater. The rate of evaporation of water is to bedetermined. SteamAssumptions 1 Steady operating conditions exist. 2 Heat losses fromthe outer surfaces of the water tank are negligible.Properties The enthalpy of vaporization of water at 100°C ishfg = 2256.4 kJ/kg (Table A-4).Analysis Noting that the enthalpy of vaporization representsthe amount of energy needed to vaporize a unit mass of aliquid at a specified temperature, the rate of evaporation of Waterwater is determined to be 100°C & W e, boiling 3 kJ/s m evaporation = & = Heater h fg 2256.4 kJ/kg = 0.00133 kg/s = 4.79 kg/hPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 136. 5-1365-164 Two streams of same ideal gas at different states are mixed in a mixing chamber. The simplestexpression for the mixture temperature in a specified format is to be obtained.Analysis The energy balance for this steady-flow system can be expressed in the rate form as & & Ein − Eout = ΔEsystem 0 (steady) & =0 1 24 4 3 1442443 Rate of net energy transfer Rate of change in internal, kinetic, by heat, work, and mass potential, etc. energies & & Ein = Eout m1, T1 & & & & & m1h1 + m2 h2 = m3h3 (since Q = W = 0) Mixing m3, T3 m1c pT1 + m2c pT2 = m3c pT3 & & & deviceand, m2, T2 m3 = m1 + m 2 & & &Solving for final temperature, we find & m1 m& T3 = T1 + 2 T2 & m3 & m35-165 An ideal gas expands in a turbine. The volume flow rate at the inlet for a power output of 200 kW isto be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potentialenergy changes are negligible. 3 The device is adiabatic and thus heat transfer is negligible.Properties The properties of the ideal gas are given as R = 0.30 kPa.m3/kg.K, cp = 1.13 kJ/kg·°C, cv = 0.83kJ/kg·°C.Analysis We take the turbine as the system, which is a control volume since mass crosses the boundary.The energy balance for this steady-flow system can be expressed in the rate form as & & Ein − Eout = ΔEsystem 0 (steady) & =0 1 24 4 3 1442443 Rate of net energy transfer Rate of change in internal, kinetic, by heat, work, and mass potential, etc. energies P1 = 600 kPa T1 = 1200 K & & ⎯→ & & & Ein = Eout ⎯ mh1 = Wout + mh2 (since Q ≅ Δke = Δpe ≅ 0) &which can be rearranged to solve for mass flow rate & Wout & Wout 200 kWm=& = = = 0.354 kg/s Ideal gas h1 − h2 c p (T1 − T2 ) (1.13 kJ/kg.K)(1200 - 700)K 200 kWThe inlet specific volume and the volume flow rate are v1 = RT1 = ( 0.3 kPa ⋅ m 3 /kg ⋅ K (1200 K ) ) = 0.6 m 3 /kg T2 = 700 K P1 600 kPaThus, V& = mv 1 = (0.354 kg/s)(0.6 m 3 /kg) = 0.212 m 3 /s &PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 137. 5-1375-166 Chilled air is to cool a room by removing the heat generated in a large insulated classroom by lightsand students. The required flow rate of air that needs to be supplied to the room is to be determined.Assumptions 1 The moisture produced by the bodies leave the room as vapor without any condensing, andthus the classroom has no latent heat load. 2 Heat gain through the walls and the roof is negligible. 4 Air isan ideal gas with constant specific heats at room temperature. 5 Steady operating conditions exist.Properties The specific heat of air at room temperature is 1.005 kJ/kg⋅°C (Table A-2). The average rate ofsensible heat generation by a person is given to be 60 W.Analysis The rate of sensible heat generation by the people in the room and the total rate of sensibleinternal heat generation are & Qgen, sensible = q gen, sensible (No. of people) = (60 W/person)(150 persons) = 9000 W & & & & Q total, sensible = Qgen, sensible + Qlighting = 9000 + 6000 = 15,000 WBoth of these effects can be viewed as heat gain forthe chilled air stream, which can be viewed as a Chilled air Return airsteady stream of cool air that is heated as it flows in 15°C 25°Can imaginary duct passing through the room. Theenergy balance for this imaginary steady-flow systemcan be expressed in the rate form as & & Ein − Eout = ΔEsystem 0 (steady) & =0 1 24 4 3 1442443 Rate of net energy transfer Rate of change in internal, kinetic, by heat, work, and mass potential, etc. energies & & Ein = Eout & Qin + mh1 = mh2 (since Δke ≅ Δpe ≅ 0) & & & & Qin = Qtotal, sensible = mc p (T2 − T1 ) &Then the required mass flow rate of chilled air becomes & Q total, sensible 15 kJ/s m air = & = = 1.49 kg/s c p ΔT (1.005 kJ/kg ⋅ °C)(25 − 15)°CDiscussion The latent heat will be removed by the air-conditioning system as the moisture condensesoutside the cooling coils.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 138. 5-1385-167 Chickens are to be cooled by chilled water in an immersion chiller. The rate of heat removal fromthe chicken and the mass flow rate of water are to be determined.Assumptions 1 Steady operating conditions exist. 2 The thermal properties of chickens and water areconstant.Properties The specific heat of chicken are given to be 3.54 kJ/kg.°C. The specific heat of water is 4.18kJ/kg.°C (Table A-3).Analysis (a) Chickens are dropped into the chiller at a rate of 500 per hour. Therefore, chickens can beconsidered to flow steadily through the chiller at a mass flow rate of mchicken = (500 chicken / h)(2.2 kg / chicken) = 1100 kg / h = 0.3056 kg / s &Taking the chicken flow stream in the chiller as thesystem, the energy balance for steadily flowing Immersionchickens can be expressed in the rate form as chilling, 0.5°C & & Ein − Eout = ΔEsystem 0 (steady) & =0 1 24 4 3 1442443Rate of net energy transfer Rate of change in internal, kinetic,by heat, work, and mass potential, etc. energies & & Ein = Eout & & mh1 = Qout + mh2 (since Δke ≅ Δpe ≅ 0) & Chicken & & 15°C Q =Q out =m& c (T − T ) chicken chicken p 1 2Then the rate of heat removal from the chickens asthey are cooled from 15°C to 3ºC becomes & Q chicken =( mc p ΔT ) chicken = (0.3056 kg/s)(3.54 kJ/kg.º C)(15 − 3)º C = 13.0 kW &The chiller gains heat from the surroundings at a rate of 200 kJ/h = 0.0556 kJ/s. Then the total rate of heatgain by the water is & & & Qwater = Qchicken + Qheat gain = 13.0 + 0.056 = 13.056 kWNoting that the temperature rise of water is not to exceed 2ºC as it flows through the chiller, the mass flowrate of water must be at least & Q water 13.056 kW m water = & = = 1.56 kg/s (c p ΔT ) water (4.18 kJ/kg.º C)(2º C)If the mass flow rate of water is less than this value, then the temperature rise of water will have to be morethan 2°C.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 139. 5-1395-168 Chickens are to be cooled by chilled water in an immersion chiller. The rate of heat removal fromthe chicken and the mass flow rate of water are to be determined.Assumptions 1 Steady operating conditions exist. 2 The thermal properties of chickens and water areconstant. 3 Heat gain of the chiller is negligible.Properties The specific heat of chicken are given to be 3.54 kJ/kg.°C. The specific heat of water is 4.18kJ/kg.°C (Table A-3).Analysis (a) Chickens are dropped into the chiller at a rate of 500 per hour. Therefore, chickens can beconsidered to flow steadily through the chiller at a mass flow rate of mchicken = (500 chicken / h)(2.2 kg / chicken) = 1100 kg / h = 0.3056 kg / s &Taking the chicken flow stream in the chiller as thesystem, the energy balance for steadily flowingchickens can be expressed in the rate form as Immersion chilling, 0.5°C & & Ein − Eout = ΔEsystem 0 (steady) & =0 1 24 4 3 1442443 Rate of net energy transfer Rate of change in internal, kinetic, by heat, work, and mass potential, etc. energies & & Ein = Eout Chicken & mh1 = Qout + mh2 (since Δke ≅ Δpe ≅ 0) & & 15°C & Q =Q & =m & c (T − T ) out chicken chicken p 1 2Then the rate of heat removal from the chickens as they arecooled from 15°C to 3ºC becomes & Q chicken =( mc p ΔT ) chicken = (0.3056 kg/s)(3.54 kJ/kg.º C)(15 − 3)º C = 13.0 kW &Heat gain of the chiller from the surroundings is negligible. Then the total rate of heat gain by the water is & & Q water = Qchicken = 13.0 kWNoting that the temperature rise of water is not to exceed 2ºC as it flows through the chiller, the massflow rate of water must be at least & Q water 13.0 kW m water = & = = 1.56 kg/s (c p ΔT ) water (4.18 kJ/kg.º C)(2º C)If the mass flow rate of water is less than this value, then the temperature rise of water will have to be morethan 2°C.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 140. 5-1405-169 A regenerator is considered to save heat during the cooling of milk in a dairy plant. The amounts offuel and money such a generator will save per year are to be determined.Assumptions 1 Steady operating conditions exist. 2 The properties of the milk are constant.Properties The average density and specific heat of milk can be taken to be ρmilk ≅ ρ water = 1 kg/L andcp, milk = 3.79 kJ/kg.°C (Table A-3).Analysis The mass flow rate of the milk is m milk = ρV&milk & Hot milk = (1 kg/L)(12 L/s) = 12 kg/s 72°C = 43,200 kg/hTaking the pasteurizing section as the system,the energy balance for this steady-flow systemcan be expressed in the rate form as & Q & & Ein − Eout = ΔEsystem 0 (steady) & =0 4°C 1 24 4 3 1442443Rate of net energy transfer Rate of change in internal, kinetic,by heat, work, and mass potential, etc. energies Cold milk & & Ein = Eout Regenerator & Qin + mh1 = mh2 (since Δke ≅ Δpe ≅ 0) & & & Qin = mmilk c p (T2 − T1 ) &Therefore, to heat the milk from 4 to 72°C as being done currently, heat must be transferred to the milk at arate of & Qcurrent = [mcp (Tpasturization − Trefrigeration )]milk & = (12 kg/s)(3.79 kJ/kg.ëC)(72 − 4)°C = 3093 kJ/sThe proposed regenerator has an effectiveness of ε = 0.82, and thus it will save 82 percent of this energy.Therefore, & & Qsaved = εQcurrent = (0.82)(3093 kJ / s) = 2536 kJ / sNoting that the boiler has an efficiency of ηboiler = 0.82, the energy savings above correspond to fuelsavings of & Qsaved (2536 kJ / s) (1therm) Fuel Saved = = = 0.02931therm / s η boiler (0.82) (105,500 kJ)Noting that 1 year = 365×24=8760 h and unit cost of natural gas is $1.10/therm, the annual fuel and moneysavings will be Fuel Saved = (0.02931 therms/s)(8760×3600 s) = 924,400 therms/yr Money saved = (Fuel saved)(Unit cost of fuel) = (924,400 therm/yr)($1.10/therm) = $1,016,800 /yrPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 141. 5-1415-170E A refrigeration system is to cool eggs by chilled air at a rate of 10,000 eggs per hour. The rate ofheat removal from the eggs, the required volume flow rate of air, and the size of the compressor of therefrigeration system are to be determined.Assumptions 1 Steady operating conditions exist. 2 The eggs are at uniform temperatures before and aftercooling. 3 The cooling section is well-insulated. 4 The properties of eggs are constant. 5 The localatmospheric pressure is 1 atm.Properties The properties of the eggs are given to ρ = 67.4 lbm/ft3 and cp = 0.80 Btu/lbm.°F. The specificheat of air at room temperature cp = 0.24 Btu/lbm. °F (Table A-2E). The gas constant of air is R = 0.3704psia.ft3/lbm.R (Table A-1E).Analysis (a) Noting that eggs are cooled at a rate of 10,000 eggs per hour, eggs can be considered to flowsteadily through the cooling section at a mass flow rate of megg = (10,000 eggs/h)(0.14 lbm/egg) = 1400 lbm/h = 0.3889 lbm/s &Taking the egg flow stream in the cooler as the system, the energybalance for steadily flowing eggs can be expressed in the rate form as Egg 0.14 lbm & & Ein − Eout = ΔEsystem 0 (steady) & =0 1 24 4 3 1442443 Air Rate of net energy transfer Rate of change in internal, kinetic, by heat, work, and mass potential, etc. energies 34°F & & Ein = Eout & & mh1 = Qout + mh2 (since Δke ≅ Δpe ≅ 0) & & = Q = m c (T − T ) Qout & & egg p 1 2 eggThen the rate of heat removal from the eggs as they are cooled from 90°F to 50ºF at this rate becomes & Qegg = ( mc p ΔT ) egg = (1400 lbm/h)(0.8 0 Btu/lbm.°F)(90 − 50)°F = 44,800 Btu/h &(b) All the heat released by the eggs is absorbed by the refrigerated air since heat transfer through he wallsof cooler is negligible, and the temperature rise of air is not to exceed 10°F. The minimum mass flow andvolume flow rates of air are determined to be & Qair 44,800 Btu/h mair = & = = 18,667 lbm/h (c p ΔT )air (0.24 Btu/lbm.°F)(10°F) P 14.7 psia ρ air = = = 0.0803 lbm/ft3 RT (0.3704 psia.ft 3/lbm.R)(34 + 460)R m& 18,667 lbm/h V&air = air = = 232,500 ft³/h ρ air 0.0803 lbm/ft³PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 142. 5-1425-171 Glass bottles are washed in hot water in an uncovered rectangular glass washing bath. The rates ofheat and water mass that need to be supplied to the water are to be determined.Assumptions 1 Steady operating conditions exist. 2 The entire water body is maintained at a uniformtemperature of 55°C. 3 Heat losses from the outer surfaces of the bath are negligible. 4 Water is anincompressible substance with constant properties.Properties The specific heat of water at room temperature is cp = 4.18 kJ/kg.°C. Also, the specific heat ofglass is 0.80 kJ/kg.°C (Table A-3).Analysis (a) The mass flow rate of glass bottles through the water bath in steady operation is mbottle = mbottle × Bottle flow rate = (0.150 kg / bottle)(800 bottles / min) = 120 kg / min = 2 kg / s &Taking the bottle flow section as the system, which is a steady-flow control volume, the energy balance for this steady-flow Water bathsystem can be expressed in the rate form as 55°C & & Ein − Eout = ΔEsystem 0 (steady) & =0 1 24 4 3 1442443 Rate of net energy transfer Rate of change in internal, kinetic, by heat, work, and mass potential, etc. energies & & Ein = Eout & Qin + mh1 = mh2 (since Δke ≅ Δpe ≅ 0) & & & & Qin = Qbottle = mwater c p (T2 − T1 ) &Then the rate of heat removal by the bottles as they are heated & Qfrom 20 to 55°C is Qbottle = mbottle c p ΔT = (2 kg/s )(0.8 kJ/kg.º C )(55 − 20 )º C = 56,000 W & &The amount of water removed by the bottles is m water,out = (Flow rate of bottles )(Water removed per bottle ) & = (800 bottles / min )(0.2 g/bottle ) = 160 g/min = 2.67×10 − 3 kg/sNoting that the water removed by the bottles is made up by fresh water enteringat 15°C, the rate of heat removal by the water that sticks to the bottles is Q water removed = m water removed c p ΔT = ( 2.67 ×10 −3 kg/s )(4180 J/kg ⋅º C)(55 − 15)º C= 446 W & &Therefore, the total amount of heat removed by the wet bottles is & & & Q total, removed = Q glass removed + Q water removed = 56,000 + 446 = 56,446 WDiscussion In practice, the rates of heat and water removal will be much larger since the heat losses fromthe tank and the moisture loss from the open surface are not considered.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 143. 5-1435-172 Glass bottles are washed in hot water in an uncovered rectangular glass washing bath. The rates ofheat and water mass that need to be supplied to the water are to be determined.Assumptions 1 Steady operating conditions exist. 2 The entire water body is maintained at a uniformtemperature of 50°C. 3 Heat losses from the outer surfaces of the bath are negligible. 4 Water is anincompressible substance with constant properties.Properties The specific heat of water at room temperature is cp = 4.18 kJ/kg.°C. Also, the specific heat ofglass is 0.80 kJ/kg.°C (Table A-3).Analysis (a) The mass flow rate of glass bottles through the water bath in steady operation is mbottle = mbottle × Bottle flow rate = (0.150 kg / bottle)(800 bottles / min) = 120 kg / min = 2 kg / s &Taking the bottle flow section as the system, which is asteady-flow control volume, the energy balance for this Water bathsteady-flow system can be expressed in the rate form as 50°C & & Ein − Eout = ΔEsystem 0 (steady) & =0 1 24 4 3 1442443 Rate of net energy transfer Rate of change in internal, kinetic, by heat, work, and mass potential, etc. energies & & Ein = Eout & Qin + mh1 = mh2 (since Δke ≅ Δpe ≅ 0) & & & & Qin = Qbottle = mwater c p (T2 − T1 ) &Then the rate of heat removal by the bottles as they are & Qheated from 20 to 50°C is Q bottle = m bottle c p ΔT = (2 kg/s )(0.8 kJ/kg.º C )(50 − 20 )º C= 48,000 W & &The amount of water removed by the bottles is m water,out = (Flow rate of bottles )(Water removed per bottle ) & = (800 bottles / min )(0.2 g/bottle )= 160 g/min = 2.67×10 − 3 kg/sNoting that the water removed by the bottles is made up by fresh water entering at 15°C, the rate of heatremoval by the water that sticks to the bottles is Qwater removed = mwater removedc p ΔT = ( 2.67 ×10 −3 kg/s )( 4180 J/kg ⋅º C)(50 − 15)º C = 391 W & &Therefore, the total amount of heat removed by the wet bottles is & & & Q total, removed = Q glass removed + Q water removed = 48,000 + 391 = 48,391 WDiscussion In practice, the rates of heat and water removal will be much larger since the heat losses fromthe tank and the moisture loss from the open surface are not considered.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 144. 5-1445-173 Long aluminum wires are extruded at a velocity of 10 m/min, and are exposed to atmospheric air.The rate of heat transfer from the wire is to be determined.Assumptions 1 Steady operating conditions exist. 2 The thermal properties of the wire are constant.Properties The properties of aluminum are given to be ρ = 2702 kg/m3 and cp = 0.896 kJ/kg.°C.Analysis The mass flow rate of the extruded wire through the air is m = ρV& = ρ (πr02 )V = (2702 kg/m 3 )π (0.0015 m) 2 (10 m/min) = 0.191 kg/min &Taking the volume occupied by the extruded wire as the system, which is a steady-flow control volume, theenergy balance for this steady-flow system can be expressed in the rate form as & & Ein − Eout = ΔEsystem 0 (steady) & =0 1 24 4 3 1442443Rate of net energy transfer Rate of change in internal, kinetic,by heat, work, and mass potential, etc. energies 350°C 10 m/min & & Ein = Eout & & mh1 = Qout + mh2 (since Δke ≅ Δpe ≅ 0) & Aluminum wire & & Qout = Qwire = mwirec p (T1 − T2 ) &Then the rate of heat transfer from the wire to the air becomes & & Q = mc p [T (t ) − T∞ ] = (0.191 kg/min )(0.896 kJ/kg. °C)(350 − 50)°C = 51.3 kJ/min = 0.856 kW5-174 Long copper wires are extruded at a velocity of 10 m/min, and are exposed to atmospheric air. Therate of heat transfer from the wire is to be determined.Assumptions 1 Steady operating conditions exist. 2 The thermal properties of the wire are constant.Properties The properties of copper are given to be ρ = 8950 kg/m3 and cp = 0.383 kJ/kg.°C.Analysis The mass flow rate of the extruded wire through the air is m = ρV& = ρ (πr02 )V = (8950 kg/m 3 )π (0.0015 m) 2 (10 m/min) = 0.633 kg/min &Taking the volume occupied by the extruded wire as the system, which is a steady-flow control volume, theenergy balance for this steady-flow system can be expressed in the rate form as & & Ein − Eout = ΔEsystem 0 (steady) & =0 1 24 4 3 1442443Rate of net energy transfer Rate of change in internal, kinetic,by heat, work, and mass potential, etc. energies 350°C 10 m/min & & Ein = Eout & & mh1 = Qout + mh2 (since Δke ≅ Δpe ≅ 0) & Copper wire & & Qout = Qwire = mwirec p (T1 − T2 ) &Then the rate of heat transfer from the wire to the air becomes & & Q = mc p [T (t ) − T∞ ] = (0.633 kg/min )(0.383 kJ/kg. °C)(350 − 50)°C = 72.7 kJ/min = 1.21 kWPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 145. 5-1455-175 Steam at a saturation temperature of Tsat = 40°C condenses on the outside of a thin horizontal tube.Heat is transferred to the cooling water that enters the tube at 25°C and exits at 35°C. The rate ofcondensation of steam is to be determined.Assumptions 1 Steady operating conditions exist. 2 Water is an incompressible substance with constantproperties at room temperature. 3 The changes in kinetic and potential energies are negligible.Properties The properties of water at room temperature are ρ = 997 kg/m3 and cp = 4.18 kJ/kg.°C (TableA-3). The enthalpy of vaporization of water at 40°C is hfg = 2406.0 kJ/kg (Table A-4).Analysis The mass flow rate of water through the tube is m water = ρVAc = (997 kg/m 3 )(2 m/s)[π (0.03 m) 2 / 4] = 1.409 kg/s &Taking the volume occupied by the cold water in the tube as the system, which is a steady-flow controlvolume, the energy balance for this steady-flow system can be expressed in the rate form as & & Ein − Eout = ΔEsystem 0 (steady) & =0 Steam 1 24 4 3 1442443Rate of net energy transfer Rate of change in internal, kinetic, 40°Cby heat, work, and mass potential, etc. energies & & 35°C Ein = Eout Cold & Qin + mh1 = mh2 (since Δke ≅ Δpe ≅ 0) & & Water & & Qin = Qwater = mwater c p (T2 − T1 ) & 25°CThen the rate of heat transfer to the water and the rate of condensation become & & Q = mc p (Tout − Tin ) = (1.409 kg/s)(4.18 kJ/kg ⋅ °C)(35 − 25)°C = 58.9 kW Q& 58.9 kJ/s & & Q = mcond h fg → mcond = & = = 0.0245 kg/s h fg 2406.0 kJ/kgPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 146. 5-1465-176E Steam is mixed with water steadily in an adiabatic device. The temperature of the water leavingthis device is to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potentialenergy changes are negligible. 3 There are no work and heat interactions. 4 There is no heat transferbetween the mixing device and the surroundings.Analysis We take the mixing device as the system, which is a control volume. The energy balance for thissteady-flow system can be expressed in the rate form as & & E in − E out = ΔE system 0 (steady) & =0 1 24 4 3 1442443 4 Rate of net energy transfer Rate of change in internal, kinetic, by heat, work, and mass potential, etc. energies & & E in = E out Steam m1 h1 + m 2 h2 = m3 h3 & & & 60 psia 350°FFrom a mass balance 0.05 lbm/s m 3 = m1 + m 2 = 0.05 + 1 = 1.05 lbm/s & & & 1The enthalpies of steam and water are (Table A-6E 3 60 psiaand A-4E) Water 2 P1 = 60 psia ⎫ 60 psia ⎬ h1 = 1208.3 Btu/lbm 40°F T1 = 350°F ⎭ 1 lbm/s P2 = 60 psia ⎫ ⎬ h2 ≅ h f @ 40° F = 8.032 Btu/lbm T2 = 40°F ⎭Substituting into the energy balance equation solving for the exit enthalpy gives m1 h1 + m 2 h2 (0.05 lbm/s)(1208.3 Btu/lbm) + (1 lbm/s)(8.032 Btu/lbm) & & h3 = = = 65.19 Btu/lbm & m3 1.05 lbm/sAt the exit state P3=60 psia and h3=65.19 kJ/kg. An investigation of Table A-5E reveals that this iscompressed liquid state. Approximating this state as saturated liquid at the specified temperature, thetemperature may be determined from Table A-4E to be P3 = 60 psia ⎫ ⎬ T3 ≅ T f @ h = 65.19 Btu/lbm = 97.2°F h3 = 65.19 Btu/lbm ⎭Discussion The exact answer is determined at the compressed liquid state using EES to be 97.0°F,indicating that the saturated liquid approximation is a reasonable one.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 147. 5-1475-177 A constant-pressure R-134a vapor separation unit separates the liquid and vapor portions of asaturated mixture into two separate outlet streams. The flow power needed to operate this unit and the massflow rate of the two outlet streams are to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potentialenergy changes are negligible. 3 There are no work and heat interactions.Analysis The specific volume at the inlet is (Table A-12) P1 = 200 kPa ⎫ 3 ⎬ v 1 = v f + x 1v fg = 0.0007533 + (0.70)(0.099867 − 0.0007533) = 0.070133 m /kg x1 = 0.70 ⎭The mass flow rate at the inlet is then R-134a V& 1 3 0.003 m /s 200 kPa Saturatedm1 =& = = 0.042776 kg/s x = 0.7 Vapor separation vapor v1 0.070133 m /kg3 3 L/s unitFor each kg of mixture processed, 0.7 kg ofvapor are processed. Therefore, m 2 = 0.7 m1 = 0.7 × 0.042776 = 0.02994 kg/s & & Saturated liquid m 3 = m1 − m 2 = 0.3m1 = 0.3 × 0.042776 = 0.01283 kg/s & & & &The flow power for this unit is & Wflow = m 2 P2v 2 + m3 P3v 3 − m1 P1v 1 & & & = (0.02994 kg/s)(200 kPa)(0.099867 m 3 /kg) + (0.01283 kg/s)(200 kPa)(0.0007533 m 3 /kg) − (0.042776 kg/s)(200 kPa)(0.070133 m 3 /kg) = 0 kWPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 148. 5-1485-178E A small positioning control rocket in a satellite is driven by a container filled with R-134a atsaturated liquid state. The number of bursts this rocket experience before the quality in the container is90% or more is to be determined.Analysis The initial and final specific volumes are T = −10°F ⎫ 3 ⎬ v 1 = 0.01171 ft /lbm (Table A-11E) x1 = 0 ⎭ T2 = −10°F ⎫ 3 ⎬ v 2 = v f + x 2v fg = 0.01171 + (0.90)(2.7091 − 0.01171) = 2.4394 ft /lbm x 2 = 0.90 ⎭The initial and final masses in the container are V 2 ft 3 m1 = = = 170.8 lbm v 1 0.01171 m 3 /kg V 2 ft 3 m2 = = = 0.8199 lbm v 2 2.4394 ft 3 /lbmThen, Δm = m1 − m 2 = 170.8 − 0.8199 = 170.0 lbmThe amount of mass released during each control burst is Δm b = mΔt = (0.05 lbm/s)(5 s) = 0.25 lbm &The number of bursts that can be executed is then Δm 170.0 lbm Nb = = = 680 bursts Δm b 0.25 lbm/burstPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 149. 5-1495-179E The relationships between the mass flow rate and the time for the inflation and deflation of an airbag are given. The volume of this bag as a function of time are to be plotted.Assumptions Uniform flow exists at the inlet and outlet.Properties The specific volume of air during inflation and deflation are given to be 15 and 13 ft3/lbm,respectively.Analysis The volume of the airbag at any time is given by V (t ) = ∫ ( mv ) & in flow time in dt − ∫ ( mv ) & out flow time out dApplying at different time periods as given in problem statement give t 20 lbm/s ⎛ 1 s ⎞ ∫ V (t ) = (15 ft 3 /lbm) 0 ⎜ ⎟tdt 10 ms ⎝ 1000 ms ⎠ 0 ≤ t ≤ 10 ms t ∫ V (t ) = 0.015t 2 ft 3 /ms 2 0 0 ≤ t ≤ 10 ms t ⎛ 1s ⎞ V (t ) = V (10 ms) + 10 ms ∫(15 ft 3 /lbm)(20 lbm/s)⎜ ⎟tdt ⎝ 1000 ms ⎠ 10 < t ≤ 12 ms V (t ) = V (10 ms) + 0.03 ft 3 /ms 2 (t − 10 ms) 10 < t ≤ 12 ms V (t ) = V (12 ms) + 0.03 ft 3 /ms 2 (t − 12 ms) t 16 lbm/s ⎛ 1 s ⎞ − ∫ 12 ms (13 ft 3 /lbm) ⎜ ⎟(t − 12 ms)dt (30 - 12) ms ⎝ 1000 ms ⎠ 12 < t ≤ 25 ms V (t ) = V (12 ms) + 0.03 ft 3 /ms 2 (t − 12 ms) t ∫ 0.011556 ft 3 − /ms 2 (t − 12 ms)dt 12 < t ≤ 25 ms 12 ms V (t ) = V (12 ms) + 0.03 ft 3 /ms 2 (t − 12 ms) 0.011556 2 − (t − 144 ms 2 ) + 0.13867(t − 12 ms) 12 < t ≤ 25 ms 2 0.011556 2 V (t ) = V (25 ms) − (t − 625 ms 2 ) + 0.13867(t − 25 ms) 25 < t ≤ 30 ms 2 t ⎛ 1s ⎞ V (t ) = V (30 ms) − ∫ 12 ms (13 ft 3 /lbm)(16 lbm/s)⎜ ⎟dt ⎝ 1000 ms ⎠ 30 < t ≤ 50 ms V (t ) = V (30 ms) − (0.208 ft 3 /ms)(t − 30 ms) 30 < t ≤ 50 msPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 150. 5-150The results with some suitable time intervals are 6 Time, ms V, ft3 0 0 5 2 0.06 4 0.24 6 0.54 Volume [ft ] 4 3 8 0.96 10 1.50 12 2.10 3 15 2.95 20 4.13 25 5.02 2 27 4.70 30 4.13 1 40 2.05 46 0.80 49.85 0 0 0 10 20 30 40 50 Time [ms]Alternative solutionThe net volume flow rate is obtained from V& = (mv ) in − (mv ) out & &which is sketched on the figure below. The volume of the airbag is given by ∫ V = V&dtThe results of a graphical interpretation of the volume is also given in the figure below. Note that theevaluation of the above integral is simply the area under the process curve. 400 6 300 5 5.03 200 150 100 4 4.14 V [ft /s] V [ft ]3 3 0 3 -100 -150 2 2.10 -200 -208 -300 1 -400 0 10 20 30 40 50 60 0 time [mlllisec] 0 10 20 30 40 50 60 time [millisec]PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 151. 5-1515-180 A rigid container filled with an ideal gas is heated while gas is released so that the temperature of thegas remaining in the container stays constant. An expression for the mass flow rate at the outlet as afunction of the rate of pressure change in the container is to be derived.Analysis At any instant, the mass in the control volume may be expressed as V V mCV = = P v RTSince there are no inlets to this control volume, the conservation of mass principle becomes dm CV m in − m out = & & dt dm CV V dP & m out =− =− dt RT dtPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 152. 5-1525-181E A winterizing project is to reduce the infiltration rate of a house from 2.2 ACH to 1.1 ACH. Theresulting cost savings are to be determined.Assumptions 1 The house is maintained at 72°F at all times. 2 The latent heat load during the heatingseason is negligible. 3 The infiltrating air is heated to 72°F before it exfiltrates. 4 Air is an ideal gas withconstant specific heats at room temperature. 5 The changes in kinetic and potential energies are negligible.6 Steady flow conditions exist.Properties The gas constant of air is 0.3704 psia.ft3/lbm⋅R (Table A-1E). The specific heat of air at roomtemperature is 0.24 Btu/lbm⋅°F (Table A-2E).Analysis The density of air at the outdoor conditions is Po 13.5 psia ρo = = 3 = 0.0734 lbm/ft 3 RTo (0.3704 psia.ft /lbm.R)(496.5 R)The volume of the house is V building = ( Floor area)(Height) = (3000 ft 2 )(9 ft) = 27,000 ft 3We can view infiltration as a steady stream of air thatis heated as it flows in an imaginary duct passingthrough the house. The energy balance for thisimaginary steady-flow system can be expressed inthe rate form as & & Ein − Eout = ΔEsystem 0 (steady) & =0 Warm 1 24 4 3 1442443 airRate of net energy transfer Rate of change in internal, kinetic, Cold air Warm airby heat, work, and mass potential, etc. energies 36.5°F 72°F & & Ein = Eout 72°F & Qin + mh1 = mh2 (since Δke ≅ Δpe ≅ 0) & & Qin = mc p (T2 − T1 ) = ρV&c p (T2 − T1 ) & &The reduction in the infiltration rate is 2.2 – 1.1 = 1.1 ACH. The reduction in the sensible infiltration heatload corresponding to it is & Qinfiltration, saved = ρ o c p ( ACH saved )(V building )(Ti − To ) = (0.0734 lbm/ft 3 )(0.24 Btu/lbm.°F)(1.1/h)(27,000 ft 3 )(72 - 36.5)°F = 18,573 Btu/h = 0.18573 therm/hsince 1 therm = 100,000 Btu. The number of hours during a six month period is 6×30×24 = 4320 h. Notingthat the furnace efficiency is 0.65 and the unit cost of natural gas is $1.24/therm, the energy and moneysaved during the 6-month period are & Energy savings = (Qinfiltration, saved )( No. of hours per year)/Efficiency = (0.18573 therm/h)(4320 h/year)/0.65 = 1234 therms/year Cost savings = (Energy savings)( Unit cost of energy) = (1234 therms/year)($1.24/therm) = $1530/yearTherefore, reducing the infiltration rate by one-half will reduce the heating costs of this homeowner by$1530 per year.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 153. 5-1535-182 Outdoors air at -5°C and 90 kPa enters the building at a rate of 35 L/s while the indoors ismaintained at 20°C. The rate of sensible heat loss from the building due to infiltration is to be determined.Assumptions 1 The house is maintained at 20°C at all times. 2 The latent heat load is negligible. 3 Theinfiltrating air is heated to 20°C before it exfiltrates. 4 Air is an ideal gas with constant specific heats atroom temperature. 5 The changes in kinetic and potential energies are negligible. 6 Steady flow conditionsexist.Properties The gas constant of air is R = 0.287 kPa.m3/kg⋅K. The specific heat of air at room temperatureis cp = 1.005 kJ/kg⋅°C (Table A-2).Analysis The density of air at the outdoor conditions is Po 90 kPa ρo = = = 1.17 kg/m 3 RTo (0.287 kPa.m 3 /kg.K)(-5 + 273 K)We can view infiltration as a steady stream of air that isheated as it flows in an imaginary duct passing through thebuilding. The energy balance for this imaginary steady-flow system can be expressed in the rate form as & & Ein − Eout = ΔEsystem 0 (steady) & =0 Warm 1 24 4 3 1442443Rate of net energy transfer airby heat, work, and mass Rate of change in internal, kinetic, potential, etc. energies Cold air Warm air & & -5°C 20°C Ein = Eout 20°C 90 kPa & Qin + mh1 = mh2 (since Δke ≅ Δpe ≅ 0) & & 35 L/s & = mc (T − T ) Qin & p 2 1Then the sensible infiltration heat load corresponding to an infiltration rate of 35 L/s becomes Qinfiltration = ρ oV&air c p (Ti − To ) & = (1.17 kg/m 3 )(0.035 m 3 /s)(1.005 kJ/kg.°C)[20 - (-5)]°C = 1.029 kWTherefore, sensible heat will be lost at a rate of 1.029 kJ/s due to infiltration.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 154. 5-1545-183 A fan is powered by a 0.5 hp motor, and delivers air at a rate of 85 m3/min. The highest possible airvelocity at the fan exit is to be determined.Assumptions 1 This is a steady-flow process since there is no change with time at any point and thusΔmCV = 0 and ΔE CV = 0 . 2 The inlet velocity and the change in potential energy are negligible,V1 ≅ 0 and Δpe ≅ 0 . 3 There are no heat and work interactions other than the electrical power consumedby the fan motor. 4 The efficiencies of the motor and the fan are 100% since best possible operation isassumed. 5 Air is an ideal gas with constant specific heats at room temperature.Properties The density of air is given to be ρ = 1.18 kg/m3. The constant pressure specific heat of air atroom temperature is cp = 1.005 kJ/kg.°C (Table A-2).Analysis We take the fan-motor assembly as the system. This is a control volume since mass crosses thesystem boundary during the process. We note that there is only one inlet and one exit, and thusm1 = m2 = m . & & & The velocity of air leaving the fan will be highest when all of the entire electrical energy drawn bythe motor is converted to kinetic energy, and the friction between the air layers is zero. In this best possiblecase, no energy will be converted to thermal energy, and thus the temperature change of air will be zero, T2 = T1 . Then the energy balance for this steady-flow system can be expressed in the rate form as & & Ein − Eout = ΔEsystem 0 (steady) & =0 1 24 4 3 1442443 Rate of net energy transfer Rate of change in internal, kinetic, by heat, work, and mass potential, etc. energies & & Ein = Eout & We,in + mh1 = m(h2 + V22 /2) (since V1 ≅ 0 and Δpe ≅ 0) & &Noting that the temperature and thus enthalpy remains constant, therelation above simplifies further to & We,in = mV 22 /2 & 0.5 hpwhere 85 m3/min m = ρV& = (1.18 kg/m 3 )(85 m 3 /min) = 100.3 kg/min = 1.67 kg/s &Solving for V2 and substituting gives & 2We,in 2(0.5 hp) ⎛ 745.7 W ⎞⎛ 1 m 2 / s 2 ⎞ V2 = = ⎜ ⎟⎜ ⎟ = 21.1 m/s & m 1.67 kg/s ⎝ 1 hp ⎟⎜ 1 W ⎜ ⎠⎝ ⎟ ⎠Discussion In reality, the velocity will be less because of the inefficiencies of the motor and the fan.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 155. 5-1555-184 The average air velocity in the circular duct of an air-conditioning system is not to exceed 10 m/s. Ifthe fan converts 70 percent of the electrical energy into kinetic energy, the size of the fan motor needed andthe diameter of the main duct are to be determined.Assumptions 1 This is a steady-flow process since there is no change with time at any point and thus ΔmCV = 0 and ΔE CV = 0 . 2 The inlet velocity is negligible, V1 ≅ 0. 3 There are no heat and workinteractions other than the electrical power consumed by the fan motor. 4 Air is an ideal gas with constantspecific heats at room temperature.Properties The density of air is given to be ρ = 1.20 kg/m3. The constant pressure specific heat of air atroom temperature is cp = 1.005 kJ/kg.°C (Table A-2).Analysis We take the fan-motor assembly as the system. This is a control volume since mass crosses thesystem boundary during the process. We note that there is only one inlet and one exit, and thusm1 = m2 = m . The change in the kinetic energy of air as it is accelerated from zero to 10 m/s at a rate of & & &180 m3/s is m = ρV& = (1.20 kg/m 3 )(180 m 3 /min) = 216 kg/min = 3.6 kg/s & & & V 22 − V12 (10 m/s) 2 − 0 ⎛ 1 kJ/kg ⎞ ΔKE = m = (3.6 kg/s ) ⎜ ⎟ = 0.18 kW 2 2 ⎝ 1000 m 2 / s 2 ⎠It is stated that this represents 70% of the electrical energy consumed by themotor. Then the total electrical power consumed by the motor is determined to be & ΔKE 018 kW . & & 0.7Wmotor = ΔKE → & Wmotor = = = 0.257 kW 0.7 0.7 10 m/sThe diameter of the main duct is 180 m3/min 4V& 4(180 m 3 / min) ⎛ 1 min ⎞ V& = VA = V (πD 2 / 4) → D = = ⎜ ⎟ = 0.618 m πV π (10 m/s) ⎝ 60 s ⎠Therefore, the motor should have a rated power of at least 0.257 kW, and the diameter of the duct shouldbe at least 61.8 cmPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 156. 5-1565-185 An evacuated bottle is surrounded by atmospheric air. A valve is opened, and air is allowed to fillthe bottle. The amount of heat transfer through the wall of the bottle when thermal and mechanicalequilibrium is established is to be determined.Assumptions 1 This is an unsteady process since the conditions within the device are changing during theprocess, but it can be analyzed as a uniform-flow process since the state of fluid at the inlet remainsconstant. 2 Air is an ideal gas. 3 Kinetic and potential energies are negligible. 4 There are no workinteractions involved. 5 The direction of heat transfer is to the air in the bottle (will be verified).Analysis We take the bottle as the system. It is a control volume since mass crosses the boundary. Notingthat the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internalenergy u, respectively, the mass and energy balances for this uniform-flow system can be expressed asMass balance: min − mout = Δmsystem → mi = m2 (since mout = minitial = 0) P0Energy balance: T0 Ein − Eout = ΔEsystem 1 24 4 3 123 4 4 Net energy transfer by heat, work, and mass Change in internal, kinetic, V potential, etc. energies Evacuated Qin + mi hi = m2u2 (since W ≅ Eout = Einitial = ke ≅ pe ≅ 0)Combining the two balances: Qin = m2 (u2 − hi ) = m2 (cv T2 − c pTi )but Ti = T2 = T0and cp - cv = R.Substituting, ( ) Qin = m2 cv − c p T0 = −m2 RT0 = − P0V RT0 RT0 = − P0VTherefore, Qout = P0V (Heat is lost from the tank)PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 157. 5-1575-186 An adiabatic air compressor is powered by a direct-coupled steam turbine, which is also driving agenerator. The net power delivered to the generator is to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potentialenergy changes are negligible. 3 The devices are adiabatic and thus heat transfer is negligible. 4 Air is anideal gas with variable specific heats.Properties From the steam tables (Tables A-4 through 6) P3 = 12.5 MPa ⎫ ⎬ h3 = 3343.6 kJ/kg T3 = 500°C ⎭and P4 = 10 kPa ⎫ ⎬ h4 = h f + x4 h fg = 191.81 + (0.92)(2392.1) = 2392.5 kJ/kg x4 = 0.92 ⎭From the air table (Table A-17), T1 = 295 K ⎯ ⎯→ h1 = 295.17 kJ/kg 2 12.5 MPa 3 T2 = 620 K ⎯ ⎯→ h2 = 628.07 kJ/kg 1 MPa 500°C 620 KAnalysis There is only one inlet and one exit for eitherdevice, and thus min = mout = m . We take either the & & & Air Steamturbine or the compressor as the system, which is a control comp turbinevolume since mass crosses the boundary. The energybalance for either steady-flow system can be expressed inthe rate form as 4 1 98 kPa & & Ein − Eout = ΔEsystem 0 (steady) & =0 295 K 1 24 4 3 1442443 10 kPa Rate of net energy transfer Rate of change in internal, kinetic, by heat, work, and mass potential, etc. energies & & Ein = EoutFor the turbine and the compressor it becomesCompressor: & Wcomp, in + mair h1 = mair h2 & & → & Wcomp, in = mair (h2 − h1 ) &Turbine: & & msteam h3 = Wturb, out + msteam h4 & → & Wturb, out = msteam (h3 − h4 ) &Substituting, Wcomp,in = (10 kg/s )(628.07 − 295.17 ) kJ/kg = 3329 kW & Wturb, out = (25 kg/s )(3343.6 − 2392.5 ) kJ/kg = 23,777 kW &Therefore, & & & W net,out = W turb,out − Wcomp,in = 23,777 − 3329 = 20,448 kWPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 158. 5-1585-187 Helium is compressed by a compressor. The power required is to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potentialenergy changes are negligible. 3 Helium is an ideal gas with constant specific heats.Properties The constant pressure specific heat of helium is cp = 5.1926 kJ/kg·K (Table A-2b). The gasconstant of air is R = 2.0769 kPa⋅m3/kg⋅K (Table A-1).Analysis There is only one inlet and one exit, and thus m1 = m2 = m . We take the compressor as the & & &system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as & & E in − E out = ΔE system 0 (steady) & =0 1 24 4 3 1442443 4 Rate of net energy transfer Rate of change in internal, kinetic, 400 kPa by heat, work, and mass potential, etc. energies 200°C & & E in = E out & Compressor Win + mh1 = mh2 (since Δke ≅ Δpe ≅ 0) & & & = m(h − h ) = mc (T − T ) Win & 2 1 & p 2 1 HeliumThe specific volume of air at the inlet and the mass flow rate are 150 kPa 20°C RT1 (2.0769 kPa ⋅ m 3 /kg ⋅ K)(20 + 273 K) v1 = = = 4.0569 m 3 /kg P1 150 kPa A1V1 (0.1 m 2 )(15 m/s) m= & = = 0.3697 kg/s v1 4.0569 m 3 /kgThen the power input is determined from the energy balance equation to be & Win = mc p (T2 − T1 ) = (0.3697 kg/s)(5.19 26 kJ/kg ⋅ K)(200 − 20)K = 345.5 kW &PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 159. 5-1595-188 A submarine that has an air-ballast tank originally partially filled with air is considered. Air ispumped into the ballast tank until it is entirely filled with air. The final temperature and mass of the air inthe ballast tank are to be determined.Assumptions 1 Air is an ideal gas with constant specific heats. 2 The process is adiabatic. 3 There are nowork interactions.Properties The gas constant of air is R = 0.287 kPa⋅m3/kg⋅K (Table A-1). The specific heat ratio of air atroom temperature is k = 1.4 (Table A-2a). The specific volume of water is taken 0.001 m3/kg.Analysis The conservation of mass principle applied to the air gives dm a = m in & dtand as applied to the water becomes dm w = −m out & dtThe first law for the ballast tank produces d (mu ) a d (mu ) w 0= + + h w m w − ha m a & & dt dtCombining this with the conservation of mass expressions, rearranging and canceling the common dt termproduces d (mu ) a + d (mu ) w = ha dm a + hw dm wIntegrating this result from the beginning to the end of the process gives [(mu ) 2 − (mu )1 ]a + [(mu ) 2 − (mu )1 ]w = ha (m 2 − m1 ) a + hw (m 2 − m1 ) wSubstituting the ideal gas equation of state and the specific heat models for the enthalpies and internalenergies expands this to PV 2 PV 1 ⎛ PV 2 PV1 ⎞ cv T2 − cv T1 − m w,1 c wTw = c p Tin ⎜ ⎜ RT − RT ⎟ − m w,1c w Tw ⎟ RT2 RT1 ⎝ 2 1 ⎠When the common terms are cancelled, this result becomes V2 1000 T2 = = = 393.5 K V1 1 100 1 + (V 2 −V 1 ) + (900) T1 kTin 288 (1.4)(293)The final mass from the ideal gas relation is PV 2 (2000 kPa)(1000 m 3 ) m2 = = = 17,710 kg RT2 (0.287 kPa ⋅ m 3 /kg ⋅ K)(393.5 K)PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 160. 5-1605-189 A submarine that has an air-ballast tank originally partially filled with air is considered. Air ispumped into the ballast tank in an isothermal manner until it is entirely filled with air. The final mass of theair in the ballast tank and the total heat transfer are to be determined.Assumptions 1 Air is an ideal gas with constant specific heats. 2 There are no work interactions.Properties The gas constant of air is R = 0.287 kPa⋅m3/kg⋅K (Table A-1). The specific heats of air at roomtemperature are cp = 1.005 kJ/kg·K and cv = 0.718 kJ/kg·K (Table A-2a). The specific volume of water istaken 0.001 m3/kg.Analysis The initial air mass is P1V1 (2000 kPa)(100 m 3 ) m1 = = = 2420 kg RT1 (0.287 kPa ⋅ m 3 /kg ⋅ K)(288 K)and the initial water mass is V1 900 m 3 m w,1 = = = 900,000 kg v 1 0.001 m 3 /kgand the final mass of air in the tank is P2V 2 (2000 kPa)(1000 m 3 ) m2 = = = 24,200 kg RT2 (0.287 kPa ⋅ m 3 /kg ⋅ K)(288 K)The first law when adapted to this system gives Qin + mi hi − m e he = m 2 u 2 − m1u1 Qin = m 2 u 2 − m1u1 + m e he − mi hi Qin = m 2 cv T − (m a ,1cv T + m w u w ) + m w hw − (m 2 − m1 )c p TNoting that u w ≅ h w = 62.98 kJ/kgSubstituting, Qin = 24,200 × 0.718 × 288 − (2420 × 0.718 × 288 + 900,000 × 62.98) + 900,000 × 62.98 − (24200 − 2420) × 1.005 × 288 = −1.800 × 10 6 kJThe negative sign shows that the direction of heat is from the tank.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 161. 5-1615-190 A cylindrical tank is charged with nitrogen from a supply line. The final mass of nitrogen in the tankand final temperature are to be determined for two cases.Assumptions 1 This is an unsteady process since the conditions within the device are changing during theprocess, but it can be analyzed as a uniform-flow process since the state of fluid at the inlet remainsconstant. 2 Air is an ideal gas with constant specific heats. 3 Kinetic and potential energies are negligible. 4There are no work interactions involved.Properties The gas constant of nitrogen is 0.2968 kPa·m3/kg·K (Table A-1). The specific heats of nitrogenat room temperature are cp = 1.039 kJ/kg·K and cv = 0.743 kJ/kg·K (Table A-2a).Analysis (a) We take the tank as the system, which is a control volume since mass crosses the boundary.Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h andinternal energy u, respectively, the mass and energy balances for this uniform-flow system can beexpressed asMass balance: min − mout = Δmsystem → mi = m 2 − m1Energy balance: Air 800 kPa, 25°C E in − E out = ΔE system 1 24 4 3 1 24 4 3 Net energy transfer Change in internal, kinetic, by heat, work, and mass potential, etc. energies mi hi = m 2 u 2 − m1u1 200 kPa m i c p Ti = m 2 cv T2 − m1cv T1 25°CCombining the two balances: 0.1 m3 (m 2 − m1 )c p Ti = m 2 cv T2 − m1cv T1The initial and final masses are given by PV (200 kPa )(0.1 m 3 ) m1 = 1 = = 0.2261 kg RT1 (0.2968 kPa ⋅ m 3 /kg ⋅ K )(298 K ) P2V (800 kPa )(0.1 m 3 ) 278.7 m2 = = = RT2 (0.287 kPa ⋅ m 3 /kg ⋅ K )T2 T2Substituting, ⎛ 278.7 ⎞ 278.7 ⎜ ⎜ T − 0.2261⎟(1.039)(298) = T (0.743)T2 − (0.2261)(0.743)(298) ⎟ ⎝ 2 ⎠ 2whose solution is T2 = 380.1 K = 107.1 °CThe final mass is then 278.7 278.7 m2 = = = 0.7332 kg T2 380.1(b) When there is rapid heat transfer between the nitrogen and tank such that the cylinder and nitrogenremain in thermal equilibrium during the process, the energy balance equation may be written as (m 2 − m1 )c p Ti = (m nit , 2 cv T2 + mt c t T2 ) − (m nit ,1cv T1 + mt c t T1 )Substituting, ⎛ 278.7 ⎞ ⎡ 278.7 ⎤ ⎜ T − 0.2261⎟(1.039)(298) = ⎢ T (0.743)T2 + (50)(0.43)T2 ⎥ − [(0.2261)(0.743)(298) + (50)(0.43)(298)] ⎜ ⎟ ⎝ 2 ⎠ ⎣ 2 ⎦whose solution is T2 = 300.8 K = 27.8 °CThe final mass is then 278.7 278.7 m2 = = = 0.9266 kg 300.8 300.8PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 162. 5-1625-191 The air in a tank is released until the pressure in the tank reduces to a specified value. The masswithdrawn from the tank is to be determined for three methods of analysis.Assumptions 1 This is an unsteady process since the conditions within the device are changing during theprocess, but it can be analyzed as a uniform-flow process. 2 Air is an ideal gas with constant specific heats.3 Kinetic and potential energies are negligible. 4 There are no work or heat interactions involved.Properties The gas constant of air is 0.287 kPa·m3/kg·K (Table A-1). The specific heats of air at roomtemperature are cp = 1.005 kJ/kg·K and cv = 0.718 kJ/kg·K. Also k =1.4 (Table A-2a).Analysis (a) We take the tank as the system, which is a control volume since mass crosses the boundary.Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h andinternal energy u, respectively, the mass and energy balances for this uniform-flow system can beexpressed asMass balance: min − m out = Δmsystem − m e = m 2 − m1 m e = m1 − m 2Energy balance: Air E in − E out = ΔE system 800 kPa 1 24 4 3 1 24 4 3 Net energy transfer by heat, work, and mass Change in internal, kinetic, 25°C, 1 m3 potential, etc. energies − m e he = m 2 u 2 − m1u1 0 = m 2 u 2 − m1u1 + m e he 0 = m 2 cv T2 − m1cv T1 + m e c p TeCombining the two balances: 0 = m 2 cv T2 − m1cv T1 + (m1 − m 2 )c p TeThe initial and final masses are given by P1V (800 kPa )(1 m 3 ) m1 = = = 9.354 kg RT1 (0.287 kPa ⋅ m 3 /kg ⋅ K )(25 + 273 K ) P2V (150 kPa )(1 m 3 ) 522.6 m2 = = 3 = RT2 (0.287 kPa ⋅ m /kg ⋅ K )T2 T2The temperature of air leaving the tank changes from the initial temperature in the tank to the finaltemperature during the discharging process. We assume that the temperature of the air leaving the tank isthe average of initial and final temperatures in the tank. Substituting into the energy balance equation gives 0 = m 2 cv T2 − m1 cv T1 + (m1 − m 2 )c p Te 522.6 ⎛ 522.6 ⎞ ⎛ 298 + T2 ⎞ 0= (0.718)T2 − (9.354)(0.718)(298) + ⎜ 9.354 − ⎜ ⎟(1.005)⎜ ⎟ ⎜ ⎟ ⎟ T2 ⎝ T2 ⎠ ⎝ 2 ⎠whose solution is T2 = 191.0 KSubstituting, the final mass is 522.6 m2 = = 2.736 kg 191PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 163. 5-163and the mass withdrawn is m e = m1 − m 2 = 9.354 − 2.736 = 6.618 kg(b) Considering the process in two parts, first from 800 kPa to 400 kPa and from 400 kPa to 150 kPa, thesolution will be as follows:From 800 kPa to 400 kPa: P2V (400 kPa )(1 m 3 ) 1394 m2 = = 3 = RT2 (0.287 kPa ⋅ m /kg ⋅ K )T2 T2 1394 ⎛ 1394 ⎞ ⎛ 298 + T2 ⎞ 0= (0.718)T2 − (9.354)(0.718)(298) + ⎜ 9.354 − ⎜ ⎟(1.005)⎜ ⎜ ⎟ ⎟ T2 ⎝ T2 ⎟⎠ ⎝ 2 ⎠ T2 = 245.1 K 1394 m2 = = 5.687 kg 245.1 m e ,1 = m1 − m 2 = 9.354 − 5.687 = 3.667 kgFrom 400 kPa to 150 kPa: 522.6 ⎛ 522.6 ⎞ ⎛ 298 + T2 ⎞ 0= (0.718)T2 − (5.687)(0.718)(298) + ⎜ 5.687 − ⎜ ⎟(1.005)⎜ ⎟ ⎜ ⎟ ⎟ T2 ⎝ T2 ⎠ ⎝ 2 ⎠ T2 = 215.5 K 522.6 m2 = = 2.425 kg 215.5 m e, 2 = m1 − m 2 = 5.687 − 2.425 = 3.262 kgThe total mass withdrawn is m e = m e,1 + m e, 2 = 3.667 + 3.262 = 6.929 kg(c) The mass balance may be written as dm = −m e & dtWhen this is combined with the ideal gas equation of state, it becomes V d (P / T ) = −me & R dtsince the tank volume remains constant during the process. An energy balance on the tank givesPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 164. 5-164 d (mu ) = − he m e & dt d (mT ) dm cv = c pT dt dt V dP V d (P / T ) cv = c pT R dt R dt dP ⎛ dP P dT ⎞ cv = cp⎜ − ⎟ dt ⎝ dt T dt ⎠ dP dT ( c p − cv ) = cp P dtWhen this result is integrated, it gives ( k −1) / k 0.4 / 1.4 ⎛P ⎞ ⎛ 150 kPa ⎞ T2 = T1 ⎜ 2 ⎜P ⎟ ⎟ = (298 K )⎜ ⎟ = 184.7 K ⎝ 1 ⎠ ⎝ 800 kPa ⎠The final mass is P2V (150 kPa )(1 m 3 ) m2 = = = 2.830 kg RT2 (0.287 kPa ⋅ m 3 /kg ⋅ K )(184.7 K )and the mass withdrawn is m e = m1 − m 2 = 9.354 − 2.830 = 6.524 kgDiscussion The result in first method is in error by 1.4% while that in the second method is in error by6.2%.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 165. 5-1655-192 A tank initially contains saturated mixture of R-134a. A valve is opened and R-134a vapor only isallowed to escape slowly such that temperature remains constant. The heat transfer necessary with thesurroundings to maintain the temperature and pressure of the R-134a constant is to be determined.Assumptions 1 This is an unsteady process since the conditions within the device are changing during theprocess, but it can be analyzed as a uniform-flow process since the state of fluid at the exit remainsconstant. 2 Kinetic and potential energies are negligible. 3 There are no work interactions involved.Analysis We take the tank as the system, which is a control volume since mass crosses the boundary.Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h andinternal energy u, respectively, the mass and energy balances for this uniform-flow system can beexpressed asMass balance: min − m out = Δmsystem − m e = m 2 − m1 m e = m1 − m 2Energy balance: R-134a E in − E out = ΔE system 0.4 kg, 1 L 1 24 4 3 1 24 4 3 Net energy transfer Change in internal, kinetic, 25°C by heat, work, and mass potential, etc. energies Qin − m e he = m 2 u 2 − m1u1 Qin = m 2 u 2 − m1u1 + m e heCombining the two balances: Qin = m 2 u 2 − m1u1 + (m1 − m 2 )heThe specific volume at the initial state is V 0.001 m 3 v1 = = = 0.0025 m 3 /kg m1 0.4 kgThe initial state properties of R-134a in the tank are v1 −v f 0.0025 − 0.0008313 T1 = 26°C ⎫ ⎪ x1 = = = 0.05726 3 ⎬ v fg 0.029976 − 0.0008313 (Table A-11) v 1 = 0.0025 m /kg ⎪ ⎭ u = u + x u = 87.26 + (0.05726)(156.87) = 96.24 kJ/kg 1 f 1 fgThe enthalpy of saturated vapor refrigerant leaving the bottle is he = h g @ 26°C = 264.68 kJ/kgThe specific volume at the final state is V 0.001 m 3 v2 = = = 0.01 m 3 /kg m2 0.1 kgThe internal energy at the final state is v −v f 0.01 − 0.0008313 T2 = 26°C ⎫ x2 = 2 ⎪ = = 0.3146 3 ⎬ v fg 0.029976 − 0.0008313 (Table A-11) ⎪ v 2 = 0.01 m /kg ⎭ u 2 = u f + x1u fg = 87.26 + (0.3146)(156.87) = 136.61 kJ/kgSubstituting into the energy balance equation, Qin = m 2 u 2 − m1u1 + (m1 − m 2 )he = (0.1 kg )(136.61 kJ/kg ) − (0.4 kg )(96.24 kJ/kg ) + (0.4 − 0.1 kg )(264.68 kJ/kg ) = 54.6 kJPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 166. 5-1665-193 [Also solved by EES on enclosed CD] Steam expands in a turbine steadily. The mass flow rate of thesteam, the exit velocity, and the power output are to be determined.Assumptions 1 This is a steady-flow process since there is no 1 30 kJ/kgchange with time. 2 Potential energy changes are negligible.Properties From the steam tables (Tables A-4 through 6) P1 = 10 MPa ⎫ v 1 = 0.035655 m 3 /kg H2O ⎬ T1 = 550°C ⎭ h1 = 3502.0 kJ/kgand P2 = 25 kPa ⎫ v 2 = v f + x 2v fg = 0.00102 + (0.95)(6.2034 - 0.00102 ) = 5.8933 m 3 /kg ⎬ x 2 = 0.95 ⎭ h2 = h f + x 2 h fg = 271.96 + (0.95)(2345.5) = 2500.2 kJ/kg 2Analysis (a) The mass flow rate of the steam is m= & 1 V1 A1 = 1 (60 m/s )(0.015 m 2 ) = 25.24 kg/s v1 0.035655 m 3 /kg(b) There is only one inlet and one exit, and thus m1 = m2 = m . Then the exit velocity is determined from & & & 1 mv 2 (25.24 kg/s)(5.8933 m 3 /kg ) & m= & V 2 A2 ⎯ ⎯→ V 2 = = = 1063 m/s v2 A2 0.14 m 2(c) We take the turbine as the system, which is a control volume since mass crosses the boundary. Theenergy balance for this steady-flow system can be expressed in the rate form as & & Ein − Eout = ΔEsystem 0 (steady) & =0 1 24 4 3 1442443 Rate of net energy transfer Rate of change in internal, kinetic, by heat, work, and mass potential, etc. energies & & Ein = Eout & & m(h1 + V12 / 2) = Wout + Qout + m(h2 + V 22 /2) (since Δpe ≅ 0) & & ⎛ V 2 − V12 ⎞ Wout = −Qout − m⎜ h2 − h1 + 2 & & & ⎟ ⎜ 2 ⎟ ⎝ ⎠Then the power output of the turbine is determined by substituting to be & ⎛ Wout = −(25.24 × 30 ) kJ/s − (25.24 kg/s )⎜ 2500.2 − 3502.0 + (1063 m/s )2 − (60 m/s )2 ⎛ 1 kJ/kg ⎞ ⎞ ⎜ ⎟⎟ ⎜ 2 ⎜ 1000 m 2 /s 2 ⎟ ⎟ ⎝ ⎝ ⎠⎠ = 10,330 kWPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 167. 5-1675-194 EES Problem 5-193 is reconsidered. The effects of turbine exit area and turbine exit pressure on theexit velocity and power output of the turbine as the exit pressure varies from 10 kPa to 50 kPa (with thesame quality), and the exit area to varies from 1000 cm2 to 3000 cm2 is to be investigated. The exit velocityand the power output are to be plotted against the exit pressure for the exit areas of 1000, 2000, and 3000cm2.Analysis The problem is solved using EES, and the results are tabulated and plotted below.Fluid$=Steam_IAPWSA[1]=150 [cm^2]T[1]=550 [C]P[1]=10000 [kPa]Vel[1]= 60 [m/s]A[2]=1400 [cm^2]P[2]=25 [kPa]q_out = 30 [kJ/kg]m_dot = A[1]*Vel[1]/v[1]*convert(cm^2,m^2)v[1]=volume(Fluid$, T=T[1], P=P[1]) "specific volume of steam at state 1"Vel[2]=m_dot*v[2]/(A[2]*convert(cm^2,m^2))v[2]=volume(Fluid$, x=0.95, P=P[2]) "specific volume of steam at state 2"T[2]=temperature(Fluid$, P=P[2], v=v[2]) "[C]" "not required, but good to know""[conservation of Energy for steady-flow:""Ein_dot - Eout_dot = DeltaE_dot" "For steady-flow, DeltaE_dot = 0"DELTAE_dot=0 "[kW]""For the turbine as the control volume, neglecting the PE of each flow steam:"E_dot_in=E_dot_outh[1]=enthalpy(Fluid$,T=T[1], P=P[1])E_dot_in=m_dot*(h[1]+ Vel[1]^2/2*Convert(m^2/s^2, kJ/kg))h[2]=enthalpy(Fluid$,x=0.95, P=P[2])E_dot_out=m_dot*(h[2]+ Vel[2]^2/2*Convert(m^2/s^2, kJ/kg))+ m_dot *q_out+ W_dot_outPower=W_dot_outQ_dot_out=m_dot*q_out Power [kW] P2 [kPa] Vel2 [m/s] -54208 10 2513 -14781 14.44 1778 750.2 18.89 1382 8428 23.33 1134 12770 27.78 962.6 15452 32.22 837.6 17217 36.67 742.1 18432 41.11 666.7 19299 45.56 605.6 19935 50 555PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 168. 5-168 25000 20000 Power [kW] 15000 A2 =1000 cm^2 A2 =2000 cm^2 10000 A2 =3000 cm^2 5000 0 10 15 20 25 30 35 40 45 50 P[2] [kPa] 4000 A2 = 1000 cm^2 3500 A2 = 2000 cm^2 3000 A2 = 3000 cm^2 Vel[2] [m/s] 2500 2000 1500 1000 500 0 10 15 20 25 30 35 40 45 50 P[2] [kPa]PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 169. 5-1695-195E Refrigerant-134a is compressed steadily by a compressor. The mass flow rate of the refrigerant andthe exit temperature are to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potentialenergy changes are negligible. 3 The device is adiabatic and thus heat transfer is negligible.Properties From the refrigerant tables (Tables A-11E through A-13E) P1 = 15 psia ⎫ v 1 = 3.2551 ft 3 /lbm ⎬ 2 T1 = 20°F ⎭ h1 = 107.52 Btu/lbmAnalysis (a) The mass flow rate of refrigerant is V&1 10 ft 3 /s R-134a m= & = = 3.072 lbm/s v 1 3.2551 ft 3 /lbm(b) There is only one inlet and one exit, and thus m1 = m2 = m . We take & & &the compressor as the system, which is a control volume since mass 1crosses the boundary. The energy balance for this steady-flow systemcan be expressed in the rate form as & & Ein − Eout = ΔEsystem 0 (steady) & =0 1 24 4 3 1442443 Rate of net energy transfer Rate of change in internal, kinetic, by heat, work, and mass potential, etc. energies & & Ein = Eout & & Win + mh1 = mh2 (since Q ≅ Δke ≅ Δpe ≅ 0) & & & Win = m(h2 − h1 ) &Substituting, ⎛ 0.7068 Btu/s ⎞ (45 hp)⎜ ⎜ ⎟ = (3.072 lbm/s)(h2 − 107.52)Btu/lbm ⎟ ⎝ 1 hp ⎠ h2 = 117.87 Btu/lbmThen the exit temperature becomes P2 = 100 psia ⎫ ⎬ T2 = 95.7°F h2 = 117.87 Btu/lbm ⎭PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 170. 5-1705-196 Air is preheated by the exhaust gases of a gas turbine in a regenerator. For a specified heat transferrate, the exit temperature of air and the mass flow rate of exhaust gases are to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potentialenergy changes are negligible. 3 There are no work interactions. 4 Heat loss from the regenerator to thesurroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the coldfluid. 5 Exhaust gases can be treated as air. 6 Air is an ideal gas with variable specific heats.Properties The gas constant of air is 0.287 kPa.m3/kg.K (Table A-1). The enthalpies of air are (Table A-17) T1 = 550 K → h1 = 555.74 kJ/kg T3 = 800 K → h3 = 821.95 kJ/kg AIR T4 = 600 K → h4 = 607.02 kJ/kg 1 ExhaustAnalysis (a) We take the air side of the heat exchanger as Gasesthe system, which is a control volume since mass crossesthe boundary. There is only one inlet and one exit, and 3thus m1 = m2 = m . The energy balance for this steady- & & &flow system can be expressed in the rate form as 4 2 & & Ein − Eout = ΔEsystem 0 (steady) & =0 1 24 4 3 1442443 Rate of net energy transfer Rate of change in internal, kinetic, by heat, work, and mass potential, etc. energies & & Ein = Eout & & Qin + mair h1 = mair h2 (since W = Δke ≅ Δpe ≅ 0) & & & Qin = mair (h2 − h1 ) &Substituting, 3200 kJ/s = (800/60 kg/s )(h2 − 554.71 kJ/kg ) → h2 = 794.71 kJ/kgThen from Table A-17 we read T2 = 775.1 K(b) Treating the exhaust gases as an ideal gas, the mass flow rate of the exhaust gases is determined fromthe steady-flow energy relation applied only to the exhaust gases, & & Ein = Eout & & mexhaust h3 = Qout + mexhaust h4 (since W = Δke ≅ Δpe ≅ 0) & & & Q =m & (h − h ) out exhaust 3 4 3200 kJ/s = mexhaust (821.95 - 607.02 ) kJ/kg &It yields mexhaust = 14.9 kg/s &PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 171. 5-1715-197 Water is to be heated steadily from 20°C to 55°C by an electrical resistor inside an insulated pipe.The power rating of the resistance heater and the average velocity of the water are to be determined.Assumptions 1 This is a steady-flow process since there is no change with time at any point within thesystem and thus ΔmCV = 0 and ΔE CV = 0 . 2 Water is an incompressible substance with constant specificheats. 3 The kinetic and potential energy changes are negligible, Δke ≅ Δpe ≅ 0 . 4 The pipe is insulatedand thus the heat losses are negligible.Properties The density and specific heat of water at room temperature are ρ = 1000 kg/m3 and c = 4.18kJ/kg·°C (Table A-3).Analysis (a) We take the pipe as the system. This is a control volume since mass crosses the systemboundary during the process. Also, there is only one inlet and one exit and thus m1 = m2 = m . The energy & & &balance for this steady-flow system can be expressed in the rate form as & & Ein − Eout = ΔEsystem 0 (steady) & =0 1 24 4 3 1442443 WATERRate of net energy transfer Rate of change in internal, kinetic,by heat, work, and mass potential, etc. energies 30 L/min D = 5 cm & & Ein = Eout & & & & We,in + mh1 = mh2 (since Qout ≅ Δke ≅ Δpe ≅ 0) · We & We,in = m(h2 − h1 ) = m[c(T2 − T1 ) + vΔP & & 0 ] = mc(T2 − T1 ) &The mass flow rate of water through the pipe is m = ρV&1 = (1000 kg/m 3 )(0.030 m 3 /min) = 30 kg/min &Therefore, & We,in = mc(T2 − T1 ) = (30/60 kg/s)(4.18 kJ/kg⋅ o C)(55 − 20) o C = 73.2 kW &(b) The average velocity of water through the pipe is determined from V&1 V& 0.030 m 3 /min V1 = = = = 15.3 m/min A1 πr 2 π(0.025 m) 2PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 172. 5-1725-198 [Also solved by EES on enclosed CD] An insulated cylinder equipped with an external springinitially contains air. The tank is connected to a supply line, and air is allowed to enter the cylinder until itsvolume doubles. The mass of the air that entered and the final temperature in the cylinder are to bedetermined.Assumptions 1 This is an unsteady process since the conditions within the device are changing during theprocess, but it can be analyzed as a uniform-flow process since the state of fluid at the inlet remainsconstant. 2 The expansion process is quasi-equilibrium. 3 Kinetic and potential energies are negligible. 4The spring is a linear spring. 5 The device is insulated and thus heat transfer is negligible. 6 Air is an idealgas with constant specific heats.Properties The gas constant of air is R = 0.287 kJ/kg·K (Table A-1). The specific heats of air at roomtemperature are cv = 0.718 and cp = 1.005 kJ/kg·K (Table A-2a). Also, u = cvT and h = cpT.Analysis We take the cylinder as the system, which is acontrol volume since mass crosses the boundary. Noting Fspringthat the microscopic energies of flowing and nonflowingfluids are represented by enthalpy h and internal energy u,respectively, the mass and energy balances for thisuniform-flow system can be expressed asMass balance: Air P = 200 kPa min − mout = Δmsystem → mi = m2 − m1 T1 = 22°C V1 = 0.2 m3 Pi = 0.8 MPaEnergy balance: Ti = 22°C Ein − Eout = ΔEsystem 1 24 4 3 123 4 4 Net energy transfer Change in internal, kinetic, by heat, work, and mass potential, etc. energies mi hi = Wb,out + m2u2 − m1u1 (since Q ≅ ke ≅ pe ≅ 0)Combining the two relations, (m 2 − m1 )hi = Wb ,out + m 2 u 2 − m1u1or, (m 2 − m1 )c p Ti = Wb,out + m 2 cv T2 − m1cv T1The initial and the final masses in the tank are m1 = P1V 1 = (200 kPa ) 0.2 m 3 ( ) = 0.472 kg RT1 ( 0.287 kPa ⋅ m 3 /kg ⋅ K (295 K ) ) m2 = P2V 2 = (600 kPa ) 0.4 m ( = 3 836.2 ) RT2 ( 0.287 kPa ⋅ m 3 /kg ⋅ K T2 T2 ) 836.2Then from the mass balance becomes mi = m2 − m1 = − 0.472 T2The spring is a linear spring, and thus the boundary work for this process can be determined from (V 2 −V1 ) = (200 + 600)kPa (0.4 − 0.2)m 3 = 80 kJ P1 + P2 Wb = Area = 2 2Substituting into the energy balance, the final temperature of air T2 is determined to be ⎛ 836.2 ⎞ ⎛ 836.2 ⎞ ⎜ T − 0.472 ⎟(1.005)(295) + ⎜ T ⎟(0.718)(T2 ) − (0.472 )(0.718)(295) − 80 = −⎜ ⎟ ⎜ ⎟ ⎝ 2 ⎠ ⎝ 2 ⎠It yields T2 = 344.1 K 836.2 836.2Thus, m2 = = = 2.430 kg T2 344.1and mi = m2 - m1 = 2.430 - 0.472 = 1.96 kgPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 173. 5-1735-199 R-134a is allowed to leave a piston-cylinder device with a pair of stops. The work done and the heattransfer are to be determined.Assumptions 1 This is an unsteady process since the conditions within the device are changing during theprocess, but it can be analyzed as a uniform-flow process since the state of fluid leaving the device isassumed to be constant. 2 Kinetic and potential energies are negligible.Properties The properties of R-134a at various states are(Tables A-11 through A-13) 3 P1 = 800 kPa ⎫v 1 = 0.032659 m /kg ⎬u1 = 290.84 kJ/kg R-134a T1 = 80°C ⎭ h1 = 316.97 kJ/kg 2 kg Q 3 800 kPa P2 = 500 kPa ⎫v 2 = 0.042115 m /kg 80°C ⎬u 2 = 242.40 kJ/kg T2 = 20°C ⎭ h2 = 263.46 kJ/kgAnalysis (a) We take the tank as the system, which is a control volume since mass crosses the boundary.Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h andinternal energy u, respectively, the mass and energy balances for this uniform-flow system can beexpressed asMass balance: min − mout = Δmsystem → me = m1 − m2Energy balance: Ein − Eout = ΔEsystem 1 24 4 3 123 4 4 Net energy transfer Change in internal, kinetic, by heat, work, and mass potential, etc. energies Wb,in − Qout − me he = m2u2 − m1u1 (since ke ≅ pe ≅ 0)The volumes at the initial and final states and the mass that has left the cylinder are V1 = m1v 1 = (2 kg)(0.032659 m 3 /kg) = 0.06532 m 3 V 2 = m 2v 2 = (1 / 2)m1v 2 = (1/2)(2 kg)(0.042115 m 3 /kg) = 0.04212 m 3 m e = m1 − m 2 = 2 − 1 = 1 kgThe enthalpy of the refrigerant withdrawn from the cylinder is assumed to be the average of initial andfinal enthalpies of the refrigerant in the cylinder he = (1 / 2)(h1 + h2 ) = (1 / 2)(316.97 + 263.46) = 290.21 kJ/kgNoting that the pressure remains constant after the piston starts moving, the boundary work is determinedfrom W b,in = P2 (V1 −V 2 ) = (500 kPa)(0.06532 − 0.04212)m 3 = 11.6 kJ(b) Substituting, 11.6 kJ − Qout − (1 kg)(290.21 kJ/kg) = (1 kg)(242.40 kJ/kg) − ( 2 kg)(290.84 kJ/kg) Qout = 60.7 kJPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 174. 5-1745-200 The pressures across a pump are measured. The mechanical efficiency of the pump and thetemperature rise of water are to be determined.Assumptions 1 The flow is steady and incompressible. 2 The pump is driven by an external motor so thatthe heat generated by the motor is dissipated to the atmosphere. 3 The elevation difference between theinlet and outlet of the pump is negligible, z1 = z2. 4 The inlet and outlet diameters are the same and thus theinlet and exit velocities are equal, V1 = V2.Properties We take the density of water to be 1 kg/L = 1000 15 kWkg/m3 and its specific heat to be 4.18 kJ/kg · °C (Table A-3).Analysis (a) The mass flow rate of water through the pump is m = ρV& = (1 kg/L)(50 L/s) = 50 kg/s & PUMP MotorThe motor draws 15 kW of power and is 90 percent efficient.Thus the mechanical (shaft) power it delivers to the pump is Pump & & inlet W pump,shaft = η motor W electric = (0.90)(15 kW) = 13.5 kWTo determine the mechanical efficiency of the pump, we need to know the increase in the mechanicalenergy of the fluid as it flows through the pump, which is ⎛P V2 ⎞ ⎛P V2 ⎞ ΔE mech,fluid = E mech,out − E mech,in = m⎜ 2 + 2 + gz 2 ⎟ − m⎜ 1 + 1 + gz1 ⎟ & & & & & ⎜ ρ 2 ⎟ ⎜ρ 2 ⎟ ⎝ ⎠ ⎝ ⎠Simplifying it for this case and substituting the given values, ⎛ P − P1 ⎞ ⎛ (300 − 100)kPa ⎞⎛ 1 kJ ⎞ & ΔE mech,fluid = m⎜ 2 ⎟ = (50 kg/s)⎜ ⎟ = 10 kW ⎜ 1000 kg/m 3 ⎟⎜ 1 kPa ⋅ m 3 ⎟ &⎜ ⎟ ⎝ ρ ⎠ ⎝ ⎠⎝ ⎠Then the mechanical efficiency of the pump becomes & ΔE mech,fluid 10 kW η pump = = = 0.741 = 74.1% & Wpump,shaft 13.5 kW(b) Of the 13.5-kW mechanical power supplied by the pump, only 10 kW is imparted to the fluid asmechanical energy. The remaining 3.5 kW is converted to thermal energy due to frictional effects, and this“lost” mechanical energy manifests itself as a heating effect in the fluid, & & & E mech,loss = W pump,shaft − ΔE mech,fluid = 13.5 − 10 = 3.5 kWThe temperature rise of water due to this mechanical inefficiency is determined from the thermal energybalance, & E mech,loss = m(u 2 − u1 ) = mcΔT & &Solving for ∆T, & E mech,loss 3.5 kW ΔT = = = 0.017 °C & mc (50 kg/s)(4.18 kJ/kg ⋅ K)Therefore, the water will experience a temperature rise of 0.017°C, which is very small, as it flows throughthe pump.Discussion In an actual application, the temperature rise of water will probably be less since part of theheat generated will be transferred to the casing of the pump and from the casing to the surrounding air. Ifthe entire pump motor were submerged in water, then the 1.5 kW dissipated to the air due to motorinefficiency would also be transferred to the surrounding water as heat. This would cause the watertemperature to rise more.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 175. 5-1755-201 Heat is lost from the steam flowing in a nozzle. The exit velocity and the mass flow rate are to bedetermined.Assumptions 1 This is a steady-flow process sincethere is no change with time. 2 Potential energychange is negligible. 3 There are no workinteractions. 150°C STEAM 75 kPaAnalysis (a) We take the steam as the system, which 200 kPa Sat. vap.is a control volume since mass crosses the boundary.The energy balance for this steady-flow system can qbe expressed in the rate form asEnergy balance: & & Ein − Eout = ΔEsystem 0 (steady) & =0 1 24 4 3 1442443 Rate of net energy transfer Rate of change in internal, kinetic, by heat, work, and mass potential, etc. energies & & Ein = Eout ⎛ V2 ⎞ ⎛ V2 ⎞ & m⎜ h1 + 1 ⎟ = m⎜ h2 + 2 ⎟ + Qout & & & since W ≅ Δpe ≅ 0) ⎜ ⎟ 2 ⎠ ⎜ 2 ⎟ ⎝ ⎝ ⎠or V 2 = 2( h1 − h 2 − q out )The properties of steam at the inlet and exit are (Table A-6) P1 = 200 kPa ⎫ ⎬h1 = 2769.1 kJ/kg T1 = 150°C ⎭ P2 = 75 kPa ⎫v 2 = 2.2172 m 3 /kg ⎬ sat. vap. ⎭ h2 = 2662.4 kJ/kgSubstituting, ⎛ 1 kJ/kg ⎞ V 2 = 2(h1 − h2 − q out ) = 2( 2769.1 − 2662.4 − 26) kJ/kg ⎜ ⎟ = 401.7 m/s ⎝ 1000 m 2 /s 2 ⎠(b) The mass flow rate of the steam is 1 1 m= & A2V 2 = (0.001 m 2 )(401.7 m/s) = 0.181 kg/s v2 2.2172 m 3 /kgPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 176. 5-1765-202 The turbocharger of an internal combustion engine consisting of a turbine, a compressor, and anaftercooler is considered. The temperature of the air at the compressor outlet and the minimum flow rate ofambient air are to be determined.Assumptions 1 All processes are steady since there is Airno change with time. 2 Kinetic and potential energychanges are negligible. 3 Air properties are used forexhaust gases. 4 Air is an ideal gas with constant Turbine Compressorspecific heats. 5 The mechanical efficiency between theturbine and the compressor is 100%. 6 All devices areadiabatic. 7 The local atmospheric pressure is 100 kPa. ExhaustProperties The constant pressure specific heats of gasesexhaust gases, warm air, and cold ambient air are taken Aftercoolerto be cp = 1.063, 1.008, and 1.005 kJ/kg·K, respectively Cold air(Table A-2b).Analysis (a) An energy balance on turbine gives W T = m exh c p ,exh (Texh,1 − Texh,2 ) = (0.02 kg/s)(1.06 3 kJ/kg ⋅ K)(400 − 350)K = 1.063 kW & &This is also the power input to the compressor since the mechanical efficiency between the turbine and thecompressor is assumed to be 100%. An energy balance on the compressor gives the air temperature at thecompressor outlet & W C = m a c p ,a (Ta,2 − Ta,1 ) & 1.063 kW = (0.018 kg/s)(1.008 kJ/kg ⋅ K)(Ta,2 − 50)K ⎯ Ta,2 = 108.6 °C ⎯→(b) An energy balance on the aftercooler gives the mass flow rate of cold ambient air ma c p ,a (Ta,2 − Ta,3 ) = mca c p ,ca (Tca,2 − Tca,1 ) & & (0.018 kg/s)(1.008 kJ/kg ⋅ °C)(108.6 − 80)°C = mca (1.005 kJ/kg ⋅ °C)(40 − 30)°C & mca = 0.05161 kg/s &The volume flow rate may be determined if we first calculate specific volume of cold ambient air at theinlet of aftercooler. That is, RT (0.287 kJ/kg ⋅ K)(30 + 273 K) v ca = = = 0.8696 m 3 /kg P 100 kPa V&ca = mv ca = (0.05161 kg/s)(0.8696 m 3 /kg) = 0.0449 m 3 /s = 44.9 L/s &PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 177. 5-177Fundamentals of Engineering (FE) Exam Problems5-203 Steam is accelerated by a nozzle steadily from a low velocity to a velocity of 210 m/s at a rate of 3.2kg/s. If the temperature and pressure of the steam at the nozzle exit are 400°C and 2 MPa, the exit area ofthe nozzle is(a) 24.0 cm2 (b) 8.4 cm2 (c) 10.2 cm2 (d) 152 cm2 (e) 23.0 cm2Answer (e) 23.0 cm2Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines ona blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numericalvalues).Vel_1=0 "m/s"Vel_2=210 "m/s"m=3.2 "kg/s"T2=400 "C"P2=2000 "kPa""The rate form of energy balance is E_dot_in - E_dot_out = DELTAE_dot_cv"v2=VOLUME(Steam_IAPWS,T=T2,P=P2)m=(1/v2)*A2*Vel_2 "A2 in m^2""Some Wrong Solutions with Common Mistakes:"R=0.4615 "kJ/kg.K"P2*v2ideal=R*(T2+273)m=(1/v2ideal)*W1_A2*Vel_2 "assuming ideal gas"P1*v2ideal=R*T2m=(1/v2ideal)*W2_A2*Vel_2 "assuming ideal gas and using C"m=W3_A2*Vel_2 "not using specific volume"PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 178. 5-1785-204 Steam enters a diffuser steadily at 0.5 MPa, 300°C, and 122 m/s at a rate of 3.5 kg/s. The inlet areaof the diffuser is(a) 15 cm2 (b) 50 cm2 (c) 105 cm2 (d) 150 cm2 (e) 190 cm2Answer (d) 150 cm2Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines ona blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numericalvalues).Vel_1=122 "m/s"m=3.5 "kg/s"T1=300 "C"P1=500 "kPa""The rate form of energy balance is E_dot_in - E_dot_out = DELTAE_dot_cv"v1=VOLUME(Steam_IAPWS,T=T1,P=P1)m=(1/v1)*A*Vel_1 "A in m^2""Some Wrong Solutions with Common Mistakes:"R=0.4615 "kJ/kg.K"P1*v1ideal=R*(T1+273)m=(1/v1ideal)*W1_A*Vel_1 "assuming ideal gas"P1*v2ideal=R*T1m=(1/v2ideal)*W2_A*Vel_1 "assuming ideal gas and using C"m=W3_A*Vel_1 "not using specific volume"PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 179. 5-1795-205 An adiabatic heat exchanger is used to heat cold water at 15°C entering at a rate of 5 kg/s by hot airat 90°C entering also at rate of 5 kg/s. If the exit temperature of hot air is 20°C, the exit temperature ofcold water is(a) 27°C (b) 32°C (c) 52°C (d) 85°C (e) 90°CAnswer (b) 32°CSolution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines ona blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numericalvalues).C_w=4.18 "kJ/kg-C"Cp_air=1.005 "kJ/kg-C"Tw1=15 "C"m_dot_w=5 "kg/s"Tair1=90 "C"Tair2=20 "C"m_dot_air=5 "kg/s""The rate form of energy balance for a steady-flow system is E_dot_in = E_dot_out"m_dot_air*Cp_air*(Tair1-Tair2)=m_dot_w*C_w*(Tw2-Tw1)"Some Wrong Solutions with Common Mistakes:"(Tair1-Tair2)=(W1_Tw2-Tw1) "Equating temperature changes of fluids"Cv_air=0.718 "kJ/kg.K"m_dot_air*Cv_air*(Tair1-Tair2)=m_dot_w*C_w*(W2_Tw2-Tw1) "Using Cv for air"W3_Tw2=Tair1 "Setting inlet temperature of hot fluid = exit temperature of cold fluid"W4_Tw2=Tair2 "Setting exit temperature of hot fluid = exit temperature of cold fluid"PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 180. 5-1805-206 A heat exchanger is used to heat cold water at 15°C entering at a rate of 2 kg/s by hot air at 100°Centering at rate of 3 kg/s. The heat exchanger is not insulated, and is loosing heat at a rate of 40 kJ/s. If theexit temperature of hot air is 20°C, the exit temperature of cold water is(a) 44°C (b) 49°C (c) 39°C (d) 72°C (e) 95°CAnswer (c) 39°CSolution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines ona blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numericalvalues).C_w=4.18 "kJ/kg-C"Cp_air=1.005 "kJ/kg-C"Tw1=15 "C"m_dot_w=2 "kg/s"Tair1=100 "C"Tair2=20 "C"m_dot_air=3 "kg/s"Q_loss=40 "kJ/s""The rate form of energy balance for a steady-flow system is E_dot_in = E_dot_out"m_dot_air*Cp_air*(Tair1-Tair2)=m_dot_w*C_w*(Tw2-Tw1)+Q_loss"Some Wrong Solutions with Common Mistakes:"m_dot_air*Cp_air*(Tair1-Tair2)=m_dot_w*C_w*(W1_Tw2-Tw1) "Not considering Q_loss"m_dot_air*Cp_air*(Tair1-Tair2)=m_dot_w*C_w*(W2_Tw2-Tw1)-Q_loss "Taking heat loss asheat gain"(Tair1-Tair2)=(W3_Tw2-Tw1) "Equating temperature changes of fluids"Cv_air=0.718 "kJ/kg.K"m_dot_air*Cv_air*(Tair1-Tair2)=m_dot_w*C_w*(W4_Tw2-Tw1)+Q_loss "Using Cv for air"PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 181. 5-1815-207 An adiabatic heat exchanger is used to heat cold water at 15°C entering at a rate of 5 kg/s by hotwater at 90°C entering at rate of 4 kg/s. If the exit temperature of hot water is 50°C, the exit temperature ofcold water is(a) 42°C (b) 47°C (c) 55°C (d) 78°C (e) 90°CAnswer (b) 47°CSolution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines ona blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numericalvalues).C_w=4.18 "kJ/kg-C"Tcold_1=15 "C"m_dot_cold=5 "kg/s"Thot_1=90 "C"Thot_2=50 "C"m_dot_hot=4 "kg/s"Q_loss=0 "kJ/s""The rate form of energy balance for a steady-flow system is E_dot_in = E_dot_out"m_dot_hot*C_w*(Thot_1-Thot_2)=m_dot_cold*C_w*(Tcold_2-Tcold_1)+Q_loss"Some Wrong Solutions with Common Mistakes:"Thot_1-Thot_2=W1_Tcold_2-Tcold_1 "Equating temperature changes of fluids"W2_Tcold_2=90 "Taking exit temp of cold fluid=inlet temp of hot fluid"5-208 In a shower, cold water at 10°C flowing at a rate of 5 kg/min is mixed with hot water at 60°Cflowing at a rate of 2 kg/min. The exit temperature of the mixture will be(a) 24.3°C (b) 35.0°C (c) 40.0°C (d) 44.3°C (e) 55.2°CAnswer (a) 24.3°CSolution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines ona blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numericalvalues).C_w=4.18 "kJ/kg-C"Tcold_1=10 "C"m_dot_cold=5 "kg/min"Thot_1=60 "C"m_dot_hot=2 "kg/min""The rate form of energy balance for a steady-flow system is E_dot_in = E_dot_out"m_dot_hot*C_w*Thot_1+m_dot_cold*C_w*Tcold_1=(m_dot_hot+m_dot_cold)*C_w*Tmix"Some Wrong Solutions with Common Mistakes:"W1_Tmix=(Tcold_1+Thot_1)/2 "Taking the average temperature of inlet fluids"PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 182. 5-1825-209 In a heating system, cold outdoor air at 10°C flowing at a rate of 6 kg/min is mixed adiabaticallywith heated air at 70°C flowing at a rate of 3 kg/min. The exit temperature of the mixture is(a) 30°C (b) 40°C (c) 45°C (d) 55°C (e) 85°CAnswer (a) 30°CSolution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines ona blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numericalvalues).C_air=1.005 "kJ/kg-C"Tcold_1=10 "C"m_dot_cold=6 "kg/min"Thot_1=70 "C"m_dot_hot=3 "kg/min""The rate form of energy balance for a steady-flow system is E_dot_in = E_dot_out"m_dot_hot*C_air*Thot_1+m_dot_cold*C_air*Tcold_1=(m_dot_hot+m_dot_cold)*C_air*Tmix"Some Wrong Solutions with Common Mistakes:"W1_Tmix=(Tcold_1+Thot_1)/2 "Taking the average temperature of inlet fluids"5-210 Hot combustion gases (assumed to have the properties of air at room temperature) enter a gas turbineat 1 MPa and 1500 K at a rate of 0.1 kg/s, and exit at 0.2 MPa and 900 K. If heat is lost from the turbine tothe surroundings at a rate of 15 kJ/s, the power output of the gas turbine is(a) 15 kW (b) 30 kW (c) 45 kW (d) 60 kW (e) 75 kWAnswer (c) 45 kWSolution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines ona blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numericalvalues).Cp_air=1.005 "kJ/kg-C"T1=1500 "K"T2=900 "K"m_dot=0.1 "kg/s"Q_dot_loss=15 "kJ/s""The rate form of energy balance for a steady-flow system is E_dot_in = E_dot_out"W_dot_out+Q_dot_loss=m_dot*Cp_air*(T1-T2)"Alternative: Variable specific heats - using EES data"W_dot_outvariable+Q_dot_loss=m_dot*(ENTHALPY(Air,T=T1)-ENTHALPY(Air,T=T2))"Some Wrong Solutions with Common Mistakes:"W1_Wout=m_dot*Cp_air*(T1-T2) "Disregarding heat loss"W2_Wout-Q_dot_loss=m_dot*Cp_air*(T1-T2) "Assuming heat gain instead of loss"PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 183. 5-1835-211 Steam expands in a turbine from 4 MPa and 500°C to 0.5 MPa and 250°C at a rate of 1350 kg/h.Heat is lost from the turbine at a rate of 25 kJ/s during the process. The power output of the turbine is(a) 157 kW (b) 207 kW (c) 182 kW (d) 287 kW (e) 246 kWAnswer (a) 157 kWSolution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines ona blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numericalvalues).T1=500 "C"P1=4000 "kPa"T2=250 "C"P2=500 "kPa"m_dot=1350/3600 "kg/s"Q_dot_loss=25 "kJ/s"h1=ENTHALPY(Steam_IAPWS,T=T1,P=P1)h2=ENTHALPY(Steam_IAPWS,T=T2,P=P2)"The rate form of energy balance for a steady-flow system is E_dot_in = E_dot_out"W_dot_out+Q_dot_loss=m_dot*(h1-h2)"Some Wrong Solutions with Common Mistakes:"W1_Wout=m_dot*(h1-h2) "Disregarding heat loss"W2_Wout-Q_dot_loss=m_dot*(h1-h2) "Assuming heat gain instead of loss"u1=INTENERGY(Steam_IAPWS,T=T1,P=P1)u2=INTENERGY(Steam_IAPWS,T=T2,P=P2)W3_Wout+Q_dot_loss=m_dot*(u1-u2) "Using internal energy instead of enthalpy"W4_Wout-Q_dot_loss=m_dot*(u1-u2) "Using internal energy and wrong direction for heat"PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 184. 5-1845-212 Steam is compressed by an adiabatic compressor from 0.2 MPa and 150°C to 0.8 MPa and 350°C ata rate of 1.30 kg/s. The power input to the compressor is(a) 511 kW (b) 393 kW (c) 302 kW (d) 717 kW (e) 901 kWAnswer (a) 511 kWSolution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines ona blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numericalvalues).P1=200 "kPa"T1=150 "C"P2=800 "kPa"T2=350 "C"m_dot=1.30 "kg/s"Q_dot_loss=0 "kJ/s"h1=ENTHALPY(Steam_IAPWS,T=T1,P=P1)h2=ENTHALPY(Steam_IAPWS,T=T2,P=P2)"The rate form of energy balance for a steady-flow system is E_dot_in = E_dot_out"W_dot_in-Q_dot_loss=m_dot*(h2-h1)"Some Wrong Solutions with Common Mistakes:"W1_Win-Q_dot_loss=(h2-h1)/m_dot "Dividing by mass flow rate instead of multiplying"W2_Win-Q_dot_loss=h2-h1 "Not considering mass flow rate"u1=INTENERGY(Steam_IAPWS,T=T1,P=P1)u2=INTENERGY(Steam_IAPWS,T=T2,P=P2)W3_Win-Q_dot_loss=m_dot*(u2-u1) "Using internal energy instead of enthalpy"W4_Win-Q_dot_loss=u2-u1 "Using internal energy and ignoring mass flow rate"PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 185. 5-1855-213 Refrigerant-134a is compressed by a compressor from the saturated vapor state at 0.14 MPa to 1.2MPa and 70°C at a rate of 0.108 kg/s. The refrigerant is cooled at a rate of 1.10 kJ/s during compression.The power input to the compressor is(a) 5.54 kW (b) 7.33 kW (c) 6.64 kW (d) 7.74 kW (e) 8.13 kWAnswer (d) 7.74 kWSolution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines ona blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numericalvalues).P1=140 "kPa"x1=1P2=1200 "kPa"T2=70 "C"m_dot=0.108 "kg/s"Q_dot_loss=1.10 "kJ/s"h1=ENTHALPY(R134a,x=x1,P=P1)h2=ENTHALPY(R134a,T=T2,P=P2)"The rate form of energy balance for a steady-flow system is E_dot_in = E_dot_out"W_dot_in-Q_dot_loss=m_dot*(h2-h1)"Some Wrong Solutions with Common Mistakes:"W1_Win+Q_dot_loss=m_dot*(h2-h1) "Wrong direction for heat transfer"W2_Win =m_dot*(h2-h1) "Not considering heat loss"u1=INTENERGY(R134a,x=x1,P=P1)u2=INTENERGY(R134a,T=T2,P=P2)W3_Win-Q_dot_loss=m_dot*(u2-u1) "Using internal energy instead of enthalpy"W4_Win+Q_dot_loss=u2-u1 "Using internal energy and wrong direction for heat transfer"PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 186. 5-1865-214 Refrigerant-134a expands in an adiabatic turbine from 1.2 MPa and 100°C to 0.18 MPa and 50°C ata rate of 1.25 kg/s. The power output of the turbine is(a) 46.3 kW (b) 66.4 kW (c) 72.7 kW (d) 89.2 kW (e) 112.0 kWAnswer (a) 46.3 kWSolution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines ona blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numericalvalues).P1=1200 "kPa"T1=100 "C"P2=180 "kPa"T2=50 "C"m_dot=1.25 "kg/s"Q_dot_loss=0 "kJ/s"h1=ENTHALPY(R134a,T=T1,P=P1)h2=ENTHALPY(R134a,T=T2,P=P2)"The rate form of energy balance for a steady-flow system is E_dot_in = E_dot_out"-W_dot_out-Q_dot_loss=m_dot*(h2-h1)"Some Wrong Solutions with Common Mistakes:"-W1_Wout-Q_dot_loss=(h2-h1)/m_dot "Dividing by mass flow rate instead of multiplying"-W2_Wout-Q_dot_loss=h2-h1 "Not considering mass flow rate"u1=INTENERGY(R134a,T=T1,P=P1)u2=INTENERGY(R134a,T=T2,P=P2)-W3_Wout-Q_dot_loss=m_dot*(u2-u1) "Using internal energy instead of enthalpy"-W4_Wout-Q_dot_loss=u2-u1 "Using internal energy and ignoring mass flow rate"PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 187. 5-1875-215 Refrigerant-134a at 1.4 MPa and 90°C is throttled to a pressure of 0.6 MPa. The temperature of therefrigerant after throttling is(a) 22°C (b) 56°C (c) 82°C (d) 80°C (e) 90.0°CAnswer (d) 80°CSolution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines ona blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numericalvalues).P1=1400 "kPa"T1=90 "C"P2=600 "kPa"h1=ENTHALPY(R134a,T=T1,P=P1)T2=TEMPERATURE(R134a,h=h1,P=P2)"Some Wrong Solutions with Common Mistakes:"W1_T2=T1 "Assuming the temperature to remain constant"W2_T2=TEMPERATURE(R134a,x=0,P=P2) "Taking the temperature to be the saturationtemperature at P2"u1=INTENERGY(R134a,T=T1,P=P1)W3_T2=TEMPERATURE(R134a,u=u1,P=P2) "Assuming u=constant"v1=VOLUME(R134a,T=T1,P=P1)W4_T2=TEMPERATURE(R134a,v=v1,P=P2) "Assuming v=constant"5-216 Air at 20°C and 5 atm is throttled by a valve to 2 atm. If the valve is adiabatic and the change inkinetic energy is negligible, the exit temperature of air will be(a) 10°C (b) 14°C (c) 17°C (d) 20°C (e) 24°CAnswer (d) 20°CSolution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines ona blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numericalvalues)."The temperature of an ideal gas remains constant during throttling, and thus T2=T1"T1=20 "C"P1=5 "atm"P2=2 "atm"T2=T1 "C""Some Wrong Solutions with Common Mistakes:"W1_T2=T1*P1/P2 "Assuming v=constant and using C"W2_T2=(T1+273)*P1/P2-273 "Assuming v=constant and using K"W3_T2=T1*P2/P1 "Assuming v=constant and pressures backwards and using C"W4_T2=(T1+273)*P2/P1 "Assuming v=constant and pressures backwards and using K"PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 188. 5-1885-217 Steam at 1 MPa and 300°C is throttled adiabatically to a pressure of 0.4 MPa. If the change inkinetic energy is negligible, the specific volume of the steam after throttling will be(a) 0.358 m3/kg (b) 0.233 m3/kg (c) 0.375 m3/kg (d) 0.646 m3/kg (e) 0.655 m3/kgAnswer (d) 0.646 m3/kgSolution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines ona blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numericalvalues).P1=1000 "kPa"T1=300 "C"P2=400 "kPa"h1=ENTHALPY(Steam_IAPWS,T=T1,P=P1)v2=VOLUME(Steam_IAPWS,h=h1,P=P2)"Some Wrong Solutions with Common Mistakes:"W1_v2=VOLUME(Steam_IAPWS,T=T1,P=P2) "Assuming the volume to remain constant"u1=INTENERGY(Steam,T=T1,P=P1)W2_v2=VOLUME(Steam_IAPWS,u=u1,P=P2) "Assuming u=constant"W3_v2=VOLUME(Steam_IAPWS,T=T1,P=P2) "Assuming T=constant"5-218 Air is to be heated steadily by an 8-kW electric resistance heater as it flows through an insulatedduct. If the air enters at 50°C at a rate of 2 kg/s, the exit temperature of air will be(a) 46.0°C (b) 50.0°C (c) 54.0°C (d) 55.4°C (e) 58.0°CAnswer (c) 54.0°CSolution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines ona blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numericalvalues).Cp=1.005 "kJ/kg-C"T1=50 "C"m_dot=2 "kg/s"W_dot_e=8 "kJ/s"W_dot_e=m_dot*Cp*(T2-T1)"Checking using data from EES table"W_dot_e=m_dot*(ENTHALPY(Air,T=T_2table)-ENTHALPY(Air,T=T1))"Some Wrong Solutions with Common Mistakes:"Cv=0.718 "kJ/kg.K"W_dot_e=Cp*(W1_T2-T1) "Not using mass flow rate"W_dot_e=m_dot*Cv*(W2_T2-T1) "Using Cv"W_dot_e=m_dot*Cp*W3_T2 "Ignoring T1"PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 189. 5-1895-219 Saturated water vapor at 50°C is to be condensed as it flows through a tube at a rate of 0.35 kg/s.The condensate leaves the tube as a saturated liquid at 50°C. The rate of heat transfer from the tube is(a) 73 kJ/s (b) 980 kJ/s (c) 2380 kJ/s (d) 834 kJ/s (e) 907 kJ/sAnswer (d) 834 kJ/sSolution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines ona blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numericalvalues).T1=50 "C"m_dot=0.35 "kg/s"h_f=ENTHALPY(Steam_IAPWS,T=T1,x=0)h_g=ENTHALPY(Steam_IAPWS,T=T1,x=1)h_fg=h_g-h_fQ_dot=m_dot*h_fg"Some Wrong Solutions with Common Mistakes:"W1_Q=m_dot*h_f "Using hf"W2_Q=m_dot*h_g "Using hg"W3_Q=h_fg "not using mass flow rate"W4_Q=m_dot*(h_f+h_g) "Adding hf and hg"5-220 … 5-224 Design and Essay ProblemsPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.

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