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# Solution manual of Thermodynamics-Ch.8

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• 1. 8-1 Chapter 8 EXERGY – A MEASURE OF WORK POTENTIAL Exergy, Irreversibility, Reversible Work, and Second-Law Efficiency 8-1C Reversible work differs from the useful work by irreversibilities. For reversible processes both are identical. Wu = Wrev − I. 8-2C Reversible work and irreversibility are identical for processes that involve no actual useful work. 8-3C The dead state. 8-4C Yes; exergy is a function of the state of the surroundings as well as the state of the system. 8-5C Useful work differs from the actual work by the surroundings work. They are identical for systems that involve no surroundings work such as steady-flow systems. 8-6C Yes. 8-7C No, not necessarily. The well with the higher temperature will have a higher exergy. 8-8C The system that is at the temperature of the surroundings has zero exergy. But the system that is at a lower temperature than the surroundings has some exergy since we can run a heat engine between these two temperature levels. 8-9C They would be identical. 8-10C The second-law efficiency is a measure of the performance of a device relative to its performance under reversible conditions. It differs from the first law efficiency in that it is not a conversion efficiency. 8-11C No. The power plant that has a lower thermal efficiency may have a higher second-law efficiency. 8-12C No. The refrigerator that has a lower COP may have a higher second-law efficiency. 8-13C A processes with Wrev = 0 is reversible if it involves no actual useful work. Otherwise it is irreversible. PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 2. 8-2 8-14C Yes. 8-15 Windmills are to be installed at a location with steady winds to generate power. The minimum number of windmills that need to be installed is to be determined. Assumptions Air is at standard conditions of 1 atm and 25°C Properties The gas constant of air is 0.287 kPa.m3/kg.K (Table A-1). Analysis The exergy or work potential of the blowing air is the kinetic energy it possesses, Exergy = ke = V 2 (8 m/s) 2 ⎛ 1 kJ/kg ⎞ = ⎜ ⎟ = 0.032 kJ/kg 2 2 ⎝ 1000 m 2 / s 2 ⎠ At standard atmospheric conditions (25°C, 101 kPa), the density and the mass flow rate of air are ρ= P 101 kPa = = 118 m 3 / kg . RT (0.287 kPa ⋅ m 3 / kg ⋅ K)(298 K) and & m = ρAV1 = ρ π D2 4 V1 = (1.18 kg/m 3 )(π / 4)(10 m) 2 (8 m/s) = 742 kg/s Thus, & Available Power = mke = (742 kg/s)(0.032 kJ/kg) = 23.74 kW The minimum number of windmills that needs to be installed is N= & W total 600 kW = = 25.3 ≅ 26 windmills & 23.74 kW W PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 3. 8-3 8-16E Saturated steam is generated in a boiler by transferring heat from the combustion gases. The wasted work potential associated with this heat transfer process is to be determined. Also, the effect of increasing the temperature of combustion gases on the irreversibility is to be discussed. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis The properties of water at the inlet and outlet of the boiler and at the dead state are (Tables A-4E through A-6E) ⎫ h1 = h f = 355.46 Btu/lbm ⎬ x1 = 0 (sat. liq.) ⎭ s1 = s f = 0.54379 Btu/lbm ⋅ R P2 = 200 psia ⎫ h2 = h g = 1198.8 Btu/lbm ⎬ x 2 = 1 (sat. vap.) ⎭ s 2 = s g = 1.5460 Btu/lbm ⋅ R P1 = 200 psia q Water 200 psia sat. liq. 200 psia sat. vap. ⎫ h0 ≅ h f @ 80°F = 48.07 Btu/lbm ⎬ P0 = 14.7 psia ⎭ s 0 ≅ s f @ 80°F = 0.09328 Btu/lbm ⋅ R T0 = 80°F The heat transfer during the process is qin = h2 − h1 = 1198.8 − 355.46 = 843.3 Btu/lbm The entropy generation associated with this process is sgen = Δsw + ΔsR = ( s2 − s1 ) − qin TR = (1.5460 − 0.54379)Btu/lbm ⋅ R − 843.3 Btu/lbm (500 + 460)R = 0.12377 Btu/lbm ⋅ R The wasted work potential (exergy destruction is) xdest = T0 sgen = (80 + 460 R)(0.12377 Btu/lbm ⋅ R) = 66.8 Btu/lbm The work potential (exergy) of the steam stream is Δψ w = h2 − h1 − T0 ( s2 − s1 ) = (1198.8 − 355.46)Btu/lbm − (540 R )(1.5460 − 0.54379)Btu/lbm ⋅ R = 302.1 Btu/lbm Increasing the temperature of combustion gases does not effect the work potential of steam stream since it is determined by the states at which water enters and leaves the boiler. Discussion This problem may also be solved as follows: Exergy transfer by heat transfer: ⎛ T ⎞ ⎛ 540 ⎞ xheat = q⎜1 − 0 ⎟ = (843.3)⎜1 − ⎟ = 368.9 Btu/lbm ⎜ T ⎟ ⎝ 960 ⎠ R ⎠ ⎝ Exergy increase of steam: Δψ w = 302.1 Btu/lbm The net exergy destruction: xdest = xheat − Δψ w = 368.9 − 302.1 = 66.8 Btu/lbm PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 4. 8-4 8-17 Water is to be pumped to a high elevation lake at times of low electric demand for use in a hydroelectric turbine at times of high demand. For a specified energy storage capacity, the minimum amount of water that needs to be stored in the lake is to be determined. Assumptions The evaporation of water from the lake is negligible. Analysis The exergy or work potential of the water is the potential energy it possesses, 75 m Exergy = PE = mgh Thus, m= PE 5 × 10 6 kWh ⎛ 3600 s ⎞⎛ 1000 m 2 / s 2 = ⎜ ⎟⎜ gh (9.8 m/s 2 )(75 m) ⎝ 1 h ⎠⎜ 1 kW ⋅ s/kg ⎝ ⎞ ⎟ = 2.45 × 10 10 kg ⎟ ⎠ 8-18 A heat reservoir at a specified temperature can supply heat at a specified rate. The exergy of this heat supplied is to be determined. Analysis The exergy of the supplied heat, in the rate form, is the amount of power that would be produced by a reversible heat engine, T0 298 K = 1− = 0.8013 TH 1500 K & & = W rev,out = η th, rev Qin η th,max = η th, rev = 1 − & Exergy = W max,out = (0.8013)(150,000 / 3600 kJ/s) 1500 K & Wrev HE 298 K = 33.4 kW 8-19 [Also solved by EES on enclosed CD] A heat engine receives heat from a source at a specified temperature at a specified rate, and rejects the waste heat to a sink. For a given power output, the reversible power, the rate of irreversibility, and the 2nd law efficiency are to be determined. Analysis (a) The reversible power is the power produced by a reversible heat engine operating between the specified temperature limits, η th, max = η th,rev = 1 − TL 320 K = 1− = 0.787 TH 1500 K 1500 K & & W rev,out = η th, rev Qin = (0.787)(700 kJ/s) = 550.7 kW (b) The irreversibility rate is the difference between the reversible power and the actual power output: 700 kJ/s HE 320 kW & & & I = W rev,out − W u,out = 550.7 − 320 = 230.7 kW (c) The second law efficiency is determined from its definition, η II = W u,out W rev,out = 320 K 320 kW = 58.1% 550.7 kW PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 5. 8-5 8-20 EES Problem 8-19 is reconsidered. The effect of reducing the temperature at which the waste heat is rejected on the reversible power, the rate of irreversibility, and the second law efficiency is to be studied and the results are to be plotted. Analysis The problem is solved using EES, and the solution is given below. "Input Data" T_H= 1500 [K] Q_dot_H= 700 [kJ/s] {T_L=320 [K]} W_dot_out = 320 [kW] T_Lsurr =25 [C] "The reversible work is the maximum work done by the Carnot Engine between T_H and T_L:" Eta_Carnot=1 - T_L/T_H W_dot_rev=Q_dot_H*Eta_Carnot "The irreversibility is given as:" I_dot = W_dot_rev-W_dot_out "The thermal efficiency is, in percent:" Eta_th = Eta_Carnot*Convert(, %) "The second law efficiency is, in percent:" Eta_II = W_dot_out/W_dot_rev*Convert(, %) ηII [%] 68.57 67.07 65.63 64.25 62.92 61.65 60.43 59.26 58.13 57.05 I [kJ/s] 146.7 157.1 167.6 178.1 188.6 199 209.5 220 230.5 240.9 Wrev [kJ/s] 466.7 477.1 487.6 498.1 508.6 519 529.5 540 550.5 560.9 TL [K] 500 477.6 455.1 432.7 410.2 387.8 365.3 342.9 320.4 298 PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 6. 8-6 570 Wrev [kJ/s] 548 526 504 482 460 275 320 365 410 455 500 TL [K] 70 68 ηII [%] 66 64 62 60 58 56 275 320 365 410 455 500 TL [K] 250 I [kJ/s] 228 206 184 162 140 275 320 365 410 455 500 TL [K] PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 7. 8-7 8-21E The thermal efficiency and the second-law efficiency of a heat engine are given. The source temperature is to be determined. Analysis From the definition of the second law efficiency, η II = TH η th η 0.36 ⎯ ⎯→ η th, rev = th = = 0.60 η II 0.60 η th, rev HE Thus, η th,rev = 1 − TL ⎯ ⎯→ T H = T L /(1 − η th, rev ) = (530 R)/0.40 = 1325 R TH η th = 36% η II = 60% 530 R 8-22 A body contains a specified amount of thermal energy at a specified temperature. The amount that can be converted to work is to be determined. Analysis The amount of heat that can be converted to work is simply the amount that a reversible heat engine can convert to work, η th,rev = 1 − 800 K T0 298 K = 1− = 0.6275 TH 800 K 100 kJ HE W max,out = W rev,out = η th, rev Qin = (0.6275)(100 kJ) = 62.75 kJ 298 K 8-23 The thermal efficiency of a heat engine operating between specified temperature limits is given. The second-law efficiency of a engine is to be determined. Analysis The thermal efficiency of a reversible heat engine operating between the same temperature reservoirs is η th,rev = 1 − T0 293 K = 1− = 0.801 TH 1200 + 273 K 1200°C HE η th = 0.40 Thus, η II = η th 0.40 = = 49.9% η th, rev 0.801 20°C PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 8. 8-8 8-24 A house is maintained at a specified temperature by electric resistance heaters. The reversible work for this heating process and irreversibility are to be determined. Analysis The reversible work is the minimum work required to accomplish this process, and the irreversibility is the difference between the reversible work and the actual electrical work consumed. The actual power input is & & & Win = Qout = Q H = 80,000 kJ/h = 22.22 kW 80,000 kJ/h The COP of a reversible heat pump operating between the specified temperature limits is COPHP, rev = 1 1 = = 42.14 1 − T L / T H 1 − 288 / 295 House 22 °C 15 °C Thus, & W rev,in = & QH 22.22 kW = = 0.53 kW COPHP, rev 42.14 and · W & & & I = W u,in − W rev,in = 22.22 − 0.53 = 21.69 kW 8-25E A freezer is maintained at a specified temperature by removing heat from it at a specified rate. The power consumption of the freezer is given. The reversible power, irreversibility, and the second-law efficiency are to be determined. Analysis (a) The reversible work is the minimum work required to accomplish this task, which is the work that a reversible refrigerator operating between the specified temperature limits would consume, COPR, rev = & W rev,in = 1 1 = = 8.73 T H / T L − 1 535 / 480 − 1 & QL 1 hp 75 Btu/min ⎛ ⎞ = ⎜ ⎟ = 0.20 hp COPR, rev 8.73 42.41 Btu/min ⎠ ⎝ (b) The irreversibility is the difference between the reversible work and the actual electrical work consumed, & & & I = W u,in − W rev,in = 0.70 − 0.20 = 0.50 hp (c) The second law efficiency is determined from its definition, η II = 75°F R 0.70 hp 75 Btu/min Freezer 20°F & W rev 0.20 hp = = 28.9% & 0.7 hp Wu PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 9. 8-9 8-26 It is to be shown that the power produced by a wind turbine is proportional to the cube of the wind velocity and the square of the blade span diameter. Analysis The power produced by a wind turbine is proportional to the kinetic energy of the wind, which is equal to the product of the kinetic energy of air per unit mass and the mass flow rate of air through the blade span area. Therefore, Wind power = (Efficiency)(Kinetic energy)(Mass flow rate of air) V2 V 2 π D2 ( ρAV ) = η wind ρ V 2 2 4 πV 3 D 2 = η wind ρ = (Constant )V 3 D 2 8 = η wind which completes the proof that wind power is proportional to the cube of the wind velocity and to the square of the blade span diameter. 8-27 A geothermal power produces 14 MW power while the exergy destruction in the plant is 18.5 MW. The exergy of the geothermal water entering to the plant, the second-law efficiency of the plant, and the exergy of the heat rejected from the plant are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3 Water properties are used for geothermal water. Analysis (a) The properties of geothermal water at the inlet of the plant and at the dead state are (Tables A4 through A-6) T1 = 160°C ⎫ h1 = 675.47 kJ/kg ⎬ x1 = 0 ⎭ s1 = 1.9426 kJ/kg.K T0 = 25°C⎫ h0 = 104.83 kJ/kg ⎬ x0 = 0 ⎭ s 0 = 0.36723 kJ/kg.K The exergy of geothermal water entering the plant is & & X in = m[h1 − h0 − T0 ( s1 − s 0 ] = (440 kg/s)[(675.47 − 104.83) kJ/kg + 0 − (25 + 273 K )(1.9426 − 0.36723)kJ/kg.K ] = 44,525 kW = 44.53 MW (b) The second-law efficiency of the plant is the ratio of power produced to the exergy input to the plant η II = & Wout 14,000 kW = = 0.314 & 44,525 kW X in (c) The exergy of the heat rejected from the plant may be determined from an exergy balance on the plant & & & & X heat,out = X in − Wout − X dest = 44,525 − 14,000 − 18,500 = 12,025 kW = 12.03 MW PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 10. 8-10 Second-Law Analysis of Closed Systems 8-28C Yes. 8-29C Yes, it can. For example, the 1st law efficiency of a reversible heat engine operating between the temperature limits of 300 K and 1000 K is 70%. However, the second law efficiency of this engine, like all reversible devices, is 100%. PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 11. 8-11 8-30E Air is expanded in an adiabatic closed system with an isentropic efficiency of 95%. The second law efficiency of the process is to be determined. Assumptions 1 Kinetic and potential energy changes are negligible. 2 The process is adiabatic, and thus there is no heat transfer. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are cp = 0.240 Btu/lbm·R, cv = 0.171 Btu/lbm·R, k = 1.4, and R = 0.06855 Btu/lbm·R (Table A-2Ea). Analysis We take the air as the system. This is a closed system since no mass crosses the boundaries of the system. The energy balance for this system can be expressed as E in − E out = ΔE system 1 24 4 3 1 24 4 3 Net energy transfer by heat, work, and mass Change in internal, kinetic, potential, etc. energies Air 150 psia 100°F − Wb,out = ΔU = mcv (T2 − T1 ) The final temperature for the isentropic case is ( k −1) / k 0.4 / 1.4 ⎛P ⎞ ⎛ 15 psia ⎞ = (560 R)⎜ T2 s = T1 ⎜ 2 ⎟ = 290.1 R ⎜ 150 psia ⎟ ⎟ ⎜P ⎟ ⎝ ⎠ ⎝ 1⎠ The actual exit temperature from the isentropic relation is T −T η= 1 2 T1 − T2s T T2 = T1 − η (T1 − T2 s ) = 560 − (0.95)(560 − 290.1) = 303.6 R The boundary work output is wb,out = cv (T1 − T2 ) = (0.171 Btu/lbm ⋅ R)(560 − 303.6)R = 43.84 Btu/lbm 150 psia 1 15 psia 2 2s s The entropy change of air is T P Δs air = c p ln 2 − R ln 2 T1 P1 = (0.240 Btu/lbm ⋅ R)ln 15 psia 303.6 R − (0.06855 Btu/lbm ⋅ R)ln 560 R 150 psia = 0.01091 Btu/lbm ⋅ R The exergy difference between states 1 and 2 is φ1 − φ2 = u1 − u2 + P0 (v1 − v 2 ) − T0 (s1 − s2 ) ⎛T T ⎞ = cv (T1 − T2 ) + P0 R⎜ 1 − 2 ⎟ − T0 ( s1 − s2 ) ⎜P P ⎟ 2⎠ ⎝ 1 ⎛ 560 R 303.6 R ⎞ = 43.84 Btu/lbm + (14.7 psia)(0.06855 Btu/lbm ⋅ R)⎜ ⎜ 150 psia − 15 psia ⎟ − (537 R)(−0.01091 Btu/lbm ⋅ R) ⎟ ⎝ ⎠ = 33.07 Btu/lbm The useful work is determined from ⎛T T ⎞ wu = wb,out − wsurr = cv (T1 − T2 ) − P0 (v 2 − v1 ) = cv (T1 − T2 ) − P0 R⎜ 2 − 1 ⎟ ⎜P P ⎟ 1⎠ ⎝ 2 ⎛ 303.6 R 560 R ⎞ = 43.84 Btu/lbm − (14.7 psia)(0.06855 Btu/lbm ⋅ R)⎜ ⎜ 15 psia − 150 psia ⎟ ⎟ ⎝ ⎠ = 27.21 Btu/lbm The second law efficiency is then w 27.21 Btu/lbm = 0.823 η II = u = Δφ 33.07 Btu/lbm PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 12. 8-12 8-31E Air and helium at specified states are considered. The gas with the higher exergy content is to be identified. Assumptions 1 Kinetic and potential energy changes are negligible. 2 Air and helium are ideal gases with constant specific heats. Properties The properties of air at room temperature are cp = 0.240 Btu/lbm·R, cv = 0.171 Btu/lbm·R, k = 1.4, and R = 0.06855 Btu/lbm·R = 0.3704 psia⋅ft3/lbm·R. For helium, cp = 1.25 Btu/lbm·R, cv = 0.753 Btu/lbm·R, k = 1.667, and R = 0.4961 Btu/lbm·R= 2.6809 psia⋅ft3/lbm·R. (Table A-2E). Analysis The mass of air in the system is (100 psia)(10 ft 3 ) PV = = 3.552 lbm RT (0.3704 psia ⋅ ft 3 /lbm ⋅ R)(760 R) m= The entropy change of air between the given state and the dead state is s − s 0 = c p ln T P − R ln T0 P0 = (0.240 Btu/lbm ⋅ R)ln Air 10 ft3 100 psia 300°F 100 psia 760 R − (0.06855 Btu/lbm ⋅ R)ln 537 R 14.7 psia = −0.04808 Btu/lbm ⋅ R The air’s specific volumes at the given state and dead state are v= RT (0.3704 psia ⋅ ft 3 /lbm ⋅ R)(760 R) = = 2.815 ft 3 /lbm P 100 psia v0 = RT0 (0.3704 psia ⋅ ft 3 /lbm ⋅ R)(537 R) = = 13.53 ft 3 /lbm P0 14.7 psia The specific closed system exergy of the air is then φ = u − u 0 + P0 (v − v 0 ) − T0 ( s − s 0 ) = cv (T − T0 ) + P0 (v − v 0 ) − T0 ( s − s 0 ) ⎛ 1 Btu = (0.171 Btu/lbm ⋅ R )(300 − 77)R + (14.7 psia)(2.815 − 13.53)ft 3 /lbm⎜ ⎜ 5.404 psia ⋅ ft 3 ⎝ − (537 R)( −0.04808) Btu/lbm ⋅ R ⎞ ⎟ ⎟ ⎠ = 34.80 Btu/lbm The total exergy available in the air for the production of work is then Φ = mφ = (3.552 lbm)(34.80 Btu/lbm) = 123.6 Btu We now repeat the calculations for helium: m= (80 psia)(20 ft 3 ) PV = = 0.9043 lbm RT (2.6809 psia ⋅ ft 3 /lbm ⋅ R)(660 R) s − s 0 = c p ln T P − R ln T0 P0 = (1.25 Btu/lbm ⋅ R)ln Helium 20 ft3 80 psia 200°F 80 psia 660 R − (0.4961 Btu/lbm ⋅ R)ln 537 R 14.7 psia = −0.5827 Btu/lbm ⋅ R PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 13. 8-13 v= RT (2.6809 psia ⋅ ft 3 /lbm ⋅ R)(660 R) = = 22.12 ft 3 /lbm P 80 psia v0 = RT0 (2.6809 psia ⋅ ft 3 /lbm ⋅ R)(537 R) = = 97.93 ft 3 /lbm P0 14.7 psia φ = u − u 0 + P0 (v − v 0 ) − T0 ( s − s 0 ) = cv (T − T0 ) + P0 (v − v 0 ) − T0 ( s − s 0 ) ⎛ 1 Btu = (0.753 Btu/lbm ⋅ R )(200 − 77)R + (14.7 psia)(22.12 − 97.93)ft 3 /lbm⎜ ⎜ 5.404 psia ⋅ ft 3 ⎝ − (537 R)(−0.5827) Btu/lbm ⋅ R ⎞ ⎟ ⎟ ⎠ = 199.3 Btu/lbm Φ = mφ = (0.9043 lbm)(199.3 Btu/lbm) = 180.2 Btu Comparison of two results shows that the helium system has a greater potential for the production of work. PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 14. 8-14 8-32 Steam and R-134a at the same states are considered. The fluid with the higher exergy content is to be identified. Assumptions Kinetic and potential energy changes are negligible. Analysis The properties of water at the given state and at the dead state are u = 2594.7 kJ/kg P = 800 kPa ⎫ 3 ⎬ v = 0.24720 m /kg T = 180°C ⎭ s = 6.7155 kJ/kg ⋅ K Steam 1 kg 800 kPa 180°C (Table A - 6) u 0 ≅ u f @ 25°C = 104.83 kJ/kg ⎫ 3 ⎬ v 0 ≅ v f @ 25°C = 0.001003 m /kg P0 = 100 kPa ⎭ s 0 ≅ s f @ 25°C = 0.3672 kJ/kg ⋅ K T0 = 25°C (Table A - 4) The exergy of steam is Φ = m[u − u 0 + P0 (v − v 0 ) − T0 ( s − s 0 )] ⎡ ⎛ 1 kJ 3 ⎢(2594.7 − 104.83)kJ/kg + (100 kPa)(0.24720 − 0.001003)m /kg⎜ = (1 kg) ⎢ ⎝ 1 kPa ⋅ m 3 ⎢− (298 K)(6.7155 − 0.3672)kJ/kg ⋅ K ⎣ ⎞⎤ ⎟⎥ ⎠⎥ ⎥ ⎦ = 622.7 kJ For R-134a; u = 386.99 kJ/kg P = 800 kPa ⎫ 3 ⎬ v = 0.044554 m /kg T = 180°C ⎭ s = 1.3327 kJ/kg ⋅ K (Table A - 13) u 0 ≅ u f @ 25°C = 85.85 kJ/kg ⎫ v 0 ≅ v f @ 25°C = 0.0008286 m 3 /kg ⎬ P0 = 100 kPa ⎭ s 0 ≅ s f @ 25°C = 0.32432 kJ/kg ⋅ K T0 = 25°C (Table A - 11) R-134a 1 kg 800 kPa 180°C Φ = m[u − u 0 + P0 (v − v 0 ) − T0 ( s − s 0 )] ⎡ ⎛ 1 kJ 3 ⎢(386.99 − 85.85)kJ/kg + (100 kPa)(0.044554 − 0.0008286)m /kg⎜ = (1 kg) ⎢ ⎝ 1 kPa ⋅ m 3 ⎢− (298 K)(1.3327 − 0.32432)kJ/kg ⋅ K ⎣ ⎞⎤ ⎟⎥ ⎠⎥ ⎥ ⎦ = 5.02 kJ The steam can therefore has more work potential than the R-134a. PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 15. 8-15 8-33 A cylinder initially contains air at atmospheric conditions. Air is compressed to a specified state and the useful work input is measured. The exergy of the air at the initial and final states, and the minimum work input to accomplish this compression process, and the second-law efficiency are to be determined Assumptions 1 Air is an ideal gas with constant specific heats. 2 The kinetic and potential energies are negligible. Properties The gas constant of air is R = 0.287 kPa.m3/kg.K (Table A-1). The specific heats of air at the average temperature of (298+423)/2=360 K are cp = 1.009 kJ/kg·K and cv = 0.722 kJ/kg·K (Table A-2). Analysis (a) We realize that X1 = Φ1 = 0 since air initially is at the dead state. The mass of air is P1V1 (100 kPa)(0.002 m 3 ) = = 0.00234 kg RT1 (0.287 kPa ⋅ m 3 / kg ⋅ K)(298 K) m= Also, P2V 2 P1V1 PT (100 kPa)(423 K) = ⎯ ⎯→ V 2 = 1 2 V 1 = (2 L) = 0.473 L (600 kPa)(298 K) T2 T1 P2 T1 AIR V1 = 2 L and s 2 − s 0 = c p ,avg P1 = 100 kPa T1 = 25°C T P ln 2 − R ln 2 T0 P0 = (1.009 kJ/kg ⋅ K) ln 423 K 600 kPa − (0.287 kJ/kg ⋅ K) ln 298 K 100 kPa = −0.1608 kJ/kg ⋅ K Thus, the exergy of air at the final state is [ ] X 2 = Φ 2 = m cv ,avg (T2 − T0 ) − T0 ( s 2 − s 0 ) + P0 (V 2 −V 0 ) = (0.00234 kg)[(0.722 kJ/kg ⋅ K)(423 - 298)K - (298 K)(-0.1608 kJ/kg ⋅ K)] + (100 kPa)(0.000473 - 0.002)m 3 [kJ/m 3 ⋅ kPa] = 0.171 kJ (b) The minimum work input is the reversible work input, which can be determined from the exergy balance by setting the exergy destruction equal to zero, X in − X out 1 24 4 3 Net exergy tra nsfer by heat, work,and mass − X destroyed 0 (reversible) = ΔX system 144 2444 4 3 123 4 4 Exergy destructio n Change in exergy Wrev,in = X 2 − X1 = 0.171− 0 = 0.171kJ (c) The second-law efficiency of this process is ηII = Wrev,in Wu,in = 0.171 kJ = 14.3% 1.2 kJ PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 16. 8-16 8-34 A cylinder is initially filled with R-134a at a specified state. The refrigerant is cooled and condensed at constant pressure. The exergy of the refrigerant at the initial and final states, and the exergy destroyed during this process are to be determined. Assumptions The kinetic and potential energies are negligible. Properties From the refrigerant tables (Tables A-11 through A-13), v = 0.034875 m 3 / kg P1 = 0.7 MPa ⎫ 1 ⎬ u1 = 274.01 kJ/kg T1 = 60°C ⎭ s = 1.0256 kJ/kg ⋅ K 1 v ≅ v f @ 24°C = 0.0008261 m 3 / kg P2 = 0.7 MPa ⎫ 2 ⎬ u 2 ≅ u f @ 24°C = 84.44 kJ/kg T2 = 24°C ⎭s ≅s 2 f @ 24°C = 0.31958 kJ/kg ⋅ K R-134a 0.7 MPa P = const. Q v = 0.23718 m 3 / kg P0 = 0.1 MPa ⎫ 0 ⎬ u 0 = 251.84 kJ/kg T0 = 24°C ⎭ s 0 = 1.1033 kJ/kg ⋅ K Analysis (a) From the closed system exergy relation, X 1 = Φ 1 = m{(u1 − u 0 ) − T0 ( s1 − s 0 ) + P0 (v 1 − v 0 )} = (5 kg){(274.01 − 251.84) kJ/kg − (297 K)(1.0256 − 1.1033) kJ/kg ⋅ K ⎛ 1 kJ ⎞ + (100 kPa)(0.034875 − 0.23718)m 3 /kg⎜ ⎟} ⎝ 1 kPa ⋅ m 3 ⎠ = 125.1 kJ and, X 2 = Φ 2 = m{(u 2 − u 0 ) − T0 ( s 2 − s 0 ) + P0 (v 2 − v 0 )} = (5 kg){(84.44 − 251.84) kJ/kg - (297 K)(0.31958 − 1.1033) kJ/kg ⋅ K ⎛ 1 kJ ⎞ + (100 kPa)(0.0008261 − 0.23718)m 3 /kg⎜ ⎟} ⎝ 1 kPa ⋅ m 3 ⎠ = 208.6 kJ (b) The reversible work input, which represents the minimum work input Wrev,in in this case can be determined from the exergy balance by setting the exergy destruction equal to zero, X in − X out 1 24 4 3 Net exergy tra nsfer by heat, work,and mass − X destroyed 0 (reversible) = ΔX system 144 2444 4 3 123 4 4 Exergy destructio n Change in exergy Wrev,in = X 2 − X1 = 208.6 − 125.1 = 83.5 kJ Noting that the process involves only boundary work, the useful work input during this process is simply the boundary work in excess of the work done by the surrounding air, W u,in = Win − Wsurr,in = Win − P0 (V1 −V 2 ) = P(V1 −V 2 ) − P0 m(v 1 − v 2 ) = m( P − P0 )(v 1 − v 2 ) ⎛ 1 kJ ⎞ = (5 kg)(700 - 100 kPa)(0.034875 − 0.0008261 m 3 / kg)⎜ ⎟ = 102.1 kJ ⎝ 1 kPa ⋅ m 3 ⎠ Knowing both the actual useful and reversible work inputs, the exergy destruction or irreversibility that is the difference between the two is determined from its definition to be X destroyed = I = W u,in − W rev,in = 102.1 − 83.5 = 18.6 kJ PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 17. 8-17 8-35E An insulated rigid tank contains saturated liquid-vapor mixture of water at a specified pressure. An electric heater inside is turned on and kept on until all the liquid is vaporized. The exergy destruction and the second-law efficiency are to be determined. Assumptions Kinetic and potential energies are negligible. Properties From the steam tables (Tables A-4 through A-6) v = v f + x1v fg = 0.01708 + 0.25 × (11.901 − 0.01708) = 2.9880 ft 3 / lbm P1 = 35 psia ⎫ 1 ⎬ u1 = u f + x1u fg = 227.92 + 0.25 × 862.19 = 443.47 Btu / lbm x1 = 0.25 ⎭ s1 = s f + x1 s fg = 0.38093 + 0.25 × 1.30632 = 0.70751 Btu / lbm ⋅ R v 2 = v 1 ⎫ u 2 = u g @ v g = 2.9880 ft 3 /lbm = 1110.9 Btu/lbm ⎬ sat. vapor ⎭ s 2 = s g @ v g = 2.9880 ft 3 /lbm = 1.5692 Btu/lbm ⋅ R Analysis (a) The irreversibility can be determined from its definition Xdestroyed = T0Sgen where the entropy generation is determined from an entropy balance on the tank, which is an insulated closed system, S in − S out 1 24 4 3 Net entropy transfer by heat and mass H2O 35 psia We + S gen = ΔS system { 123 4 4 Entropy generation Change in entropy S gen = ΔS system = m( s 2 − s1 ) Substituting, X destroyed = T0 S gen = mT0 ( s 2 − s1 ) = (6 lbm)(535 R)(1.5692 - 0.70751)Btu/lbm ⋅ R = 2766 Btu (b) Noting that V = constant during this process, the W and Wu are identical and are determined from the energy balance on the closed system energy equation, E in − E out 1 24 4 3 Net energy transfer by heat, work, and mass = ΔE system 1 24 4 3 Change in internal, kinetic, potential, etc. energies We,in = ΔU = m(u 2 − u1 ) or, We,in = (6 lbm)(1110.9 - 443.47)Btu/lbm = 4005 Btu Then the reversible work during this process and the second-law efficiency become W rev,in = W u,in − X destroyed = 4005 − 2766 = 1239 Btu Thus, ηII = Wrev 1239 Btu = = 30.9% Wu 4005 Btu PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 18. 8-18 8-36 A rigid tank is divided into two equal parts by a partition. One part is filled with compressed liquid while the other side is evacuated. The partition is removed and water expands into the entire tank. The exergy destroyed during this process is to be determined. Assumptions Kinetic and potential energies are negligible. Analysis The properties of the water are (Tables A-4 through A-6) v 1 ≅ v f @ 60°C = 0.001017 m 3 / kg P1 = 300 kPa ⎫ ⎬ u1 ≅ u f @ 60°C = 251.16 kJ/kg T1 = 60°C ⎭ s1 ≅ s f @ 60°C = 0.8313 kJ/kg ⋅ K 1.5 kg 300 kPa 60°C WATER Vacuum Noting that v 2 = 2v 1 = 2 × 0.001017 = 0.002034 m 3 / kg , v 2 −v f 0.002034 − 0.001014 = 0.0001017 10.02 − 0.001014 v fg P2 = 15 kPa ⎫ ⎬ u 2 = u f + x 2 u fg = 225.93 + 0.0001017 × 2222.1 = 226.15 kJ/kg v 2 = 0.002034 ⎭ s 2 = s f + x 2 s fg = 0.7549 + 0.0001017 × 7.2522 = 0.7556 kJ/kg ⋅ K x2 = = Taking the direction of heat transfer to be to the tank, the energy balance on this closed system becomes E in − E out 1 24 4 3 = Net energy transfer by heat, work, and mass ΔE system 1 24 4 3 Change in internal, kinetic, potential, etc. energies Qin = ΔU = m(u 2 − u1 ) or, Qin = (1.5 kg)(226.15 - 251.16)kJ/kg = -37.51 kJ → Qout = 37.51 kJ The irreversibility can be determined from its definition Xdestroyed = T0Sgen where the entropy generation is determined from an entropy balance on an extended system that includes the tank and its immediate surroundings so that the boundary temperature of the extended system is the temperature of the surroundings at all times, S in − S out 1 24 4 3 Net entropy transfer by heat and mass − + S gen = ΔS system { 123 4 4 Entropy generation Change in entropy Qout + S gen = ΔS system = m( s 2 − s1 ) Tb,out S gen = m( s 2 − s1 ) + Qout Tsurr Substituting, ⎛ Q ⎞ X destroyed = T0 S gen = T0 ⎜ m( s 2 − s1 ) + out ⎟ ⎜ Tsurr ⎟ ⎠ ⎝ 37.51 kJ ⎤ ⎡ = (298 K) ⎢(1.5 kg)(0.7556 - 0.8313)kJ/kg ⋅ K + 298 K ⎥ ⎣ ⎦ = 3.67 kJ PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 19. 8-19 8-37 EES Problem 8-36 is reconsidered. The effect of final pressure in the tank on the exergy destroyed during the process is to be investigated. Analysis The problem is solved using EES, and the solution is given below. T_1=60 [C] P_1=300 [kPa] m=1.5 [kg] P_2=15 [kPa] T_o=25 [C] P_o=100 [kPa] T_surr = T_o "Conservation of energy for closed system is:" E_in - E_out = DELTAE DELTAE = m*(u_2 - u_1) E_in=0 E_out= Q_out u_1 =intenergy(steam_iapws,P=P_1,T=T_1) v_1 =volume(steam_iapws,P=P_1,T=T_1) s_1 =entropy(steam_iapws,P=P_1,T=T_1) v_2 = 2*v_1 u_2 = intenergy(steam_iapws, v=v_2,P=P_2) s_2 = entropy(steam_iapws, v=v_2,P=P_2) S_in -S_out+S_gen=DELTAS_sys S_in=0 [kJ/K] S_out=Q_out/(T_surr+273) DELTAS_sys=m*(s_2 - s_1) X_destroyed = (T_o+273)*S_gen Xdestroyed [kJ] 3.666 2.813 1.974 1.148 0.336 -0.4629 -1.249 -2.022 -2.782 -3.531 Qout [kJ] 37.44 28.07 19.25 10.89 2.95 -4.612 -11.84 -18.75 -25.39 -31.77 4 3 Xdestroyed [kJ] P2 [kPa] 15 16.11 17.22 18.33 19.44 20.56 21.67 22.78 23.89 25 2 1 0 -1 -2 -3 -4 15 17.2 19.4 21.6 23.8 P2 [kPa] PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 26
• 20. 8-20 8-38 An insulated cylinder is initially filled with saturated liquid water at a specified pressure. The water is heated electrically at constant pressure. The minimum work by which this process can be accomplished and the exergy destroyed are to be determined. Assumptions 1 The kinetic and potential energy changes are negligible. 2 The cylinder is well-insulated and thus heat transfer is negligible. 3 The thermal energy stored in the cylinder itself is negligible. 4 The compression or expansion process is quasi-equilibrium. Analysis (a) From the steam tables (Tables A-4 through A-6), u1 = u f @ 150 kPa = 466.97 kJ / kg 3 P1 = 150 kPa ⎫ v 1 = v f @ 150 kPa = 0.001053 m /kg ⎬ sat. liquid ⎭ h1 = h f @ 150 kPa = 467.13 kJ/kg s1 = s f @ 150 kPa = 1.4337 kJ/kg ⋅ K The mass of the steam is m= V 0.002 m 3 = = 1.899 kg v 1 0.001053 m 3 / kg Saturated Liquid H2O P = 150 kPa We We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed as E in − E out 1 24 4 3 = Net energy transfer by heat, work, and mass ΔE system 1 24 4 3 Change in internal, kinetic, potential, etc. energies We,in − W b,out = ΔU We,in = m(h2 − h1 ) since ΔU + Wb = ΔH during a constant pressure quasi-equilibrium process. Solving for h2, h2 = h1 + We,in m = 467.13 + 2200 kJ = 1625.1 kJ/kg 1.899 kg Thus, h2 − h f 1625.1 − 467.13 = 0.5202 2226.0 h fg P2 = 150 kPa ⎫ ⎬ s 2 = s f + x 2 s fg = 1.4337 + 0.5202 × 5.7894 = 4.4454 kJ/kg ⋅ K h2 = 1625.1 kJ/kg ⎭ u 2 = u f + x 2 u fg = 466.97 + 0.5202 × 2052.3 = 1534.6 kJ/kg x2 = = v 2 = v f + x 2v fg = 0.001053 + 0.5202 × (1.1594 − 0.001053) = 0.6037 m 3 /kg The reversible work input, which represents the minimum work input Wrev,in in this case can be determined from the exergy balance by setting the exergy destruction equal to zero, X in − X out 1 24 4 3 Net exergy tra nsfer by heat, work,and mass − X destroyed 0 (reversible) = ΔX system → Wrev,in = X 2 − X1 144 2444 4 3 123 4 4 Exergy destructio n Change in exergy Substituting the closed system exergy relation, the reversible work input during this process is determined to be PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 21. 8-21 W rev,in = −m[(u1 − u 2 ) − T0 ( s1 − s 2 ) + P0 (v 1 − v 2 )] = −(1.899 kg){(466.97 − 1534.6) kJ/kg − (298 K)(1.4337 − 4.4454) kJ/kg ⋅ K + (100 kPa)(0.001053 − 0.6037)m 3 / kg[1 kJ/1 kPa ⋅ m 3 ]} = 437.7 kJ (b) The exergy destruction (or irreversibility) associated with this process can be determined from its definition Xdestroyed = T0Sgen where the entropy generation is determined from an entropy balance on the cylinder, which is an insulated closed system, S in − S out 1 24 4 3 Net entropy transfer by heat and mass + S gen = ΔS system { 123 4 4 Entropy generation Change in entropy S gen = ΔS system = m( s 2 − s1 ) Substituting, X destroyed = T0 S gen = mT0 ( s 2 − s1 ) = (298 K)(1.899 kg)(4.4454 − 1.4337)kJ/kg ⋅ K = 1705 kJ PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 22. 8-22 8-39 EES Problem 8-38 is reconsidered. The effect of the amount of electrical work on the minimum work and the exergy destroyed is to be investigated. Analysis The problem is solved using EES, and the solution is given below. x_1=0 P_1=150 [kPa] V=2 [L] P_2=P_1 {W_Ele = 2200 [kJ]} T_o=25 [C] P_o=100 [kPa] "Conservation of energy for closed system is:" E_in - E_out = DELTAE DELTAE = m*(u_2 - u_1) E_in=W_Ele E_out= W_b W_b = m*P_1*(v_2-v_1) u_1 =intenergy(steam_iapws,P=P_1,x=x_1) v_1 =volume(steam_iapws,P=P_1,x=x_1) s_1 =entropy(steam_iapws,P=P_1,x=x_1) u_2 = intenergy(steam_iapws, v=v_2,P=P_2) s_2 = entropy(steam_iapws, v=v_2,P=P_2) m=V*convert(L,m^3)/v_1 W_rev_in=m*(u_2 - u_1 -(T_o+273.15) *(s_2-s_1)+P_o*(v_2-v_1)) 400 350 300 Wrev,in [kJ] "Entropy Balance:" S_in - S_out+S_gen = DELTAS_sys DELTAS_sys = m*(s_2 - s_1) S_in=0 [kJ/K] S_out= 0 [kJ/K] 450 250 200 150 100 "The exergy destruction or irreversibility is:" X_destroyed = (T_o+273.15)*S_gen 50 0 0 450 900 1350 1800 2250 WEle [kJ] Wrev,in [kJ] 0 48.54 97.07 145.6 194.1 242.7 291.2 339.8 388.3 436.8 Xdestroyed [kJ] 0 189.5 379.1 568.6 758.2 947.7 1137 1327 1516 1706 1800 1600 Xdestroyed [kJ/K] WEle [kJ] 0 244.4 488.9 733.3 977.8 1222 1467 1711 1956 2200 1400 1200 1000 800 600 400 200 0 0 450 900 1350 1800 2250 WEle [kJ] PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 23. 8-23 8-40 An insulated cylinder is initially filled with saturated R-134a vapor at a specified pressure. The refrigerant expands in a reversible manner until the pressure drops to a specified value. The change in the exergy of the refrigerant during this process and the reversible work are to be determined. Assumptions 1 The kinetic and potential energy changes are negligible. 2 The cylinder is well-insulated and thus heat transfer is negligible. 3 The thermal energy stored in the cylinder itself is negligible. 4 The process is stated to be reversible. Analysis This is a reversible adiabatic (i.e., isentropic) process, and thus s2 = s1. From the refrigerant tables (Tables A-11 through A-13), v 1 = v g @ 0.8 MPa = 0.02562 m 3 / kg P1 = 0.8 MPa ⎫ ⎬ u1 = u g @ 0.8 MPa = 246.79 kJ/kg sat. vapor ⎭s =s 1 g @ 0.8 MPa = 0.9183 kJ/kg ⋅ K The mass of the refrigerant is m= R-134a 0.8 MPa Reversible 3 V 0.05 m = = 1.952 kg v 1 0.02562 m 3 / kg x2 = s2 − s f = 0.9183 − 0.15457 = 0.9753 0.78316 s fg P2 = 0.2 MPa ⎫ 3 ⎬ v 2 = v f + x 2v fg = 0.0007533 + 0.099867 × (0.099867 − 0.0007533) = 0.09741 m /kg s 2 = s1 ⎭ u = u + x u = 38.28 + 0.9753 × 186.21 = 219.88 kJ/kg 2 f 2 fg The reversible work output, which represents the maximum work output Wrev,out can be determined from the exergy balance by setting the exergy destruction equal to zero, X in − X out 1 24 4 3 Net exergy tra nsfer by heat, work,and mass − X destroyed 0 (reversible) = ΔX system 144 2444 4 3 123 4 4 Exergy destructio n Change in exergy - Wrev,out = X 2 − X1 Wrev,out = X1 − X 2 = Φ1 − Φ2 Therefore, the change in exergy and the reversible work are identical in this case. Using the definition of the closed system exergy and substituting, the reversible work is determined to be [ W rev,out = Φ 1 − Φ 2 = m (u1 − u 2 ) − T0 ( s1 − s 2 ) 0 ] + P0 ( v 1 − v 2 ) = m[(u1 − u 2 ) + P0 (v 1 − v 2 )] = (1.952 kg)[(246.79 − 219.88) kJ/kg + (100 kPa)(0.02562 − 0.09741)m 3 / kg[kJ/kPa ⋅ m 3 ] = 38.5 kJ PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 24. 8-24 8-41E Oxygen gas is compressed from a specified initial state to a final specified state. The reversible work and the increase in the exergy of the oxygen during this process are to be determined. Assumptions At specified conditions, oxygen can be treated as an ideal gas with constant specific heats. Properties The gas constant of oxygen is R = 0.06206 Btu/lbm.R (Table A-1E). The constant-volume specific heat of oxygen at the average temperature is Tavg = (T1 + T2 ) / 2 = (75 + 525) / 2 = 300°F ⎯ ⎯→ cv ,avg = 0.164 Btu/lbm ⋅ R Analysis The entropy change of oxygen is ⎛T ⎞ ⎛v ⎞ s 2 − s1 = cv, avg ln⎜ 2 ⎟ + R ln⎜ 2 ⎟ ⎜T ⎟ ⎜v ⎟ ⎝ 1⎠ ⎝ 1⎠ ⎛ 1.5 ft 3 /lbm ⎞ ⎛ 985 R ⎞ ⎟ = (0.164 Btu/lbm ⋅ R) ln⎜ ⎟ + (0.06206 Btu/lbm ⋅ R) ln⎜ ⎜ 12 ft 3 /lbm ⎟ ⎝ 535 R ⎠ ⎝ ⎠ = −0.02894 Btu/lbm ⋅ R O2 12 ft3/lbm 75°F The reversible work input, which represents the minimum work input Wrev,in in this case can be determined from the exergy balance by setting the exergy destruction equal to zero, X in − X out 1 24 4 3 Net exergy tra nsfer by heat, work,and mass − X destroyed 0 (reversible) = ΔX system → Wrev,in = X 2 − X1 144 2444 4 3 123 4 4 Exergy destructio n Change in exergy Therefore, the change in exergy and the reversible work are identical in this case. Substituting the closed system exergy relation, the reversible work input during this process is determined to be w rev,in = φ 2 − φ1 = −[(u1 − u 2 ) − T0 ( s1 − s 2 ) + P0 (v 1 − v 2 )] = −{(0.164 Btu/lbm ⋅ R)(535 - 985)R − (535 R)(0.02894 Btu/lbm ⋅ R) + (14.7 psia)(12 − 1.5)ft 3 /lbm[Btu/5.4039 psia ⋅ ft 3 ]} = 60.7 Btu/lbm Also, the increase in the exergy of oxygen is φ 2 − φ1 = wrev,in = 60.7 Btu/lbm PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 25. 8-25 8-42 An insulated tank contains CO2 gas at a specified pressure and volume. A paddle-wheel in the tank stirs the gas, and the pressure and temperature of CO2 rises. The actual paddle-wheel work and the minimum paddle-wheel work by which this process can be accomplished are to be determined. Assumptions 1 At specified conditions, CO2 can be treated as an ideal gas with constant specific heats at the average temperature. 2 The surroundings temperature is 298 K. Analysis (a) The initial and final temperature of CO2 are T1 = P1V1 (100 kPa)(1.2 m 3 ) = = 298.2 K mR (2.13 kg)(0.1889 kPa ⋅ m 3 / kg ⋅ K ) T2 = 1.2 m3 2.13 kg CO2 100 kPa P2V 2 (120 kPa)(1.2 m 3 ) = = 357.9 K mR (2.13 kg)(0.1889 kPa ⋅ m 3 / kg ⋅ K ) Wpw Tavg = (T1 + T2 ) / 2 = (298.2 + 357.9) / 2 = 328 K ⎯ ⎯→ cv ,avg = 0.684 kJ/kg ⋅ K The actual paddle-wheel work done is determined from the energy balance on the CO gas in the tank, We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed as E in − E out 1 24 4 3 = Net energy transfer by heat, work, and mass ΔE system 1 24 4 3 Change in internal, kinetic, potential, etc. energies W pw,in = ΔU = mcv (T2 − T1 ) or, W pw,in = (2.13 kg)(0.684 kJ/kg ⋅ K)(357.9 − 298.2)K = 87.0 kJ (b) The minimum paddle-wheel work with which this process can be accomplished is the reversible work, which can be determined from the exergy balance by setting the exergy destruction equal to zero, X in − X out 1 24 4 3 Net exergy tra nsfer by heat, work,and mass − X destroyed 0 (reversible) = ΔX system → Wrev,in = X 2 − X1 144 2444 4 3 123 4 4 Exergy destruction Change in exergy Substituting the closed system exergy relation, the reversible work input for this process is determined to be [ 0 W rev,in = m (u 2 − u1 ) − T0 ( s 2 − s1 ) + P0 (v 2 − v 1 ) [ = m cv ,avg (T2 − T1 ) − T0 ( s 2 − s1 ) ] ] = (2.13 kg)[(0.684 kJ/kg ⋅ K)(357.9 − 298.2)K − (298.2)(0.1253 kJ/kg ⋅ K)] = 7.74 kJ since s 2 − s1 = cv ,avg ln T2 v + R ln 2 T1 v1 0 ⎛ 357.9 K ⎞ = (0.684 kJ/kg ⋅ K) ln⎜ ⎟ = 0.1253 kJ/kg ⋅ K ⎝ 298.2 K ⎠ PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 26. 8-26 8-43 An insulated cylinder initially contains air at a specified state. A resistance heater inside the cylinder is turned on, and air is heated for 15 min at constant pressure. The exergy destruction during this process is to be determined. Assumptions Air is an ideal gas with variable specific heats. Properties The gas constant of air is R = 0.287 kJ/kg.K (Table A-1). Analysis The mass of the air and the electrical work done during this process are PV (120 kPa)(0.03 m 3 ) m= 1 1 = = 0.0418 kg RT1 (0.287 kPa ⋅ m 3 /kg ⋅ K)(300 K) AIR 120 kPa P = const We & We = We Δt = (−0.05 kJ / s)(5 × 60 s) = −15 kJ Also, T1 = 300 K ⎯→ ⎯ h1 = 30019 kJ / kg . and o s1 = 1.70202 kJ / kg ⋅ K The energy balance for this stationary closed system can be expressed as E in − E out 1 24 4 3 = Net energy transfer by heat, work, and mass ΔE system 1 24 4 3 Change in internal, kinetic, potential, etc. energies We,in − W b,out = ΔU We,in = m(h2 − h1 ) since ΔU + Wb = ΔH during a constant pressure quasi-equilibrium process. Thus, h2 = h1 + We,in m = 300.19 + T2 = 650 K 15 kJ = 659.04 kJ/kg ⎯ ⎯→ o s 2 = 2.49364 kJ/kg ⋅ K 0.0418 kg Also, s 2 − s1 = o s2 − o s1 ⎛P − R ln⎜ 2 ⎜P ⎝ 1 ⎞ ⎟ ⎟ ⎠ 0 o o = s 2 − s1 = 2.49364 − 1.70202 = 0.79162 kJ/kg ⋅ K The exergy destruction (or irreversibility) associated with this process can be determined from its definition Xdestroyed = T0Sgen where the entropy generation is determined from an entropy balance on the cylinder, which is an insulated closed system, S in − S out 1 24 4 3 Net entropy transfer by heat and mass + S gen = ΔS system { 123 4 4 Entropy generation Change in entropy S gen = ΔS system = m( s 2 − s1 ) Substituting, X destroyed = T0 S gen = mT0 ( s 2 − s1 ) == (300 K)(0.0418 kg)(0.79162 kJ/kg ⋅ K) = 9.9 kJ PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 27. 8-27 8-44 A fixed mass of helium undergoes a process from a specified state to another specified state. The increase in the useful energy potential of helium is to be determined. Assumptions 1 At specified conditions, helium can be treated as an ideal gas. 2 Helium has constant specific heats at room temperature. Properties The gas constant of helium is R = 2.0769 kJ/kg.K (Table A-1). The constant volume specific heat of helium is cv = 3.1156 kJ/kg.K (Table A-2). Analysis From the ideal-gas entropy change relation, s 2 − s1 = cv ,avg ln v T2 + R ln 2 v1 T1 = (3.1156 kJ/kg ⋅ K) ln He 8 kg 288 K 0.5 m 3 /kg 353 K + (2.0769 kJ/kg ⋅ K) ln = −3.087 kJ/kg ⋅ K 288 K 3 m 3 /kg The increase in the useful potential of helium during this process is simply the increase in exergy, Φ 2 − Φ 1 = −m[(u1 − u 2 ) − T0 ( s1 − s 2 ) + P0 (v 1 − v 2 )] = −(8 kg){(3.1156 kJ/kg ⋅ K)(288 − 353) K − (298 K)(3.087 kJ/kg ⋅ K) + (100 kPa)(3 − 0.5)m 3 / kg[kJ/kPa ⋅ m 3 ]} = 6980 kJ PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 28. 8-28 8-45 One side of a partitioned insulated rigid tank contains argon gas at a specified temperature and pressure while the other side is evacuated. The partition is removed, and the gas fills the entire tank. The exergy destroyed during this process is to be determined. Assumptions Argon is an ideal gas with constant specific heats, and thus ideal gas relations apply. Properties The gas constant of argon is R = 0.208 kJ/kg.K (Table A-1). Analysis Taking the entire rigid tank as the system, the energy balance can be expressed as Ein − Eout 1 24 4 3 = Net energy transfer by heat, work, and mass ΔEsystem 1 24 4 3 Change in internal, kinetic, potential, etc. energies 0 = ΔU = m(u2 − u1 ) u2 = u1 → T2 = T1 Argon 300 kPa 70°C Vacuum since u = u(T) for an ideal gas. The exergy destruction (or irreversibility) associated with this process can be determined from its definition Xdestroyed = T0Sgen where the entropy generation is determined from an entropy balance on the entire tank, which is an insulated closed system, S in − S out 1 24 4 3 Net entropy transfer by heat and mass + S gen = ΔS system { 123 4 4 Entropy generation Change in entropy S gen = ΔS system = m( s 2 − s1 ) where ⎛ T 0 V ⎞ V ΔS system = m( s 2 − s1 ) = m⎜ cv ,avg ln 2 + R ln 2 ⎟ = mR ln 2 ⎜ T1 V1 ⎟ V1 ⎝ ⎠ = (3 kg)(0.208 kJ/kg ⋅ K) ln (2) = 0.433 kJ/K Substituting, X destroyed = T0 S gen = mT0 ( s 2 − s1 ) = (298 K)(0.433 kJ/K) = 129 kJ PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 29. 8-29 8-46E A hot copper block is dropped into water in an insulated tank. The final equilibrium temperature of the tank and the work potential wasted during this process are to be determined. Assumptions 1 Both the water and the copper block are incompressible substances with constant specific heats at room temperature. 2 The system is stationary and thus the kinetic and potential energies are negligible. 3 The tank is well-insulated and thus there is no heat transfer. Properties The density and specific heat of water at the anticipated average temperature of 90°F are ρ = 62.1 lbm/ft3 and cp = 1.00 Btu/lbm.°F. The specific heat of copper at the anticipated average temperature of 100°F is cp = 0.0925 Btu/lbm.°F (Table A-3E). Analysis We take the entire contents of the tank, water + copper block, as the system, which is a closed system. The energy balance for this system can be expressed as E in − E out 1 24 4 3 = Net energy transfer by heat, work, and mass ΔE system 1 24 4 3 Change in internal, kinetic, potential, etc. energies 0 = ΔU or, ΔU Cu + ΔU water = 0 Water 75°F Copper 250°F [mc(T2 − T1 )] Cu + [mc(T2 − T1 )] water = 0 where m w = ρV = (62.1 lbm/ft 3 )(1.5 ft 3 ) = 93.15 lbm Substituting, 0 = (70 lbm)(0.0925 Btu/lbm ⋅ °F)(T2 − 250°F) + (93.15 lbm)(1.0 Btu/lbm ⋅ °F)(T2 − 75°F) T2 = 86.4°F = 546.4 R The wasted work potential is equivalent to the exergy destruction (or irreversibility), and it can be determined from its definition Xdestroyed = T0Sgen where the entropy generation is determined from an entropy balance on the system, which is an insulated closed system, S in − S out 1 24 4 3 Net entropy transfer by heat and mass + S gen = ΔS system { 123 4 4 Entropy generation Change in entropy S gen = ΔS system = ΔS water + ΔS copper where ⎛T ⎞ ⎛ 546.4 R ⎞ ΔS copper = mc avg ln⎜ 2 ⎟ = (70 lbm)(0.092 Btu/lbm ⋅ R) ln⎜ ⎟ = −1.696 Btu/R ⎜T ⎟ ⎝ 710 R ⎠ ⎝ 1⎠ ⎛T ⎞ ⎛ 546.4 R ⎞ ΔS water = mc avg ln⎜ 2 ⎟ = (93.15 lbm)(1.0 Btu/lbm ⋅ R) ln⎜ ⎟ = 1.960 Btu/R ⎜T ⎟ ⎝ 535 R ⎠ ⎝ 1⎠ Substituting, X destroyed = (535 R)(−1.696 + 1.960)Btu/R = 140.9 Btu PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 30. 8-30 8-47 A hot iron block is dropped into water in an insulated tank that is stirred by a paddle-wheel. The mass of the iron block and the exergy destroyed during this process are to be determined. √ Assumptions 1 Both the water and the iron block are incompressible substances with constant specific heats at room temperature. 2 The system is stationary and thus the kinetic and potential energies are negligible. 3 The tank is well-insulated and thus there is no heat transfer. Properties The density and specific heat of water at 25°C are ρ = 997 kg/m3 and cp = 4.18 kJ/kg.°F. The specific heat of iron at room temperature (the only value available in the tables) is cp = 0.45 kJ/kg.°C (Table A-3). Analysis We take the entire contents of the tank, water + iron block, as the system, which is a closed system. The energy balance for this system can be expressed as E in − E out 1 24 4 3 = Net energy transfer by heat, work, and mass ΔE system 1 24 4 3 Change in internal, kinetic, potential, etc. energies W pw,in = ΔU = ΔU iron + ΔU water W pw,in = [mc(T2 − T1 )] iron + [mc(T2 − T1 )] water 100 L 20°C Iron 85°C where m water = ρV = (997 kg/m 3 )(0.1 m 3 ) = 99.7 kg & W pw = W pw,in Δt = (0.2 kJ/s)(20 × 60 s) = 240 kJ Wpw Water Substituting, 240 kJ = miron (0.45 kJ/kg ⋅ °C)(24 − 85)°C + (99.7 kg)(4.18 kJ/kg ⋅ °C)(24 − 20)°C miron = 52.0 kg (b) The exergy destruction (or irreversibility) can be determined from its definition Xdestroyed = T0Sgen where the entropy generation is determined from an entropy balance on the system, which is an insulated closed system, S in − S out 1 24 4 3 Net entropy transfer by heat and mass + S gen = ΔS system { 123 4 4 Entropy generation Change in entropy S gen = ΔS system = ΔS iron + ΔS water where ⎛T ⎞ ⎛ 297 K ⎞ ΔS iron = mc avg ln⎜ 2 ⎟ = (52.0 kg)(0.45 kJ/kg ⋅ K) ln⎜ ⎟ = −4.371 kJ/K ⎜T ⎟ ⎝ 358 K ⎠ ⎝ 1⎠ ⎛T ⎞ ⎛ 297 K ⎞ ΔS water = mc avg ln⎜ 2 ⎟ = (99.7 kg)(4.18 kJ/kg ⋅ K) ln⎜ ⎟ = 5.651 kJ/K ⎜T ⎟ ⎝ 293 K ⎠ ⎝ 1⎠ Substituting, X destroyed = T0 S gen = (293 K)(−4.371 + 5.651) kJ/K = 375.0 kJ PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 31. 8-31 8-48 An iron block and a copper block are dropped into a large lake where they cool to lake temperature. The amount of work that could have been produced is to be determined. Assumptions 1 The iron and copper blocks and water are incompressible substances with constant specific heats at room temperature. 2 Kinetic and potential energies are negligible. Properties The specific heats of iron and copper at room temperature are cp, iron = 0.45 kJ/kg.°C and cp,copper = 0.386 kJ/kg.°C (Table A-3). Analysis The thermal-energy capacity of the lake is very large, and thus the temperatures of both the iron and the copper blocks will drop to the lake temperature (15°C) when the thermal equilibrium is established. We take both the iron and the copper blocks as the system, which is a closed system. The energy balance for this system can be expressed as E in − E out 1 24 4 3 = Net energy transfer by heat, work, and mass ΔE system 1 24 4 3 Lake 15°C Change in internal, kinetic, potential, etc. energies − Qout = ΔU = ΔU iron + ΔU copper or, Iron 85°C Copper Iron Qout = [mc(T1 − T2 )] iron + [mc(T1 − T2 )] copper Substituting, Qout = (50 kg )(0.45 kJ/kg ⋅ K )(353 − 288)K + (20 kg )(0.386 kJ/kg ⋅ K )(353 − 288)K = 1964 kJ The work that could have been produced is equal to the wasted work potential. It is equivalent to the exergy destruction (or irreversibility), and it can be determined from its definition Xdestroyed = T0Sgen . The entropy generation is determined from an entropy balance on an extended system that includes the blocks and the water in their immediate surroundings so that the boundary temperature of the extended system is the temperature of the lake water at all times, S in − S out 1 24 4 3 Net entropy transfer by heat and mass − + S gen = ΔS system { 123 4 4 Entropy generation Change in entropy Qout + S gen = ΔS system = ΔS iron + ΔS copper Tb,out S gen = ΔS iron + ΔS copper + Qout Tlake where ⎛T ⎞ ⎛ 288 K ⎞ ΔS iron = mc avg ln⎜ 2 ⎟ = (50 kg )(0.45 kJ/kg ⋅ K ) ln⎜ ⎜ 353 K ⎟ = −4.579 kJ/K ⎟ ⎜T ⎟ ⎝ ⎠ ⎝ 1⎠ ⎛T ΔS copper = mc avg ln⎜ 2 ⎜T ⎝ 1 ⎞ ⎛ 288 K ⎞ ⎟ = (20 kg )(0.386 kJ/kg ⋅ K ) ln⎜ ⎜ 353 K ⎟ = −1.571 kJ/K ⎟ ⎟ ⎝ ⎠ ⎠ Substituting, ⎛ 1964 kJ ⎞ ⎟kJ/K = 196 kJ X destroyed = T0 S gen = (293 K)⎜ − 4.579 − 1.571 + ⎜ 288 K ⎟ ⎝ ⎠ PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 32. 8-32 8-49E A rigid tank is initially filled with saturated mixture of R-134a. Heat is transferred to the tank from a source until the pressure inside rises to a specified value. The amount of heat transfer to the tank from the source and the exergy destroyed are to be determined. Assumptions 1 The tank is stationary and thus the kinetic and potential energy changes are zero. 2 There is no heat transfer with the environment. Properties From the refrigerant tables (Tables A-11E through A-13E), u1 = u f + x1u fg = 21.246 + 0.55 × 77.307 = 63.76 Btu / lbm P1 = 40 psia ⎫ ⎬ s1 = s f + x1 s fg = 0.04688 + 0.55 × 0.17580 = 0.1436 Btu / lbm ⋅ R x1 = 0.55 ⎭ v 1 = v f + x1 v fg = 0.01232 + 0.55 × 1.16368 = 0.65234 ft 3 / lbm v 2 −v f 0.65234 − 0.01270 = 0.8191 0.79361 − 0.01270 P2 = 60 psia ⎫ ⎬ s 2 = s f + x 2 s fg = 0.06029 + 0.8191× 0.16098 = 0.1922 Btu/lbm ⋅ R (v 2 = v 1 ) ⎭ u 2 = u f + x 2 u fg = 27.939 + 0.8191× 73.360 = 88.03 Btu/lbm x2 = v fg = Analysis (a) The mass of the refrigerant is m= 12 ft 3 V = = 18.40 lbm v 1 0.65234 ft 3 / lbm We take the tank as the system, which is a closed system. The energy balance for this stationary closed system can be expressed as E in − E out 1 24 4 3 = Net energy transfer by heat, work, and mass Source 120°C R-134a 40 psia Q ΔE system 1 24 4 3 Change in internal, kinetic, potential, etc. energies Qin = ΔU = m(u 2 − u1 ) Substituting, Qin = m(u 2 − u1 ) = (18.40 lbm)(88.03 - 63.76) Btu/lbm = 446.3 Btu (b) The exergy destruction (or irreversibility) can be determined from its definition Xdestroyed = T0Sgen . The entropy generation is determined from an entropy balance on an extended system that includes the tank and the region in its immediate surroundings so that the boundary temperature of the extended system where heat transfer occurs is the source temperature, S in − S out 1 24 4 3 Net entropy transfer by heat and mass + S gen = ΔS system { 123 4 4 Entropy generation Change in entropy Qin + S gen = ΔS system = m( s 2 − s1 ) , Tb,in S gen = m( s 2 − s1 ) − Qin Tsource Substituting, 446.3 Btu ⎤ ⎡ X destroyed = T0 S gen = (535 R) ⎢(18.40 lbm)(0.1922 − 0.1436)Btu/lbm ⋅ R − = 66.5 Btu 580 R ⎥ ⎣ ⎦ PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 33. 8-33 8-50 Chickens are to be cooled by chilled water in an immersion chiller that is also gaining heat from the surroundings. The rate of heat removal from the chicken and the rate of exergy destruction during this process are to be determined. Assumptions 1 Steady operating conditions exist. 2 Thermal properties of chickens and water are constant. 3 The temperature of the surrounding medium is 25°C. Properties The specific heat of chicken is given to be 3.54 kJ/kg.°C. The specific heat of water at room temperature is 4.18 kJ/kg.°C (Table A-3). Analysis (a) Chickens are dropped into the chiller at a rate of 500 per hour. Therefore, chickens can be considered 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 the system, the energy balance for steadily flowing chickens can be expressed in the rate form as & & & & & E in − E out = ΔE system 0 (steady) = 0 → E in = E out 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 & & & mh1 = Qout + mh2 (since Δke ≅ Δpe ≅ 0) & & & Qout = Qchicken = m chicken c p (T1 − T2 ) Then the rate of heat removal from the chickens as they are cooled from 15°C to 3ºC becomes & & Qchicken =(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 as a rate of 200 kJ/h = 0.0556 kJ/s. Then the total rate of heat gain by the water is & & & Qwater = Qchicken + Qheat gain = 13.0 + 0.056 = 13.056 kW Noting that the temperature rise of water is not to exceed 2ºC as it flows through the chiller, the mass flow rate 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) (b) The exergy destruction can be determined from its definition Xdestroyed = T0Sgen. The rate of entropy generation during this chilling process is determined by applying the rate form of the entropy balance on an extended system that includes the chiller and the immediate surroundings so that the boundary temperature is the surroundings temperature: & & & & + = ΔS system 0 (steady) S in − S out S gen 1 24 4 3 { 1442443 Rate of net entropy transfer by heat and mass Rate of entropy generation & & & & m1 s1 + m 3 s 3 − m 2 s 2 − m 3 s 4 + Qin & + S gen = 0 Tsurr & & & & m chicken s1 + m water s 3 − m chicken s 2 − m water s 4 + Rate of change of entropy Qin & + S gen = 0 Tsurr Q & & & S gen = m chicken ( s 2 − s1 ) + m water ( s 4 − s 3 ) − in Tsurr Noting that both streams are incompressible substances, the rate of entropy generation is determined to be & Q T T & & & S gen = m chicken c p ln 2 + m water c p ln 4 − in T1 T3 Tsurr = (0.3056 kg/s)(3.54 kJ/kg.K) ln 276 275.5 0.0556 kW + (1.56 kg/s)(4.18 kJ/kg.K) ln − = 0.00128 kW/K 288 273.5 298 K & & Finally, X destroyed = T0 S gen = (298 K)(0.00128 kW/K) = 0.381 kW PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 34. 8-34 8-51 Carbon steel balls are to be annealed at a rate of 2500/h by heating them first and then allowing them to cool slowly in ambient air at a specified rate. The total rate of heat transfer from the balls to the ambient air and the rate of exergy destruction due to this heat transfer are to be determined. Assumptions 1 The thermal properties of the balls are constant. 2 There are no changes in kinetic and potential energies. 3 The balls are at a uniform temperature at the end of the process. Properties The density and specific heat of the balls are given to be ρ = 7833 kg/m3 and cp = 0.465 kJ/kg.°C. Analysis (a) We take a single ball as the system. The energy balance for this closed system can be expressed as E in − E out 1 24 4 3 = Net energy transfer by heat, work, and mass ΔE system 1 24 4 3 Change in internal, kinetic, potential, etc. energies − Qout = ΔU ball = m(u 2 − u1 ) Qout = mc(T1 − T2 ) The amount of heat transfer from a single ball is m = ρV = ρ Qout πD 3 = (7833 kg/m 3 ) π (0.008 m) 3 = 0.00210 kg 6 6 = mc p (T1 − T2 ) = (0.0021 kg )(0.465 kJ/kg.°C)(900 − 100)°C = 781 J = 0.781 kJ (per ball) Then the total rate of heat transfer from the balls to the ambient air becomes & & Qout = n ball Qout = (1200 balls/h) × (0.781 kJ/ball) = 936 kJ/h = 260 W (b) The exergy destruction (or irreversibility) can be determined from its definition Xdestroyed = T0Sgen. The entropy generated during this process can be determined by applying an entropy balance on an extended system that includes the ball and its immediate surroundings so that the boundary temperature of the extended system is at 35°C at all times: S in − S out 1 24 4 3 Net entropy transfer by heat and mass − + S gen = ΔS system { 123 4 4 Entropy generation Change in entropy Qout Q + S gen = ΔS system → S gen = out + ΔS system Tb Tb where ΔS system = m( s 2 − s1 ) = mc avg ln T2 100 + 273 = (0.00210 kg )(0.465 kJ/kg.K)ln = −0.00112 kJ/K T1 900 + 273 Substituting, S gen = Qout 0.781 kJ + ΔS system = − 0.00112 kJ/K = 0.00142 kJ/K (per ball) Tb 308 K Then the rate of entropy generation becomes & & S gen = S gen n ball = (0.00142 kJ/K ⋅ ball)(1200 balls/h) = 1.704 kJ/h.K = 0.000473 kW/K Finally, & & X destroyed = T0 S gen = (308 K)(0.000473 kW/K) = 0.146 kW = 146 W PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 35. 8-35 8-52 A tank containing hot water is placed in a larger tank. The amount of heat lost to the surroundings and the exergy destruction during the process are to be determined. Assumptions 1 Kinetic and potential energy changes are negligible. 2 Air is an ideal gas with constant specific heats. 3 The larger tank is well-sealed. Properties The properties of air at room temperature are R = 0.287 kPa.m3/kg.K, cp = 1.005 kJ/kg.K, cv = 0.718 kJ/kg.K (Table A-2). The properties of water at room temperature are ρ = 1000 kg/m3, cw = 4.18 kJ/kg.K. Analysis (a) The final volume of the air in the tank is V a 2 = V a1 −V w = 0.04 − 0.015 = 0.025 m 3 Air, 22°C The mass of the air in the room is Water 85°C 15 L PV (100 kPa)(0.04 m 3 ) m a = 1 a1 = = 0.04724 kg RTa1 (0.287 kPa ⋅ m 3 /kg ⋅ K)(22 + 273 K) Q The pressure of air at the final state is Pa 2 = m a RTa 2 V a2 = (0.04724 kg)(0.287 kPa ⋅ m 3 /kg ⋅ K)(44 + 273 K) 0.025 m 3 = 171.9 kPa The mass of water is m w = ρ wV w = (1000 kg/m 3 )(0.015 m 3 ) = 14.53 kg An energy balance on the system consisting of water and air is used to determine heat lost to the surroundings Qout = −[m w c w (T2 − Tw1 ) + m a cv (T2 − Ta1 )] = −(14.53 kg)(4.18 kJ/kg.K)(44 − 85) − (0.04724 kg)(0.718 kJ/kg.K)(44 − 22) = 2489 kJ (b) An exergy balance written on the (system + immediate surroundings) can be used to determine exergy destruction. But we first determine entropy and internal energy changes ΔS w = m w c w ln Tw1 (85 + 273) K = 7.3873 kJ/K = (14.53 kg)(4.18 kJ/kg.K)ln (44 + 273) K T2 ⎡ P ⎤ T ΔS a = m a ⎢c p ln a1 − R ln a1 ⎥ P2 ⎦ T2 ⎣ ⎡ (22 + 273) K 100 kPa ⎤ − (0.287 kJ/kg.K)ln = (0.04724 kg) ⎢(1.005 kJ/kg.K)ln ⎥ 171.9 kPa ⎦ (44 + 273) K ⎣ = 0.003931 kJ/K ΔU w = m w c w (T1w − T2 ) = (14.53 kg)(4.18 kJ/kg.K)(85 - 44)K = 2490 kJ ΔU a = m a cv (T1a − T2 ) = (0.04724 kg)(0.718 kJ/kg.K)(22 - 44)K = −0.7462 kJ X dest = ΔX w + ΔX a = ΔU w − T0 ΔS w + ΔU a − T0 ΔS a = 2490 kJ − (295 K)(7.3873 kJ/K) + (-0.7462 kJ) − (295 K)(0.003931 kJ/K) = 308.8 kJ PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 36. 8-36 8-53 Heat is transferred to a piston-cylinder device with a set of stops. The work done, the heat transfer, the exergy destroyed, and the second-law efficiency are to be determined. Assumptions 1 The device is stationary and kinetic and potential energy changes are zero. 2 There is no friction between the piston and the cylinder. Analysis (a) The properties of the refrigerant at the initial and final states are (Tables A-11 through A-13) 3 P1 = 120 kPa ⎫ v 1 = 0.16544 m /kg ⎬ T1 = 20°C ⎭ u1 = 248.22 kJ/kg s1 = 1.0624 kJ/kg.K 3 P2 = 180 kPa ⎫ v 2 = 0.17563 m /kg ⎬ T2 = 120°C ⎭ u 2 = 331.96 kJ/kg s 2 = 1.3118 kJ/kg.K R-134a 1.4 kg 140 kPa 20°C Q The boundary work is determined to be Wb,out = mP2 (v 2 − v1 ) = (1.4 kg)(180 kPa)(0.17563 − 0.16544)m3/kg = 2.57 kJ (b) The heat transfer can be determined from an energy balance on the system Qin = m(u 2 − u1 ) + W b,out = (1.4 kg)(331.96 − 248.22)kJ/kg + 2.57 kJ = 119.8 kJ (c) The exergy difference between the inlet and exit states is ΔX = m[u2 − u1 − T0 ( s2 − s1 ) + P0 (v 2 − v1 )] [ = (1.4 kg) (331.96 − 248.22)kJ/kg − (298 K)(1.3118 − 1.0624)kg.K + (100 kPa)(0.17563 − 0.16544)m3 /kg = 14.61 kJ The useful work output for the process is Wu,out = Wb,out − mP0 (v 2 − v1 ) = 2.57 kJ − (1.4 kg)(100 kPa)(0.17563 − 0.16544) m3/kg = 1.14 kJ The exergy destroyed is the difference between the exergy difference and the useful work output X dest = ΔX − Wu,out = 14.61 − 1.14 = 13.47 kJ (d) The second-law efficiency for this process is η II = W u,out ΔX = 1.14 kJ = 0.078 14.61 kJ PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. ]
• 37. 8-37 Second-Law Analysis of Control Volumes 8-54 Steam is throttled from a specified state to a specified pressure. The wasted work potential during this throttling process is to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 The temperature of the surroundings is given to be 25°C. 4 Heat transfer is negligible. Properties The properties of steam before and after the throttling process are (Tables A-4 through A-6) P1 = 8 MPa ⎫ h1 = 3273.3 kJ/kg ⎬ T1 = 450°C ⎭ s1 = 6.5579 kJ/kg ⋅ K P2 = 6 MPa ⎫ ⎬ s 2 = 6.6806 kJ/kg ⋅ K h2 = h1 ⎭ Analysis The wasted work potential is equivalent to the exergy destruction (or irreversibility). It can be determined from an exergy balance or directly from its definition Xdestroyed = T0Sgen where the entropy generation is determined from an entropy balance on the device, which is an adiabatic steady-flow system, & & S in − S out 1 24 4 3 + Rate of net entropy transfer by heat and mass & S gen { Rate of entropy generation Steam 1 2 & = ΔS system 0 = 0 14 4 2 3 Rate of change of entropy & & & & & ms1 − ms 2 + S gen = 0 → S gen = m( s 2 − s1 ) or s gen = s 2 − s1 Substituting, x destroyed = T0 s gen = T0 ( s 2 − s1 ) = (298 K)(6.6806 − 6.5579)kJ/kg ⋅ K = 36.6 kJ/kg Discussion Note that 36.6 kJ/kg of work potential is wasted during this throttling process. PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 38. 8-38 8-55 [Also solved by EES on enclosed CD] Air is compressed steadily by an 8-kW compressor from a specified state to another specified state. The increase in the exergy of air and the rate of exergy destruction are to be determined. Assumptions 1 Air is an ideal gas with variable specific heats. 2 Kinetic and potential energy changes are negligible. Properties The gas constant of air is R = 0.287 kJ/kg.K (Table A-1). From the air table (Table A-17) T1 = 290 K ⎯→ ⎯ 600 kPa 167°C h1 = 29016 kJ / kg . o s1 = 1.66802 kJ / kg ⋅ K T2 = 440 K ⎯→ ⎯ h2 = 441.61 kJ / kg AIR o s2 = 2.0887 kJ / kg ⋅ K 8 kW Analysis The increase in exergy is the difference between the exit and inlet flow exergies, Increase in exergy = ψ 2 −ψ 1 = [(h2 − h1 ) + Δke 0 + Δpe 0 100 kPa 17°C − T0 ( s 2 − s1 )] = (h2 − h1 ) − T0 ( s 2 − s1 ) where o o s 2 − s1 = ( s 2 − s1 ) − R ln P2 P1 = (2.0887 − 1.66802)kJ/kg ⋅ K - (0.287 kJ/kg ⋅ K) ln = −0.09356 kJ/kg ⋅ K 600 kPa 100 kPa Substituting, Increase in exergy = ψ 2 −ψ 1 = [(441.61 − 290.16)kJ/kg - (290 K)(−0.09356 kJ/kg ⋅ K)] = 178.6 kJ/kg Then the reversible power input becomes & & W rev,in = m(ψ 2 −ψ 1 ) = (2.1 / 60 kg/s)(178.6 kJ/kg) = 6.25 kW (b) The rate of exergy destruction (or irreversibility) is determined from its definition, & & & X destroyed = Win − W rev,in = 8 − 6.25 = 1.75 kW Discussion Note that 1.75 kW of power input is wasted during this compression process. PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 39. 8-39 8-56 EES Problem 8-55 is reconsidered. The problem is to be solved and the actual heat transfer, its direction, the minimum power input, and the compressor second-law efficiency are to be determined. Analysis The problem is solved using EES, and the solution is given below. Function Direction\$(Q) If Q<0 then Direction\$='out' else Direction\$='in' end Function Violation\$(eta) If eta>1 then Violation\$='You have violated the 2nd Law!!!!!' else Violation\$='' end {"Input Data from the Diagram Window" T_1=17 [C] P_1=100 [kPa] W_dot_c = 8 [kW] P_2=600 [kPa] S_dot_gen=0 Q_dot_net=0} {"Special cases" T_2=167 [C] m_dot=2.1 [kg/min]} T_o=T_1 P_o=P_1 m_dot_in=m_dot*Convert(kg/min, kg/s) "Steady-flow conservation of mass" m_dot_in = m_dot_out "Conservation of energy for steady-flow is:" E_dot_in - E_dot_out = DELTAE_dot DELTAE_dot = 0 E_dot_in=Q_dot_net + m_dot_in*h_1 +W_dot_c "If Q_dot_net < 0, heat is transferred from the compressor" E_dot_out= m_dot_out*h_2 h_1 =enthalpy(air,T=T_1) h_2 = enthalpy(air, T=T_2) W_dot_net=-W_dot_c W_dot_rev=-m_dot_in*(h_2 - h_1 -(T_1+273.15)*(s_2-s_1)) "Irreversibility, entropy generated, second law efficiency, and exergy destroyed:" s_1=entropy(air, T=T_1,P=P_1) s_2=entropy(air,T=T_2,P=P_2) s_2s=entropy(air,T=T_2s,P=P_2) s_2s=s_1"This yields the isentropic T_2s for an isentropic process bewteen T_1, P_1 and P_2"I_dot=(T_o+273.15)*S_dot_gen"Irreversiblility for the Process, KW" S_dot_gen=(-Q_dot_net/(T_o+273.15) +m_dot_in*(s_2-s_1)) "Entropy generated, kW" Eta_II=W_dot_rev/W_dot_net"Definition of compressor second law efficiency, Eq. 7_6" h_o=enthalpy(air,T=T_o) s_o=entropy(air,T=T_o,P=P_o) Psi_in=h_1-h_o-(T_o+273.15)*(s_1-s_o) "availability function at state 1" Psi_out=h_2-h_o-(T_o+273.15)*(s_2-s_o) "availability function at state 2" X_dot_in=Psi_in*m_dot_in X_dot_out=Psi_out*m_dot_out DELTAX_dot=X_dot_in-X_dot_out "General Exergy balance for a steady-flow system, Eq. 7-47" (1-(T_o+273.15)/(T_o+273.15))*Q_dot_net-W_dot_net+m_dot_in*Psi_in - m_dot_out*Psi_out =X_dot_dest "For the Diagram Window" PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 40. 8-40 Text\$=Direction\$(Q_dot_net) Text2\$=Violation\$(Eta_II) ηII 0.7815 0.8361 0.8908 0.9454 1 I [kW] 1.748 1.311 0.874 0.437 1.425E-13 Xdest [kW] 1.748 1.311 0.874 0.437 5.407E-15 T2s [C] 209.308 209.308 209.308 209.308 209.308 T2 [C] 167 200.6 230.5 258.1 283.9 Qnet [kW] -2.7 -1.501 -0.4252 0.5698 1.506 How can entropy decrease? 250 2s 200 2 ideal T [C] 150 100 kPa 100 600 kPa actual 50 1 0 5.0 5.5 6.0 6.5 s [kJ/kg-K] 300 2.0 280 T2 240 1.0 Xdest 1.5 260 220 200 0.5 180 160 0.75 0.80 0.85 0.90 0.95 0.0 1.00 ηII 2 2.0 1 0 1.0 X dest Q net 1.5 -1 0.5 -2 -3 0.75 0.80 0.85 0.90 0.95 0.0 1.00 ηII PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 41. 8-41 8-57 Air expands in an adiabatic turbine from a specified state to another specified state. The actual and maximum work outputs are to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 The device is adiabatic and thus heat transfer is negligible. 3 Air is an ideal gas with constant specific heats. 4 Potential energy changes are negligible. Properties At the average temperature of (425 + 325)/2 = 375 K, the constant pressure specific heat of air is cp = 1.011 kJ/kg.K (Table A-2b). The gas constant of air is R = 0.287 kJ/kg.K (Table A-1). & & & 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 system can be expressed in the rate form as & & E in − E out 1 24 4 3 & ΔE system 0 (steady) 1442443 4 = Rate of net energy transfer by heat, work, and mass =0 Rate of change in internal, kinetic, potential, etc. energies & & E in = E out 550 kPa 425 K 150 m/s & & & m(h1 + V12 / 2) = Wout + m(h2 + V 22 / 2) ⎡ V 2 − V 22 ⎤ & & Wout = m ⎢h1 − h2 + 1 ⎥ 2 ⎢ ⎥ ⎣ ⎦ wout = c p (T1 − T2 ) + V12 − V 22 2 Substituting, wout = c p (T1 − T2 ) + Air 110 kPa 325 K 50 m/s V12 − V 22 2 = (1.011 kJ/kg ⋅ K)(425 − 325)K + (150 m/s) 2 − (50 m/s) 2 ⎛ 1 kJ/kg ⎞ ⎜ ⎟ 2 ⎝ 1000 m 2 /s 2 ⎠ = 111.1 kJ/kg The entropy change of air is s 2 − s1 = c p ln P T2 − R ln 2 P1 T1 = (1.011 kJ/kg ⋅ K) ln = 0.1907 kJ/kg ⋅ K 110 kPa 325 K − (0.287 kJ/kg ⋅ K) ln 550 kPa 425 K The maximum (reversible) work is the exergy difference between the inlet and exit states V12 − V 22 − T0 ( s1 − s 2 ) 2 − T0 ( s1 − s 2 ) w rev,out = c p (T1 − T2 ) + = wout = 111.1 kJ/kg − (298 K)(−0.1907 kJ/kg ⋅ K) = 167.9 kJ/kg Irreversibilites occurring inside the turbine cause the actual work production to be less than the reversible (maximum) work. The difference between the reversible and actual works is the irreversibility. PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 42. 8-42 8-58E Air is compressed in a compressor from a specified state to another specified state. The minimum work input is to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. 4 The inlet state of air is taken as the dead state. Properties The properties of air at room temperature are cp = 0.240 Btu/lbm⋅R and R = 0.06855 Btu/lbm⋅R (Table A-2Ea). Analysis The entropy change of air is 140 psia P T 200°F s 2 − s1 = c p ln 2 − R ln 2 P1 T1 140 psia 660 R = (0.240 Btu/lbm ⋅ R) ln − (0.06855 Btu/lbm ⋅ R) ln 14.7 psia 537 R Air = −0.1050 Btu/lbm ⋅ R The minimum (reversible) work is the exergy difference between the inlet and exit states wrev,in = c p (T2 − T1 ) − T0 ( s 2 − s1 ) = (0.240 Btu/lbm ⋅ R)(200 − 77)R − (537 R)( −0.1050 Btu/lbm ⋅ R) 14.7 psia 77°F = 85.9 Btu/lbm & & & 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 & & & = ΔE system 0 (steady) =0 E in − E out 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 & & E in = E out h1 + win = h2 + q out win = h2 − h1 + q out Inspection of this result reveals that for the same inlet and exit states, any rejection of heat will increase the actual work that must be supplied to the compressor. The reversible (or minimum) power input is determined from the exergy balance applied on the compressor, and setting the exergy destruction term equal to zero, & & X in − X out 1 24 4 3 Rate of net exergy tra nsfer by heat, work,and mass & & − X destroyed 0 (reversible) = ΔX system 0 (steady) = 0 144 2444 4 3 1442443 Rate of exergy destructio n Rate of change of exergy & & X in = X out & & & & mψ1 + Wrev,in = mψ 2 + X heat, out & ⎛ T ⎞ & & Wrev,in = m(ψ 2 −ψ1) + Qout ⎜1 − 0 ⎟ ⎝ T⎠ & ⎛ T ⎞ & & Wrev,in = m[(h2 − h1) − T0 (s2 − s1)] + Qout ⎜1 − 0 ⎟ ⎝ T⎠ ⎛ T ⎞ wrev,in = c p (T2 − T1) − T0 (s2 − s1) + qout ⎜1 − 0 ⎟ ⎝ T⎠ Inspection of this result reveals that for the same inlet and exit states, any rejection of heat will increase the reversible (minimum) work that must be supplied to the compressor. PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 43. 8-43 8-59 Refrigerant-134a is compressed by an adiabatic compressor from a specified state to another specified state. The minimum power required is to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 The device is adiabatic and thus heat transfer is negligible. 800 kPa 50°C Properties From the refrigerant tables (Tables A-11 through A-13) P1 = 160 kPa ⎫ ⎬ x1 = 1 (sat. vap.) ⎭ P2 = 800 kPa ⎫ h2 ⎬ T2 = 50°C ⎭ s 2 h1 = 241.11 kJ/kg s1 = 0.94190 kJ/kg ⋅ K R-134a = 286.69 kJ/kg = 0.9802 kJ/kg ⋅ K The minimum (reversible) power is & & W rev,in = m[h2 − h1 − T0 ( s 2 − s1 )] 160 kPa sat. vapor 0.1 kg/s = (0.1 kg/s)[(286.69 − 241.11)kJ/kg − (298 K)(0.9802 − 0.94190) kJ/kg ⋅ K ] = 3.42 kW PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 44. 8-44 8-60 Steam is decelerated in a diffuser. The second law efficiency of the diffuser is to be determined. Assumptions 1 The diffuser operates steadily. 2 The changes in potential energies are negligible. Properties The properties of steam at the inlet and the exit of the diffuser are (Tables A-4 through A-6) P1 = 500 kPa ⎫ h1 = 2855.8 kJ/kg ⎬ T1 = 200°C ⎭ s1 = 7.0610 kJ/kg ⋅ K P2 = 200 kPa ⎫ h2 = 2706.3 kJ/kg ⎬ x 2 = 1 (sat. vapor) ⎭ s 2 = 7.1270 kJ/kg ⋅ K Analysis We take the diffuser 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 & & E in − E out 1 24 4 3 & ΔE system 0 (steady) 1442443 4 4 = Rate of net energy transfer by heat, work, and mass =0 Rate of change in internal, kinetic, potential, etc. energies & & E in = E out 500 kPa 200°C 30 m/s & & m(h1 + V12 / 2) = m(h2 + V 22 /2) V 22 − V12 = h1 − h2 = Δke actual 2 H2O 200 kPa sat. vapor Substituting, Δke actual = h1 − h2 = 2855.8 − 2706.3 = 149.5 kJ/kg An exergy balance on the diffuser gives & & X in − X out 1 24 4 3 Rate of net exergy tra nsfer by heat, work,and mass & & − X destroyed 0 (reversible) = ΔX system 0 (steady) = 0 144 2444 4 3 1442443 Rate of exergy destructio n Rate of change of exergy & & X in = X out & & mψ1 = mψ 2 h1 − h0 + V12 V2 − T0 (s1 − s0 ) = h2 − h0 + 2 − T0 (s2 − s0 ) 2 2 2 2 V2 − V1 = h1 − h2 − T0 (s1 − s2 ) 2 Δkerev = h1 − h2 − T0 (s1 − s2 ) Substituting, Δke rev = h1 − h2 − T0 ( s1 − s 2 ) = (2855.8 − 2706.3)kJ/kg − (298 K)(7.0610 − 7.1270) kJ/kg ⋅ K = 169.2 kJ/kg The second law efficiency is then η II = Δke actual 149.5 kJ/kg = = 0.884 Δke rev 169.2 kJ/kg PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 45. 8-45 8-61 Air is accelerated in a nozzle while losing some heat to the surroundings. The exit temperature of air and the exergy destroyed during the process are to be determined. Assumptions 1 Air is an ideal gas with variable specific heats. 2 The nozzle operates steadily. Properties The gas constant of air is R = 0.287 kJ/kg.K (Table A-1). The properties of air at the nozzle inlet are (Table A-17) T1 = 360 K ⎯→ ⎯ h1 = 360.58 kJ / kg 4 kJ/kg o s1 = 1.88543 kJ / kg ⋅ K Analysis (a) We take the nozzle as 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 1 24 4 3 & ΔEsystem 0 (steady) 1442443 = Rate of net energy transfer by heat, work, and mass 50 m/s AIR 300 m/s =0 Rate of change in internal, kinetic, potential, etc. energies & & Ein = Eout 2 & & & m(h1 + V12 / 2) = m( h2 + V2 /2) + Qout or 0 = q out + h2 − h1 + V 22 − V12 2 Therefore, h2 = h1 − q out − V 22 − V12 (300 m/s) 2 − (50 m/s) 2 ⎛ 1 kJ/kg ⎞ = 360.58 − 4 − ⎜ ⎟ = 312.83 kJ/kg 2 2 2 2 ⎝ 1000 m / s ⎠ 0 T2 = 312.5 K = 39.5°C and s 2 = 1.74302 kJ/kg ⋅ K At this h2 value we read, from Table A-17, (b) The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0S gen where the entropy generation Sgen is determined from an entropy balance on an extended system that includes the device and its immediate surroundings so that the boundary temperature of the extended system is Tsurr at all times. It gives & & S in − S out 1 24 4 3 Rate of net entropy transfer by heat and mass & ms1 − ms 2 − & S gen { + Rate of entropy generation & = ΔS system 0 = 0 14 4 2 3 Rate of change of entropy & Qout & + S gen = 0 Tb,surr & Q & & S gen = m(s 2 − s1 ) + out Tsurr where o o Δs air = s 2 − s1 − R ln P2 P1 = (1.74302 − 1.88543)kJ/kg ⋅ K − (0.287 kJ/kg ⋅ K) ln 95 kPa = 0.1876 kJ/kg ⋅ K 300 kPa Substituting, the entropy generation and exergy destruction per unit mass of air are determined to be PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 46. 8-46 x destroyed = T0 s gen = Tsurr s gen ⎛ q = T0 ⎜ s 2 − s1 + surr ⎜ Tsurr ⎝ ⎞ 4 kJ/kg ⎞ ⎛ ⎟ = (290 K)⎜ 0.1876 kJ/kg ⋅ K + ⎟ = 58.4 kJ/kg ⎟ 290 K ⎠ ⎝ ⎠ Alternative solution The exergy destroyed during a process can be determined from an exergy balance applied on the extended system that includes the device and its immediate surroundings so that the boundary temperature of the extended system is environment temperature T0 (or Tsurr) at all times. Noting that exergy transfer with heat is zero when the temperature at the point of transfer is the environment temperature, the exergy balance for this steady-flow system can be expressed as & & X in − X out 1 24 4 3 Rate of net exergy tra nsfer by heat, work,and mass & & & & & & & & − X destroyed = ΔX system 0 (steady) = 0 → X destroyed = X in − X out = mψ1 − mψ 2 = m(ψ1 −ψ 2 ) 1 24 4 3 1442443 Rate of exergy destructio n Rate of change of exergy & & = m[(h1 − h2 ) − T0 (s1 − s2 ) − Δke − Δpe 0 ] = m[T0 (s2 − s1) − (h2 − h1 + Δke)] & = m[T0 (s2 − s1) + qout ] since, from energy balance, − qout = h2 − h1 + Δke & ⎛ Q ⎞ & & = T0 ⎜ m(s2 − s1) + out ⎟ = T0Sgen ⎜ T0 ⎟ ⎝ ⎠ Therefore, the two approaches for the determination of exergy destruction are identical. PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 47. 8-47 8-62 EES Problem 8-61 is reconsidered. The effect of varying the nozzle exit velocity on the exit temperature and exergy destroyed is to be investigated. Analysis The problem is solved using EES, and the solution is given below. "Knowns:" WorkFluid\$ = 'Air' P[1] = 300 [kPa] T[1] =87 [C] P[2] = 95 [kPa] Vel[1] = 50 [m/s] {Vel[2] = 300 [m/s]} T_o = 17 [C] T_surr = T_o q_loss = 4 [kJ/kg] "Conservation of Energy - SSSF energy balance for nozzle -- neglecting the change in potential energy:" h[1]=enthalpy(WorkFluid\$,T=T[1]) s[1]=entropy(WorkFluid\$,P=P[1],T=T[1]) ke[1] = Vel[1]^2/2 ke[2]=Vel[2]^2/2 h[1]+ke[1]*convert(m^2/s^2,kJ/kg) = h[2] + ke[2]*convert(m^2/s^2,kJ/kg)+q_loss T[2]=temperature(WorkFluid\$,h=h[2]) s[2]=entropy(WorkFluid\$,P=P[2],h=h[2]) "The entropy generated is detemined from the entropy balance:" s[1] - s[2] - q_loss/(T_surr+273) + s_gen = 0 x_destroyed = (T_o+273)*s_gen 80 75 Vel2 [m/s] 79.31 74.55 68.2 60.25 50.72 39.6 xdestroyed [kJ/kg] 93.41 89.43 84.04 77.17 68.7 58.49 100 140 180 220 260 300 70 T[2] [C] T2 [C] 65 60 55 50 45 40 35 100 95 140 220 260 Vel[2] 90 xdestroyed [kJ/kg] 180 85 80 75 70 65 60 55 100 140 180 220 260 300 Vel[2] PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 300
• 48. 8-48 8-63 Steam is decelerated in a diffuser. The mass flow rate of steam and the wasted work potential during the process are to be determined. Assumptions 1 The diffuser operates steadily. 2 The changes in potential energies are negligible. Properties The properties of steam at the inlet and the exit of the diffuser are (Tables A-4 through A-6) P1 = 10 kPa ⎫ h1 = 2592.0 kJ/kg ⎬ T1 = 50°C ⎭ s1 = 8.1741 kJ/kg ⋅ K h2 = 2591.3 kJ/kg T2 = 50°C ⎫ ⎬ s 2 = 8.0748 kJ/kg ⋅ K sat.vapor ⎭ v 2 = 12.026 m 3 /kg 300 m/s H2O 70 m/s Analysis (a) The mass flow rate of the steam is & m= 1 v2 A2V 2 = 1 3 12.026 m / kg (3 m 2 )(70 m/s) = 17.46 kg/s (b) We take the diffuser to be the system, which is a control volume. Assuming the direction of heat transfer to be from the stem, the energy balance for this steady-flow system can be expressed in the rate form as & & E in − E out 1 24 4 3 = Rate of net energy transfer by heat, work, and mass & ΔE system 0 (steady) 1442443 4 4 =0 Rate of change in internal, kinetic, potential, etc. energies & & E in = E out 2 & & & m(h1 + V12 / 2) = m(h2 + V2 /2) + Qout ⎛ V 2 − V12 & & Qout = − m⎜ h2 − h1 + 2 ⎜ 2 ⎝ ⎞ ⎟ ⎟ ⎠ Substituting, ⎡ (70 m/s) 2 − (300 m/s) 2 ⎛ 1 kJ/kg & Qout = −(17.46 kg/s) ⎢2591.3 − 2592.0 + ⎜ 2 ⎝ 1000 m 2 / s 2 ⎢ ⎣ ⎞⎤ ⎟⎥ = 754.8 kJ/s ⎠⎥ ⎦ The wasted work potential is equivalent to exergy destruction. The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0S gen where the entropy generation Sgen is determined from an entropy balance on an extended system that includes the device and its immediate surroundings so that the boundary temperature of the extended system is Tsurr at all times. It gives & & S in − S out 1 24 4 3 Rate of net entropy transfer by heat and mass & & ms1 − ms 2 − + & S gen { Rate of entropy generation & = ΔS system 0 = 0 14 4 2 3 Rate of change of entropy & & Qout Q & & & + S gen = 0 → S gen = m(s 2 − s1 ) + out Tb,surr Tsurr Substituting, the exergy destruction is determined to be & ⎛ Q & & & X destroyed = T0 S gen = T0 ⎜ m( s 2 − s1 ) + out ⎜ T0 ⎝ ⎞ ⎟ ⎟ ⎠ 754.8 kW ⎞ ⎛ = (298 K)⎜ (17.46 kg/s)(8.0748 - 8.1741)kJ/kg ⋅ K + ⎟ = 238.3 kW 298 K ⎠ ⎝ PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 49. 8-49 8-64E Air is compressed steadily by a compressor from a specified state to another specified state. The minimum power input required for the compressor is to be determined. Assumptions 1 Air is an ideal gas with variable specific heats. 2 Kinetic and potential energy changes are negligible. Properties The gas constant of air is R = 0.06855 Btu/lbm.R (Table A-1E). From the air table (Table A17E) ⎯→ h1 = 124.27 Btu/lbm T1 = 520 R ⎯ o s1 = 0.59173 Btu/lbm ⋅ R 100 psia 480°F ⎯→ h2 = 226.11 Btu/lbm T2 = 940 R ⎯ o s 2 = 0.73509 Btu/lbm ⋅ R Analysis The reversible (or minimum) power input is determined from the rate form of the exergy balance applied on the compressor and setting the exergy destruction term equal to zero, & & X in − X out 1 24 4 3 Rate of net exergy tra nsfer by heat, work,and mass & & − X destroyed 0 (reversible) = ΔX system 0 (steady) = 0 144 2444 4 3 1442443 Rate of exergy destructio n Rate of change of exergy AIR 15 lbm/min 14.7 psia 60°F & & X in = X out & & & mψ1 + Wrev,in = mψ 2 & & & Wrev,in = m(ψ 2 −ψ1) = m[(h2 − h1) − T0 (s2 − s1) + Δke 0 + Δpe 0 ] where o o Δs air = s 2 − s1 − R ln P2 P1 = (0.73509 − 0.59173)Btu/lbm ⋅ R − (0.06855 Btu/lbm ⋅ R) ln 100 psia 14.7 psia = 0.01193 Btu/lbm ⋅ R Substituting, & W rev,in = (22/60 lbm/s)[(226.11 − 124.27)Btu/lbm − (520 R)(0.01193 Btu/lbm ⋅ R)] = 35.1 Btu/s = 49.6 hp Discussion Note that this is the minimum power input needed for this compressor. PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 50. 8-50 8-65 Steam expands in a turbine from a specified state to another specified state. The actual power output of the turbine is given. The reversible power output and the second-law efficiency are to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 The temperature of the surroundings is given to be 25°C. Properties From the steam tables (Tables A-4 through A-6) P1 = 6 MPa ⎫ h1 = 3658.8 kJ/kg ⎬ T1 = 600°C ⎭ s1 = 7.1693 kJ/kg ⋅ K P2 = 50 kPa ⎫ h2 = 2682.4 kJ/kg ⎬ T2 = 100°C ⎭ s 2 = 7.6953 kJ/kg ⋅ K & & & Analysis (b) 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 system can be expressed in the rate form as & & E in − E out 1 24 4 3 = Rate of net energy transfer by heat, work, and mass & ΔE system 0 (steady) 1442443 4 =0 80 m/s 6 MPa 600°C Rate of change in internal, kinetic, potential, etc. energies & & E in = E out & & & m( h1 + V12 / 2) = Wout + m( h2 + V 22 / 2) STEAM ⎡ V 2 − V 22 ⎤ & & Wout = m ⎢h1 − h2 + 1 ⎥ 2 ⎢ ⎥ ⎣ ⎦ 5 MW Substituting, ⎛ (80 m/s) 2 − (140 m/s) 2 & 5000 kJ/s = m⎜ 3658.8 − 2682.4 + ⎜ 2 ⎝ & m = 5.156 kg/s ⎛ ⎞⎞ ⎜ ⎟⎟ 2 2 ⎟ ⎝ 1000 m / s ⎠ ⎠ 1 kJ/kg 50 kPa 100°C 140 m/s The reversible (or maximum) power output is determined from the rate form of the exergy balance applied on the turbine and setting the exergy destruction term equal to zero, & & X in − X out 1 24 4 3 Rate of net exergy tra nsfer by heat, work,and mass & & − X destroyed 0 (reversible) = ΔX system 0 (steady) = 0 144 2444 4 3 1442443 Rate of exergy destructio n Rate of change of exergy & & X in = X out & & mψ = W 1 rev,out & + mψ 2 & & & Wrev,out = m(ψ1 −ψ 2 ) = m[(h1 − h2 ) − T0 (s1 − s2 ) − Δke 0 − Δpe 0 ] Substituting, & & W rev,out = m[(h1 − h2 ) − T0 ( s1 − s 2 )] = (5.156 kg/s)[3658.8 − 2682.4 − (298 K)(7.1693 − 7.6953) kJ/kg ⋅ K ] = 5842 kW (b) The second-law efficiency of a turbine is the ratio of the actual work output to the reversible work, η II = & Wout & W rev,out = 5 MW = 85.6% 5.842 MW Discussion Note that 14.4% percent of the work potential of the steam is wasted as it flows through the turbine during this process. PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 51. 8-51 8-66 Steam is throttled from a specified state to a specified pressure. The decrease in the exergy of the steam during this throttling process is to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 The temperature of the surroundings is given to be 25°C. 4 Heat transfer is negligible. Properties The properties of steam before and after throttling are (Tables A-4 through A-6) P1 = 9 MPa ⎫ h1 = 3387.4 kJ/kg ⎬ T1 = 500°C ⎭ s1 = 6.6603 kJ/kg ⋅ K Steam P2 = 7 MPa ⎫ ⎬ s 2 = 6.7687 kJ/kg ⋅ K ⎭ 1 h2 = h1 Analysis The decrease in exergy is of the steam is the difference between the inlet and exit flow exergies, Decrease in exergy = ψ 1 − ψ 2 = −[Δh 0 − Δke 0 − Δpe 0 − T0 ( s1 − s 2 )] = T0 ( s 2 − s1 ) 2 = (298 K)(6.7687 − 6.6603)kJ/kg ⋅ K = 32.3 kJ/kg Discussion Note that 32.3 kJ/kg of work potential is wasted during this throttling process. PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 52. 8-52 8-67 Combustion gases expand in a turbine from a specified state to another specified state. The exergy of the gases at the inlet and the reversible work output of the turbine are to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Potential energy changes are negligible. 3 The temperature of the surroundings is given to be 25°C. 4 The combustion gases are ideal gases with constant specific heats. Properties The constant pressure specific heat and the specific heat ratio are given to be cp = 1.15 kJ/kg.K and k = 1.3. The gas constant R is determined from R = c p − cv = c p − c p / k = c p (1 − 1 / k ) = (1.15 kJ/kg ⋅ K)(1 − 1/1.3) = 0.265 kJ/kg ⋅ K Analysis (a) The exergy of the gases at the turbine inlet is simply the flow exergy, ψ 1 = h1 − h0 − T0 ( s1 − s 0 ) + V12 2 + gz1 800 kPa 900°C 0 where T P s1 − s 0 = c p ln 1 − R ln 1 T0 P0 800 kPa 1173 K = (1.15 kJ/kg ⋅ K)ln − (0.265 kJ/kg ⋅ K)ln 100 kPa 298 K = 1.025 kJ/kg ⋅ K GAS TURBINE 400 kPa 650°C Thus, ψ 1 = (1.15 kJ/kg.K)(900 − 25)°C − (298 K)(1.025 kJ/kg ⋅ K) + (100 m/s) 2 ⎛ 1 kJ/kg ⎞ ⎜ ⎟ = 705.8 kJ/kg 2 ⎝ 1000 m 2 / s 2 ⎠ (b) The reversible (or maximum) work output is determined from an exergy balance applied on the turbine and setting the exergy destruction term equal to zero, & & X in − X out 1 24 4 3 Rate of net exergy tra nsfer by heat, work,and mass & & − X destroyed 0 (reversible) = ΔX system 0 (steady) = 0 144 2444 4 3 1442443 Rate of exergy destructio n Rate of change of exergy & & X in = X out & & mψ = W 1 rev,out & + mψ 2 & & & Wrev,out = m(ψ1 −ψ 2 ) = m[(h1 − h2 ) − T0 (s1 − s2 ) − Δke − Δpe 0 ] where Δke = V 22 − V12 (220 m/s) 2 − (100 m/s) 2 ⎛ 1 kJ/kg ⎞ = ⎜ ⎟ = 19.2 kJ/kg 2 2 ⎝ 1000 m 2 / s 2 ⎠ and s 2 − s1 = c p ln T2 P − R ln 2 T1 P1 400 kPa 923 K − (0.265 kJ/kg ⋅ K)ln 800 kPa 1173 K = −0.09196 kJ/kg ⋅ K = (1.15 kJ/kg ⋅ K)ln Then the reversible work output on a unit mass basis becomes wrev,out = h1 − h2 + T0 ( s 2 − s1 ) − Δke = c p (T1 − T2 ) + T0 ( s 2 − s1 ) − Δke = (1.15 kJ/kg ⋅ K)(900 − 650)°C + (298 K)(−0.09196 kJ/kg ⋅ K) − 19.2 kJ/kg = 240.9 kJ/kg PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 53. 8-53 8-68E Refrigerant-134a enters an adiabatic compressor with an isentropic efficiency of 0.80 at a specified state with a specified volume flow rate, and leaves at a specified pressure. The actual power input and the second-law efficiency 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 potential energy 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) h1 = h g @ 30 psia = 105.32 Btu / lbm P2 = 70 psia ⎫ P1 = 30 psia ⎫ ⎬ h2 s = 112.80 Btu/lbm ⎬ s1 = s g @ 30 psia = 0.2238 Btu/lbm ⋅ R s 2 s = s1 sat.vapor ⎭ ⎭ 3 v 1 = v g @ 30 psia = 1.5492 ft /lbm Analysis From the isentropic efficiency relation, h −h ⎯→ h2 a = h1 + (h2 s − h1 ) / η c η c = 2s 1 ⎯ h2 a − h1 = 105.32 + (112.80 − 105.32) / 0.80 = 114.67 Btu/lbm 70 psia s2 = s1 R-134a 20 ft3/min Then, P2 = 70 psia ⎫ ⎬ s 2 = 0.2274 Btu/lbm h2 a = 114.67 ⎭ Also, & m= 30 psia sat. vapor V&1 20 / 60 ft 3 / s = = 0.2152 lbm/s v 1 1.5492 ft 3 / lbm & & & There is only one inlet and one exit, and thus m1 = m2 = m . We take the actual compressor as the system, which is a control volume. The energy balance for this steady-flow system can be expressed 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 & & E in = E out & & & & Wa,in + mh1 = mh2 (since Q ≅ Δke ≅ Δpe ≅ 0) & & Wa,in = m(h2 − h1 ) Substituting, the actual power input to the compressor becomes 1 hp ⎞ ⎛ & Wa,in = (0.2152 lbm/s)(114.67 − 105.32) Btu/lbm⎜ ⎟ = 2.85 hp 0.7068 Btu/s ⎠ ⎝ (b) The reversible (or minimum) power input is determined from the exergy balance applied on the compressor and setting the exergy destruction term equal to zero, & & & & − X destroyed 0 (reversible) = ΔX system 0 (steady) = 0 X in − X out 1 24 4 3 144 2444 4 3 1442443 Rate of net exergy tra nsfer by heat, work,and mass Rate of exergy destructio n Rate of change of exergy & & X in = X out & & & Wrev,in + mψ1 = mψ 2 & & & Wrev,in = m(ψ 2 −ψ1) = m[(h2 − h1) − T0 (s2 − s1) + Δke Substituting, & W rev,in Thus, 0 + Δpe 0 ] = (0.2152 lbm/s)[(114.67 − 105.32)Btu/lbm − (535 R)(0.2274 − 0.2238)Btu/lbm ⋅ R ] = 1.606 Btu/s = 2.27 hp & W rev,in 2.27 hp η II = = = 79.8% & 2.85 hp W (since 1 hp = 0.7068 Btu/s) act,in PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 54. 8-54 8-69 Refrigerant-134a is compressed by an adiabatic compressor from a specified state to another specified state. The isentropic efficiency and the second-law efficiency of the compressor are to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy 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) h1 = 246.36 kJ/kg P1 = 140 kPa ⎫ ⎬ s1 = 0.97236 kJ/kg ⋅ K T1 = −10°C ⎭ v 1 = 0.14605 m 3 /kg 700 kPa 60°C P2 = 700 kPa ⎫ h2 = 298.42 kJ/kg ⎬ T2 = 60°C ⎭ s 2 = 1.0256 kJ/kg ⋅ K R-134a P2 s = 700 kPa ⎫ ⎬ h2 s = 281.16 kJ/kg s 2 s = s1 ⎭ 0.5 kW Analysis (a) The isentropic efficiency is ηc = h2 s − h1 281.16 − 246.36 = = 0.668 = 66.8% h2 a − h1 298.42 − 246.36 140 kPa -10°C & & & (b) There is only one inlet and one exit, and thus m1 = m2 = m . We take the actual compressor as the system, which is a control volume. The energy balance for this steady-flow system can be expressed as & & E in − E out 1 24 4 3 & ΔE system 0 (steady) 1442443 4 = Rate of net energy transfer by heat, work, and mass =0 Rate of change in internal, kinetic, potential, etc. energies & & E in = E out & & & & Wa,in + mh1 = mh2 (since Q ≅ Δke ≅ Δpe ≅ 0) & & Wa,in = m(h2 − h1 ) Then the mass flow rate of the refrigerant becomes & m= & Wa,in h2 a − h1 = 0.5 kJ/s = 0.009603 kg/s (298.42 − 246.36)kJ/kg The reversible (or minimum) power input is determined from the exergy balance applied on the compressor and setting the exergy destruction term equal to zero, & & X in − X out 1 24 4 3 Rate of net exergy tra nsfer by heat, work,and mass & & − X destroyed 0 (reversible) = ΔX system 0 (steady) = 0 144 2444 4 3 1442443 Rate of exergy destructio n Rate of change of exergy & & X in = X out & & & Wrev,in + mψ1 = mψ 2 & & & Wrev,in = m(ψ 2 −ψ1) = m[(h2 − h1) − T0 (s2 − s1) + Δke 0 + Δpe 0 ] Substituting, & W rev, in = (0.009603 kg/s) [(298.42 − 246.36 ) kJ/kg − (300 K)(1.0256 − 0.97236 ) kJ/kg ⋅ K ] = 0 .347 kW and & W 0.347 kW rev, ηII = & in = = 69.3% Wa,in 0.5 kW PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 55. 8-55 8-70 Air is compressed steadily by a compressor from a specified state to another specified state. The increase in the exergy of air and the rate of exergy destruction are to be determined. Assumptions 1 Air is an ideal gas with variable specific heats. 2 Kinetic and potential energy changes are negligible. Properties The gas constant of air is R = 0.287 kJ/kg.K (Table A-1). From the air table (Table A-17) T1 = 300 K ⎯ ⎯→ h1 = 300.19 kJ/kg o s1 = 1.702 kJ/kg ⋅ K 600 kPa 277°C AIR 0.06 kg/s T2 = 550 K ⎯ ⎯→ h2 = 555.74 kJ/kg o s 2 = 2.318 kJ/kg ⋅ K 95 kPa 27°C Analysis The reversible (or minimum) power input is determined from the rate form of the exergy balance applied on the compressor and setting the exergy destruction term equal to zero, & & X in − X out 1 24 4 3 Rate of net exergy tra nsfer by heat, work,and mass & & − X destroyed 0 (reversible) = ΔX system 0 (steady) = 0 144 2444 4 3 1442443 Rate of exergy destructio n Rate of change of exergy & & X in = X out & & & mψ1 + Wrev,in = mψ 2 & & & Wrev,in = m(ψ 2 −ψ1) = m[(h2 − h1) − T0 (s2 − s1) + Δke 0 + Δpe 0 ] where o o s2 − s1 = s2 − s1 − R ln P2 P 1 = (2.318 − 1.702) kJ / kg ⋅ K − (0.287 kJ / kg ⋅ K)ln = 0.0870 kJ / kg ⋅ K 600 kPa 95 kPa Substituting, & W rev,in = (0.06 kg/s)[(555.74 − 300.19)kJ/kg - (298 K)(0.0870 kJ/kg ⋅ K)] = 13.7 kW Discussion Note that a minimum of 13.7 kW of power input is needed for this compression process. PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 56. 8-56 8-71 EES Problem 8-70 is reconsidered. The effect of compressor exit pressure on reversible power is to be investigated. Analysis The problem is solved using EES, and the solution is given below. T_1=27 [C] P_1=95 [kPa] m_dot = 0.06 [kg/s] {P_2=600 [kPa]} T_2=277 [C] T_o=25 [C] P_o=100 [kPa] m_dot_in=m_dot "Steady-flow conservation of mass" m_dot_in = m_dot_out h_1 =enthalpy(air,T=T_1) h_2 = enthalpy(air, T=T_2) W_dot_rev=m_dot_in*(h_2 - h_1 -(T_1+273.15)*(s_2-s_1)) s_1=entropy(air, T=T_1,P=P_1) s_2=entropy(air,T=T_2,P=P_2) P2 [kPa] 200 250 300 350 400 450 500 550 600 Wrev [kW] 8.025 9.179 10.12 10.92 11.61 12.22 12.76 13.25 13.7 14 Wrev [kW] 13 12 11 10 9 8 200 250 300 350 400 450 500 550 600 P2 [kPa] PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 57. 8-57 8-72 An expression is to be derived for the work potential of the single-phase contents of a rigid adiabatic container when the initially empty container is filled through a single opening from a source of working fluid whose properties remain fixed. Analysis The conservation of mass principle for this system reduces to dm CV & = mi dt where the subscript i stands for the inlet state. When the entropy generation is set to zero (for calculating work potential) and the combined first and second law is reduced to fit this system, it becomes d (U − T0 S ) & & W rev = − + ( h − T0 S ) i m i dt When these are combined, the result is d (U − T0 S ) dm CV & W rev = − + ( h − T0 S ) i dt dt Recognizing that there is no initial mass in the system, integration of the above equation produces W rev = (h − T0 s ) i m 2 − m 2 (h2 − T0 s 2 ) W rev = (hi − h2 ) − T0 ( s i − s 2 ) m2 where the subscript 2 stands for the final state in the container. PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 58. 8-58 8-73 A rigid tank initially contains saturated liquid of refrigerant-134a. R-134a is released from the vessel until no liquid is left in the vessel. The reversible work associated with this process is to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process. It can be analyzed as a uniform-flow process since the state of fluid leaving the device remains constant. 2 Kinetic and potential energies are negligible. 3 There are no work interactions involved. Properties The properties of R-134a are (Tables A-11 through A-13) v 1 = v f @ 20°C = 0.0008161 m 3 / kg T1 = 20°C ⎫ ⎬ u1 = u f @ 20°C = 78.86 kJ/kg sat. liquid ⎭ s1 = s f @ 20°C = 0.30063 kJ/kg ⋅ K 3 v 2 = v g @ 20°C = 0.035969 m / kg T2 = 20°C ⎫ u 2 = u g @ 20°C = 241.02 kJ/kg ⎬ sat. vapor ⎭ s 2 = s e = s g @ 20°C = 0.92234 kJ/kg ⋅ K he = h g @ 20°C = 261.59 kJ/kg Analysis The volume of the container is R-134a 1 kg 20°C sat. liq. me 3 V = m1v 1 = (1 kg )(0.0008161 m /kg ) = 0.0008161 m 3 The mass in the container at the final state is m2 = V 0.0008161 m 3 = = 0.02269 kg v 2 0.035969 m 3 /kg The amount of mass leaving the container is m e = m1 − m 2 = 1 − 0.02269 = 0.9773 kg The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 S gen . The entropy generation Sgen in this case is determined from an entropy balance on the system: S in − S out 1 24 4 3 Net entropy transfer by heat and mass + S gen = ΔS system { 123 4 4 Entropy generation Change in entropy − m e s e + S gen = ΔS tank = (m 2 s 2 − m1 s1 ) tank S gen = m 2 s 2 − m1 s1 + m e s e Substituting, X destroyed = T0 S gen = T0 (m 2 s 2 − m1 s1 + m e s e ) = (293 K)(0.02269 × 0.92234 − 1× 0.30063 + 0.9773 × 0.92234) = 182.2 kJ PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 59. 8-59 8-74 An adiabatic rigid tank that is initially evacuated is filled by air from a supply line. The work potential associated with this process is to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process. It can be analyzed as a uniform-flow process since the state of fluid entering the device remains constant. 2 Kinetic and potential energies are negligible. 3 There are no work interactions involved. 4 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are cp = 0.240 Btu/lbm⋅R and R = 0.06855 Btu/lbm⋅R = 0.3704 kPa⋅m3/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 and internal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as Mass balance: min − m out = Δmsystem → mi = m 2 Air 200 psia, 100°F Energy balance: E in − E out 1 24 4 3 Net energy transfer by heat, work, and mass = ΔE system 1 24 4 3 Change in internal, kinetic, potential, etc. energies m i hi = m 2 u 2 10 ft3 Combining the two balances: hi = u 2 ⎯ ⎯→ c p Ti = cv T2 ⎯ T2 = ⎯→ cp cv Ti = kTi Substituting, T2 = kTi = (1.4)(560 R ) = 784 R The final mass in the tank is m 2 = mi = (200 psia )(10 ft 3 ) PV = = 6.887 lbm RT2 (0.3704 psia ⋅ ft 3 /lbm ⋅ R )(784 R ) The work potential associated with this process is equal to the exergy destroyed during the process. The exergy destruction during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 S gen . The entropy generation Sgen in this case is determined from an entropy balance on the system: S in − S out 1 24 4 3 Net entropy transfer by heat and mass + S gen = ΔS system { 123 4 4 Entropy generation Change in entropy mi s i + S gen = ΔS tank = m 2 s 2 S gen = m 2 s 2 − mi s i S gen = m 2 ( s 2 − s i ) Substituting, ⎛ T ⎞ W rev = X destyroyed = m 2 T0 ( s 2 − s i ) = m 2 T0 ⎜ c p ln 2 ⎟ ⎜ T1 ⎟ ⎝ ⎠ 784 R ⎤ ⎡ = (6.887 lbm)(540 R) ⎢(0.240 Btu/lbm ⋅ R)ln 560 R ⎥ ⎣ ⎦ = 300.3 Btu PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 60. 8-60 8-75 An rigid tank that is initially evacuated is filled by air from a supply line. The work potential associated with this process is to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process. It can be analyzed as a uniform-flow process since the state of fluid entering the device remains constant. 2 Kinetic and potential energies are negligible. 3 There are no work interactions involved. 4 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are cp = 0.240 Btu/lbm⋅R and R = 0.06855 Btu/lbm⋅R = 0.3704 kPa⋅m3/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 and internal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as Mass balance: min − m out = Δmsystem → mi = m 2 Air Energy balance: E in − E out 1 24 4 3 Net energy transfer by heat, work, and mass = 200 psia, 100°F ΔE system 1 24 4 3 Change in internal, kinetic, potential, etc. energies mi hi − Qout = m 2 u 2 10 ft3 Qout = mi hi − m 2 u 2 Combining the two balances: Qout = m 2 (hi − u 2 ) The final mass in the tank is (200 psia )(10 ft 3 ) PV m 2 = mi = = = 9.642 lbm RT2 (0.3704 psia ⋅ ft 3 /lbm ⋅ R )(560 R ) Substituting, Qout = m 2 ( hi − u 2 ) = m 2 (c p Ti − c v Ti ) = m 2 Ti (c p − c v ) = m 2 Ti R = (9.642 lbm)(560 R)(0.06855 Btu/lbm ⋅ R) = 370.1 Btu The work potential associated with this process is equal to the exergy destroyed during the process. The exergy destruction during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 S gen . The entropy generation Sgen in this case is determined from an entropy balance on the system: S in − S out + S gen = ΔS system 1 24 4 3 { 123 4 4 Net entropy transfer by heat and mass mi s i − Entropy generation Change in entropy Qout + S gen = ΔS tank = m 2 s 2 T0 S gen = m 2 s 2 − mi s i + Qout T0 Qout T0 Noting that both the temperature and pressure in the tank is same as those in the supply line at the final state, substituting gives, ⎡ Q ⎤ W rev = X destroyed = T0 ⎢m 2 ( s 2 − s i ) + out ⎥ T0 ⎦ ⎣ S gen = m 2 ( s 2 − s i ) + ⎛ Q = T0 ⎜ 0 + out ⎜ T0 ⎝ ⎞ ⎛Q ⎟ = T0 ⎜ out ⎟ ⎜ T ⎠ ⎝ 0 ⎞ ⎟ = Qout = 370.1 Btu ⎟ ⎠ PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 61. 8-61 8-76 Steam expands in a turbine steadily at a specified rate from a specified state to another specified state. The power potential of the steam at the inlet conditions and the reversible power output are to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 The temperature of the surroundings is given to be 25°C. Properties From the steam tables (Tables A-4 through 6) 8 MPa 450°C P1 = 8 MPa ⎫ h1 = 3273.3 kJ/kg ⎬ T1 = 450°C ⎭ s1 = 6.5579 kJ/kg ⋅ K P2 = 50 kPa ⎫ h2 = 2645.2 kJ/kg ⎬ sat. vapor ⎭ s 2 = 7.5931 kJ/kg ⋅ K STEAM 15,000 kg/h P0 = 100 kPa ⎫ h0 ≅ h f @ 25°C = 104.83 kJ/kg ⎬ T0 = 25°C ⎭ s 0 ≅ s f @ 25°C = 0.36723 kJ/kg ⋅ K Analysis (a) The power potential of the steam at the inlet conditions is equivalent to its exergy at the inlet state, 50 kPa sat. vapor 0 ⎛ ⎞ 0 V2 & & & & Ψ = mψ 1 = m⎜ h1 − h0 − T0 ( s1 − s 0 ) + 1 + gz1 ⎟ = m(h1 − h0 − T0 ( s1 − s 0 ) ) ⎜ ⎟ 2 ⎜ ⎟ ⎝ ⎠ = (15,000 / 3600 kg/s)[(3273.3 − 104.83)kJ/kg − (298 K)(6.5579 - 0.36723)kJ/kg ⋅ K ] = 5515 kW (b) The power output of the turbine if there were no irreversibilities is the reversible power, is determined from the rate form of the exergy balance applied on the turbine and setting the exergy destruction term equal to zero, & & X in − X out 1 24 4 3 Rate of net exergy tra nsfer by heat, work,and mass & & − X destroyed 0 (reversible) = ΔX system 0 (steady) = 0 144 2444 4 3 1442443 Rate of exergy destructio n Rate of change of exergy & & X in = X out & & mψ = W 1 rev,out & + mψ 2 & & & Wrev,out = m(ψ1 −ψ 2 ) = m[(h1 − h2 ) − T0 (s1 − s2 ) − Δke 0 − Δpe 0 ] Substituting, & & W rev,out = m[(h1 − h2 ) − T0 ( s1 − s 2 )] = (15,000/3600 kg/s)[(3273.3 − 2645.2) kJ/kg − (298 K)(6.5579 − 7.5931) kJ/kg ⋅ K ] = 3902 kW PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 62. 8-62 8-77E Air is compressed steadily by a 400-hp compressor from a specified state to another specified state while being cooled by the ambient air. The mass flow rate of air and the part of input power that is used to just overcome the irreversibilities are to be determined. Assumptions 1 Air is an ideal gas with variable specific heats. 2 Potential energy changes are negligible. 3 The temperature of the surroundings is given to be 60°F. Properties The gas constant of air is R = 0.06855 350 ft/s Btu/lbm.R (Table A-1E). From the air table (Table A-17E) 150 psia 620°F T1 = 520 R ⎫ h1 = 124.27 Btu/lbm ⎬ o P1 = 15 psia ⎭ s1 = 0.59173 Btu/lbm ⋅ R 1500 Btu/min T2 = 1080 R ⎫ h2 = 260.97 Btu/lbm ⎬ o AIR P2 = 150 psia ⎭ s1 = 0.76964 Btu/lbm ⋅ R 400 hp & & & Analysis (a) There is only one inlet and one exit, and thus m1 = m2 = m . We take the actual compressor as the system, which is a control volume. The energy balance for this steady-flow system can be expressed as 15 psia 0 (steady) & −E & & 60°F E in = ΔE system =0 1 24 4 out 3 1442443 4 Rate of net energy transfer by heat, work, and mass Rate of change in internal, kinetic, potential, etc. energies & & E in = E out ⎛ V 2 − V12 & & & & & & & W a ,in + m(h1 + V12 / 2) = m(h2 + V 22 / 2) + Qout → W a ,in − Qout = m⎜ h2 − h1 + 2 ⎜ 2 ⎝ Substituting, the mass flow rate of the refrigerant becomes ⎞ ⎟ ⎟ ⎠ ⎛ ⎛ 0.7068 Btu/s ⎞ (350 ft/s) 2 − 0 1 Btu/lbm & ⎟ − (1500 / 60 Btu/s) = m⎜ 260.97 − 124.27 + (400 hp)⎜ ⎜ ⎟ ⎜ 1 hp 2 25,037 ft 2 / s 2 ⎝ ⎠ ⎝ ⎞ ⎟ ⎟ ⎠ & It yields m = 1.852 lbm / s (b) The portion of the power output that is used just to overcome the irreversibilities is equivalent to exergy destruction, which can be determined from an exergy balance or directly from its definition X destroyed = T0 S gen where the entropy generation Sgen is determined from an entropy balance on an extended system that includes the device and its immediate surroundings. It gives & & & & S in − S out + S gen = ΔS system 0 = 0 1 24 4 3 { 14 4 2 3 Rate of net entropy transfer by heat and mass & & ms1 − ms 2 − Rate of entropy generation Rate of change of entropy & & Qout Q & & & + S gen = 0 → S gen = m(s 2 − s1 ) + out Tb,surr T0 where P2 150 psia = (0.76964 − 0.59173) Btu/lbm − (0.06855 Btu/lbm.R) ln P1 15 psia = 0.02007 Btu/lbm.R Substituting, the exergy destruction is determined to be & ⎛ Q ⎞ & & & X destroyed = T0 S gen = T0 ⎜ m( s 2 − s1 ) + out ⎟ ⎜ T0 ⎟ ⎝ ⎠ 1 hp 1500 / 60 Btu/s ⎞⎛ ⎞ ⎛ = (520 R)⎜ (1.852 lbm/s)(0.02007 Btu/lbm ⋅ R) + ⎟ = 62.72 hp ⎟⎜ 520 R 0.7068 Btu/s ⎠ ⎠⎝ ⎝ 0 0 s 2 − s1 = s 2 − s1 − R ln PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 63. 8-63 8-78 Hot combustion gases are accelerated in an adiabatic nozzle. The exit velocity and the decrease in the exergy of the gases are to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Potential energy changes are negligible. 3 The device is adiabatic and thus heat transfer is negligible. 4 The combustion gases are ideal gases with constant specific heats. Properties The constant pressure specific heat and the specific heat ratio are given to be cp = 1.15 kJ/kg.K and k = 1.3. The gas constant R is determined from R = c p − cv = c p − c p / k = c p (1 − 1 / k ) = (1.15 kJ/kg ⋅ K)(1 − 1/1.3) = 0.2654 kJ/kg ⋅ K & & & Analysis (a) There is only one inlet and one exit, and thus m1 = m2 = m . We take the nozzle as the system, which is a control volume. The energy balance for this steady-flow system can be expressed as & & E in − E out 1 24 4 3 = Rate of net energy transfer by heat, work, and mass & ΔE system 0 (steady) 1442443 4 =0 Rate of change in internal, kinetic, potential, etc. energies & & E in = E out 2 & & & & m(h1 + V12 / 2) = m(h2 + V2 /2) (since W = Q ≅ Δpe ≅ 0) h2 = h1 − V 22 − V12 2 260 kPa 747°C 80 m/s Comb. gases 70 kPa 500°C Then the exit velocity becomes V 2 = 2c p (T1 − T2 ) + V12 ⎛ 1000 m 2 /s 2 = 2(1.15 kJ/kg ⋅ K)(747 − 500)K⎜ ⎜ 1 kJ/kg ⎝ = 758 m/s ⎞ ⎟ + (80 m/s) 2 ⎟ ⎠ (b) The decrease in exergy of combustion gases is simply the difference between the initial and final values of flow exergy, and is determined to be ψ 1 −ψ 2 = w rev = h1 − h2 − Δke − Δpe 0 + T0 ( s 2 − s1 ) = c p (T1 − T2 ) + T0 ( s 2 − s1 ) − Δke where Δke = V 22 − V12 (758 m/s) 2 − (80 m/s) 2 ⎛ 1 kJ/kg = ⎜ 2 2 ⎝ 1000 m 2 / s 2 ⎞ ⎟ = 284.1 kJ/kg ⎠ and s 2 − s1 = c p ln T2 P − R ln 2 T1 P1 = (1.15 kJ/kg ⋅ K) ln = 0.02938 kJ/kg ⋅ K 773 K 70 kPa − (0.2654 kJ/kg ⋅ K) ln 1020 K 260 kPa Substituting, Decrease in exergy = ψ 1 − ψ 2 = (1.15 kJ/kg ⋅ K)(747 − 500)°C + (293 K)(0.02938 kJ/kg ⋅ K) − 284.1 kJ/kg = 8.56 kJ/kg PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 64. 8-64 8-79 Steam is accelerated in an adiabatic nozzle. The exit velocity of the steam, the isentropic efficiency, and the exergy destroyed within the nozzle are to be determined. Assumptions 1 The nozzle operates steadily. 2 The changes in potential energies are negligible. Properties The properties of steam at the inlet and the exit of the nozzle are (Tables A-4 through A-6) P1 = 7 MPa ⎫ h1 = 3411.4 kJ/kg ⎬ T1 = 500°C ⎭ s1 = 6.8000 kJ/kg ⋅ K 7 MPa 500°C 70 m/s P2 = 5 MPa ⎫ h2 = 3317.2 kJ/kg ⎬ T2 = 450°C ⎭ s 2 = 6.8210 kJ/kg ⋅ K STEAM 5 MPa 450°C P2 s = 5 MPa ⎫ ⎬ h2 s = 3302.0 kJ/kg ⎭ s 2 s = s1 Analysis (a) We take the nozzle 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 & & E in − E out 1 24 4 3 & ΔE system 0 (steady) 1442443 4 = Rate of net energy transfer by heat, work, and mass =0 Rate of change in internal, kinetic, potential, etc. energies & & E in = E out 2 & & & & m(h1 + V12 / 2) = m(h2 + V2 /2) (since W = Q ≅ Δpe ≅ 0) 0 = h2 − h1 + V 22 − V12 2 Then the exit velocity becomes ⎛ 1000 m 2 /s 2 V 2 = 2(h1 − h2 ) + V12 = 2(3411.4 − 3317.2) kJ/kg⎜ ⎜ 1 kJ/kg ⎝ ⎞ ⎟ + (70 m/s) 2 = 439.6 m/s ⎟ ⎠ (b) The exit velocity for the isentropic case is determined from ⎛ 1000 m 2 /s 2 V 2 s = 2(h1 − h2 s ) + V12 = 2(3411.4 − 3302.0) kJ/kg⎜ ⎜ 1 kJ/kg ⎝ ⎞ ⎟ + (70 m/s) 2 = 472.9 m/s ⎟ ⎠ Thus, ηN = V 22 / 2 V 22s / 2 = (439.6 m/s) 2 / 2 (472.9 m/s) 2 / 2 = 86.4% (c) The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0S gen where the entropy generation Sgen is determined from an entropy balance on the actual nozzle. It gives & & S in − S out 1 24 4 3 + Rate of net entropy transfer by heat and mass & S gen { Rate of entropy generation & = ΔS system 0 = 0 14 4 2 3 Rate of change of entropy & & & & & ms1 − ms 2 + S gen = 0 → S gen = m(s 2 − s1 ) or s gen = s 2 − s1 Substituting, the exergy destruction in the nozzle on a unit mass basis is determined to be x destroyed = T0 s gen = T0 ( s 2 − s1 ) = (298 K)(6.8210 − 6.8000)kJ/kg ⋅ K = 6.28 kJ/kg PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 65. 8-65 8-80 CO2 gas is compressed steadily by a compressor from a specified state to another specified state. The power input to the compressor if the process involved no irreversibilities is to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 The device is adiabatic and thus heat transfer is negligible. 4 CO2 is an ideal gas with constant specific heats. Properties At the average temperature of (300 + 450)/2 = 375 K, the constant pressure specific heat and the specific heat ratio of CO2 are k = 1.261 and cp = 0.917 kJ/kg.K (Table A-2). 600 kPa 450 K Analysis The reversible (or minimum) power input is determined from the exergy balance applied on the compressor, and setting the exergy destruction term equal to zero, & & X in − X out 1 24 4 3 0 (reversibl e) Rate of net exergy tra nsfer by heat, work,and mass CO2 0.2 kg/s 0 (steady) & & − X destroyed = ΔX system =0 144 2444 4 3 1442443 Rate of exergy destructio n Rate of change of exergy 100 kPa 300 K & & X in = X out & & & mψ1 + Wrev,in = mψ 2 & & Wrev,in = m(ψ 2 −ψ1) & = m[(h2 − h1) − T0 (s2 − s1) + Δke 0 + Δpe 0 ] where s 2 − s1 = c p ln T2 P − R ln 2 T1 P1 = (0.9175 kJ/kg ⋅ K) ln = 0.03335 kJ/kg ⋅ K 450 K 600 kPa − (0.1889 kJ/kg ⋅ K) ln 300 K 100 kPa Substituting, & W rev,in = (0.2 kg/s)[(0.917 kJ/kg ⋅ K)(450 − 300)K − (298 K)(0.03335 kJ/kg ⋅ K)] = 25.5 kW Discussion Note that a minimum of 25.5 kW of power input is needed for this compressor. PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 66. 8-66 8-81 Liquid water is heated in a chamber by mixing it with superheated steam. For a specified mixing temperature, the mass flow rate of the steam and the rate of exergy destruction are to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 There are no work interactions. Properties Noting that T < Tsat @ 200 kPa = 120.23°C, the cold water and the exit mixture streams exist as a compressed liquid, which can be approximated as a saturated liquid at the given temperature. From Tables A-4 through A-6, 600 kJ/min P1 = 200 kPa ⎫ h1 ≅ h f @ 20o C = 83.91 kJ/kg ⎬s ≅ s = 0.29649 kJ/kg ⋅ K T1 = 20°C f @ 20o C ⎭ 1 20°C 1 P2 = 200 kPa ⎫ h2 = 3072.1 kJ/kg MIXING 2.5 kg/s ⎬ CHAMBER T2 = 300°C ⎭ s 2 = 7.8941 kJ/kg ⋅ K 60°C 3 200 kPa P3 = 200 kPa ⎫ h3 ≅ h f @60o C = 251.18 kJ/kg 2 300°C ⎬s ≅ s = 0.83130 kJ/kg ⋅ K o T3 = 60°C 3 f @ 60 C ⎭ Analysis (a) We take the mixing chamber as the system, which is a control volume. The mass and energy balances for this steady-flow system can be expressed in the rate form as & & & min − m out = Δmsystem Mass balance: 0 (steady) & & & =0 ⎯ ⎯→ m1 + m 2 = m3 Energy balance: & & E in − E out 1 24 4 3 Rate of net energy transfer by heat, work, and mass = & ΔE system 0 (steady) 1442443 4 =0 Rate of change in internal, kinetic, potential, etc. energies & & E in = E out & & & & mh1 + m 2 h2 = Qout + m3 h3 Combining the two relations gives & & & & & & & Qout = m1 h1 + m 2 h2 − (m1 + m 2 )h3 = m1 (h1 − h3 ) + m 2 (h2 − h3 ) & Solving for m2 and substituting, the mass flow rate of the superheated steam is determined to be & m2 = Also, & & Qout − m1 (h1 − h3 ) (600/60 kJ/s) − (2.5 kg/s )(83.91 − 251.18)kJ/kg = = 0.148 kg/s (3072.1 − 251.18)kJ/kg h2 − h3 & & & m3 = m1 + m 2 = 2.5 + 0.148 = 2.648 kg/s (b) The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 S gen where the entropy generation Sgen is determined from an entropy balance on an extended system that includes the mixing chamber and its immediate surroundings. It gives & & & & S in − S out + S gen = ΔS system 0 = 0 1 24 4 3 { 14 4 2 3 Rate of net entropy transfer by heat and mass & & & m1 s1 + m 2 s 2 − m3 s 3 − Rate of entropy generation Rate of change of entropy & & Qout Q & & & & & + S gen = 0 → S gen = m3 s 3 − m1 s1 − m 2 s 2 + out Tb,surr T0 Substituting, the exergy destruction is determined to be & ⎛ ⎞ Q & & & & & X destroyed = T0 S gen = T0 ⎜ m3 s 3 − m 2 s 2 − m1 s1 + out ⎟ ⎜ Tb, surr ⎟ ⎝ ⎠ = (298 K)(2.648 × 0.83130 − 0.148 × 7.8941 − 2.5 × 0.29649 + 10 / 298)kW/K = 96.4 kW PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 67. 8-67 8-82 A rigid tank initially contains saturated R-134a vapor. The tank is connected to a supply line, and R134a is allowed to enter the tank. The mass of the R-134a that entered the tank and the exergy destroyed during this process are to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process since the state of fluid at the inlet remains constant. 2 Kinetic and potential energies are negligible. 3 There are no work interactions involved. 4 The direction of heat transfer is to the tank (will be verified). Properties The properties of refrigerant are (Tables A-11 through A-13) v 1 = v g @ 1.2 MPa = 0.01672 m 3 / kg P1 = 1.2 MPa ⎫ ⎬ u1 = u g @ 1.2 MPa = 253.81 kJ/kg sat. vapor ⎭s =s 1 g @ 1.2 MPa = 0.91303 kJ/kg ⋅ K R-134a v 2 = v f @ 1.4 MPa = 0.0009166 m 3 / kg T2 = 1.4 MPa ⎫ ⎬ u 2 = u f @ 1.4 MPa = 125.94 kJ/kg sat. liquid ⎭s =s 2 f @ 1.4 MPa = 0.45315 kJ/kg ⋅ K Pi = 1.6 MPa ⎫ hi = 93.56 kJ/kg ⎬ Ti = 30°C ⎭ s i = 0.34554 kJ/kg ⋅ K 1.6 MPa 30°C R-134a 0.1 m3 1.2 MPa Sat. vapor Q Analysis We take the tank as the system, which is a control volume. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as Mass balance: min − m out = Δmsystem → mi = m 2 − m1 Energy balance: E in − E out 1 24 4 3 = Net energy transfer by heat, work, and mass ΔE system 1 24 4 3 Change in internal, kinetic, potential, etc. energies Qin + mi hi = m 2 u 2 − m1u1 (since W ≅ ke ≅ pe ≅ 0) (a) The initial and the final masses in the tank are m1 = V1 0.1 m 3 = = 5.983 kg v 1 0.01672 m 3 /kg m2 = V2 0.1 m 3 = = 109.10 kg v 2 0.0009166 m 3 /kg Then from the mass balance mi = m 2 − m1 = 109.10 − 5.983 = 103.11 kg The heat transfer during this process is determined from the energy balance to be Qin = −mi hi + m 2 u 2 − m1u1 = −(103.11 kg )(93.56 kJ/kg) + (109.10)(125.94 kJ/kg ) − (5.983 kg )(253.81 kJ/kg ) = 2573 kJ PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 68. 8-68 (b) The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 S gen . The entropy generation Sgen in this case is determined from an entropy balance on an extended system that includes the tank and its immediate surroundings so that the boundary temperature of the extended system is the surroundings temperature Tsurr at all times. It gives S in − S out 1 24 4 3 Net entropy transfer by heat and mass + S gen = ΔS system { 123 4 4 Entropy generation Change in entropy Qin + mi s i + S gen = ΔS tank = (m 2 s 2 − m1 s1 ) tank Substituting, the Tb,in S gen = m 2 s 2 − m1 s1 − mi s i − Qin T0 exergy destruction is determined to be ⎡ Q ⎤ X destroyed = T0 S gen = T0 ⎢m 2 s 2 − m1 s1 − mi s i − in ⎥ T0 ⎦ ⎣ = (318 K)[109.10 × 0.45315 − 5.983 × 0.91303 − 103.11× 0.34554 − (2573 kJ)/(318 K)] = 80.3 kJ PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 69. 8-69 8-83 A rigid tank initially contains saturated liquid water. A valve at the bottom of the tank is opened, and half of mass in liquid form is withdrawn from the tank. The temperature in the tank is maintained constant. The amount of heat transfer, the reversible work, and the exergy destruction during this process are to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process since the state of fluid leaving the device remains constant. 2 Kinetic and potential energies are negligible. 3 There are no work interactions involved. 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) 3 v 1 = v f @170o C = 0.001114 m /kg T1 = 170°C ⎫ ⎬ u1 = u f @170o C = 718.20 kJ/kg sat. liquid ⎭ s1 = s f @170o C = 2.0417 kJ/kg ⋅ K Te = 170°C ⎫ he = h f @170o C = 719.08 kJ/kg ⎬ sat. liquid ⎭ s e = s f @170o C = 2.0417 kJ/kg ⋅ K H2O 0.6 m3 170°C T = const. Q me 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 and internal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as Mass balance: m in − m out = Δm system → m e = m1 − m 2 Energy balance: E in − E out 1 24 4 3 = Net energy transfer by heat, work, and mass ΔE system 1 24 4 3 Change in internal, kinetic, potential, etc. energies Qin = m e he + m 2 u 2 − m1u1 (since W ≅ ke ≅ pe ≅ 0) The initial and the final masses in the tank are m1 = 0.6 m 3 V = = 538.47 kg v 1 0.001114 m 3 /kg m2 = 1 1 m1 = (538.47 kg ) = 269.24 kg = m e 2 2 Now we determine the final internal energy and entropy, v2 = x2 = V m2 = 0.6 m 3 = 0.002229 m 3 /kg 269.24 kg v 2 −v f v fg = 0.002229 − 0.001114 = 0.004614 0.24260 − 0.001114 ⎫ u 2 = u f + x 2 u fg = 718.20 + (0.004614)(1857.5) = 726.77 kJ/kg ⎬ x 2 = 0.004614 ⎭ s 2 = s f + x 2 s fg = 2.0417 + (0.004614 )(4.6233) = 2.0630 kJ/kg ⋅ K T2 = 170°C The heat transfer during this process is determined by substituting these values into the energy balance equation, PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 70. 8-70 Qin = m e he + m 2 u 2 − m1u1 = (269.24 kg )(719.08 kJ/kg ) + (269.24 kg )(726.77 kJ/kg ) − (538.47 kg )(718.20 kJ/kg ) = 2545 kJ (b) The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 S gen . The entropy generation Sgen in this case is determined from an entropy balance on an extended system that includes the tank and the region between the tank and the source so that the boundary temperature of the extended system at the location of heat transfer is the source temperature Tsource at all times. It gives S in − S out 1 24 4 3 Net entropy transfer by heat and mass + S gen = ΔS system { 123 4 4 Entropy generation Change in entropy Qin − m e s e + S gen = ΔS tank = (m 2 s 2 − m1 s1 ) tank Tb,in S gen = m 2 s 2 − m1 s1 + m e s e − Qin Tsource Substituting, the exergy destruction is determined to be ⎡ Q ⎤ X destroyed = T0 S gen = T0 ⎢m 2 s 2 − m1 s1 + m e s e − in ⎥ Tsource ⎦ ⎣ = (298 K)[269.24 × 2.0630 − 538.47 × 2.0417 + 269.24 × 2.0417 − (2545 kJ)/(523 K)] = 141.2 kJ For processes that involve no actual work, the reversible work output and exergy destruction are identical. Therefore, X destroyed = W rev,out − Wact,out → W rev,out = X destroyed = 141.2 kJ PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 71. 8-71 8-84E An insulated rigid tank equipped with an electric heater initially contains pressurized air. A valve is opened, and air is allowed to escape at constant temperature until the pressure inside drops to 30 psia. The amount of electrical work done and the exergy destroys are to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process since the exit temperature (and enthalpy) of air remains constant. 2 Kinetic and potential energies are negligible. 3 The tank is insulated and thus heat transfer is negligible. 4 Air is an ideal gas with variable specific heats. 5 The environment temperature is given to be 70°F. 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) ⎯→ he = 143.47 Btu/lbm Te = 600 R ⎯ ⎯→ T1 = 600 R ⎯ u1 = 102.34 Btu/lbm ⎯→ u 2 = 102.34 Btu/lbm T2 = 600 R ⎯ Analysis We take the tank as the system, which is a control volume. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as Mass balance: min − m out = Δmsystem → m e = m1 − m 2 Energy balance: E in − E out 1 24 4 3 Net energy transfer by heat, work, and mass = ΔE system 1 24 4 3 Change in internal, kinetic, potential, etc. energies AIR 150 ft3 75 psia 140°F We We,in − m e he = m 2 u 2 − m1u1 (since Q ≅ ke ≅ pe ≅ 0) The initial and the final masses of air in the tank are m1 = P1V (75 psia)(150 ft 3 ) = = 50.62 lbm RT1 (0.3704 psia ⋅ ft 3 /lbm ⋅ R)(600 R ) m2 = P2V (30 psia)(150 ft 3 ) = = 20.25 lbm RT2 (0.3704 psia ⋅ ft 3 /lbm ⋅ R)(600 R ) Then from the mass and energy balances, me = m1 − m2 = 50.62 − 20.25 = 30.37 lbm We,in = m e he + m 2 u 2 − m1u1 = (30.37 lbm)(143.47 Btu/lbm) + (20.25 lbm)(102.34 Btu/lbm) − (50.62 lbm)(102.34 Btu/lbm) = 1249 Btu PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 72. 8-72 (b) The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 S gen where the entropy generation Sgen is determined from an entropy balance on the insulated tank. It gives S in − S out 1 24 4 3 Net entropy transfer by heat and mass + S gen = ΔS system { 123 4 4 Entropy generation Change in entropy − m e s e + S gen = ΔS tank = (m 2 s 2 − m1 s1 ) tank S gen = m 2 s 2 − m1 s1 + m e s e = m 2 s 2 − m1 s1 + (m1 − m 2 ) s e = m 2 ( s 2 − s e ) − m1 ( s1 − s e ) Assuming a constant average pressure of (75 + 30) / 2 = 52.5 psia for the exit stream, the entropy changes are determined to be 0 T P 30 psia s 2 − s e = c p ln 2 − R ln 2 = −(0.06855 Btu/lbm ⋅ R) ln = 0.03836 Btu/lbm ⋅ R Te 52.5 psia Pe 0 s1 − s e = c p ln T1 P 75 psia − R ln 1 = −(0.06855 Btu/lbm ⋅ R) ln = −0.02445 Btu/lbm ⋅ R Te 52.5 psia Pe Substituting, the exergy destruction is determined to be X destroyed = T0 S gen = T0 [m 2 ( s 2 − s e ) − m1 ( s1 − s e )] = (530 R)[(20.25 lbm)(0.03836 Btu/lbm ⋅ R) − (50.62 lbm)(−0.02445 Btu/lbm ⋅ R)] = 1068 Btu PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 73. 8-73 8-85 A cylinder initially contains helium gas at a specified pressure and temperature. A valve is opened, and helium is allowed to escape until its volume decreases by half. The work potential of the helium at the initial state and the exergy destroyed during the process are to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process by using constant average properties for the helium leaving the tank. 2 Kinetic and potential energies are negligible. 3 There are no work interactions involved other than boundary work. 4 The tank is insulated and thus heat transfer is negligible. 5 Helium is an ideal gas with constant specific heats. Properties The gas constant of helium is R = 2.0769 kPa.m3/kg.K = 2.0769 kJ/kg.K. The specific heats of helium are cp = 5.1926 kJ/kg.K and cv = 3.1156 kJ/kg.K (Table A-2). Analysis (a) From the ideal gas relation, the initial and the final masses in the cylinder are determined to be m1 = P1V (300 kPa)(0.1 m 3 ) = = 0.0493 kg RT1 (2.0769 kPa ⋅ m 3 /kg ⋅ K)(293 K ) m e = m 2 = m1 / 2 = 0.0493 / 2 = 0.0247 kg The work potential of helium at the initial state is simply the initial exergy of helium, and is determined from the closed-system exergy relation, Φ 1 = m1φ = m1 [(u1 − u 0 ) − T0 ( s1 − s 0 ) + P0 (v 1 − v 0 )] where v1 = RT1 (2.0769 kPa ⋅ m 3 /kg ⋅ K)(293 K ) = = 2.0284 m 3 /kg P1 300 kPa 3 RT (2.0769 kPa ⋅ m /kg ⋅ K)(293 K ) v0 = 0 = = 6.405 m 3 /kg P0 95 kPa HELIUM 300 kPa 0.1 m3 20°C Q and s1 − s 0 = c p ln T1 P − R ln 1 T0 P0 = (5.1926 kJ/kg ⋅ K) ln = −2.28 kJ/kg ⋅ K 293 K 300 kPa − (2.0769 kJ/kg ⋅ K) ln 293 K 100 kPa Thus, Φ 1 = (0.0493 kg){(3.1156 kJ/kg ⋅ K)(20 − 20)°C − (293 K)(−2.28 kJ/kg ⋅ K) + (95 kPa)(2.0284 − 6.405)m 3 /kg[kJ/kPa ⋅ m 3 ]} = 12.44 kJ (b) We take the cylinder as the system, which is a control volume. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as Mass balance: min − m out = Δmsystem → m e = m1 − m 2 Energy balance: PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 74. 8-74 E in − E out 1 24 4 3 = Net energy transfer by heat, work, and mass ΔE system 1 24 4 3 Change in internal, kinetic, potential, etc. energies Qin − m e he + W b,in = m 2 u 2 − m1u1 Combining the two relations gives Qin = (m1 − m 2 )he + m 2 u 2 − m1u1 − W b,in = (m1 − m 2 )he + m 2 h2 − m1 h1 = (m1 − m 2 + m 2 − m1 )h1 =0 since the boundary work and ΔU combine into ΔH for constant pressure expansion and compression processes. The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 S gen where the entropy generation Sgen can be determined from an entropy balance on the cylinder. Noting that the pressure and temperature of helium in the cylinder are maintained constant during this process and heat transfer is zero, it gives S in − S out 1 24 4 3 Net entropy transfer by heat and mass + S gen = ΔS system { 123 4 4 Entropy generation Change in entropy − m e s e + S gen = ΔS cylinder = (m 2 s 2 − m1 s1 ) cylinder S gen = m 2 s 2 − m1 s1 + m e s e = m 2 s 2 − m1 s1 + (m1 − m 2 ) s e = (m 2 − m1 + m1 − m 2 ) s1 =0 since the initial, final, and the exit states are identical and thus se = s2 = s1. Therefore, this discharge process is reversible, and X destroyed = T0 Sgen = 0 PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 75. 8-75 8-86 A rigid tank initially contains saturated R-134a vapor at a specified pressure. The tank is connected to a supply line, and R-134a is allowed to enter the tank. The amount of heat transfer with the surroundings and the exergy destruction are to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process since the state of fluid at the inlet remains constant. 2 Kinetic and potential energies are negligible. 3 There are no work interactions involved. 4 The direction of heat transfer is from the tank (will be verified). Properties The properties of refrigerant are (Tables A-11 through A-13) 1.4 MPa 60°C R-134a u1 = u g @ 1 MPa = 250.68 kJ/kg P1 = 1 MPa ⎫ ⎬ s1 = s g @ 1 MPa = 0.91558 kJ/kg ⋅ K sat.vapor ⎭ v 1 = v g @ 1 MPa = 0.020313 m 3 / kg R-134a 0.2 m3 1 MPa Sat. vapor Pi = 1.4 MPa ⎫ hi = 285.47 kJ/kg ⎬ Ti = 60°C ⎭ s i = 0.93889 kJ/kg ⋅ K Q 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 and internal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as Mass balance: m in − m out = Δm system → m i = m 2 − m1 Energy balance: E in − E out 1 24 4 3 Net energy transfer by heat, work, and mass = ΔE system 1 24 4 3 Change in internal, kinetic, potential, etc. energies mi hi − Qout = m 2 u 2 − m1u1 (since W ≅ ke ≅ pe ≅ 0) The initial and the final masses in the tank are m1 = 0.2 m 3 V = = 9.846 kg v 1 0.020313 m 3 / kg m2 = m f + m g = Vf vf + Vg vg = 0.1 m 3 0.0008934 m 3 / kg + 0.1 m 3 0.016715 m 3 / kg = 111.93 + 5.983 = 117.91 kg U 2 = m2 u 2 = m f u f + m g u g = 111.93 × 116.70 + 5.983 × 253.81 = 14,581 kJ S 2 = m 2 s 2 = m f s f + m g s g = 111.93 × 0.42441 + 5.983 × 0.91303 = 52.967 kJ/K Then from the mass and energy balances, mi = m 2 − m1 = 117.91 − 9.846 = 108.06 kg The heat transfer during this process is determined from the energy balance to be Qout = mi hi − m 2 u 2 + m1u1 = 108.06 × 285.47 − 14,581 + 9.846 × 250.68 = 18,737 kJ PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 76. 8-76 (b) The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 S gen . The entropy generation Sgen in this case is determined from an entropy balance on an extended system that includes the cylinder and its immediate surroundings so that the boundary temperature of the extended system is the surroundings temperature Tsurr at all times. It gives S in − S out 1 24 4 3 Net entropy transfer by heat and mass − + S gen = ΔS system { 123 4 4 Entropy generation Change in entropy Qout + mi s i + S gen = ΔS tank = (m 2 s 2 − m1 s1 ) tank Tb,out S gen = m 2 s 2 − m1 s1 − mi s i + Qout T0 Substituting, the exergy destruction is determined to be ⎡ Q ⎤ X destroyed = T0 S gen = T0 ⎢m 2 s 2 − m1 s1 − mi s i + out ⎥ T0 ⎦ ⎣ = (298 K)[52.967 − 9.846 × 0.91558 − 108.06 × 0.93889 + 18,737 / 298] = 1599 kJ PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 77. 8-77 8-87 An insulated cylinder initially contains saturated liquid-vapor mixture of water. The cylinder is connected to a supply line, and the steam is allowed to enter the cylinder until all the liquid is vaporized. The amount of steam that entered the cylinder and the exergy destroyed are to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process since the state of fluid at the inlet remains constant. 2 The expansion process is quasi-equilibrium. 3 Kinetic and potential energies are negligible. 4 The device is insulated and thus heat transfer is negligible. Properties The properties of steam are (Tables A-4 through A-6) P1 = 200 kPa ⎫ h1 = h f + x1 h fg = 504.71 + 0.6 × 2201.6 = 1825.6 kJ/kg ⎬ x1 = 9 / 15 = 0.6 ⎭ s1 = s f + x1 s fg = 1.5302 + 0.6 × 5.5968 = 4.8883 kJ/kg ⋅ K P2 = 200 kPa ⎫ h2 = h g @ 200 kPa = 2706.3 kJ/kg ⎬ sat.vapor ⎭ s 2 = s g @ 200 kPa = 7.1270 kJ/kg ⋅ K Pi = 1 MPa ⎫ hi = 3264.5 kJ/kg ⎬ Ti = 400°C ⎭ s i = 7.4670 kJ/kg ⋅ K Analysis (a) We take the cylinder as the system, which is a control volume. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this unsteady-flow system can be expressed as Mass balance: min − m out = Δmsystem → mi = m 2 − m1 Energy balance: E in − E out 1 24 4 3 = Net energy transfer by heat, work, and mass H2O 200 kPa P = const. 1 MPa 400°C ΔE system 1 24 4 3 Change in internal, kinetic, potential, etc. energies mi hi = W b,out + m 2 u 2 − m1u1 (since Q ≅ ke ≅ pe ≅ 0) Combining the two relations gives 0 = W b,out − (m 2 − m1 )hi + m 2 u 2 − m1u1 0 = −(m 2 − m1 )hi + m 2 h2 − m1 h1 or, since the boundary work and ΔU combine into ΔH for constant pressure expansion and compression processes. Solving for m2 and substituting, h − h1 (3264.5 − 1825.6)kJ/kg (15 kg) = 38.66 kg m2 = i m1 = (3264.5 − 2706.3)kJ/kg hi − h2 Thus, mi = m 2 − m1 = 38.66 − 15 = 23.66 kg (b) The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 S gen where the entropy generation Sgen is determined from an entropy balance on the insulated cylinder, S in − S out + S gen = ΔS system 1 24 4 3 { 123 4 4 Net entropy transfer by heat and mass Entropy generation Change in entropy mi s i + S gen = ΔS system = m 2 s 2 − m1 s1 S gen = m 2 s 2 − m1 s1 − mi s i Substituting, the exergy destruction is determined to be X destroyed = T0 S gen = T0 [m 2 s 2 − m1 s1 − mi s i ] = (298 K)(38.66 × 7.1270 − 15 × 4.8883 − 23.66 × 7.4670) = 7610 kJ PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 78. 8-78 8-88 Each member of a family of four takes a shower every day. The amount of exergy destroyed by this family per year is to be determined. Assumptions 1 Steady operating conditions exist. 2 The kinetic and potential energies are negligible. 3 & Heat losses from the pipes, mixing section are negligible and thus Q ≅ 0. 4 Showers operate at maximum flow conditions during the entire shower. 5 Each member of the household takes a shower every day. 6 Water is an incompressible substance with constant properties at room temperature. 7 The efficiency of the electric water heater is 100%. Properties The density and specific heat of water are at room temperature are ρ = 1 kg/L = 1000 kg/3 and c = 4.18 kJ/kg.°C (Table A-3). Analysis The mass flow rate of water at the shower head is & m = ρV& = (1 kg/L)(10 L/min) = 10 kg/min The mass balance for the mixing chamber can be expressed in the rate form as & & & min − m out = Δmsystem 0 (steady) & & & & & = 0 → min = m out → m1 + m 2 = m3 where the subscript 1 denotes the cold water stream, 2 the hot water stream, and 3 the mixture. The rate of entropy generation during this process can be determined by applying the rate form of the entropy balance on a system that includes the electric water heater and the mixing chamber (the Telbow). Noting that there is no entropy transfer associated with work transfer (electricity) and there is no heat transfer, the entropy balance for this steady-flow system can be expressed as & & S in − S out 1 24 4 3 Rate of net entropy transfer by heat and mass + & S gen { Rate of entropy generation & = ΔS system 0 (steady) 1442443 Rate of change of entropy & & & & m1 s1 + m 2 s 2 − m3 s 3 + S gen = 0 (since Q = 0 and work is entropy free) & & & & S gen = m3 s 3 − m1 s1 − m 2 s 2 & & & Noting from mass balance that m1 + m2 = m3 and s2 = s1 since hot water enters the system at the same temperature as the cold water, the rate of entropy generation is determined to be T & & & & & & S gen = m3 s 3 − (m1 + m 2 ) s1 = m3 ( s 3 − s1 ) = m3 c p ln 3 T1 = (10 kg/min)(4.18 kJ/kg.K) ln 42 + 273 = 3.746 kJ/min.K 15 + 273 Noting that 4 people take a 6-min shower every day, the amount of entropy generated per year is & S gen = ( S gen )Δt ( No. of people)(No. of days) = (3.746 kJ/min.K)(6 min/person ⋅ day)(4 persons)(365 days/year) = 32,815 kJ/K (per year) The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 S gen , X destroyed = T0 S gen = (298 K)(32,815 kJ/K ) = 9,779,000 kJ Discussion The value above represents the exergy destroyed within the water heater and the T-elbow in the absence of any heat losses. It does not include the exergy destroyed as the shower water at 42°C is discarded or cooled to the outdoor temperature. Also, an entropy balance on the mixing chamber alone (hot water entering at 55°C instead of 15°C) will exclude the exergy destroyed within the water heater. PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 79. 8-79 8-89 Air is compressed in a steady-flow device isentropically. The work done, the exit exergy of compressed air, and the exergy of compressed air after it is cooled to ambient temperature are to be determined. Assumptions 1 Air is an ideal gas with constant specific heats at room temperature. 2 The process is given to be reversible and adiabatic, and thus isentropic. Therefore, isentropic relations of ideal gases apply. 3 The environment temperature and pressure are given to be 300 K and 100 kPa. 4 The kinetic and potential energies are negligible. Properties The gas constant of air is R = 0.287 kJ/kg.K (Table A-1). The constant pressure specific heat and specific heat ratio of air at room temperature are cp = 1.005 kJ/kg.K and k = 1.4 (Table A-2). Analysis (a) From the constant specific heats ideal gas isentropic relations, ⎛P T2 = T1 ⎜ 2 ⎜P ⎝ 1 ⎞ ⎟ ⎟ ⎠ ( k −1) / k ⎛ 1000 kPa ⎞ = (300 K )⎜ ⎜ 100 kPa ⎟ ⎟ ⎝ ⎠ 0.4 / 1.4 1 MPa s2 = s1 = 579.2 K For a steady-flow isentropic compression process, the work input is determined from wcomp,in { AIR } kRT1 (P2 P1 )(k −1) / k − 1 = k −1 (1.4)(0.287kJ/kg ⋅ K )(300K ) (1000/100) 0.4/1.4 − 1 = 1.4 − 1 = 280.5 kJ/kg { } 100 kPa 300 K (b) The exergy of air at the compressor exit is simply the flow exergy at the exit state, ψ 2 = h 2 − h 0 − T0 ( s 2 − s 0 ) 0 V2 + 2 2 0 0 + gz 2 (since the proccess 0 - 2 is isentropic) = c p (T2 − T0 ) = (1.005 kJ/kg.K)(579.2 - 300)K = 280.6 kJ/kg which is the same as the compressor work input. This is not surprising since the compression process is reversible. (c) The exergy of compressed air at 1 MPa after it is cooled to 300 K is again the flow exergy at that state, V2 ψ 3 = h3 − h0 − T0 ( s 3 − s 0 ) + 3 2 = c p (T3 − T0 ) 0 0 + gz 3 0 − T0 ( s 3 − s 0 ) (since T3 = T0 = 300 K) = −T0 ( s 3 − s 0 ) where T s 3 − s 0 = c p ln 3 T0 0 − R ln P3 P 1000 kPa = − R ln 3 = −(0.287 kJ/kg ⋅ K)ln = −0.661 kJ/kg.K P0 P0 100 kPa Substituting, ψ 3 = −(300 K)(−0.661 kJ / kg.K) = 198 kJ / kg Note that the exergy of compressed air decreases from 280.6 to 198 as it is cooled to ambient temperature. PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 80. 8-80 8-90 Air is preheated by hot exhaust gases in a cross-flow heat exchanger. The rate of heat transfer and the rate of exergy destruction in the heat exchanger are to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Fluid properties are constant. Properties The specific heats of air and combustion gases are given to be 1.005 and 1.10 kJ/kg.°C, respectively. The gas constant of air is R = 0.287 kJ/kg.K (Table A-1). Analysis We take the exhaust pipes as the system, which is a control volume. The energy balance for this steady-flow system can be expressed in the rate form as Air & & & 95 kPa E in − E out = ΔE system 0 (steady) =0 1 24 4 3 1442443 4 20°C Rate of net energy transfer Rate of change in internal, kinetic, by heat, work, and mass potential, etc. energies 0.8 m3/s & & E =E in out & & & mh1 = Qout + mh2 (since Δke ≅ Δpe ≅ 0) & & Qout = mC p (T1 − T2 ) Exhaust gases 1.1 kg/s, 95°C Then the rate of heat transfer from the exhaust gases becomes & & Q = [mc (T − T )] = (1.1 kg/s)(1.1 kJ/kg.°C)(180°C − 95°C) = 102.85 kW p in out gas. The mass flow rate of air is (95 kPa)(0.8 m 3 /s) PV& & m= = = 0.904 kg/s RT (0.287 kPa.m 3 /kg.K) × 293 K Noting that heat loss by exhaust gases is equal to the heat gain by the air, the air exit temperature becomes & Q 102.85 kW & & Q = mC p (Tout − Tin ) air → Tout = Tin + = 20°C + = 133.2°C & (0.904 kg/s)(1.005 kJ/kg.°C) mc p [ ] The rate of entropy generation within the heat exchanger is determined by applying the rate form of the entropy balance on the entire heat exchanger: & & S in − S out 1 24 4 3 Rate of net entropy transfer by heat and mass + & S gen { Rate of entropy generation & = ΔS system 0 (steady) 1442443 Rate of change of entropy & & & & & m1 s1 + m 3 s 3 − m 2 s 2 − m 3 s 4 + S gen = 0 (since Q = 0) & & & & & m exhaust s1 + m air s 3 − m exhaust s 2 − m air s 4 + S gen = 0 & & & S gen = m exhaust ( s 2 − s1 ) + m air ( s 4 − s 3 ) Noting that both fluid streams are liquids (incompressible substances), the rate of entropy generation is T T & & & S gen = m exhaust c p ln 2 + m air c p ln 4 T3 T1 = (1.1 kg/s)(1.1 kJ/kg.K)ln 133.2 + 273 95 + 273 + (0.904 kg/s)(1.005 kJ/kg.K)ln = 0.0453 kW/K 20 + 273 180 + 273 The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 S gen , & & X destroyed = T0 S gen = (293 K)(0.0453 kW/K ) = 13.3 kW PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 81. 8-81 8-91 Water is heated by hot oil in a heat exchanger. The outlet temperature of the oil and the rate of exergy destruction within the heat exchanger are to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Fluid properties 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, which is a control volume. The energy balance for this steady-flow system can be expressed in the rate form as & & E in − E out 1 24 4 3 = Rate of net energy transfer by heat, work, and mass & ΔE system 0 (steady) 1442443 4 =0 Oil 170°C 10 kg/s 70°C Rate of change in internal, kinetic, potential, etc. energies & & E in = E out Water 20°C 4.5 kg/s & & & Qin + mh1 = mh2 (since Δke ≅ Δpe ≅ 0) & = mc (T − T ) Qin & p 2 1 (12 tube passes) Then the rate of heat transfer to the cold water in this heat 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 water is 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) (b) The rate of entropy generation within the heat exchanger is determined by applying the rate form of the entropy balance on the entire heat exchanger: & & S in − S out 1 24 4 3 Rate of net entropy transfer by heat and mass + & S gen { Rate of entropy generation & = ΔS system 0 (steady) 1442443 Rate of change of entropy & & & & & m1 s1 + m 3 s 3 − m 2 s 2 − m 3 s 4 + S gen = 0 (since Q = 0) & & & & & m water s1 + m oil s 3 − m water s 2 − m oil s 4 + S gen = 0 & & & S gen = m water ( s 2 − s1 ) + m oil ( s 4 − s 3 ) Noting that both fluid streams are liquids (incompressible substances), the rate of entropy generation is determined to be T T & & & S gen = m water c p ln 2 + m oil c p ln 4 T1 T3 = (4.5 kg/s)(4.18 kJ/kg.K) ln 70 + 273 129.1 + 273 + (10 kg/s)(2.3 kJ/kg.K) ln = 0.736 kW/K 20 + 273 170 + 273 The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 S gen , & & X destroyed = T0 S gen = (298 K)(0.736 kW/K ) = 219 kW PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 82. 8-82 8-92E Steam is condensed by cooling water in a condenser. The rate of heat transfer and the rate of exergy destruction within the heat exchanger are to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Fluid properties are constant. 5 The temperature of the environment is 77°F. Properties The specific heat of water is 1.0 Btu/lbm.°F (Table A-3E). The enthalpy and entropy of vaporization of water at 120°F are 1025.2 Btu/lbm and sfg = 1.7686 Btu/lbm.R (Table A-4E). Analysis We take the tube-side of the heat exchanger where cold water is flowing as the system, which is a control volume. The energy balance for this steady-flow system can be expressed in the rate form as & & E in − E out 1 24 4 3 & ΔE system 0 (steady) 1442443 4 = Rate of net energy transfer by heat, work, and mass =0 Steam 120°F Rate of change in internal, kinetic, potential, etc. energies & & E in = E out 73°F & & & Qin + mh1 = mh2 (since Δke ≅ Δpe ≅ 0) & & Qin = mc p (T2 − T1 ) Then the rate of heat transfer to the cold water in this heat exchanger becomes & & Q = [mc (T − T )] p out in 60°F water Water = (115.3 lbm/s)(1.0 Btu/lbm.°F)(73°F − 60°F) = 1499 Btu/s Noting that heat gain by the water is equal to the heat loss by the condensing steam, the rate of condensation of the steam in the heat exchanger is determined from & Q 1499 Btu/s & & & Q = (mh fg ) steam = ⎯ ⎯→ msteam = = = 1.462 lbm/s h fg 1025.2 Btu/lbm 120°F (b) The rate of entropy generation within the heat exchanger is determined by applying the rate form of the entropy balance on the entire heat exchanger: & & S in − S out 1 24 4 3 + Rate of net entropy transfer by heat and mass & S gen { Rate of entropy generation & = ΔS system 0 (steady) 1442443 Rate of change of entropy & & & & & m1 s1 + m 3 s 3 − m 2 s 2 − m 4 s 4 + S gen = 0 (since Q = 0) & & & & & m water s1 + m steam s 3 − m water s 2 − msteam s 4 + S gen = 0 & & & S gen = m water ( s 2 − s1 ) + msteam ( s 4 − s 3 ) Noting that water is an incompressible substance and steam changes from saturated vapor to saturated liquid, the rate of entropy generation is determined to be T T & & & & & S gen = m water c p ln 2 + msteam ( s f − s g ) = m water c p ln 2 − msteam s fg T1 T1 = (115.3 lbm/s)(1.0 Btu/lbm.R)ln 73 + 460 − (1.462 lbm/s)(1.7686 Btu/lbm.R) = 0.2613 Btu/s.R 60 + 460 The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 S gen , & & X destroyed = T0 S gen = (537 R)(0.2613 Btu/s.R ) = 140.3 Btu/s PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 83. 8-83 8-93 Steam expands in a turbine, which is not insulated. The reversible power, the exergy destroyed, the second-law efficiency, and the possible increase in the turbine power if the turbine is well insulated are to be determined. Assumptions 1 Steady operating conditions exist. 2 Potential energy change is negligible. Analysis (a) The properties of the steam at the inlet and exit of the turbine are (Tables A-4 through A-6) P1 = 12 MPa ⎫ h1 = 3481.7 kJ/kg ⎬ T1 = 550°C ⎭ s1 = 6.6554 kJ/kg.K P2 = 20 kPa ⎫ h2 = 2491.1 kJ/kg ⎬ x 2 = 0.95 ⎭ s 2 = 7.5535 kJ/kg.K The enthalpy at the dead state is T0 = 25°C ⎫ ⎬h0 = 104.83 kJ/kg x=0 ⎭ The mass flow rate of steam may be determined from an energy balance on the turbine ⎛ V2 & m⎜ h1 + 1 ⎜ 2 ⎝ Steam 12 MPa 550°C, 60 m/s Turbine Q 20 kPa 130 m/s x = 0.95 2 ⎞ ⎞ ⎛ & & ⎟ = m⎜ h2 + V 2 ⎟ + Qout + Wa & ⎟ ⎜ 2 ⎟ ⎠ ⎝ ⎠ ⎡ ⎡ (60 m/s) 2 ⎛ 1 kJ/kg ⎞⎤ (130 m/s) 2 ⎛ 1 kJ/kg & & m ⎢3481.7 kJ/kg + ⎜ ⎟⎥ = m ⎢2491.1 kJ/kg + ⎜ 2 2 ⎝ 1000 m 2 /s 2 ⎠⎥ ⎝ 1000 m 2 /s 2 ⎢ ⎢ ⎣ ⎦ ⎣ ⎞⎤ ⎟⎥ ⎠⎥ ⎦ & + 150 kW + 2500 kW ⎯ ⎯→ m = 2.693 kg/s The reversible power may be determined from ⎡ V 2 - V 22 ⎤ & & W rev = m ⎢h1 − h2 − T0 ( s1 − s 2 ) + 1 ⎥ 2 ⎥ ⎢ ⎣ ⎦ ⎡ (60 m/s) 2 − (130 m/s) 2 ⎛ 1 kJ/kg = (2.693) ⎢(3481.7 − 2491.1) − (298)(6.6554 - 7.5535) + ⎜ 2 ⎝ 1000 m 2 /s 2 ⎢ ⎣ ⎞⎤ ⎟⎥ ⎠⎥ ⎦ = 3371 kW (b) The exergy destroyed in the turbine is & & & X = W − W = 3371 − 2500 = 871 kW dest rev a (c) The second-law efficiency is & W 2500 kW η II = a = = 0.742 & Wrev 3371 kW (d) The energy of the steam at the turbine inlet in the given dead state is & & Q = m(h − h ) = (2.693 kg/s)(3481.7 - 104.83)kJ/kg = 9095 kW 1 0 The fraction of energy at the turbine inlet that is converted to power is & W 2500 kW f = a = = 0.2749 & Q 9095 kW Assuming that the same fraction of heat loss from the turbine could have been converted to work, the possible increase in the power if the turbine is to be well-insulated becomes & & W = fQ = (0.2749)(150 kW) = 41.2 kW increase out PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 84. 8-84 8-94 Air is compressed in a compressor that is intentionally cooled. The actual and reversible power inputs, the second law efficiency, and the mass flow rate of cooling water are to be determined. Assumptions 1 Steady operating conditions exist. 2 Potential energy change is negligible. 3 Air is an ideal gas with constant specific heats. 900 kPa 60°C 80 m/s Properties The gas constant of air is R = 0.287 kJ/kg.K and the specific heat of air at room is cp = 1.005 kJ/kg.K. the specific heat of water at room temperature is cw = 4.18 kJ/kg.K (Tables A-2, A-3). Analysis (a) The mass flow rate of air is Compressor P (100 kPa) & (4.5 m 3 /s) = 5.351 kg/s m = ρV&1 = 1 V&1 = (0.287 kJ/kg.K)(20 + 273 K) RT1 The power input for a reversible-isothermal process is given by Q Air 100 kPa 20°C P ⎛ 900 kPa ⎞ & & Wrev = mRT1 ln 2 = (5.351 kg/s)(0.287 kJ/kg.K)(20 + 273 K)ln⎜ ⎟ = 988.8 kW P1 ⎝ 100 kPa ⎠ Given the isothermal efficiency, the actual power may be determined from & W 988.8 kW & Wactual = rev = = 1413 kW ηT 0.70 (b) The given isothermal efficiency is actually the second-law efficiency of the compressor η II = η T = 0.70 (c) An energy balance on the compressor gives ⎡ V 2 − V 22 ⎤ & & & Qout = m ⎢C p (T1 − T2 ) + 1 ⎥ + Wactual,in 2 ⎢ ⎥ ⎣ ⎦ ⎡ 0 − (80 m/s) 2 ⎛ 1 kJ/kg = (5.351 kg/s) ⎢(1.005 kJ/kg.°C)(20 − 60)°C + ⎜ 2 ⎝ 1000 m 2 /s 2 ⎢ ⎣ = 1181 kW ⎞⎤ ⎟⎥ + 1413 kW ⎠⎥ ⎦ The mass flow rate of the cooling water is & mw = & Qout 1181 kW = = 28.25 kg/s c w ΔT (4.18 kJ/kg.°C)(10°C) PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 85. 8-85 8-95 Water is heated in a chamber by mixing it with saturated steam. The temperature of the steam entering the chamber, the exergy destruction, and the second-law efficiency are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3 Heat loss from the chamber is negligible. Analysis (a) The properties of water are (Tables A-4 through A-6) T1 = 15°C⎫ h1 = h0 = 62.98 kJ/kg ⎬ x1 = 0 ⎭ s1 = s 0 = 0.22447 kJ/kg.K T3 = 45°C⎫ h3 = 188.44 kJ/kg ⎬ x1 = 0 ⎭ s 3 = 0.63862 kJ/kg.K Water 15°C 4.6 kg/s Mixing chamber Mixture 45°C Sat. vap. 0.23 kg/s An energy balance on the chamber gives & & & & & m1h1 + m2 h2 = m3h3 = (m1 + m2 )h3 (4.6 kg/s)(62.98 kJ/kg) + (0.23 kg/s) h2 = ( 4.6 + 0.23 kg/s )(188.44 kJ/kg) h2 = 2697.5 kJ/kg The remaining properties of the saturated steam are h2 = 2697.5 kJ/kg ⎫ T2 = 114.3°C ⎬ x2 = 1 ⎭ s 2 = 7.1907 kJ/kg.K (b) The specific exergy of each stream is ψ1 = 0 ψ 2 = h2 − h0 − T0 ( s2 − s0 ) = (2697.5 − 62.98)kJ/kg − (15 + 273 K)(7.1907 − 0.22447)kJ/kg.K = 628.28 kJ/kg ψ 3 = h3 − h0 − T0 ( s3 − s0 ) = (188.44 − 62.98)kJ/kg − (15 + 273 K)(0.63862 − 0.22447)kJ/kg.K = 6.18 kJ/kg The exergy destruction is determined from an exergy balance on the chamber to be & & & & & X dest = m1ψ 1 + m 2ψ 2 − ( m1 + m 2 )ψ 3 = 0 + (0.23 kg/s)(628.28 kJ/kg) − (4.6 + 0.23 kg/s)(6.18 kJ/kg) = 114.7 kW (c) The second-law efficiency for this mixing process may be determined from ηII = & & (m1 + m2 )ψ 3 (4.6 + 0.23 kg/s)(6.18 kJ/kg) = = 0.207 & & m1ψ 1 + m2ψ 2 0 + (0.23 kg/s)(628.28 kJ/kg) PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 86. 8-86 Review Problems 8-96 Refrigerant-134a is expanded adiabatically in an expansion valve. The work potential of R-134a at the inlet, the exergy destruction, and the second-law efficiency are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) The properties of the refrigerant at the inlet and exit of the valve and at dead state are (Tables A-11 through A-13) P1 = 1.2 MPa ⎫ h1 = 108.23 kJ/kg ⎬ T1 = 40°C ⎭ s1 = 0.39424 kJ/kg.K P2 = 180 kPa ⎫ ⎬s 2 = 0.42271 kJ/kg.K h2 = h1 = 108.23 kJ/kg ⎭ R-134a 1.2 MPa 40°C 180 kPa P0 = 100 kPa ⎫ h0 = 272.17 kJ/kg ⎬ T0 = 20°C ⎭ s 0 = 1.0918 kJ/kg.K The specific exergy of the refrigerant at the inlet and exit of the valve are ψ 1 = h1 − h0 − T0 ( s1 − s0 ) = (108.23 − 272.17)kJ/kg - (20 + 273.15 K)(0.39424 - 1.0918)kJ/kg.K = 40.55 kJ/kg ψ 2 = h2 − h0 − T0 ( s2 − s0 ) = (108.23 − 272.17)kJ/kg − (20 + 273.15 K)(0.42271 − 1.0918 kJ/kg.K = 32.20 kJ/kg (b) The exergy destruction is determined to be xdest = T0 ( s2 − s1 ) = (20 + 273.15 K)(0.42271 - 0.39424)kJ/kg.K = 8.34 kJ/kg (c) The second-law efficiency for this process may be determined from η II = ψ 2 32.20 kJ/kg = = 0.794 ψ 1 40.55 kJ/kg PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 87. 8-87 8-97 Steam is accelerated in an adiabatic nozzle. The exit velocity, the rate of exergy destruction, and the second-law efficiency are to be determined. Assumptions 1 Steady operating conditions exist. 2 Potential energy changes are negligible. Analysis (a) The properties of the steam at the inlet and exit of the turbine and at the dead state are (Tables A-4 through A-6) P1 = 3.5 MPa ⎫ h1 = 2978.4 kJ/kg ⎬ T1 = 300°C ⎭ s1 = 6.4484 kJ/kg.K P2 = 1.6 kPa ⎫ h2 = 2919.9 kJ/kg ⎬ T2 = 250°C ⎭ s 2 = 6.6753 kJ/kg.K T0 = 18°C⎫ h0 = 75.54 kJ/kg ⎬ x=0 ⎭ s 0 = 0.2678 kJ/kg.K 1.6 MPa 250°C V2 Steam 3.5 MPa 300°C The exit velocity is determined from an energy balance on the nozzle h1 + 2978.4 kJ/kg + V12 V2 = h2 + 2 2 2 V 2 ⎛ 1 kJ/kg ⎞ (0 m/s) 2 ⎛ 1 kJ/kg ⎞ ⎜ ⎟ = 2919.9 kJ/kg + 2 ⎜ ⎟ 2 2 ⎝ 1000 m 2 /s 2 ⎠ ⎝ 1000 m 2 /s 2 ⎠ V 2 = 342.0 m/s (b) The rate of exergy destruction is the exergy decrease of the steam in the nozzle ⎡ ⎤ V 2 − V12 & & − T0 ( s 2 − s1 ⎥ X dest = m ⎢h2 − h1 + 2 2 ⎢ ⎥ ⎣ ⎦ ⎡ (342 m/s) 2 − 0 ⎛ 1 kJ/kg ⎜ ⎢(2919.9 − 2978.4)kJ/kg + = (0.4 kg/s) ⎢ 2 ⎝ 1000 m 2 /s 2 ⎢− (291 K )(6.6753 − 6.4484)kJ/kg.K ⎣ = 26.41 kW ⎞⎤ ⎟⎥ ⎠⎥ ⎥ ⎦ (c) The exergy of the refrigerant at the inlet is ⎡ ⎤ V2 & & X 1 = m ⎢h1 − h0 + 1 − T0 ( s1 − s0 ⎥ 2 ⎢ ⎥ ⎣ ⎦ = (0.4 kg/s)[(2978.4 − 75.54) kJ/kg + 0 − (291 K )(6.4484 − 0.2678)kJ/kg.K ] = 441.72 kW The second-law efficiency for this device may be defined as the exergy output divided by the exergy input: η II = & & X X2 26.41 kW = 1 − dest = 1 − = 0.940 & & 441.72 kW X1 X1 PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 88. 8-88 8-98 R-134a is expanded in an adiabatic process with an isentropic efficiency of 0.85. The second law efficiency is to be determined. Assumptions 1 Kinetic and potential energy changes are negligible. 2 The device is adiabatic and thus heat transfer is negligible. Analysis We take the R-134a as the system. This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed as E in − E out 1 24 4 3 Net energy transfer by heat, work, and mass = ΔE system 1 24 4 3 Change in internal, kinetic, potential, etc. energies − Wout = ΔU = m(u 2 − u1 ) T From the R-134a tables (Tables A-11 through A-13), 1 1.6 MPa 3 v = 0.014362 m /kg P1 = 1600 kPa ⎫ 1 ⎬ u1 = 282.09 kJ/kg T1 = 80°C ⎭ s = 0.9875 kJ/kg ⋅ K 1 100 kPa 2s 2 s P2 = 100 kPa ⎫ ⎬ u 2 s = 223.16 kJ/kg s 2 s = s1 ⎭ The actual work input is wa ,out = η T w s ,out = η T (u1 − u 2 s ) = (0.85)(282.09 − 223.16)kJ/kg = 50.09 kJ/kg The actual internal energy at the end of the expansion process is wa ,out = (u1 − u 2 ) ⎯ u 2 = u1 − wa ,out = 282.09 − 50.09 = 232.00 kJ/kg ⎯→ Other actual properties at the final state are (Table A-13) P2 = 100 kPa ⎫ v 2 = 0.2139 m 3 /lbm ⎬ u 2 = 232.00 kJ/kg ⎭ s 2 = 1.0251 kJ/kg ⋅ K The useful work is determined from wu = wa ,out − wsurr = wa ,out − P0 (v 2 − v 1 ) ⎛ 1 kJ ⎞ = 50.09 kJ/kg − (100 kPa )(0.2139 − 0.014362) m 3 /kg⎜ ⎟ ⎝ 1 kPa ⋅ m 3 ⎠ = 30.14 kJ/kg The exergy change between initial and final states is φ1 − φ 2 = u1 − u 2 + P0 (v 1 − v 2 ) − T0 ( s1 − s 2 ) ⎛ 1 kJ ⎞ = (282.09 − 232.00)kJ/kg + (100 kPa )(0.014362 − 0.2139) m 3 /kg⎜ ⎟ ⎝ 1 kPa ⋅ m 3 ⎠ − (298 K )(0.9875 − 1.0251)kJ/kg ⋅ K = 41.34 kJ/kg The second law efficiency is then η II = wu 30.14 kJ/kg = = 0.729 Δφ 41.34 kJ/kg PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 89. 8-89 8-99 Steam is condensed in a closed system at a constant pressure from a saturated vapor to a saturated liquid by rejecting heat to a thermal energy reservoir. The second law efficiency is to be determined. Assumptions 1 Kinetic and potential energy changes are negligible. Analysis We take the steam as the system. This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed as E in − E out 1 24 4 3 = Net energy transfer by heat, work, and mass ΔE system 1 24 4 3 Change in internal, kinetic, potential, etc. energies Steam 40 kPa Sat. vapor Wb,in − Qout = ΔU = m(u 2 − u1 ) From the steam tables (Table A-5), v 1 = v g = 3.9933 m 3 /kg P1 = 40 kPa ⎫ ⎬ u1 = u g = 2476.3 kJ/kg Sat. vapor ⎭ s1 = s g = 7.6691 kJ/kg ⋅ K q T v 2 = v f = 0.001026 m 3 /kg P2 = 40 kPa ⎫ ⎬ u 2 = u f = 317.58 kJ/kg Sat. liquid ⎭ s 2 = s f = 1.0261 kJ/kg ⋅ K 2 40 kPa 1 s The boundary work during this process is ⎛ 1 kJ ⎞ wb,in = P(v 1 − v 2 ) = (40 kPa )(3.9933 − 0.001026) m 3 /kg⎜ ⎟ = 159.7 kJ/kg ⎝ 1 kPa ⋅ m 3 ⎠ The heat transfer is determined from the energy balance: q out = wb,in − (u 2 − u1 ) = 159.7 kJ/kg − (317.58 − 2476.3)kJ/kg = 2318.4 kJ/kg The exergy change between initial and final states is ⎛ φ1 − φ 2 = u1 − u 2 + P0 (v 1 − v 2 ) − T0 ( s1 − s 2 ) − q out ⎜1 − ⎜ ⎝ T0 TR ⎞ ⎟ ⎟ ⎠ ⎛ 1 kJ ⎞ = (2476.3 − 317.58)kJ/kg + (100 kPa )(3.9933 − 0.001026) m 3 /kg⎜ ⎟ ⎝ 1 kPa ⋅ m 3 ⎠ ⎛ 298 K ⎞ − (298 K )(7.6691 − 1.0261)kJ/kg ⋅ K − (2318.4 kJ/kg)⎜1 − ⎟ ⎝ 303 K ⎠ = 540.1 kJ/kg The second law efficiency is then η II = wb,in Δφ = 159.7 kJ/kg = 0.296 540.1 kJ/kg PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 90. 8-90 8-100 R-134a is vaporized in a closed system at a constant pressure from a saturated liquid to a saturated vapor by transferring heat from a reservoir at two pressures. The pressure that is more effective from a second-law point of view is to be determined. Assumptions 1 Kinetic and potential energy changes are negligible. Analysis We take the R-134a as the system. This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed as E in − E out 1 24 4 3 Net energy transfer by heat, work, and mass = ΔE system 1 24 4 3 Change in internal, kinetic, potential, etc. energies R-134a 100 kPa sat. liquid Qin − Wb,out = ΔU = m(u 2 − u1 ) Qin = Wb,out + ΔU Qin = ΔH = m(h2 − h1 ) q At 100 kPa: T From the R-134a tables (Table A-12), u fg @ 100 kPa = 197.98 kJ/kg h fg @ 100 kPa = 217.16 kJ/kg 1 100 kPa s fg @ 100 kPa = 0.87995 kJ/kg ⋅ K v fg @ 100 kPa = v g − v f = 0.19254 − 0.0007259 = 0.19181 m 3 /kg 2 s The boundary work during this process is ⎛ 1 kJ ⎞ wb,out = P(v 2 − v 1 ) = Pv fg = (100 kPa )(0.19181) m 3 /kg⎜ ⎟ = 19.18 kJ/kg ⎝ 1 kPa ⋅ m 3 ⎠ The useful work is determined from wu = wb,out − wsurr = P(v 2 − v 1 ) − P0 (v 2 − v 1 ) = 0 kJ/kg since P = P0 = 100 kPa. The heat transfer from the energy balance is q in = h fg = 217.16 kJ/kg The exergy change between initial and final states is ⎛ φ1 − φ 2 = u1 − u 2 + P0 (v 1 − v 2 ) − T0 ( s1 − s 2 ) + q in ⎜1 − ⎜ ⎝ ⎛ T = −u fg − P0v fg + T0 s fg + q in ⎜1 − 0 ⎜ T R ⎝ T0 TR ⎞ ⎟ ⎟ ⎠ ⎞ ⎟ ⎟ ⎠ ⎛ 1 kJ ⎞ = −197.98 kJ/kg − (100 kPa )(0.19181 m 3 /kg)⎜ ⎟ + (298 K )(0.87995 kJ/kg ⋅ K) ⎝ 1 kPa ⋅ m 3 ⎠ ⎛ 298 K ⎞ + (217.16 kJ/kg)⎜1 − ⎟ ⎝ 273 K ⎠ = 25.18 kJ/kg The second law efficiency is then η II = wu 0 kJ/kg = =0 Δφ 25.18 kJ/kg PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 91. 8-91 At 200 kPa: u fg @ 200 kPa = 186.21 kJ/kg h fg @ 200 kPa = 206.03 kJ/kg s fg @ 200 kPa = 0.78316 kJ/kg ⋅ K v fg @ 200 kPa = v g − v f = 0.099867 − 0.0007533 = 0.099114 m 3 /kg ⎛ 1 kJ ⎞ wb,out = P(v 2 − v 1 ) = Pv fg = (200 kPa )(0.099114) m 3 /kg⎜ ⎟ = 19.82 kJ/kg ⎝ 1 kPa ⋅ m 3 ⎠ wu = wb,out − wsurr = P (v 2 − v 1 ) − P0 (v 2 − v 1 ) ⎛ 1 kJ = ( P − P0 )v fg = (200 − 100) kPa (0.099114) m 3 /kg⎜ ⎝ 1 kPa ⋅ m 3 ⎞ ⎟ = 9.911 kJ/kg ⎠ q in = h fg = 206.03 kJ/kg ⎛ φ1 − φ 2 = u1 − u 2 + P0 (v 1 − v 2 ) − T0 ( s1 − s 2 ) + q in ⎜1 − ⎜ ⎝ ⎛ T = −u fg − P0v fg + T0 s fg + q in ⎜1 − 0 ⎜ T R ⎝ T0 TR ⎞ ⎟ ⎟ ⎠ ⎞ ⎟ ⎟ ⎠ ⎛ 1 kJ ⎞ = −186.21 kJ/kg − (100 kPa )(0.099114 m 3 /kg)⎜ ⎟ + (298 K )(0.78316 kJ/kg ⋅ K) ⎝ 1 kPa ⋅ m 3 ⎠ ⎛ 298 K ⎞ + (206.03 kJ/kg)⎜1 − ⎟ ⎝ 273 K ⎠ = 18.39 kJ/kg η II = wu 9.911 kJ/kg = = 0.539 Δφ 18.39 kJ/kg The process at 200 kPa is more effective from a work production standpoint. PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 92. 8-92 8-101E Steam is expanded in a two-stage turbine. Six percent of the inlet steam is bled for feedwater heating. The isentropic efficiencies for the two stages of the turbine are given. The second-law efficiency of the turbine is to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 The turbine is wellinsulated, and there is no heat transfer from the turbine. Analysis There is one inlet and two exits. 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 & & & = ΔE system 0 (steady) =0 E in − E out 500 psia 1 24 4 3 1442443 4 Rate of net energy transfer Rate of change in internal, kinetic, 600°F by heat, work, and mass potential, etc. energies & & E in = E out Turbine & & & & m1 h1 = m 2 h2 + m3 h3 + Wout & & & & Wout = m1 h1 − m 2 h2 − m3 h3 100 psia wout = h1 − 0.06h2 − 0.94h3 wout = (h1 − h2 ) + 0.94(h2 − h3 ) 5 psia T The isentropic and actual enthalpies at three states are determined using steam tables as follows: P1 = 500 psia ⎫ h1 = 1298.6 Btu/lbm ⎬ T1 = 600°F ⎭ s1 = 1.5590 Btu/lbm ⋅ R 1 500 psia 100 psia 5 psia 2 P2 = 100 psia ⎫ x 2 s = 0.9609 3s 3 ⎬ s s 2 s = s1 = 1.5590 Btu/lbm ⋅ R ⎭ h2 s = 1152.7 Btu/lbm h −h η T ,1 = 1 2 ⎯ ⎯→ h2 = h1 − η T ,1 (h1 − h2 s ) = 1298.6 − (0.97)(1298.6 − 1152.7) = 1157.1 kJ/kg h1 − h2 s P2 = 100 psia ⎫ x 2 = 0.9658 ⎬ h2 = 1157.1 Btu/lbm ⎭ s 2 = 1.5646 Btu/lbm ⋅ R P3 = 5 psia ⎫ x 3s = 0.8265 ⎬ s 3 = s 2 = 1.5646 kJ/kg ⋅ K ⎭ h3s = 957.09 Btu/lbm h −h ηT ,2 = 2 3 ⎯ h3 = h2 − ηT ,2 (h2 − h3s ) = 1157.1 − (0.95)(1157.1 − 957.09) = 967.09 kJ/kg ⎯→ h2 − h3s P3 = 5 psia ⎫ x 3 = 0.8364 ⎬ h3 = 967.09 Btu/lbm ⎭ s 3 = 1.5807 Btu/lbm ⋅ R Substituting into the energy balance per unit mass flow at the inlet of the turbine, we obtain wout = (h1 − h2 ) + 0.94(h2 − h3 ) = (1298.6 − 1157.1) + 0.94(1157.1 − 967.09) = 320.1 Btu/lbm The reversible work output per unit mass flow at the turbine inlet is wrev = h1 − h2 − T0 ( s1 − s 2 ) + 0.94[h2 − h3 − T0 ( s 2 − s 3 )] = 1298.6 − 1157.1 − (537)(1.5590 − 1.5646) + 0.94[(1157.1 − 967.09 − (537)(1.5646 − 1.5807)] = 331.2 Btu/lbm The second law efficiency is then w 320.1 Btu/lbm η II = out = = 0.966 wrev 331.2 Btu/lbm PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 93. 8-93 8-102 An electrical radiator is placed in a room and it is turned on for a period of time. The time period for which the heater was on, the exergy destruction, and the second-law efficiency are to be determined. Assumptions 1 Kinetic and potential energy changes are negligible. 2 Air is an ideal gas with constant specific heats. 3 The room is well-sealed. 4 Standard atmospheric pressure of 101.3 kPa is assumed. Properties The properties of air at room temperature are R = 0.287 kPa.m3/kg.K, cp = 1.005 kJ/kg.K, cv = 0.718 kJ/kg.K (Table A-2). The properties of oil are given to be ρ = 950 kg/m3, coil = 2.2 kJ/kg.K. Analysis (a) The masses of air and oil are ma = P1V (101.3 kPa)(50 m 3 ) = = 62.36 kg RT1 (0.287 kPa ⋅ m 3 /kg ⋅ K)(10 + 273 K) 3 10°C 3 m oil = ρ oilV oil = (950 kg/m )(0.030 m ) = 28.50 kg Room Q Radiator An energy balance on the system can be used to determine time period for which the heater was kept on & & (W − Q )Δt = [mc (T − T )] + [mc(T − T )] in v out 2 1 a 2 1 oil (1.8 − 0.35 kW)Δt = [(62.36 kg)(0.718 kJ/kg.°C)(20 − 10)°C] + [(28.50 kg)(2.2 kJ/kg.°C)(50 − 10)°C] Δt = 2038 s = 34 min (b) The pressure of the air at the final state is Pa 2 = m a RTa 2 V = (62.36 kg)(0.287 kPa ⋅ m 3 /kg ⋅ K)(20 + 273 K) 50 m 3 The amount of heat transfer to the surroundings is & Q = Q Δt = (0.35 kJ/s)(2038 s) = 713.5 kJ out = 104.9 kPa out The entropy generation is the sum of the entropy changes of air, oil, and the surroundings ⎡ T P ⎤ ΔS a = m ⎢c p ln 2 − R ln 2 ⎥ T1 P1 ⎦ ⎣ ⎡ (20 + 273) K 104.9 kPa ⎤ = (62.36 kg) ⎢(1.005 kJ/kg.K)ln − (0.287 kJ/kg.K)ln ⎥ (10 + 273) K 101.3 kPa ⎦ ⎣ = 1.5548 kJ/K T (50 + 273) K ΔS oil = mc ln 2 = (28.50 kg)(2.2 kJ/kg.K)ln = 8.2893 kJ/K T1 (10 + 273) K ΔS surr = Qout 713.5 kJ = = 2.521 kJ/K Tsurr (10 + 273) K S gen = ΔS a + ΔS oil + ΔS surr = 1.5548 + 8.2893 + 2.521 = 12.365 kJ/K The exergy destruction is determined from X dest = T0 S gen = (10 + 273 K)(12.365 kJ/K) = 3500 kJ (c) The second-law efficiency may be defined in this case as the ratio of the exergy recovered to the exergy input. That is, X a ,2 = m[cv (T2 − T1 )] − T0 ΔS a = (62.36 kg)[(0.718 kJ/kg.°C)(20 - 10)°C] − (10 + 273 K)(1.5548 kJ/K) = 7.729 kJ X oil, 2 = m[C (T2 − T1 )] − T0 ΔS a = (28.50 kg)[(2.2 kJ/kg.°C)(50 - 10)°C] − (10 + 273 K)(8.2893 kJ/K) = 162.13 kJ η II = X recovered X a ,2 + X oil,2 (7.729 + 162.13) kJ = = = 0.046 = 4.6% & (1.8 kJ/s)(2038 s) X supplied Win Δt PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 94. 8-94 8-103 Hot exhaust gases leaving an internal combustion engine is to be used to obtain saturated steam in an adiabatic heat exchanger. The rate at which the steam is obtained, the rate of exergy destruction, and the second-law efficiency are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3 Air properties are used for exhaust gases. 4 Pressure drops in the heat exchanger are negligible. Properties The gas constant of air is R = 0.287 kJkg.K. The specific heat of air at the average temperature of exhaust gases (650 K) is cp = 1.063 kJ/kg.K (Table A-2). Analysis (a) We denote the inlet and exit states of exhaust gases by (1) and (2) and that of the water by (3) and (4). The properties of water are (Table A-4) T3 = 20°C⎫ h3 = 83.91 kJ/kg ⎬ x3 = 0 ⎭ s 3 = 0.29649 kJ/kg.K T4 = 200°C⎫ h4 = 2792.0 kJ/kg ⎬ x4 = 1 ⎭ s 4 = 6.4302 kJ/kg.K An energy balance on the heat exchanger gives Exh. gas 400°C 150 kPa Heat Exchanger Sat. vap. 200°C 350°C Water 20°C & & & & ma h1 + mwh3 = ma h2 + mwh4 & & ma c p (T1 − T2 ) = mw (h4 − h3 ) & (0.8 kg/s)(1.063 kJ/kg°C)(400 − 350)°C = mw (2792.0 − 83.91)kJ/kg & w = 0.01570 kg/s m (b) The specific exergy changes of each stream as it flows in the heat exchanger is Δs a = c p ln T2 (350 + 273) K = (0.8 kg/s)(1.063 kJ/kg.K)ln = −0.08206 kJ/kg.K T1 (400 + 273) K Δψ a = c p (T2 − T1 ) − T0Δsa = (1.063 kJ/kg.°C)(350 - 400)°C − (20 + 273 K)(-0.08206 kJ/kg.K) = −29.106 kJ/kg Δψ w = h4 − h3 − T0 ( s4 − s3 ) = (2792.0 − 83.91)kJ/kg − (20 + 273 K)(6.4302 − 0.29649)kJ/kg.K = 910.913 kJ/kg The exergy destruction is determined from an exergy balance on the heat exchanger to be & & & − X dest = ma Δψ a + mwΔψ w = (0.8 kg/s)(-29.106 kJ/kg) + (0.01570 kg/s)(910.913) kJ/kg = −8.98 kW or & X dest = 8.98 kW (c) The second-law efficiency for a heat exchanger may be defined as the exergy increase of the cold fluid divided by the exergy decrease of the hot fluid. That is, η II = & mwΔψ w (0.01570 kg/s)(910.913 kJ/kg) = = 0.614 & − ma Δψ a − (0.8 kg/s)(-29.106 kJ/kg) PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 95. 8-95 8-104 The inner and outer surfaces of a window glass are maintained at specified temperatures. The amount of heat loss and the amount of exergy destruction in 5 h are to be determined Assumptions Steady operating conditions exist since the surface temperatures of the glass remain constant at the specified values. Analysis We take the glass to be the system, which is a closed system. The amount of heat loss is determined from & Q = QΔt = (4.4 kJ/s)(5 × 3600 s) = 79,200 k J Glass Under steady conditions, the rate form of the entropy balance for the glass simplifies to & & S in − S out 1 24 4 3 Rate of net entropy transfer by heat and mass + & S gen { Rate of entropy generation & = ΔS system 0 = 0 14 4 2 3 Rate of change of entropy & & Q Qin & − out + S gen,glass = 0 Tb,in Tb,out 10°C 3°C 4400 W 4400 W & & − + S gen, wall = 0 → S gen,glass = 0.3943 W/K 283 K 276 K Then the amount of entropy generation over a period of 5 h becomes & S gen,glass = S gen,glass Δt = (0.3943 W/K)(5 × 3600 s) = 7098 J/K The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 S gen , X destroyed = T0 S gen = (278 K)(7.098 kJ/K ) = 1973 kJ Discussion The total entropy generated during this process can be determined by applying the entropy balance on an extended system that includes the glass and its immediate surroundings on both sides so that the boundary temperature of the extended system is the room temperature on one side and the environment temperature on the other side at all times. Using this value of entropy generation will give the total exergy destroyed during the process, including the temperature gradient zones on both sides of the window. PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 96. 8-96 8-105 Heat is transferred steadily to boiling water in the pan through its bottom. The inner and outer surface temperatures of the bottom of the pan are given. The rate of exergy destruction within the bottom plate is to be determined. Assumptions Steady operating conditions exist since the surface temperatures of the pan remain constant at the specified values. Analysis We take the bottom of the pan to be the system, which is a closed system. Under steady conditions, the rate form of the entropy balance for this system can be expressed as & & S in − S out 1 24 4 3 + Rate of net entropy transfer by heat and mass & S gen { Rate of entropy generation & = ΔS system 0 = 0 14 4 2 3 Rate of change of entropy & & Q Qin & − out + S gen,system = 0 Tb,in Tb,out 800 W 800 W & − + S gen,system = 0 378 K 377 K & S gen,system = 0.00561 W/K 104°C 800 W 105°C The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 S gen , & & X destroyed = T0 S gen = (298 K)(0.00561 W/K ) = 1.67 W 8-106 A elevation, base area, and the depth of a crater lake are given. The maximum amount of electricity that can be generated by a hydroelectric power plant is to be determined. Assumptions The evaporation of water from the lake is negligible. Analysis The exergy or work potential of the water is the potential energy it possesses relative to the ground level, Exergy = PE = mgh Therefore, 12 m ∫ ∫ dz ∫ Exergy = PE = dPE = gz dm = gz ( ρAdz ) = ρAg ∫ z2 z1 2 2 zdz = ρAg ( z 2 − z1 ) / 2 = 0.5(1000 kg/m3 )(2 × 104 m 2 )(9.81 m/s 2 ) ( ) 140 m z ⎛ 1 h ⎞⎛ 1 kJ/kg ⎞ × (152 m)2 − (140 m 2 ) ⎜ ⎟⎜ ⎟ 2 2 ⎝ 3600 s ⎠⎝ 1000 m / s ⎠ = 9.55 × 104 kWh PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 97. 8-97 8-107E The 2nd-law efficiency of a refrigerator and the refrigeration rate are given. The power input to the refrigerator is to be determined. Analysis From the definition of the second law efficiency, the COP of the refrigerator is determined to be COPR, rev = η II 1 1 = = 12.375 T H / T L − 1 535 / 495 − 1 75°F COPR = ⎯ ⎯→ COPR = η II COPR, rev = 0.45 × 12.375 = 5.57 COPR, rev η II = 0.45 R Thus the power input is 200 Btu/min & QL 1 hp 200 Btu/min ⎛ ⎞ & Win = = ⎟ = 0.85 hp ⎜ COPR 5.57 ⎝ 42.41 Btu/min ⎠ 35°F 8-108 Writing energy and entropy balances, a relation for the reversible work is to be obtained for a closed system that exchanges heat with surroundings at T0 in the amount of Q0 as well as a heat reservoir at temperature TR in the amount QR. Assumptions Kinetic and potential changes are negligible. Analysis We take the direction of heat transfers to be to the system (heat input) and the direction of work transfer to be from the system (work output). The result obtained is still general since quantities wit opposite directions can be handled the same way by using negative signs. The energy and entropy balances for this stationary closed system can be expressed as ⎯→ W = U 1 − U 2 + Q0 + Q R (1) Energy balance: E in − E out = ΔE system → Q0 + Q R − W = U 2 − U 1 ⎯ Entropy balance: S in − S out + S gen = ΔS system → S gen = ( S 2 − S1 ) + −Q R −Q0 + TR T0 (2) Solving for Q0 from (2) and substituting in (1) yields ⎛ T W = (U 1 − U 2 ) − T0 ( S1 − S 2 ) − Q R ⎜1 − 0 ⎜ T R ⎝ ⎞ ⎟ − T0 S gen ⎟ ⎠ Source TR System The useful work relation for a closed system is obtained from QR Wu = W − Wsurr ⎛ T = (U 1 − U 2 ) − T0 ( S1 − S 2 ) − Q R ⎜1 − 0 ⎜ T R ⎝ ⎞ ⎟ − T0 S gen − P0 (V 2 −V1 ) ⎟ ⎠ Then the reversible work relation is obtained by substituting Sgen = 0, ⎛ T W rev = (U 1 − U 2 ) − T0 ( S1 − S 2 ) + P0 (V1 −V 2 ) − Q R ⎜1 − 0 ⎜ T R ⎝ ⎞ ⎟ ⎟ ⎠ A positive result for Wrev indicates work output, and a negative result work input. Also, the QR is a positive quantity for heat transfer to the system, and a negative quantity for heat transfer from the system. PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 98. 8-98 8-109 Writing energy and entropy balances, a relation for the reversible work is to be obtained for a & steady-flow system that exchanges heat with surroundings at T0 at a rate of Q0 as well as a heat reservoir & at temperature TR in the amount Q . R Analysis We take the direction of heat transfers to be to the system (heat input) and the direction of work transfer to be from the system (work output). The result obtained is still general since quantities wit opposite directions can be handled the same way by using negative signs. The energy and entropy balances for this stationary closed system can be expressed as & & & & & Energy balance: E in − E out = ΔE system → E in = E out V2 V2 & & & & & Q0 + Q R − W = ∑ m e (he + e + gz e ) − ∑ mi (hi + i + gz i ) 2 2 or V2 V2 & & & & & W = ∑ mi (hi + i + gz i ) − ∑ m e (he + e + gz e ) + Q0 + Q R (1) 2 2 System Entropy balance: & & & & S in − S out + S gen = ΔS system = 0 & & & S gen = S out − S in & − QR − Q0 & & & + Sgen = ∑ me se − ∑ mi si + TR T0 (2) & Solving for Q0 from (2) and substituting in (1) yields V2 V2 & & ⎛ T & & & W = ∑ mi (hi + i + gz i − T0 s i ) − ∑ m e (he + e + gz e − T0 s e ) − T0 S gen − Q R ⎜1 − 0 ⎜ T 2 2 R ⎝ ⎞ ⎟ ⎟ ⎠ Then the reversible work relation is obtained by substituting Sgen = 0, V2 V2 & ⎛ T & & & W rev = ∑ mi (hi + i + gz i − T0 s i ) − ∑ m e (he + e + gz e − T0 s e ) − Q R ⎜1 − 0 ⎜ T 2 2 R ⎝ ⎞ ⎟ ⎟ ⎠ A positive result for Wrev indicates work output, and a negative result work input. Also, the QR is a positive quantity for heat transfer to the system, and a negative quantity for heat transfer from the system. PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 99. 8-99 8-110 Writing energy and entropy balances, a relation for the reversible work is to be obtained for a uniform-flow system that exchanges heat with surroundings at T0 in the amount of Q0 as well as a heat reservoir at temperature TR in the amount QR. Assumptions Kinetic and potential changes are negligible. Analysis We take the direction of heat transfers to be to the system (heat input) and the direction of work transfer to be from the system (work output). The result obtained is still general since quantities wit opposite directions can be handled the same way by using negative signs. The energy and entropy balances for this stationary closed system can be expressed as Energy balance: E in − E out = ΔE system Q 0 + Q R − W = ∑ m e ( he + or, W = ∑ mi (hi + Ve2 V2 + gz e ) − ∑ mi (hi + i + gz i ) + (U 2 − U 1 ) cv 2 2 Vi 2 V2 + gz i ) − ∑ m e (he + e + gz e ) − (U 2 − U 1 ) cv + Q0 + Q R 2 2 (1) Entropy balance: Sin − Sout + Sgen = ΔSsystem Sgen = ( S2 − S1 )cv + ∑ me se − ∑ mi si + −QR −Q0 + TR T0 Source TR (2) System Solving for Q0 from (2) and substituting in (1) yields Vi 2 V2 + gz i − T0 s i ) − ∑ m e (he + e + gz e − T0 s e ) 2 2 ⎛ T ⎞ +[(U 1 − U 2 ) − T0 ( S1 − S 2 )]cv − T0 S gen − Q R ⎜1 − 0 ⎟ ⎜ T ⎟ R ⎠ ⎝ Q W = ∑ mi (hi + me The useful work relation for a closed system is obtained from Vi 2 V2 + gz i − T0 s i ) − ∑ m e (he + e + gz e − T0 s e ) 2 2 ⎛ T ⎞ +[(U 1 − U 2 ) − T0 ( S1 − S 2 )]cv − T0 S gen − Q R ⎜1 − 0 ⎟ − P0 (V 2 −V 1 ) ⎜ T ⎟ R ⎠ ⎝ Wu = W − Wsurr = ∑ mi (hi + Then the reversible work relation is obtained by substituting Sgen = 0, Vi 2 V2 + gz i − T0 s i ) − ∑ m e (he + e + gz e − T0 s e ) 2 2 ⎛ T ⎞ +[(U 1 − U 2 ) − T0 ( S1 − S 2 ) + P0 (V1 −V 2 )]cv − Q R ⎜1 − 0 ⎟ ⎜ T ⎟ R ⎠ ⎝ W rev = ∑ mi (hi + A positive result for Wrev indicates work output, and a negative result work input. Also, the QR is a positive quantity for heat transfer to the system, and a negative quantity for heat transfer from the system. PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 100. 8-100 8-111 An electric resistance heater is immersed in water. The time it will take for the electric heater to raise the water temperature to a specified temperature, the minimum work input, and the exergy destroyed during this process are to be determined. Assumptions 1 Water is an incompressible substance with constant specific heats. 2 The energy stored in the container itself and the heater is negligible. 3 Heat loss from the container is negligible. 4 The environment temperature is given to be T0 = 20°C. Properties The specific heat of water at room temperature is c = 4.18 kJ/kg·°C (Table A-3). Analysis Taking the water in the container as the system, which is a closed system, the energy balance can be expressed as E in − E out 1 24 4 3 = Net energy transfer by heat, work, and mass ΔE system 1 24 4 3 Change in internal, kinetic, potential, etc. energies We,in = (ΔU ) water Water 40 kg & We,in Δt = mc(T2 − T1 ) water Heater Substituting, (800 J/s)Δt = (40 kg)(4180 J/kg·°C)(80 - 20)°C Solving for Δt gives Δt = 12,544 s = 209.1 min = 3.484 h Again we take the water in the tank to be the system. Noting that no heat or mass crosses the boundaries of this system and the energy and entropy contents of the heater are negligible, the entropy balance for it can be expressed as S in − S out 1 24 4 3 Net entropy transfer by heat and mass + S gen = ΔS system { 123 4 4 Entropy generation Change in entropy 0 + S gen = ΔS water Therefore, the entropy generated during this process is S gen = ΔS water = mc ln T2 353 K = (40 kg )(4.184 kJ/kg ⋅ K ) ln = 31.18 kJ/K T1 293 K The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 S gen , X destroyed = T0 S gen = (293 K)(31.18 kJ/K ) = 9136 kJ The actual work input for this process is & Wact,in = Wact,in Δt = (0.8 kJ/s)(12,552 s) = 10,042 kJ Then the reversible (or minimum required )work input becomes W rev,in = Wact,in − X destroyed = 10,042 − 9136 = 906 kJ PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 101. 8-101 8-112 A hot water pipe at a specified temperature is losing heat to the surrounding air at a specified rate. The rate at which the work potential is wasted during this process is to be determined. Assumptions Steady operating conditions exist. Analysis We take the air in the vicinity of the pipe (excluding the pipe) as our system, which is a closed system.. The system extends from the outer surface of the pipe to a distance at which the temperature drops to the surroundings temperature. In steady operation, the rate form of the entropy balance for this system can be expressed as & & S in − S out 1 24 4 3 Rate of net entropy transfer by heat and mass + & S gen { Rate of entropy generation 80°C & = ΔS system 0 = 0 14 4 2 3 Rate of change of entropy & & Q Qin & − out + S gen,system = 0 Tb,in Tb,out D = 5 cm L = 10 m 1175 W 1175 W & & − + S gen,system = 0 → S gen,system = 0.8980 W/K 353 K 278 K Air, 5°C The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 S gen , & & X destroyed = T0 S gen = (278 K)(0.8980 W/K ) = 250 W PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 102. 8-102 8-113 Air expands in an adiabatic turbine from a specified state to another specified state. The second-law efficiency is to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 The device is adiabatic and thus heat transfer is negligible. 3 Air is an ideal gas with constant specific heats. 4 Kinetic and potential energy changes are negligible. Properties At the average temperature of (425 + 325)/2 = 375 K, the constant pressure specific heat of air is cp = 1.011 kJ/kg.K (Table A-2b). The gas constant of air is R = 0.287 kJ/kg.K (Table A-1). & & & 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 system can be expressed in the rate form as & & E in − E out 1 24 4 3 = Rate of net energy transfer by heat, work, and mass & ΔE system 0 (steady) 1442443 4 =0 Rate of change in internal, kinetic, potential, etc. energies & & E in = E out & & & mh1 = Wout + mh2 & & Wout = m(h1 − h2 ) 550 kPa 425 K Air wout = c p (T1 − T2 ) Substituting, wout = c p (T1 − T2 ) = (1.011 kJ/kg ⋅ K)(425 − 325)K = 101.1 kJ/kg 110 kPa 325 K The entropy change of air is s 2 − s1 = c p ln P T2 − R ln 2 P1 T1 = (1.011 kJ/kg ⋅ K) ln = 0.1907 kJ/kg ⋅ K 110 kPa 325 K − (0.287 kJ/kg ⋅ K) ln 550 kPa 425 K The maximum (reversible) work is the exergy difference between the inlet and exit states wrev,out = c p (T1 − T2 ) − T0 ( s1 − s 2 ) = wout − T0 ( s1 − s 2 ) = 101.1 kJ/kg − (298 K)(−0.1907 kJ/kg ⋅ K) = 157.9 kJ/kg The second law efficiency is then η II = wout 101.1 kJ/kg = = 0.640 wrev,out 157.9 kJ/kg PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 103. 8-103 8-114 Steam is accelerated in a nozzle. The actual and maximum outlet velocities are to be determined. Assumptions 1 The nozzle operates steadily. 2 The changes in potential energies are negligible. Properties The properties of steam at the inlet and the exit of the nozzle are (Tables A-4 through A-6) P1 = 500 kPa ⎫ h1 = 2855.8 kJ/kg ⎬ T1 = 200°C ⎭ s1 = 7.0610 kJ/kg ⋅ K P2 = 200 kPa ⎫ h2 = 2706.3 kJ/kg ⎬ x 2 = 1 (sat. vapor) ⎭ s 2 = 7.1270 kJ/kg ⋅ K Analysis We take the nozzle 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 & & E in − E out 1 24 4 3 = Rate of net energy transfer by heat, work, and mass & ΔE system 0 (steady) 1442443 4 4 =0 Rate of change in internal, kinetic, potential, etc. energies & & E in = E out 2 & & m(h1 + V12 / 2) = m(h2 + V2 /2) V 22 − V12 = h1 − h2 = Δke actual 2 500 kPa 200°C 30 m/s 200 kPa sat. vapor H2O Substituting, Δke actual = h1 − h2 = 2855.8 − 2706.3 = 149.5 kJ/kg The actual velocity at the exit is then V 22 − V12 = Δke actual 2 ⎛ 1000 m 2 /s 2 V 2 = V12 + 2Δke actual = (30 m/s) 2 + 2(149.5 kJ/kg)⎜ ⎜ 1 kJ/kg ⎝ ⎞ ⎟ = 547.6 m/s ⎟ ⎠ The maximum kinetic energy change is determined from Δke max = h1 − h2 − T0 ( s1 − s 2 ) = 2855.8 − 2706.3 − (298)(7.0610 − 7.1270) = 169.2 kJ/kg The maximum velocity at the exit is then V 22,max − V12 2 = Δke max ⎛ 1000 m 2 /s 2 V 2, max = V12 + 2Δke max = (30 m/s) 2 + 2(169.2 kJ/kg)⎜ ⎜ 1 kJ/kg ⎝ = 582.5 m/s ⎞ ⎟ ⎟ ⎠ PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 104. 8-104 8-115 A throttle valve is placed in the steam line supplying the turbine inlet in order to control an isentropic steam turbine. The second-law efficiency of this system when the valve is partially open to when it is fully open is to be compared. Assumptions 1 This is a steady-flow process since there is no change with time. 2 The turbine is wellinsulated, and there is no heat transfer from the turbine. Analysis Valve is fully open: The properties of steam at various states are T 6 MPa P1 = P2 = 6 MPa ⎫ h1 = h2 = 3894.3 kJ/kg ⎬ T1 = T2 = 700°C ⎭ s1 = s 2 = 7.4247 kJ/kg ⋅ K P3 = 70 kPa ⎫ x 3 = 0.9914 ⎬ s 2 = s1 ⎭ h3 = 2639.7 kJ/kg 1 2 70 kPa 3 The second law efficiency of the entire system is then η II = 3 MPa 3p s wout h1 − h3 h −h = = 1 3 = 1.0 wrev h1 − h3 − T0 ( s1 − s 3 ) h1 − h3 since s1 = s3 for this system. Valve is partly open: P2 = 3 MPa ⎫ ⎬ s 2 = 7.7405 kJ/kg ⋅ K h2 = h1 = 3894.3 kJ/kg ⎭ (from EES) P3 = 70 kPa ⎫ ⎬ h3 = 2760.8 kJ/kg (from EES) s3 = s 2 ⎭ η II = wout h1 − h3 3894.3 − 2760.8 = = = 0.923 wrev h1 − h3 − T0 ( s1 − s 3 ) 3894.3 − 2760.8 − (298)(7.4247 − 7.7405) PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 105. 8-105 8-116 Two rigid tanks that contain water at different states are connected by a valve. The valve is opened and steam flows from tank A to tank B until the pressure in tank A drops to a specified value. Tank B loses heat to the surroundings. The final temperature in each tank and the work potential wasted during this process are to be determined. Assumptions 1 Tank A is insulated and thus heat transfer is negligible. 2 The water that remains in tank A undergoes a reversible adiabatic process. 3 The thermal energy stored in the tanks themselves is negligible. 4 The system is stationary and thus kinetic and potential energy changes are negligible. 5 There are no work interactions. Analysis (a) The steam in tank A undergoes a reversible, adiabatic process, and thus s2 = s1. From the steam tables (Tables A-4 through A-6), Tank A : v 1, A = v f + x1v fg = 0.001084 + (0.8)(0.46242 − 0.001084 ) = 0.37015 m 3 /kg P1 = 400 kPa ⎫ ⎬ u1, A = u f + x1u fg = 604.22 + (0.8)(1948.9 ) = 2163.3 kJ/kg x1 = 0.8 ⎭ s = s + x s = 1.7765 + (0.8)(5.1191) = 5.8717 kJ/kg ⋅ K 1, A 1 fg f T2, A = Tsat @300kPa = 133.52 °C P2 = 300 kPa ⎫ x = s 2, A − s f = 5.8717 − 1.6717 = 0.7895 ⎪ 2, A 5.3200 s fg s 2 = s1 ⎬ (sat. mixture) ⎪ v 2, A = v f + x 2, Av fg = 0.001073 + (0.7895)(0.60582 − 0.001073) = 0.47850 m 3 /kg ⎭ u 2, A = u f + x 2, A u fg = 561.11 + (0.7895)(1982.1 kJ/kg ) = 2125.9 kJ/kg Tank B : v = 1.1989 m 3 /kg P1 = 200 kPa ⎫ 1, B 900 kJ ⎬ u1, B = 2731.4 kJ/kg T1 = 250°C ⎭ s1, B = 7.7100 kJ/kg ⋅ K A B × 3 The initial and the final masses in tank A are V = 0.2 m m = 3 kg steam steam 3 V 0.2 m P = 400 kPa m1, A = A = = 0.5403 kg T = 250°C v 1, A 0.37015 m 3 /kg x = 0.8 P = 200 kPa and V 0.2m 3 m 2, A = A = = 0.4180 kg v 2, A 0.479m 3 /kg Thus, 0.540 - 0.418 = 0.122 kg of mass flows into tank B. Then, m2, B = m1, B − 0122 = 3 + 0122 = 3.122 kg . . The final specific volume of steam in tank B is determined from v 2, B = VB m 2, B = (m1v1 )B m 2, B = (3 kg )(1.1989 m 3 /kg ) = 1.152 m 3 /kg 3.122 m 3 We take the entire contents of both tanks as the system, which is a closed system. The energy balance for this stationary closed system can be expressed as E in − E out 1 24 4 3 Net energy transfer by heat, work, and mass = ΔE system 1 24 4 3 Change in internal, kinetic, potential, etc. energies − Qout = ΔU = (ΔU ) A + (ΔU ) B (since W = KE = PE = 0) − Qout = (m 2 u 2 − m1u1 ) A + (m 2 u 2 − m1u1 ) B PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 106. 8-106 Substituting, { } − 900 = {(0.418)(2125.9) − (0.5403)(2163.3)} + (3.122)u 2, B − (3)(2731.4) u 2, B = 2425.9 kJ/kg Thus, v 2, B = 1.152 m 3 /kg ⎫ ⎪ u 2, B T2, B = 110.1°C ⎬ = 2425.9 kJ/kg ⎪ s 2, B = 6.9772 kJ/kg ⋅ K ⎭ (b) The total entropy generation during this process is determined by applying the entropy balance on an extended system that includes both tanks and their immediate surroundings so that the boundary temperature of the extended system is the temperature of the surroundings at all times. It gives S in − S out 1 24 4 3 Net entropy transfer by heat and mass − + S gen = ΔS system { 123 4 4 Entropy generation Change in entropy Qout + S gen = ΔS A + ΔS B Tb,surr Rearranging and substituting, the total entropy generated during this process is determined to be S gen = ΔS A + ΔS B + Qout Q = (m 2 s 2 − m1 s1 ) A + (m 2 s 2 − m1 s1 ) B + out Tb,surr Tb,surr = {(0.418)(5.8717 ) − (0.5403)(5.8717 )} + {(3.122 )(6.9772) − (3)(7.7100 )} + 900 kJ 273 K = 1.234 kJ/K The work potential wasted is equivalent to the exergy destroyed during a process, which can be determined from an exergy balance or directly from its definition X destroyed = T0 S gen , X destroyed = T0 S gen = (273 K)(1.234 kJ/K ) = 337 kJ PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 107. 8-107 8-117E A cylinder initially filled with helium gas at a specified state is compressed polytropically to a specified temperature and pressure. The actual work consumed and the minimum useful work input needed are to be determined. Assumptions 1 Helium is an ideal gas with constant specific heats. 2 The cylinder is stationary and thus the kinetic and potential energy changes are negligible. 3 The thermal energy stored in the cylinder itself is negligible. 4 The compression or expansion process is quasi-equilibrium. 5 The environment temperature is 70°F. Properties The gas constant of helium is R = 2.6805 psia.ft3/lbm.R = 0.4961 Btu/lbm.R (Table A-1E). The specific heats of helium are cv = 0.753 and cv = 1.25 Btu/lbm.R (Table A-2E). Analysis (a) Helium at specified conditions can be treated as an ideal gas. The mass of helium is m= P1V1 (25 psia )(15 ft 3 ) = = 0.264 lbm RT1 (2.6805 psia ⋅ ft 3 /lbm ⋅ R )(530 R ) The exponent n and the boundary work for this polytropic process are determined to be HELIUM 15 ft3 PV n = const P1V1 P2V 2 T P (760 R )(25 psia ) = ⎯ ⎯→ V 2 = 2 1 V1 = (15 ft 3 ) = 7.682 ft 3 T1 T2 T1 P2 (530 R )(70 psia ) P2V 2n = P1V1n ⎛P ⎯ ⎯→ ⎜ 2 ⎜P ⎝ 1 ⎞ ⎛ V1 ⎟=⎜ ⎟ ⎜V ⎠ ⎝ 2 n ⎞ ⎛ 70 ⎞ ⎛ 15 ⎞ ⎟ ⎯ ⎟ ⎯→ ⎜ 25 ⎟ = ⎜ 7.682 ⎟ ⎝ ⎠ ⎝ ⎠ ⎠ Q n ⎯ ⎯→ n = 1.539 Then the boundary work for this polytropic process can be determined from ∫ 2 W b,in = − P dV = − 1 =− P2V 2 − P1V1 mR(T2 − T1 ) =− 1− n 1− n (0.264 lbm)(0.4961 Btu/lbm ⋅ R )(760 − 530)R = 55.9 Btu 1 − 1.539 Also, ⎛ 1 Btu Wsurr,in = − P0 (V 2 −V1 ) = −(14.7 psia)(7.682 − 15)ft 3 ⎜ ⎜ 5.4039 psia ⋅ ft 3 ⎝ ⎞ ⎟ = 19.9 Btu ⎟ ⎠ Thus, W u,in = W b,in − Wsurr,in = 55.9 − 19.9 = 36.0 Btu (b) We take the helium in the cylinder as the system, which is a closed system. Taking the direction of heat transfer to be from the cylinder, the energy balance for this stationary closed system can be expressed as E in − E out 1 24 4 3 Net energy transfer by heat, work, and mass = ΔE system 1 24 4 3 Change in internal, kinetic, potential, etc. energies − Qout + W b,in = ΔU = m(u 2 − u1 ) − Qout = m(u 2 − u1 ) − W b,in Qout = W b,in − mcv (T2 − T1 ) Substituting, Qout = 55.9 Btu − (0.264 lbm )(0.753 Btu/lbm ⋅ R )(760 − 530)R = 10.2 Btu PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 108. 8-108 The total entropy generation during this process is determined by applying the entropy balance on an extended system that includes the cylinder and its immediate surroundings so that the boundary temperature of the extended system is the temperature of the surroundings at all times. It gives S in − S out 1 24 4 3 Net entropy transfer by heat and mass − + S gen = ΔS system { 123 4 4 Entropy generation Change in entropy Qout + S gen = ΔS sys Tb,surr where the entropy change of helium is ⎛ T P ⎞ ΔS sys = ΔS helium = m⎜ c p ,avg ln 2 − R ln 2 ⎟ ⎜ T1 P1 ⎟ ⎝ ⎠ ⎡ 760 R 70 psia ⎤ = (0.264 lbm) ⎢(1.25 Btu/lbm ⋅ R )ln − (0.4961 Btu/lbm ⋅ R )ln ⎥ 530 R 25 psia ⎦ ⎣ = −0.0159 Btu/R Rearranging and substituting, the total entropy generated during this process is determined to be S gen = ΔS helium + Qout 10.2 Btu = (−0.0159 Btu/R) + = 0.003345 Btu/R T0 530 R The work potential wasted is equivalent to the exergy destroyed during a process, which can be determined from an exergy balance or directly from its definition X destroyed = T0 S gen , X destroyed = T0 S gen = (530 R)(0.003345 Btu/R ) = 1.77 Btu The minimum work with which this process could be accomplished is the reversible work input, Wrev, in. which can be determined directly from W rev,in = Wact,in − X destroyed = 36.0 − 1.77 = 34.23 Btu Discussion The reversible work input, which represents the minimum work input Wrev,in in this case can be determined from the exergy balance by setting the exergy destruction term equal to zero, X in − X out 1 24 4 3 Net exergy tra nsfer by heat, work,and mass − X destroyed 0 (reversible) = ΔX system → Wrev,in = X 2 − X1 144 2444 4 3 123 4 4 Exergy destructio n Change in exergy Substituting the closed system exergy relation, the reversible work input during this process is determined to be W rev = (U 2 − U 1 ) − T0 ( S 2 − S1 ) + P0 (V 2 −V1 ) = (0.264 lbm)(0.753 Btu/lbm ⋅ R)(300 − 70)°F − (530 R)(-0.0159 Btu/R) + (14.7 psia)(7.682 − 15)ft 3 [Btu/5.4039 psia ⋅ ft 3 ] = 34.24 Btu PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 109. 8-109 8-118 Steam expands in a two-stage adiabatic turbine from a specified state to specified pressure. Some steam is extracted at the end of the first stage. The wasted power potential is to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 The turbine is adiabatic and thus heat transfer is negligible. 4 The environment temperature is given to be T0 = 25°C. Analysis The wasted power potential is equivalent to the rate of exergy destruction during a process, which can be determined from an exergy balance or directly from its definition X destroyed = T0 S gen . The total rate of entropy generation during this process is determined by taking the entire turbine, which is a control volume, as the system and applying the entropy balance. Noting that this is a steady-flow process and there is no heat transfer, & & S in − S out 1 24 4 3 Rate of net entropy transfer by heat and mass + & S gen { Rate of entropy generation & = ΔS system 0 = 0 14 4 2 3 Rate of change of entropy & & & & m1 s1 − m 2 s 2 − m3 s 3 + S gen = 0 & & & & & & m1 s1 − 0.1m1 s 2 − 0.9m1 s 3 + S gen = 0 → S gen = m1 [0.9 s 3 + 0.1s 2 − s1 ] and & X destroyed = T0 Sgen = T0m1[0.9 s3 + 0.1s2 − s1 ] From the steam tables (Tables A-4 through 6) P1 = 9 MPa ⎫ h1 = 3387.4 kJ / kg ⎬ T1 = 500°C ⎭ s1 = 6.6603 kJ / kg ⋅ K 9 MPa 500°C STEAM 13.5 kg/s STEAM 15 kg/s P2 = 1.4 MPa ⎫ ⎬ h2 s = 2882.4 kJ / kg s 2 s = s1 ⎭ I II 1.4 MPa and, h − h2 ηT = 1 ⎯ ⎯→ h2 = h1 − ηT (h1 − h2 s ) h1 − h2 s = 3387.4 − 0.88(3387.4 − 2882.4) = 2943.0 kJ/kg 50 kPa 10% 90% P2 = 1.4 MPa ⎫ ⎬ s 2 = 6.7776 kJ / kg ⋅ K h2 = 2943.0 kJ/kg ⎭ s 3s − s f 6.6603 − 1.0912 P3 = 50 kPa ⎫ x 3s = = = 0.8565 s fg 6.5019 ⎬ s 3s = s1 ⎭ h = h + x h = 340.54 + 0.8565 × 2304.7 = 2314.6 kJ/kg 3s f 3 s fg and ηT = h1 − h3 ⎯ ⎯→ h3 = h1 − ηT (h1 − h3s ) h1 − h3s = 3387.4 − 0.88(3387.4 − 2314.6) = 2443.3 kJ/kg h3 − h f 2443.3 − 340.54 = = 0.9124 ⎫ x3 = h fg 2304.7 ⎬ h3 = 2443.3 kJ/kg ⎭ s 3 = s f + x3 s fg = 1.0912 + 0.9124 × 6.5019 = 7.0235 kJ/kg ⋅ K P3 = 50 kPa Substituting, the wasted work potential is determined to be & & X destroyed = T0 S gen = (298 K)(15 kg/s)(0.9 × 7.0235 + 0.1× 6.7776 − 6.6603)kJ/kg = 1514 kW PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 110. 8-110 8-119 Steam expands in a two-stage adiabatic turbine from a specified state to another specified state. Steam is reheated between the stages. For a given power output, the reversible power output and the rate of exergy destruction are to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 The turbine is adiabatic and thus heat transfer is negligible. 4 The environment temperature is given to be T0 = 25°C. Heat 2 MPa 350°C Properties From the steam tables (Tables A4 through 6) Stage I 2 MPa 500°C Stage II 5 MW P1 = 8 MPa ⎫ h1 = 3399.5 kJ / kg ⎬ T1 = 500°C ⎭ s1 = 6.7266 kJ / kg ⋅ K 8 MPa 500°C P2 = 2 MPa ⎫ h2 = 3137.7 kJ / kg ⎬ T2 = 350°C ⎭ s 2 = 6.9583 kJ / kg ⋅ K 30 kPa x = 97% P3 = 2 MPa ⎫ h3 = 3468.3 kJ / kg ⎬ T3 = 500°C ⎭ s 3 = 7.4337 kJ / kg ⋅ K P4 = 30 kPa ⎫ h4 = h f + x 4 h fg = 289.27 + 0.97 × 2335.3 = 2554.5 kJ/kg ⎬ x 4 = 0.97 ⎭ s 4 = s f + x 4 s fg = 0.9441 + 0.97 × 6.8234 = 7.5628 kJ/kg ⋅ K Analysis We take the entire turbine, excluding the reheat section, as the system, which is a control volume. The energy balance for this steady-flow system can be expressed in the rate form as & & E in − E out 1 24 4 3 = Rate of net energy transfer by heat, work, and mass & ΔE system 0 (steady) 1442443 4 =0 Rate of change in internal, kinetic, potential, etc. energies & & E in = E out & & & & & mh1 + mh3 = mh2 + mh4 + Wout & & Wout = m[(h1 − h2 ) + (h3 − h4 )] Substituting, the mass flow rate of the steam is determined from the steady-flow energy equation applied to the actual process, & m= & Wout 5000 kJ/s = = 4.253 kg/s h1 − h2 + h3 − h4 (3399.5 − 3137.7 + 3468.3 − 2554.5)kJ/kg The reversible (or maximum) power output is determined from the rate form of the exergy balance applied on the turbine and setting the exergy destruction term equal to zero, PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 111. 8-111 & & X in − X out 1 24 4 3 Rate of net exergy tra nsfer by heat, work,and mass & & − X destroyed 0 (reversible) = ΔX system 0 (steady) = 0 144 2444 4 3 1442443 Rate of exergy destructio n Rate of change of exergy & & X in = X out & & & & & mψ1 + mψ 3 = mψ 2 + mψ 4 + Wrev,out & & & Wrev,out = m(ψ1 −ψ 2 ) + m(ψ 3 −ψ 4 ) & = m[(h1 − h2 ) + T0 (s2 − s1) − Δke & + m[(h3 − h4 ) + T0 (s4 − s3 ) − Δke 0 − Δpe 0 ] 0 − Δpe 0 ] Then the reversible power becomes & & W rev,out = m[h1 − h2 + h3 − h4 + T0 ( s 2 − s1 + s 4 − s 3 )] = (4.253 kg/s)[(3399.5 − 3137.7 + 3468.3 − 2554.5)kJ/kg +(298 K)(6.9583 − 6.7266 + 7.5628 − 7.4337)kJ/kg ⋅ K] = 5457 kW Then the rate of exergy destruction is determined from its definition, & & & X destroyed = W rev,out − Wout = 5457 − 5000 = 457 kW PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 112. 8-112 8-120 One ton of liquid water at 80°C is brought into a room. The final equilibrium temperature in the room and the entropy generated are to be determined. Assumptions 1 The room is well insulated and well sealed. 2 The thermal properties of water and air are constant at room temperature. 3 The system is stationary and thus the kinetic and potential energy changes are zero. 4 There are no work interactions involved. Properties The gas constant of air is R = 0.287 kPa.m3/kg.K (Table A-1). The constant volume specific heat of water at room temperature is cv = 0.718 kJ/kg⋅°C (Table A-2). The specific heat of water at room temperature is c = 4.18 kJ/kg⋅°C (Table A-3). Analysis The volume and the mass of the air in the room are V = 4x5x6 = 120 m³ 4m×5m×6m P1V 1 (100 kPa )(120 m 3 ) m air = = = 141.74 kg ROOM RT1 (0.2870 kPa ⋅ m 3 /kg ⋅ K )(295 K ) 22°C Taking the contents of the room, including the water, as 100 kPa our system, the energy balance can be written as Heat E in − E out = ΔE system 1 24 4 3 1 24 4 3 Water Net energy transfer Change in internal, kinetic, by heat, work, and mass potential, etc. energies 80°C 0 = ΔU = (ΔU ) water + (ΔU )air [mc(T2 − T1 )]water + [mcv (T2 − T1 )]air = 0 Substituting, (1000 kg )(4.18 kJ/kg⋅o C )(T f − 80)o C + (141.74 kg )(0.718 kJ/kg⋅o C )(T f or ) − 22 o C = 0 It gives the final equilibrium temperature in the room to be Tf = 78.6°C (b) We again take the room and the water in it as the system, which is a closed system. Considering that the system is well-insulated and no mass is entering and leaving, the entropy balance for this system can be expressed as S in − S out 1 24 4 3 Net entropy transfer by heat and mass + S gen = ΔS system { 123 4 4 Entropy generation Change in entropy 0 + S gen = ΔS air + ΔS water where ΔS air = mcv ln ΔS water = mc ln T2 V + mR ln 2 T1 V1 0 = (141.74 kg )(0.718 kJ/kg ⋅ K )ln 351.6 K = 17.87 kJ/K 295 K T2 351.6 K = (1000 kg )(4.18 kJ/kg ⋅ K ) ln = −16.36 kJ/K T1 353 K Substituting, the entropy generation is determined to be Sgen = 17.87 - 16.36 = 1.51 kJ/K The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 S gen , X destroyed = T0 S gen = (283 K)(1.51 kJ/K) = 427 kJ (c) The work potential (the maximum amount of work that can be produced) during a process is simply the reversible work output. Noting that the actual work for this process is zero, it becomes X destroyed = W rev,out − Wact,out → W rev,out = X destroyed = 427 kJ PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 113. 8-113 8-121 An insulated cylinder is divided into two parts. One side of the cylinder contains N2 gas and the other side contains He gas at different states. The final equilibrium temperature in the cylinder and the wasted work potential are to be determined for the cases of piston being fixed and moving freely. Assumptions 1 Both N2 and He are ideal gases with constant specific heats. 2 The energy stored in the container itself is negligible. 3 The cylinder is well-insulated and thus heat transfer is negligible. Properties The gas constants and the constant volume specific heats are R = 0.2968 kPa.m3/kg.K is cv = 0.743 kJ/kg·°C for N2, and R = 2.0769 kPa.m3/kg.K is cv = 3.1156 kJ/kg·°C for He (Tables A-1 and A-2) Analysis The mass of each gas in the cylinder is ( ) ( ) (500 kPa )(1 m ) = (2.0769 kPa ⋅ m /kg ⋅ K )(298 K ) = 0.808 kg ⎛ PV ⎞ (500 kPa ) 1 m 3 m N2 = ⎜ 1 1 ⎟ = = 4.77 kg ⎜ RT ⎟ 0.2968 kPa ⋅ m 3 /kg ⋅ K (353 K ) ⎝ 1 ⎠ N2 m He ⎛ PV ⎞ =⎜ 1 1 ⎟ ⎜ RT ⎟ ⎝ 1 ⎠ He 3 Taking the entire contents of the cylinder as our system, the 1st law relation can be written as E in − E out 1 24 4 3 = Net energy transfer by heat, work, and mass He 1 m3 500 kPa 25°C N2 1 m3 500 kPa 80°C 3 ΔE system 1 24 4 3 Change in internal, kinetic, potential, etc. energies 0 = ΔU = (ΔU ) N 2 + (ΔU )He 0 = [mcv (T2 − T1 )] N 2 + [mcv (T2 − T1 )] He Substituting, (4.77 kg )(0.743 kJ/kg⋅o C)(T f ) ( )( ) − 80 o C + (0.808 kg ) 3.1156 kJ/kg⋅ o C T f − 25 o C = 0 It gives Tf = 57.2°C where Tf is the final equilibrium temperature in the cylinder. The answer would be the same if the piston were not free to move since it would effect only pressure, and not the specific heats. (b) We take the entire cylinder as our system, which is a closed system. Noting that the cylinder is wellinsulated and thus there is no heat transfer, the entropy balance for this closed system can be expressed as S in − S out 1 24 4 3 Net entropy transfer by heat and mass + S gen = ΔS system { 123 4 4 Entropy generation Change in entropy 0 + S gen = ΔS N 2 + ΔS He But first we determine the final pressure in the cylinder: 4.77 kg 0.808 kg ⎛m⎞ ⎛m⎞ N total = N N 2 + N He = ⎜ ⎟ + ⎜ ⎟ = + = 0.372 kmol ⎝ M ⎠ N 2 ⎝ M ⎠ He 28 kg/kmol 4 kg/kmol P2 = ( ) N total Ru T (0.372 kmol) 8.314 kPa ⋅ m 3 /kmol ⋅ K (330.2 K ) = = 510.6 kPa V total 2 m3 Then, PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 114. 8-114 ⎛ T P ⎞ ΔS N 2 = m⎜ c p ln 2 − R ln 2 ⎟ ⎜ T1 P1 ⎟ N ⎝ ⎠ 2 ⎡ 330.2 K 510.6 kPa ⎤ = (4.77 kg )⎢(1.039 kJ/kg ⋅ K ) ln − (0.2968 kJ/kg ⋅ K ) ln ⎥ = −0.361 kJ/K 353 K 500 kPa ⎦ ⎣ ⎛ T P ⎞ ΔS He = m⎜ c p ln 2 − R ln 2 ⎟ ⎜ T1 P1 ⎟ He ⎝ ⎠ ⎡ 330.2 K 510.6 kPa ⎤ = (0.808 kg )⎢(5.1926 kJ/kg ⋅ K ) ln − (2.0769 kJ/kg ⋅ K ) ln ⎥ = 0.395 kJ/K 298 K 500 kPa ⎦ ⎣ S gen = ΔS N 2 + ΔS He = −0.361 + 0.395 = 0.034 kJ/K The wasted work potential is equivalent to the exergy destroyed during a process, and it can be determined from an exergy balance or directly from its definition X destroyed = T0 S gen , X destroyed = T0 S gen = (298 K)(0.034 kJ/K) = 10.1 kJ If the piston were not free to move, we would still have T2 = 330.2 K but the volume of each gas would remain constant in this case: ⎛ V T ΔS N 2 = m⎜ cv ln 2 − R ln 2 ⎜ V1 T1 ⎝ 0 ⎞ ⎟ = (4.77 kg )(0.743 kJ/kg ⋅ K ) ln 330.2 K = −0.237 kJ/K ⎟ 353 K ⎠ N2 ⎛ V T ΔS He = m⎜ cv ln 2 − R ln 2 ⎜ V1 T1 ⎝ 0 ⎞ ⎟ = (0.808 kg )(3.1156 kJ/kg ⋅ K ) ln 330.2 K = 0.258 kJ/K ⎟ 298 K ⎠ He S gen = ΔS N 2 + ΔS He = −0.237 + 0.258 = 0.021 kJ/K and X destroyed = T0 S gen = (298 K)(0.021 kJ/K) = 6.26 kJ PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 115. 8-115 8-122 An insulated cylinder is divided into two parts. One side of the cylinder contains N2 gas and the other side contains He gas at different states. The final equilibrium temperature in the cylinder and the wasted work potential are to be determined for the cases of piston being fixed and moving freely. √ Assumptions 1 Both N2 and He are ideal gases with constant specific heats. 2 The energy stored in the container itself, except the piston, is negligible. 3 The cylinder is well-insulated and thus heat transfer is negligible. 4 Initially, the piston is at the average temperature of the two gases. Properties The gas constants and the constant volume specific heats are R = 0.2968 kPa.m3/kg.K is cv = 0.743 kJ/kg·°C for N2, and R = 2.0769 kPa.m3/kg.K is cv = 3.1156 kJ/kg·°C for He (Tables A-1 and A2). The specific heat of copper piston is c = 0.386 kJ/kg·°C (Table A-3). Analysis The mass of each gas in the cylinder is ( ) ( ) (500 kPa )(1 m ) = (2.0769 kPa ⋅ m /kg ⋅ K )(353 K ) = 0.808 kg ⎛ PV ⎞ (500 kPa ) 1 m 3 m N2 = ⎜ 1 1 ⎟ = = 4.77 kg ⎜ RT ⎟ 0.2968 kPa ⋅ m 3 /kg ⋅ K (353 K ) ⎝ 1 ⎠ N2 ⎛ PV ⎞ m He = ⎜ 1 1 ⎟ ⎜ RT ⎟ ⎝ 1 ⎠ He 3 3 He 1 m3 500 kPa 25°C N2 1 m3 500 kPa 80°C Taking the entire contents of the cylinder as our system, the 1st law relation can be written as E in − E out 1 24 4 3 = Net energy transfer by heat, work, and mass Copper ΔE system 1 24 4 3 Change in internal, kinetic, potential, etc. energies 0 = ΔU = (ΔU ) N 2 + (ΔU )He + (ΔU )Cu 0 = [mcv (T2 − T1 )] N 2 + [mcv (T2 − T1 )] He + [mc(T2 − T1 )] Cu where T1, Cu = (80 + 25) / 2 = 52.5°C Substituting, (4.77 kg )(0.743 kJ/kg⋅ o C)(T f ) ) ( )( + (5.0 kg )(0.386 kJ/kg⋅ C )(T − 52.5) C = 0 − 80 o C + (0.808 kg ) 3.1156 kJ/kg⋅ o C T f − 25 o C o f o It gives Tf = 56.0°C where Tf is the final equilibrium temperature in the cylinder. The answer would be the same if the piston were not free to move since it would effect only pressure, and not the specific heats. (b) We take the entire cylinder as our system, which is a closed system. Noting that the cylinder is wellinsulated and thus there is no heat transfer, the entropy balance for this closed system can be expressed as S in − S out 1 24 4 3 Net entropy transfer by heat and mass + S gen = ΔS system { 123 4 4 Entropy generation Change in entropy 0 + S gen = ΔS N 2 + ΔS He + ΔS piston But first we determine the final pressure in the cylinder: PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 116. 8-116 4.77 kg 0.808 kg ⎛m⎞ ⎛m⎞ N total = N N 2 + N He = ⎜ ⎟ + ⎜ ⎟ = + = 0.372 kmol ⎝ M ⎠ N 2 ⎝ M ⎠ He 28 kg/kmol 4 kg/kmol P2 = N total Ru T V total = (0.372 kmol)(8.314 kPa ⋅ m 3 /kmol ⋅ K )(329 K ) = 508.8 kPa 2 m3 Then, ⎛ T P ⎞ ΔS N 2 = m⎜ c p ln 2 − R ln 2 ⎟ ⎜ T1 P1 ⎟ N ⎝ ⎠ 2 ΔS He ⎡ 329 K 508.8 kPa ⎤ = (4.77 kg )⎢(1.039 kJ/kg ⋅ K )ln − (0.2968 kJ/kg ⋅ K ) ln ⎥ = −0.374 kJ/K 353 K 500 kPa ⎦ ⎣ ⎛ T P ⎞ = m⎜ c p ln 2 − R ln 2 ⎟ ⎜ T1 P1 ⎟ He ⎝ ⎠ ⎡ 329 K 508.8 kPa ⎤ = (0.808 kg )⎢(5.1926 kJ/kg ⋅ K ) ln − (2.0769 kJ/kg ⋅ K ) ln ⎥ = 0.386 kJ/K 353 K 500 kPa ⎦ ⎣ ⎛ T ⎞ 329 K ΔS piston = ⎜ mc ln 2 ⎟ = (5 kg )(0.386 kJ/kg ⋅ K ) ln = 0.021 kJ/K ⎜ ⎟ 325.5 K T1 ⎠ piston ⎝ S gen = ΔS N 2 + ΔS He + ΔS piston = −0.374 + 0.386 + 0.021 = 0.0334 kJ/K The wasted work potential is equivalent to the exergy destroyed during a process, and it can be determined from an exergy balance or directly from its definition X destroyed = T0 S gen , X destroyed = T0 S gen = (298 K)(0.033 kJ/K) = 9.83 kJ If the piston were not free to move, we would still have T2 = 330.2 K but the volume of each gas would remain constant in this case: ⎛ V T ΔS N 2 = m⎜ cv ln 2 − R ln 2 ⎜ V1 T1 ⎝ 0 ⎞ ⎟ = (4.77 kg )(0.743 kJ/kg ⋅ K ) ln 329 K = −0.250 kJ/K ⎟ 353 K ⎠ N2 ⎛ V 0⎞ T ⎟ = (0.808 kg )(3.1156 kJ/kg ⋅ K ) ln 329 K = 0.249 kJ/K ΔS He = m⎜ cv ln 2 − R ln 2 ⎜ V1 ⎟ T1 353 K ⎝ ⎠ He S gen = ΔS N 2 + ΔS He + ΔS piston = −0.250 + 0.249 + 0.021 = 0.020 kJ/K and X destroyed = T0 S gen = (298 K)(0.020 kJ/K) = 6.0 kJ PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 117. 8-117 8-123E Argon enters an adiabatic turbine at a specified state with a specified mass flow rate, and leaves at a specified pressure. The isentropic efficiency of turbine is to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 The device is adiabatic and thus heat transfer is negligible. 4 Argon is an ideal gas with constant specific heats. Properties The specific heat ratio of argon is k = 1.667. The constant pressure specific heat of argon is cp = 0.1253 Btu/lbm.R (Table A-2E). & & & Analysis There is only one inlet and one exit, and thus m1 = m2 = m . We take the isentropic turbine as the system, which is a control volume since mass crosses the boundary. The energy balance for this steadyflow system can be expressed in the rate form as & & E in − E out 1 24 4 3 & ΔE system 0 (steady) 1442443 4 = Rate of net energy transfer by heat, work, and mass =0 1 Rate of change in internal, kinetic, potential, etc. energies & & E in = E out & & & mh1 = W s ,out + mh2 s & (since Q ≅ Δke ≅ Δpe ≅ 0) 370 kW Ar ηT & & W s ,out = m(h1 − h2 s ) From the isentropic relations, ⎛P ⎞ T2 s = T1⎜ 2 s ⎟ ⎜ P ⎟ ⎝ 1 ⎠ ( k −1) / k ⎛ 30 psia ⎞ = (1960 R)⎜ ⎜ 200 psia ⎟ ⎟ ⎝ ⎠ 2 0.667 / 1.667 = 917.5 R Then the power output of the isentropic turbine becomes 1 hp ⎛ ⎞ & & W s ,out = mc p (T1 − T2 s ) = (40 lbm/min)(0.1253 Btu/lbm ⋅ R)(1960 − 917.5)R ⎜ ⎟ = 123.2 hp ⎝ 42.41 Btu/min ⎠ Then the isentropic efficiency of the turbine is determined from ηT = & W a ,out 95 hp = = 0.771 = 77.1% & W s ,out 123.2 hp & & (b) Using the steady-flow energy balance relation W a ,out = mc p (T1 − T2 ) above, the actual turbine exit temperature is determined to be T2 = T1 − & W a ,out & mc p = 1500 − ⎛ 42.41 Btu/min ⎞ 95 hp ⎜ ⎟ = 696.1°F = 1156.1 R ⎟ (40 lbm/min)(0.1253 Btu/lbm ⋅ R) ⎜ 1 hp ⎝ ⎠ The entropy generation during this process can be determined from an entropy balance on the turbine, & & S in − S out 1 24 4 3 + Rate of net entropy transfer by heat and mass & S gen { Rate of entropy generation & = ΔS system 0 = 0 14 4 2 3 Rate of change of entropy & & & ms1 − ms 2 + S gen = 0 & & S gen = m( s 2 − s1 ) where PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 118. 8-118 s 2 − s1 = c p ln T2 P − R ln 2 T1 P1 = (0.1253 Btu/lbm ⋅ R) ln 30 psia 1156.1 R − (0.04971 Btu/lbm ⋅ R) ln 1960 R 200 psia = 0.02816 Btu/lbm.R The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 S gen , & & & X destroyed = T0 Sgen = mT0 ( s2 − s1 ) 1 hp ⎛ ⎞ = (40 lbm/min)(537 R)(0.02816 Btu/lbm ⋅ R)⎜ ⎟ ⎝ 42.41 Btu/min ⎠ = 14.3 hp Then the reversible power and second-law efficiency become & & & W rev,out = W a ,out + X destroyed = 95 + 14.3 = 109.3 hp and η II = & 95 hp W = = 86.9% & 109.3 hp W rev PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 119. 8-119 8-124 The feedwater of a steam power plant is preheated using steam extracted from the turbine. The ratio of the mass flow rates of the extracted steam and the feedwater are to be determined. Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 Heat loss from the device to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. Properties The properties of steam and feedwater are (Tables A-4 through A-6) P1 = 1 MPa ⎫ h1 = 2828.3kJ/kg 1 ⎬ 1 MPa T1 = 200°C ⎭ s1 = 6.6956 kJ/kg ⋅ K Steam 200°C from h2 = h f @1 MPa = 762.51 kJ/kg Feedwater turbine P2 = 1 MPa ⎫ ⎬ s 2 = s f @1 MPa = 2.1381 kJ/kg ⋅ K 3 sat. liquid ⎭ T2 = 179.88°C 2.5 MPa 50°C P3 = 2.5 MPa ⎫ h3 ≅ h f @50o C = 209.34 kJ/kg ⎬s ≅ s = 0.7038 kJ/kg ⋅ K T3 = 50°C f @ 50o C ⎭ 3 4 P4 = 2.5 MPa ⎫ h4 ≅ h f @170o C = 719.08 kJ/kg ⎬ T4 = T2 − 10°C ≅ 170°C ⎭ s 4 ≅ s f @170o C = 2.0417 kJ/kg ⋅ K Analysis (a) We take the heat exchanger as the system, which is 2 a control volume. The mass and energy balances for this steadysat. liquid flow system can be expressed in the rate form as follows: Mass balance (for each fluid stream): & & & & & & & & & & & min − m out = Δmsystem 0 (steady) = 0 → min = m out → m1 = m 2 = m s and m3 = m 4 = m fw Energy balance (for the heat exchanger): & & & Ein − Eout = ΔEsystem 0 (steady) 1 24 4 3 1442443 Rate of net energy transfer by heat, work, and mass & & =0⎯ ⎯→ Ein = Eout Rate of change in internal, kinetic, potential, etc. energies & & & & & & m1h1 + m3h3 = m2h2 + m4h4 (since Q = W = Δke ≅ Δpe ≅ 0) & & Combining the two, m s (h2 − h1 ) = m fw (h3 − h4 ) & Dividing by m fw and substituting, & ms h − h4 (209.34 − 719.08)kJ/kg = 3 = = 0.247 & m fw h2 − h1 (762.51 − 2828.3)kJ/kg (b) The entropy generation during this process per unit mass of feedwater can be determined from an entropy balance on the feedwater heater expressed in the rate form as & & & & S in − S out + S gen = ΔS system 0 = 0 1 24 4 3 { 14 4 2 3 Rate of net entropy transfer by heat and mass Rate of entropy generation Rate of change of entropy & & & & & m1 s1 − m 2 s 2 + m3 s 3 − m 4 s 4 + S gen = 0 & & & m s ( s1 − s 2 ) + m fw ( s 3 − s 4 ) + S gen = 0 & S gen & m fw = & ms (s 2 − s1 ) + (s 4 − s 3 ) = (0.247 )(2.1381 − 6.6956) + (2.0417 − 0.7038) = 0.213 kJ/K ⋅ kg fw & m fw Noting that this process involves no actual work, the reversible work and exergy destruction become equivalent since X destroyed = W rev,out − Wact,out → W rev,out = X destroyed . The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 S gen , X destroyed = T0 S gen = (298 K)(0.213 kJ/K ⋅ kg fw) = 63.5 kJ/kgfeedwater PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 120. 8-120 8-125 EES Problem 8-124 is reconsidered. The effect of the state of the steam at the inlet of the feedwater heater on the ratio of mass flow rates and the reversible power is to be investigated. Analysis Using EES, the problem is solved as follows: "Input Data" "Steam (let st=steam data):" Fluid\$='Steam_IAPWS' T_st[1]=200 [C] {P_st[1]=1000 [kPa]} P_st[2] = P_st[1] x_st[2]=0 "saturated liquid, quality = 0%" T_st[2]=temperature(steam, P=P_st[2], x=x_st[2]) "Feedwater (let fw=feedwater data):" T_fw[1]=50 [C] P_fw[1]=2500 [kPa] P_fw[2]=P_fw[1] "assume no pressure drop for the feedwater" T_fw[2]=T_st[2]-10 "Surroundings:" T_o = 25 [C] P_o = 100 [kPa] "Assumed value for the surrroundings pressure" "Conservation of mass:" "There is one entrance, one exit for both the steam and feedwater." "Steam: m_dot_st[1] = m_dot_st[2]" "Feedwater: m_dot_fw[1] = m_dot_fw[2]" "Let m_ratio = m_dot_st/m_dot_fw" "Conservation of Energy:" "We write the conservation of energy for steady-flow control volume having two entrances and two exits with the above assumptions. Since neither of the flow rates is know or can be found, write the conservation of energy per unit mass of the feedwater." E_in - E_out =DELTAE_cv DELTAE_cv=0 "Steady-flow requirement" E_in = m_ratio*h_st[1] + h_fw[1] h_st[1]=enthalpy(Fluid\$, T=T_st[1], P=P_st[1]) h_fw[1]=enthalpy(Fluid\$,T=T_fw[1], P=P_fw[1]) E_out = m_ratio*h_st[2] + h_fw[2] h_fw[2]=enthalpy(Fluid\$, T=T_fw[2], P=P_fw[2]) h_st[2]=enthalpy(Fluid\$, x=x_st[2], P=P_st[2]) "The reversible work is given by Eq. 7-47, where the heat transfer is zero (the feedwater heater is adiabatic) and the Exergy destroyed is set equal to zero" W_rev = m_ratio*(Psi_st[1]-Psi_st[2]) +(Psi_fw[1]-Psi_fw[2]) Psi_st[1]=h_st[1]-h_st_o -(T_o + 273)*(s_st[1]-s_st_o) s_st[1]=entropy(Fluid\$,T=T_st[1], P=P_st[1]) h_st_o=enthalpy(Fluid\$, T=T_o, P=P_o) s_st_o=entropy(Fluid\$, T=T_o, P=P_o) Psi_st[2]=h_st[2]-h_st_o -(T_o + 273)*(s_st[2]-s_st_o) s_st[2]=entropy(Fluid\$,x=x_st[2], P=P_st[2]) Psi_fw[1]=h_fw[1]-h_fw_o -(T_o + 273)*(s_fw[1]-s_fw_o) h_fw_o=enthalpy(Fluid\$, T=T_o, P=P_o) s_fw[1]=entropy(Fluid\$,T=T_fw[1], P=P_fw[1]) s_fw_o=entropy(Fluid\$, T=T_o, P=P_o) PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 121. 8-121 Psi_fw[2]=h_fw[2]-h_fw_o -(T_o + 273)*(s_fw[2]-s_fw_o) s_fw[2]=entropy(Fluid\$,T=T_fw[2], P=P_fw[2]) mratio [kg/kg] 0.06745 0.1067 0.1341 0.1559 0.1746 0.1912 0.2064 0.2204 0.2335 0.246 Wrev [kJ/kg] 12.9 23.38 31.24 37.7 43.26 48.19 52.64 56.72 60.5 64.03 Pst,1 [kPa] 100 200 300 400 500 600 700 800 900 1000 0.28 mratio [kgst/kg,fw] 0.24 0.2 0.16 0.12 0.08 0.04 100 200 300 400 500 600 700 800 900 1000 700 800 900 1000 Pst[1] [kPa] 70 Wrev [kJ/kgfw] 60 50 40 30 20 10 100 200 300 400 500 600 Pst[1] [kPa] PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 122. 8-122 8-126 A 1-ton (1000 kg) of water is to be cooled in a tank by pouring ice into it. The final equilibrium temperature in the tank and the exergy destruction are to be determined. Assumptions 1 Thermal properties of the ice and water are constant. 2 Heat transfer to the water tank is negligible. 3 There is no stirring by hand or a mechanical device (it will add energy). Properties The specific heat of water at room temperature is c = 4.18 kJ/kg·°C, and the specific heat of ice at about 0°C is c = 2.11 kJ/kg·°C (Table A-3). The melting temperature and the heat of fusion of ice at 1 atm are 0°C and 333.7 kJ/kg.. Analysis (a) We take the ice and the water as the system, and disregard ice any heat transfer between the system and the surroundings. Then the -5°C energy balance for this process can be written as 80 kg E in − E out = ΔE system 1 24 4 3 1 24 4 3 Net energy transfer by heat, work, and mass Change in internal, kinetic, potential, etc. energies WATER 1 ton 0 = ΔU 0 = ΔU ice + ΔU water o [mc(0 C − T1 ) solid + mhif + mc(T2 −0 o C) liquid ] ice + [mc(T2 − T1 )] water = 0 Substituting, (80 kg){(2.11 kJ / kg⋅o C)[0 − (-5)]o C + 333.7 kJ / kg + (4.18 kJ / kg⋅o C)(T2 − 0) o C} + (1000 kg)(4.18 kJ / kg⋅o C)(T2 − 20)o C = 0 It gives T2 = 12.42°C which is the final equilibrium temperature in the tank. (b) We take the ice and the water as our system, which is a closed system .Considering that the tank is well-insulated and thus there is no heat transfer, the entropy balance for this closed system can be expressed as S in − S out 1 24 4 3 Net entropy transfer by heat and mass + S gen = ΔS system { 123 4 4 Entropy generation Change in entropy 0 + S gen = ΔS ice + ΔS water where ⎛ T ⎞ 285.42 K = (1000 kg )(4.18 kJ/kg ⋅ K )ln = −109.590 kJ/K ΔS water = ⎜ mc ln 2 ⎟ ⎟ ⎜ 293 K T1 ⎠ water ⎝ ΔS ice = ΔS solid + ΔS melting + ΔS liquid ice ( ⎛⎛ Tmelting = ⎜ ⎜ mc ln ⎜ ⎜ T1 ⎝⎝ ) mhig ⎞ ⎛ T ⎟ + + ⎜ mc ln 2 ⎜ ⎟ T1 ⎠ solid Tmelting ⎝ ⎞ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ liquid ⎠ ice ⎛ 273 K 333.7 kJ/kg 285.42 K ⎞ ⎟ = (80 kg )⎜ (2.11 kJ/kg ⋅ K )ln + + (4.18 kJ/kg ⋅ K )ln ⎜ 268 K 273 K 273 K ⎟ ⎝ ⎠ = 115.783 kJ/K Then, S gen = ΔS water + ΔS ice = −109.590 + 115.783 = 6.193 kJ/K The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 S gen , X destroyed = T0 S gen = (293 K)(6.193 kJ/K) = 1815 kJ PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 123. 8-123 8-127 An evacuated bottle is surrounded by atmospheric air. A valve is opened, and air is allowed to fill the bottle. The amount of heat transfer through the wall of the bottle when thermal and mechanical equilibrium is established and the amount of exergy destroyed are to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process, but it can be analyzed as a uniform-flow process since the state of fluid at the inlet remains constant. 2 Air is an ideal gas. 3 Kinetic and potential energies are negligible. 4 There are no work interactions involved. 5 The direction of heat transfer is to the air in the bottle (will be verified). 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 and internal energy u, respectively, the mass and energy balances can be expressed as Mass balance: m in − m out = Δm system → m i = m 2 (since m out = m initial = 0) Energy balance: E in − E out 1 24 4 3 = Net energy transfer by heat, work, and mass 100 kPa 17°C ΔE system 1 24 4 3 Change in internal, kinetic, potential, etc. energies Qin + mi hi = m 2 u 2 (since W ≅ E out = E initial = ke ≅ pe ≅ 0) Combining the two balances: Qin = m 2 (u 2 − hi ) where ( 12 L Evacuated ) (100 kPa ) 0.012 m3 P2V = = 0.0144 kg RT2 0.287 kPa ⋅ m3 /kg ⋅ K (290 K ) h = 290.16 kJ/kg Table A -17 Ti = T2 = 290 K ⎯⎯ ⎯ ⎯ → i ⎯ u2 = 206.91 kJ/kg m2 = ( ) Substituting, Qin = (0.0144 kg)(206.91 - 290.16) kJ/kg = - 1.2 kJ → Qout = 1.2 kJ Note that the negative sign for heat transfer indicates that the assumed direction is wrong. Therefore, we reversed the direction. The entropy generated during this process is determined by applying the entropy balance on an extended system that includes the bottle and its immediate surroundings so that the boundary temperature of the extended system is the temperature of the surroundings at all times. The entropy balance for it can be expressed as S in − S out 1 24 4 3 Net entropy transfer by heat and mass mi s i − + S gen = ΔS system { 123 4 4 Entropy generation Change in entropy Qout + S gen = ΔS tank = m 2 s 2 − m1 s1 Tb,in 0 = m2 s 2 Therefore, the total entropy generated during this process is S gen = − mi s i + m 2 s 2 + Qout = m 2 (s 2 − s i ) Tb,out 0 + Qout Q 1.2 kJ = out = = 0.00415 kJ/K Tb,out Tsurr 290 K The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 S gen , X destroyed = T0 S gen = (290 K)(0.00415 kJ/K) = 1.2 kJ PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 124. 8-124 8-128 A heat engine operates between two tanks filled with air at different temperatures. The maximum work that can be produced and the final temperatures of the tanks are to be determined. Assumptions 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 volume specific heat of air at room temperature is cv = 0.718 kJ/kg.K (Table A-2). Analysis For maximum power production, the entropy generation must be zero. We take the two tanks (the heat source and heat sink) and the heat engine as the system. Noting that the system involves no heat and mass transfer and that the entropy change for cyclic devices is zero, the entropy balance can be expressed as S in − S out 1 24 4 3 Net entropy transfer by heat and mass + S gen 0 = ΔS system 123 1 2 3 4 4 Entropy generation 0 + S gen 0 Change in entropy = ΔS tank,source + ΔS tank,sink + ΔS heat engine 0 AIR 30 kg 900 K QH ΔS tank,source + ΔS tank,sink = 0 ⎛ ⎜ mcv ln T2 + mR ln V 2 ⎜ V1 T1 ⎝ ln 0 ⎞ ⎛ V T ⎟ + ⎜ mcv ln 2 + mR ln 2 ⎟ ⎜ V1 T1 ⎠ source ⎝ 0 ⎞ ⎟ =0 ⎟ ⎠ sink T 2 T2 =0 ⎯ ⎯→ T22 = T1 AT1B T1 A T1B where T1A and T1B are the initial temperatures of the source and the sink, respectively, and T2 is the common final temperature. Therefore, the final temperature of the tanks for maximum power production is W HE QL AIR 30 kg 300 K T2 = T1 AT1B = (900 K)(300 K) = 519.6 K The energy balance Ein − Eout = ΔEsystem for the source and sink can be expressed as follows: Source: −Qsource,out = ΔU = mcv (T2 − T1 A ) → Qsource,out = mcv (T1 A − T2 ) Qsource,out = mcv (T1 A − T2 ) = (30 kg)(0.718 kJ/kg ⋅ K)(900 − 519.6)K = 8193 kJ Sink: Qsink,in = mcv (T2 − T1B ) = (30 kg)(0.718 kJ/kg ⋅ K)(519.6 − 300)K = 4731 kJ Then the work produced in this case becomes W max,out = Q H − Q L = Qsource,out − Qsink,in = 8193 − 4731 = 3463 kJ Therefore, a maximum of 3463 kJ of work can be produced during this process. PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 125. 8-125 8-129 A heat engine operates between two constant-pressure cylinders filled with air at different temperatures. The maximum work that can be produced and the final temperatures of the cylinders are to be determined. Assumptions 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 of air at room temperature is cp = 1.005 kJ/kg.K (Table A-2). Analysis For maximum power production, the entropy generation must be zero. We take the two cylinders (the heat source and heat sink) and the heat engine as the system. Noting that the system involves no heat and mass transfer and that the entropy change for cyclic devices is zero, the entropy balance can be expressed as S in − S out 1 24 4 3 Net entropy transfer by heat and mass + S gen 0 = ΔS system 123 1 2 3 4 4 Entropy generation 0 + S gen 0 Change in entropy = ΔS cylinder,source + ΔS cylinder,sink + ΔS heat engine 0 AIR 30 kg 900 K QH ΔS cylinder,source + ΔS cylinder,sink = 0 ⎛ ⎜ mc ln T2 − mR ln P2 ⎜ p T1 P1 ⎝ ln 0 ⎞ ⎛ T P ⎟ + 0 + ⎜ mc p ln 2 − mR ln 2 ⎟ ⎜ T1 P1 ⎠ source ⎝ 0 ⎞ ⎟ =0 ⎟ ⎠ sink T2 T2 =0 ⎯ ⎯→ T22 = T1 AT1B T1 A T1B where T1A and T1B are the initial temperatures of the source and the sink, respectively, and T2 is the common final temperature. Therefore, the final temperature of the tanks for maximum power production is W HE QL AIR 30 kg 300 K T2 = T1 AT1B = (900 K)(300 K) = 519.6 K The energy balance Ein − Eout = ΔEsystem for the source and sink can be expressed as follows: Source: −Qsource,out + Wb,in = ΔU → Qsource,out = ΔH = mc p (T1 A − T2 ) Qsource,out = mc p (T1 A − T2 ) = (30 kg)(1.005 kJ/kg ⋅ K)(900 − 519.6)K = 11,469 kJ Sink: Qsink,in − Wb,out = ΔU → Qsink,in = ΔH = mc p (T2 − T1 A ) Qsink,in = mc p (T2 − T1B ) = (30 kg)(1.005 kJ/kg ⋅ K)(519.6 − 300)K = 6621 kJ Then the work produced becomes Wmax,out = Q H − Q L = Qsource,out − Qsink,in = 11,469 − 6621 = 4847 kJ Therefore, a maximum of 4847 kJ of work can be produced during this process PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 126. 8-126 8-130 A heat engine operates between a nitrogen tank and an argon cylinder at different temperatures. The maximum work that can be produced and the final temperatures are to be determined. Assumptions Nitrogen and argon are ideal gases with constant specific heats at room temperature. Properties The constant volume specific heat of nitrogen at room temperature is cv = 0.743 kJ/kg.K. The constant pressure specific heat of argon at room temperature is cp = 0.5203 kJ/kg.K (Table A-2). Analysis For maximum power production, the entropy generation must be zero. We take the tank, the cylinder (the heat source and the heat sink) and the heat engine as the system. Noting that the system involves no heat and mass transfer and that the entropy change for cyclic devices is zero, the entropy balance can be expressed as S in − S out 1 24 4 3 Net entropy transfer by heat and mass + S gen 0 = ΔS system 123 1 2 3 4 4 Entropy generation 0 + S gen 0 Change in entropy = ΔS tank,source + ΔS cylinder,sink + ΔS heat engine 0 N2 20 kg 1000 K QH (ΔS ) source + (ΔS ) sink = 0 ⎛ ⎜ mcv ln T2 − mR ln V 2 ⎜ T1 V1 ⎝ 0 ⎞ ⎛ T P ⎟ + 0 + ⎜ mc p ln 2 − mR ln 2 ⎟ ⎜ T1 P1 ⎠ source ⎝ 0 ⎞ ⎟ =0 ⎟ ⎠ sink Substituting, T2 T2 (20 kg)(0.743 kJ / kg ⋅ K) ln + (10 kg)(0.5203 kJ / kg ⋅ K) ln =0 1000 K 300 K Solving for T2 yields T2 = 731.8 K W HE QL Ar 10 kg 300 K where T2 is the common final temperature of the tanks for maximum power production. The energy balance E in − E out = ΔE system for the source and sink can be expressed as follows: Source: −Qsource,out = ΔU = mcv (T2 − T1 A ) → Qsource,out = mcv (T1 A − T2 ) Qsource,out = mcv (T1 A − T2 ) = (20 kg)(0.743 kJ/kg ⋅ K)(1000 − 731.8)K = 3985 kJ Sink: Qsink,in − W b,out = ΔU → Qsink,in = ΔH = mc p (T2 − T1 A ) Qsink,in = mcv (T2 − T1 A ) = (10 kg)(0.5203 kJ/kg ⋅ K)(731.8 − 300)K = 2247 kJ Then the work produced becomes Wmax,out = Q H − Q L = Qsource,out − Qsink,in = 3985 − 2247 = 1739 kJ Therefore, a maximum of 1739 kJ of work can be produced during this process PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 127. 8-127 8-131 A rigid tank containing nitrogen is considered. Heat is now transferred to the nitrogen from a reservoir and nitrogen is allowed to escape until the mass of nitrogen becomes one-half of its initial mass. The change in the nitrogen's work potential is to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process. 2 Kinetic and potential energies are negligible. 3 There are no work interactions involved. 4 Nitrogen is an ideal gas with constant specific heats. Properties The properties of nitrogen at room temperature are cp = 1.039 kJ/kg⋅K, cv = 0.743 kJ/kg⋅K, and R = 0.2968 kJ/kg⋅K (Table A-2a). Analysis The initial and final masses in the tank are m1 = (1000 kPa )(0.100 m 3 ) PV = = 1.150 kg RT1 (0.2968 kPa ⋅ m 3 /kg ⋅ K )(293 K ) m 2 = me = m1 1.150 kg = = 0.575 kg 2 2 Nitrogen 100 L 1000 kPa 20°C Q The final temperature in the tank is T2 = (1000 kPa )(0.100 m 3 ) PV = = 586 K m 2 R (0.575 kg)(0.2968 kPa ⋅ m 3 /kg ⋅ K ) me 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 and internal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as Mass balance: min − m out = Δmsystem → m e = m 2 Energy balance: E in − E out 1 24 4 3 = Net energy transfer by heat, work, and mass ΔE system 1 24 4 3 Change in internal, kinetic, potential, etc. energies Qin − m e he = m 2 u 2 − m1u1 Qout = m e he + m 2 u 2 − m1u1 Using the average of the initial and final temperatures for the exiting nitrogen, Te = 0.5(T1 + T2 ) = 0.5((293 + 586) = 439.5 K this energy balance equation becomes Qout = me he + m 2 u 2 − m1u1 = m e c p Te + m 2 cv T2 − m1cv T1 = (0.575)(1.039)(439.5) + (0.575)(0.743)(586) − (1.150)(0.743)(293) = 262.6 kJ The work potential associated with this process is equal to the exergy destroyed during the process. The exergy destruction during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 S gen . The entropy generation Sgen in this case is determined from an entropy balance on the system: PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 128. 8-128 S in − S out 1 24 4 3 Net entropy transfer by heat and mass + S gen = ΔS system { 123 4 4 Entropy generation Change in entropy Qin − m e s e + S gen = ΔS tank = m 2 s 2 − m1 s1 TR S gen = m 2 s 2 − m1 s1 + m e s e − Qin TR Noting that pressures are same, rearranging and substituting gives S gen = m 2 s 2 − m1 s1 + m e s e − Qin TR = m 2 c p ln T2 − m1c p ln T1 + m e c p ln Te − Qin TR = (0.575)(1.039) ln(586) − (1.150)(1.039) ln(293) + (0.575)(1.039) ln(439.5) − 262.6 673 = 0.2661 kJ/K Then, W rev = X destroyed = T0 S gen = (293 K)(0.2661 kJ/K) = 78.0 kJ Alternative More Accurate Solution This problem may also be solved by considering the variation of gas temperature at the outlet of the tank. The mass and energy balances are dm dt & d (mu ) − h dm = c v d (mT ) − c T dm Q= p dt dt dt dt & me = − Combining these expressions and replacing T in the last term gives d (mT ) c p PV dm & − Q = cv dt Rm dt Integrating this over the time required to release one-half the mass produces Q = c v (m 2 T2 − m1T1 ) − c p PV R ln m2 m1 The reduced combined first and second law becomes &⎛ T & W rev = Q⎜1 − 0 ⎜ T R ⎝ ⎞ d (U − T0 S ) dm ⎟− + ( h − T0 s ) ⎟ dt dt ⎠ when the mass balance is substituted and the entropy generation is set to zero (for maximum work production). Expanding the system time derivative gives PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 129. 8-129 &⎛ T & W rev = Q⎜1 − 0 ⎜ T R ⎝ &⎛ T = Q⎜1 − 0 ⎜ T R ⎝ &⎛ T = Q ⎜1 − 0 ⎜ T R ⎝ ⎞ d ( mu − T0 ms) dm ⎟− + ( h − T0 s) ⎟ dt dt ⎠ ⎞ d ( mu ) ds dm dm ⎟− + T0 m + T0 s + ( h − T0 s ) ⎟ dt dt dt dt ⎠ ⎞ d (mu ) T dh dm ⎟− +h +m 0 ⎟ dt dt T dt ⎠ & Substituting Q from the first law, dm ⎤⎛ T0 ⎡ d ( mu ) & ⎜1 − −h W rev = ⎢ dt ⎥⎜ TR ⎣ dt ⎦⎝ ⎞ ⎡ d ( mu ) T0 dh dm ⎤ ⎟−⎢ −h ⎥ + m T dt ⎟ dt ⎦ ⎠ ⎣ dt T ⎡ d ( mu ) dm dh ⎤ =− 0 ⎢ −h −m ⎥ TR ⎣ dt dt dt ⎦ =− T0 TR dm dT ⎤ ⎡ d (mT ) ⎢c v dt − c p T dt − mc p dt ⎥ ⎣ ⎦ At any time, T= PV mR which further reduces this result to ⎛ c p dT R dP ⎞ T PV dm & ⎟ W rev = 0 c p + T0 m⎜ ⎜ T dt − P dt ⎟ TR mR dt ⎝ ⎠ When this integrated over the time to complete the process, the result is c p PV T0 c p PV m ln 2 + T0 TR R m1 R ⎛ 1 1 ⎞ ⎜ − ⎟ ⎜T T ⎟ 2 ⎠ ⎝ 1 (1.039)(1000)(0.100) ⎛ 1 293 (1.039)(1000)(0.100) 1 1 ⎞ = − ln + (293) ⎜ ⎟ 673 0.2968 2 0.2968 ⎝ 293 586 ⎠ = 69.4 kJ W rev = PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 130. 8-130 8-132 A rigid tank containing nitrogen is considered. Nitrogen is allowed to escape until the mass of nitrogen becomes one-half of its initial mass. The change in the nitrogen's work potential is to be determined. Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process. 2 Kinetic and potential energies are negligible. 3 There are no work interactions involved. 4 Nitrogen is an ideal gas with constant specific heats. Properties The properties of nitrogen at room temperature are cp = 1.039 kJ/kg⋅K, cv = 0.743 kJ/kg⋅K, k = 1.4, and R = 0.2968 kJ/kg⋅K (Table A-2a). Nitrogen 100 L 1000 kPa 20°C Analysis The initial and final masses in the tank are m1 = (1000 kPa )(0.100 m 3 ) PV = = 1.150 kg RT1 (0.2968 kPa ⋅ m 3 /kg ⋅ K )(293 K ) m 2 = me = m1 1.150 kg = = 0.575 kg 2 2 me 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 and internal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as Mass balance: min − m out = Δmsystem → m e = m 2 Energy balance: E in − E out 1 24 4 3 = Net energy transfer by heat, work, and mass ΔE system 1 24 4 3 Change in internal, kinetic, potential, etc. energies − m e he = m 2 u 2 − m1u1 Using the average of the initial and final temperatures for the exiting nitrogen, this energy balance equation becomes − m e he = m 2 u 2 − m1u1 − m e c p Te = m 2 cv T2 − m1 cv T1 − (0.575)(1.039)(0.5)(293 + T2 ) = (0.575)(0.743)T2 − (1.150)(0.743)(293) Solving for the final temperature, we get T2 = 224.3 K The final pressure in the tank is P2 = m 2 RT2 V = (0.575 kg)(0.2968 kPa ⋅ m 3 /kg ⋅ K )(224.3 K) 0.100 m 3 = 382.8 kPa The average temperature and pressure for the exiting nitrogen is Te = 0.5(T1 + T2 ) = 0.5(293 + 224.3) = 258.7 K Pe = 0.5(T1 + T2 ) = 0.5(1000 + 382.8) = 691.4 kPa The work potential associated with this process is equal to the exergy destroyed during the process. The exergy destruction during a process can be determined from an exergy balance or directly from its PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 131. 8-131 definition X destroyed = T0 S gen . The entropy generation Sgen in this case is determined from an entropy balance on the system: S in − S out 1 24 4 3 Net entropy transfer by heat and mass + S gen = ΔS system { 123 4 4 Entropy generation Change in entropy − m e s e + S gen = ΔS tank = m 2 s 2 − m1 s1 S gen = m 2 s 2 − m1 s1 + m e s e Rearranging and substituting gives S gen = m 2 s 2 − m1 s1 + m e s e = m 2 (c p ln T2 − R ln P2 ) − m1 (c p ln T1 − R ln P1 ) + m e (c p ln Te − R ln Pe ) = (0.575)[1.039 ln(224.3) − (0.2968) ln(382.8)] − (1.15)[1.039 ln(293) − (0.2968) ln(1000)] + (0.575)[1.039 ln(258.7) − (0.2968) ln(691.4)] = 2.2188 − 4.4292 + 2.2032 = −0.007152 kJ/K Then, W rev = X destroyed = T0 S gen = (293 K)(−0.007152 kJ/K) = −2.10 kJ The entropy generation cannot be negative for a thermodynamically possible process. This result is probably due to using average temperature and pressure values for the exiting gas and using constant specific heats for nitrogen. This sensitivity occurs because the entropy generation is very small in this process. Alternative More Accurate Solution This problem may also be solved by considering the variation of gas temperature and pressure at the outlet of the tank. The mass balance in this case is & me = − dm dt which when combined with the reduced first law gives d (mu ) dm =h dt dt Using the specific heats and the ideal gas equation of state reduces this to cv V dP R dt = c pT dm dt which upon rearrangement and an additional use of ideal gas equation of state becomes 1 dP c p 1 dm = P dt c v m dt When this is integrated, the result is ⎛m P2 = P1 ⎜ 2 ⎜m ⎝ 1 k ⎞ ⎛1⎞ ⎟ = 1000⎜ ⎟ ⎟ ⎝2⎠ ⎠ 1.4 = 378.9 kPa The final temperature is then PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 132. 8-132 T2 = P2V (378.9 kPa )(0.100 m 3 ) = = 222.0 K m 2 R (0.575 kg)(0.2968 kPa ⋅ m 3 /kg ⋅ K ) The process is then one of mk = const or P m k −1 = const T The reduced combined first and second law becomes d (U − T0 S ) dm & + ( h − T0 s ) W rev = − dt dt when the mass balance is substituted and the entropy generation is set to zero (for maximum work production). Replacing the enthalpy term with the first law result and canceling the common dU/dt term reduces this to d (ms) dm & − T0 s W rev = T0 dt dt Expanding the first derivative and canceling the common terms further reduces this to ds & W rev = T0 m dt k k Letting a = P1 / m1 and b = T1 / m1 −1 , the pressure and temperature of the nitrogen in the system are related to the mass by P = am k and T = bm k −1 according to the first law. Then, dP = akm k −1 dm and dT = b(k − 1)m k − 2 dm The entropy change relation then becomes ds = c p [ ] dT dP dm −R = (k − 1)c p − Rk T P m Now, multiplying the combined first and second laws by dt and integrating the result gives 2 2 ∫ ∫ [ = T [(k − 1)c − Rk ](m ] W rev = T0 mds = T0 mds (k − 1)c p − Rk dm 1 0 1 p 2 − m1 ) = (293)[(1.4 − 1)(1.039) − (0.2968)(1.4)](0.575 − 1.15) = −0.0135 kJ Once again the entropy generation is negative, which cannot be the case for a thermodynamically possible process. This is probably due to using constant specific heats for nitrogen. This sensitivity occurs because the entropy generation is very small in this process. PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 133. 8-133 8-133 A system consisting of a compressor, a storage tank, and a turbine as shown in the figure is considered. The change in the exergy of the air in the tank and the work required to compress the air as the tank was being filled are to be determined. Assumptions 1 Changes in the kinetic and potential energies are negligible. 4 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are R = 0.287 kPa⋅m3/kg⋅K, cp = 1.005 kJ/kg⋅K, cv = 0.718 kJ/kg⋅K, k = 1.4 (Table A-2a). Analysis The initial mass of air in the tank is minitial = PinitialV (100 kPa)(5 × 10 5 m 3 ) = = 0.5946 × 10 6 kg RTinitial (0.287 kPa ⋅ m 3 /kg ⋅ K)(293 K) and the final mass in the tank is m final = PfinalV (600 kPa)(5 × 10 5 m 3 ) = = 3.568 × 10 6 kg RTfinal (0.287 kPa ⋅ m 3 /kg ⋅ K)(293 K) Since the compressor operates as an isentropic device, ⎛P T2 = T1 ⎜ 2 ⎜P ⎝ 1 ⎞ ⎟ ⎟ ⎠ ( k −1) / k The conservation of mass applied to the tank gives dm & = min dt while the first law gives & d (mu ) − h dm Q= dt dt Employing the ideal gas equation of state and using constant specific heats, expands this result to & Vc dP − c T V dP Q= v p 2 R dt RT dt Using the temperature relation across the compressor and multiplying by dt puts this result in the form ⎛P⎞ Vc & Qdt = v dP − c p T1 ⎜ ⎟ ⎜P ⎟ R ⎝ 1⎠ ( k −1) / k V RT dP When this integrated, it yields (i and f stand for initial and final states) Q= = Vcv R ( P f − Pi ) − ⎡ k c pV ⎢ Pf 2k − 1 R ⎢ ⎣ ⎛ Pf ⎜ ⎜P ⎝ i ⎞ ⎟ ⎟ ⎠ ( k −1) / k ⎤ − Pi ⎥ ⎥ ⎦ (5 × 10 5 )(0.718) (1.005)(5 × 10 5 ) ⎡ ⎛ 600 ⎞ 1.4 (600 − 100) − ⎢600⎜ ⎟ 0.287 2(1.4) − 1 0.287 ⎢ ⎝ 100 ⎠ ⎣ 0.4 / 1.4 ⎤ − 100⎥ ⎥ ⎦ = −6.017 × 10 8 kJ The negative result show that heat is transferred from the tank. Applying the first law to the tank and compressor gives PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 134. 8-134 & & (Q − Wout )dt = d (mu ) − h1 dm which integrates to Q − Wout = (m f u f − m i u i ) − h1 (m f − m i ) Upon rearrangement, Wout = Q + (c p − cv )T (m f − mi ) = −6.017 × 10 8 + (1.005 − 0.718)(293)[(3.568 − 0.5946) × 10 6 ] = −3.516 × 10 8 kJ The negative sign shows that work is done on the compressor. When the combined first and second laws is reduced to fit the compressor and tank system and the mass balance incorporated, the result is &⎛ T & W rev = Q⎜1 − 0 ⎜ T R ⎝ ⎞ d (U − T0 S ) dm ⎟− + ( h − T0 s ) ⎟ dt dt ⎠ which when integrated over the process becomes [ ⎞ ⎟ + mi [(u i − h1 ) − T0 ( s i − s1 )] − m f (u f − h1 ) − T0 (s f − s1 ) ⎟ ⎠ ⎛ 293 ⎞ 6 = −6.017 × 10 8 ⎜1 − ⎟ + 0.5946 × 10 [(0.718 − 1.005)293] ⎝ 293 ⎠ ⎛ T W rev = Q⎜1 − 0 ⎜ T R ⎝ ] 600 ⎤ ⎡ − 3.568 × 10 6 [(0.718 − 1.005)293]⎢(0.718 − 1.005)293 + 293(0.287) ln 100 ⎥ ⎣ ⎦ = −2.876 × 10 8 kJ This is the exergy change of the air stored in the tank. PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 135. 8-135 8-134 The air stored in the tank of the system shown in the figure is released through the isentropic turbine. The work produced and the change in the exergy of the air in the tank are to be determined. Assumptions 1 Changes in the kinetic and potential energies are negligible. 4 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are R = 0.287 kPa⋅m3/kg⋅K, cp = 1.005 kJ/kg⋅K, cv = 0.718 kJ/kg⋅K, k = 1.4 (Table A-2a). Analysis The initial mass of air in the tank is minitial = PinitialV (600 kPa)(5 × 10 5 m 3 ) = 3.568 × 10 6 kg = RTinitial (0.287 kPa ⋅ m 3 /kg ⋅ K)(293 K) and the final mass in the tank is mfinal = (100 kPa)(5 × 105 m3 ) PfinalV = 0.5946 × 106 kg = RTfinal (0.287 kPa ⋅ m3/kg ⋅ K)(293 K) The conservation of mass is dm & = min dt while the first law gives & d (mu ) − h dm Q= dt dt Employing the ideal gas equation of state and using constant specific heats, expands this result to & Vc dP − c T V dP Q= v p R dt RT dt cv − c p dP V = R dt dP = −V dt When this is integrated over the process, the result is (i and f stand for initial and final states) Q = −V ( P f − Pi ) = −5 × 10 5 (100 − 600) = 2.5 × 10 8 kJ Applying the first law to the tank and compressor gives & & (Q − Wout ) dt = d (mu ) − hdm which integrates to PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 136. 8-136 Q − Wout = (m f u f − m i u i ) + h(mi − m f ) − Wout = −Q + m f u f − mi u i + h(mi − m f ) Wout = Q − m f u f + mi u i − h(mi − m f ) = Q − m f c v T + m i c p T − c p T (m i − m f ) = 2.5 × 10 8 − (0.5946 × 10 6 )(0.718)(293) + (3.568 × 10 6 )(1.005)(293) − (1.005)(293)(3.568 × 10 6 − 0.5946 × 10 6 ) = 3.00 × 10 8 kJ This is the work output from the turbine. When the combined first and second laws is reduced to fit the turbine and tank system and the mass balance incorporated, the result is &⎛ T & W rev = Q⎜1 − 0 ⎜ T R ⎝ ⎞ d (U − T0 S ) dm ⎟− + ( h − T0 s ) ⎟ dt dt ⎠ &⎛ T = Q ⎜1 − 0 ⎜ T R ⎝ &⎛ T = Q ⎜1 − 0 ⎜ T R ⎝ ⎞ d (u − T0 s ) dm dm ⎟ − (u − T0 s ) −m + ( h − T0 s ) ⎟ dt dt dt ⎠ ⎞ dm ds ⎟ + (c p − c v )T + mT0 ⎟ dt dt ⎠ &⎛ T = Q ⎜1 − 0 ⎜ T R ⎝ ⎞ T dm ⎟ + (c p − c v )T −V 0 ( P f − Pi ) ⎟ dt T ⎠ where the last step uses entropy change equation. When this is integrated over the process it becomes ⎞ T ⎟ + (c p − c v )T (m f − mi ) −V 0 ( P f − Pi ) ⎟ T ⎠ ⎛ 293 ⎞ 6 5 293 (100 − 600) = 3.00 × 10 8 ⎜1 − ⎟ + (1.005 − 0.718)(293)(0.5946 − 3.568) × 10 − 5 × 10 293 ⎝ 293 ⎠ ⎛ T W rev = Q⎜1 − 0 ⎜ T R ⎝ = 0 − 2.500 × 10 8 − 2.5 × 10 8 = −5.00 × 10 8 kJ This is the exergy change of the air in the storage tank. PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 137. 8-137 8-135 A heat engine operates between a tank and a cylinder filled with air at different temperatures. The maximum work that can be produced and the final temperatures are to be determined. Assumptions Air is an ideal gas with constant specific heats at room temperature. Properties The specific heats of air are cv = 0.718 kJ/kg.K and cp = 1.005 kJ/kg.K (Table A-2). Analysis For maximum power production, the entropy generation must be zero. We take the tank, the cylinder (the heat source and the heat sink) and the heat engine as the system. Noting that the system involves no heat and mass transfer and that the entropy change for cyclic devices is zero, the entropy balance can be expressed as S in − S out 1 24 4 3 Net entropy transfer by heat and mass + S gen 0 = ΔS system 123 1 2 3 4 4 Entropy generation 0 + S gen 0 Change in entropy 0 = ΔS tank,source + ΔS cylinder,sink + ΔS heat engine Air 20 kg 800 K (ΔS ) source + (ΔS ) sink = 0 ⎛ ⎜ mc ln T2 − mR ln V 2 ⎜ v T1 V1 ⎝ ln 0 QH ⎞ ⎛ P T ⎟ + 0 + ⎜ mc p ln 2 − mR ln 2 ⎟ ⎜ P1 T1 ⎠ source ⎝ T2 c p T T ⎛T ln 2 = 0 ⎯ + ⎯→ 2 ⎜ 2 T1 A cv T1B T1 A ⎜ T1B ⎝ k 0 ( ⎞ ⎟ =0 ⎟ ⎠ sink ⎞ ⎟ =1 ⎯ ⎯→ T2 = T1 AT1k B ⎟ ⎠ ) 1 /( k +1) where T1A and T1B are the initial temperatures of the source and the sink, respectively, and T2 is the common final temperature. Therefore, the final temperature of the tanks for maximum power production is ( T2 = (800 K)(290 K)1.4 ) 1 2.4 W HE QL Air 20 kg 290 K = 442.6 K Source: −Qsource,out = ΔU = mcv (T2 − T1 A ) → Qsource,out = mcv (T1 A − T2 ) Qsource,out = mcv (T1 A − T2 ) = (20 kg)(0.718 kJ/kg ⋅ K)(800 − 442.6)K = 5132 kJ Sink: Qsink,in − W b,out = ΔU → Qsink,in = ΔH = mc p (T2 − T1 A ) Qsink,in = mcv (T2 − T1 A ) = (20 kg)(1.005 kJ/kg ⋅ K)(442.6 − 290)K = 3068 kJ Then the work produced becomes Wmax,out = Q H − Q L = Qsource,out − Qsink,in = 5132 − 3068 = 2064 kJ Therefore, a maximum of 2064 kJ of work can be produced during this process. PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 138. 8-138 8-136 Using an incompressible substance as an example, it is to be demonstrated if closed system and flow exergies can be negative. Analysis The availability of a closed system cannot be negative. However, the flow availability can be negative at low pressures. A closed system has zero availability at dead state, and positive availability at any other state since we can always produce work when there is a pressure or temperature differential. To see that the flow availability can be negative, consider an incompressible substance. The flow availability can be written as ψ = h − h0 + T0 (s − s0 ) = (u − u0 ) + v ( P − P0 ) + T0 (s − s0 ) = ξ + v ( P − P0 ) The closed system availability ξ is always positive or zero, and the flow availability can be negative when P << P0. 8-137 A relation for the second-law efficiency of a heat engine operating between a heat source and a heat sink at specified temperatures is to be obtained. Analysis The second-law efficiency is defined as the ratio of the availability recovered to availability supplied during a process. The work W produced is the availability recovered. The decrease in the availability of the heat supplied QH is the availability supplied or invested. Source TH QH Therefore, η II = W ⎛ T0 ⎞ ⎛ T ⎜1 − 0 ⎟Q H − ⎜1 − ⎜ T ⎜ T ⎟ H ⎠ L ⎝ ⎝ W HE ⎞ ⎟(Q H − W ) ⎟ ⎠ Note that the first term in the denominator is the availability of heat supplied to the heat engine whereas the second term is the availability of the heat rejected by the heat engine. The difference between the two is the availability consumed during the process. QL TL Sink PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 139. 8-139 8-138E Large brass plates are heated in an oven at a rate of 300/min. The rate of heat transfer to the plates in the oven and the rate of exergy destruction associated with this heat transfer process are to be determined. Assumptions 1 The thermal properties of the plates are constant. 2 The changes in kinetic and potential energies are negligible. 3 The environment temperature is 75°F. Properties The density and specific heat of the brass are given to be ρ = 532.5 lbm/ft3 and cp = 0.091 Btu/lbm.°F. Analysis We take the plate to be the system. The energy balance for this closed system can be expressed as E in − E out 1 24 4 3 = Net energy transfer by heat, work, and mass ΔE system 1 24 4 3 Change in internal, kinetic, potential, etc. energies Qin = ΔU plate = m(u 2 − u1 ) = mc(T2 − T1 ) The mass of each plate and the amount of heat transfer to each plate is m = ρV = ρLA = (532.5 lbm/ft 3 )[(1.2 / 12 ft )(2 ft)(2 ft)] = 213 lbm Qin = mc(T2 − T1 ) = (213 lbm/plate)(0.091 Btu/lbm.°F)(1000 − 75)°F = 17,930 Btu/plate Then the total rate of heat transfer to the plates becomes & & Q total = n plate Qin, per plate = (300 plates/min) × (17,930 Btu/plate) = 5,379,000 Btu/min = 89,650 Btu/s We again take a single plate as the system. The entropy generated during this process can be determined by applying an entropy balance on an extended system that includes the plate and its immediate surroundings so that the boundary temperature of the extended system is at 1300°F at all times: S in − S out 1 24 4 3 Net entropy transfer by heat and mass + S gen = ΔS system { 123 4 4 Entropy generation Change in entropy Q Qin + S gen = ΔS system → S gen = − in + ΔS system Tb Tb where ΔS system = m( s 2 − s1 ) = mc avg ln T2 (1000 + 460) R = (213 lbm)(0.091 Btu/lbm.R) ln = 19.46 Btu/R T1 (75 + 460) R Substituting, S gen = − Qin 17,930 Btu + ΔS system = − + 19.46 Btu/R = 9.272 Btu/R (per plate) Tb 1300 + 460 R Then the rate of entropy generation becomes & & S gen = S gen n ball = (9.272 Btu/R ⋅ plate)(300 plates/min) = 2781 Btu/min.R = 46.35 Btu/s.R The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 S gen , & & X destroyed = T0 S gen = (535 R)(46.35 Btu/s.R) = 24,797 Btu/s PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 140. 8-140 8-139 Long cylindrical steel rods are heat-treated in an oven. The rate of heat transfer to the rods in the oven and the rate of exergy destruction associated with this heat transfer process are to be determined. Assumptions 1 The thermal properties of the rods are constant. 2 The changes in kinetic and potential energies are negligible. 3 The environment temperature is 30°C. Properties The density and specific heat of the steel rods are given to be ρ = 7833 kg/m3 and cp = 0.465 kJ/kg.°C. Analysis Noting that the rods enter the oven at a velocity of 3 m/min and exit at the same velocity, we can say that a 3-m long section of the rod is heated in the oven in 1 min. Then the mass of the rod heated in 1 minute is m = ρV = ρLA = ρL(πD 2 / 4 ) = (7833 kg / m 3 )(3 m)[π (01 m) 2 / 4 ] = 184.6 kg . We take the 3-m section of the rod in the oven as the system. The energy balance for this closed system can be expressed as E in − E out 1 24 4 3 = Net energy transfer by heat, work, and mass ΔE system 1 24 4 3 Change in internal, kinetic, potential, etc. energies Qin = ΔU rod = m(u 2 − u1 ) = mc(T2 − T1 ) Substituting, Qin = mc(T2 − T1 ) = (184.6 kg )(0.465 kJ/kg.°C)(700 − 30)°C = 57,512 kJ Noting that this much heat is transferred in 1 min, the rate of heat transfer to the rod becomes & Qin = Qin / Δt = (57,512 kJ)/(1 min) = 57,512 kJ/min = 958.5 kW We again take the 3-m long section of the rod as the system The entropy generated during this process can be determined by applying an entropy balance on an extended system that includes the rod and its immediate surroundings so that the boundary temperature of the extended system is at 900°C at all times: S in − S out 1 24 4 3 Net entropy transfer by heat and mass + S gen = ΔS system { 123 4 4 Entropy generation Change in entropy Q Qin + S gen = ΔS system → S gen = − in + ΔS system Tb Tb where ΔS system = m( s 2 − s1 ) = mc avg ln T2 700 + 273 = (184.6 kg)(0.465 kJ/kg.K) ln = 100.1 kJ/K T1 30 + 273 Substituting, S gen = − Qin 57,512 kJ + ΔS system = − + 100.1 kJ/K = 51.1 kJ/K Tb (900 + 273) R Noting that this much entropy is generated in 1 min, the rate of entropy generation becomes & S gen = S gen Δt = 51.1 kJ/K = 51.1 kJ/min.K = 0.852 kW/K 1 min The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 S gen , & & X destroyed = T0 S gen = (298 K)(0.852 kW/K) = 254 kW PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 141. 8-141 8-140 Steam is condensed by cooling water in the condenser of a power plant. The rate of condensation of steam and the rate of exergy destruction are to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Fluid properties are constant. Properties The enthalpy and entropy of vaporization of water at 60°C are hfg =2357.7 kJ/kg and sfg= 7.0769 kJ/kg.K (Table A-4). The specific heat of water at room temperature is cp = 4.18 kJ/kg.°C (Table A-3). Analysis (a) We take the cold water tubes as the system, which is a control volume. The energy balance for this steady-flow system can be expressed in the rate form as & & E in − E out 1 24 4 3 = Rate of net energy transfer by heat, work, and mass & ΔE system 0 (steady) 1442443 4 =0 Steam 60°C Rate of change in internal, kinetic, potential, etc. energies & & E in = E out 25°C & & & Qin + mh1 = mh2 (since Δke ≅ Δpe ≅ 0) & = mC (T − T ) Qin & p 2 1 Then the heat transfer rate to the cooling water in the condenser becomes & & Q = [mC (T − T )] p out in 15°C cooling water = (140 kg/s)(4.18 kJ/kg.°C)(25°C − 15°C) = 5852 kJ/s Water The rate of condensation of steam is determined to be & Q 5852 kJ/s & & & ⎯→ msteam = = = 2.482 kg/s Q = (mh fg ) steam ⎯ h fg 2357.7 kJ/kg 60°C (b) The rate of entropy generation within the condenser during this process can be determined by applying the rate form of the entropy balance on the entire condenser. Noting that the condenser is well-insulated and thus heat transfer is negligible, the entropy balance for this steady-flow system can be expressed as & & S in − S out 1 24 4 3 + Rate of net entropy transfer by heat and mass & S gen { Rate of entropy generation & = ΔS system 0 (steady) 1442443 Rate of change of entropy & & & & & m1 s1 + m3 s 3 − m 2 s 2 − m 4 s 4 + S gen = 0 (since Q = 0) & & & & & m water s1 + msteam s 3 − m water s 2 − msteam s 4 + S gen = 0 & & & S gen = m water ( s 2 − s1 ) + msteam ( s 4 − s 3 ) Noting that water is an incompressible substance and steam changes from saturated vapor to saturated liquid, the rate of entropy generation is determined to be T T & & & & & S gen = m water c p ln 2 + msteam ( s f − s g ) = m water c p ln 2 − msteam s fg T1 T1 = (140 kg/s)(4.18 kJ/kg.K)ln 25 + 273 − (2.482 kg/s)(7.0769 kJ/kg.K) = 2.409 kW/K 15 + 273 Then the exergy destroyed can be determined directly from its definition X destroyed = T0 S gen to be & & X destroyed = T0 S gen = (288 K)(2.409 kW/K) = 694 kW PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 142. 8-142 8-141 Water is heated in a heat exchanger by geothermal water. The rate of heat transfer to the water and the rate of exergy destruction within the heat exchanger are to be determined. Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Fluid properties are constant. 5 The environment temperature is 25°C. Properties The specific heats of water and geothermal fluid are given to be 4.18 and 4.31 kJ/kg.°C, respectively. Analysis (a) We take the cold water tubes as the system, which is a control volume. The energy balance for this steady-flow system can be expressed in the rate form as & & E in − E out 1 24 4 3 = Rate of net energy transfer by heat, work, and mass & ΔE system 0 (steady) 1442443 4 60°C Brine 140°C =0 Rate of change in internal, kinetic, potential, etc. energies & & E in = E out & & & Qin + mh1 = mh2 (since Δke ≅ Δpe ≅ 0) & & Qin = mc p (T2 − T1 ) Water 25°C Then the rate of heat transfer to the cold water in the heat exchanger becomes & & Qin, water = [mc p (Tout − Tin )] water = (0.4 kg/s)(4.18 kJ/kg.°C)(60°C − 25°C) = 58.52 kW Noting that heat transfer to the cold water is equal to the heat loss from the geothermal water, the outlet temperature of the geothermal water is determined from & Q 58.52 kW & & ⎯→ = 94.7°C Qout = [mc p (Tin − Tout )] geo ⎯ Tout = Tin − out = 140°C − & mc p (0.3 kg/s)(4.31 kJ/kg.°C) (b) The rate of entropy generation within the heat exchanger is determined by applying the rate form of the entropy balance on the entire heat exchanger: & & S in − S out 1 24 4 3 + Rate of net entropy transfer by heat and mass & S gen { Rate of entropy generation & = ΔS system 0 (steady) 1442443 Rate of change of entropy & & & & & m1 s1 + m3 s 3 − m 2 s 2 − m 4 s 4 + S gen = 0 (since Q = 0) & & & & & m water s1 + m geo s 3 − m water s 2 − m geo s 4 + S gen = 0 & & & S gen = m water ( s 2 − s1 ) + m geo ( s 4 − s 3 ) Noting that both fresh and geothermal water are incompressible substances, the rate of entropy generation is determined to be T T & & & S gen = m water c p ln 2 + m geo c p ln 4 T1 T3 94.7 + 273 60 + 273 + (0.3 kg/s)(4.31 kJ/kg.K)ln = 0.0356 kW/K 140 + 273 25 + 273 The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 S gen , = (0.4 kg/s)(4.18 kJ/kg.K)ln & & X destroyed = T0 S gen = (298 K)(0.0356 kW/K) = 10.61 kW PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 143. 8-143 8-142 A regenerator is considered to save heat during the cooling of milk in a dairy plant. The amounts of fuel and money such a generator will save per year and the rate of exergy destruction within the regenerator are to be determined. Assumptions 1 Steady operating conditions exist. 2 The properties of the milk are constant. 5 The environment temperature is 18°C. Properties The average density and specific heat of milk can be taken to be ρmilk ≅ ρ water = 1 kg/L and cp,milk= 3.79 kJ/kg.°C (Table A-3). Analysis The mass flow rate of the milk is & m milk = ρV&milk = (1 kg/L)(12 L/s) = 12 kg/s = 43,200 kg/h Taking the pasteurizing section as the system, 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 → E in = E out 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 & & & Qin + mh1 = mh2 (since Δke ≅ Δpe ≅ 0) & = m c (T − T ) Qin & milk p 2 1 Therefore, to heat the milk from 4 to 72°C as being done currently, heat must be transferred to the milk at a rate of & & Qcurrent = [mc p (Tpasturization − Trefrigeration )] milk = (12 kg/s)(3.79 kJ/kg.°C)(72 − 4)°C = 3093 kJ/s The 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 / s Noting that the boiler has an efficiency of ηboiler = 0.82, the energy savings above correspond to fuel savings of & Q (2536 kJ / s) (1therm) Fuel Saved = saved = = 0.02931therm / s (0.82) (105,500 kJ) η boiler Noting that 1 year = 365×24=8760 h and unit cost of natural gas is \$1.04/therm, the annual fuel and money savings will be Fuel Saved = (0.02931 therms/s)(8760×3600 s) = 924,450 therms/yr Money saved = (Fuel saved)(Unit cost of fuel) = (924,450 therm/yr)(\$1.04/therm) = \$961,430/y r The rate of entropy generation during this process is determined by applying the rate form of the entropy balance on an extended system that includes the regenerator and the immediate surroundings so that the boundary temperature is the surroundings temperature, which we take to be the cold water temperature of 18°C.: & & & & & & & S in − S out + S gen = ΔS system 0 (steady) → S gen = S out − S in 1 24 4 3 { 1442443 Rate of net entropy transfer by heat and mass Rate of entropy generation Rate of change of entropy Disregarding entropy transfer associated with fuel flow, the only significant difference between the two cases is the reduction is the entropy transfer to water due to the reduction in heat transfer to water, and is determined to be & & Qout, reduction Qsaved 2536 kJ/s & & = = = 8.715 kW/K S gen, reduction = S out, reduction = Tsurr Tsurr 18 + 273 & S =S Δt = (8.715 kJ/s.K)(8760 × 3600 s/year) = 2.75 × 10 8 kJ/K (per year) gen, reduction gen, reduction The exergy destroyed during a process can be determined from an exergy balance or directly from its definition X destroyed = T0 S gen , X destroyed, reduction = T0 Sgen, reduction = (291 K)(2.75 × 108 kJ/K) = 8.00 × 1010 kJ (per year) PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 144. 8-144 8-143 Exhaust gases are expanded in a turbine, which is not well-insulated. Tha actual and reversible power outputs, the exergy destroyed, and the second-law efficiency are to be determined. Assumptions 1 Steady operating conditions exist. 2 Potential energy change is negligible. 3 Air is an ideal gas with constant specific heats. Exh. gas Properties The gas constant of air is R = 0.287 kJ/kg.K and the 750°C specific heat of air at the average temperature of (750+630)/2 = 1.2 MPa 690ºC is cp = 1.134 kJ/kg.ºC (Table A-2). Analysis (a) The enthalpy and entropy changes of air across the turbine are Δh = c p (T1 − T2 ) = (1.134 kJ/kg.°C)(750 − 630)°C = 136.08 kJ/kg Δs = c p ln P T1 − R ln 1 P2 T2 (750 + 273) K 1200 kPa − (0.287 kJ/kg.K) ln (630 + 273) K 500 kPa = −0.005354 kJ/kg.K Turbine Q 630°C 500 kPa = (1.134 kJ/kg.K)ln The actual and reversible power outputs from the turbine are & & & Wa = mΔh − Qout = (3.4 kg/s)(136.08 kJ/kg) − 30 kW = 432.7 kW & & Wrev = m(Δh − T0 Δs ) = (3.4 kg/s)(136.08 kJ/kg) − (25 + 273 K)(−0.005354 kJ/kg.K) = 516.9 kW (b) The exergy destroyed in the turbine is & & & X dest = W rev − Wa = 516.9 − 432.7 = 84.2 kW (c) The second-law efficiency is η II = & Wa 432.7 kW = = 0.837 & W rev 516.9 kW PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 145. 8-145 8-144 Refrigerant-134a is compressed in an adiabatic compressor, whose second-law efficiency is given. The actual work input, the isentropic efficiency, and the exergy destruction are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) The properties of the refrigerant at the inlet of the compressor are (Tables A-11 through A-13) Tsat@160 kPa = −15.60°C P1 = 160 kPa ⎫ h1 = 243.60 kJ/kg ⎬ T1 = (−15.60 + 3)°C⎭ s1 = 0.95153 kJ/kg.K The enthalpy at the exit for if the process was isentropic is P2 = 1 MPa ⎫ ⎬h2 s = 282.41 kJ/kg s 2 = s1 = 0.95153 kJ/kg.K ⎭ The expressions for actual and reversible works are 1 MPa Compressor R-134a 160 kPa wa = h2 − h1 = (h2 − 243.60)kJ/kg wrev = h2 − h1 − T0 ( s 2 − s1 ) = (h2 − 243.60)kJ/kg − (25 + 273 K)(s 2 − 0.95153)kJ/kg.K Substituting these into the expression for the second-law efficiency η II = wrev h − 243.60 − (298)( s 2 − 0.95153) ⎯ ⎯→ 0.80 = 2 wa h2 − 243.60 The exit pressure is given (1 MPa). We need one more property to fix the exit state. By a trial-error approach or using EES, we obtain the exit temperature to be 60ºC. The corresponding enthalpy and entropy values satisfying this equation are h2 = 293.36 kJ/kg s 2 = 0.98492 kJ/kg.K Then, wa = h2 − h1 = 293.36 − 243.60 = 49.76kJ/kg wrev = h2 − h1 − T0 ( s2 − s1 ) = (293.36 − 243.60)kJ/kg − (25 + 273 K)(0.98492 − 0.9515)kJ/kg ⋅ K = 39.81 kJ/kg (b) The isentropic efficiency is determined from its definition ηs = h2s − h1 (282.41 − 243.60)kJ/kg = 0.780 = h2 − h1 (293.36 − 243.60)kJ/kg (b) The exergy destroyed in the compressor is x dest = wa − wrev = 49.76 − 39.81 = 9.95 kJ/kg PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 146. 8-146 8-145 The isentropic efficiency of a water pump is specified. The actual power output, the rate of frictional heating, the exergy destruction, and the second-law efficiency are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) Using saturated liquid properties at the given temperature for the inlet state (Table A-4) h = 125.82 kJ/kg T1 = 30°C⎫ s1 = 0.4367 kJ/kg.K ⎬ 1 x1 = 0 ⎭ v = 0.001004 m 3 /kg 1 Water 100 kPa 30°C 1.35 kg/s 4 MPa PUMP The power input if the process was isentropic is & & Ws = mv1 ( P2 − P ) = (1.35 kg/s)(0.001004 m3/kg)(4000 − 100)kPa = 5.288 kW 1 Given the isentropic efficiency, the actual power may be determined to be & W 5.288 kW & Wa = s = = 7.554 kW ηs 0.70 (b) The difference between the actual and isentropic works is the frictional heating in the pump & & & Qfrictional = W a − W s = 7.554 − 5.288 = 2.266 kW (c) The enthalpy at the exit of the pump for the actual process can be determined from & & Wa = m(h2 − h1 ) ⎯ 7.554 kW = (1.35 kg/s)(h2 − 125.82)kJ/kg ⎯ h2 = 131.42 kJ/kg ⎯→ ⎯→ The entropy at the exit is P2 = 4 MPa ⎫ ⎬s 2 = 0.4423 kJ/kg.K h2 = 131.42 kJ/kg ⎭ The reversible power and the exergy destruction are & & Wrev = m[h2 − h1 − T0 ( s2 − s1 )] = (1.35 kg/s)[(131.42 − 125.82)kJ/kg − (20 + 273 K)(0.4423 − 0.4367)kJ/kg.K ] = 5.362 kW & & & X dest = Wa − W rev = 7.554 − 5.362 = 2.193 kW (d) The second-law efficiency is η II = & W rev 5.362 kW = = 0.710 & 7.554 kW Wa PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 147. 8-147 8-146 Argon gas is expanded adiabatically in an expansion valve. The exergy of argon at the inlet, the exergy destruction, and the second-law efficiency are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are zero. 3 Argon is an ideal gas with constant specific heats. Properties The properties of argon gas are R = 0.2081 kJ/kg.K, cp = 0.5203 kJ/kg.ºC (Table A-2). Analysis (a) The exergy of the argon at the inlet is x1 = h1 − h0 − T0 ( s1 − s 0 ) Argon 3.5 MPa 100°C 500 kPa ⎡ P ⎤ T = c p (T1 − T0 ) − T0 ⎢c p ln 1 − R ln 1 ⎥ P0 ⎦ T0 ⎣ 373 K 3500 kPa ⎤ ⎡ = (0.5203 kJ/kg.K)(100 − 25)°C − (298 K) ⎢(0.5203 kJ/kg.K)ln − (0.2081 kJ/kg.K)ln 298 K 100 kPa ⎥ ⎣ ⎦ = 224.7 kJ/kg (b) Noting that the temperature remains constant in a throttling process, the exergy destruction is determined from x dest = T0 s gen = T0 ( s 2 − s1 ) ⎛ P ⎞ ⎡ ⎛ 500 kPa ⎞⎤ = T0 ⎜ − R ln 1 ⎟ = (298 K) ⎢− (0.2081 kJ/kg.K)ln⎜ ⎟⎥ ⎟ ⎜ P0 ⎠ ⎝ 3500 kPa ⎠⎦ ⎣ ⎝ = 120.7 kJ/kg (c) The second-law efficiency is η II = x1 − x dest (224.7 − 120.7)kJ/kg = = 0.463 224.7 kJ/kg x1 PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 148. 8-148 8-147 Heat is lost from the air flowing in a diffuser. The exit temperature, the rate of exergy destruction, and the second law efficiency are to be determined. Assumptions 1 Steady operating conditions exist. 2 Potential energy change is negligible. 3 Nitrogen is an ideal gas with variable specific heats. Properties The gas constant of nitrogen is R = 0.2968 kJ/kg.K. Analysis (a) For this problem, we use the properties from EES software. Remember that for an ideal gas, enthalpy is a function of temperature only whereas entropy is functions of both temperature and pressure. At the inlet of the diffuser and at the dead state, we have T1 = 15°C = 423 K ⎫ h1 = 130.08 kJ/kg ⎬ P1 = 100 kPa ⎭ s1 = 7.2006 kJ/kg ⋅ K T1 = 300 K ⎫ h0 = 1.93 kJ/kg ⎬ P1 = 100 kPa ⎭ s 0 = 6.8426 kJ/kg ⋅ K An energy balance on the diffuser gives q Nitrogen 100 kPa 150°C 180 m/s 110 kPa 25 m/s V12 V2 = h2 + 2 + q out 2 2 (180 m/s) 2 ⎛ 1 kJ/kg ⎞ (25 m/s) 2 ⎛ 1 kJ/kg ⎞ 130.08 kJ/kg + ⎜ ⎟ = h2 + ⎜ ⎟ + 4.5 kJ/kg 2 2 ⎝ 1000 m 2 /s 2 ⎠ ⎝ 1000 m 2 /s 2 ⎠ h1 + ⎯ ⎯→ h2 = 141.47 kJ/kg The corresponding properties at the exit of the diffuser are h2 = 141.47 kJ/kg ⎫ T2 = 160.9 °C = 433.9 K ⎬ P1 = 110 kPa ⎭ s 2 = 7.1989 kJ/kg ⋅ K (b) The mass flow rate of the nitrogen is determined to be & m = ρ 2 A2V 2 = P2 110 kPa A2V 2 = (0.06 m 2 )(25 m/s) = 1.281 kg/s RT2 (0.2968 kJ/kg.K)(433.9 K) The exergy destruction in the nozzle is the exergy difference between the inlet and exit of the diffuser ⎤ ⎡ V 2 − V 22 & & − T0 ( s1 − s 2 )⎥ X dest = m ⎢h1 − h2 + 1 2 ⎥ ⎢ ⎦ ⎣ ⎡ (180 m/s) 2 − (25 m/s) 2 ⎛ 1 kJ/kg ⎜ ⎢(130.08 − 141.47)kJ/kg + = (1.281 kg/s) ⎢ 2 ⎝ 1000 m 2 /s 2 ⎢− (300 K )(7.2006 − 7.1989)kJ/kg.K ⎣ ⎞⎤ ⎟⎥ ⎠⎥ = 5.11 kW ⎥ ⎦ (c) The second-law efficiency for this device may be defined as the exergy output divided by the exergy input: ⎤ ⎡ V2 & & X 1 = m ⎢h1 − h0 + 1 − T0 ( s1 − s0 )⎥ 2 ⎥ ⎢ ⎦ ⎣ ⎡ ⎤ (180 m/s) 2 ⎛ 1 kJ/kg ⎞ = (1.281 kg/s) ⎢(130.08 − 1.93) kJ/kg + ⎜ ⎟ − (300 K )(7.2006 − 6.8426)kJ/kg.K ⎥ 2 2 2 ⎝ 1000 m /s ⎠ ⎢ ⎥ ⎣ ⎦ = 47.35 kW & & X X 5.11 kW η II = 2 = 1 − dest = 1 − = 0.892 & & 47.35 kW X1 X1 PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 149. 8-149 Fundamentals of Engineering (FE) Exam Problems 8-148 Heat is lost through a plane wall steadily at a rate of 800 W. If the inner and outer surface temperatures of the wall are 20°C and 5°C, respectively, and the environment temperature is 0°C, the rate of exergy destruction within the wall is (a) 40 W (b) 17,500 W (c) 765 W (d) 32,800 W (e) 0 W Answer (a) 40 W Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). Q=800 "W" T1=20 "C" T2=5 "C" To=0 "C" "Entropy balance S_in - S_out + S_gen= DS_system for the wall for steady operation gives" Q/(T1+273)-Q/(T2+273)+S_gen=0 "W/K" X_dest=(To+273)*S_gen "W" "Some Wrong Solutions with Common Mistakes:" Q/T1-Q/T2+Sgen1=0; W1_Xdest=(To+273)*Sgen1 "Using C instead of K in Sgen" Sgen2=Q/((T1+T2)/2); W2_Xdest=(To+273)*Sgen2 "Using avegage temperature in C for Sgen" Sgen3=Q/((T1+T2)/2+273); W3_Xdest=(To+273)*Sgen3 "Using avegage temperature in K" W4_Xdest=To*S_gen "Using C for To" PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 150. 8-150 8-149 Liquid water enters an adiabatic piping system at 15°C at a rate of 5 kg/s. It is observed that the water temperature rises by 0.5°C in the pipe due to friction. If the environment temperature is also 15°C, the rate of exergy destruction in the pipe is (a) 8.36 kW (b) 10.4 kW (c) 197 kW (d) 265 kW (e) 2410 kW Answer (b) 10.4 kW Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). Cp=4.18 "kJ/kg.K" m=5 "kg/s" T1=15 "C" T2=15.5 "C" To=15 "C" S_gen=m*Cp*ln((T2+273)/(T1+273)) "kW/K" X_dest=(To+273)*S_gen "kW" "Some Wrong Solutions with Common Mistakes:" W1_Xdest=(To+273)*m*Cp*ln(T2/T1) "Using deg. C in Sgen" W2_Xdest=To*m*Cp*ln(T2/T1) "Using deg. C in Sgen and To" W3_Xdest=(To+273)*Cp*ln(T2/T1) "Not using mass flow rate with deg. C" W4_Xdest=(To+273)*Cp*ln((T2+273)/(T1+273)) "Not using mass flow rate with K" 8-150 A heat engine receives heat from a source at 1500 K at a rate of 600 kJ/s and rejects the waste heat to a sink at 300 K. If the power output of the engine is 400 kW, the second-law efficiency of this heat engine is (a) 42% (b) 53% (c) 83% (d) 67% (e) 80% Answer (c) 83% Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). Qin=600 "kJ/s" W=400 "kW" TL=300 "K" TH=1500 "K" Eta_rev=1-TL/TH Eta_th=W/Qin Eta_II=Eta_th/Eta_rev "Some Wrong Solutions with Common Mistakes:" W1_Eta_II=Eta_th1/Eta_rev; Eta_th1=1-W/Qin "Using wrong relation for thermal efficiency" W2_Eta_II=Eta_th "Taking second-law efficiency to be thermal efficiency" W3_Eta_II=Eta_rev "Taking second-law efficiency to be reversible efficiency" W4_Eta_II=Eta_th*Eta_rev "Multiplying thermal and reversible efficiencies instead of dividing" PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 151. 8-151 8-151 A water reservoir contains 100 tons of water at an average elevation of 60 m. The maximum amount of electric power that can be generated from this water is (a) 8 kWh (b) 16 kWh (c) 1630 kWh (d) 16,300 kWh (e) 58,800 kWh Answer (b) 16 kWh Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). m=100000 "kg" h=60 "m" g=9.81 "m/s^2" "Maximum power is simply the potential energy change," W_max=m*g*h/1000 "kJ" W_max_kWh=W_max/3600 "kWh" "Some Wrong Solutions with Common Mistakes:" W1_Wmax =m*g*h/3600 "Not using the conversion factor 1000" W2_Wmax =m*g*h/1000 "Obtaining the result in kJ instead of kWh" W3_Wmax =m*g*h*3.6/1000 "Using worng conversion factor" W4_Wmax =m*h/3600"Not using g and the factor 1000 in calculations" 8-152 A house is maintained at 25°C in winter by electric resistance heaters. If the outdoor temperature is 2°C, the second-law efficiency of the resistance heaters is (a) 0% (b) 7.7% (c) 8.7% (d) 13% (e) 100% Answer (b) 7.7% Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). TL=2+273 "K" TH=25+273 "K" To=TL COP_rev=TH/(TH-TL) COP=1 Eta_II=COP/COP_rev "Some Wrong Solutions with Common Mistakes:" W1_Eta_II=COP/COP_rev1; COP_rev1=TL/(TH-TL) "Using wrong relation for COP_rev" W2_Eta_II=1-(TL-273)/(TH-273) "Taking second-law efficiency to be reversible thermal efficiency with C for temp" W3_Eta_II=COP_rev "Taking second-law efficiency to be reversible COP" W4_Eta_II=COP_rev2/COP; COP_rev2=(TL-273)/(TH-TL) "Using C in COP_rev relation instead of K, and reversing" PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 152. 8-152 8-153 A 10-kg solid whose specific heat is 2.8 kJ/kg.°C is at a uniform temperature of -10°C. For an environment temperature of 25°C, the exergy content of this solid is (a) Less than zero (b) 0 kJ (c) 22.3 kJ (d) 62.5 kJ (e) 980 kJ Answer (d) 62.5 kJ Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). m=10 "kg" Cp=2.8 "kJ/kg.K" T1=-10+273 "K" To=25+273 "K" "Exergy content of a fixed mass is x1=u1-uo-To*(s1-so)+Po*(v1-vo)" ex=m*(Cp*(T1-To)-To*Cp*ln(T1/To)) "Some Wrong Solutions with Common Mistakes:" W1_ex=m*Cp*(To-T1) "Taking the energy content as the exergy content" W2_ex=m*(Cp*(T1-To)+To*Cp*ln(T1/To)) "Using + for the second term instead of -" W3_ex=Cp*(T1-To)-To*Cp*ln(T1/To) "Using exergy content per unit mass" W4_ex=0 "Taking the exergy content to be zero" 8-154 Keeping the limitations imposed by the second-law of thermodynamics in mind, choose the wrong statement below: (a) A heat engine cannot have a thermal efficiency of 100%. (b) For all reversible processes, the second-law efficiency is 100%. (c) The second-law efficiency of a heat engine cannot be greater than its thermal efficiency. (d) The second-law efficiency of a process is 100% if no entropy is generated during that process. (e) The coefficient of performance of a refrigerator can be greater than 1. Answer (c) The second-law efficiency of a heat engine cannot be greater than its thermal efficiency. PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 153. 8-153 8-155 A furnace can supply heat steadily at a 1600 K at a rate of 800 kJ/s. The maximum amount of power that can be produced by using the heat supplied by this furnace in an environment at 300 K is (a) 150 kW (b) 210 kW (c) 325 kW (d) 650 kW (e) 984 kW Answer (d) 650 kW Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). Q_in=800 "kJ/s" TL=300 "K" TH=1600 "K" W_max=Q_in*(1-TL/TH) "kW" "Some Wrong Solutions with Common Mistakes:" W1_Wmax=W_max/2 "Taking half of Wmax" W2_Wmax=Q_in/(1-TL/TH) "Dividing by efficiency instead of multiplying by it" W3_Wmax =Q_in*TL/TH "Using wrong relation" W4_Wmax=Q_in "Assuming entire heat input is converted to work" 8-156 Air is throttled from 50°C and 800 kPa to a pressure of 200 kPa at a rate of 0.5 kg/s in an environment at 25°C. The change in kinetic energy is negligible, and no heat transfer occurs during the process. The power potential wasted during this process is (a) 0 (b) 0.20 kW (c) 47 kW (d) 59 kW (e) 119 kW Answer (d) 59 kW Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). R=0.287 "kJ/kg.K" Cp=1.005 "kJ/kg.K" m=0.5 "kg/s" T1=50+273 "K" P1=800 "kPa" To=25 "C" P2=200 "kPa" "Temperature of an ideal gas remains constant during throttling since h=const and h=h(T)" T2=T1 ds=Cp*ln(T2/T1)-R*ln(P2/P1) X_dest=(To+273)*m*ds "kW" "Some Wrong Solutions with Common Mistakes:" W1_dest=0 "Assuming no loss" W2_dest=(To+273)*ds "Not using mass flow rate" W3_dest=To*m*ds "Using C for To instead of K" W4_dest=m*(P1-P2) "Using wrong relations" PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
• 154. 8-154 8-157 Steam enters a turbine steadily at 4 MPa and 400°C and exits at 0.2 MPa and 150°C in an environment at 25°C. The decrease in the exergy of the steam as it flows through the turbine is (a) 58 kJ/kg (b) 445 kJ/kg (c) 458 kJ/kg (d) 518 kJ/kg (e) 597 kJ/kg Answer (e) 597 kJ/kg Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=4000 "kPa" T1=400 "C" P2=200 "kPa" T2=150 "C" To=25 "C" h1=ENTHALPY(Steam_IAPWS,T=T1,P=P1) s1=ENTROPY(Steam_IAPWS,T=T1,P=P1) h2=ENTHALPY(Steam_IAPWS,T=T2,P=P2) s2=ENTROPY(Steam_IAPWS,T=T2,P=P2) "Exergy change of s fluid stream is Dx=h2-h1-To(s2-s1)" -Dx=h2-h1-(To+273)*(s2-s1) "Some Wrong Solutions with Common Mistakes:" -W1_Dx=0 "Assuming no exergy destruction" -W2_Dx=h2-h1 "Using enthalpy change" -W3_Dx=h2-h1-To*(s2-s1) "Using C for To instead of K" -W4_Dx=(h2+(T2+273)*s2)-(h1+(T1+273)*s1) "Using wrong relations for exergy" 8- 158 … 8- 162 Design and Essay Problems PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.