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- 1. 11-1 Chapter 11 REFRIGERATION CYCLESThe Reversed Carnot Cycle11-1C Because the compression process involves the compression of a liquid-vapor mixture whichrequires a compressor that will handle two phases, and the expansion process involves the expansion ofhigh-moisture content refrigerant.11-2 A steady-flow Carnot refrigeration cycle with refrigerant-134a as the working fluid is considered. Thecoefficient of performance, the amount of heat absorbed from the refrigerated space, and the net work inputare to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) Noting that TH = 30°C = 303 K and TL = Tsat @ 160 kPa = -15.60°C = 257.4 K, the COP of thisCarnot refrigerator is determined from 1 1 T COPR,C = = = 5.64 TH / TL − 1 (303 K ) / (257.4 K ) − 1(b) From the refrigerant tables (Table A-11), QH 4 3 h3 = h g @30°C = 266.66 kJ/kg 30°C h4 = h f @30°C = 93.58 kJ/kgThus, 160 kPa q H = h3 − h4 = 266.66 − 93.58 = 173.08 kJ/kg 1 QL 2and s ⎛ 257.4 K ⎞ ⎜ 303 K ⎟(173.08 kJ/kg ) = 147.03 kJ/kg q H TH T = ⎯→ q L = L q H = ⎜ ⎯ ⎟ q L TL TH ⎝ ⎠(c) The net work input is determined from wnet = q H − q L = 173.08 − 147.03 = 26.05 kJ/kgPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 2. 11-211-3E A steady-flow Carnot refrigeration cycle with refrigerant-134a as the working fluid is considered.The coefficient of performance, the quality at the beginning of the heat-absorption process, and the network input are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) Noting that TH = Tsat @ 90 psia = 72.78°F = 532.8 R and TL = Tsat @ 30 psia = 15.37°F = 475.4 R. 1 1 COPR,C = = = 8.28 TH / TL − 1 (532.8 R )/ (475.4 R ) − 1 T(b) Process 4-1 is isentropic, and thus ( s1 = s 4 = s f + x 4 s fg ) @ 90 psia = 0.07481 + (0.05)(0.14525) QH 4 3 = 0.08207 Btu/lbm ⋅ R ⎛ s1 − s f ⎞ 0.08207 − 0.03793 x1 = ⎜ ⎟ = = 0.2374 ⎜ s fg ⎟ 0.18589 ⎝ ⎠ @ 30 psia 1 QL 2(c) Remembering that on a T-s diagram the area enclosed srepresents the net work, and s3 = sg @ 90 psia = 0.22006 Btu/lbm·R, w net,in = (T H − T L )(s 3 − s 4 ) = (72.78 − 15.37)(0.22006 − 0.08207 ) Btu/lbm ⋅ R = 7.92 Btu/lbmPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 3. 11-3Ideal and Actual Vapor-Compression Cycles11-4C Yes; the throttling process is an internally irreversible process.11-5C To make the ideal vapor-compression refrigeration cycle more closely approximate the actualcycle.11-6C No. Assuming the water is maintained at 10°C in the evaporator, the evaporator pressure will bethe saturation pressure corresponding to this pressure, which is 1.2 kPa. It is not practical to designrefrigeration or air-conditioning devices that involve such extremely low pressures.11-7C Allowing a temperature difference of 10°C for effective heat transfer, the condensation temperatureof the refrigerant should be 25°C. The saturation pressure corresponding to 25°C is 0.67 MPa. Therefore,the recommended pressure would be 0.7 MPa.11-8C The area enclosed by the cyclic curve on a T-s diagram represents the net work input for thereversed Carnot cycle, but not so for the ideal vapor-compression refrigeration cycle. This is because thelatter cycle involves an irreversible process for which the process path is not known.11-9C The cycle that involves saturated liquid at 30°C will have a higher COP because, judging from theT-s diagram, it will require a smaller work input for the same refrigeration capacity.11-10C The minimum temperature that the refrigerant can be cooled to before throttling is the temperatureof the sink (the cooling medium) since heat is transferred from the refrigerant to the cooling medium.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 4. 11-411-11 A commercial refrigerator with refrigerant-134a as the working fluid is considered. The quality ofthe refrigerant at the evaporator inlet, the refrigeration load, the COP of the refrigerator, and the theoreticalmaximum refrigeration load for the same power input to the compressor are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) From refrigerant-134a tables (Tables A-11 through A-13) P1 = 60 kPa ⎫ ⎬ h1 = 230.03 kJ/kg T1 = −34°C ⎭ 26°C Water 18°C P2 = 1200 kPa ⎫ ⎬ h2 = 295.16 kJ/kg QH T2 = 65°C ⎭ 1.2 MPa 42°C P3 = 1200 kPa ⎫ 65°C Condenser ⎬ h3 = 111.23 kJ/kg T3 = 42°C ⎭ 3 2 h4 = h3 = 111.23 kJ/kg Expansion Win P4 = 60 kPa ⎫ valve Compressor ⎬ x 4 = 0.4795 h4 = 111.23 kJ/kg ⎭ 4 1 60 kPaUsing saturated liquid enthalpy at the giventemperature, for water we have (Table A-4) Evaporator -34°C hw1 = h f @ 18°C = 75.47 kJ/kg QL hw 2 = h f @ 26°C = 108.94 kJ/kg(b) The mass flow rate of the refrigerant may be determined from an energy balance on the compressor mR (h2 − h3 ) = mw (hw 2 − hw1 ) & & mR (295.16 − 111.23)kJ/kg = (0.25 kg/s)(108.94 − 75.47)kJ/kg & ⎯ mR = 0.0455 kg/s ⎯→ &The waste heat transferred from the refrigerant, the compressor power input, and the refrigeration load are & Q H = m R (h2 − h3 ) = (0.0455 kg/s)(295.16 − 111.23)kJ/kg = 8.367 kW & & & Win = m R (h2 − h1 ) − Qin = (0.0455 kg/s)(295.16 − 230.03)kJ/kg − 0.45 kW = 2.513 kW & & & & Q L = Q H − Win = 8.367 − 2.513 = 5.85 kW (c) The COP of the refrigerator is determined T 2from its definition 2 · & QH · Q 5.85 Win COP = L = = 2.33 & Win 2.513 3(d) The reversible COP of the refrigerator for thesame temperature limits is 4 · 1 1 1 COPmax = = = 5.063 QL T H / T L − 1 (18 + 273) /(−30 + 273) − 1 sThen, the maximum refrigeration load becomes & & Q L,max = COPmax Win = (5.063)(2.513 kW) = 12.72 kWPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 5. 11-511-12 An ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid isconsidered. The COP and the power requirement are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis In an ideal vapor-compression refrigeration cycle, the compression process is isentropic, therefrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenseras saturated liquid at the condenser pressure. From the refrigerant tables (Tables A-11, A-12, and A-13), T1 = 4°C ⎫ h1 = h g @ 4°C = 252.77 kJ/kg T ⎬ sat. vapor ⎭ s1 = s g @ 4°C = 0.92927 kJ/kg ⋅ K · QH P2 = 1 MPa ⎫ 2 ⎬ h2 = 275.29 kJ/kg 3 1 MPa s 2 = s1 ⎭ · Win P3 = 1 MPa ⎫ ⎬ h = hf = 107.32 kJ/kg sat. liquid ⎭ 3 @ 1 MPa 4°C h4 ≅ h3 = 107.32 kJ/kg ( throttling) 4s 4 · 1 QLThe mass flow rate of the refrigerant is & s & QL 400 kJ/s Q L = m(h1 − h4 ) ⎯ & ⎯→ m = & = = 2.750 kg/s h1 − h4 (252.77 − 107.32) kJ/kgThe power requirement is & Win = m(h2 − h1 ) = (2.750 kg/s)(275.29 − 252.77) kJ/kg = 61.93 kW &The COP of the refrigerator is determined from its definition, & QL 400 kW COPR = = = 6.46 & Win 61.93 kWPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 6. 11-611-13 An ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid isconsidered. The mass flow rate of the refrigerant and the power requirement are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis In an ideal vapor-compression refrigeration cycle, the compression process is isentropic, therefrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenseras saturated liquid at the condenser pressure. From the refrigerant tables (Tables A-11, A-12, and A-13), P1 = 400 kPa ⎫ h1 = h g @ 400 kPa = 255.55 kJ/kg T ⎬ s =s sat. vapor ⎭ 1 g @ 400 kPa = 0.92691 kJ/kg ⋅ K · QH 2 P2 = 800 kPa ⎫ ⎬ h2 = 269.90 kJ/kg 3 0.8 MPa s 2 = s1 ⎭ · Win P3 = 800 kPa ⎫ ⎬ h3 = h f @ 800 kPa = 95.47 kJ/kg sat. liquid ⎭ 0.4 MPa h4 ≅ h3 = 95.47 kJ/kg ( throttling ) 4s 4 · 1 QLThe mass flow rate of the refrigerant is determined from & s & QL 10 kJ/s Q L = m(h1 − h4 ) ⎯ & ⎯→ m = & = = 0.06247 kg/s h1 − h4 (255.55 − 95.47) kJ/kgThe power requirement is & Win = m(h2 − h1 ) = (0.06247 kg/s)(269.90 − 255.55) kJ/kg = 0.8964 kW &PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 7. 11-711-14E An ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid isconsidered. The mass flow rate of the refrigerant and the power requirement are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis In an ideal vapor-compression refrigeration cycle, the compression process is isentropic, therefrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenseras saturated liquid at the condenser pressure. From the refrigerant tables (Tables A-11E, A-12E, and A-13E), T1 = −10°F ⎫ h1 = h g @ −10° F = 101.61 Btu/lbm T ⎬ sat. vapor ⎭ s1 = s g @ −10°F = 0.22660 Btu/lbm ⋅ R · QH 2 P2 = 100 psia ⎫ 3 100 psia ⎬ h2 = 117.57 Btu/lbm · s 2 = s1 ⎭ Win P3 = 100 psia ⎫ ⎬ h3 = h f @ 100 psia = 37.869 Btu/lbm -10°F sat. liquid ⎭ 4s 4 · 1 h4 ≅ h3 = 37.869 Btu/lbm ( throttling ) QLThe mass flow rate of the refrigerant is s & QL 24,000 Btu/h & Q L = m(h1 − h4 ) ⎯ & ⎯→ m = & = = 376.5 lbm/h h1 − h4 (101.61 − 37.869) Btu/lbmThe power requirement is & ⎛ 1 kW ⎞ Win = m(h2 − h1 ) = (376.5 lbm/h)(117.57 − 101.61) Btu/lbm⎜ & ⎟ = 1.761 kW ⎝ 3412.14 Btu/h ⎠PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 8. 11-811-15E Problem 11-14E is to be repeated if ammonia is used as the refrigerant.Analysis The problem is solved using EES, and the solution is given below."Given"x[1]=1T[1]=-10 [F]x[3]=0P[3]=100 [psia]Q_dot_L=24000 [Btu/h]"Analysis"Fluid$=ammonia"compressor"h[1]=enthalpy(Fluid$, T=T[1], x=x[1])s[1]=entropy(Fluid$, T=T[1], x=x[1])s[2]=s[1]P[2]=P[3]h[2]=enthalpy(Fluid$, P=P[2], s=s[2])"expansion valve"h[3]=enthalpy(Fluid$, P=P[3], x=x[3])h[4]=h[3]"cycle"m_dot_R=Q_dot_L/(h[1]-h[4])W_dot_in=m_dot_R*(h[2]-h[1])*Convert(Btu/h, kW)Solution for ammoniaFluid$=ammonia COP_R=5.847 h[1]=615.92 [Btu/lb_m]h[2]=701.99 [Btu/lb_m] h[3]=112.67 [Btu/lb_m] h[4]=112.67 [Btu/lb_m]m_dot_R=47.69 [lbm/h] P[2]=100 [psia] P[3]=100 [psia]Q_dot_L=24000 [Btu/h] s[1]=1.42220 [Btu/lb_m-R] s[2]=1.42220 [Btu/lb_m-R]T[1]=-10 [F] W_dot_in=1.203 [kW] x[1]=1x[3]=0Solution for R-134aFluid$=R134a COP_R=3.993 h[1]=101.62 [Btu/lb_m]h[2]=117.58 [Btu/lb_m] h[3]=37.87 [Btu/lb_m] h[4]=37.87 [Btu/lb_m]m_dot_R=376.5 [lbm/h] P[2]=100 [psia] P[3]=100 [psia]Q_dot_L=24000 [Btu/h] s[1]=0.22662 [Btu/lb_m-R] s[2]=0.22662 [Btu/lb_m-R]T[1]=-10 [F] W_dot_in=1.761 [kW] x[1]=1x[3]=0PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 9. 11-911-16 An ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid isconsidered. The rate of heat removal from the refrigerated space, the power input to the compressor, therate of heat rejection to the environment, and the COP are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) In an ideal vapor-compression refrigeration cycle, the compression process is isentropic, therefrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenseras saturated liquid at the condenser pressure. From the refrigerant tables (Tables A-12 and A-13), P1 = 120 kPa ⎫ h1 = h g @ 120 kPa = 236.97 kJ/kg ⎬s = s g @ 120 kPa = 0.94779 kJ/kg ⋅ K sat. vapor ⎭ 1 T P2 = 0.7 MPa ⎫ · ⎬ h2 = 273.50 kJ/kg (T2 = 34.95°C ) QH 2 s 2 = s1 ⎭ 3 0.7 MPa · P3 = 0.7 MPa ⎫ Win ⎬ h3 = h f @ 0.7 MPa = 88.82 kJ/kg sat. liquid ⎭ h4 ≅ h3 = 88.82 kJ/kg (throttling ) 0.12 4s · 1Then the rate of heat removal from the refrigerated 4 QLspace and the power input to the compressor aredetermined from s Q L = m(h1 − h4 ) = (0.05 kg/s )(236.97 − 88.82) kJ/kg = 7.41 kW & &and Win = m(h2 − h1 ) = (0.05 kg/s )(273.50 − 236.97 ) kJ/kg = 1.83 kW & &(b) The rate of heat rejection to the environment is determined from & & & Q H = Q L + Win = 7.41 + 1.83 = 9.23 kW(c) The COP of the refrigerator is determined from its definition, & QL 7.41 kW COPR = = = 4.06 & Win 1.83 kWPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 10. 11-1011-17 An ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid isconsidered. The rate of heat removal from the refrigerated space, the power input to the compressor, therate of heat rejection to the environment, and the COP are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) In an ideal vapor-compression refrigeration cycle, the compression process is isentropic, therefrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenseras saturated liquid at the condenser pressure. From the refrigerant tables (Tables A-12 and A-13), P1 = 120 kPa ⎫ h1 = h g @ 120 kPa = 236.97 kJ/kg ⎬s = s sat. vapor ⎭ 1 g @ 120 kPa = 0.94779 kJ/kg ⋅ K T P2 = 0.9 MPa ⎫ · ⎬ h2 = 278.93 kJ/kg (T2 = 44.45°C ) QH 2 s 2 = s1 ⎭ 3 0.9 MPa · P3 = 0.9 MPa ⎫ Win ⎬ h3 = h f @ 0.9 MPa = 101.61 kJ/kg sat. liquid ⎭ h4 ≅ h3 = 101.61 kJ/kg (throttling ) 0.12 MPa 4s · 1Then the rate of heat removal from the 4 QLrefrigerated space and the power input to thecompressor are determined from s Q L = m(h1 − h4 ) = (0.05 kg/s )(236.97 − 101.61) kJ/kg = 6.77 kW & &and Win = m(h2 − h1 ) = (0.05 kg/s )(278.93 − 236.97 ) kJ/kg = 2.10 kW & &(b) The rate of heat rejection to the environment is determined from & & & Q H = Q L + Win = 6.77 + 2.10 = 8.87 kW(c) The COP of the refrigerator is determined from its definition, & QL 6.77 kW COPR = = = 3.23 & Win 2.10 kWPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 11. 11-1111-18 An ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid isconsidered. The throttling valve in the cycle is replaced by an isentropic turbine. The percentage increasein the COP and in the rate of heat removal from the refrigerated space due to this replacement are to bedetermined.Assumptions 1 Steady operating conditions exist. T2 Kinetic and potential energy changes are ·negligible. QH 2Analysis If the throttling valve in the previous 3 0.7 MPa ·problem is replaced by an isentropic turbine, we Winwould have s4s = s3 = sf @ 0.7 MPa = 0.33230kJ/kg·K, and the enthalpy at the turbine exit 0.12 MPawould be 4s 4 · 1 ⎛ s3 − s f ⎞ QL 0.33230 − 0.09275x4s = ⎜ ⎟ = = 0.2802 ⎜ ⎟ s ⎝ s fg ⎠ @ 120 kPa 0.85503 (h4 s = h f + x 4 s h fg )@ 120 kPa = 22.49 + (0.2802)(214.48) = 82.58 kJ/kgThen, Q L = m(h1 − h4 s ) = (0.05 kg/s )(236.97 − 82.58) kJ/kg = 7.72 kW & &and & QL 7.72 kW COPR = = = 4.23 & Win 1.83 kW &Then the percentage increase in Q and COP becomes & ΔQ L 7.72 − 7.41 & Increase in Q L = = = 4.2% & QL 7.41 ΔCOPR 4.23 − 4.06 Increase in COPR = = = 4.2% COPR 4.06PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 12. 11-1211-19 A refrigerator with refrigerant-134a as the working fluid is considered. The rate of heat removalfrom the refrigerated space, the power input to the compressor, the isentropic efficiency of the compressor,and the COP of the refrigerator are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) From the refrigerant tables (Tables A-12 and A-13), P1 = 0.14 MPa ⎫ h1 = 246.36 kJ/kg T1 = −10°C ⎬ s = 0.97236 kJ/kg ⋅ K T ⎭ 1 2 0.7 MPa 0.65 2 50°C P2 = 0.7 MPa ⎫ MPa · T2 = 50°C ⎬ h2 = 288.53 kJ/kg QH · ⎭ Win P2 s = 0.7 MPa ⎫ ⎬ h2 s = 281.16 kJ/kg 3 s 2 s = s1 ⎭ P3 = 0.65 MPa ⎫ 0.15 0.14 T3 = 24°C ⎬ h3 = h f @ 24°C = 84.98 kJ/kg ⎭ MP 4 · 1 MPa h4 ≅ h3 = 84.98 kJ/kg (throttling ) QL sThen the rate of heat removal from the refrigerated spaceand the power input to the compressor are determined from Q L = m(h1 − h4 ) = (0.12 kg/s )(246.36 − 84.98) kJ/kg = 19.4 kW & &and Win = m(h2 − h1 ) = (0.12 kg/s )(288.53 − 246.36) kJ/kg = 5.06 kW & &(b) The adiabatic efficiency of the compressor is determined from h2 s − h1 281.16 − 246.36 ηC = = = 82.5% h2 − h1 288.53 − 246.36(c) The COP of the refrigerator is determined from its definition, & Q L 19.4 kW COPR = = = 3.83 & Win 5.06 kWPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 13. 11-1311-20 An air conditioner operating on the ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The COP of the system is to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis In an ideal vapor-compression refrigeration cycle, the compression process is isentropic, therefrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenseras saturated liquid at the condenser pressure. The evaporating temperature will be 22-2=20°C. From therefrigerant tables (Tables A-11, A-12, and A-13), T1 = 20°C ⎫ h1 = h g @ 20°C = 261.59 kJ/kg T ⎬ sat. vapor ⎭ s1 = s g @ 20°C = 0.92234 kJ/kg ⋅ K · QH P2 = 1 MPa ⎫ 2 ⎬ h2 = 273.11 kJ/kg s 2 = s1 ⎭ 3 1 MPa · Win P3 = 1 MPa ⎫ ⎬ h = hf = 107.32 kJ/kg sat. liquid ⎭ 3 @ 1 MPa 20°C h4 ≅ h3 = 107.32 kJ/kg ( throttling) 1 4s 4 · QLThe COP of the air conditioner is determined from its definition, qL h -h 261.59 − 107.32 s COPAC = = 1 4 = = 13.39 win h2 -h1 273.11 − 261.59PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 14. 11-1411-21E A refrigerator operating on the ideal vapor-compression refrigeration cycle with refrigerant-134aas the working fluid is considered. The increase in the COP if the throttling process were replaced by anisentropic expansion is to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis In an ideal vapor-compression refrigeration cycle, the compression process is isentropic, therefrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenseras saturated liquid at the condenser pressure. From the refrigerant tables (Tables A-11E, A-12E, and A-13E), T1 = 20°F ⎫ h1 = h g @ 20° F = 105.98 Btu/lbm T ⎬ sat. vapor ⎭ s1 = s g @ 20°F = 0.22341 Btu/lbm ⋅ R · QH P2 = 300 psia ⎫ 2 ⎬ h2 = 125.68 Btu/lbm 3 300 psia s 2 = s1 ⎭ · Win P3 = 300 psia ⎫ h3 = h f @ 300 psia = 66.339 Btu/lbm ⎬ s =s sat. liquid ⎭ 3 f @ 300 psia = 0.12715 Btu/lbm ⋅ R 20°F h4 ≅ h3 = 66.339 Btu/lbm ( throttling) 4s 4 · 1 QL T4 = 15°F ⎫ ⎬ h = 59.80 Btu/lbm (isentropic expansion) s 4 = s3 ⎭ 4s sThe COP of the refrigerator for the throttling case is qL h -h 105.98 − 66.339 COPR = = 1 4 = = 2.012 win h2 -h1 125.68 − 105.98The COP of the refrigerator for the isentropic expansion case is qL h -h 105.98 − 59.80 COPR = = 1 4s = = 2.344 win h2 -h1 125.68 − 105.98The increase in the COP by isentropic expansion is 16.5%.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 15. 11-1511-22 An ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid isconsidered. The COP of the system and the cooling load are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis In an ideal vapor-compression refrigeration cycle, the compression process is isentropic, therefrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenseras saturated liquid at the condenser pressure. From the refrigerant tables (Tables A-11, A-12, and A-13), T1 = −10°C ⎫ h1 = h g @ −10°C = 244.51 kJ/kg ⎬ sat. vapor ⎭ s1 = s g @ −10°C = 0.93766 kJ/kg ⋅ K T P2 = 600 kPa ⎫ · ⎬ h2 = 267.12 kJ/kg QH s 2 = s1 ⎭ 2 3 600 kPa · P3 = 600 kPa ⎫ Win ⎬ h3 = h f @ 600 kPa = 81.51 kJ/kg sat. liquid ⎭ h4 ≅ h3 = 81.51 kJ/kg ( throttling) -10°C 4s 4 · 1The COP of the air conditioner is determined from its definition, QL qL h -h 244.51 − 81.51 COPAC = = 1 4 = = 7.21 s win h2 -h1 267.12 − 244.51The cooling load is & Q L = m(h1 − h4 ) = (7 kg/s)(244.51 − 81.51) kJ/kg = 1141 kW &PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 16. 11-1611-23 A refrigerator with refrigerant-134a as the working fluid is considered. The power input to thecompressor, the rate of heat removal from the refrigerated space, and the pressure drop and the rate of heatgain in the line between the evaporator and the compressor are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) From the refrigerant tables (Tables A-12 and A-13), h1 = 246.36 kJ/kg P1 = 140 kPa ⎫ T 2 s = 0.97236 kJ/kg ⋅ K T1 = −10°C ⎬ 1 ⎭ 0.95 2s 1 MPa v 1 = 0.14605 m 3 /kg MPa · QH · P2 = 1.0 MPa ⎫ Win ⎬ h2 s = 289.20 kJ/kg s 2 s = s1 ⎭ 3 P3 = 0.95 MPa ⎫ T3 = 30°C ⎬ h3 ≅ h f @ 30 °C = 93.58 kJ/kg ⎭ 0.14 MPa 0.15 MPa h4 ≅ h3 = 93.58 kJ/kg (throttling ) 1 -10°C 4 · QL -18.5°C T5 = −18.5°C ⎫ P5 = 0.14165 MPa ⎬ h = 239.33 kJ/kg sat. vapor ⎭ 5 sThen the mass flow rate of the refrigerant and the power input becomes V&1 0.3/60 m3/s m= & = = 0.03423 kg/s v1 0.14605 m3/kg Win = m(h2 s − h1 ) /ηC = (0.03423 kg/s )[(289.20 − 246.36) kJ/kg ]/ (0.78) = 1.88 kW & &(b) The rate of heat removal from the refrigerated space is Q L = m(h5 − h4 ) = (0.03423 kg/s )(239.33 − 93.58) kJ/kg = 4.99 kW & &(c) The pressure drop and the heat gain in the line between the evaporator and the compressor are ΔP = P5 − P1 = 141.65 − 140 = 1.65and Qgain = m(h1 − h5 ) = (0.03423 kg/s )(246.36 − 239.33) kJ/kg = 0.241 kW & &PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 17. 11-1711-24 EES Problem 11-23 is reconsidered. The effects of the compressor isentropic efficiency and thecompressor inlet volume flow rate on the power input and the rate of refrigeration are to be investigated.Analysis The problem is solved using EES, and the solution is given below."Input Data""T[5]=-18.5 [C] P[1]=140 [kPa]T[1] = -10 [C]}V_dot[1]=0.1 [m^3/min]P[2] = 1000 [kPa]P[3]=950 [kPa]T[3] = 30 [C]Eta_c=0.78Fluid$=R134a""Compressor"h[1]=enthalpy(Fluid$,P=P[1],T=T[1]) "properties for state 1"s[1]=entropy(Fluid$,P=P[1],T=T[1])v[1]=volume(Fluid$,P=P[1],T=T[1])"[m^3/kg]"m_dot=V_dot[1]/v[1]*convert(m^3/min,m^3/s)"[kg/s]"h2s=enthalpy(Fluid$,P=P[2],s=s[1]) "Identifies state 2s as isentropic"h[1]+Wcs=h2s "energy balance on isentropic compressor"Wc=Wcs/Eta_c"definition of compressor isentropic efficiency"h[1]+Wc=h[2] "energy balance on real compressor-assumed adiabatic"s[2]=entropy(Fluid$,h=h[2],P=P[2]) "properties for state 2"T[2]=temperature(Fluid$,h=h[2],P=P[2])W_dot_c=m_dot*Wc"Condenser"h[3]=enthalpy(Fluid$,P=P[3],T=T[3]) "properties for state 3"s[3]=entropy(Fluid$,P=P[3],T=T[3])h[2]=q_out+h[3] "energy balance on condenser"Q_dot_out=m_dot*q_out"Throttle Valve"h[4]=h[3] "energy balance on throttle - isenthalpic"x[4]=quality(Fluid$,h=h[4],P=P[4]) "properties for state 4"s[4]=entropy(Fluid$,h=h[4],P=P[4])T[4]=temperature(Fluid$,h=h[4],P=P[4])"Evaporator"P[4]=pressure(Fluid$,T=T[5],x=0)"pressure=Psat at evaporator exit temp."P[5] = P[4]h[5]=enthalpy(Fluid$,T=T[5],x=1) "properties for state 5"q_in + h[4]=h[5] "energy balance on evaporator"Q_dot_in=m_dot*q_inCOP=Q_dot_in/W_dot_c "definition of COP"COP_plot = COPW_dot_in = W_dot_cQ_dot_line5to1=m_dot*(h[1]-h[5])"[kW]"PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 18. 11-18 COPplot Win[kW] Qin [kW] ηc 2.041 0.8149 1.663 0.6 2.381 0.6985 1.663 0.7 2.721 0.6112 1.663 0.8 3.062 0.5433 1.663 0.9 3.402 0.4889 1.663 1 9 3 8 V 1 m /m in 7 1.0 0.5 0.1 6 5 W in 4 3 2 1 0 0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1 ηc 4 3.5 3 2.5 COP plot 2 3 V 1 m /m in 1.5 1.0 0.5 0.1 1 0.5 0 0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1 ηc 18 3 14.4 V 1 m /m in 1.0 0.5 Q in [kW ] 10.8 0.1 7.2 3.6 0 0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1 ηcPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 19. 11-1911-25 A refrigerator uses refrigerant-134a as the working fluid and operates on the ideal vapor-compression refrigeration cycle. The mass flow rate of the refrigerant, the condenser pressure, and theCOP of the refrigerator are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) (b) From the refrigerant-134a tables (Tables A-11 through A-13) P4 = 120 kPa ⎫ . ⎬ h4 = 86.83 kJ/kg QH x 4 = 0.30 ⎭ 60°C h3 = h4 Condenser h3 = 86.83 kJ/kg ⎫ 3 ⎬ P3 = 671.8 kPa 2 . x 3 = 0 (sat. liq.) ⎭ Expansion Win P2 = P3 valve Compressor P2 = 671.8 kPa ⎫ ⎬ h2 = 298.87 kJ/kg T2 = 60°C ⎭ 4 1 P1 = P4 = 120 kPa ⎫ Evaporator ⎬ h1 = 236.97 kJ/kg 120 kPa x1 = 1 (sat. vap.) ⎭ . x=0.3 QLThe mass flow rate of the refrigerant isdetermined from T & W in 0.45 kW · m= & = = 0.00727 kg/s QH h2 − h1 (298.87 − 236.97)kJ/kg 2 3 ·(c) The refrigeration load and the COP are Win & Q L = m(h1 − h4 ) & = (0.0727 kg/s)(236.97 − 86.83)kJ/kg 120 kPa = 1.091 kW 4s · 1 4 QL & Q L 1.091 kW COP = = = 2.43 s & Win 0.45 kWPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 20. 11-20Selecting the Right Refrigerant11-26C The desirable characteristics of a refrigerant are to have an evaporator pressure which is above theatmospheric pressure, and a condenser pressure which corresponds to a saturation temperature above thetemperature of the cooling medium. Other desirable characteristics of a refrigerant include being nontoxic,noncorrosive, nonflammable, chemically stable, having a high enthalpy of vaporization (minimizes themass flow rate) and, of course, being available at low cost.11-27C The minimum pressure that the refrigerant needs to be compressed to is the saturation pressure ofthe refrigerant at 30°C, which is 0.771 MPa. At lower pressures, the refrigerant will have to condense attemperatures lower than the temperature of the surroundings, which cannot happen.11-28C Allowing a temperature difference of 10°C for effective heat transfer, the evaporation temperatureof the refrigerant should be -20°C. The saturation pressure corresponding to -20°C is 0.133 MPa.Therefore, the recommended pressure would be 0.12 MPa.11-29 A refrigerator that operates on the ideal vapor-compression cycle with refrigerant-134a isconsidered. Reasonable pressures for the evaporator and the condenser are to be selected.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis Allowing a temperature difference of 10°C for effective heat transfer, the evaporation andcondensation temperatures of the refrigerant should be -20°C and 35°C, respectively. The saturationpressures corresponding to these temperatures are 0.133 MPa and 0.888 MPa. Therefore, the recommendedevaporator and condenser pressures are 0.133 MPa and 0.888 MPa, respectively.11-30 A heat pump that operates on the ideal vapor-compression cycle with refrigerant-134a is considered.Reasonable pressures for the evaporator and the condenser are to be selected.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis Allowing a temperature difference of 10°C for effective heat transfer, the evaporation andcondensation temperatures of the refrigerant should be 0°C and 32°C, respectively. The saturationpressures corresponding to these temperatures are 0.293 MPa and 0.816 MPa. Therefore, the recommendedevaporator and condenser pressures are 0.293 MPa and 0.816 MPa, respectively.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 21. 11-21Heat Pump Systems11-31C A heat pump system is more cost effective in Miami because of the low heating loads and highcooling loads at that location.11-32C A water-source heat pump extracts heat from water instead of air. Water-source heat pumps havehigher COPs than the air-source systems because the temperature of water is higher than the temperature ofair in winter.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 22. 11-2211-33 An actual heat pump cycle with R-134a as the refrigerant is considered. The isentropic efficiency ofthe compressor, the rate of heat supplied to the heated room, the COP of the heat pump, and the COP andthe rate of heat supplied to the heated room if this heat pump operated on the ideal vapor-compressioncycle between the same pressure limits are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) The properties of refrigerant-134a are (Tables A-11 through A-13) P2 = 800 kPa ⎫ ⎬ h2 = 291.76 kJ/kg . T2 = 55°C ⎭ QH 800 kPa T3 = Tsat@750 kPa = 29.06°C 750 kPa 55°C Condenser P3 = 750 kPa ⎫ ⎬ h3 = 87.91 kJ/kg 3 2 T3 = (29.06 − 3)°C⎭ . Expansion Win h4 = h3 = 87.91 kJ/kg valve Compressor Tsat@200 kPa = −10.09°C P1 = 200 kPa ⎫h1 = 247.87 kJ/kg ⎬ 4 1 T1 = (−10.09 + 4)°C⎭ s1 = 0.9506 kJ/kg Evaporator P2 = 800 kPa ⎫ . ⎬h2 s = 277.26 s 2 = s1 ⎭ QLThe isentropic efficiency of the compressor is h2 s − h1 277.26 − 247.87 T 2 ηC = = = 0.670 h2 − h1 291.76 − 247.87 · 2 Q ·(b) The rate of heat supplied to the room is Win & Q H = m(h2 − h3 ) = (0.018 kg/s)(291.76 − 87.91)kJ/kg = 3.67 kW & 3(c) The power input and the COP are 4 · 1 & Win = m(h2 − h1 ) = (0.018 kg/s)(291.76 − 247.87)kJ/kg = 0.790 kW & QL & s QH 3.67 COP = = = 4.64 & Win 0.790(d) The ideal vapor-compression cycle analysis of thecycle is as follows: T h1 = h g @ 200 kPa = 244.46 kJ/kg · s1 = s g @ 200 kPa = 0.9377 kJ/kg.K QH 2 3 0.8 MPa · P2 = 800 MPa ⎫ Win ⎬ h2 = 273.25 kJ/kg s 2 = s1 ⎭ h3 = h f @ 800 kPa = 95.47 kJ/kg 0.2 MPa h4 = h3 4s · 1 4 QL h2 − h3 273.25 − 95.47 s COP = = = 6.18 h2 − h1 273.25 − 244.46 & Q H = m(h2 − h3 ) = (0.018 kg/s)(273.25 − 95.47)kJ/kg = 3.20 kW &PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 23. 11-2311-34 A heat pump operating on the ideal vapor-compression refrigeration cycle with refrigerant-134a asthe working fluid is considered. The COP and the rate of heat supplied to the evaporator are to bedetermined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis In an ideal vapor-compression refrigeration cycle, the compression process is isentropic, therefrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenseras saturated liquid at the condenser pressure. From the refrigerant tables (Tables A-11, A-12, and A-13), P1 = 280 kPa ⎫ h1 = h g @ 280 kPa = 249.72 kJ/kg ⎬ s =s T sat. vapor ⎭ 1 g @ 280 kPa = 0.93210 kJ/kg ⋅ K · P2 = 1200 kPa ⎫ QH 2 ⎬ h2 = 280.00 kJ/kg s 2 = s1 ⎭ 3 1.2 MPa · Win P3 = 1200 kPa ⎫ ⎬ h3 = h f @ 1200 kPa = 117.77 kJ/kg sat. liquid ⎭ 280 kPa h4 ≅ h3 = 117.77 kJ/kg ( throttling) 4s 4 · 1 QLThe mass flow rate of the refrigerant is determined from & s & Win 20 kJ/s Win = m(h2 − h1 ) ⎯ & ⎯→ m = & = = 0.6605 kg/s h2 − h1 (280.00 − 249.72) kJ/kgThen the rate of heat supplied to the evaporator is & Q L = m(h1 − h4 ) = (0.6605 kg/s)(249.72 − 117.77) kJ/kg = 87.15 kW &The COP of the heat pump is determined from its definition, qH h -h 280.00 − 117.77 COPHP = = 2 3 = = 5.36 win h2 -h1 280.00 − 249.72PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 24. 11-2411-35 A heat pump operating on a vapor-compression refrigeration cycle with refrigerant-134a as theworking fluid is considered. The effect of compressor irreversibilities on the COP of the cycle is to bedetermined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis In this cycle, the refrigerant enters the compressor as a saturated vapor at the evaporator pressure,and leaves the condenser as saturated liquid at the condenser pressure. The compression process is notisentropic. The saturation pressure of refrigerant at −1.25°C is 280 kPa. From the refrigerant tables (TablesA-11, A-12, and A-13), P1 = 280 kPa ⎫ h1 = h g @ 280 kPa = 249.72 kJ/kg T ⎬ s =s sat. vapor ⎭ 1 g @ 280 kPa = 0.93210 kJ/kg ⋅ K · QH 2s 2 P2 = 800 kPa ⎫ ⎬ h2 s = 271.50 kJ/kg 3 0.8 MPa · s 2 = s1 ⎭ Win P3 = 800 kPa ⎫ ⎬ h3 = h f @ 800 kPa = 95.47 kJ/kg sat. liquid ⎭ -1.25°C 1 h4 ≅ h3 = 95.47 kJ/kg ( throttling) 4s 4 · QLThe actual enthalpy at the compressor exit is determined byusing the compressor efficiency: s h2 s − h1 h − h1 271.50 − 249.72 ηC = ⎯ ⎯→ h2 = h1 + 2 s = 249.72 + = 275.34 kJ/kg h2 − h1 ηC 0.85The COPs of the heat pump for isentropic and irreversible compression cases are qH h − h3 271.50 − 95.47 COPHP, ideal = = 2s = = 8.082 win h2 s − h1 271.50 − 249.72 qH h − h3 275.34 − 95.47 COPHP, actual = = 2 = = 7.021 win h2 − h1 275.34 − 249.72The irreversible compressor decreases the COP by 13.1%.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 25. 11-2511-36 A heat pump operating on a vapor-compression refrigeration cycle with refrigerant-134a as theworking fluid is considered. The effect of superheating at the compressor inlet on the COP of the cycle isto be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis In this cycle, the compression process is isentropic and leaves the condenser as saturated liquid atthe condenser pressure. The refrigerant entering the compressor is superheated by 2°C. The saturationpressure of refrigerant at −1.25°C is 280 kPa. From the refrigerant tables (Tables A-11, A-12, and A-13), P1 = 280 kPa ⎫ h1 = 251.96 kJ/kg ⎬ T1 = −1.25 + 2 = 1.25°C ⎭ s1 = 0.9403 kJ/kg ⋅ K T P2 = 800 kPa ⎫ ⎬ h2 = 274.04 kJ/kg · s 2 = s1 ⎭ QH 2 P3 = 800 kPa ⎫ 3 0.8 MPa · ⎬ h3 = h f @ 800 kPa = 95.47 kJ/kg Win sat. liquid ⎭ h4 ≅ h3 = 95.47 kJ/kg ( throttling) -1.25°CThe states at the inlet and exit of the compressor when the 4s 4 · 1refrigerant enters the compressor as a saturated vapor are QL P1 = 280 kPa ⎫ h1 = h g @ 280 kPa = 249.72 kJ/kg s ⎬ s =s sat. vapor ⎭ 1 g @ 280 kPa = 0.93210 kJ/kg ⋅ K P2 = 800 kPa ⎫ ⎬ h2 s = 271.50 kJ/kg s 2 = s1 ⎭The COPs of the heat pump for the two cases are qH h − h3 271.50 − 95.47 COPHP, ideal = = 2s = = 8.082 win h2 s − h1 271.50 − 249.72 qH h − h3 274.04 − 95.47 COPHP, actual = = 2 = = 8.087 win h2 − h1 274.04 − 251.96The effect of superheating on the COP is negligible.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 26. 11-2611-37E A heat pump operating on the vapor-compression refrigeration cycle with refrigerant-134a as theworking fluid is considered. The effect of subcooling at the exit of the condenser on the power requirementis to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis In this cycle, the compression process is isentropic, the refrigerant enters the compressor as asaturated vapor at the evaporator pressure, and leaves the condenser as subcooled liquid. From therefrigerant tables (Tables A-11E, A-12E, and A-13E), P1 = 50 psia ⎫ h1 = h g @ 50 psia = 108.81 Btu/lbm ⎬ sat. vapor ⎭ s1 = s g @ 50 psia = 0.22188 kJ/kg ⋅ K T P2 = 160 psia ⎫ ⎬ h2 = 119.19 Btu/lbm · QH s 2 = s1 ⎭ 2 3 160 psia T3 = Tsat @ 160 psia − 9.5 = 109.5 − 9.5 = 100°F · Win P3 = 160 psia ⎫ ⎬ h ≅ hf @ 100° F = 45.124 Btu/lbm T3 = 100°F ⎭ 3 50 psia h4 ≅ h3 = 45.124 Btu/lbm ( throttling) 4s 4 · 1 QLThe states at the inlet and exit of the expansion valve whenthe refrigerant is saturated liquid at the condenser exit are s P3 = 160 psia ⎫ ⎬ h3 = h f @ 160 psia = 48.519 Btu/lbm sat. liquid ⎭ h4 ≅ h3 = 48.519 Btu/lbm ( throttling)The mass flow rate of the refrigerant in the ideal case is & QH 100,000 Btu/h & Q L = m(h1 − h4 ) ⎯ & ⎯→ m = & = = 1415.0 lbm/h h2 − h3 (119.19 − 48.519) Btu/lbmThe power requirement is & ⎛ 1 kW ⎞ Win = m(h2 − h1 ) = (1415.0 lbm/h)(119.19 − 108.81) Btu/lbm⎜ & ⎟ = 4.305 kW ⎝ 3412.14 Btu/h ⎠With subcooling, the mass flow rate and the power input are & QH 100,000 Btu/h & Q L = m(h1 − h4 ) ⎯ & ⎯→ m = & = = 1350.1 lbm/h h2 − h3 (119.19 − 45.124) Btu/lbm & ⎛ 1 kW ⎞ Win = m(h2 − h1 ) = (1350.1 lbm/h)(119.19 − 108.81) Btu/lbm⎜ & ⎟ = 4.107 kW ⎝ 3412.14 Btu/h ⎠Subcooling decreases the power requirement by 4.6%.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 27. 11-2711-38E A heat pump operating on the vapor-compression refrigeration cycle with refrigerant-134a as theworking fluid is considered. The effect of superheating at the compressor inlet on the power requirement isto be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis In this cycle, the compression process is isentropic and leaves the condenser as saturated liquid atthe condenser pressure. The refrigerant entering the compressor is superheated by 10°F. The saturationtemperature of the refrigerant at 50 psia is 40.23°F. From the refrigerant tables (Tables A-11E, A-12E, andA-13E), P1 = 50 psia ⎫ h1 = 110.99 Btu/lbm ⎬ T1 = 40.23 + 10 = 50.2°F ⎭ s1 = 0.2262 kJ/kg ⋅ K T P2 = 160 psia ⎫ · ⎬ h2 = 121.71 Btu/lbm QH 2 s 2 = s1 ⎭ 3 160 psia · P3 = 160 psia ⎫ Win ⎬ h3 = h f @ 160 psia = 48.519 Btu/lbm sat. liquid ⎭ h4 ≅ h3 = 48.519 Btu/lbm ( throttling) 50 psia 4s 4 · 1The states at the inlet and exit of the compressor when the QLrefrigerant enters the compressor as a saturated vapor are P1 = 50 psia ⎫ h1 = h g @ 50 psia = 108.81 Btu/lbm s ⎬ sat. vapor ⎭ s1 = s g @ 50 psia = 0.22188 kJ/kg ⋅ K P2 = 160 psia ⎫ ⎬ h2 = 119.19 Btu/lbm s 2 = s1 ⎭The mass flow rate of the refrigerant in the ideal case is & QH 100,000 Btu/h & Q L = m(h1 − h4 ) ⎯ & ⎯→ m = & = = 1415.0 lbm/h h2 − h3 (119.19 − 48.519) Btu/lbmThe power requirement is & ⎛ 1 kW ⎞ Win = m(h2 − h1 ) = (1415.0 lbm/h)(119.19 − 108.81) Btu/lbm⎜ & ⎟ = 4.305 kW ⎝ 3412.14 Btu/h ⎠With superheating, the mass flow rate and the power input are & QH 100,000 Btu/h & Q L = m(h1 − h4 ) ⎯ & ⎯→ m = & = = 1366.3 lbm/h h2 − h3 (121.71 − 48.519) Btu/lbm & ⎛ 1 kW ⎞ Win = m(h2 − h1 ) = (1366.3 lbm/h)(121.71 − 110.99) Btu/lbm⎜ & ⎟ = 4.293 kW ⎝ 3412.14 Btu/h ⎠Superheating decreases the power requirement slightly.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 28. 11-2811-39 A geothermal heat pump is considered. The degrees of subcooling done on the refrigerant in thecondenser, the mass flow rate of the refrigerant, the heating load, the COP of the heat pump, the minimumpower input are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) From the refrigerant-134a tables .(Tables A-11 through A-13) QH T4 = 20°C⎫ P4 = 572.1 kPa ⎬ Condenser x 4 = 0.23 ⎭ h4 = 121.24 kJ/kg 1.4 MPa 3 s2 = s1 h3 = h4 2 . Expansion Win P1 = 572.1 kPa ⎫ h1 = 261.59 kJ/kg valve Compressor ⎬ x1 = 1 (sat. vap.)⎭ s1 = 0.9223 kJ/kg P2 = 1400 kPa ⎫ 4 1 ⎬ h2 = 280.00 kJ/kg Evaporator s 2 = s1 ⎭ 20°C sat. vap.From the steam tables (Table A-4) x=0.23 hw1 = h f @ 50°C = 209.34 kJ/kg 40°C Water hw 2 = h f @ 40°C = 167.53 kJ/kg 50°CThe saturation temperature at the condenser Tpressure of 1400 kPa and the actual temperatureat the condenser outlet are · QH 2 Tsat @ 1400 kPa = 52.40°C 1.4 MPa · P3 = 1400 kPa ⎫ 3 Win ⎬T3 = 48.59°C (from EES) h3 = 121.24 kJ ⎭Then, the degrees of subcooling is 1 ΔTsubcool = Tsat − T3 = 52.40 − 48.59 = 3.81°C 4s 4 · QL(b) The rate of heat absorbed from the sgeothermal water in the evaporator is & Q = m (h − h ) = (0.065 kg/s)(209.34 − 167.53)kJ/kg = 2.718 kW & L w w1 w2This heat is absorbed by the refrigerant in the evaporator & QL 2.718 kW mR = & = = 0.01936 kg/s h1 − h4 (261.59 − 121.24)kJ/kg(c) The power input to the compressor, the heating load and the COP are & & W = m (h − h ) + Q = (0.01936 kg/s)(280.00 − 261.59)kJ/kg = 0.6564 kW & in R 2 1 out & Q H = m R (h2 − h3 ) = (0.01936 kg/s)(280.00 − 121.24)kJ/kg = 3.074 kW & & QH 3.074 kW COP = = = 4.68 & Win 0.6564 kW(d) The reversible COP of the cycle is 1 1 COPrev = = = 12.92 1 − T L / T H 1 − (25 + 273) /(50 + 273)The corresponding minimum power input is & QH 3.074 kW & Win, min = = = 0.238 kW COPrev 12.92PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 29. 11-29Innovative Refrigeration Systems11-40C Performing the refrigeration in stages is called cascade refrigeration. In cascade refrigeration, twoor more refrigeration cycles operate in series. Cascade refrigerators are more complex and expensive, butthey have higher COPs, they can incorporate two or more different refrigerants, and they can achievemuch lower temperatures.11-41C Cascade refrigeration systems have higher COPs than the ordinary refrigeration systems operatingbetween the same pressure limits.11-42C The saturation pressure of refrigerant-134a at -32°C is 77 kPa, which is below the atmosphericpressure. In reality a pressure below this value should be used. Therefore, a cascade refrigeration systemwith a different refrigerant at the bottoming cycle is recommended in this case.11-43C We would favor the two-stage compression refrigeration system with a flash chamber since it issimpler, cheaper, and has better heat transfer characteristics.11-44C Yes, by expanding the refrigerant in stages in several throttling devices.11-45C To take advantage of the cooling effect by throttling from high pressures to low pressures.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 30. 11-3011-46 [Also solved by EES on enclosed CD] A two-stage compression refrigeration system withrefrigerant-134a as the working fluid is considered. The fraction of the refrigerant that evaporates as it isthrottled to the flash chamber, the rate of heat removed from the refrigerated space, and the COP are to bedetermined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3The flash chamber is adiabatic.Analysis (a) The enthalpies of the refrigerant at several states are determined from the refrigerant tables(Tables A-11, A-12, and A-13) to be h1 = 239.16 kJ/kg, h2 = 265.31 kJ/kg T 1 MPa h3 = 259.30 kJ/kg, 4 h5 = 107.32 kJ/kg, h6 = 107.32 kJ/kg 0.5 MPa 5 h7 = 73.33 kJ/kg, h8 = 73.33 kJ/kg 2 AThe fraction of the refrigerant that evaporates as 7 0.14 MPait is throttled to the flash chamber is simply the 6 3 9 Bquality at state 6, · 1 8 h6 − h f 107.32 − 73.33 QL x6 = = = 0.1828 h fg 185.98 s(b) The enthalpy at state 9 is determined from an energy balance on the mixing chamber: & & & E in − E out = ΔE system 0 (steady) =0 & & E in = E out ∑m h = ∑m h & & e e i i (1)h9= x6 h3 + (1 − x6 )h2 h9 = (0.1828)(259.30 ) + (1 − 0.1828)(265.31) = 264.21 kJ/kgalso, P4 = 1 MPa ⎫ ⎬ h4 = 278.97 kJ/kg s 4 = s 9 = 0.94083 kJ/kg ⋅ K ⎭Then the rate of heat removed from the refrigerated space and the compressor work input per unit mass ofrefrigerant flowing through the condenser are m B = (1 − x 6 )m A = (1 − 0.1828)(0.25 kg/s ) = 0.2043 kg/s & & Q L = m B (h1 − h8 ) = (0.2043 kg/s )(239.16 − 73.33) kJ/kg = 33.88 kW & & Win = WcompI,in + WcompII,in = m A (h4 − h9 ) + m B (h2 − h1 ) & & & & & = (0.25 kg/s )(278.97 − 264.21) kJ/kg + (0.2043 kg/s )(265.31 − 239.16) kJ/kg = 9.03 kW(c) The coefficient of performance is determined from & QL 33.88 kW COPR = = = 3.75 & Wnet,in 9.03 kWPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 31. 11-3111-47 EES Problem 11-46 is reconsidered. The effects of the various refrigerants in EES data bank forcompressor efficiencies of 80, 90, and 100 percent is to be investigated.Analysis The problem is solved using EES, and the results are tabulated and plotted below."Input Data""P[1]=140 [kPa] P[4] = 1000 [kPa]P[6]=500 [kPa]Eta_compB =1.0Eta_compA =1.0"m_dot_A=0.25 [kg/s]"High Pressure Compressor A"P[9]=P[6]h4s=enthalpy(R134a,P=P[4],s=s[9]) "State 4s is the isentropic value of state 4"h[9]+w_compAs=h4s "energy balance on isentropic compressor"w_compA=w_compAs/Eta_compA"definition of compressor isentropic efficiency"h[9]+w_compA=h[4] "energy balance on real compressor-assumed adiabatic"s[4]=entropy(R134a,h=h[4],P=P[4]) "properties for state 4"T[4]=temperature(R134a,h=h[4],P=P[4])W_dot_compA=m_dot_A*w_compA"Condenser"P[5]=P[4] "neglect pressure drops across condenser"T[5]=temperature(R134a,P=P[5],x=0) "properties for state 5, assumes sat. liq. at cond. exit"h[5]=enthalpy(R134a,T=T[5],x=0) "properties for state 5"s[5]=entropy(R134a,T=T[5],x=0)h[4]=q_out+h[5] "energy balance on condenser"Q_dot_out = m_dot_A*q_out"Throttle Valve A"h[6]=h[5] "energy balance on throttle - isenthalpic"x6=quality(R134a,h=h[6],P=P[6]) "properties for state 6"s[6]=entropy(R134a,h=h[6],P=P[6])T[6]=temperature(R134a,h=h[6],P=P[6])"Flash Chamber"m_dot_B = (1-x6) * m_dot_AP[7] = P[6]h[7]=enthalpy(R134a, P=P[7], x=0)s[7]=entropy(R134a,h=h[7],P=P[7])T[7]=temperature(R134a,h=h[7],P=P[7])"Mixing Chamber"x6*m_dot_A*h[3] + m_dot_B*h[2] =(x6* m_dot_A + m_dot_B)*h[9]P[3] = P[6]h[3]=enthalpy(R134a, P=P[3], x=1) "properties for state 3"s[3]=entropy(R134a,P=P[3],x=1)T[3]=temperature(R134a,P=P[3],x=x1)s[9]=entropy(R134a,h=h[9],P=P[9]) "properties for state 9"T[9]=temperature(R134a,h=h[9],P=P[9])"Low Pressure Compressor B"x1=1 "assume flow to compressor inlet to be saturated vapor"PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 32. 11-32h[1]=enthalpy(R134a,P=P[1],x=x1) "properties for state 1"T[1]=temperature(R134a,P=P[1], x=x1)s[1]=entropy(R134a,P=P[1],x=x1)P[2]=P[6]h2s=enthalpy(R134a,P=P[2],s=s[1]) " state 2s is isentropic state at comp. exit"h[1]+w_compBs=h2s "energy balance on isentropic compressor"w_compB=w_compBs/Eta_compB"definition of compressor isentropic efficiency"h[1]+w_compB=h[2] "energy balance on real compressor-assumed adiabatic"s[2]=entropy(R134a,h=h[2],P=P[2]) "properties for state 2"T[2]=temperature(R134a,h=h[2],P=P[2])W_dot_compB=m_dot_B*w_compB"Throttle Valve B"h[8]=h[7] "energy balance on throttle - isenthalpic"x8=quality(R134a,h=h[8],P=P[8]) "properties for state 8"s[8]=entropy(R134a,h=h[8],P=P[8])T[8]=temperature(R134a,h=h[8],P=P[8])"Evaporator"P[8]=P[1] "neglect pressure drop across evaporator"q_in + h[8]=h[1] "energy balance on evaporator"Q_dot_in=m_dot_B*q_in"Cycle Statistics"W_dot_in_total = W_dot_compA + W_dot_compBCOP=Q_dot_in/W_dot_in_total "definition of COP" ηcompA ηcompB Qout COP 0,8 0,8 45.32 2.963 0,8333 0,8333 44.83 3.094 0,8667 0,8667 44.39 3.225 0,9 0,9 43.97 3.357 0,9333 0,9333 43.59 3.488 0,9667 0,9667 43.24 3.619 1 1 42.91 3.751 R134a 125 100 75 50 5 4T [C] 1000 kPa 25 2 7 500 kPa 9 6 3 0 140 kPa 1 -25 8 -50 0,0 0,2 0,4 0,6 0,8 1,0 1,2 s [kJ/kg-K]PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
- 33. 11-33 45,5 3,8 3,7 45 3,6 44,5 3,5 Qout [kW] 3,4 COP 44 3,3 43,5 3,2 3,1 43 3 42,5 2,9 0,8 0,84 0,88 0,92 0,96 1 ηcomp COP vs Flash Cham ber Pressure, P 6 3.80 3.75 3.70 3.65 COP 3.60 3.55 3.50 3.45 200 300 400 500 600 700 800 900 1000 1100 P[6] [kPa]PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.

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