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Ch.11

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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.
  • 34. 11-3411-48 [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 = 255.90 kJ/kg h3 = 251.88 kJ/kg, T 1 MPa h5 = 107.32 kJ/kg, h6 = 107.32 kJ/kg 4 h7 = 55.16 kJ/kg, h8 = 55.16 kJ/kg 0.32 MPaThe fraction of the refrigerant that evaporates 5 2as it is throttled to the flash chamber is simply Athe quality at state 6, 7 0.14 MPa 6 3 9 h6 − h f 107.32 − 55.16 B x6 = = = 0.2651 h fg 196.71 1 8 ·(b) The enthalpy at state 9 is determined from an QLenergy balance on the mixing chamber: s & & & E in − E out = ΔEsystem 0 (steady) =0 & & E in = E out ∑m h = ∑m h & & e e i i (1)h9= x 6 h3 + (1 − x 6 )h2 h9 = (0.2651)(251.88) + (1 − 0.2651)(255.90 ) = 254.84 kJ/kgand P9 = 0.32 MPa ⎫ ⎬ s 9 = 0.94074 kJ/kg ⋅ K h9 = 254.84 kJ/kg ⎭ P4 = 1 MPa ⎫also, ⎬ h4 = 278.94 kJ/kg s 4 = s 9 = 0.94074 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 − x6 )m A = (1 − 0.2651)(0.25 kg/s ) = 0.1837 kg/s & & QL = m B (h1 − h8 ) = (0.1837 kg/s )(239.16 − 55.16 ) kJ/kg = 33.80 kW & & Win = WcompI,in + WcompII,in = m A (h4 − h9 ) + m B (h2 − h1 ) & & & & & = (0.25 kg/s )(278.94 − 254.84) kJ/kg + (0.1837 kg/s )(255.90 − 239.16) kJ/kg = 9.10 kW(c) The coefficient of performance is determined from & QL 33.80 kW COPR = = = 3.71 & Wnet,in 9.10 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.
  • 35. 11-3511-49 A two-stage cascade refrigeration cycle is considered. The mass flow rate of the refrigerant throughthe upper cycle, the rate of heat removal from the refrigerated space, and the COP of the refrigerator are tobe determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) The properties are to be obtained from the refrigerant tables (Tables A-11 through A-13): h1 = h g @ 200 kPa = 244.46 kJ/kg . s1 = s g @ 200 kPa = 0.9377 kJ/kg.K QH P2 = 500 kPa ⎫ Condenser ⎬ h2 s = 263.30 kJ/kg s 2 = s1 ⎭ 7 6 . h −h Expansion Win η C = 2s 1 valve h2 − h1 Compressor 263.30 − 244.46 0.80 = ⎯ ⎯→ h2 = 268.01 kJ/kg h2 − 244.46 8 5 Evaporator h3 = h f @ 500 kPa = 73.33 kJ/kg h4 = h3 = 73.33 kJ/kg Condenser 3 h5 = h g @ 400 kPa = 255.55 kJ/kg 2 . Expansion s 5 = s g @ 400 kPa = 0.9269 kJ/kg.K Win valve Compressor P6 = 1200 kPa ⎫ ⎬ h6 s = 278.33 kJ/kg s6 = s5 ⎭ 4 1 Evaporator h 6 s − h5 ηC = . h6 − h5 QL 278.33 − 255.55 0.80 = ⎯ ⎯→ h6 = 284.02 kJ/kg h6 − 255.55 h7 = h f @ 1200 kPa = 117.77 kJ/kg h8 = h7 = 117.77 kJ/kgThe mass flow rate of the refrigerant through the upper cycle is determined from an energy balance on theheat exchanger m A (h5 − h8 ) = m B (h2 − h3 ) & & m A (255.55 − 117.77)kJ/kg = (0.15 kg/s)(268.01 − 73.33)kJ/kg ⎯ & ⎯→ m A = 0.212 kg/s &(b) The rate of heat removal from the refrigerated space is & Q L = m B (h1 − h4 ) = (0.15 kg/s)(244.46 − 73.33)kJ/kg = 25.67 kW &(c) The power input and the COP are & Win = m A (h6 − h5 ) + m B (h2 − h1 ) & & = (0.15 kg/s)(284.02 − 255.55)kJ/kg + (0.212 kg/s)(268.01 − 244.46)kJ/kg = 9.566 kW & Q L 25.67 COP = = = 2.68 & Win 9.566PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
  • 36. 11-3611-50 A two-evaporator compression refrigeration cycle with refrigerant-134a as the working fluid isconsidered. The cooling rate of the high-temperature evaporator, the power required by the compressor,and the COP of the system are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Condenser 2 T 3 · Compressor QH 2 Expansion 3 800 kPa · Expansion Win 1 valve valve Evaporator 1 4 0°C 5 5 4 -26.4°C Expansion 6 · 7 1 QL valve 7 Evaporator 2 s 6Analysis From the refrigerant tables (Tables A-11, A-12, and A-13), P3 = 800 kPa ⎫ ⎬ h3 = h f @ 800 kPa = 95.47 kJ/kg sat. liquid ⎭ h4 = h6 ≅ h3 = 95.47 kJ/kg ( throttling) T5 = 0°C ⎫ ⎬ h = h g @ 0°C = 250.45 kJ/kg sat. vapor ⎭ 5 T7 = −26.4°C ⎫ ⎬ h7 = h g @ − 26.4°C = 234.44 kJ/kg sat. vapor ⎭The mass flow rate through the low-temperature evaporator is found by & QL 8 kJ/s & Q L = m 2 ( h7 − h6 ) ⎯ & ⎯→ m 2 = & = = 0.05757 kg/s h7 − h6 (234.44 − 95.47) kJ/kgThe mass flow rate through the warmer evaporator is then m1 = m − m 2 = 0.1 − 0.05757 = 0.04243 kg/s & & &Applying an energy balance to the point in the system where the two evaporator streams are recombinedgives m1 h5 + m 2 h7 (0.04243)(250.45) + (0.05757)(234.44) & & m1 h5 + m 2 h7 = mh1 ⎯ h1 = & & & ⎯→ = = 241.23 kJ/kg & m 0.1Then,PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
  • 37. 11-37 P1 = Psat @ − 26.4°C ≅ 100 kPa ⎫ ⎬ s1 = 0.9789 kJ/kg ⋅ K h1 = 241.23 kJ/kg ⎭ P2 = 800 kPa ⎫ ⎬ h2 = 286.26 kJ/kg s 2 = s1 ⎭The cooling rate of the high-temperature evaporator is & Q L = m1 (h5 − h4 ) = (0.04243 kg/s )(250.45 − 95.47) kJ/kg = 6.58 kW &The power input to the compressor is & Win = m(h2 − h1 ) = (0.1 kg/s)(286.26 − 241.23) kJ/kg = 4.50 kW &The COP of this refrigeration system is determined from its definition, & Q L (8 + 6.58) kW COPR = = = 3.24 & Win 4.50 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.
  • 38. 11-3811-51E A two-evaporator compression refrigeration cycle with refrigerant-134a as the working fluid isconsidered. The power required by the compressor and the COP of the system are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Condenser T 2 3 · Compressor QH 2 Expansion Expansion 3 140 psia · 1 valve Win valve Evaporator 1 30 psia 5 5 4 4 15 psia Expansion 6 · 7 1 valve QL 7 Evaporator 2 6 sAnalysis From the refrigerant tables (Tables A-11E, A-12E, and A-13E), P3 = 140 psia ⎫ ⎬ h3 = h f @ 140 psia = 45.304 Btu/lbm sat. liquid ⎭ h4 = h6 ≅ h3 = 45.304 Btu/lbm ( throttling) P5 = 30 psia ⎫ ⎬ h = h g @ 30 psia = 105.32 Btu/lbm sat. vapor ⎭ 5 P7 = 15 psia ⎫ ⎬ h = h g @ 15 psia = 100.99 Btu/lbm sat. vapor ⎭ 7The mass flow rates through the high-temperature and low-temperature evaporators are found by & Q L,1 3000 Btu/h & Q L,1 = m1 (h5 − h4 ) ⎯ & ⎯→ m1 = & = = 49.99 lbm/h h5 − h4 (105.32 − 45.304) Btu/lbm & Q L, 2 10,000 Btu/h & Q L , 2 = m 2 ( h7 − h6 ) ⎯ & ⎯→ m 2 = & = = 179.58 lbm/h h7 − h6 (100.99 − 45.304) Btu/lbmApplying an energy balance to the point in the system where the two evaporator streams are recombinedgives m h + m2h7 (49.99)(105.32) + (179.58)(100.99) & & m1h5 + m2h7 = (m1 + m2 )h1 ⎯ h1 = 1 5 & & & & ⎯→ = = 101.93 Btu/lbm m1 + m2 & & 49.99 + 179.58Then, P1 = 15 psia ⎫ ⎬ s = 0.2293 Btu/lbm ⋅ R h1 = 101.93 Btu/lbm ⎭ 1 P2 = 140 psia ⎫ ⎬ h2 = 122.24 Btu/lbm s 2 = s1 ⎭The power input to the compressor is & ⎛ 1 kW ⎞ Win = (m1 + m2 )(h2 − h1) = (49.99 + 179.58) lbm/h(122.24 − 101.93) Btu/lbm⎜ & & ⎟ = 1.366 kW ⎝ 3412.14 Btu/h ⎠The COP of this refrigeration system is determined from its definition, Q& (10,000 + 3000) Btu/h ⎛ 1 kW ⎞ COPR = L = ⎜ ⎟ = 2.79 & Win 1.366 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.
  • 39. 11-3911-52E A two-evaporator compression refrigeration cycle with refrigerant-134a as the working fluid isconsidered. The power required by the compressor and the COP of the system are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Condenser T 2 3 · Compressor QH 2 Expansion Expansion 3 140 psia · 1 valve Win valve Evaporator II 60 psia 5 4 4 5 15 psia Expansion 6 · 7 1 QL 7 valve Evaporator I s 6Analysis From the refrigerant tables (Tables A-11E, A-12E, and A-13E), P3 = 140 psia ⎫ ⎬ h3 = h f @ 140 psia = 45.304 Btu/lbm sat. liquid ⎭ h4 = h6 ≅ h3 = 45.304 Btu/lbm ( throttling ) P5 = 60 psia ⎫ ⎬ h = h g @ 60 psia = 110.11 Btu/lbm sat. vapor ⎭ 5 P7 = 15 psia ⎫ ⎬ h = h g @ 15 psia = 100.99 Btu/lbm sat. vapor ⎭ 7The mass flow rates through the high-temperature and low-temperature evaporators are found by & Q L,1 48,000 Btu/h & Q L,1 = m1 (h5 − h4 ) ⎯ & ⎯→ m1 = & = = 740.7 lbm/h h5 − h4 (110.11 − 45.304) Btu/lbm & Q L, 2 10,000 Btu/h & Q L , 2 = m 2 ( h7 − h6 ) ⎯ & ⎯→ m 2 = & = = 179.6 lbm/h h7 − h6 (100.99 − 45.304) Btu/lbmApplying an energy balance to the point in the system where the two evaporator streams are recombinedgives m h + m2h7 (740.7)(110.11) + (179.58)(100.99) & & m1h5 + m2h7 = (m1 + m2 )h1 ⎯ h1 = 1 5 & & & & ⎯→ = = 108.33 Btu/lbm m1 + m2 & & 740.7 + 179.58Then, P1 = 15 psia ⎫ ⎬ s = 0.2430 Btu/lbm ⋅ R h1 = 108.33 Btu/lbm ⎭ 1 P2 = 140 psia ⎫ ⎬ h2 = 130.45 Btu/lbm s 2 = s1 ⎭The power input to the compressor is & ⎛ 1 kW ⎞ Win = (m1 + m2 )(h2 − h1) = (740.7 + 179.6) lbm/h(130.45 − 108.33) Btu/lbm⎜ & & ⎟ = 5.966 kW ⎝ 3412.14 Btu/h ⎠The COP of this refrigeration system is determined from its definition, & Q (48,000 + 10,000) Btu/h ⎛ 1 kW ⎞ COPR = L = ⎜ ⎟ = 2.85 & Win 5.966 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.
  • 40. 11-4011-53 A two-stage compression refrigeration system with a separation unit is considered. The mass flowrate through the two compressors, the power used by the compressors, and the system’s COP are to bedetermined.Assumptions 1 Steady operating conditionsexist. 2 Kinetic and potential energy 2 Condenserchanges are negligible.Analysis From the refrigerant tables (Tables Compressor 3A-11, A-12, and A-13),T1 = −10.1°C ⎫ h1 = h g @ −10.1°C = 244.46 kJ/kg 1 Expansion ⎬ s =s Separator g @ −10.1°C = 0.93773 kJ/kg ⋅ Ksat. vapor ⎭ 1 8 valveP2 = 800 kPa ⎫ ⎬ h2 = 273.24 kJ/kg 4s 2 = s1 ⎭ Compressor 5 ExpansionP3 = 800 kPa ⎫ valve ⎬ h3 = h f @800 kPa = 95.47 kJ/kg 7sat. liquid ⎭ Evaporator 6h4 ≅ h3 = 95.47 kJ/kg ( throttling)T5 = −10.1°C ⎫ ⎬ h5 = h f @ −10.1° C = 38.43 kJ/kg Tsat. liquid ⎭ 2h6 ≅ h5 = 38.43 kJ/kg ( throttling) 0.8 MPa 3T7 = −40°C ⎫ h7 = h g @ − 40°C = 225.86 kJ/kg 8 ⎬ -10.1°Csat. vapor ⎭ s 7 = s g @ − 40°C = 0.96866 kJ/kg ⋅ K 5 4 1P8 = Psat @ −10.1°C = 200 kPa ⎫ ⎬ h8 = 252.74 kJ/kg -40°Cs8 = s 7 ⎭ 6 · 7 QLThe mass flow rate through the evaporator isdetermined from s & QL 30 kJ/s & Q L = m 6 ( h7 − h6 ) ⎯ & ⎯→ m 6 = & = = 0.1601 kg/s h7 − h6 (225.86 − 38.43) kJ/kgAn energy balance on the separator gives h8 − h5 252.74 − 38.43 m 6 (h8 − h5 ) = m 2 (h1 − h4 ) ⎯ & & ⎯→ m 2 = m 6 & & = (0.1601) = 0.2303 kg/s h1 − h4 244.46 − 95.47The total power input to the compressors is & Win = m 6 (h8 − h7 ) + m 2 (h2 − h1 ) & & = (0.1601 kg/s)(252.74 − 225.86) kJ/kg + (0.2303 kg/s)(273.24 − 244.46) kJ/kg = 10.93 kWThe COP of this refrigeration system is determined from its definition, & QL 30 kW COPR = = = 2.74 & Win 10.93 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.
  • 41. 11-4111-54E A two-stage compression refrigeration system with a separation unit is considered. The coolingload and the system’s COP are to be determined.Assumptions 1 Steady operating conditionsexist. 2 Kinetic and potential energy Condenser 2changes are negligible. Compressor 3Analysis From the refrigerant tables (TablesA-11E, A-12E, and A-13E), 1 ExpansionP1 = 120 psia ⎫ h1 = h g @ 120 psia = 115.16 Btu/lbm 8 Separator valve ⎬ s =ssat. vapor ⎭ 1 g @ 120 psia = 0.21924 Btu/lbm ⋅ R 4P2 = 300 psia ⎫ 5 ⎬ h2 = 123.06 Btu/lbm Compressors 2 = s1 ⎭ Expansion valveP3 = 300 psia ⎫ ⎬ h3 = h f @ 300 psia = 66.339 Btu/lbm 7 Evaporator 6sat. liquid ⎭h 4 ≅ h3 = 66.339 Btu/lbm ( throttling) TP5 = 120 psia ⎫ ⎬ h5 = h f @ 120 psia = 41.787 Btu/lbm 2sat. liquid ⎭ 300 psia 3h6 ≅ h5 = 41.787 kJ/kg ( throttling) 8 120 psiaP7 = 60 psia ⎫ h7 = h g @ 60 psia = 110.11 Btu/lbm 5 ⎬ 4 1sat. vapor ⎭ s 7 = s g @ 60 psia = 0.22127 Btu/lbm ⋅ R 60 psiaP8 = 120 psia ⎫ 6 · 7 ⎬ h8 = 116.28 Btu/lbm QLs8 = s 7 ⎭An energy balance on the separator gives s h8 − h5 116.28 − 41.787 m 6 (h8 − h5 ) = m 2 (h1 − h4 ) ⎯ & & ⎯→ m 2 = m 6 & & = m6 & = 1.5258m 6 & h1 − h4 115.16 − 66.339The total power input to the compressors is given by & Win = m 6 (h8 − h7 ) + m 2 (h2 − h1 ) & & = m 6 (h8 − h7 ) + 1.5258m 6 (h2 − h1 ) & & &Solving for m 6 , & Win (25 × 3412.14) Btu/h m6 = & = = 4681 lbm/h (h8 − h7 ) + 1.5258(h2 − h1 ) (116.28 − 110.11) + 1.5258(123.06 − 115.16) Btu/lbmThe cooling effect produced by this system is then & Q L = m 6 ( h7 − h6 ) = ( 4681 lbm/h)(110.11 − 41.787) Btu/lbm = 319,800 Btu/h &The COP of this refrigeration system is determined from its definition, & Q L (319,800/3412.14) kW COPR = = = 3.75 & Win 25 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.
  • 42. 11-4211-55 A two-stage cascade refrigeration system is considered. Each stage operates on the ideal vapor-compression cycle with upper cycle using water and lower cycle using refrigerant-134a as the workingfluids. The mass flow rate of R-134a and water in their respective cycles and the overall COP of thissystem are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3The heat exchanger is adiabatic.Analysis From the water and refrigerant tables (Tables A-4, A-5, A-6, A-11, A-12, and A-13), T1 = 5°C ⎫ h1 = h g @ 5°C = 2510.1 kJ/kg ⎬ sat. vapor ⎭ s1 = s g @ 5°C = 9.0249 kJ/kg ⋅ K T P2 = 1.6 MPa ⎫ ⎬ h2 = 5083.4 kJ/kg s 2 = s1 ⎭ 2 1.6 MPa 3 P3 = 1.6 MPa ⎫ 6 ⎬ h3 = h f @ 1.6 MPa = 858.44 kJ/kg sat. liquid ⎭ 5°C 7 h4 ≅ h3 = 858.44 kJ/kg ( throttling) 4 400 kPa 1 T5 = −40°C ⎫ h5 = h g @ − 40°C = 225.86 kJ/kg ⎬ 8 -40°C · 5 sat. vapor ⎭ s 5 = s g @ − 40°C = 0.96866 kJ/kg ⋅ K QL P6 = 400 kPa ⎫ ⎬ h6 = 267.59 kJ/kg s s6 = s5 ⎭ P7 = 400 kPa ⎫ ⎬ h7 = h f @ 400 kPa = 63.94 kJ/kg sat. liquid ⎭ h8 ≅ h7 = 63.94 kJ/kg ( throttling)The mass flow rate of R-134a is determined from & QL 20 kJ/s & Q L = m R (h5 − h8 ) ⎯ & ⎯→ m R = & = = 0.1235 kg/s h5 − h8 (225.86 − 63.94) kJ/kgAn energy balance on the heat exchanger gives the mass flow rate of water m R (h6 − h7 ) = m w (h1 − h4 ) & & h6 − h7 267.59 − 63.94 ⎯ ⎯→ m w = m R & & = (0.1235 kg/s) = 0.01523 kg/s h1 − h4 2510.1 − 858.44The total power input to the compressors is & Win = m R (h6 − h5 ) + m w (h2 − h1 ) & & = (0.1235 kg/s)(267.59 − 225.86) kJ/kg + (0.01523 kg/s)(5083.4 − 2510.1) kJ/kg = 44.35 kJ/sThe COP of this refrigeration system is determined from its definition, & QL 20 kJ/s COPR = = = 0.451 & Win 44.35 kJ/sPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
  • 43. 11-4311-56 A two-stage vapor-compression refrigeration system with refrigerant-134a as the working fluid isconsidered. The process with the greatest exergy destruction is to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis From Prob. 11-55 and the water and refrigerant tables (Tables A-4, A-5, A-6, A-11, A-12, andA-13), s1 = s 2 = 9.0249 kJ/kg ⋅ K s 3 = 2.3435 kJ/kg ⋅ K T s 4 = 3.0869 kJ/kg ⋅ K 2 s 5 = s 6 = 0.96866 kJ/kg ⋅ K 1.6 MPa s 7 = 0.24757 kJ/kg ⋅ K 3 6 s 8 = 0.27423 kJ/kg ⋅ K 5°C m R = 0.1235 kg/s & 7 m w = 0.01751 kg/s & 4 400 kPa 1 q L = h5 − h8 = 161.92 kJ/kg q H = h2 − h3 = 4225.0 kJ/kg 8 -40°C · 5 QL T L = −30°C = 243 K T H = 30°C = 303 K s T0 = 30°C = 303 KThe exergy destruction during a process of a stream from an inlet state to exit state is given by ⎛ q q ⎞ x dest = T0 s gen = T0 ⎜ s e − s i − in + out ⎜ ⎟ ⎟ ⎝ Tsource Tsink ⎠Application of this equation for each process of the cycle gives & ⎛ q ⎞ X destroyed, 23 = m w T0 ⎜ s 3 − s 2 + H ⎟ & ⎜ ⎝ TH ⎟⎠ ⎛ 4225.0 ⎞ = (0.01751)(303 K )⎜ 2.3435 − 9.0249 + ⎟ = 38.53 kJ/s ⎝ 303 ⎠ & X destroyed, 34 = m w T0 ( s 4 − s 3 ) = (0.01751)(303)(3.0869 − 2.3435) = 3.94 kJ/s & & X destroyed, 78 = m R T0 ( s8 − s 7 ) = (0.1235)(303)(0.27423 − 0.24757) = 1.00 kJ/s & & ⎛ q ⎞ X destroyed, 85 = m R T0 ⎜ s 5 − s8 − L ⎟ & ⎜ ⎝ TL ⎟ ⎠ ⎛ 161.92 ⎞ = (0.1235)(303)⎜ 0.96866 − 0.27423 − ⎟ = 1.05 kJ/s ⎝ 243 ⎠ X destroyed, heat exch = T0 [m w ( s1 − s 4 ) + m R ( s 7 − s 6 )] & & & = (303)[(0.01751)(9.0249 − 3.0869) + (0.1235)(0.24757 − 0.96866)] = 4.52 kJ/sFor isentropic processes, the exergy destruction is zero: & X destroyed,12 = 0 & X destroyed, 56 = 0Note that heat is absorbed from a reservoir at -30°C (243 K) and rejected to a reservoir at 30°C (303 K),which is also taken as the dead state temperature. Alternatively, one may use the standard 25°C (298 K) asthe dead state temperature, and perform the calculations accordingly. The greatest exergy destructionoccurs in the condenser.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.
  • 44. 11-44Gas Refrigeration Cycles11-57C The ideal gas refrigeration cycle is identical to the Brayton cycle, except it operates in the reverseddirection.11-58C The reversed Stirling cycle is identical to the Stirling cycle, except it operates in the reverseddirection. Remembering that the Stirling cycle is a totally reversible cycle, the reversed Stirling cycle isalso totally reversible, and thus its COP is 1 COPR, Stirling = TH / TL − 111-59C In the ideal gas refrigeration cycle, the heat absorption and the heat rejection processes occur atconstant pressure instead of at constant temperature.11-60C In aircraft cooling, the atmospheric air is compressed by a compressor, cooled by the surroundingair, and expanded in a turbine. The cool air leaving the turbine is then directly routed to the cabin.11-61C No; because h = h(T) for ideal gases, and the temperature of air will not drop during a throttling(h1 = h2) process.11-62C By regeneration.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.
  • 45. 11-4511-63 [Also solved by EES on enclosed CD] An ideal-gas refrigeration cycle with air as the working fluidis considered. The rate of refrigeration, the net power input, and the COP are to be determined.Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with variable specific heats. 3Kinetic and potential energy changes are negligible.Analysis (a) We assume both the turbine and the compressor to be isentropic, the turbine inlet temperatureto be the temperature of the surroundings, and the compressor inlet temperature to be the temperature ofthe refrigerated space. From the air table (Table A-17), T1 = 285 K ⎯→ ⎯ h1 = 28514 kJ / kg . T Pr 1 = 11584 . · 2 QH T1 = 320 K ⎯→ ⎯ h3 = 320.29 kJ / kg Pr 3 = 1.7375 3 47°C 12°CThus, 1 · ⎛ 250 ⎞ QRefrig ⎟(1.1584 ) = 5.792 ⎯ T2 = 450.4 K P2 4 Pr2 = Pr = ⎜ ⎯→ P 1 ⎝ 50 ⎠ 1 h2 = 452.17 kJ/kg s ⎛ 50 ⎞ ⎟(1.7375) = 0.3475 ⎯ T4 = 201.8 K P4 Pr4 = Pr = ⎜ ⎯→ P3 3 ⎝ 250 ⎠ h4 = 201.76 kJ/kgThen the rate of refrigeration is Qrefrig = m(qL ) = m (h1 − h4 ) = (0.08 kg/s)(285.14 − 201.76) kJ/kg = 6.67 kW & & &(b) The net power input is determined from & & & Wnet, in = Wcomp, in − Wturb, outwhere Wcomp,in = m(h2 − h1 ) = (0.08 kg/s)(452.17 − 285.14 ) kJ/kg = 13.36 kW & & Wturb,out = m (h3 − h4 ) = (0.08 kg/s)(320.29 − 201.76 ) kJ/kg = 9.48 kW & &Thus, & Wnet, in = 13.36 − 9.48 = 3.88 kW(c) The COP of this ideal gas refrigeration cycle is determined from & Q 6.67 kW COPR = & L = = 1.72 Wnet, in 3.88 kWPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
  • 46. 11-4611-64 EES Problem 11-63 is reconsidered. The effects of compressor and turbine isentropic efficiencies onthe rate of refrigeration, the net power input, and the COP are to be investigated.Analysis The problem is solved using EES, and the solution is given below."Input data"T[1] = 12 [C]P[1]= 50 [kPa]T[3] = 47 [C]P[3]=250 [kPa]m_dot=0.08 [kg/s]Eta_comp = 1.00Eta_turb = 1.0"Compressor anaysis"s[1]=ENTROPY(Air,T=T[1],P=P[1])s2s=s[1] "For the ideal case the entropies are constant across the compressor"P[2] = P[3]s2s=ENTROPY(Air,T=Ts2,P=P[2])"Ts2 is the isentropic value of T[2] at compressor exit"Eta_comp = W_dot_comp_isen/W_dot_comp "compressor adiabatic efficiency,W_dot_comp > W_dot_comp_isen"m_dot*h[1] + W_dot_comp_isen = m_dot*hs2"SSSF First Law for the isentropic compressor,assuming: adiabatic, ke=pe=0, m_dot is the mass flow rate in kg/s"h[1]=ENTHALPY(Air,T=T[1])hs2=ENTHALPY(Air,T=Ts2)m_dot*h[1] + W_dot_comp = m_dot*h[2]"SSSF First Law for the actual compressor,assuming: adiabatic, ke=pe=0"h[2]=ENTHALPY(Air,T=T[2])s[2]=ENTROPY(Air,h=h[2],P=P[2])"Heat Rejection Process 2-3, assumed SSSF constant pressure process"m_dot*h[2] + Q_dot_out = m_dot*h[3]"SSSF First Law for the heat exchanger,assuming W=0, ke=pe=0"h[3]=ENTHALPY(Air,T=T[3])"Turbine analysis"s[3]=ENTROPY(Air,T=T[3],P=P[3])s4s=s[3] "For the ideal case the entropies are constant across the turbine"P[4] = P[1]s4s=ENTROPY(Air,T=Ts4,P=P[4])"Ts4 is the isentropic value of T[4] at turbine exit"Eta_turb = W_dot_turb /W_dot_turb_isen "turbine adiabatic efficiency, W_dot_turb_isen >W_dot_turb"m_dot*h[3] = W_dot_turb_isen + m_dot*hs4"SSSF First Law for the isentropic turbine, assuming:adiabatic, ke=pe=0"hs4=ENTHALPY(Air,T=Ts4)m_dot*h[3] = W_dot_turb + m_dot*h[4]"SSSF First Law for the actual compressor, assuming:adiabatic, ke=pe=0"h[4]=ENTHALPY(Air,T=T[4])s[4]=ENTROPY(Air,h=h[4],P=P[4])"Refrigeration effect:"m_dot*h[4] + Q_dot_Refrig = m_dot*h[1]"Cycle analysis"W_dot_in_net=W_dot_comp-W_dot_turb"External work supplied to compressor"COP= Q_dot_Refrig/W_dot_in_net"The following is for plotting data only:"Ts[1]=Ts2ss[1]=s2sTs[2]=Ts4ss[2]=s4sPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
  • 47. 11-47 COP ηcomp ηturb QRefrig Winnet [kW] [kW] 0.6937 0.7 1 6.667 9.612 0.9229 0.8 1 6.667 7.224 1.242 0.9 1 6.667 5.368 1.717 1 1 6.667 3.882 7 6 5 QRefrig [kW] 4 3 η turb 2 0.7 0.85 1 1.0 0 0,7 0,75 0,8 0,85 0,9 0,95 1 ηcomp 12 10 Win;net [kW] 8 6 η turb 4 0.7 2 0.85 1.0 0 0,7 0,75 0,8 0,85 0,9 0,95 1 ηcomp 2 η turb 1,5 0.7 0.85 1 1.0 COP 0,5 0 0,7 0,75 0,8 0,85 0,9 0,95 1 ηcompPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
  • 48. 11-4811-65 [Also solved by EES on enclosed CD] An ideal-gas refrigeration cycle with air as the workingfluid is considered. The rate of refrigeration, the net power input, and the COP are to be determined.Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with variable specific heats. 3Kinetic and potential energy changes are negligible.Analysis (a) We assume the turbine inlet temperature to be the temperature of the surroundings, and thecompressor inlet temperature to be the temperature of the refrigerated space. From the air table (Table A-17), T1 = 285 K ⎯→ ⎯ h1 = 28514 kJ / kg . 2 Pr 1 = 11584 . T 2 · T1 = 320 K ⎯→ ⎯ h3 = 320.29 kJ / kg QH Pr 3 = 1.7375 3 47°CThus, 12°C 1 ⎛ 250 ⎞ ⎟(1.1584 ) = 5.792 ⎯ T2 s = 450.4 K P2 · Pr2 = Pr = ⎜ ⎯→ 4 QRefrig P 1 ⎝ 50 ⎠ 4s 1 h2 s = 452.17 kJ/kg s ⎛ 50 ⎞ ⎟(1.7375) = 0.3475 ⎯ T4 s = 201.8 K P Pr4 = 4 Pr3 = ⎜ ⎯→ P3 ⎝ 250 ⎠ h4 s = 201.76 kJ/kgAlso, h3 − h 4 ηT = ⎯→ h4 = h3 − ηT (h3 − h4 s ) ⎯ h3 − h4 s = 320.29 − (0.85)(320.29 − 201.76) = 219.54 kJ/kgThen the rate of refrigeration is Qrefrig = m(qL ) = m (h1 − h4 ) = (0.08 kg/s)(285.14 − 219.54 ) kJ/kg = 5.25 kW & & &(b) The net power input is determined from & & & Wnet, in = Wcomp, in − Wturb, outwhere Wcomp,in = m(h2 − h1 ) = m(h2 s − h1 ) / ηC & & & = (0.08 kg/s)[(452.17 − 285.14 ) kJ/kg]/ (0.80) = 16.70 kW Wturb,out = m(h3 − h4 ) = (0.08 kg/s)(320.29 − 219.54 ) kJ/kg = 8.06 kW & &Thus, & Wnet, in = 16.70 − 8.06 = 8.64 kW(c) The COP of this ideal gas refrigeration cycle is determined from & QL 5.25 kW COPR = = = 0.61 & Wnet, in 8.64 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.
  • 49. 11-4911-66 A gas refrigeration cycle with helium as the working fluid is considered. The minimum temperaturein the cycle, the COP, and the mass flow rate of the helium are to be determined.Assumptions 1 Steady operating conditions exist. 2 Helium is an ideal gas with constant specific heats. 3Kinetic and potential energy changes are negligible.Properties The properties of helium are cp = 5.1926 kJ/kg·K 2and k = 1.667 (Table A-2). T 2Analysis (a) From the isentropic relations, · QH (k −1) / k ⎛P ⎞ T2 s = T1 ⎜ 2 ⎜P ⎟ ⎟ = (263K )(3)0.667 / 1.667 = 408.2K 50°C 3 ⎝ 1 ⎠ (k −1) / k -10°C ⎛P ⎞ 0.667 / 1.667 ⎛1⎞ 1 T4 s = T3 ⎜ 4 ⎜P ⎟ ⎟ = (323K )⎜ ⎟ = 208.1K · ⎝3⎠ 4 QRefrig ⎝ 3 ⎠ 4sand s h3 − h4 T − T4 ηT = = 3 ⎯→ T4 = T3 − ηT (T3 − T4 s ) = 323 − (0.80 )(323 − 208.1) ⎯ h3 − h4 s T3 − T4 s = 231.1 K = T min h − h1 T2 s − T1 ηC = 2s = ⎯→ T2 = T1 + (T2 s − T1 ) / η C = 263 + (408.2 − 263) / (0.80 ) ⎯ h2 − h1 T2 − T1 = 444.5 K(b) The COP of this gas refrigeration cycle is determined from qL qL COPR = = wnet,in wcomp,in − wturb,out h1 − h4 = (h2 − h1 ) − (h3 − h4 ) T1 − T4 = (T2 − T1 ) − (T3 − T4 ) 263 − 231.1 = = 0.356 (444.5 − 263) − (323 − 231.1)(c) The mass flow rate of helium is determined from & Q refrig & Q refrig & Q refrig 18 kJ/s m= & = = = = 0.109 kg/s qL h1 − h4 c p (T1 − T4 ) (5.1926 kJ/kg ⋅ K )(263 − 231.1) KPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
  • 50. 11-5011-67 A regenerative gas refrigeration cycle using air as the working fluid is considered. The effectivenessof the regenerator, the rate of heat removal from the refrigerated space, the COP of the cycle, and therefrigeration load and the COP if this system operated on the simple gas refrigeration cycle are to bedetermined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3Air is an ideal gas with constant specific heats.Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2).Analysis (a) From the isentropic relations, (k −1) / k ⎛P ⎞ T2 s = T1 ⎜ 2 ⎜P ⎟ ⎟ = (273.2 K )(5)0.4 / 1.4 = 432.4 K ⎝ 1 ⎠ . QL h2 s − h1 T2 s − T1 ηC = = Heat h2 − h1 T2 − T1 Exch. 432.4 − 273.2 6 0.80 = ⎯ T2 = 472.5 K ⎯→ T2 − 273.2 Regenerator HeatThe temperature at state 4 can be Exch. 3determined by solving the following two 5 1equations simultaneously: 4 . QH 2 (k −1) / k ⎛P ⎞ 0.4 / 1.4 ⎛1⎞T5 s = T4 ⎜ 5 ⎜P ⎟ ⎟ = T4 ⎜ ⎟ ⎝ 4 ⎠ ⎝5⎠ Turbine h −h T − 193.2η T = 4 5 → 0.85 = 4 h4 − h5 s T4 − T5 s CompressorUsing EES, we obtain T4 = 281.3 K. T 2An energy balance on the regenerator may be written as · 2s QH mc p (T3 − T4 ) = mc p (T1 − T6 ) ⎯ & & ⎯→ T3 − T4 = T1 − T6 3or, 35°C T6 = T1 − T3 + T4 = 273.2 − 308.2 + 281.3 = 246.3 K 0°C Qrege 1The effectiveness of the regenerator is 4 h3 − h4 T3 − T4 308.2 − 281.3 ε regen = = = = 0.434 -80°C ·6 h3 − h6 T3 − T6 308.2 − 246.3 5 5 QRefrig s(b) The refrigeration load is & Q L = mc p (T6 − T5 ) = (0.4 kg/s)(1.005 kJ/kg.K)(246.3 − 193.2)K = 21.36 kW &(c) The turbine and compressor powers and the COP of the cycle are & WC,in = mc p (T2 − T1 ) = (0.4 kg/s)(1.005 kJ/kg.K)(472.5 − 273.2)K = 80.13 kW & & WT,out = mc p (T4 − T5 ) = (0.4 kg/s)(1.005 kJ/kg.K)(281.3 − 193.2)kJ/kg = 35.43 kW & & QL & QL 21.36 COP = = = = 0.478 W& & & WC,in − WT,out 80.13 − 35.43 net,inPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
  • 51. 11-51(d) The simple gas refrigeration cycle analysis is as follows: (k −1) / k 0.4 / 1.4 2 ⎛1⎞ ⎛1⎞ T4 s = T3 ⎜ ⎟ = (308.2 K )⎜ ⎟ = 194.6 K T 2 ⎝r⎠ ⎝5⎠ · QH T3 − T4 308.2 − T4 ηT = ⎯ ⎯→ 0.85 = ⎯ T4 = 211.6 K ⎯→ 3 T3 − T4 s 308.2 − 194.6 35°C 0°C & Q L = mc p (T1 − T4 ) & 1 · 4 QRefrig = (0.4 kg/s)(1.005 kJ/kg.K)(273.2 − 211.6)kJ/kg 4s = 24.74 kW s &W net,in = mc p (T2 − T1 ) − mc p (T3 − T4 ) & & = (0.4 kg/s)(1.005 kJ/kg.K)[(472.5 − 273.2) − (308.2 − 211.6)kJ/kg ] = 41.32 kW & QL 24.74 COP = = = 0.599 W& 41.32 net,inPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
  • 52. 11-5211-68E An ideal gas refrigeration cycle with air as the working fluid has a compression ratio of 4. TheCOP of the cycle is to be determined.Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant specific heats. 3Kinetic and potential energy changes are negligible.Properties The properties of air at room temperature are cp = 0.240 Btu/lbm·R and k = 1.4 (Table A-2Ea).Analysis From the isentropic relations, ( k −1) / k ⎛P ⎞ T2 = T1 ⎜ 2 ⎜P ⎟ ⎟ = (450 R)(4) 0.4 / 1.4 = 668.7 R T 2 · ⎝ 1 ⎠ QH ( k −1) / k ⎛P ⎞ 0.4 / 1.4 ⎛1⎞ 3 T4 = T3 ⎜ 4 ⎜P ⎟ ⎟ = (560 R)⎜ ⎟ = 376.8 R 100°F ⎝ 3 ⎠ ⎝4⎠ -10°F 1The COP of this ideal gas refrigeration cycle is determined from · qL qL 4 QRefrig COPR = = wnet,in wcomp,in − w turb,out s h1 − h4 = (h2 − h1 ) − (h3 − h4 ) T1 − T4 = (T2 − T1 ) − (T3 − T4 ) 450 − 376.8 = = 2.06 (668.7 − 450) − (560 − 376.8)PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
  • 53. 11-5311-69E An gas refrigeration cycle with air as the working fluid has a compression ratio of 4. The COP ofthe cycle is to be determined.Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant specific heats. 3Kinetic and potential energy changes are negligible.Properties The properties of air at room temperature are 2cp = 0.240 Btu/lbm·R and k = 1.4 (Table A-2Ea). 2s T ·Analysis From the isentropic relations, QH ( k −1) / k ⎛P ⎞ 3 T2 s = T1 ⎜ 2 ⎜P ⎟ ⎟ = (450 R)(4) 0.4 / 1.4 = 668.7 R 100°F ⎝ 1 ⎠ -10°F ( k −1) / k 0.4 / 1.4 1 ⎛P ⎞ ⎛ 6 psia ⎞ · T4 s = T3 ⎜ 4 ⎜P ⎟ ⎟ = (560 R)⎜ ⎜ 19 psia ⎟ ⎟ = 402.9 R 4 4s QRefrig ⎝ 3 ⎠ ⎝ ⎠and s h3 − h4 T − T4 ηT = = 3 ⎯→ T4 = T3 − η T (T3 − T4 s ) = 560 − (0.94)(560 − 402.9) ⎯ h3 − h4 s T3 − T4 s = 412.3 R h −h T −T η C = 2s 1 = 2s 1 ⎯ ⎯→ T2 = T1 + (T2 s − T1 ) / η C = 450 + (668.7 − 450) /(0.87) h2 − h1 T2 − T1 = 701.4 RThe COP of this gas refrigeration cycle is determined from qL qL COPR = = wnet,in wcomp,in − w turb,out h1 − h4 = (h2 − h1 ) − (h3 − h4 ) T1 − T4 = (T2 − T1 ) − (T3 − T4 ) 450 − 412.3 = = 0.364 (701.4 − 450) − (560 − 412.3)PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
  • 54. 11-5411-70 An ideal gas refrigeration cycle with air as the working fluid provides 15 kW of cooling. The massflow rate of air and the rates of heat addition and rejection are to be determined.Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant specific heats. 3Kinetic and potential energy changes are negligible.Properties The properties of air at room temperature arecp = 1.005 kJ/kg·K and k = 1.4 (Table A-2a). T 2 · QHAnalysis From the isentropic relations, ( k −1) / k 3 ⎛P ⎞ 0.4 / 1.4 30°C ⎛ 500 kPa ⎞ ⎜ T2 = T1 ⎜ 2 ⎟ ⎟ = (293 K)⎜ ⎟ = 464.1 K 20°C ⎝ P1 ⎠ ⎝ 100 kPa ⎠ 1 ( k −1) / k · ⎛P ⎞ 0.4 / 1.4 ⎛ 100 kPa ⎞ 4 QRefrig T4 = T3 ⎜ 4 ⎜P ⎟ ⎟ = (303 K)⎜ ⎟ = 191.3 K ⎝ 3 ⎠ ⎝ 500 kPa ⎠ sThe mass flow rate of the air is determined from & Q Refrig & 15 kJ/s Q Refrig = mc p (T1 − T4 ) ⎯ & ⎯→ m = & = = 0.1468 kg/s c p (T1 − T4 ) (1.005 kJ/kg ⋅ K)(293 − 191.3) KThe rate of heat addition to the cycle is the same as the rate of cooling, & & Qin = Q Refrig = 15 kWThe rate of heat rejection from the cycle is & Q H = mc p (T2 − T3 ) = (0.1468 kg/s)(1.005 kJ/kg ⋅ K)(464.1 − 303)K = 23.8 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.
  • 55. 11-5511-71 An ideal gas refrigeration cycle with air as the working fluid is considered. The minimum pressureratio for this system to operate properly is to be determined.Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant specific heats. 3Kinetic and potential energy changes are negligible.Properties The properties of air at room temperature arecp = 1.005 kJ/kg·K and k = 1.4 (Table A-2a). T 2 · QHAnalysis An energy balance on process 4-1 gives q Refrig = c p (T1 − T4 ) 3 20°C q Refrig 20 kJ/kg -5°C T4 = T1 − = 268 K − = 248.1 K 1 cp 1.005 kJ/kg ⋅ K · 4 QRefrigThe minimum temperature at the turbine inlet would be thesame as that to which the heat is rejected. That is, s T3 = 293 KThen the minimum pressure ratio is determined from the isentropic relation to be k /( k −1) P3 ⎛ T3 ⎞ 1.4 / 0.4 ⎛ 293 K ⎞ =⎜ ⎟ =⎜ ⎟ = 1.79 P4 ⎜ T4 ⎟ ⎝ ⎠ ⎝ 248.1 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.
  • 56. 11-5611-72 An ideal gas refrigeration cycle with with two stages of compression with intercooling using air asthe working fluid is considered. The COP of this system and the mass flow rate of air are to be determined.Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant specific heats. 3Kinetic and potential energy changes are negligible.Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2a). . . Q Q 4 T 2 Heat exch. Heat exch. 3 5 5 10°C 4 3 2 -18°C 1 Turbine 6 6 s Heat exch. 1 . QAnalysis From the isentropic relations, ( k −1) / k ⎛P ⎞ T2 = T1 ⎜ 2 ⎜P ⎟ ⎟ = (255 K)(4) 0.4 / 1.4 = 378.9 K ⎝ 1 ⎠ ( k −1) / k ⎛P ⎞ T4 = T3 ⎜ 4 ⎜P ⎟ ⎟ = (283 K)(4) 0.4 / 1.4 = 420.5 K ⎝ 3 ⎠ ( k −1) / k ⎛P ⎞ ⎛1⎞ 0.4 / 1.4 T6 = T5 ⎜ 6 ⎜ ⎟ ⎟ = (283 K)⎜ ⎟ = 128.2 K ⎝ P5 ⎠ ⎝ 16 ⎠The COP of this ideal gas refrigeration cycle is determined from qL qL COPR = = w net,in wcomp,in − w turb,out h1 − h6 = ( h2 − h1 ) + ( h4 − h3 ) − ( h5 − h6 ) T1 − T6 = (T2 − T1 ) + (T4 − T3 ) − (T5 − T6 ) 255 − 128.2 = = 1.19 (378.9 − 255) + (420.5 − 283) − (283 − 128.2)The mass flow rate of the air is determined from & Q Refrig (75,000 / 3600) kJ/s & Q Refrig = mc p (T1 − T6 ) ⎯ & ⎯→ m = & = = 0.163 kg/s c p (T1 − T6 ) (1.005 kJ/kg ⋅ K)(255 − 128.2) KPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
  • 57. 11-5711-73 A gas refrigeration cycle with with two stages of compression with intercooling using air as theworking fluid is considered. The COP of this system and the mass flow rate of air are to be determined.Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant specific heats. 3Kinetic and potential energy changes are negligible.Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2a). . . Q Q 4 4s Heat exch. Heat exch. T 2s 2 3 5 5 4 10°C 2 3 Turbine -18°C 1 6s 6 6 Heat exch. 1 s . QAnalysis From the isentropic relations, ( k −1) / k ⎛P ⎞ T2 s = T1 ⎜ 2 ⎜P ⎟ ⎟ = (255 K)(4) 0.4 / 1.4 = 378.9 K ⎝ 1 ⎠ ( k −1) / k ⎛P ⎞ T4 s = T3 ⎜ 4 ⎜ ⎟ ⎟ = (283 K)(4) 0.4 / 1.4 = 420.5 K ⎝ P3 ⎠ ( k −1) / k ⎛P ⎞ ⎛1⎞ 0.4 / 1.4 T6 s = T5 ⎜ 6 ⎜P ⎟ ⎟ = (283 K)⎜ ⎟ = 128.2 K ⎝ 5 ⎠ ⎝ 16 ⎠and h2 s − h1 T2 s − T1 ηC = = ⎯→ T2 = T1 + (T2 s − T1 ) / η C = 255 + (378.9 − 255) / 0.85 = 400.8 K ⎯ h2 − h1 T2 − T1 h4 s − h3 T4 s − T3 ηC = = ⎯→ T4 = T3 + (T4 s − T3 ) / η C = 283 + (420.5 − 283) / 0.85 = 444.8 K ⎯ h 4 − h3 T4 − T3 h5 − h 6 T − T6 ηT = = 5 ⎯⎯→ T6 = T5 − η T (T5 − T6 s ) = 283 − (0.95)(283 − 128.2) = 135.9 K h5 − h6 s T5 − T6 sThe COP of this ideal gas refrigeration cycle is determined from qL qL COPR = = wnet,in wcomp,in − w turb,out h1 − h6 = ( h2 − h1 ) + ( h4 − h3 ) − ( h5 − h6 ) T1 − T6 = (T2 − T1 ) + (T4 − T3 ) − (T5 − T6 ) 255 − 135.9 = = 0.742 (400.8 − 255) + (444.8 − 283) − (283 − 135.9)The mass flow rate of the air is determined from & Q Refrig (75,000 / 3600) kJ/s & Q Refrig = mc p (T1 − T6 ) ⎯ & ⎯→ m =& = = 0.174 kg/s c p (T1 − T6 ) (1.005 kJ/kg ⋅ K)(255 − 135.9) KPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
  • 58. 11-58Absorption Refrigeration Systems11-74C Absorption refrigeration is the kind of refrigeration that involves the absorption of the refrigerantduring part of the cycle. In absorption refrigeration cycles, the refrigerant is compressed in the liquidphase instead of in the vapor form.11-75C The main advantage of absorption refrigeration is its being economical in the presence of aninexpensive heat source. Its disadvantages include being expensive, complex, and requiring an externalheat source.11-76C In absorption refrigeration, water can be used as the refrigerant in air conditioning applicationssince the temperature of water never needs to fall below the freezing point.11-77C The fluid in the absorber is cooled to maximize the refrigerant content of the liquid; the fluid inthe generator is heated to maximize the refrigerant content of the vapor.11-78C The coefficient of performance of absorption refrigeration systems is defined as desiredoutput QL Q COPR = = ≅ L requiredinput Qgen + Wpump,in Qgen11-79C The rectifier separates the water from NH3 and returns it to the generator. The regeneratortransfers some heat from the water-rich solution leaving the generator to the NH3-rich solution leaving thepump.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.
  • 59. 11-5911-80 The COP of an absorption refrigeration system that operates at specified conditions is given. It is tobe determined whether the given COP value is possible.Analysis The maximum COP that this refrigeration system can have is ⎛ T ⎞⎛ TL ⎞ ⎛ 300 K ⎞⎛ 268 ⎞ COPR,max = ⎜1 − 0 ⎟⎜ ⎜ T ⎟⎜ T − T ⎟ = ⎜1 − 403 K ⎟⎜ 300 − 268 ⎟ = 2.14 ⎟ ⎝ s ⎠⎝ 0 L ⎠ ⎝ ⎠⎝ ⎠which is slightly greater than 2. Thus the claim is possible, but not probable.11-81 The conditions at which an absorption refrigeration system operates are specified. The maximumCOP this absorption refrigeration system can have is to be determined.Analysis The maximum COP that this refrigeration system can have is ⎛ T ⎞⎛ TL ⎞ ⎛ 298 K ⎞⎛ 273 ⎞ COPR,max = ⎜1 − 0 ⎜ T ⎟⎜ ⎟⎜ T − T ⎟ ⎟ = ⎜1 − 393 K ⎟⎜ 298 − 273 ⎟ = 2.64 ⎝ s ⎠⎝ 0 L ⎠ ⎝ ⎠⎝ ⎠11-82 The conditions at which an absorption refrigeration system operates are specified. The maximumrate at which this system can remove heat from the refrigerated space is to be determined.Analysis The maximum COP that this refrigeration system can have is ⎛ T ⎞⎛ TL ⎞ ⎛ 298 K ⎞⎛ 243 ⎞ COPR,max = ⎜1 − 0 ⎟⎜ ⎜ ⎟⎜ ⎟ = ⎜1 − ⎟ ⎜ ⎟⎜ ⎟ ⎟ = 1.15 ⎝ Ts ⎠⎝ T0 − TL ⎠ ⎝ 403 K ⎠⎝ 298 − 243 ⎠Thus, ( ) QL,max = COPR, maxQgen = (1.15) 5 × 105 kJ/h = 5.75 × 105 kJ/h & &11-83E The conditions at which an absorption refrigeration system operates are specified. The COP is alsogiven. The maximum rate at which this system can remove heat from the refrigerated space is to bedetermined.Analysis For a COP = 0.55, the rate at which this system can remove heat from the refrigerated space is ( ) Q L = COPR Qgen = (0.55) 10 5 Btu/h = 0.55 × 10 5 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.
  • 60. 11-6011-84 A reversible absorption refrigerator consists of a reversible heat engine and a reversible refrigerator.The rate at which the steam condenses, the power input to the reversible refrigerator, and the second lawefficiency of an actual chiller are to be determined.Properties The enthalpy of vaporization of water at 200°C is hfg = 1939.8 kJ/kg (Table A-4).Analysis (a) The thermal efficiency of the reversible heat engine is T0 (25 + 273.15) K η th,rev = 1 − = 1− = 0.370 Ts (200 + 273.15) K Ts T0The COP of the reversible refrigerator is TL (−10 + 273.15) K COPR,rev = = = 7.52 T0 − T L (25 + 273.15) − (−10 + 273.15) KThe COP of the reversible absorption refrigerator is Rev. Rev. COPabs, rev = η th,rev COPR, rev = (0.370)(7.52) = 2.78 HE Ref.The heat input to the reversible heat engine is & QL 22 kW & Qin = = = 7.911 kW COPabs,rev 2.78 T0 TLThen, the rate at which the steam condenses becomes & Qin 7.911 kJ/s ms = & = = 0.00408 kg/s h fg 1939.8 kJ/kg(b) The power input to the refrigerator is equal to the power output from the heat engine & & & Win, R = Wout, HE = η th, rev Qin = (0.370)(7.911 kW ) = 2.93 kW(c) The second-law efficiency of an actual absorption chiller with a COP of 0.7 is COPactual 0.7 η II = = = 0.252 COPabs,rev 2.78PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
  • 61. 11-61Special Topic: Thermoelectric Power Generation and Refrigeration Systems11-85C The circuit that incorporates both thermal and electrical effects is called a thermoelectric circuit.11-86C When two wires made from different metals joined at both ends (junctions) forming a closedcircuit and one of the joints is heated, a current flows continuously in the circuit. This is called theSeebeck effect. When a small current is passed through the junction of two dissimilar wires, the junction iscooled. This is called the Peltier effect.11-87C No.11-88C No.11-89C Yes.11-90C When a thermoelectric circuit is broken, the current will cease to flow, and we can measure thevoltage generated in the circuit by a voltmeter. The voltage generated is a function of the temperaturedifference, and the temperature can be measured by simply measuring voltages.11-91C The performance of thermoelectric refrigerators improves considerably when semiconductors areused instead of metals.11-92C The efficiency of a thermoelectric generator is limited by the Carnot efficiency because athermoelectric generator fits into the definition of a heat engine with electrons serving as the working fluid.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
  • 62. 11-6211-93E A thermoelectric generator that operates at specified conditions is considered. The maximumthermal efficiency this thermoelectric generator can have is to be determined.Analysis The maximum thermal efficiency of this thermoelectric generator is the Carnot efficiency, TL 550R η th,max = η th,Carnot = 1 − = 1− = 31.3% TH 800R11-94 A thermoelectric refrigerator that operates at specified conditions is considered. The maximum COPthis thermoelectric refrigerator can have and the minimum required power input are to be determined.Analysis The maximum COP of this thermoelectric refrigerator is the COP of a Carnot refrigeratoroperating between the same temperature limits, 1 1 COPmax = COPR,Carnot = = = 10.72 (TH / T L ) − 1 (293 K ) / (268 K ) − 1Thus, & QL 130 W & Win, min = = = 12.1 W COPmax 10.7211-95 A thermoelectric cooler that operates at specified conditions with a given COP is considered. Therequired power input to the thermoelectric cooler is to be determined.Analysis The required power input is determined from the definition of COPR, & Q & QL 180 W COPR = & L ⎯→ ⎯ & Win = = = 1200 W Win COPR 0.1511-96E A thermoelectric cooler that operates at specified conditions with a given COP is considered. Therequired power input to the thermoelectric cooler is to be determined.Analysis The required power input is determined from the definition of COPR, & QL & QL 20 Btu/min COPR = ⎯ & ⎯→ Win = = = 133.3 Btu/min = 3.14 hp & Win COPR 0.15PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
  • 63. 11-6311-97 A thermoelectric refrigerator powered by a car battery cools 9 canned drinks in 12 h. The averageCOP of this refrigerator is to be determined.Assumptions Heat transfer through the walls of the refrigerator is negligible.Properties The properties of canned drinks are the same as those of water at room temperature, ρ = 1 kg/Land cp = 4.18 kJ/kg·°C (Table A-3).Analysis The cooling rate of the refrigerator is simply the rate of decrease of the energy of the canneddrinks, m = ρV = 9 × (1 kg/L)(0.350 L) = 3.15 kg Qcooling = mcΔT = (3.15 kg)(4.18 kJ/kg ⋅ °C)(25 - 3)°C = 290 kJ Qcooling 290 kJ & Qcooling = = = 0.00671 kW = 6.71 W Δt 12 × 3600 sThe electric power consumed by the refrigerator is & Win = VI = (12 V)(3 A) = 36 WThen the COP of the refrigerator becomes & Qcooling 6.71 W COP = = = 0.186 ≈ 0.20 & Win 36 WPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
  • 64. 11-6411-98E A thermoelectric cooler is said to cool a 12-oz drink or to heat a cup of coffee in about 15 min.The average rate of heat removal from the drink, the average rate of heat supply to the coffee, and theelectric power drawn from the battery of the car are to be determined.Assumptions Heat transfer through the walls of the refrigerator is negligible.Properties The properties of canned drinks are the same as those of water at room temperature, cp = 1.0Btu/lbm.°F (Table A-3E).Analysis (a) The average cooling rate of the refrigerator is simply the rate of decrease of the energy contentof the canned drinks, Qcooling = mc p ΔT = (0.771 lbm)(1.0 Btu/lbm ⋅ °F)(78 - 38)°F = 30.84 Btu & Qcooling 30.84 Btu ⎛ 1055 J ⎞ Qcooling = = ⎜ ⎟ = 36.2 W Δt 15 × 60 s ⎝ 1 Btu ⎠(b) The average heating rate of the refrigerator is simply the rate of increase of the energy content of thecanned drinks, Q heating = mc p ΔT = (0.771 lbm)(1.0 Btu/lbm ⋅ °F)(130 - 75)°F = 42.4 Btu & Q heating 42.4 Btu ⎛ 1055 J ⎞ Q heating = = ⎜ ⎟ = 49.7 W Δt 15 × 60 s ⎝ 1 Btu ⎠(c) The electric power drawn from the car battery during cooling and heating is & Qcooling 36.2 W & Win,cooling = = = 181 W COPcooling 0.2 COPheating = COPcooling + 1 = 0.2 + 1 = 1.2 & Qheating 49.7 W & Win, heating = = = 41.4 W COPheating 1.211-99 The maximum power a thermoelectric generator can produce is to be determined.Analysis The maximum thermal efficiency this thermoelectric generator can have is TL 303 K ηth,max = 1 − = 1− = 0.142 TH 353 KThus, & & Wout, max = η th, maxQin = (0.142)(106 kJ/h) = 142,000 kJ/h = 39.4 kWPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
  • 65. 11-65Review Problems11-100 A steady-flow Carnot refrigeration cycle with refrigerant-134a as the working fluid is considered.The COP, the condenser and evaporator pressures, and the net work input are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) The COP of this refrigeration cycle isdetermined from T 1 1 COPR,C = = = 5.06 (TH / TL ) − 1 (303 K )/ (253 K ) − 1 4 3(b) The condenser and evaporative pressures are 30°C(Table A-11) Pevap = Psat @ − 20°C = 132.82 kPa Pcond = Psat @30°C = 770.64 kPa -20°C 1 qL 2 s(c) The net work input is determined from ( h1 = h f + x1 h fg )@−20°C = 25.49 + (0.15)(212.91) = 57.43 kJ/kg h2 = (h f + x 2 h fg )@ − 20°C = 25.49 + (0.80 )(212.91) = 195.82 kJ/kg q L = h2 − h1 = 195.82 − 57.43 = 138.4kJ/kg qL 138.4 kJ/kg w net,in = = = 27.35 kJ/kg COPR 5.06PROPRIETARY 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.
  • 66. 11-6611-101 A heat pump that operates on the ideal vapor-compression cycle with refrigerant-134a as theworking fluid is used to heat a house. The rate of heat supply to the house, the volume flow rate of therefrigerant at the compressor inlet, and the COP of this heat pump 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), h1 = h g @ 200 kPa = 244.46 kJ/kg P1 = 200 kPa ⎫ T ⎬ s1 = s g @ 200 kPa = 0.93773 kJ/kg ⋅ K House sat. vapor ⎭ v 1 = v g @ 200 kPa = 0.099867 m 3 /kg · QH 2 P2 = 0.9 MPa ⎫ ⎬ h2 = 275.75 kJ/kg 3 0.9 MPa · s 2 = s1 ⎭ Win P3 = 0.9 MPa ⎫ ⎬ h3 = h f @ 0.9 MPa = 101.61 kJ/kg sat. liquid ⎭ 200 kPa h4 ≅ h3 = 101.61 kJ/kg (throttling ) · 1 4 QLThe rate of heat supply to the house isdetermined from s Q H = m(h2 − h3 ) = (0.32 kg/s )(275.75 − 101.61) kJ/kg = 55.73 kW & &(b) The volume flow rate of the refrigerant at the compressor inlet is ( ) V&1 = mv 1 = (0.32 kg/s ) 0.099867 m 3 /kg = 0.0320 m 3 /s &(c) The COP of t his heat pump is determined from q L h2 − h3 275.75 − 101.61 COPR = = = = 5.57 win h2 − h1 275.75 − 244.4611-102 A relation for the COP of the two-stage refrigeration system with a flash chamber shown in Fig.11-12 is to be derived.Analysis The coefficient of performance is determined from qL COPR = winwhere h6 − h f qL = (1 − x6 )(h1 − h8 ) with x6 = h fg win = wcompI,in + wcompII,in = (1 − x6 )(h2 − h1 ) + (1)(h4 − h9 )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.
  • 67. 11-6711-103 A two-stage compression refrigeration system using refrigerant-134a as the working fluid isconsidered. The fraction of the refrigerant that evaporates as it is throttled to the flash chamber, the amountof heat removed from the refrigerated space, the compressor work, and the COP are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3The flashing chamber is adiabatic.Analysis (a) The enthalpies of the refrigerant at several states are determined from the refrigerant tables tobe (Tables A-11, A-12, and A-13) h1 = 239.16 kJ/kg, h2 = 260.58 kJ/kg h3 = 255.55 kJ/kg, T 0.8 MPa h5 =95.47 kJ/kg, h6 = 95.47 kJ/kg 4 h7 = 63.94 kJ/kg, h8 = 63.94 kJ/kg 0.4 MPa 5 2The fraction of the refrigerant that evaporates as Ait is throttled to the flash chamber is simply the 7 0.14 MPaquality at state 6, 6 3 9 B h6 − h f 95.47 − 63.94 x6 = = = 0.1646 8 1 h fg 191.62 qL(b) The enthalpy at state 9 is determined from san 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 = (0.1646 )(255.55) + (1 − 0.1646 )(260.58) = 259.75 kJ/kg P9 = 0.4 MPa ⎫ ⎬ s 9 = 0.94168 kJ/kg ⋅ K h9 = 259.75 kJ/kg ⎭Also, P4 = 0.8 MPa ⎫ ⎬ h4 = 274.47 kJ/kg s 4 = s 9 = 0.94168 kJ/kg ⋅ K ⎭Then the amount of heat removed from the refrigerated space and the compressor work input per unit massof refrigerant flowing through the condenser are q L = (1 − x 6 )(h1 − h8 ) = (1 − 0.1646)(239.16 − 63.94) kJ/kg = 146.4 kJ/kg win = wcompI,in + wcompII,in = (1 − x 6 )(h2 − h1 ) + (1)(h4 − h9 ) = (1 − 0.1646)(260.58 − 239.16 ) kJ/kg + (274.47 − 259.75) kJ/kg = 32.6 kJ/kg(c) The coefficient of performance is determined from q L 146.4 kJ/kg COPR = = = 4.49 win 32.6 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.
  • 68. 11-6811-104 A refrigerator operating on a vapor-compression refrigeration cycle with refrigerant-134a as theworking fluid is considered. The process with the greatest exergy loss is to be determined.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. 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 = 700 kPa ⎫ · ⎬ h2 s = 270.38 kJ/kg QH 2s 2 s 2 = s1 ⎭ 3 0.7 MPa P3 = 700 kPa ⎫ h3 = h f @ 700 kPa = 88.82 kJ/kg · Win ⎬ s =s sat. liquid ⎭ 3 f @ 700 kPa = 0.33230 kJ/kg ⋅ K h4 ≅ h3 = 88.82 kJ/kg ( throttling) -10°C T4 = −10°C ⎫ 4s 4 · 1 ⎬ s = 0.34605 kJ/kg ⋅ K QL h4 = 88.82 kJ/kg ⎭ 4The actual enthalpy at the compressor exit is determined by susing the compressor efficiency: h2 s − h1 h − h1 270.38 − 244.51 ηC = ⎯ ⎯→ h2 = h1 + 2 s = 244.51 + = 274.95 kJ/kg h2 − h1 ηC 0.85 P2 = 700 kPa ⎫and ⎬ s = 0.95252 kJ/kg h2 = 274.95 kJ/kg ⎭ 2The heat added in the evaporator and that rejected in the condenser are q L = h1 − h4 = (244.51 − 88.82) kJ/kg = 155.69 kJ/kg q H = h2 − h3 = (274.95 − 88.82) kJ/kg = 186.13 kJ/kgThe exergy destruction during a process of a stream from an inlet state to exit state is given by ⎛ q q ⎞ x dest = T0 s gen = T0 ⎜ s e − s i − in + out ⎟ ⎜ ⎟ ⎝ Tsource Tsink ⎠Application of this equation for each process of the cycle gives x destroyed,12 = T0 ( s 2 − s1 ) = (295 K )(0.95252 − 0.93766) kJ/kg ⋅ K = 4.38 kJ/kg ⎛ q ⎞ ⎛ 186.13 kJ/kg ⎞ x destroyed, 23 = T0 ⎜ s 3 − s 2 + H ⎟ = (295 K )⎜ 0.33230 − 0.95252 + ⎜ ⎟ ⎟ = 3.17 kJ/kg ⎝ TH ⎠ ⎝ 295 K ⎠ x destroyed, 34 = T0 ( s 4 − s 3 ) = (295 K )(0.34605 − 0.33230) kJ/kg ⋅ K = 4.06 kJ/kg ⎛ q ⎞ ⎛ 155.69 kJ/kg ⎞ x destroyed, 41 = T0 ⎜ s1 − s 4 − L ⎟ = (295 K )⎜ 0.93766 − 0.34605 − ⎜ ⎟ = 6.29 kJ/kg ⎝ TL ⎟ ⎠ ⎝ 273 K ⎠The greatest exergy destruction occurs in the evaporator. Note that heat is absorbed from freezing water at0°C (273 K) and rejected to the ambient air at 22°C (295 K), which is also taken as the dead statetemperature. Alternatively, one may use the standard 25°C (298 K) as the dead state temperature, andperform the calculations accordingly.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
  • 69. 11-6911-105 A refrigerator operating on a vapor-compression refrigeration cycle with refrigerant-134a as theworking fluid is considered. The process with the greatest exergy loss is to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis 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 = 700 kPa ⎫ ⎬ h2 s = 270.38 kJ/kg · 2s 2 s 2 = s1 ⎭ QH 0.7 MPa P3 = 700 kPa ⎫ · ⎪ h3 ≅ h f @ 24°C = 84.98 kJ/kg 3 Win T3 = Tsat @ 700 kPa − 2.7⎬ s ≅s f @ 24°C = 0.31958 kJ/kg ⋅ K = 26.7 − 2.7 = 24°C ⎪ 3 ⎭ -10°C h4 ≅ h3 = 84.98 kJ/kg ( throttling) · 1 4 QL T4 = −10°C ⎫ ⎬ s = 0.33146 kJ/kg ⋅ K h4 = 84.98 kJ/kg ⎭ 4 sThe actual enthalpy at the compressor exit is determined byusing the compressor efficiency: h2 s − h1 h − h1 270.38 − 244.51 ηC = ⎯ ⎯→ h2 = h1 + 2 s = 244.51 + = 274.95 kJ/kg h2 − h1 ηC 0.85 P2 = 700 kPa ⎫and ⎬ s = 0.95252 kJ/kg h2 = 274.95 kJ/kg ⎭ 2The heat added in the evaporator and that rejected in the condenser are q L = h1 − h4 = (244.51 − 84.98) kJ/kg = 159.53 kJ/kg q H = h2 − h3 = (274.95 − 84.98) kJ/kg = 189.97 kJ/kgThe exergy destruction during a process of a stream from an inlet state to exit state is given by ⎛ q q ⎞ x dest = T0 s gen = T0 ⎜ s e − s i − in + out ⎟ ⎜ ⎟ ⎝ Tsource Tsink ⎠Application of this equation for each process of the cycle gives x destroyed,12 = T0 ( s 2 − s1 ) = (295 K )(0.95252 − 0.93766) kJ/kg ⋅ K = 4.38 kJ/kg ⎛ q ⎞ ⎛ 189.97 kJ/kg ⎞ x destroyed, 23 = T0 ⎜ s 3 − s 2 + H ⎟ = (295 K )⎜ 0.31958 − 0.95252 + ⎜ ⎟ ⎟ = 3.25 kJ/kg ⎝ TH ⎠ ⎝ 295 K ⎠ x destroyed, 34 = T0 ( s 4 − s 3 ) = (295 K )(0.33146 − 0.31958) kJ/kg ⋅ K = 3.50 kJ/kg ⎛ q ⎞ ⎛ 159.53 kJ/kg ⎞ ⎜ ⎟ x destroyed, 41 = T0 ⎜ s1 − s 4 − L ⎟ = (295 K )⎜ 0.93766 − 0.33146 − ⎟ = 6.44 kJ/kg ⎝ TL ⎠ ⎝ 273 K ⎠The greatest exergy destruction occurs in the evaporator. Note that heat is absorbed from freezing water at0°C (273 K) and rejected to the ambient air at 22°C (295 K), which is also taken as the dead statetemperature. Alternatively, one may use the standard 25°C (298 K) as the dead state temperature, andperform the calculations accordingly.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.
  • 70. 11-7011-106 A refrigerator operating on a vapor-compression refrigeration cycle with refrigerant-134a as theworking fluid is considered. The process with the greatest exergy loss is to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis In this cycle, the refrigerant leaves the condenser as saturated liquid at the condenser pressure.The compression process is not isentropic. From the refrigerant tables (Tables A-11, A-12, and A-13), P1 = Psat @ −37°C = 60 kPa ⎫ h1 = 233.09 kJ/kg ⎬ s = 0.9865 kJ/kg ⋅ K T1 = −37 + 7 = −30°C ⎭ 1 T P2 = 1.2 MPa ⎫ · 2s 2 ⎬ h2 s = 298.11 kJ/kg QH s 2 = s1 ⎭ 3 1.2 MPa · P3 = 1.2 MPa ⎫ h3 = h f @ 1.2 MPa = 117.77 kJ/kg Win ⎬ s =s sat. liquid ⎭ 3 f @ 1.2 MPa = 0.42441 kJ/kg ⋅ K h4 ≅ h3 = 117.77 kJ/kg ( throttling) -37°C 4s · 1 T4 = −37°C ⎫ x 4 = 0.5089 4 QL ⎬ h4 = 117.77 kJ/kg ⎭ s 4 = 0.4988 kJ/kg ⋅ K sThe actual enthalpy at the compressor exit is determined byusing the compressor efficiency: h2 s − h1 h − h1 298.11 − 233.09 ηC = ⎯ ⎯→ h2 = h1 + 2 s = 233.09 + = 305.33 kJ/kg h2 − h1 ηC 0.90 P2 = 1.2 MPa ⎫and ⎬ s = 1.0075 kJ/kg ⋅ K h2 = 305.33 Btu/lbm⎭ 2The heat added in the evaporator and that rejected in the condenser are q L = h1 − h4 = (233.09 − 117.77) kJ/kg = 115.32 kJ/kg q H = h2 − h3 = (305.33 − 117.77) kJ/kg = 187.56 kJ/kgThe exergy destruction during a process of a stream from an inlet state to exit state is given by ⎛ q q ⎞ x dest = T0 s gen = T0 ⎜ s e − s i − in + out ⎟ ⎜ ⎟ ⎝ Tsource Tsink ⎠Application of this equation for each process of the cycle gives x destroyed,12 = T0 ( s 2 − s1 ) = (303 K )(1.0075 − 0.9865) kJ/kg ⋅ K = 6.36 kJ/kg ⎛ q ⎞ ⎛ 187.56 kJ/kg ⎞ ⎜ ⎟ x destroyed, 23 = T0 ⎜ s 3 − s 2 + H ⎟ = (303 K )⎜ 0.42441 − 1.0075 + ⎟ = 10.88 kJ/kg ⎝ TH ⎠ ⎝ 303 K ⎠ x destroyed, 34 = T0 ( s 4 − s 3 ) = (303 K )(0.4988 − 0.42441) kJ/kg ⋅ K = 22.54 kJ/kg ⎛ q ⎞ ⎛ 115.32 kJ/kg ⎞ ⎜ x destroyed, 41 = T0 ⎜ s1 − s 4 − L ⎟ ⎟ = (303 K )⎜ 0.9865 − 0.4988 − (−34 + 273) K ⎟ = 1.57 kJ/kg ⎜ ⎟ ⎝ TL ⎠ ⎝ ⎠The greatest exergy destruction occurs in the expansion valve. Note that heat is absorbed from fruits at -34°C (239 K) and rejected to the ambient air at 30°C (303 K), which is also taken as the dead statetemperature. Alternatively, one may use the standard 25°C (298 K) as the dead state temperature, andperform the calculations accordingly.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
  • 71. 11-7111-107 A refrigerator operating on a vapor-compression refrigeration cycle with refrigerant-134a as theworking fluid is considered. The process with the greatest exergy loss is to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis From the refrigerant tables (Tables A-11, A-12, and A-13), P1 = Psat @ −37°C = 60 kPa ⎫ h1 = 233.09 kJ/kg ⎬ s = 0.9865 kJ/kg ⋅ K T1 = −37 + 7 = −30°C ⎭ 1 T P2 = 1.2 MPa ⎫ · 2s 2 ⎬ h2 s = 298.11 kJ/kg QH s 2 = s1 ⎭ 1.2 MPa · P3 = 1.2 MPa ⎫ 3 Win ⎪ h3 ≅ h f @ 40°C = 108.26 kJ/kg T3 = Tsat @ 1.2 MPa − 6.3 ⎬ s 3 ≅ s f @ 40°C = 0.39486 kJ/kg ⋅ K = 46.3 − 6.3 = 40°C ⎪ ⎭ -37°C h4 ≅ h3 = 108.26 kJ/kg ( throttling) 4 · 1 QL T4 = −37°C ⎫ x 4 = 0.4665 ⎬ h4 = 108.26 kJ/kg ⎭ s 4 = 0.4585 kJ/kg ⋅ K sThe actual enthalpy at the compressor exit is determined byusing the compressor efficiency: h2 s − h1 h − h1 298.11 − 233.09 ηC = ⎯ ⎯→ h2 = h1 + 2 s = 233.09 + = 305.33 kJ/kg h2 − h1 ηC 0.90 P2 = 1.2 MPa ⎫and ⎬ s = 1.0075 kJ/kg ⋅ K h2 = 305.33 Btu/lbm⎭ 2The heat added in the evaporator and that rejected in the condenser are q L = h1 − h4 = (233.09 − 108.26) kJ/kg = 124.83 kJ/kg q H = h2 − h3 = (305.33 − 108.26) kJ/kg = 197.07 kJ/kgThe exergy destruction during a process of a stream from an inlet state to exit state is given by ⎛ q q ⎞ x dest = T0 s gen = T0 ⎜ s e − s i − in + out ⎜ ⎟ ⎟ ⎝ Tsource Tsink ⎠Application of this equation for each process of the cycle gives x destroyed,12 = T0 ( s 2 − s1 ) = (303 K )(1.0075 − 0.9865) kJ/kg ⋅ K = 6.36 kJ/kg ⎛ q ⎞ ⎛ 197.07 kJ/kg ⎞ x destroyed, 23 = T0 ⎜ s 3 − s 2 + H ⎟ = (303 K )⎜ 0.39486 − 1.0075 + ⎜ ⎟ ⎟ = 11.44 kJ/kg ⎝ TH ⎠ ⎝ 303 K ⎠ x destroyed, 34 = T0 ( s 4 − s 3 ) = (303 K )(0.4585 − 0.39486) kJ/kg ⋅ K = 19.28 kJ/kg ⎛ q ⎞ ⎛ 124.83 kJ/kg ⎞ x destroyed, 41 = T0 ⎜ s1 − s 4 − L ⎟ = (303 K )⎜ 0.9865 − 0.4585 − ⎜ ⎟ = 1.73 kJ/kg ⎜ ⎝ TL ⎟ ⎠ ⎝ (−34 + 273) K ⎟ ⎠The greatest exergy destruction occurs in the expansion valve. Note that heat is absorbed from fruits at -34°C (239 K) and rejected to the ambient air at 30°C (303 K), which is also taken as the dead statetemperature. Alternatively, one may use the standard 25°C (298 K) as the dead state temperature, andperform the calculations accordingly.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
  • 72. 11-7211-108E A two-evaporator compression refrigeration cycle with refrigerant-134a as the working fluid isconsidered. The cooling load of both evaporators per unit of flow through the compressor and the COP ofthe system are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Condenser 2 T 3 Compressor · QH 2 Expansion Expansion 3 160 psia 1 · valve valve Win Evaporator 1 30°F 5 4 4 5 -29.5°F Expansion 6 · 7 1 QL valve 7 Evaporator 2 s 6Analysis From the refrigerant tables (Tables A-11E, A-12E, and A-13E), P3 = 160 psia ⎫ ⎬ h3 = h f @ 160 psia = 48.519 Btu/lbm sat. liquid ⎭ h4 = h6 ≅ h3 = 48.519 Btu/lbm ( throttling) T5 = 30°F ⎫ ⎬ h = h g @ 30°F = 107.40 Btu/lbm sat. vapor ⎭ 5 T7 = −29.5°F ⎫ ⎬ h7 = h g @ − 29.5°F = 98.68 Btu/lbm sat. vapor ⎭For a unit mass flowing through the compressor, the fraction of mass flowing through Evaporator II isdenoted by x and that through Evaporator I is y (y = 1-x). From the cooling loads specification, & & Q L,e vap 1 = 2Q L ,e vap 2 x(h5 − h4 ) = 2 y (h7 − h6 )where x = 1− yCombining these results and solving for y gives h5 − h 4 107.40 − 48.519 y= = = 0.3698 2(h7 − h6 ) + (h5 − h4 ) 2(98.68 − 48.519) + (107.40 − 48.519)Then, x = 1 − y = 1 − 0.3698 = 0.6302Applying an energy balance to the point in the system where the two evaporator streams are recombinedgivesPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
  • 73. 11-73 xh5 + yh7 (0.6302)(107.40) + (0.3698)(98.68) xh5 + yh7 = h1 ⎯ ⎯→ h1 = = = 104.18 Btu/lbm 1 1Then, P1 = Psat @ − 29.5° F ≅ 10 psia ⎫ ⎬ s1 = 0.2418 Btu/lbm ⋅ R h1 = 104.18 Btu/lbm ⎭ P2 = 160 psia ⎫ ⎬ h2 = 131.14 Btu/lbm s 2 = s1 ⎭The cooling load of both evaporators per unit mass through the compressor is q L = x ( h5 − h 4 ) + y ( h 7 − h 6 ) = (0.6302)(107.40 − 48.519) Btu/lbm + (0.3698)(98.68 − 48.519) Btu/lbm = 55.66 Btu/lbmThe work input to the compressor is win = h2 − h1 = (131.14 − 104.18) Btu/lbm = 26.96 Btu/lbmThe COP of this refrigeration system is determined from its definition, qL 55.66 Btu/lbm COPR = = = 2.06 win 26.96 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.
  • 74. 11-7411-109E A two-evaporator compression refrigeration cycle with refrigerant-134a as the working fluid isconsidered. The process with the greatest exergy destruction is to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis From Prob. 11-107E and the refrigerant tables (Tables A-11E, A-12E, and A-13E), s1 = s 2 = 0.2418 Btu/lbm ⋅ R s 3 = 0.09774 Btu/lbm ⋅ R T s 4 = 0.10238 Btu/lbm ⋅ R s 5 = 0.22260 Btu/lbm ⋅ R · QH 2 s 6 = 0.11286 Btu/lbm ⋅ R s 7 = 0.22948 Btu/lbm ⋅ R 3 160 psia · Win x = 0.6302 y = 1 − x = 0.3698 30°F 5 4 q L, 45 = h5 − h4 = 58.881 Btu/lbm -29.5°F q L,67 = h7 − h6 = 50.161 Btu/lbm 6 · 7 1 q H = 82.621 Btu/lbm QLThe exergy destruction during a process of a stream from an sinlet state to exit state is given by ⎛ q q ⎞ x dest = T0 s gen = T0 ⎜ s e − s i − in + out ⎜ ⎟ ⎟ ⎝ Tsource Tsink ⎠Application of this equation for each process of the cycle gives the exergy destructions per unit massflowing through the compressor: ⎛ q ⎞ ⎛ 82.621 Btu/lbm ⎞ x destroyed, 23 = T0 ⎜ s 3 − s 2 + H ⎟ = (555 R )⎜ 0.09774 − 0.2418 + ⎜ ⎟ = 2.67 Btu/lbm ⎝ TH ⎟ ⎠ ⎝ 555 R ⎠ x destroyed, 346 = T0 ( xs 4 + ys 6 − s 3 ) = (555 R )(0.6302 × 0.10238 + 0.3698 × 0.11286 − 0.09774) Btu/lbm ⋅ R = 4.73 Btu/lbm ⎛ q L , 45 ⎞ x destroyed, 45 = xT0 ⎜ s 5 − s 4 − ⎜ ⎟ ⎟ ⎝ TL ⎠ ⎛ 58.881 Btu/lbm ⎞ = (0.6302)(555 R )⎜ 0.22260 − 0.10238 − ⎟ = 0.44 Btu/lbm ⎝ 495 R ⎠ ⎛ q L ,67 ⎞ x destroyed, 67 = yT0 ⎜ s 7 − s 6 − ⎜ ⎟ ⎝ TL ⎟ ⎠ ⎛ 50.161 Btu/lbm ⎞ = (0.3698)(555 R )⎜ 0.22948 − 0.11286 − ⎟ = 0.54 Btu/lbm ⎝ 440 R ⎠&X destroyed, mixing = T0 ( s1 − xs 5 − ys 6 ) = (555 R )[0.2418 − (0.6302)(0.22260) − (0.3698)(0.11286)] = 3.18 Btu/lbmFor isentropic processes, the exergy destruction is zero: & X destroyed,12 = 0The greatest exergy destruction occurs during the mixing process. Note that heat is absorbed in evaporator2 from a reservoir at -20°F (440 R), in evaporator 1 from a reservoir at 35°F (495 R), and rejected to areservoir at 95°F (555 R), which is also taken as the dead state temperature.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.
  • 75. 11-7511-110 A two-stage compression refrigeration system with a separation unit is considered. The rate ofcooling and the power requirement are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 2 Condenser T Compressor 3 2 1.4 MPa 3 Expansion 8 1 Separator valve 8.9°C 8 5 4 1 4 -32°C Compressor 5 6 · 7 Expansion QL valve 7 s Evaporator 6Analysis From the refrigerant tables (Tables A-11, A-12, and A-13), T1 = 8.9°C ⎫ h1 = h g @ 8.9°C = 255.55 kJ/kg ⎬ sat. vapor ⎭ s1 = s g @ 8.9°C = 0.92691 kJ/kg ⋅ K P2 = 1400 kPa ⎫ ⎬ h2 = 281.49 kJ/kg s 2 = s1 ⎭ P3 = 1400 kPa ⎫ ⎬ h3 = h f @ 1400 kPa = 127.22 kJ/kg sat. liquid ⎭ h4 ≅ h3 = 127.22 kJ/kg ( throttling) T5 = 8.9°C ⎫ ⎬ h = hf @ 8.9°C = 63.94 kJ/kg sat. liquid ⎭ 5 h6 ≅ h5 = 63.94 kJ/kg ( throttling) T7 = −32°C ⎫ h7 = h g @ −32°C = 230.91 kJ/kg ⎬ sat. vapor ⎭ s 7 = s g @ −32°C = 0.95813 kJ/kg ⋅ K P8 = Psat @ 8.9°C = 400 kPa ⎫ ⎬ h8 = 264.51 kJ/kg s8 = s 7 ⎭An energy balance on the separator gives h1 − h4 255.55 − 127.22 m6 (h8 − h5 ) = m 2 (h1 − h4 ) ⎯ & & ⎯→ m 6 = m 2 & & = (2 kg/s) = 1.280 kg/s h8 − h5 264.51 − 63.94The rate of cooling produced by this system is then & Q = m (h − h ) = (1.280 kg/s)(230.91 − 63.94) kJ/kg = 213.7 kJ/s & L 6 7 6The total power input to the compressors is & W = m (h − h ) + m (h − h ) & & in 6 8 7 2 2 1 = (1.280 kg/s)(264.51 − 230.91) kJ/kg + (2 kg/s)(281.49 − 255.55) kJ/kg = 94.89 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.
  • 76. 11-7611-111 A two-stage vapor-compression refrigeration system with refrigerant-134a as the working fluid isconsidered. The process with the greatest exergy destruction is to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis From Prob. 11-109 and the refrigerant tables (Tables A-11, A-12, and A-13), s1 = s 2 = 0.92691 kJ/kg ⋅ K s 3 = 0.45315 kJ/kg ⋅ K s 4 = 0.4720 kJ/kg ⋅ K T s 5 = 0.24761 kJ/kg ⋅ K 2 s 6 = 0.2658 kJ/kg ⋅ K 3 1.4 MPa s 7 = s 8 = 0.95813 kJ/kg ⋅ K 8 m upper = 2 kg/s & 8.9°C 5 m lower = 1.280 kg/s & 4 1 q L = h7 − h6 = 166.97 kJ/kg -32°C q H = h2 − h3 = 154.27 kJ/kg 6 · 7 T L = −18 + 273 = 255 K QL T H = T0 = 25 + 273 = 298 K sThe exergy destruction during a process of a stream from aninlet state to exit state is given by ⎛ q q ⎞ x dest = T0 s gen = T0 ⎜ s e − s i − in + out ⎜ ⎟ ⎟ ⎝ Tsource Tsink ⎠Application of this equation for each process of the cycle gives & ⎛ q ⎞ X destroyed, 23 = m upper T0 ⎜ s 3 − s 2 + H ⎟ & ⎜ ⎝ TH ⎟⎠ ⎛ 154.27 kJ/kg ⎞ = (2 kg/s)(298 K )⎜ 0.45315 − 0.92691 + ⎟ = 26.18 kW ⎝ 298 K ⎠ & X destroyed, 34 = m upper T0 ( s 4 − s 3 ) = (2 kg/s)(298 K )(0.4720 − 0.45315) kJ/kg ⋅ K = 11.23 kW & & X destroyed, 56 = m lower T0 ( s 6 − s 5 ) = (1.280 kg/s)(298 K )(0.2658 − 0.24761) kJ/kg ⋅ K = 6.94 kW & & ⎛ q ⎞ X destroyed, 67 = m lower T0 ⎜ s 7 − s 6 − L ⎟ & ⎜ ⎝ TL ⎟⎠ ⎛ 166.97 kJ/kg ⎞ = (1.280 kg/s)(298 K )⎜ 0.95813 − 0.2658 − ⎟ = 14.32 kW ⎝ 255 K ⎠ & [ ] X destroyed, separator = T0 m lower ( s 5 − s 8 ) − m upper ( s1 − s 4 ) & & = (298 K )[(1.280 kg/s)(0.24761 − 0.95813) + (2 kg/s)(0.92691 − 0.4720)] = 0.11 kWFor isentropic processes, the exergy destruction is zero: & X destroyed,12 = 0 & X destroyed, 78 = 0Note that heat is absorbed from a reservoir at 0°F (460 R) and rejected to the standard ambient air at 77°F(537 R), which is also taken as the dead state temperature. The greatest exergy destruction occurs duringthe condensation process.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
  • 77. 11-7711-112 A regenerative gas refrigeration cycle with helium as the working fluid is considered. Thetemperature of the helium at the turbine inlet, the COP of the cycle, and the net power input required are tobe determined.Assumptions 1 Steady operating conditions exist. 2 Helium is an ideal gas with constant specific heats atroom temperature. 3 Kinetic and potential energy changes are negligible.Properties The properties of helium are cp = 5.1926kJ/kg·K and k = 1.667 (Table A-2). T · QH 2Analysis (a) The temperature of the helium at theturbine inlet is determined from an energy balance 3on the regenerator, 20°C -10°C 1 & & & E in − E out = ΔE system 0 (steady) =0 & & E in = E out 4 · Qregen ∑m h = ∑m h & & e e i i ⎯→ m(h3 − h4 ) = m(h1 − h6 ) ⎯ & & -25°C ·6 5 QRefrigor, s mc p (T3 − T4 ) = mc p (T1 − T6 ) ⎯ & & ⎯→ T3 − T4 = T1 − T6Thus, T4 = T3 − T1 + T6 = 20°C − (− 10°C ) + (− 25°C ) = 5°C = 278 K(b) From the isentropic relations, (k −1) / k ⎛P ⎞ T2 = T1 ⎜ 2 ⎟ ⎜P ⎟ = (263 K )(3)0.667 / 1.667 = 408.2 K = 135.2°C ⎝ 1⎠ (k −1) / k ⎛P ⎞ 0.667 / 1.667 ⎛1⎞ T5 = T4 ⎜ 5 ⎟ ⎜P ⎟ = (278 K )⎜ ⎟ = 179.1 K = −93.9°C ⎝ 4⎠ ⎝ 3⎠Then the COP of this ideal gas refrigeration cycle is determined from qL qL h6 − h5 COPR = = = wnet,in wcomp,in − w turb,out (h2 − h1 ) − (h4 − h5 ) T6 − T5 − 25°C − (− 93.9°C ) = = = 1.49 (T2 − T1 ) − (T4 − T5 ) [135.2 − (− 10)]°C − [5 − (− 93.9)]°C(c) The net power input is determined from W net,in = Wcomp,in − W turb,out = m[(h2 − h1 ) − (h4 − h5 )] & & & & = mc p [(T2 − T1 ) − (T4 − T5 )] & = (0.45 kg/s )(5.1926 kJ/kg ⋅ °C )([135.2 − (− 10)] − [5 − (− 93.9 )]) = 108.2 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.
  • 78. 11-7811-113 An absorption refrigeration system operating at specified conditions is considered. The minimumrate of heat supply required is to be determined.Analysis The maximum COP that this refrigeration system can have is ⎛ T ⎞⎛ T L ⎞ ⎛ 298K ⎞⎛ 263 ⎞ COPR, max = ⎜1 − 0 ⎜ T ⎟⎜ ⎟⎜ T − T ⎟ = ⎜1 − ⎟ ⎜ 358K ⎟⎜ 298 − 263 ⎟ = 1.259 ⎟ ⎝ s ⎠⎝ 0 L ⎠ ⎝ ⎠⎝ ⎠Thus, & QL 12 kW & Qgen,min = = = 9.53 kW COPR, max 1.25911-114 EES Problem 11-113 is reconsidered. The effect of the source temperature on the minimum rate ofheat supply is to be investigated.Analysis The problem is solved using EES, and the solution is given below."Input Data:"T_L = -10 [C]T_0 = 25 [C]T_s = 85 [C]Q_dot_L = 8 [kW]"The maximum COP that this refrigeration system can have is:"COP_R_max = (1-(T_0+273)/(T_s+273))*((T_L+273)/(T_0 - T_L))"The minimum rate of heat supply is:"Q_dot_gen_min = Q_dot_L/COP_R_max Qgenmin [kW] Ts [C] 14 13.76 50 8.996 65 12 6.833 80 Qgen;min [kW] 5.295 100 10 4.237 125 3.603 150 8 2.878 200 2.475 250 6 4 2 50 90 130 170 210 250 Ts [C]PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
  • 79. 11-7911-115 A room is cooled adequately by a 5000 Btu/h window air-conditioning unit. The rate of heat gainof the room when the air-conditioner is running continuously is to be determined.Assumptions 1 The heat gain includes heat transfer through the walls and the roof, infiltration heat gain,solar heat gain, internal heat gain, etc. 2 Steady operating conditions exist.Analysis The rate of heat gain of the room in steady operation is simply equal to the cooling rate of the air-conditioning system, & & Qheat gain = Qcooling = 5,000 Btu / hPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
  • 80. 11-8011-116 A regenerative gas refrigeration cycle using air as the working fluid is considered. Theeffectiveness of the regenerator, the rate of heat removal from the refrigerated space, the COP of the cycle,and the refrigeration load and the COP if this system operated on the simple gas refrigeration cycle are tobe determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3Air is an ideal gas with variable specific heats.Analysis (a) For this problem, we use the properties of air from EES:T1 = 0°C ⎯ ⎯→ h1 = 273.40 kJ/kg . QLP1 = 100 kPa ⎫ ⎬ s1 = 5.6110 kJ/kg.KT1 = 0°C ⎭ Heat Exch.P2 = 500 kPa ⎫ 6 ⎬ h2 s = 433.50 kJ/kgs 2 = s1 ⎭ Regenerator Heat h2 s − h1 Exch. ηC = 3 h2 − h1 5 1 4 . 433.50 − 273.40 20.80 = QH h2 − 273.40 h2 = 473.52 kJ/kg T3 = 35°C ⎯ ⎯→ h3 = 308.63 kJ/kg TurbineFor the turbine inlet and exit we have CompressorT5 = −80°C ⎯ h5 = 193.45 kJ/kg ⎯→ T 2T4 = ? ⎯ h4 = ⎯→ · 2s QH h4 − h5ηT = h4 − h5 s 35°C 3P1 = 100 kPa ⎫ 0°C ⎬ s1 = 5.6110 kJ/kg.K Qregen 1T1 = 0°C ⎭ 4P4 = 500 kPa ⎫ ⎬ s4 =T4 = ? ⎭ ·6 -80°C 5 QRefrigP5 = 500 kPa ⎫ 5s s ⎬ h5 s =s5 = s 4 ⎭We can determine the temperature at the turbine inlet from EES using the above relations. A hand solutionwould require a trial-error approach. T4 = 281.8 K, h4 = 282.08 kJ/kgAn energy balance on the regenerator gives h6 = h1 − h3 + h4 = 273.40 − 308.63 + 282.08 = 246.85 kJ/kgThe effectiveness of the regenerator is determined from h3 − h4 308.63 − 282.08 ε regen = = = 0.430 h3 − h6 308.63 − 246.85PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
  • 81. 11-81(b) The refrigeration load is & Q L = m(h6 − h5 ) = (0.4 kg/s)(246.85 − 193.45)kJ/kg = 21.36 kW &(c) The turbine and compressor powers and the COP of the cycle are & WC,in = m(h2 − h1 ) = (0.4 kg/s)(473.52 − 273.40)kJ/kg = 80.05 kW & & WT,out = m(h4 − h5 ) = (0.4 kg/s)(282.08 − 193.45)kJ/kg = 35.45 kW & & QL & QL 21.36 COP = = = = 0.479 W& & & WC,in − WT, out 80.05 − 35.45 net,in(d) The simple gas refrigeration cycle analysis is as follows: 2 h1 = 273.40 kJ/kg T 2 h2 = 473.52 kJ/kg · QH h3 = 308.63 kJ/kg 3 35°C P3 = 500 kPa ⎫ ⎬ s 3 = 5.2704 kJ/kg 0°C T3 = 35°C ⎭ 1 · 4 QRefrig P1 = 100 kPa ⎫ 4s ⎬ h4 s = 194.52 kJ/kg.K s s 4 = s3 ⎭ h3 − h4 308.63 − h4 ηT = ⎯ ⎯→ 0.85 = ⎯ ⎯→ h4 = 211.64 kJ/kg h3 − h4 s 308.63 − 194.52 & Q L = m(h1 − h4 ) = (0.4 kg/s)(273.40 − 211.64)kJ/kg = 24.70 kW & W net,in = m(h2 − h1 ) − m(h3 − h4 ) = (0.4 kg/s)[(473.52 − 273.40) − (308.63 − 211.64)kJ/kg ] = 41.25 kW & & & & QL 24.70 COP = = = 0.599 W& 41.25 net,inPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
  • 82. 11-8211-117 A heat pump water heater has a COP of 2.2 and consumes 2 kW when running. It is to bedetermined if this heat pump can be used to meet the cooling needs of a room by absorbing heat from it.Assumptions The COP of the heat pump remains constant whether heat is absorbed from the outdoor airor room air.Analysis The COP of the heat pump is given to be 2.2. Then the COP of the air-conditioning systembecomes COPair-cond = COPheat pump − 1 = 2.2 − 1 = 1.2Then the rate of cooling (heat absorption from the air) becomes & & Qcooling = COPair-cond Win = (1.2)(2 kW) = 2.4 kW = 8640 kJ / hsince 1 kW = 3600 kJ/h. We conclude that this heat pump can meet the cooling needs of the room since itscooling rate is greater than the rate of heat gain of the room.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.
  • 83. 11-8311-118 An innovative vapor-compression refrigeration system with a heat exchanger is considered. Thesystem’s COP is to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 2 Condenser T Compressor 3 · QH 2 1 3 800 kPa · Heat 4 Win exchanger 4 -10.1°C · 6 1 6 Throttle 5 QL valve Evaporator 5 sAnalysis From the refrigerant tables (Tables A-11, A-12, and A-13), P3 = 800 kPa ⎫ ⎬ h3 = h f @ 800 kPa = 95.47 kJ/kg sat. liquid ⎭ T4 = Tsat @ 800 kPa − 11.3 ⎫ ⎪ = 31.3 − 11.3 = 20°C ⎬ h4 ≅ h f @ 20°C = 79.32 kJ/kg P4 = 800 kPa ⎪ ⎭ h5 ≅ h4 = 79.32 kJ/kg ( throttling) T6 = −10.1°C ⎫ h6 = h g @ −10.1°C = 244.46 kJ/kg ⎬ sat. vapor ⎭ P6 = Psat @ −10.1°C = 200 kPaAn energy balance on the heat exchanger gives m(h1 − h6 ) = m(h3 − h4 ) ⎯ & & ⎯→ h1 = h3 − h4 + h6 = 95.47 − 79.32 + 244.46 = 260.61 kJ/kgThen, P1 = 200 kPa ⎫ ⎬ s = 0.9970 kJ/kg ⋅ K h1 = 260.61 kJ/kg ⎭ 1 P2 = 800 kPa ⎫ ⎬ h2 = 292.17 kJ/kg s 2 = s1 ⎭The COP of this refrigeration system is determined from its definition, qL h − h5 244.46 − 79.32 COPR = = 6 = = 5.23 win h2 − h1 292.17 − 260.61PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
  • 84. 11-8411-119 An innovative vapor-compression refrigeration system with a heat exchanger is considered. Thesystem’s COP is to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 2 Condenser T Compressor 3 · QH 2 1 3 800 kPa · Heat 4 Win exchanger 4 200 kPa · 6 1 5 6 Throttle QL valve Evaporator 5 sAnalysis From the refrigerant tables (Tables A-11, A-12, and A-13), P3 = 800 kPa ⎫ ⎬ h3 = h f @ 800 kPa = 95.47 kJ/kg sat. liquid ⎭ T4 = Tsat @ 800 kPa − 21.3 ⎫ ⎪ = 31.3 − 21.3 = 10°C ⎬ h4 ≅ h f @ 10°C = 65.43 kJ/kg P4 = 800 kPa ⎪ ⎭ h5 ≅ h4 = 65.43 kJ/kg ( throttling) T6 = −10.1°C ⎫ h6 = h g @ −10.1°C = 244.46 kJ/kg ⎬ sat. vapor ⎭ P6 = Psat @ −10.1°C = 200 kPaAn energy balance on the heat exchanger gives m(h1 − h6 ) = m(h3 − h4 ) ⎯ & & ⎯→ h1 = h3 − h4 + h6 = 95.47 − 65.43 + 244.46 = 274.50 kJ/kgThen, P1 = 200 kPa ⎫ ⎬ s = 1.0449 kJ/kg ⋅ K h1 = 274.50 kJ/kg ⎭ 1 P2 = 800 kPa ⎫ ⎬ h2 = 308.28 kJ/kg s 2 = s1 ⎭The COP of this refrigeration system is determined from its definition, qL h − h5 244.46 − 65.43 COPR = = 6 = = 5.30 win h2 − h1 308.28 − 274.50PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
  • 85. 11-8511-120 An ideal gas refrigeration cycle with with three stages of compression with intercooling using air asthe working fluid is considered. The COP of this system is to be determined.Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant specific heats. 3Kinetic and potential energy changes are negligible.Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2a).Analysis From the isentropic relations, ( k −1) / k ⎛P ⎞ 6 4 T2 = T1 ⎜ 2 ⎜ ⎟ ⎟ = (253 K)(5) 0.4 / 1.4 = 400.7 K T ⎝ P1 ⎠ 2 ( k −1) / k 7 ⎛P ⎞ 15°C T4 = T6 = T3 ⎜ 4 ⎜P ⎟ ⎟ = (288 K)(5) 0.4 / 1.4 = 456.1 K 5 3 ⎝ 3 ⎠ -20°C ( k −1) / k 1 ⎛P ⎞ 0.4 / 1.4 ⎛ 1 ⎞ T8 = T7 ⎜ 8 ⎜ ⎟ ⎟ = (288 K)⎜ ⎟ = 72.5 K ⎝ P7 ⎠ ⎝ 5× 5× 5 ⎠ 8 sThe COP of this ideal gas refrigeration cycle is determined from qL qL COPR = = w net,in wcomp,in − w turb,out h1 − h8 = (h2 − h1 ) + ( h4 − h3 ) + ( h6 − h5 ) − ( h7 − h8 ) T1 − T8 = (T2 − T1 ) + 2(T4 − T3 ) − (T7 − T8 ) 253 − 72.5 = = 0.673 (400.7 − 253) + 2(456.1 − 288) − (288 − 72.5)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.
  • 86. 11-8611-121 An ideal vapor-compression refrigeration cycle with refrigerant-22 as the working fluid isconsidered. The evaporator is located inside the air handler of building. The hardware and the T-s diagramfor this heat pump application are to be sketched. The COP of the unit and the ratio of volume flow rate ofair entering the air handler to mass flow rate of R-22 through the air handler 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-22 data from the problem statement, Air T 45°C Condenser · QH 2 3 2 3 45°C · Expansion Win Win valve Compressor -5°C 4 1 1 Evaporator 4s 4 · QL -5°C sat. vap. QL s T1 = −5°C ⎫ h1 = h g @ −5°C = 248.1 kJ/kg ⎬ sat. vapor ⎭ s1 = s g @ −5°C = 0.9344 kJ/kg ⋅ K P2 = 1728 kPa ⎫ ⎬ h2 = 283.7 kJ/kg s 2 = s1 ⎭ P3 = 1728 kPa ⎫ ⎬ h3 = h f @ 1728 kPa = 101 kJ/kg sat. liquid ⎭ h4 ≅ h3 = 101 kJ/kg ( throttling)(b) The COP of the heat pump is determined from its definition, qH h − h3 283.7 − 101 COPHP = = 2 = = 5.13 win h2 − h1 283.7 − 248.1(c) An energy balance on the condenser gives & V& QH = mR (h2 − h3 ) = ma c p ΔT = a c p ΔT & & vaRearranging, we obtain the ratio of volume flow rate of air entering the air handler to mass flow rate of R-22 through the air handler V&a h 2 − h3 (283.7 − 101) kJ/kg = = & mR (1 / v )c p ΔT (1 / 0.847 m 3 /kg)(1.005 kJ/kg ⋅ K))(20 K ) = 7.699 (m 3 air/s)/(kg R22/s) = 462 (m 3 air/min)/(kg R22/s)Note that the specific volume of air is obtained from ideal gas equation taking the pressure of air to be 101kPa and using the room temperature of air (25°C = 298 K) to be 0.847 m3/kg.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
  • 87. 11-8711-122 An ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid isconsidered. Cooling water flows through the water jacket surrounding the condenser. To produce ice,potable water is supplied to the chiller section of the refrigeration cycle. The hardware and the T-s diagramfor this refrigerant-ice making system are to be sketched. The mass flow rates of the refrigerant and thepotable water 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-134a data from the problem statement, Water 200 kg/s T 1.2 MPa Condenser · QH 2 3 2 3 1.2 MPa Expansion · Win Win valve Compressor 140 kPa 4 1 Evaporator 4s · 1 4 sat. vap. QL 140 kPa Potable s water T1 = 140 kPa ⎫ h1 = h g @ 140 kPa = 239.16 kJ/kg ⎬ s =s sat. vapor ⎭ 1 g @ 140 kPa = 0.94456 kJ/kg ⋅ K P2 = 1200 kPa ⎫ ⎬ h2 = 284.07 kJ/kg s 2 = s1 ⎭ P3 = 1200 kPa ⎫ ⎬ h3 = h f @ 1200 kPa = 117.77 kJ/kg sat. liquid ⎭ h4 ≅ h3 = 117.77 kJ/kg ( throttling)(b) An energy balance on the condenser gives & Q H = m R (h2 − h3 ) = m w c p ΔT & &Solving for the mass flow rate of the refrigerant m w c p ΔT & (200 kg/s)(4.18 kJ/kg ⋅ K))(10 K ) mR = & = = 50.3 kg/s h2 − h3 (284.07 − 117.77)kJ/kg(c) An energy balance on the evaporator gives & Q L = m R (h1 − h4 ) = m w hif & &Solving for the mass flow rate of the potable water m R (h1 − h4 ) (50.3 kg/s)(239.16 − 117.77)kJ/kg & mw = & = = 18.3 kg/s hif 333 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.
  • 88. 11-8811-123 A vortex tube receives compressed air at 500 kPa and 300 K, and supplies 25 percent of it as coldair and the rest as hot air. The COP of the vortex tube is to be compared to that of a reversed Brayton cyclefor the same pressure ratio; the exit temperature of the hot fluid stream and the COP are to be determined;and it is to be shown if this process violates the second law.Assumptions 1 The vortex tube is adiabatic. 2 Air is an ideal gas with constant specific heats at roomtemperature. 3 Steady operating conditions exist.Properties The gas constant of air is 0.287 kJ/kg.K (Table A-1). The specific heat of air at roomtemperature is cp = 1.005 kJ/kg.K (Table A-2). The enthalpy of air at absolute temperature T can beexpressed in terms of specific heats as h = cpT.Analysis (a) The COP of the vortex tube is much lower than the COP of a reversed Brayton cycle of thesame pressure ratio since the vortex tube involves vortices, which are highly irreversible. Owing to thisirreversibility, the minimum temperature that can be obtained by the vortex tube is not as low as the onethat can be obtained by the revered Brayton cycle.(b) We take the vortex tube as the system. This is a steady flow system with one inlet and two exits, and itinvolves no heat or work interactions. Then the steady-flow energy balance equation for this system & & Ein = Eout for a unit mass flow rate at the inlet (m1 = 1 kg / s) can be expressed as & m1 h1 = m 2 h2 + m 3 h3 & & & m1 c p T1 = m 2 c p T2 + m3 c p T3 & & & Compressed 1 air 1c p T1 = 0.25c p T2 + 0.75c p T3Canceling cp and solving for T3 gives Cold air Warm air T − 0.25T2 T3 = 1 0.75 2 3 300 − 0.25 × 278 = = 307.3 K 0.75Therefore, the hot air stream will leave the vortex tube at an average temperature of 307.3 K. & & &(c) The entropy balance for this steady flow system Sin − Sout + Sgen = 0 can be expressed as with one inletand two exits, and it involves no heat or work interactions. Then the steady-flow entropy balance equationfor this system for a unit mass flow rate at the inlet (m1 = 1 kg / s) can be expressed & & & & S gen = S out − S in = m 2 s 2 + m3 s 3 − m1 s1 = m 2 s 2 + m3 s 3 − (m 2 + m3 ) s1 & & & & & & & = m 2 ( s 2 − s1 ) + m3 ( s 3 − s1 ) & & = 0.25( s 2 − s1 ) + 0.75( s 3 − s1 ) ⎛ T P ⎞ ⎛ T P ⎞ = 0.25⎜ c p ln 2 − R ln 2 ⎟ + 0.75⎜ c p ln 3 − R ln 3 ⎟ ⎜ ⎝ T1 ⎟ P1 ⎠ ⎜ ⎝ T1 P1 ⎟ ⎠Substituting the known quantities, the rate of entropy generation is determined to be & ⎛ 278 K 100 kPa ⎞ S gen = 0.25⎜ (1.005 kJ/kg.K) ln − (0.287 kJ/kg.K) ln ⎟ ⎝ 300 K 500 kPa ⎠ ⎛ 307.3 K 100 kPa ⎞ + 0.75⎜ (1.005 kJ/kg.K) ln − (0.287 kJ/kg.K) ln ⎟ ⎝ 300 K 500 kPa ⎠ = 0.461 kW/K > 0which is a positive quantity. Therefore, this process satisfies the 2nd law of thermodynamics.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
  • 89. 11-89(d) For a unit mass flow rate at the inlet (m1 = 1 kg / s) , the cooling rate and the power input to the &compressor are determined to & Qcooling = m c (h1 − hc ) = m c c p (T1 − Tc ) & & = (0.25 kg/s)(1.005 kJ/kg.K)(300 - 278)K = 5.53 kW & m0 RT0 ⎡⎛ P ⎞ ( k −1) / k ⎤ & Wcomp,in = ⎢⎜ 1 ⎟ − 1⎥ (k − 1)η comp ⎢⎜ P0 ⎟ ⎥ ⎣⎝ ⎠ ⎦ (1 kg/s)(0.287 kJ/kg.K)(300 K) ⎡⎛ 500 kPa ⎞ ⎤ (1.4 −1) / 1.4 = ⎢⎜ ⎟ − 1⎥ = 157.1 kW (1.4 − 1)0.80 ⎢⎝ 100 kPa ⎠ ⎣ ⎥ ⎦Then the COP of the vortex refrigerator becomes & Qcooling 5.53 kW COP = = = 0.035 & Wcomp, in 157.1 kWThe COP of a Carnot refrigerator operating between the same temperature limits of 300 K and 278 K is TL 278 K COPCarnot = = = 12.6 TH − TL (300 − 278) KDiscussion Note that the COP of the vortex refrigerator is a small fraction of the COP of a Carnotrefrigerator operating between the same temperature limits.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.
  • 90. 11-9011-124 A vortex tube receives compressed air at 600 kPa and 300 K, and supplies 25 percent of it as coldair and the rest as hot air. The COP of the vortex tube is to be compared to that of a reversed Brayton cyclefor the same pressure ratio; the exit temperature of the hot fluid stream and the COP are to be determined;and it is to be shown if this process violates the second law.Assumptions 1 The vortex tube is adiabatic. 2 Air is an ideal gas with constant specific heats at roomtemperature. 3 Steady operating conditions exist.Properties The gas constant of air is 0.287 kJ/kg.K (Table A-1). The specific heat of air at roomtemperature is cp = 1.005 kJ/kg.K (Table A-2). The enthalpy of air at absolute temperature T can beexpressed in terms of specific heats as h = cp T.Analysis (a) The COP of the vortex tube is much lower than the COP of a reversed Brayton cycle of thesame pressure ratio since the vortex tube involves vortices, which are highly irreversible. Owing to thisirreversibility, the minimum temperature that can be obtained by the vortex tube is not as low as the onethat can be obtained by the revered Brayton cycle.(b) We take the vortex tube as the system. This is a steady flow system with one inlet and two exits, and itinvolves no heat or work interactions. Then the steady-flow entropy balance equation for this system & & Ein = Eout for a unit mass flow rate at the inlet (m1 = 1 kg / s) can be expressed as & m1 h1 = m 2 h2 + m3 h3 & & & m1 c p T1 = m 2 c p T2 + m3 c p T3 & & & Compressed 1 1c p T1 = 0.25c p T2 + 0.75c p T3 airCanceling cp and solving for T3 gives Cold air Warm air T1 − 0.25T2 T3 = 0.75 2 3 300 − 0.25 × 278 = = 307.3 K 0.75Therefore, the hot air stream will leave the vortex tube at an average temperature of 307.3 K. & & &(c) The entropy balance for this steady flow system Sin − Sout + Sgen = 0 can be expressed as with one inletand two exits, and it involves no heat or work interactions. Then the steady-flow energy balance equationfor this system for a unit mass flow rate at the inlet (m1 = 1 kg / s) can be expressed & & & & S gen = S out − S in = m 2 s 2 + m3 s 3 − m1 s1 = m 2 s 2 + m3 s 3 − ( m 2 + m3 ) s1 & & & & & & & =m& 2 ( s 2 − s1 ) + m 3 ( s 3 − s1 ) & = 0.25( s 2 − s1 ) + 0.75( s 3 − s1 ) ⎛ T P ⎞ ⎛ T P ⎞ = 0.25⎜ c p ln 2 − R ln 2 ⎟ + 0.75⎜ c p ln 3 − R ln 3 ⎟ ⎜ ⎝ T1 P1 ⎟ ⎠ ⎜ ⎝ T1 P1 ⎟ ⎠Substituting the known quantities, the rate of entropy generation is determined to be & ⎛ 278 K 100 kPa ⎞ S gen = 0.25⎜ (1.005 kJ/kg.K)ln − (0.287 kJ/kg.K)ln ⎟ ⎝ 300 K 600 kPa ⎠ ⎛ 307.3 K 100 kPa ⎞ + 0.75⎜ (1.005 kJ/kg.K)ln − (0.287 kJ/kg.K)ln ⎟ ⎝ 300 K 600 kPa ⎠ = 0.513 kW/K > 0which is a positive quantity. Therefore, this process satisfies the 2nd law of thermodynamics.PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
  • 91. 11-91(d) For a unit mass flow rate at the inlet (m1 = 1 kg / s) , the cooling rate and the power input to the &compressor are determined to & Qcooling = m c (h1 − hc ) = m c c p (T1 − Tc ) & & = (0.25 kg/s)(1.005 kJ/kg.K)(300 - 278)K = 5.53 kW & m0 RT0 ⎡⎛ P ⎞ ( k −1) / k ⎤ & Wcomp,in = ⎢⎜ 1 ⎟ − 1⎥ (k − 1)η comp ⎢⎜ P0 ⎟ ⎥ ⎣⎝ ⎠ ⎦ (1 kg/s)(0.287 kJ/kg.K)(300 K) ⎡⎛ 600 kPa ⎞ ⎤ (1.4 −1) / 1.4 = ⎢⎜ ⎟ − 1⎥ = 179.9 kW (1.4 − 1)0.80 ⎢⎝ 100 kPa ⎠ ⎣ ⎥ ⎦Then the COP of the vortex refrigerator becomes & Qcooling 5.53 kW COP = = = 0.031 & Wcomp, in 179.9 kWThe COP of a Carnot refrigerator operating between the same temperature limits of 300 K and 278 K is TL 278 K COPCarnot = = = 12.6 TH − TL (300 − 278) KDiscussion Note that the COP of the vortex refrigerator is a small fraction of the COP of a Carnotrefrigerator operating between the same temperature limits.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.
  • 92. 11-9211-125 EES The effect of the evaporator pressure on the COP of an ideal vapor-compression refrigerationcycle with R-134a as the working fluid is to be investigated.Analysis The problem is solved using EES, and the solution is given below."Input Data"P[1]=100 [kPa]P[2] = 1000 [kPa]Fluid$=R134aEta_c=0.7 "Compressor isentropic efficiency""Compressor"h[1]=enthalpy(Fluid$,P=P[1],x=1) "properties for state 1"s[1]=entropy(Fluid$,P=P[1],x=1)T[1]=temperature(Fluid$,h=h[1],P=P[1])h2s=enthalpy(Fluid$,P=P[2],s=s[1]) "Identifies state 2s as isentropic"h[1]+Wcs=h2s "energy balance on isentropic compressor"W_c=Wcs/Eta_c"definition of compressor isentropic efficiency"h[1]+W_c=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])"Condenser"P[3] = P[2]h[3]=enthalpy(Fluid$,P=P[3],x=0) "properties for state 3"s[3]=entropy(Fluid$,P=P[3],x=0)h[2]=Qout+h[3] "energy balance on condenser""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$,h=h[4],P=P[4])"Evaporator"P[4]= P[1]Q_in + h[4]=h[1] "energy balance on evaporator""Coefficient of Performance:"COP=Q_in/W_c "definition of COP" COP ηc P1 [kPa] 10 1.851 0.7 100 η comp 2.863 0.7 200 4.014 0.7 300 8 1.0 5.462 0.7 400 7.424 0.7 500 0.7 6 COP 4 2 0 100 150 200 250 300 350 400 450 500 P[1] [kPa]PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
  • 93. 11-9311-126 EES The effect of the condenser pressure on the COP of an ideal vapor-compression refrigerationcycle with R-134a as the working fluid is to be investigated.Analysis The problem is solved using EES, and the solution is given below."Input Data"P[1]=120 [kPa]P[2] = 400 [kPa]Fluid$=R134aEta_c=0.7 "Compressor isentropic efficiency""Compressor"h[1]=enthalpy(Fluid$,P=P[1],x=1) "properties for state 1"s[1]=entropy(Fluid$,P=P[1],x=1)T[1]=temperature(Fluid$,h=h[1],P=P[1])h2s=enthalpy(Fluid$,P=P[2],s=s[1]) "Identifies state 2s as isentropic"h[1]+Wcs=h2s "energy balance on isentropic compressor"W_c=Wcs/Eta_c"definition of compressor isentropic efficiency"h[1]+W_c=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])"Condenser"P[3] = P[2]h[3]=enthalpy(Fluid$,P=P[3],x=0) "properties for state 3"s[3]=entropy(Fluid$,P=P[3],x=0)h[2]=Qout+h[3] "energy balance on condenser""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]= P[1]Q_in + h[4]=h[1] "energy balance on evaporator""Coefficient of Performance:"COP=Q_in/W_c "definition of COP" 8 COP ηc P2 [kPa] 7 η comp 4.935 0.7 400 3.04 0.7 650 6 1.0 2.258 0.7 900 1.803 0.7 1150 5 0.7 1.492 0.7 1400 COP 4 3 2 1 0 400 600 800 1000 1200 1400 P[2] [kPa]PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
  • 94. 11-94Fundamentals of Engineering (FE) Exam Problems11-127 Consider a heat pump that operates on the reversed Carnot cycle with R-134a as the working fluidexecuted under the saturation dome between the pressure limits of 140 kPa and 800 kPa. R-134a changesfrom saturated vapor to saturated liquid during the heat rejection process. The net work input for this cycleis(a) 28 kJ/kg (b) 34 kJ/kg (c) 49 kJ/kg (d) 144 kJ/kg (e) 275 kJ/kgAnswer (a) 28 kJ/kgSolution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines ona blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numericalvalues).P1=800 "kPa"P2=140 "kPa"h_fg=ENTHALPY(R134a,x=1,P=P1)-ENTHALPY(R134a,x=0,P=P1)TH=TEMPERATURE(R134a,x=0,P=P1)+273TL=TEMPERATURE(R134a,x=0,P=P2)+273q_H=h_fgCOP=TH/(TH-TL)w_net=q_H/COP"Some Wrong Solutions with Common Mistakes:"W1_work = q_H/COP1; COP1=TL/(TH-TL) "Using COP of regrigerator"W2_work = q_H/COP2; COP2=(TH-273)/(TH-TL) "Using C instead of K"W3_work = h_fg3/COP; h_fg3= ENTHALPY(R134a,x=1,P=P2)-ENTHALPY(R134a,x=0,P=P2)"Using h_fg at P2"W4_work = q_H*TL/TH "Using the wrong relation"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.
  • 95. 11-9511-128 A refrigerator removes heat from a refrigerated space at –5°C at a rate of 0.35 kJ/s and rejects it toan environment at 20°C. The minimum required power input is(a) 30 W (b) 33 W (c) 56 W (d) 124 W (e) 350 WAnswer (b) 33 WSolution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines ona blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numericalvalues).TH=20+273TL=-5+273Q_L=0.35 "kJ/s"COP_max=TL/(TH-TL)w_min=Q_L/COP_max"Some Wrong Solutions with Common Mistakes:"W1_work = Q_L/COP1; COP1=TH/(TH-TL) "Using COP of heat pump"W2_work = Q_L/COP2; COP2=(TH-273)/(TH-TL) "Using C instead of K"W3_work = Q_L*TL/TH "Using the wrong relation"W4_work = Q_L "Taking the rate of refrigeration as power input"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.
  • 96. 11-9611-129 A refrigerator operates on the ideal vapor compression refrigeration cycle with R-134a as theworking fluid between the pressure limits of 120 kPa and 800 kPa. If the rate of heat removal from therefrigerated space is 32 kJ/s, the mass flow rate of the refrigerant is(a) 0.19 kg/s (b) 0.15 kg/s (c) 0.23 kg/s (d) 0.28 kg/s (e) 0.81 kg/sAnswer (c) 0.23 kg/sSolution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines ona blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numericalvalues).P1=120 "kPa"P2=800 "kPa"P3=P2P4=P1s2=s1Q_refrig=32 "kJ/s"m=Q_refrig/(h1-h4)h1=ENTHALPY(R134a,x=1,P=P1)s1=ENTROPY(R134a,x=1,P=P1)h2=ENTHALPY(R134a,s=s2,P=P2)h3=ENTHALPY(R134a,x=0,P=P3)h4=h3"Some Wrong Solutions with Common Mistakes:"W1_mass = Q_refrig/(h2-h1) "Using wrong enthalpies, for W_in"W2_mass = Q_refrig/(h2-h3) "Using wrong enthalpies, for Q_H"W3_mass = Q_refrig/(h1-h44); h44=ENTHALPY(R134a,x=0,P=P4) "Using wrong enthalpy h4 (atP4)"W4_mass = Q_refrig/h_fg; h_fg=ENTHALPY(R134a,x=1,P=P2) - ENTHALPY(R134a,x=0,P=P2)"Using h_fg at P2"PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
  • 97. 11-9711-130 A heat pump operates on the ideal vapor compression refrigeration cycle with R-134a as theworking fluid between the pressure limits of 0.32 MPa and 1.2 MPa. If the mass flow rate of the refrigerantis 0.193 kg/s, the rate of heat supply by the heat pump to the heated space is(a) 3.3 kW (b) 23 kW (c) 26 kW (d) 31 kW (e) 45 kWAnswer (d) 31 kWSolution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines ona blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numericalvalues).P1=320 "kPa"P2=1200 "kPa"P3=P2P4=P1s2=s1m=0.193 "kg/s"Q_supply=m*(h2-h3) "kJ/s"h1=ENTHALPY(R134a,x=1,P=P1)s1=ENTROPY(R134a,x=1,P=P1)h2=ENTHALPY(R134a,s=s2,P=P2)h3=ENTHALPY(R134a,x=0,P=P3)h4=h3"Some Wrong Solutions with Common Mistakes:"W1_Qh = m*(h2-h1) "Using wrong enthalpies, for W_in"W2_Qh = m*(h1-h4) "Using wrong enthalpies, for Q_L"W3_Qh = m*(h22-h4); h22=ENTHALPY(R134a,x=1,P=P2) "Using wrong enthalpy h2 (hg at P2)"W4_Qh = m*h_fg; h_fg=ENTHALPY(R134a,x=1,P=P1) - ENTHALPY(R134a,x=0,P=P1) "Usingh_fg at P1"PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
  • 98. 11-9811-131 An ideal vapor compression refrigeration cycle with R-134a as the working fluid operates betweenthe pressure limits of 120 kPa and 1000 kPa. The mass fraction of the refrigerant that is in the liquid phaseat the inlet of the evaporator is(a) 0.65 (b) 0.60 (c) 0.40 (d) 0.55 (e) 0.35Answer (b) 0.60Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines ona blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numericalvalues).P1=120 "kPa"P2=1000 "kPa"P3=P2P4=P1h1=ENTHALPY(R134a,x=1,P=P1)h3=ENTHALPY(R134a,x=0,P=P3)h4=h3x4=QUALITY(R134a,h=h4,P=P4)liquid=1-x4"Some Wrong Solutions with Common Mistakes:"W1_liquid = x4 "Taking quality as liquid content"W2_liquid = 0 "Assuming superheated vapor"W3_liquid = 1-x4s; x4s=QUALITY(R134a,s=s3,P=P4) "Assuming isentropic expansion"s3=ENTROPY(R134a,x=0,P=P3)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.
  • 99. 11-9911-132 Consider a heat pump that operates on the ideal vapor compression refrigeration cycle with R-134aas the working fluid between the pressure limits of 0.32 MPa and 1.2 MPa. The coefficient of performanceof this heat pump is(a) 0.17 (b) 1.2 (c) 3.1 (d) 4.9 (e) 5.9Answer (e) 5.9Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines ona blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numericalvalues).P1=320 "kPa"P2=1200 "kPa"P3=P2P4=P1s2=s1h1=ENTHALPY(R134a,x=1,P=P1)s1=ENTROPY(R134a,x=1,P=P1)h2=ENTHALPY(R134a,s=s2,P=P2)h3=ENTHALPY(R134a,x=0,P=P3)h4=h3COP_HP=qH/WinWin=h2-h1qH=h2-h3"Some Wrong Solutions with Common Mistakes:"W1_COP = (h1-h4)/(h2-h1) "COP of refrigerator"W2_COP = (h1-h4)/(h2-h3) "Using wrong enthalpies, QL/QH"W3_COP = (h22-h3)/(h22-h1); h22=ENTHALPY(R134a,x=1,P=P2) "Using wrong enthalpy h2(hg at P2)"PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.
  • 100. 11-10011-133 An ideal gas refrigeration cycle using air as the working fluid operates between the pressure limitsof 80 kPa and 280 kPa. Air is cooled to 35°C before entering the turbine. The lowest temperature of thiscycle is(a) –58°C (b) -26°C (c) 0°C (d) 11°C (e) 24°CAnswer (a) –58°CSolution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines ona blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numericalvalues).k=1.4P1= 80 "kPa"P2=280 "kPa"T3=35+273 "K""Mimimum temperature is the turbine exit temperature"T4=T3*(P1/P2)^((k-1)/k) - 273"Some Wrong Solutions with Common Mistakes:"W1_Tmin = (T3-273)*(P1/P2)^((k-1)/k) "Using C instead of K"W2_Tmin = T3*(P1/P2)^((k-1)) - 273 "Using wrong exponent"W3_Tmin = T3*(P1/P2)^k - 273 "Using wrong exponent"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.
  • 101. 11-10111-134 Consider an ideal gas refrigeration cycle using helium as the working fluid. Helium enters thecompressor at 100 kPa and –10°C and is compressed to 250 kPa. Helium is then cooled to 20°C before itenters the turbine. For a mass flow rate of 0.2 kg/s, the net power input required is(a) 9.3 kW (b) 27.6 kW (c) 48.8 kW (d) 93.5 kW (e) 119 kWAnswer (b) 27.6 kWSolution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines ona blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numericalvalues).k=1.667Cp=5.1926 "kJ/kg.K"P1= 100 "kPa"T1=-10+273 "K"P2=250 "kPa"T3=20+273 "K"m=0.2 "kg/s""Mimimum temperature is the turbine exit temperature"T2=T1*(P2/P1)^((k-1)/k)T4=T3*(P1/P2)^((k-1)/k)W_netin=m*Cp*((T2-T1)-(T3-T4))"Some Wrong Solutions with Common Mistakes:"W1_Win = m*Cp*((T22-T1)-(T3-T44)); T22=T1*P2/P1; T44=T3*P1/P2 "Using wrong relationsfor temps"W2_Win = m*Cp*(T2-T1) "Ignoring turbine work"W3_Win=m*1.005*((T2B-T1)-(T3-T4B)); T2B=T1*(P2/P1)^((kB-1)/kB); T4B=T3*(P1/P2)^((kB-1)/kB); kB=1.4 "Using air properties"W4_Win=m*Cp*((T2A-(T1-273))-(T3-273-T4A)); T2A=(T1-273)*(P2/P1)^((k-1)/k); T4A=(T3-273)*(P1/P2)^((k-1)/k) "Using C instead of 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.
  • 102. 11-10211-135 An absorption air-conditioning system is to remove heat from the conditioned space at 20°C at arate of 150 kJ/s while operating in an environment at 35°C. Heat is to be supplied from a geothermalsource at 140°C. The minimum rate of heat supply required is(a) 86 kJ/s (b) 21 kJ/s (c) 30 kJ/s (d) 61 kJ/s (e) 150 kJ/sAnswer (c) 30 kJ/sSolution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines ona blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numericalvalues).TL=20+273 "K"Q_refrig=150 "kJ/s"To=35+273 "K"Ts=140+273 "K"COP_max=(1-To/Ts)*(TL/(To-TL))Q_in=Q_refrig/COP_max"Some Wrong Solutions with Common Mistakes:"W1_Qin = Q_refrig "Taking COP = 1"W2_Qin = Q_refrig/COP2; COP2=TL/(Ts-TL) "Wrong COP expression"W3_Qin = Q_refrig/COP3; COP3=(1-To/Ts)*(Ts/(To-TL)) "Wrong COP expression, COP_HP"W4_Qin = Q_refrig*COP_max "Multiplying by COP instead of dividing"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.
  • 103. 11-10311-136 Consider a refrigerator that operates on the vapor compression refrigeration cycle with R-134a asthe working fluid. The refrigerant enters the compressor as saturated vapor at 160 kPa, and exits at 800 kPaand 50°C, and leaves the condenser as saturated liquid at 800 kPa. The coefficient of performance of thisrefrigerator is(a) 2.6 (b) 1.0 (c) 4.2 (d) 3.2 (e) 4.4Answer (d) 3.2Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines ona blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numericalvalues).P1=160 "kPa"P2=800 "kPa"T2=50 "C"P3=P2P4=P1h1=ENTHALPY(R134a,x=1,P=P1)s1=ENTROPY(R134a,x=1,P=P1)h2=ENTHALPY(R134a,T=T2,P=P2)h3=ENTHALPY(R134a,x=0,P=P3)h4=h3COP_R=qL/WinWin=h2-h1qL=h1-h4"Some Wrong Solutions with Common Mistakes:"W1_COP = (h2-h3)/(h2-h1) "COP of heat pump"W2_COP = (h1-h4)/(h2-h3) "Using wrong enthalpies, QL/QH"W3_COP = (h1-h4)/(h2s-h1); h2s=ENTHALPY(R134a,s=s1,P=P2) "Assuming isentropiccompression"11-137 ··· 11-145 Design and Essay ProblemsPROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers andeducators for course preparation. If you are a student using this Manual, you are using it without permission.

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