Thermodynamic Cycles
Content Thermodynamic Cycles Carnot Cycle Otto Cycle Rankine Cycle Refrigeration Cycle
Thermodynamic Cycles
Carnot Cycle
Derivation of the Carnot Cycle Efficiency
Otto Cycle
Otto Cycle Nomenclature for reciprocating engines TDC Top Dead Center BDC Bottom Dead Center
Otto Cycle
Otto Cycle
Otto Cycle
Otto Cycle kaire = 1.4; kco2 = 1.3; ketano = 1.667
Constant Specific Heat Otto Cycle Data: rr = vv 1 vv 2 PP 1 kpa TT 1 K qq in kj/kg RR Table A1 cc vv, kk Table A2a vv 1 = RRTT 1 PP 1
Constant Specific Heat Process 11 22 Isentropic compression Otto Cycle kk 1 kk vv 2 = vv 1 rr ; PP 2 = PP 1 rr kk PP 2 ; TT 2 = TT 1 PP 1 Process 22 33 Constant-volume heat addition oooo TT 2 = PP 2vv 2 RR vv 3 = vv 2 ; TT 3 = TT 2 + qq in TT 3 ; PP cc 3 = PP 2 vv TT 2 Process 33 44 Isentropic expansion qq in = uu = cc vv (TT 3 TT 2 ) vv 4 = vv 1 ; PP 4 = PP 3 rr kk ; TT 4 = TT 3 rr kk 1
Constant Specific Heat Otto Cycle Process 44 11 Constant-volume heat rejection qq out = cc vv (TT 4 TT 1 ) ww net = qq net = qq in qq out ηη th = ww net qq in ηη otto = 1 rr 1 kk MEP = ww nnnnnn vv 1 vv 2
Variable Specific Heat Otto Cycle Data: rr = vv 1 vv 2 PP 1 kpa TT 1 K qq in kj/kg @TT 1 (Table A17) vv rrr uu 1 RR Table A1
Variable Specific Heat Process 11 22 Isentropic compression Otto Cycle vv rrr = vv rrr rr ; @vv rrr Table A17 TT 2 uu 2 ; PP 2 = PP 1 TT 2 TT 1 rr Process 22 33 Constant-volume heat addition uu 3 = qq in +uu 2 ; @uu 3 (Table A17) TT 3 vv rrr ; PP 3 = PP 2 TT 3 TT 2 Process 33 44 Isentropic expansion vv rrr = rrrr rrr ; @vv rrr (Table A17) TT 4 TT 4 1 ; PP uu 4 = PP 3 4 TT 3 rr qq in = uu = uu 3 uu 2
Variable Specific Heat Otto Cycle Process 44 11 Constant-volume heat rejection qq out = (uu 4 uu 1 ) ww net = qq net = qq in qq out ηη th = ww net qq in ηη otto = 1 rr 1 kk vv 1 = RRTT 1 PP 1 MEP = vv 1 ww nnnnnn 1 1 rr
Example (textbook) EXAMPLE 9 2 The Ideal Otto Cycle An ideal Otto cycle has a compression ratio of 8. At the beginning of the compression process, air is at 100 kpa and 17 C, and 800 kj/kg of heat is transferred to air during the constant-volume heat-addition process. Find the thermal efficiency, and the mean effective pressure for the cycle, (a) considering constant specific heat, (b) considering variable specific heat, (c) Also, determine the power output from the cycle, in kw, for an engine speed of 4000 rpm (rev/min). Assume that this cycle is operated on an engine that has four cylinders with a total displacement volume of 1.6 L.
Example (textbook) a) Constant Specific Heat Data: rr = 8, PP 1 = 100 kpa, TT 1 = 17 C = 290 K, qq in = 800 kj/kg RR = 0.287 kpa m 3 /kgk (Table A1) cc vv = 0.718 kj/kgk, kk = 1.4 (Table A2a) vv 1 = RRTT 1 = (0.287 KPam3 /kgk)(290 K) PP 1 100 kpa Process 11 22 Isentropic compression vv 2 = vv 1 rr = 0.8323 m3 /kg = 0.10404 m 8 3 /kg PP 2 = PP 1 rr kk = 100 kpa 8 1.4 = 11111111. 9999 kkkkkk = 0.8323 m 3 /kg
Example (textbook) TT 2 = PP 2vv 2 RR = (1837.92 kpa)(0.10404 m3 /kg) 0.287 kpa m 3 /kgk = 666666 KK Process 22 33 Constant-volume heat addition vv 3 = vv 2 = 0.10404 m 3 /kg TT 3 = TT 2 + qq in 800 kj/kg = 666 K + = 11111111. 22 KK cc vv 0.718 kj/kgk PP 3 = PP 2 TT 3 TT 2 = 1837.92 kpa 1780.2 K 666 K = 44444444. 7777 kkkkkk
Example (textbook) Process 33 44 Isentropic expansion vv 4 = vv 1 = 0.8323 m 3 /kg PP 4 = PP 3 4912.71 kpa = rrkk (8) 1.4 = 222222. 33 kkkkkk TT 4 = TT 3 1780.2 K = = 777777. 8888 KK rrkk 1 (8) 1.4 1 Process 44 11 Constant-volume heat rejection qq out = cc vv TT 4 TT 1 = 0.718 kj/kgk 774.88 290 K = 333333. 1111 kkkk/kkkk
Example (textbook) ww net = qq net = qq in qq out = 800 348.14 kj/kg = 444444. 8888 kkkk/kkkk ηη th = ww net qq in = 451.86 kj/kg 800 kj/kg = 00. 55555555 = 5555. 4444 % ηη otto = 1 rr 1 kk = 1 8 1 1.4 = 00. 55555555 = 5555. 4444 % MEP = ww nnnnnn vv 1 vv 2 = 451.86 kj/kg 0.8323 0.10404 m 3 /kg 1 kpa m 3 /kg 1 kj = 666666. 4444 kkkkkk
Example (textbook) b) Variable Specific Heat Data: rr = 8, PP 1 = 100 kpa, TT 1 = 17 C = 290 K, qq in = 800 kj/kg RR = 0.287 kpa m 3 /kgk (Table A1) @TT 1 (Table A17) vv rrr = 676.1 uu 1 = 206.91 kj/kg Process 11 22 Isentropic compression vv rrr = vv rrr rr = 676.1 = 84.51 8 @vv rrr Table A17 TT 2 = 666666. 44 KK uu 2 = 475.11 kj/kg
Example (textbook) PP 2 = PP 1 TT 2 TT 1 rr = 100 kpa 652.4 K 290 K 8 = 11111111. 77 kkkkkk Process 22 33 Constant-volume heat addition uu 3 = qq in + uu 2 = 800 + 475.11 kj/kg = 1275.11 kj/kg @uu 3 (Table A17) TT 3 = 11111111. 11 KK vv rrr = 6.108 PP 3 = PP 2 TT 3 TT 2 = 1799.7 kpa 1575.1 K 652.4 K = 44444444 kkkkkk
Example (textbook) Process 33 44 Isentropic expansion vv rrr = rrrr rrr = 8 6.108 = 48.864 @vv rrr (Table A17) TT 4 = 777777. 66 KK uu 4 = 588.74 kj/kg TT 4 1 PP 4 = PP 3 TT 3 rr = 4345 kpa 795.6 K 1575.1 K 1 8 = 222222. 3333 kkkkkk Process 44 11 Constant-volume heat rejection qq out = (uu 4 uu 1 ) = (588.74 206.91)kJ/kg = 333333. 8888 kkkk/kkkk
Example (textbook) ww net = qq net = qq in qq out = 800 381.83 kj/kg = 444444. 1111 kkkk/kkkk ηη th = ww net qq in = 418.17 kj/kg 800 kj/kg = 00. 555555 = 5555. 33 % ηη otto = 1 rr 1 kk = 1 8 1 1.4 = 00. 55555555 = 5555. 4444 % vv 1 = RRTT 1 = (0.287 kpa m3 /kg)(290 K) PP 1 100 kpa = 0.8323 m 3 /kg MEP = vv 1 ww nnnnnn 1 1 rr = 418.17 kj/kg (0.8323 m 3 /kg) 1 1 8 1 kpa m 3 /kg 1 kj = 555555 kkkkkk
Example (textbook) c) Data: nn = 4000 rpm, 4 cylinders, VV dd = 1.6 L = 0.0016 m3, WW net =? (kw) WW net = mmww netnn nn rev mm = VV dd vv 1 = 0.0016 m3 0.8323 m 3 /kg = 0.001922 kg For a four-stroke engine nn rev = 2 rev/cycle (for Ideal Otto cycle nn rev = 1 rev/cycle) WW net = (0.001922 kg)(418.17 kj/kg)(4000 rev/min) 2 rev/cycle 1 min 60 s = 2222. 88 kkkk
Carnot Vapor Cycle
Rankine Cycle
Rankine Cycle qq iiii qq oooooo + ww iiii ww oooooo = h ee h ii Pump qq = 0 : ww pppppppp,iiii = h 2 h 1 = vv pp 2 pp 1 h 1 = h fffpp1 ; vv vv 1 = vv fffpp1 Boiler (ww = 0): qq 1 = h 3 h 2 Turbine (qq = 0): ww tttttttt,oooooo = h 3 h 4 Condenser ww = 0 : qq oooooo = h 4 h 1 ww nnnnnn = qq iiii qq oooooo = ww tttttttt,oooooo ww tttttttt,iiii ηη tt = ww nnnnnn qq iiii = 1 qq oooooo qq iiii
Rankine Cycle
Rankine Cycle The simple ideal Rankine cycle Data: PP 3 kpa TT 3 K PP 4 kpa Process 1 2 Isentropic compression in a pump Process 2 3 Constant pressure heat addition in a boiler Process 3 4 Isentropic expansion in a turbine Process 4 1 Constant pressure heat rejection in a condenser
Rankine Cycle Stage 1 @PP 1 = PP 4 Stage 2 Table A5 h 1 = h ff vv 1 = vv ff PP 2 = PP 3 ww pump,in = vv 1 (PP 2 PP 1 ) ss 2 = ss 1 h 2 = h 1 + ww pump,in
Rankine Cycle Stage 3 @PP 3, TT 3 Table A6 h 3 ss 3 Stage 4 ss 4 = ss 3 If @PP 4 Table A5, ss ff < ss 4 < ss gg, then: xx 4 = ss 4 ss ff ss ffff h 4 = h ff + xx 4 h ffff
Rankine Cycle qq in = h 3 h 2 qq out = h 4 h 1 ww turb,out = h 3 h 4 ww net = ww turb,out ww pump,in = qq in qq out ηη th = 1 qq out qq in = ww net qq in
Examples (textbook) EXAMPLE 10 1 The Simple Ideal Rankine Cycle Consider a steam power plant operating on the simple ideal Rankine cycle. Steam enters the turbine at 3 MPa and 350 C and is condensed in the condenser at a pressure of 75 kpa. Determine the thermal efficiency of this cycle. Data: PP 3 = 3 MPa, TT 3 = 350 C, PP 4 = 75kPa, ηη th =? Stage 1 @PP 1 = PP 4 = 75 kpa Table A5 h 1 = h ff = 333333. 4444 kkkk/kkkk vv 1 = vv ff = 0.001037 m 3 /kg
Examples (textbook) Stage 2 PP 2 = PP 3 = 3 MPa ww pump,in = vv 1 PP 2 PP 1 = 0.001037 m 3 /kg 3000 75 kpa 1 kj 1 kpa m 3 /kg = 33. 0000 kkkk/kkkk h 2 = h 1 + ww pump,in = 384.44 + 3.03 kj/kg = 333333. 4444 kkkk/kkkk Stage 3 @PP 3 = 3 MPa, TT 3 = 350 C Table A6 h 3 = 33333333. 11 kkkk/kkkk ss 3 = 6.7450 kj/kgk
Examples (textbook) Stage 4 ss 4 = ss 3 = 6.7450 kj/kgk @PP 4 = 75 kpa Table A5 ss ff = 1.2132 kj/kgk ss gg = 7.4558 kj/kgk, ss gg < ss 4 < ss ff, h ff = 384.44 kj/kg h ffff = 2278.0 kj/kg xx 4 = ss 4 ss ff ss ffff = 6.7450 1.2132 7.4558 1.2132 = 0.8861 h 4 = h ff + xx 4 h ffff = 384.44 kj/kg + 0.8861 2278.0 kj/kg = 22222222. 00 kkkk/kkkk
Examples (textbook) qq in = h 3 h 2 = 3116.1 387.47 kj/kg = 22222222. 66 kkkk/kkkk qq out = h 4 h 1 = 2403.0 384.44 kj/kg = 22222222. 66 kkkk/kkkk ww turb,out = h 3 h 4 = 3116.1 2403.0 kj/kg = 777777. 11 kkkk/kkkk ww net = ww turb,out ww pump,in = 713.1 303 kj/kg = 777777. 11 kkkk/kkkk = qq in qq out = 2728.6 2018.6 kj/kg = 777777. 00 kkkk/kkkk ηη th = 1 qq out 2018.6 kj/kg = 1 = 00. 222222 = 2222. 00 % qq in 2728.6 kj/kg = ww net 710.0 kj/kg = = 00. 222222 = 2222. 00 % qq in 2728.6 kj/kg
Examples (textbook) EXAMPLE 10 2 An Actual Steam Power Cycle A steam power plant operates on the cycle shown in Figure. If the isentropic efficiency of the turbine is 87 percent and the isentropic efficiency of the pump is 85 percent, determine (a) the thermal efficiency of the cycle and (b) the net power output of the plant for a mass flow rate of 15 kg/s.
Examples (textbook) a) ηη th =? ηη th = ww net qq in ww nnnnnn = ww turb,out ww pump,in ww turb,out = ηη TT ww ss,turb,out = ηη TT (h 5 h 6ss ) = 0.87 3583.1 2115.3 kj/kg = 11111111. 00 kkkk/kkkk ww pump,in = ww ss,pump,in = vv 1(PP 2 PP 1 ) = 0.001009 m3 /kg 16000 9 kpa ηη PP ηη PP 0.85 1 kj 1 kpam 3 = 1111. 00 kkkk/kkkk
Examples (textbook) qq in = h 4 h 3 = 3647.6 160.1 kj/kg = 33333333. 55 kkkk/kkkk ww net = ww turb,out ww pump,in = 1277.0 19.0 kj/kg = 11111111. 00 kkkk/kkkk ηη th = ww net qq in = 1258.0 kj/kg 3487.5 kj/kg = 00. 333333 = 3333. 11 % b) WW net =? WW net = mmww net = 15 kg/s 1258.0 kj/kg = 1111. 99 MMMM
Refrigeration
Coefficient of Performance Refrigeration Heat Pump CCCCCC RR = CCCCCC HHHH = CCCCCC HHHH = CCCCCC RR + 1 CCCCCC RR,CCCCCCCCCCCC = 1 TT HH TTLL 1 QQ LL WW nnnnnn,iiii QQ HH WW nnnnnn,iiii ; CCCCCC HHHH,CCCCCCCCCCCC = 1 1 TT LL TT HH
Carnot Refrigerator
Refrigerator Cycle
Refrigerator Cycle
Refrigerator Cycle
Refrigerator Cycle The Ideal Vapor-Compression Refrigeration Cycle Data: PP 1 = PP 4 = PP LL kpa PP 2 = PP 3 = PP HH kpa Process 1 2 Isentropic compression in a compressor Process 2 3 Constant pressure heat rejection in a condenser Process 3 4 Throttling in a expansion device Process 4 1 Constant pressure heat absorption in an evaporator
Refrigerator Cycle Stage 1 @PP 1 = PP LL Table A12 h 1 = h gg ss 1 = ss gg Stage 2 @PP 2 = PP HH, ss 2 = ss 1, Table A13 h 2 Stage 3 @PP 3 = PP HH, Table A12 h 3 = h ff Stage 4 h 4 h 3
Refrigerator Cycle QQ LL = WW in = QQ HH = mm(h 1 h 4 ) mm(h 2 h 1 ) mm h 2 h 3 = QQ LL + WW in COP R = QQ LL WW in
Example (textbook) EXAMPLE 11 1 The Ideal Vapor-Compression Refrigeration Cycle A refrigerator uses refrigerant-134a as the working fluid and operates on an ideal vapor-compression refrigeration cycle between 0.14 and 0.8 MPa. If the mass flow rate of the refrigerant is 0.05 kg/s, determine (a) the rate of heat removal from the refrigerated space and the power input to the compressor, (b) the rate of heat rejection to the environment, and (c) the COP of the refrigerator. Data: PP 1 = PP 4 = 0.14 MPa PP 2 = PP 3 = 0.8 Mpa mm = 0.05 kg/s
Example (textbook) a) QQ LL =?, WW in =? QQ LL = mm(h 1 h 4 ) WW in = mm(h 2 h 1 ) @PP 1 = PP LL = 0.14 MPa Table A12 h 1 = h gg = 222222. 1111 kkkk/kkkk ss 1 = ss gg = 0.94467 kj/kgk @PP 2 = PP HH = 0.8 MPa, ss 2 = ss 1, Table A13 {h 2 = 222222. 4444 kkkk/kkkk @PP 3 = PP HH = 0.8 MPa, Table A12 h 3 = h ff = 9999. 4444 kkkk/kkkk h 4 h 3 = 9999. 4444 kkkk/kkkk QQ LL = 0.05 kg/s WW in = 0.05 kg/s 239.19 95.48 kj/kg = 77. 1111 kkkk 275.40 239.19 kj/kg = 11. 8888 kkkk
Example (textbook) b) QQ HH =? QQ HH = mm h 2 h 3 = 0.05 kg/s 275.40 95.48 kj/kg = 99. 0000 kkkk = QQ LL + WW in = 7.19 kw + 1.81 kw = 99. 0000 kkkk c) COP R =? QQ COP R = LL = WW in 7.19 kw 1.81 kw = 33. 9999
Homework 4c Problems from the textbook (Thermodynamics, Yunus, 8th ed.): Choose 5 conceptual problems and answer them (those who you consider to provide better understanding to the subject seen in this section) Chapter 9, problems: 1-10, 23-30 Chapter 10, problems: 1, 6, 7 Chapter 11, problems: 1, 2, 5-10 Choose 5 problems and answer them (those who you consider to provide better understanding to the subject seen in this section) Chapter 9, problems: 11-22, 31-41 Chapter 10, problems: 2-5, -26 Chapter 11, problems: 3, 4, 11-23
Thermodynamic Cycles