Thermodynamic Cycles
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1 Thermodynamic Cycles
2 Content Thermodynamic Cycles Carnot Cycle Otto Cycle Rankine Cycle Refrigeration Cycle
3 Thermodynamic Cycles
4 Carnot Cycle
5 Derivation of the Carnot Cycle Efficiency
6 Otto Cycle
7 Otto Cycle Nomenclature for reciprocating engines TDC Top Dead Center BDC Bottom Dead Center
8 Otto Cycle
9 Otto Cycle
10 Otto Cycle
11 Otto Cycle kaire = 1.4; kco2 = 1.3; ketano = 1.667
12 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
13 Constant Specific Heat Process 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 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 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
14 Constant Specific Heat Otto Cycle Process 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
15 Variable Specific Heat Otto Cycle Data: rr = vv 1 vv 2 PP 1 kpa TT 1 K qq in 1 (Table A17) vv rrr uu 1 RR Table A1
16 Variable Specific Heat Process Isentropic compression Otto Cycle vv rrr = vv rrr rr rrr Table A17 TT 2 uu 2 ; PP 2 = PP 1 TT 2 TT 1 rr Process Constant-volume heat addition uu 3 = qq in +uu 2 3 (Table A17) TT 3 vv rrr ; PP 3 = PP 2 TT 3 TT 2 Process Isentropic expansion vv rrr = rrrr rrr rrr (Table A17) TT 4 TT 4 1 ; PP uu 4 = PP 3 4 TT 3 rr qq in = uu = uu 3 uu 2
17 Variable Specific Heat Otto Cycle Process 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
18 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.
19 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 = kpa m 3 /kgk (Table A1) cc vv = kj/kgk, kk = 1.4 (Table A2a) vv 1 = RRTT 1 = (0.287 KPam3 /kgk)(290 K) PP kpa Process Isentropic compression vv 2 = vv 1 rr = m3 /kg = m 8 3 /kg PP 2 = PP 1 rr kk = 100 kpa = kkkkkk = m 3 /kg
20 Example (textbook) TT 2 = PP 2vv 2 RR = ( kpa)( m3 /kg) kpa m 3 /kgk = KK Process Constant-volume heat addition vv 3 = vv 2 = m 3 /kg TT 3 = TT 2 + qq in 800 kj/kg = 666 K + = KK cc vv kj/kgk PP 3 = PP 2 TT 3 TT 2 = kpa K 666 K = kkkkkk
21 Example (textbook) Process Isentropic expansion vv 4 = vv 1 = m 3 /kg PP 4 = PP kpa = rrkk (8) 1.4 = kkkkkk TT 4 = TT K = = KK rrkk 1 (8) Process Constant-volume heat rejection qq out = cc vv TT 4 TT 1 = kj/kgk K = kkkk/kkkk
22 Example (textbook) ww net = qq net = qq in qq out = kj/kg = kkkk/kkkk ηη th = ww net qq in = kj/kg 800 kj/kg = = % ηη otto = 1 rr 1 kk = = = % MEP = ww nnnnnn vv 1 vv 2 = kj/kg m 3 /kg 1 kpa m 3 /kg 1 kj = kkkkkk
23 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 = kpa m 3 /kgk (Table 1 (Table A17) vv rrr = uu 1 = kj/kg Process Isentropic compression vv rrr = vv rrr rr = = rrr Table A17 TT 2 = KK uu 2 = kj/kg
24 Example (textbook) PP 2 = PP 1 TT 2 TT 1 rr = 100 kpa K 290 K 8 = kkkkkk Process Constant-volume heat addition uu 3 = qq in + uu 2 = kj/kg = (Table A17) TT 3 = KK vv rrr = PP 3 = PP 2 TT 3 TT 2 = kpa K K = kkkkkk
25 Example (textbook) Process Isentropic expansion vv rrr = rrrr rrr = = rrr (Table A17) TT 4 = KK uu 4 = kj/kg TT 4 1 PP 4 = PP 3 TT 3 rr = 4345 kpa K K 1 8 = kkkkkk Process Constant-volume heat rejection qq out = (uu 4 uu 1 ) = ( )kJ/kg = kkkk/kkkk
26 Example (textbook) ww net = qq net = qq in qq out = kj/kg = kkkk/kkkk ηη th = ww net qq in = kj/kg 800 kj/kg = = % ηη otto = 1 rr 1 kk = = = % vv 1 = RRTT 1 = (0.287 kpa m3 /kg)(290 K) PP kpa = m 3 /kg MEP = vv 1 ww nnnnnn 1 1 rr = kj/kg ( m 3 /kg) kpa m 3 /kg 1 kj = kkkkkk
27 Example (textbook) c) Data: nn = 4000 rpm, 4 cylinders, VV dd = 1.6 L = m3, WW net =? (kw) WW net = mmww netnn nn rev mm = VV dd vv 1 = m m 3 /kg = kg For a four-stroke engine nn rev = 2 rev/cycle (for Ideal Otto cycle nn rev = 1 rev/cycle) WW net = ( kg)( kj/kg)(4000 rev/min) 2 rev/cycle 1 min 60 s = kkkk
28 Carnot Vapor Cycle
29 Rankine Cycle
30 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
31 Rankine Cycle
32 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
33 Rankine Cycle Stage 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
34 Rankine Cycle Stage 3, TT 3 Table A6 h 3 ss 3 Stage 4 ss 4 = ss 3 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
35 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
36 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 4 = 75 kpa Table A5 h 1 = h ff = kkkk/kkkk vv 1 = vv ff = m 3 /kg
37 Examples (textbook) Stage 2 PP 2 = PP 3 = 3 MPa ww pump,in = vv 1 PP 2 PP 1 = m 3 /kg kpa 1 kj 1 kpa m 3 /kg = kkkk/kkkk h 2 = h 1 + ww pump,in = kj/kg = kkkk/kkkk Stage 3 = 3 MPa, TT 3 = 350 C Table A6 h 3 = kkkk/kkkk ss 3 = kj/kgk
38 Examples (textbook) Stage 4 ss 4 = ss 3 = = 75 kpa Table A5 ss ff = kj/kgk ss gg = kj/kgk, ss gg < ss 4 < ss ff, h ff = kj/kg h ffff = kj/kg xx 4 = ss 4 ss ff ss ffff = = h 4 = h ff + xx 4 h ffff = kj/kg kj/kg = kkkk/kkkk
39 Examples (textbook) qq in = h 3 h 2 = kj/kg = kkkk/kkkk qq out = h 4 h 1 = kj/kg = kkkk/kkkk ww turb,out = h 3 h 4 = kj/kg = kkkk/kkkk ww net = ww turb,out ww pump,in = kj/kg = kkkk/kkkk = qq in qq out = kj/kg = kkkk/kkkk ηη th = 1 qq out kj/kg = 1 = = % qq in kj/kg = ww net kj/kg = = = % qq in kj/kg
40 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.
41 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 ) = kj/kg = kkkk/kkkk ww pump,in = ww ss,pump,in = vv 1(PP 2 PP 1 ) = m3 /kg kpa ηη PP ηη PP kj 1 kpam 3 = kkkk/kkkk
42 Examples (textbook) qq in = h 4 h 3 = kj/kg = kkkk/kkkk ww net = ww turb,out ww pump,in = kj/kg = kkkk/kkkk ηη th = ww net qq in = kj/kg kj/kg = = % b) WW net =? WW net = mmww net = 15 kg/s kj/kg = MMMM
43 Refrigeration
44 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
45 Carnot Refrigerator
46 Refrigerator Cycle
47 Refrigerator Cycle
48 Refrigerator Cycle
49 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
50 Refrigerator Cycle Stage 1 = PP LL Table A12 h 1 = h gg ss 1 = ss gg Stage 2 = PP HH, ss 2 = ss 1, Table A13 h 2 Stage 3 = PP HH, Table A12 h 3 = h ff Stage 4 h 4 h 3
51 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
52 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
53 Example (textbook) a) QQ LL =?, WW in =? QQ LL = mm(h 1 h 4 ) WW in = mm(h 2 h 1 1 = PP LL = 0.14 MPa Table A12 h 1 = h gg = kkkk/kkkk ss 1 = ss gg = = PP HH = 0.8 MPa, ss 2 = ss 1, Table A13 {h 2 = = PP HH = 0.8 MPa, Table A12 h 3 = h ff = kkkk/kkkk h 4 h 3 = kkkk/kkkk QQ LL = 0.05 kg/s WW in = 0.05 kg/s kj/kg = kkkk kj/kg = kkkk
54 Example (textbook) b) QQ HH =? QQ HH = mm h 2 h 3 = 0.05 kg/s kj/kg = kkkk = QQ LL + WW in = 7.19 kw kw = kkkk c) COP R =? QQ COP R = LL = WW in 7.19 kw 1.81 kw =
55 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, 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, Chapter 10, problems: 2-5, -26 Chapter 11, problems: 3, 4, 11-23
56 Thermodynamic Cycles
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