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Engine Cycle AnalysisEngine Cycle Analysis
Air Standard Otto CycleAir Standard Otto Cycle
Air Standard Otto CycleAir Standard Otto Cycle
Air Standard Otto CycleAir Standard Otto Cycle
►Starting with the piston at bottom dead Starting with the piston at bottom dead center (BDC), compression proceeds center (BDC), compression proceeds isentropically from state 1 to state 2.isentropically from state 1 to state 2.
►Heat is added at constant volume from Heat is added at constant volume from state 2 to state 3.state 2 to state 3.
►Expansion occurs isentropically from Expansion occurs isentropically from state 3 to state 4.state 3 to state 4.
►Heat is rejected at constant volume Heat is rejected at constant volume from state 4 to state 1from state 4 to state 1
Air Standard Otto CycleAir Standard Otto Cycle
Air Standard Otto CycleAir Standard Otto Cycle
►Thermal Efficiency (Thermal Efficiency (ηηthth) = W) = WNetNet/Q/Qinin
►Heat Added at constant volume from state Heat Added at constant volume from state 2 to state 3 Q2 to state 3 Q2-32-3 = U = U33 – U – U2 2 = mc= mcvv(T(T33-T-T22))
►Heat Rejected at constant volume from Heat Rejected at constant volume from state 4 to state 1 Qstate 4 to state 1 Q4-14-1 = U = U11 – U – U4 4 = -mc= -mcvv(T4 (T4 – T– T11))
►Specific Heat (CSpecific Heat (Cvv) = 0.1714 Btu/lbm-) = 0.1714 Btu/lbm-°R°R►Mean Effective Pressure (mep) = WMean Effective Pressure (mep) = WNetNet / /
Displacement Volume = WDisplacement Volume = WNetNet / V / V1 1 – V– V22
Air Standard Otto CycleAir Standard Otto Cycle
►Compression ratio (r) = Vol at BDC/Vol at Compression ratio (r) = Vol at BDC/Vol at TDCTDC
►r = Vr = V11/V/V22 = V = V44/V/V33
►ηηth th = 1 – (1/(r)= 1 – (1/(r)k-1k-1))
►Specific Heat Ratio (k) = CSpecific Heat Ratio (k) = Cpp/C/Cvv (for air k = (for air k = 1.4)1.4)
►Percentage of clearance (c) = Clearance Percentage of clearance (c) = Clearance Volume/Displacement volume = VVolume/Displacement volume = V22/(V/(V1 1 – V– V22))
Air Standard Diesel CycleAir Standard Diesel Cycle
►Starting with the piston at bottom dead Starting with the piston at bottom dead center (BDC), compression occurs center (BDC), compression occurs isentropically from state 1 to state 2isentropically from state 1 to state 2
►Heat is added at constant pressure from Heat is added at constant pressure from state 2 to state 3state 2 to state 3
►Expansion occurs isentropically from Expansion occurs isentropically from state 3 to state 4state 3 to state 4
►Heat rejection occurs at constant Heat rejection occurs at constant volume from state 4 to state 1volume from state 4 to state 1
Air Standard Diesel CycleAir Standard Diesel Cycle
Air Standard Diesel CycleAir Standard Diesel Cycle
Air Standard Diesel CycleAir Standard Diesel Cycle
►Heat supplied at constant pressure for Heat supplied at constant pressure for a closed system, Qa closed system, Q2-32-3 = H = H33 – H – H22 = = mcmcpp(T(T3 3 – T– T22))
►Heat rejected at constant volume, QHeat rejected at constant volume, Q4-14-1 = U= U11 – U – U44 = mc = mcvv(T(T44 – T – T11))
►Specific Heat (CSpecific Heat (Cpp) = 0.24 Btu/lbm-) = 0.24 Btu/lbm-°R°R
►ηηthth = 1-1/k((T = 1-1/k((T44 – T – T11)/(T)/(T33 – T – T22))))
Air Standard Diesel CycleAir Standard Diesel Cycle
►Cutoff ratio (rCutoff ratio (rcc) = Vol at the end of ) = Vol at the end of heat addition/Vol at the start of heat heat addition/Vol at the start of heat addition = Vaddition = V33/V/V22
►Cutoff percentage (RCutoff percentage (Rcc) = ((V) = ((V33 – V – V22)/(V)/(V11 – – VV22)) x 100)) x 100
►ηηth th = 1 – (1/(r)= 1 – (1/(r)k-1k-1)[(r)[(rcck k – 1)/(k(r– 1)/(k(rcc – 1))] – 1))]
Polytropic Process for a Closed Polytropic Process for a Closed SystemSystem
►Pv = RTPv = RT►Specific Gas Constant (R) = for air = Specific Gas Constant (R) = for air =
53.34 ft-lb53.34 ft-lbff/lbm-/lbm-°R°R►pp11 = RT = RT11/v/v11 ►pp22 = RT = RT22/v/v22 ►PP11/P/P22 = (V = (V22/V/V11))kk
►PP22/P/P11 = (V = (V11/V/V22))kk
►TT11/T/T2 2 = (V= (V22/V/V11))k-1k-1
►TT22/T/T11 = (V = (V11/V/V22))k-1k-1
Polytropic Process for a Closed Polytropic Process for a Closed SystemSystem
►VV22/V/V11 = (T = (T11/T/T22))1/k-11/k-1
►VV11/V/V22 = (P = (P22/P/P11))1/k1/k
►TT11/T/T22 = (P = (P22/P/P11))(1-k)/k(1-k)/k = (P = (P11/P/P22))(k-1)/k(k-1)/k
►PP11/P/P22 = (T = (T11/T/T22))k/(k-1)k/(k-1)
Air Standard Otto CycleAir Standard Otto Cycle►A air standard Otto cycle has an initial A air standard Otto cycle has an initial
temperature of 100 temperature of 100 °F, a pressure of °F, a pressure of 14.7 psia, and compression pressure P14.7 psia, and compression pressure P22 = 356 psia. The pressure at the end of = 356 psia. The pressure at the end of heat addition is 1100 psia. Determine:heat addition is 1100 psia. Determine:
►(A) The compression ratio (r)(A) The compression ratio (r)►(B) The thermal efficiency (B) The thermal efficiency ηηthermalthermal
►(C) The percentage of clearance (c)(C) The percentage of clearance (c)►(D) The maximum temperature T(D) The maximum temperature Tmax max
and the remaining Pressures, and the remaining Pressures, Temperature and VolumesTemperature and Volumes
Air Standard Otto CycleAir Standard Otto Cycle
State PointTemperatur
e (°R)
Pressure(psia)
Volume(ft3/lbm)
1 560 14.7
2 356
3 1100
4
Air Standard Otto CycleAir Standard Otto Cycle
►r = Vr = V11/V/V22
►VV11/V/V22 = (P = (P22/P/P11))1/k1/k
►VV11/V/V2 2 = (356/14.7)= (356/14.7)1/1.41/1.4
►r = 9.74r = 9.74
►ηηth th = 1 – (1/(r)= 1 – (1/(r)k-1k-1))►ηηth th = 1 – (1/(9.74)= 1 – (1/(9.74)1.4-11.4-1))►ηηth th = 0.598 or 59.8 %= 0.598 or 59.8 %
Air Standard Otto CycleAir Standard Otto Cycle
►PP11 = RT = RT11/V/V11
►VV11 = RT = RT11/P/P11
►VV11 = ((53.34 ft-lb = ((53.34 ft-lbff/lbm-/lbm-°R)(560 °R)) / °R)(560 °R)) / ((14.7 lb((14.7 lbff/in/in22)(144 in)(144 in22/ft/ft22))))
►VV11 = 14.11 ft = 14.11 ft33/lbm/lbm►r = Vr = V11/V/V22
►VV22 = V = V11/r/r►VV22 = 14.11/9.74 = 14.11/9.74►VV22 = 1.449 = 1.449 ftft33/lbm/lbm
Air Standard Otto CycleAir Standard Otto Cycle
►Percentage of clearance (c) = VPercentage of clearance (c) = V22/(V/(V1 1 – – VV22))
►Percentage of clearance (c) = Percentage of clearance (c) = 1.449/(14.111.449/(14.11 – 1.449)– 1.449)
►Percentage of clearance (c) = 0.114 or Percentage of clearance (c) = 0.114 or
11.4 %11.4 %►PP22 = RT = RT22/V/V22
►TT22 = P = P22VV22/R/R
Air Standard Otto CycleAir Standard Otto Cycle
►TT22 = (((356 lb = (((356 lbff/in/in22)(144 in)(144 in22/ft/ft22)(1.449 )(1.449 ftft33/lbm)) / 53.34 ft-lb/lbm)) / 53.34 ft-lbff/lbm-/lbm-°R)°R)
►TT22 = 1393 °R = 1393 °R
Air Standard Otto CycleAir Standard Otto Cycle
Air Standard Otto CycleAir Standard Otto Cycle
►Since VSince V22 = V = V33
►VV22 = RT = RT22/P/P22
►VV33 = RT = RT33/P/P33
►RTRT22/P/P22 = RT = RT33/P/P33
►TT22/P/P22 = T = T33/P/P33
►TT33 = T = Tmaxmax = T = T22(P(P33/P/P22))►TT3 3 = 1393 = 1393 °R (1100/356)°R (1100/356)►TT33 = 4304.2 °R = 4304.2 °R
Air Standard Otto CycleAir Standard Otto Cycle
►Find PFind P44 and T and T44
►(P(P33/P/P44) = (V) = (V44/V/V33))kk
►(V(V44/V/V33) = r = 9.74) = r = 9.74
►(9.74)(9.74)1.4 1.4 = (P= (P33/P/P44) )
►(9.74)(9.74)1.4 1.4 = (1100/P= (1100/P44) )
►PP44 = 45.4 psia = 45.4 psia
Air Standard Otto CycleAir Standard Otto Cycle
►TT33/T/T4 4 = (V= (V44/V/V33))k-1k-1
►TT33/T/T4 4 = (r)= (r)k-1k-1
►4302/T4302/T4 4 = (9.74)= (9.74)1.4-11.4-1
►TT44 = 4302/(9.74) = 4302/(9.74)1.4-11.4-1
►TT44 = 1731.2 = 1731.2 °R°R
Air Standard Otto CycleAir Standard Otto Cycle
State PointTemperatur
e (°R)
Pressure(psia)
Volume(ft3/lbm)
1 560 14.7 14.11
2 1393 356 1.449
3 4302 1100 1.449
4 1731.2 45.4 14.11
Air Standard Diesel CycleAir Standard Diesel Cycle
► An air standard Diesel Cycle operates An air standard Diesel Cycle operates on 1 fton 1 ft33 of air at 14.5 psia and 140 of air at 14.5 psia and 140 °F. °F. The compression ratio (r) is 14, and the The compression ratio (r) is 14, and the cutoff is 6.2 % of the stroke. Determine:cutoff is 6.2 % of the stroke. Determine:
►(A) The temperatures, pressures and (A) The temperatures, pressures and volumes around the cyclevolumes around the cycle
►(B) The net work(B) The net work►(C) The heat added(C) The heat added►(D) The efficiency(D) The efficiency
Air Standard Diesel CycleAir Standard Diesel Cycle
Air Standard Diesel CycleAir Standard Diesel Cycle
State PointTemperatur
e (°R)
Pressure(psia)
Volume(ft3/lbm)
11 600600 14.514.5 11
22
33
44
Air Standard Diesel CycleAir Standard Diesel Cycle
►PP11 = mRT = mRT11/V/V11
►m = Pm = P11VV11/RT/RT11
►m = (((14.5 lbm = (((14.5 lbff/in/in22)(144 in)(144 in22/ft/ft22)(1 ft)(1 ft33)) / )) / ((53.34 ft-lb((53.34 ft-lbff/lbm-/lbm-°R)(600 °R)))°R)(600 °R)))
►m = 0.0652 lbmm = 0.0652 lbm►TT22/T/T11 = (V = (V11/V/V22))k-1k-1
►TT22 = T = T1 1 (V(V11/V/V22))k-1k-1
►r = (Vr = (V11/V/V22))
Air Standard Diesel CycleAir Standard Diesel Cycle
►TT22 = (600 = (600 °R)(14)°R)(14)0.40.4
►TT22 = 1724.2 °R = 1724.2 °R
►PP22/P/P11 = (V = (V11/V/V22))kk
►PP22 = P = P11(V(V11/V/V22))kk
►PP22 = (14.5 psia)(14) = (14.5 psia)(14)1.41.4
►PP22 = 583.4 psia = 583.4 psia
►r = (Vr = (V11/V/V22))
Air Standard Diesel CycleAir Standard Diesel Cycle
►VV22 = (V = (V11/r)/r)
►VV22 = (1/14) = (1/14)
►VV22 = 0.0714 ft = 0.0714 ft33
Air Standard Diesel CycleAir Standard Diesel Cycle
Air Standard Diesel CycleAir Standard Diesel Cycle
►Since the process between PSince the process between P22 to P to P33 is is constant pressure Pconstant pressure P33 = 583.4 psia. = 583.4 psia.
►Cutoff percentage (RCutoff percentage (Rcc) = ((V) = ((V33 – V – V22)/(V)/(V11 – – VV22)) x 100)) x 100
►(R(Rcc) = ((V) = ((V33 – V – V22)/(V)/(V11 – V – V22)) x 100)) x 100
►0.062 = ((V0.062 = ((V33 – 0.0714)/(1 – 0.0714)) – 0.0714)/(1 – 0.0714))
►VV33 = 0.129 ft = 0.129 ft33
Air Standard Diesel CycleAir Standard Diesel Cycle
►PP22 = RT = RT22/V/V22
►PP33 = RT = RT33/V/V33
►Since PSince P22 = P = P33
►RTRT22/V/V22 = RT = RT33/V/V33
►TT33 = T = T22(V(V33/V/V22))
►TT33 = (1724.2 = (1724.2 °R)(0.129/0.0714)°R)(0.129/0.0714)
►TT33 = 3115 °R = 3115 °R
Air Standard Diesel CycleAir Standard Diesel Cycle
►TT44/T/T33 = (V = (V33/V/V44))k-1k-1
►TT44 = T = T33(V(V33/V/V44))k-1k-1
►TT44 = (3115 = (3115 °R°R)(0.129/1.00))(0.129/1.00)0.40.4
►TT44 = 1373 = 1373 °R°R►PP44/P/P33 = (V = (V33/V/V44))kk
►PP44 = P = P33(V(V33/V/V44))kk
►PP44 = 583.4(0.129/1.0) = 583.4(0.129/1.0)1.41.4
►PP4 4 = 33.2 psia= 33.2 psia
Air Standard Diesel CycleAir Standard Diesel Cycle
State PointTemperatur
e (°R)
Pressure(psia)
Volume(ft3/lbm)
11 600600 14.514.5 11
22 1724.21724.2 583.4583.4 0.07140.0714
33 31153115 583.4583.4 0.1290.129
44 13731373 33.233.2 11
Air Standard Diesel CycleAir Standard Diesel Cycle
Air Standard Diesel CycleAir Standard Diesel Cycle
►The heat added is:The heat added is:►QQ2-32-3 = H = H33 – H – H22 = mC = mCpp(T(T33 – T – T22))
►QQ2-32-3 = (0.0652 lbm)(0.24 Btu/lbm- = (0.0652 lbm)(0.24 Btu/lbm-°R)°R)(3115 – 1742.2 °R)(3115 – 1742.2 °R)
►QQ2-32-3 = 21.76 Btu = 21.76 Btu
Air Standard Diesel CycleAir Standard Diesel Cycle
►The heat out is:The heat out is:►QQOutOut = Q = Q4-14-1 = mC = mCvv(T(T11 – T – T44))
►QQ4-1 4-1 = (0.0652)(0.1714)(600 – 1373)= (0.0652)(0.1714)(600 – 1373)
►QQ4-1 4-1 = - 8.64 Btu= - 8.64 Btu
►WWnetnet = = ∑Q = Q∑Q = Qinin + Q + Qoutout
►WWnetnet = 21.76 Btu + - 8.64 = 21.76 Btu + - 8.64
►WWnet net = 13.12 Btu= 13.12 Btu
Air Standard Diesel CycleAir Standard Diesel Cycle
►Thermal Efficiency (Thermal Efficiency (ηηthth) = W) = Wnetnet/Q/Qinin
►ηηthth = 13.12/21.76 = 13.12/21.76
►ηηthth = 0.603 = 60.3 % = 0.603 = 60.3 %