Engine Cycle Analysis. Air Standard Otto Cycle.

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Engine Cycle AnalysisEngine Cycle Analysis

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

►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

►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

►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))))

►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

State PointTemperatur

e (°R)

Pressure(psia)

Volume(ft3/lbm)

1 560 14.7

3 1100

►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 %

►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

►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

►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

►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

►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

►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

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

► 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

e (°R)

Pressure(psia)

Volume(ft3/lbm)

11 600600 14.514.5 11

►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))

►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))

►VV22 = (V = (V11/r)/r)

►VV22 = (1/14) = (1/14)

►VV22 = 0.0714 ft = 0.0714 ft33

►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

►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

►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

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

►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

►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

►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 %

Engine Cycle Analysis. Air Standard Otto Cycle.

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