Solved Problems

Solved Problems: Figure 3-2

Figure 3-4

Figure 3-6

Figure 3-5

Figure 3-8

Figure 3-11

Figure 3-15

Figure 3-16

3-1. Cylinder conditions at the start of compression in an SI engine operating at WOT on an airstandard Otto cycle are 60°C and 98 kPa. The engine has a compression ratio of 9.5:1 and uses gasoline with AF = 15.5. Combustion efficiency is 96%, and it can be assumed that there is no exhaust residual. Calculate: (a) Temperature at all states in the cycle. [deg. C], (b) Pressure at all states in the cycle. [kpa], (c) Specific work done during power stroke. [kJ/kg], (d) Heat added during combustion. [kJ/kg], (e) Net specific work done. [kJ/kg], (£) Indicated thermal efficiency. [%] Using Fig. 3-2 a)&b) T1= 60°C + 273 = 333K p1= 98kPa T2=T1rck-1=(333K)(9.5)0.35=732K=459°C p2=p1rck=98kPa(9.5)1.35=2047kPa QHVɳc=(AF+1)cv(T3-T2) (43000kJ/kg)(0.96)=(15.5+1)(0.821 kJ/kgK)(T3-732K) T3=3779K=3506°C p2=p1(T3/T2)=(2047kPa)(3779/732)=10568kPa T4=T3(1/rc ) k-1=3779K(1/9.5)0.35=1719K=1446°C p4=p3(1/rc ) k=10568kPa(1/9.5)1.35=506kPa c) w3-4=R(T4-T3)/(1-k)=(0.287kJ/kgK)(1719-3779)K/(1-1.35)=1689kJ/kg d) qin=cv(T3-T2)=(0.821kJ/kgK)(3779-732)K=2502kJ/kg e) w1-2=R(T2-T1)/(1-k)=(0.287kJ/kgK)(732-333)K/(1-1.35)=-327kJ/kg wnet= w1-2+ w3-4=-327kJ/kg+1689kJ/kg=1362kJ/kg f)ɳt=wnet/qin=1362/2502=0.545=54.5%

3-2. The engine in Problem 3-1 is a three-liter V6 engine operating at 2400 RPM. At this speed the mechanical efficiency is 84%. Calculate: (a) Brake power. [kW], (b) Torque. [N-m], (c) Brake mean effective pressure. [kPa], (d) Friction power lost. [kW], (e) Brake specific fuel consumption. [gm/kW-hr], (f) Volumetric efficiency. [%], (g) Output per displacement. [kW/L] Using Fig. 3-2

a) V4=3L/6=0.5L=0.0005m3 (for 1 cylinder) rc=9.5=(Vd+Vc)/ Vc=(0.0005m3+Vc)/Vc Vc=0.0000588m3 V1= Vd+Vc=0.0005m3+0.0000588m3=0.0005588m3 Mass at point 1: m=pV/RT=98kPa(0.0005588m3)/(0.287kJ/kgK)(333K)=0.000573kg W=mwnet=0.000573kg(1362kJ/kg)=0.780 kJ P1=WN/n=((0.780kJ/cycle)(2400/60 rev/s)/2 rev/cycle))(6cycles)=93.6 kW Pb=ɳmP1=0.84(93.6kW)=78.6 kW b) Pb=2πNτ=78.6kJ/s=(2πrad/rev)(2400/60 rev/s)τ τ=0.313kNm=313Nm c) τ=(bmep)Vd/4π= 0.313kNm=bmep(0.003m3)/ 4π bmep=1311kPa d) Pf=Pi-Pb=93.6kW-78.6kW=15kW e) m=0.000573kg=ma+mf=ma(1=FA)=ma(1+(1/15.5)) ma=0.000538kg

mf=0.000035kg

mf=(0.000035kg/cycle-cylinder)(6cylinders)(2400/60 rev/s)(2rev/cycle)=0.0042kg/s=4.2g/s=15120g/h bsfc=mf/Pb=(15120g/h)/78.6kW)=192.4g/kWh f) ɳv=ma/ρaVd=0.000538kg/(1.181kg/m3)(0.0005m3)=0.911=91.1% g) OPD=Pb/Vd=78.6kW/3L=26.2kW/L

3-3. The exhaust pressure of the engine in Problem 3-2 is 100 kPa. Calculate: (a) Exhaust temperature. [deg. C], (b) Actual exhaust residual. [%], (c) Temperature of air entering cylinders from intake manifold. [deg. C] Using Fig.3-2 a) Tex=T4(pex/p4)(k-1)/k =1719K(100/506)(1.35-1)/1.35=1129K=856°C b) xr=(1/rc) (T4/Tex) (pex/p4)= (1/9.5) (1719/1129) (100/506)=0.032=3.2%

c) T1=xrTex+(1-xr)T 333K=0.032(1129K)+(1-0.032)T T=307K=34°C

3-4. The engine of Problems 3-2 and 3-3 is operated at part throttle with intake pressure of 75 kPa. Intake manifold temperature, mechanical efficiency, exhaust residual, and air-fuel ratio all remain the same. Calculate: (a) Temperature in cylinder at start of compression stroke. [deg. C], (b) Temperature in cylinder at start of combustion. [deg. C] Using Fig.3-2 a) Tex=1129K(75/100)(1.35-1)/1.35 =1048K T1= xrTex+(1-xr)T=0.032(1048K)+(1-0.032)(307K)=331K=58°C b) T2=T1rck-1=(331K)(9.5)0.35=728K=455°C

3-5. An SI engine operating at WOT on a four-stroke air-standard cycle has cylinder conditions at the start of compression of 100°F and 14.7 psia. Compression ratio is rc = 10, and the heat added during combustion is qin = 800 BTU/lbm. During compression the temperature range is such that a value for the ratio of specific heats k = 1.4 would be correct. During the power stroke the temperature range is such that a value of k = 1.3 would be correct. Use these values for compression and expansion, respectively, when analyzing the cycle. Use a value for specific heat of Cv = 0.216 BTU/lbm-oR, which best corresponds to the temperature range during combustion. Calculate: (a) Temperature at all states in cycle. [OF], (b) Pressure at all states in cycle. [psia], (c) Average value of k which would give the same indicated thermal efficiency value as the analysis in parts (a) and (b). Solution: Using Fig 3-2 (a)(b) T1 = 100°F = 560°R P1 = 14.7 psia

given

T2 = T1 (rc)k-1= (560°R) (10)1.4-1= 1 407°R = 947 °F P2 = P1 (rc)k = (14.7 psia)(10)1.4 = 369 psia qin = cv (T3 – T2) 800 BTU/lbm = (0.216 BTU/lbm°R)(T3 – 1 407°R) T3 = 5 110°R = 4 650 °F At constant volume: P3 = P2 (T3/T2 ) = (369 psia)(5 110/1 407) = 1 340 psia

T4 = T3 (1/rc)k-1= (5110 °R)(1/10)1.3-1 = 2 101 °F P4 = P3 (1/rc)k = (1340 psia)(1/10)1.3 = 67.2 psia

(c) 𝑤1−2 = 𝑤3−4 =

𝑅 (𝑇2 −𝑇1 ) 1−𝑘

𝑅 (𝑇4 −𝑇3 ) 1−𝑘

=

=

(0.069 𝐵𝑇𝑈/𝑙𝑏𝑚°𝑅)(1407−560)°𝑅 (1−1.4)

(0.069 𝐵𝑇𝑈/𝑙𝑏𝑚°𝑅)(2561−5110)°𝑅 (1−1.3)

𝐵𝑇𝑈

= −146.1 𝑙𝑏𝑚 𝐵𝑇𝑈

= +586.3 𝑙𝑏𝑚

ηT = wNet / qin = [(+586.3) + (-146.1)]/(800) = 0.550 = 55% ηT = 0.550 = 1 – (1/rc)k-1 = 1 – (1/10)k-1 k = 1.347

3-6. A CI engine operating on the air-standard Diesel cycle has cylinder conditions at the start of compression of 65°C and 130 kPa. Light diesel fuel is used at an equivalence ratio of if> = 0.8 with a combustion efficiency Tic = 0.98. Compression ratio is rc = 19. Calculate:(a) Temperature at each state of the cycle. [0C], (b) Pressure at each state of the cycle. [kPa], (c) Cutoff ratio, (d) Indicated thermal efficiency. [%] and (e) Heat lost in exhaust. [kJ/kg] Using Fig 3-8 (a)(b) T1 = 65°C = 338 K P1 = 130 kPa T2 = T1 (rc)k-1= (338K) (19)1.35-1= 947 K = 674 °C P2 = P1 (rc)k = (130 kPa)(19)1.35 = 6 922 kPa AF = (AF)stoich/ɸ = (14.5)/(0.8) = 18.125 QHVηc = (AF+1)cp(T3 – T2) (42 500 kJ/kg)(0.98) = (18.125 + 1 )(1.108 kJ/kgK)(T3 – 947)K T3 = 2 913 K = 2 640 °C P3 = P2 = 6922 kPa v4 = v1 = RT1/P1 = (0.287)(3880/(130) = 0.7462 m3/kg v3 = RT3/P3 = (0.287)(2913)/6922 = 0.1208 m3/kg T4 = T3 (v3/v4)k-1= (2913 K)(0.1208/0.7462)1.35-1 = 1 540 K = 1 267 °C P4 = P3 (v3/v4)k = (6922 kPa)( 0.1208/0.7462)1.35 = 592 kPa (c) β = T3/T2 = 2913/947 = 3.08 1 𝑘−1

(d) (𝜼𝒕 )𝒅𝒊𝒆𝒔𝒆𝒍 = 1 − (𝑟 ) 𝑐

{𝛽 𝑘 −1}

1 1.35−1

[{𝑘(𝛽−1)}] = 1 − (19)

{3.081.35 −1}

[{1.35(3.08−1)}] = 𝟓𝟒. 𝟕%

(e) qin = cp(T3 – T2) = (1.108 kJ/kgK)(2913-947)K = 2178 kJ/kg wnet = qinηt = (2178 kJ/kg)(0.547) = 1191 kJ/kg qnet = qout = qin – wnet = 2178 – 1191 = 987 kJ/kg

3·7. A compression ignition engine for a small truck is to operate on an air-standard Dual cycle with a compression ratio of rc = 18. Due to structural limitations, maximum allowable pressure in the cycle will be 9000 kPa. Light diesel fuel is used at a fuel-air ratio of FA = 0.054. Combustion efficiency can be considered 100%. Cylinder conditions at the start of compression are 50°C and 98 kPa. Calculate: (a) Maximum indicated thermal efficiency possible with these conditions. [%], (b) Peak cycle temperature under conditions of part (a). [0C], (c) Minimum indicated thermal efficiency possible with these conditions. [%] and (d) Peak cycle temperature under conditions of part (c). [0C] Using Fig 3-11 T1 = 50 °C = 323 K P1 = 98 kPa T2 = T1 (rc)k-1= (323 K) (18)1.35-1= 888 K = 615 °C P2 = P1 (rc)k = (98 kPa)(18)1.35 = 4 851 kPa P3 = Pmax = 9 000 kPa = Px (a) highest possible thermal efficiency will be when as much of the combustion as possible is done at constant volume, i.e., as close to the Otto cycle as possible At constant volume Tx = T2 (P3/P2) = (888 K)(9000/4851) = 1 647 K (AF) = 1/(FA) = 1/0.054 = 18.52 Total heat in: (Qin)total = Q2-x + Qx-3 = mfQHVηc = (ma + mf )[cv(Tx – T2) + cp(T3 – Tx )] Let ηc = 1/mf QHV = (AF + 1)cv(Tx – T2 ) + (AF + 1)cp (T3 – Tx ) 42 500 kJ/kg = (19.52)(0.821 kJ/kgK)(1647 – 888)K + (19.52)(1.108 kJ/kgK)(T3 – 1647)K T3 = 3 050 K α = Px/P2 = 9000/4851 = 1.855 β = T3/Tx = 3050/1647 = 1.852 1 𝑘−1

(𝜂𝑡 )𝑑𝑢𝑎𝑙 = 1 − ( ) 𝑟 𝑐

{𝛼𝛽 𝑘 −1}

1 1.35−1

[{𝑘𝛼(𝛽−1)+𝛼−1}] = 1 − (18)

(𝜼𝒕 )𝒅𝒖𝒂𝒍 = 𝟎. 𝟔𝟎𝟑 = 𝟔𝟎. 𝟑% (b) Tpeak = T3 = 3050 K = 2777 °C

{1.855(1.852)1.35 −1}

[{(1.35)(1.855)(1.852−1)+1.855−1}]

(c) Minimum thermal efficiency is when combustion is at constant pressure, i.e., operate as a Diesel cycle QHVηc = (AF+1)cp(T3 – T2) = (42,500 kJ/kg)(1) = (18.52+1)(1.108 kJ/kgK)( T3 – 888K) T3 = 2853 K β = T3/T2 = 2853/888 = 3.213 1 𝑘−1

(𝜼𝒕 )𝒅𝒊𝒆𝒔𝒆𝒍 = 1 − ( ) 𝑟 𝑐

{𝛽 𝑘 −1}

1 0.35

[{𝑘(𝛽−1)}] = 1 − (18)

{3.2131.35 −1}

[{1.35(3.213−1)}] = 𝟓𝟑. 𝟑%

(d) Tpeak = T3 = 2853 K = 2580 °C

3-8. An in-line six, 3.3-liter CI engine using light diesel fuel at an air-fuel ratio of AF = 20 operates on an air-standard Dual cycle. Half the fuel can be considered burned at constant volume, and half at constant pressure with combustion efficiency Tic = 100%. Cylinder conditions at the start of compression are 60°C and 101 kPa. Compression ratio rc = 14:1. Calculate: (a) Temperature at each state of the cycle. [K], (b) Pressure at each state of the cycle. [kPa], (c) Cutoff ratio, (d) Pressure ratio, (e) Indicated thermal efficiency. [%], (f) Heat added during combustion. [kJ/kg] and (g) Net indicated work. [kJ/kg] Using Fig 3-11 (a) (b) T1 = 60 °C = 333K P1 = 101 kPa T2 = T1 (rc)k-1 = (333 K)(14)0.35 = 839 K = 566 °C P2 = P1 (rc)k = (101 kPa)(14)1.35 = 3 561 kPa QHVηc = (AF+1)cp(T3 – T2) ½(42,500 kJ/kg)(1) = (20 + 1)(1.108 kJ/kgK)(T3 – 2072 K) T3 = 2985 K = 2712 °C Px = P2(Tx/T2) = (3561 kPa)(2072/839) = 8794 kPA = P3 v4 = v1 = RT1/P1 = (0.287)(333)(101) = 0.9462 m3/kg v3 = RTy/P3 = (0.287)(2985)(8794) = 0.0574 m3/kg T4 = T3 (v3/v4)k-1= (2985 K)(0.0974/0.9462)0.35 = 1 347 K = 1 074 °C P4 = P3 (v3/v4)k = (8794 kPa)( 0.0974/0.9462)1.35 = 408 kPa (c) β = =T3/Tx = 2 985/2 072 = 1.441 (d) α = P3/P2 = 8794/3561 = 2.470 1 𝑘−1

(e) (𝜂𝑡 )𝑑𝑢𝑎𝑙 = 1 − (𝑟 ) 𝑐

{𝛼𝛽 𝑘 −1}

[{𝑘𝛼(𝛽−1)+𝛼−1}]

𝜼𝒕 = 1 – (1/14)0.35{[(2.470)(1.441)1.35 – 1]/[(1.35)(2.471)(0.441) + 2.470 – 1]} =0.589 = 58.9 % (f ) qIn = cy (Tx – T2) + cp (T3 – Tx) = (0.821 kJ/kgK)(2072 – 829)K + ((1.108 kJ/kgK)(2985 – 2072)K = 2024 kJ/kg (g ) wnet = ηtqin = (0.589)(2024 kJ/kg) = 1192 kJ/kg

3-9. The engine in Problem 3-8 produces 57 kW of brake power at 2000 RPM. Calculate: (a) Torque. [N-m], (b) Mechanical efficiency. [%], (c) Brake mean effective pressure. [kPa], (d) Indicated specific fuel consumption. [gmlkW-hr] Using Fig 3-11 (a ) Ẇ = 2𝜋𝑁𝑇 = 57 kJ/sec = (2𝜋 radians/rev)(2000/60 rev/sec)T T = 0.272 kNm = 272 Nm (b ) V4 = (0.0033 m3)/ 6 = 0.00055 m3 re = (Vd + Vc)/Vc = 14 =(0.00055 + Vc)/Vc Vc = 0.000042 m3 V1 = Vd + Vc = (0.00055 m3) + (0.000042 m3) = 0.000592 m3 m1 = P1V 1/RT1= (101)(0.000592)/(0.287)(333) = 0.000626 kg Qin = mqin = (0.000626 kg)(2024 kJ/kg)(6 cyl) = 7.602 kJ/cucle (W1)net = η1Qin = (0.589)(7.602 kJ/cycle) = 4.48 kJ/cycle Ẇ 1 = WN/n = (4.48 kJ/cycle)(2000/60 rev/sec)(2 rev/cycle) = 74.7 kW ηm = Ẇb/ Ẇ1 = 57/74.7 = 0.763 = 76.3 % (c ) r = (bmep)Vd/4𝜋 = 272 N-m = bmep(0.0033 m3)/4𝜋 bmep = 1036 kPa (d ) with AF = 20, mass of fuel will be (1/21) of total mass Mf =(0.000626 kg/cyl-cycle)(1/21)(6 cyl) = 0.00018 kg/cycle 3-10. An Otto cycle SI engine with a compression ratio of rc = 9 has peak cycle temperature and pressure of 2800 K and 9000 kPa. Cylinder pressure when the exhaust valve opens is 460 kPa, and exhaust manifold pressure is 100 kPa. Calculate: (a) Exhaust temperature during exhaust stroke. [0C], (b) Exhaust residual after each cycle. [%], (c) Velocity out of the exhaust valve

when the valve first opens. [m/sec] and (d) Theoretical momentary maximum temperature in the exhaust. [0C] a) Using Equation (3-37) and Figure 3-6 Tex = T7 = T3 (P7/P3)(k–1)/k = (2800 K)(100/9000)(1.35–1)/1.35 = 872 K = 599 °C b) Equation (3-1h) T4 = T3 (P4/P3)(k–1)/k = (2800 K)(460/9000)(1.35-1)/1.35 = 1295 K Equation (3-46) xr = (1/rc)(T4/Tex)(Pex/P4) = (1/9)(1295/872)(100/460) = 0.036 = 3.6% c) Velocity will be sonic – choked flow Equation (3-1j) Velocity = c = (kRT)1/2 = [(1.35)(287 J/kg.K)]1/2 = 708 m/sec d) As velocity is dissipated, kinetic energy will be change to an enthalpy increase V2/2gc = Δh = cpΔT (708 m/sec)2/[(2)(1 kg-m/N-sec2)] = (1.108 kJ/kg-K) * ΔT ΔT = 226 K Tmax = T7 + ΔT = 872 + 226 = 1098 K = 825 °C 3-11. An SI engine operates on an air-standard four-stroke Otto cycle with turbocharging. Airfuel enters the cylinders at 70°C and 140 kPa, and heat in by combustion equals qin = 1800 kJ/kg. Compression ratio rc = 8 and exhaust pressure Fex = 100 kPa. Calculate: (a) Temperature at each state of the cycle. [0C], (b) Pressure at each state of the cycle. [kPa], (c) Work produced during expansion stroke. [kJ/kg], (d) Work of compression stroke. [kJ/kg], (e) Net pumping work. [kJ/kg], (f) Indicated thermal efficiency. [%] and (g) Compare with Problems 3-12 and 313 a & b) Using Figure 3-5 Given:  T1 = 70 °C = 343 K  P1 = 140 kPa Equations (3-4) and (3-5) T2 = T1 (rc)k=1 = (343 K)(8)0.35 = 710 K = 437 °C P2 = P1 (rc)k = (140 kPa)(8)1.35 = 2319 kPa Equation (3-12) Qin = cv (T3 – T2) = 1800 kJ/kg = (0.821 kJ/kg-K)(T3 – 710)K T3 = 2902 K = 2629 °C At constant volume P3 = P2(T3/T2) = (2319 kPa)(2902/710) = 9479 kPa

Equation (3-16) and (3-17) T4 = T3 (1/rc)k-1 = (2902 K)(1/8)0.35 = 1402 K = 1129 °C P4 = P3 (1/rc)k = (9479 kPa)(1/8)1.35 = 572 kPa c)

Equation (3-18) w3-4 = R(T4 – T3)/(1-k) = [(0.287 kJ/kg-K)(1402-2902)K]/(1-1.35) = +1230 kJ/kg

d)

Equation (3-7) w1-2 = R(T2 – T1)/(1-k) = [(0.287 kJ/kg-K)(710-343)K]/(1-1.35) = -301 kJ/kg

e)

v1 = RT1/P1 = (0.287)(343)/(140) = 0.7032 m3/kg v2 = RT2/P2 = (0.287)(710)/(2319) = 0.0879 m3/kg Using Equation (3-35) per unit mass wpump = (P1 – Pex)(v1 – v2) = [(140-100) kPa][(0.7032-0.0879) m3/kg] = 24.6 kJ/kg

f)

wnet = (-301) + (+1230) + (+24.6) = 953.6 kJ/kg ɳt = wnet/qin = 953.6/1800 = 0.530 = 53.0%

3-12. An SI engine operates on an air-standard four-stroke Miller cycle with turbocharging. The intake valves close late, resulting in cycle 6-7-8-7-2-3-4-5-6 in Fig. 3-15. Air-fuel enters the cylinders at 70°C and 140 kPa, and heat in by combustion equals qin = 1800 kJ/kg. Compression ratio rc = 8, expansion ratio re = 10, and exhaust pressure Fex = 100 kPa. Calculate: (a) Temperature at each state of the cycle. [0C], (b) Pressure at each state of the cycle. [kPa], (c) Work produced during expansion stroke. [kJ/kg], (d) Work of compression stroke. [kJ/kg], (e) Net pumping work. [kJ/kg], (f) Indicated thermal efficiency. [%] and (g) Compare with Problems 3-11 and 3-13. a & b) Using Figure 3-15 Given:  T7 = T8 = 70 °C = 343 K  P7 = P8 = 140 kPa T2 = T7 (rc)k-1 = (343 K)(8)0.35 = 710 K = 437 °C P2 = P7 (rc)k = (140 kPa)(8)1.35 = 2319 kPa qin = cv(T3 – T2) = 1800 kJ/kg = (0.821 kJ/kg-K)(T3 – 710)K T3 = 2902 K = 2629 °C At constant volume P3 = P2(T3/T2) = (2319 kPa)(2902/710) = 9479 kPa T4 = T3 (1/rc)k-1 = (2902 K)(1/10)0.35 = 1296 K = 1023 °C

P4 = P3 (1/rc)k = (9479 kPa)(1/10)1.35 = 423 kPa At constant volume T5 = T4(P5/P4) = (1296 K)(100/423) = 306 K = 33 °C = T6 P5 = P6 = 100 kPa

given

c)

Equation (3-1i) w3-4 = R(T4 – T3)/(1-k) = (0.287 kJ/kg-K)(1296-2902)K/(1-1.35) = +1317 kJ/kg

d)

Equation (3-1i) w7-2 = (0.287 kJ/kg-K)(710-343)K/(1-1.35) = - 301 kJ/kg

e)

v5 = RT5/P5 = (0.287)(306)/(100) = 0.8790 m3/kg v7 = RT7/P7 = (0.287)(343)/(140) = 0.7032 m3/kg v6 = v7/rc = 0.7032/8 = 0.0879 m3/kg w5-6 = P(v6-v5) = (100)(0.0879-0.8790) = -79.1 kJ/kg w6-7 = P(v7-v6) = (140)(0.7032-0.0879) = +86.1 kJ/kg w7-8 cancels w8-7 wpump = (+86.1) + (-79.1) = +7.0 kJ/kg

f)

wnet = (+1317) + (-301) + (+7.0) = +1023 kJ/kg ηt = wnet/qin = 1023/1800 = 0.568 = 56.8%

3-13. An SI engine operates on an air-standard four-stroke Miller cycle with turbocharging. The intake valves close early, resulting in cycle 6-7-1-7-2-3-4-5-6 in Fig. 3-15. Air-fuel enters the cylinders at 70°C and 140 kPa, and heat in by combustion equals qin = 1800 kJ/kg. Compression ratio rc = 8, expansion ratio re = 10, and exhaust pressure Pex = 100 kPa. Calculate: (a) Temperature at each state of the cycle. [0C], (b) Pressure at each state of the cycle. [kPa], (c) Work produced during expansion stroke. [kJ/kg], (d) Work of compression stroke. [kJ/kg] (e) Net pumping work. [kJ/kg], (f) Indicated thermal efficiency. [%] and (g) Compare with Problems 3-11 and 3-12. a & b) Using Figure 3-16 Given:  T7 = 70 °C  P7 = 140 kPa T2 = T7(rc)k-1 = (343 K)(8)0.35 = 710 K = 437 °C P2 = P7(rc)k = (140 kPa)(8)1.35 = 2319 kPa qin = cv(T3 – T2) = 1800 kJ/kg = (0.821 kJ/kg-K)(T3 – 710)K T3 = 2902 K = 2629 °C At constant volume P3 = P2(T3/T2) = (2319 kPa)(2902/710) = 9479 kPa

T4 = T3(1/rc)k-1 = (2902 K)(1/10)0.35 = 1296 K = 1023 °C P4 = P3(1/rc)k = (9479 kPa)(1/10)1.35 = 423 kPa P5 = 100 kPa = P6 given At constant volume T5 = T4(P5/P4) = (1296 k)(100/423) = 306 K = 33 °C = T6 v1 = v5 = v4 = RT5/P5 = (0.287 kJ/kg-K)(306 K)/(100 kPa) = 0.8790 kg/m3 v6 = v3 = v2 = v5/rc = (0.8790 m3/kg)/10 = 0.0879 m3/kg v7 = RT7/P7 = (0.287)(343)/(140) = 0.7032 m3/kg P1 = P7(v7/v1)k = (140 kPa)(0.7032/0.8790)1.35 = 104 kPa T1 = P1v1/R = (104 kPa)(0.8790 m3/kg)/(0.287 kJ/kg-K) = 318 K = 45 °C c)

Equation (3-1i) w3-4 = R(T4-T3)/(1-k) = (0.287 kJ/kg-K)(1296-2902)K/(1-1.35) = +1317 kJ/kg

d)

Equation (3-1i) W7-2 = (0.287 kJ/kg-K)(710-343)K/(1-1.35) = -301 kJ/kg

e)

w6-7 = P(v7-v6) = (140)(0.7032-0.0879) = +86.1 kJ/kg w5-6 = P(v6-v5) = (100)(0.0879-0.890) = -79.1 kJ/kg w7-1 cancels w1-7 wpump = (+86.1) + (-79.1) = +7.0 kJ/kg

f)

wnet = (+1317) + (-301) + (+7.0) = +1023 kJ/kg ηt = wnet/qin = 1023/1800 = 0.568 = 56.8%

3.14 A six cylinder, two-stroke cycle CI ship engine with bore B = 35 cm and stroke S = 105 cm produces 3600 kW of brake power at 210 RPM. Calculate: (a) Torque at this speed. [kN-m], (b) Total displacement. [L], (c) Brake mean effective pressure. [kPa], (d) Average piston speed. [mlsec] (a) Eq. (3-4) T2 = T1 (rc)k-1 = (570 °R) (10.5)1.35-1 = 1298 °R = 838 °F (b) R = r/a = (6.64 in.)/(1.66 in.) = 4.0 Eq. (2-14) gives chamber volume intake valve closes V1/Vc = 1 + (½)(rc – 1)[R + 1 – cosƟ – √𝑅 2 − sin2 Ɵ] = (1) + (½)(10.5 – 1)[(4.0)+(1) – cos(200°) – √(4.0)2 − sin2 (200°)] = 10.283

Eq. (2-14) for volume when spark plug fires V2/Vc = 1 + (½)(10.5 – 1)[(4.0)+(1) – cos(245°) – √(4.0)2 − sin2(245°)] = 1.202 V1/V2 = (V1/Vc)/(V2/Vc) = (10.283)/(1.202) = 8.556 T2 = T1(V1/V2)k-1 = (570 °R)(8.556)1.35-1 = 1208°R = 748°F

3.15 A single-cylinder, two-stroke cycle model airplane engine with a 7.54-cm3 displacement produces 1.42 kW of brake power at 23,000 RPM using glow plug ignition. The square engine (bore = stroke) uses 31.7 gmlmin of castor oil-methanol-nitromethane fuel at an air-fuel ratio AF = 4.5. During intake scavenging, 65% of the incoming air-fuel mixture gets trapped in the cylinder, while 35% of it is lost with the exhaust before the exhaust port closes. Combustion efficiency 'TIc = 0.94. Calculate: (a) Brake specific fuel consumption. [gm/kW-hr], (b) Average piston speed. [mlsec], (c) Unburned fuel exhausted to atmosphere. [gm/min], (d) Torque. [N-m] (a) P2 = P7(rc)k = (100 Kpa)(8.2)1.35 = 1713 kPa Pmin = P1(1/rc)k = (1713 kPa)(1/10.2)1.35 = 74.5 kPa (b) Miller cycle has no pump worl Wpump = 0 (c) PEVO = P4 = P3(1/rc)k = (9197 kPa)(1/10.2)1.35 = 400 kPa

3-16. A historic single-cylinder engine with a mechanical efficiency 'TIm = 5% operates at 140 RPM on the Lenoir cycle shown in Fig. 3-20. The cylinder has a double acting piston with a 12in. bore and a 36-in. stroke. The fuel has a heating value QLHV = 12,000 BTU/lbm and is used at an air-fuel ratio AF = 18. Combustion occurs at constant volume half way through the intakepower stroke when cylinder conditions equal 70°F and 14.7 psia. Calculate: (a) Temperature at each state of cycle. [OF], (b) Pressure at each state of cycle. [psia], (c) Indicated thermal efficiency. [%], (d) Brake power. [hp] and (e) Average piston speed. [ft/sec] (a) R = r/a = (9.5 in.)/(2.5 in.) = 3.8 Eq. (2-14) V1/Vc = 1 + (½)(rc – 1)[R + 1 – cosƟ – √𝑅 2 − sin2 Ɵ] = (1) + (½)(10.5 – 1)[(3.8)+(1) – cos(110°) – √(3.8)2 − sin2 (110°)] = 7.935 T2 = T1(V1/V2)k-1 = (4660 °R)(1/7.935)1.35-1 = 2257°R = 1797°F (b) (rc)off = 7.935 (c) P7 = P6[(rc)off]k = (17.8 psia)(7.935)1.35 = 292 psia At constant volume T7 = T1(P7/P1) = (4660°R)(292/1137) = 1197°R = 737°F

3-17. Cylinder conditions at the start of compression of a four-stroke cycle SI engine are 27°C and 100 kPa. The engine has a compression ratio of rc = 8:1, and heat addition from combustion is qin = 2000 kJ/kg. Calculate: (a) Temperature and pressure at each state of the cycle, using airstandard Otto cycle analysis with constant specific heats. [OC,kPa], (b) Indicated thermal efficiency in part (a). [%], (c) Temperature and pressure at each state of the cycle, using any standard air tables which are based on variable specific heats as functions of temperature (e.g., reference [73]). [OC,kPa], and (d) Indicated thermal efficiency in part (c). [%] (a) Eq. (2 – 43) Wb = 2πNT = 3600 kK/sec = (2π radians/rev)(210/60 rev/sec)T T = 164 kN•m = 164,000 N-m (b) Eq. (2 – 9) Vd = Nc(π/4)B2S = (6 cyl)(π/4)(0.35 m)2(1.05 m) = 0.606 m3 = 606 L (c) Eq. (2 – 40) T = (bmep) Vd2π = 164 kN•m = bmep(0.606 m3)/2π bmep = 1700 kPa (d) Eq. (2 – 2) Up = 2SN = (2 strokes/rev)(1.05 m/stroke)(210/60 rev/sec) = 7.35 m/sec