8/14/2019 Fajril Ar Rahman 100401014

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NAMA : FAJRIL AR RAHMAN

NIM : 100401014

KLS : B

1. An Otto cycle SI engine with a compression ratio of rc = 9 has peak cycle temperatureand 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. [OC](b)Exhaust residual after each cycle. [%](c)Velocity out of the exhaust valve when the valve first opens. [mlsec](d)Theoretical momentary maximum temperature in the exhaust. [0C]

a. Tex = T7 = T3(P7/P3)(k-1)/k= (2800 K)(100/9000)(1,35-1)/1,35 = 872 K ( 599 oC )

b. T4= T3(P4/P3)(k-1)/k= (2800 K)(460/9000)(1,35-1)/1,35= 1295 K

xr= (1/rc)(T4/Tex)(Pex/P4) = (1/9)(1295/875)(100/60) = 0,036 = 3,6%

c. Vel = c = [kRT]1/2= [ (1,35)(287 J/kg.K)(1295 K) ]1/2= 708 m/s

d. V2/2gc= h = cpT

(708 m/s)2/[(2)(1 kg.m/Ns2)] = (1,108 kJ/kg.K)T

T = 226K

Tmax= T7+T = 872 + 226 = 1098 K = 825oC

2. An SI engine operates on an air-standard four-stroke Otto cycle with turbocharging.Air-fuel enters the cylinders at 70C 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. [oC]

(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. [%]

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a. T1= 70 oC = 343 K ; P1= 140 kPaT2= T1(rc)

k-1= (343 K)(8)0,35= 710 K = 437 oC

P2= P1(rc)k= (140 kPa)(8)1,35= 2319 kPa

Qin= cv(T3/T2) = (2319 kPa)(2902/710) = 9479 kPa

T4= T3(1/rc)k-1= (2902 K)(1/8)0,35= 1402 K = 1129 oC

P4= P3(1/rc)k= (9479 kPa)(1/8)1,35= 572 kPa

b. P2= 2319 kPa ; P4= 572 kPa

c. w3-4= R(T4T3)/(1-k) = [(0,287 kL/kg.K)(14022902)K]/(11,35)

= +1230 kJ/kg

d. w1-2= R(T2T1)/(1-k) = [(0,287 kL/kg.K)(710343)K]/(11,35) = -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

wpump= (P1Pex)(v1 - v2) = [(140100)kPa][(0,70320,0879)m3/kg]

= 24,6 kJ/kg

f. wnet= (-301) + (+1230) + (+24,6) = 953,6 kJ/kg

t= wnet/qin= 1023/1800 = 0,568 = 56,8 %

3. A CI engine operating on the air-standard Diesel cycle has cylinder conditions at thestart of compression of 65C 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. [%](e)Heat lost in exhaust. [kJ/kg]

a.T1= 65oC = 338 K ; P1= 130 kPa

T2= T1(rc)k-1= (338 K)(19)0,35= 947 K = 674 oC

P2= P1(rc)k

= (130 kPa)(19)1,35

= 6922 kPa

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AF = (AF)stolch/ = ( 14,5)/(0,8) = 18,125

QHVc = (AF + 1)cp(T3T2)

(42500 kJ/kg)(0,98) = (18,125 + 1)(1,108 kJ/kg.K)(T3947)K

T3= 2913 K = 2640oC

P3= P2= 6922 kPa

v4= v1= RT1/P1= (0,287)(338)/(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)0,35

= 1540 K = 1267o

C

P4= P3(v3/v4)k= (6922 kPa)(0,1208/0,7462)1,35= 592 kPa

b. P3= P2= 6922 kPa ; P4= 592 kPa

c. = T3/T2= 2913/947 = 3,08

d. (t)DIESEL= 1(1/rc)k-1[(k1)/{k( 1)}]

t= 1(1/19)0,35[{(3,08)1,351}/{(1,35)(3,081)}] = 0,547 = 54,7 %

e. qin= cp(T3T2) = (1,108 kJ/kg.K)(2913947)K = 2178 kJ/kg

wnet= qint= (2178 kJ/kg)(0,547) = 1191 kJ/kg

qex= qout= qinwnet= 21781191 = 987 kJ/kg

4. A compression ignition engine for a small truck is to operate on an air-standard Dualcycle 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 50C and 98 kPa. Calculate:

(a)Maximum indicated thermal efficiency possible with these conditions. [%](b)Peak cycle temperature under conditions of part (a). [oC](c)Minimum indicated thermal efficiency possible with these conditions. [%](d)Peak cycle temperature under conditions of part (c). [oC]

T1= 50oC = 323 K ;3 P1= 98 kPa

T2= T1(rc)k-1= (323 K)(18)0,35= 888 K = 615 oC

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P2= P1(rc)k= (98 kPa)(18)1,35= 4851 kPa

P3= Pmax= 9000kPa = Px

a. Tx= T2(P3/P2) = (888 K)(9000/4851) = 1647 K

(AF) = 1/(FA) = 1/0,054 = 18,52

Qin total = QHVc

QHV= (AF + 1)cv(TxT2) + (AF + 1)cp(T3T2)

42500 kJ/kg = 19,52(0,821)(1647888) + (19,52)(1,108)(T31647)

T3= 3050 K

= Px/P2= 9000/4851 = 1,855

= T3/Tx= 3050/1647 = 1,852

tdual = 1(1/rc)k-1[{k1}/{k( 1)+ 1}]

= 1(1/18)0,35[{1,855(1,852)1,351}/{(1,35)(1,8521) + 1,8551}] = 0,603

= 60,3 %

b. Tpeak= T3= 3050 K = 2777oC

c. QHVc=(AF + 1)cp(T3T2)

(42500 kJ/kg)(1) = (18,52 + 1)(1,108kJ/kg.K)(T3888 K)

T3= 2853 K

= T3/T2= 2853/888 = 3,213

tdiesel = 1(1/rc)k-1[(k1)/{k( 1)}]

= 1(1/18)0,35[{(3,213)1,351}/{1,35(3,2131)}] = 0,533 = 53,3 %

d. Tpeak= T3=2853 K =2580oC