Analysis (a) Process 1-2: isentropic compression. p = 755K. R7] h87kpa.m 3/kg.K }30~ = O.906m 3/kg = V max

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$q23,in =U3 -U2 =Cv(T3 -T2) 750kJ/kg = (0. 718kJ/kg.KXT3-689)K 2\;:::::.J; v J ~ ~ T3 = 1734K P3V3 -~--+P3 = -r;- -T2 T2 ~ = (b) Process 3-4: isentropic expansion. T4 =T3 ( ~ )k-1 =(1734K{i. 0.$

1 ~ Chapter 8 Gas Power Cycles Analysis (a) Process 1-2: isentropic compression. p - f ii, f T2=1j ( - VI = (300KX8)0.4 = 689K k-l I V2 - ~..., I ~=~ T2 Process 2-3: v = constant heat addition. q23,in =U3 -U2 =Cv(T3 -T2) 750kJ/kg = (0. 718kJ/kg.KXT3-689)K 2\;:::::.J; v J ~ ~ T3 = 1734K P3V3 -~--+P3 = -r;- -T2 T2 ~ = (b) Process 3-4: isentropic expansion. T4 =T3 ( ~ )k-1 =(1734K{i. 0.4 = 755K Proces.s 4-1: v = constant heat rejection. gout =U4 -Ul =Cv(T4-7j)=(0.718kJ/kg.KX )K=327kJ/kg Wnet.out = gin -gout = = 423kJ/kg w 'lth = ne!,qut 423 kj /k~ (c) qin - = 56.4% 750 kj /kg (d) R7] h87kpa.m 3/kg.K }30~ = O.906m 3/kg = V max VI =T= 95kPa v. - v v 2 -max -- mm- r = 423kJ/kg ( kpa.m 31 ~6m3/kgJ~ -l/8)l ~ j = 534kPa 8-14

2 8-45 An air-standard Diesel cycle with a compression ratio of 16 and a cutoff ratio of 2 is considered. The temperature after the heat addition process, the thermal efficiency, and the mean effective pressure are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats. Properties The properties of air are given in Table A-17. Analysis (a) Process 1-2: isentropic compression. p ~in '(... U1 = kj /kg 7)=300K ~ v'i=621.2 ~qout 1 V2 1 v =-v =-v =- T, TI TI.VI r 1 (621.2)=38.825~h2 T2 =862.4 =890.9kJ/kgK 16 v Process 2-3: P = constant heat addition. (b) q;n = h3 -hz = = kj Ikg Process 3-4: isentropic expansion. (c) Wnel,OUl = qin -qoul = = kJ/kg v =~= (0.287kPa.m3/kg.K}3 0~1-3 - I PI 95kPa m /kg -v max v = V2 = mm MEP= ~ ~ VI -V2 r l 3 = Wnet,out = kJ/kg kpa.m ) = 675 9kPa vi(i-i/r) (0.906m3/kgJl~I/I6) kj. 8-21

3 8-52 A four-cylinder ideal diesel engine with air as the working fluid has a compression ratio of 17 and a cutoff ratio of 2.2. The power the engine will deliver at 1500 rpm is to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are Cp = kj/kg. K, Cy = kj/kg. K, and k = 1.4 (Table A-2). p Analysis Process 1-2: isentropic compression. 2 Qin 3 (, k-t Vt T2 =Tt - = (300K Xl7 )04 = 931.8K V2 Process 2-3: p = constant heat addition. ~ T3 -~---+T3 = -T2 V2 Process 3-4: isentropic expansion. ~'4o-.,~ v T4 =T3(~ l V n-1 =T (2.2V2 Jl- V4 =T3(7,,-1 = 904.8K m=!jyl-~ - (97 kpa)(o.oo3 m3) RT1 (0.287 kpa.m3 /kg. K)(300 K) = x 10-3kg Qjn =m(h3 -h2)=mcp(t3 -T2) = (3.380 x 10-3 kg)(1.005 k1/kg. K)( )K = kj QOUl = m(u4 -UI )= mcv(t4-1; ) = ~.380x 10-3kgJo.718 kj/kg. KX )K = k1 w nel '.OUI = Q;n -QOUI = = kj/rev Wnel.OUI =riwnel.oul =(1500/60rev/sX2.330k1/rev)=58.2 kw Discussion Note that for 2-stroke engines, 1 thermodynamic cycle is equivalent to 1 mechanical cycle (and thus revolutions). 8-28

4 ~ Chapter 8 Gas Power Cycles 8-76 A gas-turbine power plant operates at specified conditions. The fraction of the turbine work output used to drive the compressor and the thermal efficiency are to be determined. Assumptions 1 Steady operating conditions exist. 2 The air-standard assumptions are applicable. 3 Kinetic and potential energy changes are negligible. 4 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are Cp = kj/kg. K and k = 1.4 (Table A-2). Analysis (a) Using constant specific heats, r =2=~=7 p PI 100 T qln =h3 -h2 =Cp(T3 -T2) ~T3 =T2 +qin/cp = 580K + (950kJ/kgy(I.OO5kJ/kg.K) 580 K- = K T4S=T3t (P }(k-1yk =(1525.3K\7 { 1)0.4/1.4 =874.8K 300 K 1 2s~ kjlkg/ ~~ i (3 \ ~ 4s 4 s WC.in =h2 -h1 =Cp(T2 -Tl)=(I.OO5kJ/kg.KX )K=281.4kJ/kg WT.oul = 17T(h3 -h4s)= 17TC P(T3 -T4s)= (0.86Xl.OO5kJ/kg.KX )K = 562.2kJ/kg Thus, WC,;n kj /kg = 50.1 % 'bw = -;- = kj /kg T,oul (b) Wnet.out = WT.out -WC.in = = kj Ikg 17th = Wnet.out = kj Ikg = 29.6% qin 950 kj Ikg 8-42

5 ~ Chapter 8 Gas Power Cycles 8.91 An ideal Bray ton cycle with regeneration is considered. The effectiveness of the regenerator is 100%. The net work output and the thermal efficiency of the cycle are to be determined. Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible. Properties The properties of air at room temperature are Cp = kj/kg.k and k = 1.4 (Table A-2a). Analysis Noting that this is an ideal cycle and thus the compression and expansion processes are isentropic, we have ( ) (k-i)1 k T2 =TI ~ =(300KXI0)04/1.4 =579.2K T 1200 K-1 ~ in ( P4 )(k-)/k 1 0.4/1.4 T4 =T3 -p; =(1200K(TO) =621.5K 300 K- 1 s & = 100% --+ T5 = T4 = 621.5K and T6 = T2 = 579.2K 1]h =1-~=I-Cp(T6-T))=I-T6-TI I "P(T3 -T5 ) T3 -Ts =1-=0.517 q L In j , ~.TI -r ) (k-i)/ k (or,'7th=l- =1- = 0.517),T3 p )(10){).4-))/ Then, Wnet =Wturb,out -Wcomp,in =(h3 -h4)-(h2 -hl) =Cp[(T3 -T4)-(T2 -TI)] = (1.005 kj/kg.k)[( ) ~ ( )]K = kj/kg or, Wner = '71hq;n = '71h(hJ -hs) = '71hCp(TJ -Ts) = (0.517)(1.005 kj /kg.k)( ) = kj / kg 8-50

6 8.92 A Bray ton cycle with regeneration using air as the working fluid is considered. The air temperature at the turbine exit, the net work output, and the thermal efficiency are to be determined. Assumptions 1 The air standard assumptions are applicable. 2 Air is an ideal gas with variable specific heats. 3 Kinetic and potential energy changes are negligible. T Properties The properties of air are given in Table A-17. A 1. ( ) Th. f K~ na ysls a e properties o air at various states are TI =310K ~ hl =310.24kJ/kg Pr. = K- 2~ &:: h2s -h] 17c =~~ h2 =h1 +(h2s -h1)/17c = ( y(O.75)=618.26kJ/kg T3 = 1150K ~ h3 = kJ/kg p'j = Thus, (b) w, net T4 = K =WT.our -WC.in ={h3 -h4)-{h2 -h1) = { )- { )= kj/kg (c) h5 -h2 &= ~ h5 =h2 +&(h4 -h2) h4 -h2 = (0.65X ) = kj/kg Then, qin = h3 -h5 = = kj/kg 1]", = Wnel q in 8-51

7 8-107 A regenerative gas-turbine cycle with two stages of compression and two stages of expansion is considered. The minimum mass flow rate of air needed to develop a specified net power output is to be determined. Assumptions 1 Argon is an ideal gas with constant specific heats. 2 Kinetic and potential energy changes are negligible. Properties The properties of argon at room temperature are Cp = kj/kg.k and k = (Table A- 2a). Analysis The mass flow rate will be a minimum when the cycle is ideal. That is, the turbine and the compressors are isentropic, the regenerator has an effectiveness of 100%, and the compression ratios across each compression or expansion stage are identical. In our case it is rp =.J9 = 3. Then the work inputs to each stage of compressor are identical, so are the work outputs of each stage of the turbine. ( l(k-i)/k T T2 =Ti ~) =(300KX3)0.667/1.667 =465.6K WC.in WT.oul lp6) (~~--~, T6=Ts-. =(1200K{~ Ps \3/ (k-i)/k,- -r-c /1667 = 773.2K = 2(h2 -hl )= 2C p(t2-1; )= 2(0.5203kJ/kg.KX )K = 172.3kJ/kg 300 K- = 2(h5 -h6)= 2C p(t5 -T6)= 2(0.5203kJ/kg.KX )K = 444.1kJ/kg / s! 1200 K- Wnel =WT,oul -WC,in = =271.8kJ/kg.w m=-!!!.!-= wnet 90,000 kj/s kj/kg = kg/s 8-60

8 8-113 A turbojet aircraft flying at an altitude of 9150 m is operating on the ideal jet propulsion cycle. The velocity of exhaust gases, the propulsive power developed, and the rate of fuel consumption are to be determined. Assumptions 1 Steady operating conditions exist. 2 The air standard assumptions are applicable. 3 Air is an ideal gas with constant specific heats at room temperature. 4 Kinetic and potential energies are negligible, except at the diffuser inlet and the nozzle exit. 5 The turbine work output is equal to the compressor work input. Properties The properties of air at room temperature are Cp = kj/kg.k and k = 1.4 (Table A-2a). Analysis (a) We assume the aircraft is stationary and the air is moving towards the aircraft at a velocity of V 1 = 320 m/s. Ideally, the air will leave the diffuser with a negligible velocity (V 2 = 0). Diffuser:...t!o Ein -Eout : Msystem (steady) Ein = Eou, h1 + V? /2 = h2 + V i /2 Compressor: PJ =P4 =~PXP2)={12X62.6kPa)=751.2kPa (p ) (k-t)/ k TJ = T2 i = {291.9KXI2)04/1.4 = 593.7K Turbine: or, T5 =T4 -T3 +T2 = =IO98.2K 8-66

$Nozzle: E, -E = L\E <:!'0 (steady) m out system Ein = EOU' h5 + V 52/2 = h6 + V 62/2 =(1400K{ ~ 1 4/1.4 \ 751.2kPa ) = 568.$

9 Nozzle: E, -E = L\E <:!'0 (steady) m out system Ein = EOU' h5 + V 52/2 = h6 + V 62/2 =(1400K{ ~ 1 4/1.4 \ 751.2kPa ) = 568.2K or, O=h6 -hs + ~ O=Cp(T6 -Ts)+Vi/2 2 = 1 032m/s (b) = {60 kg/sx )m1s{320 mls { ~ \ 1000 m2/s2 = 13,670 kw (c) Q;n =m(h4 -h3)=mcp(t4 -T3)=(60kg/SX1.005kJ/kg;KX )K = 48,620 kj/s Qj/1 ni fuel = "HV - 48,620 kj/s -- 42,700 kj/kg = 1.14 kg/s s I 8-67

..V2 l 2 2 -VI Chapter 8 Gas Power Cycles 8-117 Air enters a turbojet engine. The thrust produced by this turbojet engine is to be determined. Assumptions 1 Steady operating conditions exist.

Properties The properties of air are given in Table A-17. Analysis We assume the aircraft is stationary and the air is moving towards the aircraft at a velocity of V I = 300 mls.

10 ..V2 l 2 2 -VI Chapter 8 Gas Power Cycles Air enters a turbojet engine. The thrust produced by this turbojet engine is to be determined. Assumptions 1 Steady operating conditions exist. 2 The air standard assumptions are applicable. 3 Air is an ideal gas with variable specific heats. 4 Kinetic and potential energies are negligible, except at the diffuser inlet and the nozzle exit. Properties The properties of air are given in Table A-17. Analysis We assume the aircraft is stationary and the air is moving towards the aircraft at a velocity of V I = 300 mls. Taking the entire engine as our control volume and writing the steady-flow energy balance yield 7) = 280 K ~ hj = kj /kg T2 = 700 K --4 h2 = kj /kg Ein -out E.A = Llijsystem i- <:PO (steady) Ein = EOUl Qln + ni(hl + V( /2) = ni(h2 + V i /2) 20,000 kj/s + 7 C 300 mls +r:;;;;;;;=~ 427 C 20kg/s 1~~2 Qin =m h2-hl+~ IkJ/kg- IOOOm 2/S2 It gives V2 = 1106 mls Thus, F p = m(v2 -vi )= (20kg/sXI )mis = 16,120N 8-71

8-141 A simple ideal Bray ton cycle with air as the working fluid is considered. The changes in the net work output per unit mass and the thermal efficiency are to be determined.

$Therefore, For rp = 6, ( ~ \(k-l)/k T 3' 2" T 4=TJ-!. (P ) (k-i)/k = 779.1K PJ qin =h3 -h2 =Cp(T3 -T2) = (1.005kJ/kg.KXl300-500.6)K = 803.4kJ/kg..qoul s q out = h4 -hl = C p (T 4 -TI ) = (1.005kJ/kg.KX779.$

11 8-141 A simple ideal Bray ton cycle with air as the working fluid is considered. The changes in the net work output per unit mass and the thermal efficiency are to be determined. Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are R = kj/kg.k, Cp = kj/kg.k, Cv = kj/kg. K, and k = 1.4 (Table A-2). Analysis Processes 1-2 and 3-4 are isentropic. Therefore, For rp = 6, ( ~ \(k-l)/k T 3' 2" T 4=TJ-!. (P ) (k-i)/k = 779.1K PJ qin =h3 -h2 =Cp(T3 -T2) = (1.005kJ/kg.KXl )K = 803.4kJ/kg..qoul s q out = h4 -hl = C p (T 4 -TI ) = (1.005kJ/kg.KX )K = 481.5kJ/kg Wnel = qin -qout = = 321.9kJ/kg 1]th = =40.1% For rp = 12, 2 = TI -1.. T ( P ) (k-i)/k = (300KXI2)04/1.4 = 610.2K I PJ T 4 = T,( t 1'-1)1' : 639.2K qin =h3 -h2 =Cp(T3 -T2) = (1.005kJ/kg.KXl )K = 693.2kJ/kg qoul =h4 -hl =Cp(T4 -TI) = (1.005kJ/kg.K X )K = 340.9kJ/kg Wnel = qin -qoul = = 352.3kJ/kg = 50.8% Thus, (a) Llwnel = = 30.4 kj/kg (increase ) I. (b) Ll17lh =50.8%-40.1%=10.7% (increase) 8-89

Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. hl = IOkPa = 191.

Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. hl = IOkPa = 191. ~ Chapter 9 Vapor and Combined Power Cycles 9-16 A steam power plant that operates on a simple ideal Rankine cycle is considered. he quality of the steam at the turbine exit, the thermal efficiency of