2020-01-12T00:25:32-05:00
In a carnot cycle operating between 307 and 17 degree Celsius the maximum and minimum pressures are 60 2.4 and 1.04 bar.clclculate the cycle efficiency and the work ratio.assume air yo be working fluid.
1
2020-01-16T07:38:42-0500
The cycle efficiency:
η = 1 − T 2 T 1 = 1 − 290 580 = 0.5 \eta=1-\frac{T_2}{T_1}=1-\frac{290}{580}=0.5 η = 1 − T 1 T 2 = 1 − 580 290 = 0.5
W R = W n e t W g r o s s WR=\frac{W_{net}}{W_{gross}} W R = W g ross W n e t
W n e t = ( T 1 − T 2 ) ( S 1 − S 2 ) W_{net}=(T_1-T_2)(S_1-S_2) W n e t = ( T 1 − T 2 ) ( S 1 − S 2 ) 4-a is isothermal, so
S A − S 4 = R ln P 4 P 2 = 0.287 ln 62.4 1.04 = 1.175 k J k g S_A-S_4=R\ln{\frac{P_4}{P_2}}=0.287\ln{\frac{62.4}{1.04}}=1.175\frac{kJ}{kg} S A − S 4 = R ln P 2 P 4 = 0.287 ln 1.04 62.4 = 1.175 k g k J a-2 is isobaric, so
S 2 − S a = c p ln T 1 T 2 = 1.005 ln 580 290 = 0.6966 k J k g S_2-S_a=c_p\ln{\frac{T_1}{T_2}}=1.005\ln{\frac{580}{290}}=0.6966\frac{kJ}{kg} S 2 − S a = c p ln T 2 T 1 = 1.005 ln 290 580 = 0.6966 k g k J
W n e t = ( 580 − 290 ) ( 1.175 − 0.696 ) = 139.2 k J k g W_{net}=(580-290)(1.175-0.696)=139.2\frac{kJ}{kg} W n e t = ( 580 − 290 ) ( 1.175 − 0.696 ) = 139.2 k g k J
W g r o s s = T 1 ( S 1 − S 4 ) + c v ( T 1 − T 2 ) W_{gross}=T_1(S_1-S_4)+c_v(T_1-T_2) W g ross = T 1 ( S 1 − S 4 ) + c v ( T 1 − T 2 )
W g r o s s = 580 ( 0.48 ) + 0.718 ( 580 − 290 ) = 486.62 k J k g W_{gross}=580(0.48)+0.718(580-290)=486.62\frac{kJ}{kg} W g ross = 580 ( 0.48 ) + 0.718 ( 580 − 290 ) = 486.62 k g k J
W R = 139.2 486.62 = 0.286 WR=\frac{139.2}{486.62}=0.286 W R = 486.62 139.2 = 0.286
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