Question

Consider a Carnot cycle executed in a closed system with 0.0058 kg of air. The temperature limits of the cycle are300 K and 940 K, and the minimum and maximum pressures that occur during the cycle are20 kPa and 2,000 kPa. Assuming constant specific heats, determine the net work output per cycle.

Answer 1

Answer:0.646 KJ

Step-by-step explanation:

Using First law for cycle

`\sum Q=\sum W`

`\sum Q=Q_(1-2)+Q_(3-4)`

For adiabatic process heat transfer is zero and for isothermal process

d(Q)=d(W)

`Q_(1-2)=mRT_1\ln {(P_1)/(P_2)}`

Given
`P_1=2000KPa`

`P_3=20KPa`

`\left ((T_2)/(T_3)\right )^{(\gamma )/(\gamma -1)`=
`\left ((P_2)/(P_3)\right )}`

`P_2=1089.06K`

`Q_(1-2)=0.0058\dot 0.287\dot 940\ln (2000)/(1089.06)`=
`0.95KJ`

`Q_(3-4)=mRT_2\ln {(P_3)/(P_4)}`

`\left ((T_1)/(T_4)\right )^{(\gamma )/(\gamma -1)`=
`\left ((P_1)/(P_4)\right )}`

Now we have to find
`P_4=36.72KPa`

`Q_(3-4)=0.0058\dot 0.287\dot 300\ln (20)/(36.72)`=
`-0.30341KJ`

`Q_(net)=Q_(1-2)+Q_(3-4)`

`Q_(net)=0.95-0.303=0.646KJ`

`Q_(net)=W_(net)=0.646KJ`