Question

Air is compressed in an isentropic process from an initial pressure and temperature of P1 = 90 kPa and T1=22°C to a final pressure of P2=900 kPa. Determine: a)- The final temperature of the air. b)-The work done per kg of air during the process.

Answer 1

a)
`T_2=569.35 K`

b)Work done per kg of air=196.84 KJ/Kg

Step-by-step explanation:

Given:
`\gamma =1.4` for air.

`P_1=90 KPa ,T_=22^\circ C,P_2=900 KPa`

We know that

`(T_2)/(T_1)=\left ((P_2)/(P_1)\right )^{\frac{{\gamma-1}}{\gamma}}`

So
`(T_2)/(295)=\left ((900)/(90)\right )^{\frac{{1.4-1}}{1.4}}`

`T_2=569.35 K`

(a)
`T_2=569.35 K`

(b)Work for adiabatic process

W=
`(P_1V_1-P_2V_2)/(\gamma -1)`

We know that PV=mRT for ideal gas.

W=
`mR(T_1-T_2)/(\gamma -1)`

Now by putting values

work per kg of air=
`0.287* (295-569.35)/(1.4 -1)`

Work w=-196.84 KJ/Kg (Negative sign indicate work given to input.)

So work done per kg of air=196.84 KJ/Kg