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

Answer:

a) $T_2=569.35 K$

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

Explanation:

Given: $\gamma =1.4$ for air.

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

We know that  

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

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

$T_2=569.35 K$

(a) $T_2=569.35 K$

(b)Work for adiabatic process

  W=$\frac{P_1V_1-P_2V_2}{\gamma -1}$

We know that PV=mRT for ideal gas.

 W=$mR\frac{T_1-T_2}{\gamma -1}$

Now by putting values

work per kg of air=$0.287\times \frac{295-569.35}{1.4 -1}$

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

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