Answer to Question #289788 in Mechanical Engineering for Abhi

Question

1 kg of air at 7 bar pressure and 90 temperature undergoes a non-

flow polytropic process. The law of expansion is p

=Constant.

The pressure falls to 1.4 bar during the process.

Calculate: (1) Final Temperature (2) Change in Internal energy (3)

Work done (4) Heat exchange. Take R=287J/kg-K and = 1.4 for air

Answer 1

Solution;

Given;

m=1kg

$P_1=7bar$

$T_1=90°c=363K$

$P_2=1.4bar$

Polytropic process;

$Pv^n=Constant$

1)Final temperature;

From P,v,T relations;

$(\frac{P_1}{P_2})^{\frac{n-1}{n}}=\frac{T_1}{T_2}$

$(\frac7{1.4})^{\frac{0.4}{1.4}}=\frac{363}{T_2}$

$T_2=\frac{363}{1.583}=229.19K$

2)Change in internal energy;

$dU=mC_v(T_2-T_1)$

$C_v=\frac{R}{\gamma-1}=\frac{287}{1.4-1}=717.5$

$dU=1×717.5(229.19-363)=-96008.675J$

(Negative sign indicates a drop in internal energy)

3)Work Done;

$W=\frac{mR(T_1-T_2)}{n-1}$

$W=\frac{1×287(363-229.19)}{1.4-1}=96008.675J$

4)Heat exchange;

$Q=W+dU$ =96008.675+-96008.675=0J