Answer to Question #177932 in Mechanical Engineering for MOHAMED SADIQUE

Question

2 m3 of air at 1.1 bar and 20°C is heated in a cylinder at constant pressure until its volume is 4.5 m. It is then compressed according to the law pv#= constant till pressure and volume change to 5.5 bar and 0.6 m calculate change in entropy, work and heat transfer during the processes and the value of n.

Answer 1

Given that air at initial condition

$P_1=1.1 bar=110 kPa; T_1=20^0 C=20+273 =293 K$

$V_1=2 m^3$

Assume air is ideal gas $R=0.287 kJ/kg.K$

Now it is heated at constant pressure

$V_2=4.5 m^3 ; (P_1=P_2)=110kPa$

For 1-2

$\frac{V_1}{T_1}=\frac{V_2}{T_2} \implies T_2=\frac{4.5 \times 293}{2}=659.25 K$

Now it is compressed to $PV^n=constant$ $P_3=5.5 bar =550kPa ; V_3=0.6 m^3$

We have $P_2V_2^n=P_3V_3^n \implies \frac{550}{110}=(\frac{4.5}{0.6})^{n} \implies n=0.798 =0.8$

$\frac{T_3}{T_2}=(\frac{V_2}{V_3})^{n-1} \implies T_3=659.25 \times (\frac{4.5}{0.6})^{-0.2}=440.6 K$

Now consider process 1-2

Q-W = change in internal energy ............(1)

$W_{1-2}=P(V_2-V_1)=110(4.5-2)=275 kJ$

Change in internal energy , $mc_v(T_2-T_1)$

$Q-275 =2.616 \times 0.718 (659.25-293) \implies Q_2=962.922 kJ$

Change in entropy, $\triangle S _{1-2}=m[C_v \ln \frac{T_2}{T_1}+R\ln \frac{V_2}{V_1}]=2.616[0.718 \ln \frac{659.25}{293}+0.287\ln \frac{4.5}{2}]=2.132 kJ/K$

$W_{2-3}= \frac{P_2V_2-P_3V_3}{n-1}=\frac{(110 \times 4.5)-(550 \times 0.6)}{0.8-1}=-825 kJ$

$Q_{2-3}-W_{2-3}=mc_v(T_3-T_2)$

$Q_{2-3}-(-825)=2.616 \times0.718(440.6-659.25) \implies Q_{2-3}=-1235.7kJ$

$\triangle S _{2-3}=m[C_v \ln \frac{T_3}{T_2}+R\ln \frac{V_3}{V_2}]=2.616[0.718 \ln \frac{440.6}{659.25}+0.287\ln \frac{0.6}{4.5}]=-0.8676 kJ/K$