## Chapter 2

## Q. 2.28

3 m³ of air at a pressure of 1bar and 25°C is compressed isentropically to 5 bar. It is then expanded isothermally and restored to original volume by constant volume heat rejection process. Determine (a) pressure, volume and temperature at the end of each operation, (b) heat added during the isothermal process, (c) heat rejected during constant volume process, and (d) change in internal energy during each process.

Assume R = 0.287 kJ/kg. K and c_p = 1.005 kJ/kg. K.

## Step-by-Step

## Verified Solution

The p–v diagram is shown in Fig. 2.42.

\begin{aligned}c_v &=c_p-R=1.005-0.287 \\&=0.718 \ kJ / kg . K \\\gamma &=\frac{c_p}{c_ν}=\frac{1.005}{0.718}=1.4\end{aligned}

**Process 1–2:**

\begin{aligned}V_2 &=V_1\left\lgroup \frac{p_1}{p_2}\right\rgroup ^{\frac{1}{\gamma}} \\&=3\left(\frac{1}{5}\right)^{\frac{1}{1.4}}=0.95 \ m ^3 \\T_2 &=T_1\left\lgroup\frac{p_2}{p_1}\right\rgroup ^{\frac{\gamma-1}{\gamma}} \\&=298\left(\frac{5}{1}\right)^{\frac{(1.4-1)}{1.4}}=471.98 \ K \\W_{1-2} &=\frac{p_1 V_1-p_2 V_2}{\gamma-1} \\&=10^2 \left\lgroup\frac{1 \times 3-5 \times 0.95}{1.4-1}\right\rgroup =-437.5 \ kJ\end{aligned}

From first law of thermodynamics,

δQ – δW = dU

For isentropic process, δQ = 0

\therefore \quad \quad -W_{1-2}=\Delta U=437.5 \ kJ

**Process 2-3:**

\begin{aligned}p_2 V_2 &=p_3 V_3=p_3 V_1 \\p_3 &=\frac{5 \times 0.95}{3}=1.583 \ bar \\T_3 &=T_2=471.98 \ K\end{aligned}

For isothermal process, ΔU = 0

\therefore \quad \quad\begin{aligned}Q_{2-3} &=W_{2-3}=p_2 V_2 \ln \frac{V_3}{V_2} \\&=5 \times 10^2 \times 0.95 \ln \left(\frac{3}{0.95}\right)=546.2 \ kJ\end{aligned}

**Process 3 –1:**

\begin{aligned}Q_{3-1} &=m c_ν\left(T_1-T_3\right) \\m &=\frac{p_1 V_1}{R T_1}=\frac{1 \times 10^2 \times 3}{0.287 \times 298}=3.507 \ kg \\Q_{3-1} &=3.507 \times 0.718(298-471.98)=-438 \ kJ\end{aligned}

During constant volume process, δW = 0

\therefore \quad \quad Q_{3-1}=\Delta U=-438 \ kJ