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## Q. 2.26

A fluid is compressed reversibly from a volume of 8 m³ to 1m³ by the law $p V^n$ = const. The initial temperature and pressure of fluid is 25°C and 1 bar. Calculate the final temperature, pressure, work done, and heat transfer, if n = 1, 1.3, and 1.4. Take $c_ν$ = 0.718 kJ/kg. K, $c_p$ = 1.005 kJ/kg. K, and R = 0.287 kJ/kg.K.

## Verified Solution

(a) For

\begin{aligned}n=& 1 \\ \\p_1 V_1 &=p_2 V_2 \\ \\p_2 &=1 \times 8 \therefore p_2=8 \ bar \\ \\T_1 &=T_2=25+273=298 \ K \\ \\ W_{1-2} &=Q_{1-2}=p_1 V_1 \ln \frac{V_2}{V_1}=1 \times 10^2 \times 8 \ln \frac{1}{8}=-1663.6 \ kJ\end{aligned}

(b) For

\begin{aligned}n &=1.3 \\ \\p_1 V_1^{1.3} &=p_2 V_2^{1.3} \\ \\p_2 &=p_1\left\lgroup \frac{V_1}{V_2}\right\rgroup ^{1.3}=1 \times 8^{1.3}=14.928 \ bar \\ \\\frac{T_2}{T_1} &= \left\lgroup\frac{V_1}{V_2}\right\rgroup ^{n-1} \\ \\T_2 &=298(8)^{0.3}=556.1 \ K\end{aligned}

\begin{aligned}W_{1-2} &=\frac{p_1 V_1-p_2 V_2}{n-1} \\&=10^2\left\lgroup \frac{1 \times 8-14.928 \times 1}{1.3-1}\right\rgroup\\&=-2309.3 \ kJ \\ \\Q_{1-2} &=m C_n\left(T_2-T_1\right) \\&=m\left[\frac{c_p-n c_v}{1-n}\right] \times\left(T_2-T_1\right)\end{aligned}

\begin{aligned}m &=\frac{p_1 V_1}{R T_1}=\frac{1 \times 10^2 \times 8}{0.287 \times 298}=9.354 \ kg \\Q_{1-2} &=9.354\left[\frac{1.005-1.3 \times 0.718}{1-1.3}\right] \times(556.1-298) \\&=-576.2 \ kJ\end{aligned}

(c) For

\begin{aligned}n=\gamma &=1.4 \\ p_2 &=p_1\left\lgroup \frac{V_1}{V_2}\right\rgroup^\gamma=1 \times 8^{1.4}=18.379 \ bar \\T_2 &=T_1\left\lgroup \frac{V_1}{V_2}\right\rgroup^{\gamma-1}=298 \times 8^{0.4}=684.6 \ K\end{aligned}

\begin{aligned}W_{1-2} &=\frac{p_1 V_1-p_2 V_2}{\gamma-1} \\&=10^2\left\lgroup \frac{1 \times 8-18.379 \times 1}{1.4-1}\right\rgroup \\&=-2594.7 \ kJ \\Q_{1-2} &=0\end{aligned}