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## Q. 2.30

A reciprocating air compressor takes in air at 1 bar and 20°C and delivers at 6 bar. Calculate the work done, heat transfer, and change in internal energy per kg of air compressed if the compression process is (a) isothermal, (b) isentropic and (c) polytropic: $p ν^{1 .35}$ = C. Take $\gamma$ = 1.4 and R = 0.287 kJ/kg.K. Neglect change in potential and kinetic energies.

## Verified Solution

(a) Isothermal process :

Work done,

\quad \begin{aligned}w &=p_1 ν_1 \ln \left(\frac{v_2}{v_1}\right)=R T_1 \ln \left(\frac{p_1}{p_2}\right) \\&=0.287 \times 293 \ln \frac{1}{6}=-150.67 \ kJ / kg\end{aligned}

Change in internal energy, dU = 0

Heat transfer, q = w = – 150.67 kJ/kg

Isentropic Process :

\quad \begin{aligned}T_2 &=T_1\left(\frac{p_2}{p_1}\right)^{\frac{\gamma-1}{\gamma}}=293\left(\frac{6}{1}\right)^{\frac{1.4-1}{1.4}}=488.87 \ K \\w &=\left[\frac{\gamma}{\gamma-1}\right]\left(p_1 ν_1-p_2 ν_2\right)=\left[\frac{\gamma}{\gamma-1}\right] R\left(T_1-T_2\right) \\&=\left[\frac{1.4}{1.4-1}\right] \times 0.287(293-488.87)=-196.75 \ kJ / kg\end{aligned}

For isentropic process,     q = 0

\begin{aligned}d u=c_ν\left(T_2-T_1\right)=\left[\frac{R}{\gamma-1}\right]\left(T_2-T_1\right) \\ =\left[\frac{0.287}{1.4-1}\right](488.87-293)=140.536 \ kJ / kg \end{aligned}

Polytropic process :

\begin{aligned}T_2 &=T_1\left(\frac{p_2}{p_1}\right)^{\frac{n-1}{n}}=293 \times 6^{\frac{1.35-1}{1.35}}=466.24 \ K \\w &=\left[\frac{n}{n-1}\right]\left(p_1 ν_1-p_2 ν_2\right)=\left[\frac{n}{n-1}\right] R\left(T_1-T_2\right) \\&=\left[\frac{1.35}{1.35-1}\right] \times 0.287(293-466.24)=-191.78 \ kJ / kg\end{aligned}

\begin{aligned}h_1+q &=h_2+w \\q &=w+\left(h_2-h_1\right) \\&=w+c_p\left(T_2-T_1\right) \\&=w+\left[\frac{\gamma}{\gamma-1}\right] R\left(T_2-T_1\right) \\&=-191.78+\left[\frac{1.4}{1.4-1}\right] \times 0.287(466.24-293) \\&=-17.76 \ kJ / kg \\d u &=c_ν\left(T_2-T_1\right)=\left[\frac{R}{\gamma-1}\right]\left(T_2-T_1\right) \\&=\left(\frac{0.287}{0.4}\right)(466.24-293)=124.3 \ kJ / kg\end{aligned}