Question

# Argon gas enters an adiabatic compressor at 14 psia and 75$^\circ{}$F with a velocity of 60 ft/s, and it exits at 200 psia and 240 ft/s. If the isentropic efficiency of the compressor is 87 percent, determine (a) the exit temperature of the argon and (b) the work input to the compressor.

Solution

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First we can calculate the final temperature for an isentropic process using the initial temperature, initial and final pressures and the coefficient $k$ from table A-2E:

\begin{align*} T_{\text{2s}}&=T_{1}\Bigg(\dfrac{P_{2}}{P_{1}}\Bigg)^{(k-1)/k}\\ &=536\cdot\Bigg(\dfrac{200}{14}\Bigg)^{(1.667-1)/1.667}\:\text{R}\\ &=1553\:\text{R} \end{align*}

Next we calculate the change in the specific kinetic energy:

\begin{align*} \Delta e_{\text{k}}&=\dfrac{v_{2}^{2}-v_{1}^{2}}{2}\\ &=\dfrac{240^{2}-60^{2}}{2}\:\dfrac{\text{ft}^{2}}{\text{s}^{2}}\\ &=27000\:\dfrac{\text{ft}^{2}}{\text{s}^{2}}\\ &=1.078\:\dfrac{\text{Btu}}{\text{lbm}} \end{align*}

Now from the efficiency we can determine the actual final temperature. The specific heat is taken from A-2E

\begin{align*} \eta&=\dfrac{h_{\text{2s}}-h_{1}+\Delta e_{\text{k}}}{h_{\text{2a}}-h_{1}+\Delta e_{\text{k}}}\\ &=\dfrac{c_{\text{p}}(T_{\text{2s}}-T_{1})+\Delta e_{\text{k}}}{c_{\text{p}}(T_{\text{2a}}-T_{1})+\Delta e_{\text{k}}} \end{align*}

\begin{align*} T_{\text{2a}}&=\Bigg(1-\dfrac{1}{\eta}\Bigg)\Bigg(T_{1}-\dfrac{\Delta e_{\text{k}}}{c_{\text{p}}}\Bigg)+\dfrac{T_{\text{2s}}}{\eta}\\ &=\Bigg(1-\dfrac{1}{0.87}\Bigg)\Bigg(536-\dfrac{1.078}{0.1253}\Bigg)\:\text{R}+\dfrac{1553}{0.87}\:\text{R}\\ &=\boxed{1706\:\text{R}} \end{align*}

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