## Related questions with answers

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

VerifiedFirst 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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