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# A flow of 4 kg/s ammonia goes through a device in a polytropic process with an inlet state of 150 kPa, −20$^{\circ} \mathrm{C}$ and an exit state of 400 kPa, 60$^{\circ} \mathrm{C}$. Find the polytropic exponent n, the specific work, and the heat transfer.

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We are given following data for flow of Ammonia in a polytropic process:

$P_1=150\text{ kPa}$

$P_2=400\text{ kPa}$

$T_1=-20\text{ C}=253\text{ K}$

$T_2=60\text{ C}=333\text{ K}$

Relation between temperature and pressure in a polytropic process:

\begin{align*} \dfrac{P_2}{P_1}&=\left(\dfrac{T_2}{T_1} \right)^{\dfrac{n}{n-1}}\\ \end{align*}

We can find polytropic exponent:

\begin{align*} \dfrac{n-1}{n}&=\log_{\dfrac{P_2}{P_1}} {\dfrac{T_2}{T_1}}=\log_{\dfrac{400}{150}} {\dfrac{333}{253}}={0.28}\\\\ n&=\dfrac{1}{1-0.28}=\boxed{1.39} \end{align*}

From Properties table A.5. we can find Ammonia specific heat and gas constant:

$C_p=2.130\frac{\text{ kJ}}{\text{ kg K}}$

$R=0.4882\frac{\text{ kJ}}{\text{ kg K}}$

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