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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, −20C^{\circ} \mathrm{C} and an exit state of 400 kPa, 60C^{\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:

P1=150 kPaP_1=150\text{ kPa}

P2=400 kPaP_2=400\text{ kPa}

T1=20 C=253 KT_1=-20\text{ C}=253\text{ K}

T2=60 C=333 KT_2=60\text{ C}=333\text{ K}

Relation between temperature and pressure in a polytropic process:

P2P1=(T2T1)nn1\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:

n1n=logP2P1T2T1=log400150333253=0.28n=110.28=1.39\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:

Cp=2.130 kJ kg KC_p=2.130\frac{\text{ kJ}}{\text{ kg K}}

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

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