Answer to Question #115775 in Mechanical Engineering for Smith

Question

Air initially at 50 psia and 140 0F undergoes a polytropic process such that the temperature becomes 40 0F. The polytropic exponent for the process is equal to 1.3. Find the final pressure and specific volume.

Answer 1

$P_1=50 psia, T_1= 140 ^oF,P_2=?, T_2= 40 ^oF$ , n= 1.3

we know that for polytropic process as

$\frac{T_2}{T_1}= (\frac{P_2}{P_1})^{\frac{n-1}{n}}$

$\frac{40}{140}= (\frac{P_2}{50})^{\frac{1.3-1}{1.3}}$

0.2857 = $(\frac{P_2}{50})^{0.231}$

taking log both sides

log(0.2857) = 0.231 log $(\frac{P_2}{50})$

-2.355= log $(\frac{P_2}{50})$

$0.00441\times 50 = P_2$

$P_2=0.2205 psia$