Answer to Question #115774 in Mechanical Engineering for Smith

Question

Air initially at 75 psia and 65 0F is compressed to a final pressure of 300 psia and temperature of 320 0F. Find the polytropic exponent for the process.

Answer 1

Here in the given equation it is given that ,

$P_1=$ $75 psia, T_1= 65 ^oF,P_2=300 psia, T_2= 320 ^oF$

we know that for polytropic process

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

$\frac{320}{65}= (\frac{300}{75})^{\frac{n-1}{n}}$

4.923=$(4)^{\frac{n-1}{n}}$

On taking log both sides

log(4.923)= $(\frac{n-1}{n})log 4$

0.692n=(n-1)0.602

0.692 n = 0. 602 n - .602

0.09 n= -.602

n= - 6.68 here n is the polytropic exponent

but for real condition polytropic exponent can not be negative