Question

One kilogram per second of air initially at 101 kPa and 300K is compressed polytropically according to the process PV^1.3 = C. Calculate the power necessary to compress the air to 1380 kPa.

W =

Answer 1

$$P_1=101K ,T_1=300K,P_2=1380Kpa,\gamma=1.3$$

$∆T=T_2-T_1$

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

$$T_2=300\times(\frac{1380}{101})^{\frac{1.3-1}{1.3}}=548.50K\\W=\frac{mR∆T}{n-1}$$

$$W=\frac{1\times0.257\times(548.50-300)}{3.1-1}=237.73\frac{KJ}{sec}$$

$W={237.73KW}$