Question

A rotary type compressor compresses air from 98kPa, 27C through a compression ratio of 5 and compression efficiency of 75%, find the compressor work.

Answer 1

Given data:

P1 = 98 kPa, Ti = 300 K, r=5, n = 0.75

The compression ratio

$r=\frac{P_2}{P_1}=5$

$\therefore P_2=5\times P_1=5\times 98=490\:kPa$

Specific volume,

$\vartheta_1 =\frac{1}{\rho_1}= \frac{RT_1}{P_1}=\frac{287\times300}{98\times10^3}=0.87857\:\frac{m^3}{kg}$

The actual work transferred to air

$W_a=\vartheta_1 (P_2-P_1)$

$W_a=0.87857 (490\times10^3-98\times10^3)=344400\:J/kg$

The work consumed by the compressor

$W=\frac{W_a}{\eta }=\frac{344400}{0.75}=459200\:\frac{J}{kg}$

$\therefore\:W=459.2\:\frac{kJ}{kg}$