Question

A centrifugal air compressor of impeller diameter 0.6 m runs at a speed of 6000 rpm. Air enters the compressor at atmospheric temperature of 298 K. Slip factor = 0.8, power factor = 1, $c{p}_{air}=1.005kJ/kg,{\gamma }_{air}=\mathrm{1.4.}$

What is the pressure ratio developed if the isentropic efficiency is 90%?

• A. 1.66
• B. 1.33
• C. 2.00
• D. 2.30

Answer 1

The correct option is B 1.33


Work done/kg, $W=\psi \sigma u_2^2$

We also know that,

Work done per kg, $W=c_p(T_2-T_1)$

$u_2=\frac{\pi DN}{60}=\frac{\pi \times 0.6\times 6000}{60}=188.49m/s$

Now, from equation (1) and (2)

$C_P(T_2-T_1)=\psi \sigma u_2^2$

$T_2=T_1+\frac{\psi \sigma u_2^2}{c_p}$

$T_2=298+\frac{1\times 0.8\times {188.49}^2}{1.005\times 1000}$

$T_2=326.28K$

We know that,

Isentropic efficiency, ${\eta }_C=\frac{T_2^{\prime }-T_1}{T_2-T_1}$

$T_2^{\prime }=(326.28-298)\times 0.9+298$

$T_2=323.452K$

Now, Pressure ratio is,

${\left(\frac{T_2^{\prime }}{T_1}\right)}^{\gamma /\gamma -1}=\frac{P_2^{\prime }}{P_1}$

${\gamma }_p={\left(\frac{323.452}{298}\right)}^{1.4/0.4-1}$

Pressure ratio, ${\gamma }_p=1.33$