Question 16.4: Air at a temperature of 27°C flows into a centrifugal compre...
Air at a temperature of 27°C flows into a centrifugal compressor running at 20,000 rpm. The following data are given:
| Slip factor | 0.80 |
| Power input factor | 1 |
| Isentropic efficiency | 80% |
| Outer diameter of blade tip | 0.5 m |
Assuming the absolute velocities of air entering and leaving the compressor are same, find (i) static temperature rise of air passing through the compressor, and (ii) the static pressure ratio.
Velocity of the blade tip,
U_{2}=\frac{\pi \times 0.5 \times 20,000}{60}=523.6 \mathrm{~m} / \mathrm{s}
From Eq. (16.5),
w=\Psi \sigma U_{2}^{2}=c_{p}\left(T_{2 t}-T_{1 t}\right) (16.5)
Stagnation temperature rise \left(T_{2 t}-T_{1 t}\right)=\frac{\psi \sigma U_{2}^{2}}{c_{p}}
=\frac{0.80 \times(523.6)^{2}}{1005}=218.23^{\circ} \mathrm{C}
( c_{p} of air has been taken as 1005 J/kg K)
Since the absolute velocities at the inlet and the outlet of the stage are the same, the rise in stagnation temperature equals to that in static temperature.
The static pressure ratio can be written as
\frac{p_{2}}{p_{1}}=\left(\frac{T_{2}^{\prime}}{T_{1}}\right)^{\frac{\gamma}{\gamma-1}}=\left[1+\frac{\eta_{C}\left(T_{2}-T_{1}\right)}{T_{1}}\right]^{\frac{\gamma}{\gamma-1}}
=\left[1+\frac{0.8 \times 218.23}{300}\right]^{\frac{1.4}{0.4}}
=4.98