Question 16.1: Air at a stagnation temperature of 27º C enters the impeller...
Air at a stagnation temperature of 27º C enters the impeller of a centrifugal compressor in the axial direction. The rotor which has 15 radial vanes, rotates at 20000 rpm. The stagnation pressure ratio between the diffuser outlet and the impeller inlet is 4 and the isentropic efficiency is 85%. Determine (i) the impeller tip radius and (ii) power input to the compressor when the mass flow rate is 2 kg/s. Assume a power input factor of 1.05 and a slip factor \sigma=1-2 / n , where n is the number of vanes. For air, take \gamma=1.4, R=287 \mathrm{~J} / \mathrm{kg} \mathrm{K}
(i) From Eq (16.7), we can write
T_{3 t}-T_{1 t}=\frac{T_{1 t}\left[\left(p_{3 t} / p_{1 t}\right)^{\frac{\gamma-1}{\gamma}}-1\right]}{\eta_{c}}again with the help of Eq (16.5) and T_{2 t}=T_{3 t} it becomes
w=\Psi \sigma U_{2}^{2}=c_{p}\left(T_{2 t}-T_{1 t}\right) (16.5)
\mathrm{U}_{2}^{2}=\frac{c_{p} T_{1 t}\left[\left(p_{3 t} / p_{1 t}\right)^{\frac{\gamma-1}{\gamma}}-1\right]}{\eta_{c} \sigma \psi}Here, p_{3 t} / p_{1 t}=4
T_{\mathrm{lt}}=300 \mathrm{~K}
c_{p}=\frac{\gamma R}{\gamma-1}
=\frac{1.4 \times 287}{0.4}
=1005 \mathrm{~J} / \mathrm{kg} \mathrm{K}
\sigma=1-\frac{2}{15}
=0.867
\psi=1.05
Therefore, U_{2}^{2}=\frac{1005 \times 300 \times\left(4^{\frac{0.4}{1.4}}-1\right)}{0.85 \times 0.867 \times 1.05}
which gives U_{2}=435 \mathrm{~m} / \mathrm{s}
Thus the impeller tip radius is
r_{2}=\frac{435 \times 60}{2 \pi \times 20000}=0.21 \mathrm{~m}
(ii) Power input to the air =\frac{2 \times 1.05 \times 0.867 \times(435)^{2}}{1000} \mathrm{~kW}
=344.52 \mathrm{~kW}