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 diffuser outlet and impeller inlet is 4 and the isentropic efficiency is 85%. Determine (a) the impeller tip radius and (b) 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 σ = 1 – 2/n, where n is the number of vanes. For air, take γ = 1.4, R = 287 J/kg K.
(a) From Eq (16.7), we can write
\begin{aligned}\frac{p_{3 t}}{p_{1 t}} &=\left(\frac{T_{3 t}^{\prime}}{T_{1 t}}\right)^{\frac{\gamma}{\gamma-1}} \\&=\left[1+\frac{\eta_{c}\left(T_{3 t}-T_{1 t}\right)}{T_{1 t}}\right]^{\frac{\gamma}{\gamma-1}}\end{aligned} (16.7)
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)
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_{1 t}=300 K
c_{p}=\frac{\gamma R}{\gamma-1}
=\frac{1.4 \times 287}{0.4}
= 1005 J/kg K.
\sigma=1-\frac{2}{15}
= 0.867
\psi=1.05
Therefore, U_{2}^{2}=\frac{1005 \times 300 \times\left(\frac{0.4}{4^{1.4}}-1\right)}{0.85 \times 0.867 \times 1.05}
which gives U_{2} = 435 m/s
Thus the impeller tip radius is
r_{2}=\frac{435 \times 60}{2 \pi \times 20000}
= 0.21 m
(b) Power input to the air = \frac{2 \times 1.05 \times 0.867 \times(435)^{2}}{1000} kW
= 344.52 kW