Question 16.3: A centrifugal compressor has an impeller tip speed of 360 m/...
A centrifugal compressor has an impeller tip speed of 360 m/s. Determine (a) the absolute Mach number of flow leaving the radial vanes of the impeller and (b) the mass flow rate. The following data are given
Impeller Tip speed 360 m/s
Radial component of flow velocity at impeller exit 30 m/s
Slip factor 0.9
Flow area at impeller exit 0.1 m^{2}
Power input factor 1.0
Isentropic efficiency 0.9
Inlet stagnation temperature 300 K
Inlet stagnation pressure 100 kN/m^{2}
R (for air) 287 J/kg K
g (for air) 1.4
The absolute Mach number is the Mach number based on absolute velocity.
Therefore, M_{2}=\frac{V_{2}}{\sqrt{\gamma R T_{2}}}
Now V_{2} \text { and } T_{2} have to be determined.
From the velocity triangle at impeller exit
V_{2}=\sqrt{V_{w 2}^{2}+V_{f 2}^{2}}
In case of slip, V_{w 2}=\sigma U_{2}
Hence, V_{2}=\sqrt{\left(\sigma U_{2}\right)^{2}+V_{f 2}^{2}}
=\sqrt{(0.9 \times 360)^{2}+(30)^{2}}
= 325.38 m/s
From Eq. (16.5)
w=\Psi \sigma U_{2}^{2}=c_{p}\left(T_{2 t}-T_{1 t}\right) (16.5)
T_{2 t}=T_{1 t}+\frac{\psi \sigma U_{2}^{2}}{c_{p}}
\left[c_{p}=\frac{\gamma R}{\gamma-1}=\frac{1.4 \times 287}{0.4}=1005 J / kg K \right]
T_{2 t}=300+\frac{0.9 \times(360)^{2}}{1005}
= 416 K.
T_{2}=T_{2 t}-\frac{V_{2}^{2}}{2 c_{n}}
=416-\frac{(325.38)^{2}}{2 \times 1005}
= 363.33 K.
Therefore, M_{2}=\frac{325.28}{\sqrt{1.4 \times 287 \times 363.33}}
= 0.85
Mass flow rate \dot{m}=\rho_{2} A_{2} V_{f 2}
We have to find out \rho_{2}
With the help of 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)
\frac{p_{2 t}}{p_{1 t}}=\left[1+\frac{0.9 \times(416-300)}{300}\right]^{\frac{1.4}{0.4}}
= 2.84
again, \frac{p_{2}}{p_{2 t}}=\left(\frac{T_{2}}{T_{2 t}}\right)^{\frac{1.4}{0.4}}=\left(\frac{363.33}{416}\right)^{\frac{1.4}{0.4}}=0.623
Hence
p_{2}=0.623 p_{2 t}
=0.623 \times 2.84 p_{1 t}
=0.623 \times 2.84 \times 100 kpa
= 176.93 kPa
Therefore, \dot{m}=\left(\frac{p_{2}}{R T_{2}}\right) \cdot A_{2} V_{f 2}
=\frac{176.93 \times 10^{3}}{287 \times 363.33} \times 0.1 \times 30
= 5.09 kg/s