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 (i) the absolute Mach number of flow leaving the radial vanes of the impeller and (ii) 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 \mathrm{~m}^{2} |
| Power input factor | 1.0 |
| Isentropic efficiency | 0.9 |
| Inlet stagnation temperature | 300 K |
| Inlet stagnation pressure | 100 \mathrm{kN} / \mathrm{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} and T_{2} have to be determined.
From the velocity triangle at the impeller exit,
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 \mathrm{~m} / \mathrm{s}
From Eq. (16.5)
w=\Psi \sigma U_{2}^{2}=c_{p}\left(T_{2 t}-T_{1 t}\right)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 \mathrm{~J} / \mathrm{kg} \mathrm{K}\right]
T_{2 t}=300+\frac{0.9 \times(360)^{2}}{1005}
=416 \mathrm{~K}
T_{2}=T_{2 t}-\frac{V_{2}^{2}}{2 c_{p}}
=416-\frac{(325.38)^{2}}{2 \times 1005}
=363.33 \mathrm{~K}
Therefore, M_{2}=\frac{325.28}{\sqrt{1.4 \times 287 \times 363.33}}
=0.85Mass 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
=\left[1+\frac{\eta_{c}\left(T_{3 t}-T_{1 t}\right)}{T_{1 t}}\right]^{\gamma \gamma-1} (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 \mathrm{kPa}
=176.93 \mathrm{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 \mathrm{~kg} / \mathrm{s}