Question 12.2: At inlet of a centrifugal compressor eye, the relative Mach ......
At inlet of a centrifugal compressor eye, the relative Mach number is to be less than 0.97 to eliminate the formation of shock waves and its attendant losses. The hub to tip radius ratio of the inducer is 0.4. There is no IGV and the inlet velocity is axial and uniform over the impeller eye. The eye tip diameter is 20 cm.
Determine
a. The maximum mass flow rate throughout for a given rotational speed of 486 rps
b. The blade angle at the inducer tip corresponding to this mass flow rate (Take air inlet conditions to be 101.3 kPa, 288 K.)
The given data are
M_{1_{\text{rel}_{\text{t}}}}=0.97, \quad \zeta =\frac{r_{\text{h}}}{r_{\text{t}}} =0.4, \quad D_{\text{e}_{\text{t}}}=20 \text{ cm} \\ N=486 \text{ rps}, \quad P_{01}=101.3 \text{ kPa}, \quad T_{01}=288 \text{ K}
The rotational speed at the inducer tip is
U_{1_{\text{t}}}=\pi D_{\text{e}_{\text{t}}}N=\pi \times 0.2 \times 486=305.363 \text{ m/s}
The maximum relative Mach number is at the eye tip (refer to Figure 12.20) will be:
M_{1_{\text{rel}}}=\frac{W_{1_{\text{t}}}}{\sqrt{\gamma R T_1}} =\frac{\sqrt{C_1^2+U_{1_{\text{t}}}^2}}{\sqrt{\gamma R \left(T_{01}-\left(C_1^2/2C_{\text{P}}\right) \right) }} =\frac{\sqrt{C_1^2+U^2_{1_{\text{t}}}}}{\sqrt{ \gamma RT_{01}-\left((\gamma-1)\right)C_1^2/2 }} \\ 0.97^2=\frac{C_1^2+305.363^2}{115,718.4-0.2C_1^2} \\ C_1^2+93, \quad 246.443=108, \quad 879.44-0.19C_1^2 \\ 1.19 C_1^2 =15,633
The absolute velocity at inlet is
C_1=114.62 \text{ m/s} \\ T_1=T_{01}-\frac{C_1^2}{2C_{\text{P}}} =281.464 \text{ K} \\ \frac{P_{01}}{P_1} =\left(\frac{T_{01}}{T_1} \right) ^{\gamma/(\gamma-1)} \\ P_1=93.48 \text{ kPa} \\ \rho_1 =\frac{P_1}{RT_1} =1.157 \text{ kg/m}^3 \\ A_1 =\frac{\pi}{4} \left(D_{\text{e}_{\text{t}}}^2-D_{\text{e}_{\text{r}}}^2\right) =\frac{\pi}{4} D_{\text{e}_{\text{t}}}^2(1-\zeta^2) \\ A_1=0.0264 \text{m}^2
Since the flow is axial, C_{\text{a}1}=C_1 . The mass flow rate is calculated from the inlet conditions as
\dot{m}=\rho_1C_{\text{a}1}A_1=1.157 \times 114.62 \times 0.0264 =3.5 \text{ kg/s}
The blade inlet angle at the tip \beta_{1_{\text{t}}} is given from the relation
\tan \beta_{1_\text{t}}=\frac{C_{1}}{U_{1_{\text{t}}} } \\ \beta_{1_{\text{t}}}=20.57 ^ \circ
