Question 12.6: A single-sided centrifugal compressor is to deliver 14 kg/s ......
A single-sided centrifugal compressor is to deliver 14 kg/s of air when operating at a pressure ratio of 4:1 and a speed of 200 rps. The inlet stagnation conditions may be taken as 288 K and 0.1 MPa. The number of blades is 20, power input factor is 1.04, the compressor isentropic efficiency is 0.8. If the Mach number is not to exceed unity at the impeller tip, then 60% of the losses are assumed to occur in the impeller.
Find
1. The overall diameter of the impeller if its blades are radial.
2. The minimum possible axial depth of the diffuser for the impeller is described in (a).
3. If the radial impeller in (a) is to be replaced by a backward-leaning impeller having the same diameter, same flow coefficient at impeller outlet \phi_2=C_{\text{r2}}/U_2 and the backswept angle \beta^\prime_2=30^\circ, find the corresponding pressure ratio and temperature rise.
4. Same as in (c) but a forward-leaning impeller is employed with \beta^\prime_2=-30^\circ.
5. If the three impellers (radial-, forward-, and backward-leaning) are to have the same pressure ratio, same flow coefficient \phi_2 , same rotational speed (N = 200 rps), and same isentropic efficiency, then find the corresponding impeller diameter d_2 and tip speed U_2 .
1. For a radial impeller having 20 blades, the slip factor is
\therefore \sigma_{S}=1-\left(\frac{0.63 \pi}{N_{\text{b}}} \right) =0.901
The compressor pressure ratio is related to the impeller tip speed U_2 by the relation
\pi_{\text{c}}^{(\gamma-1)/\gamma}=\left(1+\frac{\eta_{\text{c}}\sigma \psi U^2_2}{C_{\text{P}}T_{01}} \right) \\ \therefore 4^{(1.4-1)/1.4}=1+\frac{0.8 \times 1.04 \times 0.901 \times U_2^2}{1005 \times 288} \\ \therefore U_2=435.16 \text{ m/s}
Then, the impeller tip diameter is
d_2=\frac{U_2}{\pi \times N} =\frac{435.16}{\pi \times 200} =0.69 \text{m}
The temperature rise across the impeller is
\Delta T_0=T_{02}-T_{01}=\frac{\sigma \psi U_2^2}{C_{\text{P}}} =\frac{0.901 \times 1.04 \times 18.94 \times 10^4}{1005} =175.6\text{K}.
The total temperature at the impeller outlet is thus T_{02}=T_{03}=463.6 \text{ K}. Figure 12.29 illustrates a typical T-S diagram that illustrates both of the impeller and diffuser losses.
From Equation 12.21, the impeller efficiency
\eta_{\text{imp}}=1-\lambda +\lambda \eta_{\text{c}} \\ \eta_{\text{imp}}=1-0.6+0.6 \times 0.8=0.88
The pressure ratio across the impeller is
\frac{P_{02}}{P_{01}} =\left[1+\eta_{\text{imp}}\left(\frac{T_{02}-T_{01}}{T_{01}} \right) \right] ^{\gamma/(\gamma-1)} \\ \frac{P_{02}}{P_{01}} =\left[1+0.88\left(\frac{175.6}{288} \right) \right] ^{1.4/0.4}=(1.5366)^{3.5}=4.498
The total pressure at the impeller outlet P_{02} = 4.498 bar.
2. Since the Mach number at impeller outlet must not exceed unity, the absolute velocity at outlet is taken equal to the sonic speed, then C_2 = sonic speed = \sqrt{\gamma RT_2} . The outlet flow from impeller is illustrated by Figure 12.30. The total to static temperature in this case is expressed as T_{02}/T_2=\gamma+1/2=(1.4+1)/2=1.2.
Then, the static temperature is T_2 = 386.8 K, and the absolute velocity is
C_2=\sqrt{1.4 \times 287 \times 386.8}=394 \text{ m/s}
The total and static pressure ratio is also
\frac{P_{02}}{P_2} =\left(\frac{\gamma+1}{2} \right) ^{\gamma/(\gamma-1)}=(1.2)^{3.5}=1.8929
Thus P_2=4.498/1.8929=2.376 \text{ bar}.
The air density at the impeller outlet is
\rho_2=\frac{P_2}{RT_2} =\frac{2.376 \times 10^5}{287 \times 386.8} =2.14 \text{ kg/m}^3
The mass flow rate is \dot{m}=\rho_2C_{\text{r}_2}A_2
\dot{m}=\rho_2C_{\text{r}_2}(\pi \times d_2 \times b_2) \quad \quad \quad \text{(a)} \\ \dot{m}=\rho_2\sqrt{C_2^2-(\sigma U_2)^2}(\pi d_2b_2)=\rho_2\sqrt{C_2^2-(\sigma U_2)^2}(\pi d_2 b_2)
With a constant \dot{m} and rotational speed at maximum (C_2), the impeller width b_2 will be a minimum.
C_{\text{r}_2}^2=C_2^2-(\sigma U_2)^2=(394)^2-(0.9 \times 435)^2 \\ C_{\text{r}_2}=45.5 \text{ m/s}
From Equation (a)
\therefore b_2=\frac{\dot{m}}{\rho_2C_{\text{r}_2}\pi d_2} =\frac{14}{(2.14)(45.5)(\pi \times 0.69)} =\frac{14}{211.0} =0.066 \text{ m}
3. Backward-leaning impeller (\beta^\prime_2)=30^\circ
\phi_2=\frac{C_{\text{r}_2}}{U_2} =\frac{W_{\text{r}_2}}{U_2} =\frac{45.5}{435} =0.1046
Slip factor is given by the relation
\sigma_{\text{S}}=1-\frac{0.63(\pi/N_{\text{b}})}{1-\phi_2 \tan \beta^\prime_2} =1-\frac{0.63\pi/20}{1-(0.1046)\times \tan 30^\circ} =0.8951
The total sonic speed is a_{01}=\sqrt{\gamma RT_{01}}=\sqrt{1.4 \times 287 \times 288}=340 \text{ m/s}.
The pressure ratio is determined from Equation 12.19 as follows:
The temperature ratio is
\frac{T_{02}}{T_{01}} =\frac{T_{03}}{T_{01}} =1+\frac{\pi_{\text{c}}^{(\gamma-1)/\gamma}-1}{\eta_{\text{c}}} =1+\frac{1.4405-1}{0.8}
Then, the impeller and compressor outlet temperature is T_{02} = 446.6 K.
4. Forward-leaning impeller (\beta_2^\prime)=-30^\circ
The slip factor is
\sigma_{\text{S}}=1-\frac{0.63 \pi /N_{\text{b}}}{1-\phi_2\tan \beta_2^\prime} =1-\frac{0.63\pi/20}{1-0.1046 \times (-0.5774)} =0.9067.
The pressure ratio calculated from Equation 12.19 is
The corresponding temperature ratio is
\frac{T_{03}}{T_{01}} =1+\frac{\pi_{\text{c}}^{\gamma/(\gamma-1)}-1}{\eta_{\text{c}}} =1+\frac{1.50365-1}{0.8} =1.6296
Then, the outlet temperature is T_{03} = 469.3 K. Figure 12.31 illustrates the temperature rise for different impellers having a constant diameter.
5. If the pressure rise in backward-leaning impeller = pressure rise in radial impeller
\left(\frac{P_{03}}{P_{01}} \right) ^{(\gamma-1)/\gamma}=\left[1+(\gamma-1)\eta_{\text{c}}\sigma_{\text{S}_{\text{b}}}\left(\frac{U_2}{a_{01}} \right)^2_{\text{b}} (1-\phi_2 \tan \beta^\prime_2)_{\text{b}} \right] =\left[1+(\gamma-1)\eta_{\text{c}}\sigma_{\text{S}_{\text{r}}}\left(\frac{U_2}{a_{01}} \right) _{\text{r}}\right]
Subscripts (b) and (r) stand for backward-leaning and radial impellers.
\therefore \sigma_{\text{S}_{\text{b}}}U^2_{2_\text{b}}(1-\phi_2 \tan \beta_2^\prime)_{\text{b}}=\sigma_{\text{S}_{\text{r}}}U^2_{2_{\text{r}}} \\ \left(\frac{U_{2_{\text{b}}}}{U_{2_{\text{r}}}} \right) ^2=\frac{\sigma_{\text{S}_{\text{r}}}}{\sigma_{\text{S}_{\text{b}}}(1-\phi_2 \tan \beta^\prime_2)} _{\text{b}}=\frac{0.901}{0.8951(1-0.1046 \times 0.5774)} =1.0713 \\ U_{2_{\text{b}}}=1.035, \quad U_{2_{\text{r}}}=450 \text{ m/s}
The diameter ratio between the backward and radial impellers is d_{2_\text{b}}/d_{2_\text{r}}=U_{2_\text{b}}/U_{2_\text{r}}=1.035.
Then, the diameter of the backward impeller is d_{2_\text{b}}=0.714 \text{ m}.
Moreover, \phi_2=0.1046=C_{\text{r}_2}/U_2 .
The radial velocity at the impeller outlet is C_{\text{r}_2}=47.07 \text{ m/s}.
For a forward-leaning impeller, by equating the pressure rise in forward-leaning and radial impellers, \sigma_{\text{S}_{\text{f}}}(U_2/a_{01})^2_{\text{f}}(1-\phi_2 \tan \beta_2^\prime)_{\text{f}}=\sigma_{\text{S}_{\text{r}}}(U_2/a_{01})_{\text{r}}^2.
Where subscript (f) stands for forward-leaning impeller, the rotational speed is
U_{2_{\text{f}}}=U_{2_{\text{r}}}\left\{\frac{\sigma_{\text{S}_{\text{r}}}}{\sigma_{\text{S}_{\text{f}}}(1-\phi_2 \tan \beta^\prime_2)_{\text{f}}} \right\} ^{1/2}=435\left\{\frac{0.901}{0.9067(1+0.1046 \times 0.5774)} \right\} ^{1/2}=421.09 \text{ m/s}
Then, the tip diameter is d_{2_{\text{f}}}=d_{2_{\text{r}}}(U_{2_{\text{f}}}/U_{2_{\text{r}}})=0.69 \times 0.968=0.6679 \text{ m}.
C_{\text{r}_2}=\phi_2 \times U_2=0.1046 \times 421.09=44.05 \text{ m/s}
A comparison for the sizes of three compressors are shown in Figure 12.32.
