Question 12.5: A centrifugal compressor has the following data: Deh = 0.12 ......

U_{\text{1h}}=\pi D_{\text{eh}}\frac{N}{60} =\pi \times 0.12 \times 300=113.1 \text{ m/s} \\ U_{\text{1m}}=\pi D_{\text{em}}\frac{N}{60} =\pi \times 0.18 \times 300=169.65 \text{ m/s} \\ U_{\text{1t}}=\pi \times D_{\text{et}}\frac{N}{60} =\pi \times 0.24 \times 300=226.2 \text{ m/s} \\ U_2=\pi \times D_{\text{et}}\frac{N}{60} =\pi \times 0.5 \times 300=471.24 \text{ m/s}

Case (A) No prewhirl:
Figure 12.24 illustrates the velocity triangles with no prewhirl.

T_1=T_{01}-\frac{C_1^2}{2C_{\text{P}}} =300-\frac{(100)^2}{2 \times 1005} =295 \text{K}

(i) At the hub

W_{\text{1h}}=\sqrt{U_{1\text{h}}^2+C^2_{\text{a}}} \\ W_{\text{1h}}= \sqrt{(113.1)^2+(100)^2}=150.97 \text{ m/s} \\ (M_{1_\text{rel}})_{\text{h}}=\frac{W_{\text{1h}}}{\sqrt{\gamma RT_1}} =\frac{150.97}{\sqrt{1.4 \times 287 \times 295}} =0.4385 \\ \tan \beta_{\text{1h}}=\frac{C_{\text{a}}}{U_{\text{1h}}} =\frac{100}{113.1} \\ \beta_{\text{1h}}=41.48

(ii) At the mean

W_{1\text{m}}=\sqrt{U^2_{\text{1m}}+C^2_{\text{a}}} \\ W_{1\text{m}}=\sqrt{(169.65)^2+(100)^2}=196.93 \text{ m/s} \\ (M_{1_{\text{rel}}})_{\text{m}}=\frac{W_{1\text{m}}}{\sqrt{\gamma RT_1}} =\frac{196.93}{\sqrt{1.4 \times 287 \times 295}} =0.572

(iii) At the tip

W_{1\text{t}}=\sqrt{U^2_{\text{1t}}+C_{\text{a}}^2} \\ W_{1\text{t}}=\sqrt{(226.2)^2+(100)^2}=247.32 \text{ m/s} \\ (M_{1_\text{rel}})_{\text{t}}=\frac{W_{1\text{t}}}{\sqrt{\gamma RT_1}} =\frac{247.32}{\sqrt{1.4 \times 287 \times 295}} =0.7184 \\ \pi_{\text{c}}=\frac{P_{03}}{P_{03}} =\left(1+\eta_{\text{c}}\frac{\sigma U^2_2-U_1 C_{\text{u1}}}{C_{\text{P}}T_{01}} \right) ^{\gamma/(\gamma-1)} \\ \pi_{\text{c}}=\left(1+0.85\frac{0.92 \times (471.24)^2-0}{1005 \times 300} \right) ^{1.4/0.4}=4.914

Case (B) Constant prewhirl angle (ζ = 25):
Figure 12.25 illustrates the velocity triangles with constant prewhirl angle. Since the same axial velocity is kept constant in all cases,

C_1=\frac{C_{\text{a}}}{\cos \zeta} =\frac{100}{\cos 25} =110.34 \text{ m/s}

(i) At the hub

C_{\text{u1h}}=C_{\text{a}} \tan \zeta=100 \tan 25=46.63 \text{ m/s}=C_{\text{u1m}}=C_{\text{u1h}}

Note that

C_{\text{u1h}}=C_{\text{u1m}}=C_{\text{u1t}} \\ W_{1\text{h}}=\sqrt{(U_{\text{1h}}-C_{\text{u1h}})^2+C_{\text{a}}^2} \\ W_{\text{1h}}=\sqrt{(113.1-46.63)^2+(100)^2}=120.07 \text{ m/s} \\ (M_{1_{\text{rel}}})_{\text{h}}=\frac{W_{\text{1h}}}{\sqrt{\gamma RT_1}} =\frac{120.07}{\sqrt{1.4 \times 287 \times 293.94}} =0.3494

(ii) At the mean

W_{1\text{m}}=\sqrt{(U_{\text{1m}}-C_{\text{u1m}})^2+C^2_{\text{a}}} \\ W_{\text{1h}}=\sqrt{(169.65-46.63)^2+(100)^2}=158.54 \text{ m/s} \\ (M_{1_{\text{rel}}})_{\text{m}}=\frac{W_{\text{1m}}}{\sqrt{\gamma RT_1}} =\frac{158.54}{\sqrt{1.4 \times 287 \times 293.94}} =0.4613

(iii) At the tip

W_{1\text{t}}=\sqrt{(U_{\text{1t}}-C_{\text{u1t}})^2+C^2_{\text{a}}} \\ W_{\text{1t}}=\sqrt{(226.2-46.63)^2+(100)^2}=205.54 \text{ m/s} \\ (M_{1_{\text{rel}}})_{\text{t}}=\frac{W_{\text{1t}}}{\sqrt{\gamma RT_1}} =\frac{205.54}{\sqrt{1.4 \times 287 \times 293.94}} =0.598 \\ \pi_{\text{c}}=\frac{P_{03}}{P_{03}} =\left(1+ \eta_{\text{c}}\frac{\sigma U^2_2-U_1C_{\text{u1}}}{C_{\text{P}}T_{01}} \right) ^{(\gamma/\gamma-1)} \\ \pi_{\text{c}}=\left(1+0.85\frac{0.92 \times (471.24)^2-169.65 \times 46.63}{1005 \times 300} \right) ^{(1.4/0.4)}=4.6747

Case (C) Constant inlet relative angle (\beta_1 = 41.48):
Figure 12.26 illustrates the velocity triangles with constant inlet relative angle.

(i) At the hub

T_1=T_{01}-\frac{C_1^2}{2C_{\text{P}}} =300-\frac{(100)^2}{2 \ast 1005} =295 \text{ K} \\ W_{\text{1h}}=\sqrt{U_{\text{1h}}^2+C^2_{\text{a}}} \\ W_{\text{1h}}=\sqrt{(113.1)^2+(100)^2}=150.97 \text{ m/s}=W_{\text{1m}}=W_{\text{1t}} \\ (M_{1_\text{rel}})_{\text{h}}=\frac{W_{\text{1h}}}{\sqrt{\gamma RT_1}} =\frac{150.97}{\sqrt{1.4 \times 287 \times 295}} =0.4385

(ii) At the mean

(C_{\text{u1}})_{\text{m}}=U_{\text{m}}-U_{\text{h}}=169.65-113.1=56.55 \\ C_{\text{1m}}=\sqrt{C^2_{\text{u1m}}+C_{\text{a}}^2}=\sqrt{(56.55)^2+(100)^2}=114.88 \text{ m/s} \\ T_1=T_{01}-\frac{C_1^2}{2C_{\text{P}}} =300-\frac{(114.88)^2}{2 \times 1005} =293.43 \text{ K} \\ (M_{1_\text{rel}})_{\text{m}}=\frac{W_{\text{1m}}}{\sqrt{\gamma RT_1}} =\frac{150.97}{\sqrt{1.4 \times 287 \times 293.43}} =0.4397

(iii) At the tip

(C_{\text{u1}})_{\text{t}}=U_{\text{t}}-U_{\text{h}}=226.2-113.1=113.1 \text{ m/s} \\ C_{\text{1t}}=\sqrt{C^2_{\text{u1t}}+C_{\text{a}}^2}=\sqrt{(113.1)^2+(100)^2}=150.96 \text{ m/s} \\ T_1=T_{01}-\frac{C_1^2}{2C_{\text{P}}} =300-\frac{(150.96)^2}{2 \times 1005} =288.66 \text{ K} \\ (M_{1_\text{rel}})_{\text{t}}=\frac{W_{\text{1t}}}{\sqrt{\gamma RT_1}} =\frac{150.97}{\sqrt{1.4 \times 287 \times 288.66}} =0.4433 \\ \pi_{\text{c}}=\frac{P_{03}}{P_{01}} =\left(1+\eta_{\text{c}}\frac{\sigma \text{U}^2_2-\overline{U_1C_{\text{u1}}}}{C_{\text{P}}T_{01}} \right) ^{\gamma/\gamma-1} \\ \overline {U_1C_{\text{u1}}}=\frac{\int \limits^{r_{\text{t}}}_{r_{\text{h}}}UC_{\text{u}} \ dr}{r_{\text{t}}-r_{\text{h}}} =\frac{\int \limits^{r_{\text{t}}}_{r_{\text{h}}}\omega r C_{\text{u}} \ dr}{r_{\text{t}}-r_{\text{h}}}

but

C_{\text{u}}=U-C_{\text{a}}\tan \beta \quad \text{or}\quad C_{\text{u}}=U-A \\ \overline{U_1C_{\text{u1}}}=\frac{\omega \int^{r_{\text{t}}}_{r_{\text{h}}}r(U-A)dr}{r_{\text{t}}-r_{\text{h}}} =\frac{\omega \int^{r_{\text{t}}}_{r_{\text{h}}} r(\omega r-A)dr}{r_{\text{t}}-r_{\text{h}}} \\ \overline{U_1C_{\text{u1}}}= \frac{\omega \int^{r_{\text{t}}}_{r_{\text{h}}}(\omega r^2-Ar)dr}{r_{\text{t}}-r_{\text{h}}} \\ \overline{U_1C_{\text{u1}}}=\frac{\omega}{r_{\text{t}}-r_{\text{h}}} \left[\omega \frac{r^3}{3}-A\frac{r^2}{2} \right] ^{r_{\text{t}}}_{r_{\text{h}}} \\ \overline{U_1C_{\text{u1}}}=\frac{\omega}{r_{\text{t}}-r_{\text{h}}} \left[\frac{\omega}{3}(r^3_{\text{t}}-r^3_{\text{h}}) -\frac{A}{2}(r^2_{\text{t}}-r^2_{\text{h}}) \right] \\ \overline{U_1C_{\text{u1}}}= \omega\left[\frac{\omega}{3}(r^2_{\text{t}}+r_{\text{t}}r_{\text{h}}+r^2_{\text{h}})-\frac{A}{2}(r_{\text{t}}+r_{\text{h}}) \right] \\ \overline{U_1C_{\text{u1}}}=\frac{1}{6}\left[2U_{\text{t}}^2+2U_{\text{t}}U_{\text{h}}+2U_{\text{h}}^2-3AU_{\text{t}}-3AU_{\text{h}}\right] \\ \overline{U_1C_{\text{u1}}} =\frac{1}{6} [2U_{\text{t}}(U_{\text{t}}-A)+2U_{\text{h}}(U_{\text{h}}-A)+U_{\text{t}}(U_{\text{h}}-A)+U_{\text{h}}(U_{\text{t}}-A)] \\ \overline{U_1C_{\text{u1}}}=\frac{1}{6} [2U_{\text{t}}C_{\text{ut}}+2U_{\text{h}}C_{\text{uh}}+U_{\text{t}}C_{\text{uh}}+U_{\text{h}}C_{\text{ut}}]

But C_{\text{uh}}=0

\therefore \overline{U_1C_{\text{u1}}}=\frac{1}{6} [2U_{\text{t}}C_{\text{ut}}+U_{\text{h}}C_{\text{ut}}]=\frac{1}{6}[2 \times 226.2 \times 113.1 +113.1 \times 113.1] =10659.675 \text{ (m/s)}^2 \\ \pi_{\text{c}}=\left(1+\eta_{\text{c}}\frac{\sigma U^2_2-\overline{U_1C_{\text{u1}}}}{C_{\text{P}}T_{01}} \right) ^{\gamma/(\gamma-1)} \\ \pi_{\text{c}}=\left(1+0.85\frac{0.92(471.24)^2-10,659.675}{1005 \times 300} \right) ^{1.4/0.4}=4.5936

The above results can be summarized in Table 12.1.
Table 12.1 shows that the selection will be in favor of the constant relative angle (β) as it achieves the lowest relative Mach number at tip. Moreover, the pressure ratio is close to the constant prewhirl angle (ζ ).