Question 4.DE.12: Recalculate the maximum Mach number at the impeller eye for ......
Recalculate the maximum Mach number at the impeller eye for the same data as in the previous question, assuming prewhirl angle of 20°.
Figure 4.16 shows the velocity triangle with the prewhirl angle.
From the velocity triangle:
C_1 = \frac{145} {\cos(20°)} = 154.305 m/s
Equivalent dynamic temperature:
\frac{C^2 _1} {2C_p} = \frac{154.305^2}{(2)(1005)} = 11.846 K
C_{w1} = \tan(20°) C_{a1} = (0.36)(145) = 52.78 m/s
Relative velocity at the inlet:
V^2_1 = C^2_a + (U_e – C_{w1})^2 = 145² + (220 – 52.78)², or V_1
= 221.3 m/s
Therefore the static temperature at the inlet:
T_1 = T_{01} – \frac{C^2_1} {2C_p} = 290 – 11.846 = 278.2 K
Hence,
M_1 = \frac{V_1}{\sqrt{γRT_1}} = \frac{221.3}{\sqrt{(1.4)(287)(287.2)}} = 0.662
Note the reduction in Mach number due to prewhirl.