Question 4.DE.11: The following data apply to a double-sided centrifugal compr......

(1) Let U_{er} be the impeller speed at the eye root. Then the vane angle at the eye root is:
α_{er } = \tan^{-1} \left(\frac{C_a} {U_{er}}\right)
and
U_{er} = \frac{\pi D_{er}N}{60} = \frac{\pi (0.14)(15,000)} {60} = 110 m/s
Hence, the vane angle at the impeller eye root:
α_{er } = \tan^{-1} \left(\frac{C_a} {U_{er}}\right) = \tan^{-1} \left(\frac{145} {110}\right) = 52°48′

Impeller velocity at the eye tip:
U_{et} = \frac{\pi D_{et}N} {60} = \frac{\pi (0.28)(15,000)} {60} = 220 m/s
Therefore vane angle at the eye tip:
α_{et } = \tan^{-1} \left(\frac{C_a} {U_{et}}\right) = \tan^{-1} \left(\frac{145} {220}\right) = 33°23′

(2) Work input:
W = \dot{m} ψσU_2^2 = (10)(0.819)(1.03U_2^2)
but:

U_{2} = \frac{\pi D_{2}N}{60} = \frac{\pi(0.48)(15,000)}{60} = 377.14 m/s
Hence,
W = \frac{(10)(0.89)(1.03)(377.14^2)}{1000} = 1303.86 kW
(3) The relative velocity at the eye tip:
V_1 = \sqrt{U_{et}^2 + C_a^2} = \sqrt{220^2 + 145 ^2} = 263.5 m/s
Hence, the maximum relative Mach number at the eye tip:
M_1 = \frac{V_1}{\sqrt{γRT_1}},
where T_1 is the static temperature at the inlet
T_1 = T_{01}  –  \frac{C_1^2}{2C_p} = 290 – \frac{145^2}{(2)(1005)} = 279.54 K
The Mach number at the inlet then is:
M_1 = \frac{V_1}{\sqrt{γRT_1}} = \frac{263/5}{\sqrt{(1.4)(287)(279.54)}} = 0.786