Question 4.DE.10: A centrifugal compressor is required to deliver 8 kg/s of ai......
A centrifugal compressor is required to deliver 8 kg/s of air with a stagnation pressure ratio of 4 rotating at 15,000 rpm. The air enters the compressor at 25°C and 1 bar. Assume that the air enters axially with velocity of 145 m/s and the slip factor is 0.89. If the compressor isentropic efficiency is 0.89, find the rise in stagnation temperature, impeller tip speed, diameter, work input, and area at the impeller eye.
Inlet stagnation temperature:
T_{01} = T_a + \frac{C^2 _1} {2C_p} = 298 + \frac{145^2} {(2)(1005)} = 308.46 K
Using the isentropic P–T relation for the compression process,
T_{03^′} = T_{01} \left(\frac{P_{03}}{P_{01}}\right)^{(γ-1)/γ} = (308.46)(4)^{0.286} = 458.55 K
Using the compressor efficiency,
T_{02 } – T_{01} = \frac{(T_{02^′} – T_{01})}{η_c} = \frac{(458.55 – 308.46)}{0.89} = 168.64 K
Hence, work done on the air is given by:
W = C_p (T_{02} – T_{01}) = (1.005)(168.64) = 169.48 kJ/kg
But,
W = σU_2^2 = \frac{(0.89)(U_2) } {1000}, or :169.48 = 0.89U_2^2/1000
or:
U_2 = \sqrt{\frac{(1000)(167.49)}{0.89}} = 436.38 m/s iffi
Hence, the impeller tip diameter
D = \frac{60U_2} {\pi N} = \frac{(60)(436.38)}{\pi (15,000)} = 0.555 m
The air density at the impeller eye is given by:
ρ_1 = \frac{P_1} {RT_1} =\frac{(1)(100)}{(0.287)(298)} = 1.17 kg/m³
Using the continuity equation in order to find the area at the impeller eye,
A_1 = \frac{\dot{m} }{ρ_1C_1} = \frac{8}{(1.17)(145)} = 0.047 m²
The power input is:
P = \dot{m} W = (8)(169.48) = 1355.24 kW