Question 4.9: Repeat problem 4.8, assuming the air density at the impeller......
Repeat problem 4.8, assuming the air density at the impeller tip is 1.8 kg/m³ and the axial width at the entrance to the diffuser is 12 mm. Determine the radial velocity at the impeller exit and the absolute Mach number at the impeller tip.
Slip factor: σ = \frac{C_{w2}} {U_2}, or C_{w2} = (0.8958)(449.9) = 403 m/s
Using the continuity equation,
\dot{m} = ρ_2A_2C_{r2} = ρ_22πr_2b_2C_{r2}
where:
b_2 = axial width
r_2 = radius
Therefore:
C_{r2} = \frac{2.5}{(1.8)(2\pi)(0.25)(0.012)} = 73.65 m/s
Absolute velocity at the impeller exit
C_2 = \sqrt{C_{r2}^2 + C_{w2}^2} = \sqrt{73.65^2 + 403^2} = 409.67 m/s
The temperature equivalent of work done:
T_{02} – T_{01} = 188.57/C_p = 188.57/1.005 = 187.63 K
Therefore, T_{02} = 293 + 187.63 = 480.63 K
Hence the static temperature at the impeller exit is:
T_2 = T_{02} – \frac{C_2^2} {2C_p} = 480.63 – \frac{409.67^2}{(2)(1005)} = 397 K
Now, the Mach number at the impeller exit is:
M_2 = \frac{C_2}{\sqrt{γRT_2}} = \frac{409.67}{\sqrt{(1.4)(287)(397)}} = 1.03