Question 4.DE.14: A centrifugal compressor compresses air at ambient temperatu......
A centrifugal compressor compresses air at ambient temperature and pressure of 288 K and 1 bar respectively. The impeller tip speed is 364 m/s, the radial velocity at the exit from the impeller is 28 m/s, and the slip factor is 0.89. Calculate the Mach number of the flow at the impeller tip. If the impeller total-to-total efficiency is 0.88 and the flow area from the impeller is 0.085 m², calculate the mass flow rate of air. Assume an axial entrance at the impeller eye and radial blades.
The absolute Mach number of the air at the impeller tip is:
M_2 = \frac{C_2}{\sqrt{γRT_2}}
where T_2 is the static temperature at the impeller tip. Let us first calculate C_2 and T_2.
Slip factor:
σ = \frac{C_{w2}} {U_2}
Or:
C_{w2} = σU_2 = (0.89)(364) = 323.96 m/s
From the velocity triangle,
C_2^2 = C^2_{r2} + C^2_{w2} = 28^2 + 323.96^2 = (1.06)(105) m^2/s^2
With zero whirl at the inlet
\frac{W} {m} = σU_2^2 = C_p (T_{02} – T_{01})
Hence,
T_{02} = T_{01} + \frac{σU_2^2} {C_p} = 288 + \frac{(0.89)(364^2)}{1005} = 405.33 K
Static Temperature
T_{2} = T_{02} – \frac{C_2^2} {2C_p} = 405.33 – \frac{106000}{(2)(1005)} = 352.6 K
Therefore,
M_2 = \left(\frac{(1.06)(10^5)} {(1.4)(287)(352.6)}\right)^{\frac{1}{2}} = 0.865
Using the isentropic P–T relation:
\left(\frac{P_{02}} {P_{01}}\right) = \left[1 + η_c \left(\frac{T_{02}} {T_{01}} – 1\right)\right]^{γ/(γ-1)}
= \left[1 + 0.88 \left(\frac{405.33} {288} – 1\right)\right]^{3.5} = 2.922
\left(\frac{P_{2}} {P_{02}}\right) = \left(\frac{T_{2}} {T_{02}}\right)^{3.5} = \left(\frac{352.6}{405.33}\right)^{3.5} = 0.614
Therefore,
P_2 = \left(\frac{P_{2}} {P_{02}}\right)\left(\frac{P_{02}} {P_{01}}\right)P_{01}
= (0.614)(2.922)(1)(100)
= 179.4 kPa
ρ_2 = \frac{179.4(1000)} {287(352.6)} =1.773 kg/m³
Mass flow:
\dot{m} = (1.773)(0.085)(28) = 4.22 kg/s