Question 4.DE.15: The impeller of a centrifugal compressor rotates at 15,500 r......
The impeller of a centrifugal compressor rotates at 15,500 rpm, inlet stagnation temperature of air is 290 K, and stagnation pressure at inlet is 101 kPa. The isentropic efficiency of impeller is 0.88, diameter of the impellar is 0.56 m, axial depth of the vaneless space is 38 mm, and width of the vaneless space is 43 mm. Assume ship factor as 0.9, power input factor 1.04, mass flow rate as 16 kg/s. Calculate
1. Stagnation conditions at the impeller outlet, assume no fore whirl at the inlet,
2. Assume axial velocity approximately equal to 105 m/s at the impeller outlet, calculate the Mach number and air angle at the impeller outlet,
3. The angle of the diffuser vane leading edges and the Mach number at this radius if the diffusion in the vaneless space is isentropic.
1. Impeller tip speed
U_2 = \frac{\pi D_2N} {60} = \frac{\pi × 0.56 × 15500} {60}
U_2 = 454.67 m/s
Overall stagnation temperature rise
T_{03} – T_{01} = \frac{ψσU_2^2} {1005} = \frac{1.04 × 0.9 × 454.67^2} {1005}
= 192.53 K
Since T_{03} = T_{02}
Therefore, T_{02} – T_{01} = 192.53 K and T_{02} = 192.53 + 290 = 482.53 K
Now pressure ratio for impeller
\frac{p_{02}} {p_{01}} = \left(\frac{T_{02}} {T_{01}}\right)^{3.5} = \left(\frac{482.53} {290}\right)^{3.5} = 5.94
then, p_{02} = 5.94 × 101 = 600 KPa
2.
σ = \frac{C_{w2}} {U_2}
C_{w2} = σU_2
or
C_{w2} = 0.9 × 454.67 = 409 m/s
Let C_{r2} = 105 m/s
Outlet area normal to periphery
A_2 = \pi D_2 × impeller depth
= π × 0.56 × 0.038
A_2 = 0.0669 m²
From outlet velocity triangle
C^2_2 = C^2_{r2} + C^2 _{w2}
= 105² + 409²
C_2^2 = 178306
i.e. C_{2} = 422.26 m/s
T_2 = T_{02} – \frac{C_2^2}{2C_p} = 482.53 – \frac{422.26^2} {2 × 1005}
T_2 = 393.82 K
Using isentropic P–T relations
P_2 = P_{02} \left(\frac{T_{2}} {T_{02}}\right)^{\frac{γ}{γ-1}} = 600 \left(\frac{393.82}{482.53}\right)^{3.5} = 294.69 kPa
From equation of state
ρ_2 = \frac{P_2} {RT_2} = \frac{293.69 × 10^3} {287 × 393.82} = 2.61 kg/m³
The equation of continuity gives
C_{r2} = \frac{\dot{m}} {A_2P_2} = \frac{16} {0.0669 × 2.61} = 91.63 m/s
Thus, impeller outlet radial velocity = 91.63 m/s
Impeller outlet Mach number
M_2 = \frac{C_2}{\sqrt{γRT_2}} = \frac{422.26} {(1.4 × 287 × 393.82)^{0.5}}
M_2 = 1.06
From outlet velocity triangle
Cos α_2 = \frac{C_{r2}} {C_2} = \frac{91.63} {422.26} = 0.217
i.e., α_2 = 77.47°
3. Assuming free vortex flow in the vaneless space and for convenience denoting conditions at the diffuser vane without a subscript (r = 0.28 + 0.043 = 0.323)
C_w = \frac{C_{w2}r_2} {r} = \frac{409 × 0.28} {0.323}
C_w = 354.55 m/s
The radial component of velocity can be found by trial and error. Choose as a first try, C_r = 105 m/s
\frac{C^2}{C_p} = \frac{105^2 + 354.55^2} {2 × 1005} = 68 K
T = 482.53 – 68 (since T = T_{02} in vaneless space)
T = 414.53 K
p = p_{02} \left(\frac{T_2} {T_{02}}\right)^{3.5} = 600 \left(\frac{419.53} {482.53}\right)^{3.5} = 352.58 kPa
ρ = \frac{p_2} {RT_2} = \frac{294.69} {287 × 393.82}
ρ = 2.61 kg/m³
The equation of continuity gives
A = 2πr × depth of vanes
= 2π × 0.323 × 0.038
= 0.0772 m²
C_r = \frac{16} {2.61 × 0.0772} = 79.41 m/s
Next try C_r = 79.41 m/s
\frac{C^2} {2C_p} = \frac{79.41^2 + 354.55^2} {2 × 1005} = 65.68
T = 482.53 – 65.68 = 416.85 K
p = 359.54 kPa
ρ = \frac{359.54} {416.85 × 287} = 3 kg/m³
C_r = \frac{16} {3.0 × 0.772} = 69.08 m/s
Try C_r = 69.08 m/s
\frac{C^2} {2C_p} = \frac{69.08^2 + 354.55^2} {2 × 1005} = 64.9
T = 482.53 – 64.9 = 417.63 K
p = 361.9 Pa
ρ = \frac{361.9} {417.63 × 287} = 3.02 kg/m³
C_r = \frac{16} {3.02 × 0.772} = 68.63 m/s
Taking C_r as 62.63 m/s, the vane angle
tan α = \frac{C_w} {C_r}
= \frac{354.5} {68.63} = 5.17
i.e. α = 79°
Mach number at vane
M = \left(\frac{65.68 × 2 × 1005} {1.4 × 287 × 417.63}\right)^{1/2} = 0.787