Question 4.DE.17: In a centrifugal compressor air enters at a stagnation tempe......
In a centrifugal compressor air enters at a stagnation temperature of 288 K and stagnation pressure of 1.01 bar. The impeller has 17 radial vanes and no inlet guide vanes. The following data apply:
Mass flow rate: 2.5 kg/s
Impeller tip speed: 475 m/s
Mechanical efficiency: 96%
Absolute air velocity at diffuser exit: 90 m/s
Compressor isentropic efficiency: 84%
Absolute velocity at impeller inlet: 150 m/s
Diffuser efficiency: 82%
Axial depth of impeller: 6.5 mm
Power input factor: 1.04
γ for air: 1.4
Determine:
1. shaft power
2. stagnation and static pressure at diffuser outlet
3. radial velocity, absolute Mach number and stagnation and static pressures at the impeller exit, assume reaction ratio as 0.5, and
4. impeller efficiency and rotational speed
1. Mechanical efficiency is
η_m = \frac{\text{Work transferred to air}} {\text{Work supplied to shaft}}
or shaft power = \frac{W} {η_m}
for vaned impeller, slip factor, by Stanitz formula is
σ = 1 – \frac{0.63\pi} {n} = 1 – \frac{0.63 × \pi} {17}
σ = 0.884
Work input per unit mass flow
W = ψσU_2C_{w2}
Since C_{w1} = 0
= ψσU_2^2
= 1.04 × 0.884 × 475²
Work input for 2.5 kg/s
W = 1.04 × 0.884 × 2.5 × 475²
W = 518.58 K
Hence; Shaft Power = \frac{518.58} {0.96} = 540.19 kW
2. The overall pressure ratio is
\frac{p_{03}} {p_{01}} = \left[1 +\frac{η_cψσU_2^2}{C_pT_{01}}\right]^{γ/(γ-1)}
= \left[1 + \frac{0.84 × 1.04 × 0.884 × 475^2}{1005 × 288}\right]^{3.5} = 5.2
Stagnation pressure at diffuser exit
P_{03} = p_{01} × 5.20 = 1.01 × 5.20
P_{03} = 5.25 bar
\frac{p_3} {p_{03}} = \left(\frac{T_3}{T_{03}}\right)^{γ/γ-1}
W = m × C_p (T_{03} – T_{01})
∴ T_{03} = \frac{W} {mC_p} + T_{01} = \frac{518.58 × 10^3} {2.5 × 1005} + 288 = 494.4 K
Static temperature at diffuser exit
T_3 = T_{03} – \frac{C^2_3} {2C_p} = 494.4 – \frac{90^2} {2 × 1005}
T_3 = 490.37 K
Static pressure at diffuser exit
p_3 = p_{03} \left(\frac{T_3}{T_{03}}\right)^{γ/γ-1} = 5.25 \left(\frac{490.37}{494.4}\right)^{3.5}
p_3 = 5.10 bar
3. The reaction is
0.5 = \frac{T_2 – T_1} {T_3 – T_1}
and
T_3 – T_1 = (T_{03} – T_{01}) + \left(\frac{C^2_1 – C^2_3} {2C_p}\right) = \frac{W} {mC_p} + \frac{150^2 – 90^2} {2 × 1005}
= \frac{518.58 × 10^3} {2.5 × 1005} + 7.164 = 213.56 K
Substituting
T_2 – T_1 = 0.5 × 213.56
= 106.78 K
Now
T_2 = T_{01} – \frac{C^2_1} {2C_p} + (T_2 – T_1)
= 288 – 11.19 + 106.78
T_2 = 383.59 K
At the impeller exit
T_{02} = T_2 + \frac{C^2 _2 }{2C_p}
or
T_{03} = T_2 + \frac{C^2_2} {2C_p} (Since T_{02} = T_{03})
Therefore,
C^2 _2 = 2C_p[(T_{03} – T_{01}) + (T_{01} – T_2)]
= 2 × 1005(206.4 + 288 + 383.59)
C_2 = 471.94 m/s
Mach number at impeller outlet
M_2 = \frac{C_2} {(1.4 × 287 × 383.59)^{1/2}}
M_2 = 1.20
Radial velocity at impeller outlet
C^2_{r2} = C^2_ 2 – C^2 _{w2}
= (471.94)² – (0.884 × 475)²
C^2_{r2} = 215.43 m/s
Diffuser efficiency is given by
η_D = \frac{h_{3^′} – h_2} {h_3 – h_2} = \frac{\text{isentropic enthalpy increase}} {\text{actual enthalpy increase}} = \frac{T_{3^′} – T_2} {T_3 – T_2}
= \frac{T_2 \left(\frac{T_{3^′}}{T_2} – 1\right)}{T_3 – T_2} = \frac{\left[T_2\left(\frac{p_3}{p_2}\right)^{γ-1/γ}-1\right]}{\left(T_3 – T_2\right)}
Therefore
\frac{p_3} {p_2} = \left[1 +η_D\left(\frac{T_3 – T_2}{T_2}\right)\right]^{3.5}
= \left(1 +\frac{0.821 × 106.72}{383.59}\right)^{3.5}
= 2.05
or p_2 = \frac{5.10} {2.05} = 2.49 bar
From isentropic P–T relations
p_{02} = p_2 \left(\frac{T_{02}} {T_2}\right)^{3.5} = 2.49 \left(\frac{494.4} {383.59}\right)^{3.5}
p_{02} = 6.05 bar
4. Impeller efficiency is
η_i = \frac{T_{01}\left[\left(\frac{p_3}{p_2}\right)^{\frac{γ-1}{γ}} – 1\right]}{T_{03} – T_{01}}
= \frac{288\left[\left(\frac{6.05} {1.01}\right)^{0.286} -1 \right]}{494.4 – 288}
= 0.938
ρ_2 = \frac{p_2} {RT_2} = \frac{2.49 × 10^5} {287 × 383.59}
ρ_2 = 2.27 kg/m³
\dot{m} = ρ_2A_2C_{r2}
= 2πr_2ρ_2b_2
But
U_2 = \frac{\pi ND_2} {60} = \frac{\pi N\dot{m}} {ρ_2\pi C_{r2}b_2 × 60}
N = \frac{475 × 2.27 × 246.58 × 0.0065 × 60} {2.5}
N = 41476 rpm