Question 7.DE.8: A single-stage axial flow gas turbine with equal stage inlet......
A single-stage axial flow gas turbine with equal stage inlet and outlet velocities has the following design data based on the mean diameter:
Mass flow 20 kg/s
Inlet temperature, T_{01} 1150 K
Inlet pressure 4 bar
Axial flow velocity constant through the stage 255 m/s
Blade speed, U 345 m/s
Nozzle efflux angle, α_2 60°
Gas-stage exit angle 12°
Calculate (1) the rotor-blade gas angles, (2) the degree of reaction, blade-loading coefficient, and power output and (3) the total nozzle throat area if the throat is situated at the nozzle outlet and the nozzle loss coefficient is 0.05.
(1) From the velocity triangles
C_{w2} = Ca \tanα_2
= 255 tan 60° = 441.67 m/s
C_{w3} = Ca \tanα_3 = 255 tan 12° = 55.2 m/s
V_{w2} = C_{w2} – U = 441.67 – 345 = 96.67 m/s
β_2 = \tan^{-1} \frac{V_{w2}} {Ca} = \tan ^{-1} \frac{96.67} {255} = 20.8°
Also V_{w3} = C_{w3} + U = 345 + 55.2 = 400.2 m/s
∴ β_3 = \tan {-1} \frac{V_{w3}}{Ca} = \tan {-1} \frac{400.2} {255} = 57.5°
(2) Λ = \frac{Φ} {2}(\tanβ_3 – \tanβ_2)
= \frac{255} {2 \times 345}(\tan57.5° + \tan20.8°) = 0.44
Ψ = \frac{Ca}{U}(\tanβ_2 + \tanβ_3)
= \frac{255} {345}(\tan20.8° + \tan57.5°) = 1.44
Power W = mU (C_{w2} + C_{w3})
= (20)(345)(441.67 + 54.2) = 3421.5 kW
(3) λ_N = \frac{C_p (T_2 – T^′_2)}{\frac{1}{2}C^2_2}, C_2 = Ca \text{sec} α_2 = 255 sec 60° = 510 m/s
or T_2 – T^′_2 = \frac{(0.05)(0.5)(510^2)}{1147} = 5.67
T_2 = T_{02} – \frac{C^2_2}{2C_p} = 1150 – \frac{510^2} {(2)(1147)} = 1036.6 K
T^′_2 = 1036.6 – 5.67 = 1030.93 K
\frac{p_{01}} {p_2} = \left(\frac{T_{01}}{T_2}\right)^{γ/(γ – 1)} = \left(\frac{1150}{1030.93}\right)^4 = 1.548
p_2 = \frac{4} {1.548} = 2.584 bar
ρ_2 = \frac{p_2} {RT_2} = \frac{2.584 \times 100} {0.287 \times 1036.6} = 0.869 kg/m³
m = ρ_2A_2C_2
A_2 = \frac{20} {0.869 \times 510} = 0.045 m²