Question 7.DE.8: A single-stage axial flow gas turbine with equal stage inlet......

(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