Question 7.DE.5: In a single-stage turbine, gas enters and leaves the turbine......
In a single-stage turbine, gas enters and leaves the turbine axially. Inlet stagnation temperature is 1000 K, and pressure ratio is 1.8 bar. Gas leaving the stage with velocity 270 m/s and blade speed at root is 290 m/s. Stage isentropic efficiency is 0.85 and degree of reaction is zero. Find the nozzle efflux angle and blade inlet angle at the root radius.
Since Λ = 0, therefore
Λ = \frac{T_2 – T_3}{T_1 – T_3,}
hence
T_2 = T_3
From isentropic p–T relation for expansion
T^′_{03} = \frac{T_{01}}{(p_{01}/p_{03})^{ (γ – 1)/γ}} = \frac{1000}{(1.8)^{0.249}} = 863.558 K
Using turbine efficiency
T_{03} = T_{01} – η_t (T_{01} – T^′_{03})
= 1000 – 0.85(1000 – 863.558) = 884 K
In order to find static temperature at turbine outlet, using static and stagnation temperature relation
T_3 = T_{03} – \frac{C^2_3}{2C_{pg}} = 884 – \frac{270^2} {(2)(1.147)(1000)} = 852 K = T_2
Dynamic temperature
\frac{C^2_2}{2C_{pg}} = 1000 – T_2 = 1000 – 852 = 148 K
C_2 = \sqrt{[(2)(1.147)(148)(1000)]} = 582.677 m/s
Since, C_{pg}ΔT_{os} = U (C_{w3} + C_{w2} ) = UC_{w2} (C_{w3} = 0)
Therefore, C_{w2} = \frac{(1.147)(1000)(1000 – 884)} {290} = 458.8 m/s
From velocity triangle
\sinα_2 = \frac{C_{w2}} {C_2} = \frac{458.8} {582.677} = 0.787
That is, α_2 = 51°54′
\tanβ_2 = \frac{C_{w2} – U} {C_{a2}} = \frac{458.8 – 290} {C_2 \cosα_2}
= \frac{458.8 – 290} {582.677 \cos 51.90°} = 0.47
i.e., β_2 = 25°9′