Question 7.DE.6: In a single-stage axial flow gas turbine, gas enters the tur......
In a single-stage axial flow gas turbine, gas enters the turbine at a stagnation temperature and pressure of 1150 K and 8 bar, respectively.
Isentropic efficiency of stage is equal to 0.88, mean blade speed is 300 m/s, and rotational speed is 240 rps. The gas leaves the stage with velocity 390 m/s.
Assuming inlet and outlet velocities are same and axial, find the blade height at the outlet conditions when the mass flow of gas is 34 kg/s, and temperature drop in the stage is 145 K.
Annulus area A is given by
A = 2 πr_mh
where h = blade height
r_m = mean radius
As we have to find the blade height from the outlet conditions, in this case
annulus area is A_3.
∴ h = \frac{A_3} {2\pi r_m}
U_m = \pi D_mN
or D_m = \frac{(U_m)} {\pi N} = \frac{300} {(\pi)(240)} = 0.398
i.e., r_m = 0.199 m
Temperature drop in the stage is given by
T_{01} – T_{03} = 145 K
Hence T_{03} = 1150 – 145 = 1005 K
T_3 = T_{03} – \frac{C^2_3} {2C_{pg}} = 1005 – \frac{390^2} {(2)(1.147)(1000)} = 938.697 K
Using turbine efficiency to find isentropic temperature drop
T^{′}_{03} = 1150 – \frac{145}{0.88} = 985.23 K
Using isentropic p–T relation for expansion process
p_{03} = \frac{p_{01}} {(T_{01}/T^′_{03})}^{γ/(γ – 1)} = \frac{8 }{(1150/985.23)^4} = \frac{8} {1.856}
i.e., p_{03} = 4.31 bar
Also from isentropic relation
p_3 = \frac{p_{03}} {(T^′_{03}/T_{3})}^{γ/(γ – 1)} = \frac{4.31}{(985.23/938.697)^4} = \frac{4.31}{1.214} = 3.55 bar
ρ_3 = \frac{p_3} {RT_3} = \frac{(3.55)(100)} {(0.287)(938.697)} = 1.32 kg/m³
A_3 = \frac{m}{ρ_3Ca_3} = \frac{34} {(1.32)(390)} = 0.066 m²
Finally,
h = \frac{A_3} {2 \pi r_m} = \frac{0.066} {(2)(0.199)} = 0.053 m