Question 7.DE.3: In a single-stage axial flow gas turbine gas enters at stagn......
In a single-stage axial flow gas turbine gas enters at stagnation temperature of 1100 K and stagnation pressure of 5 bar. Axial velocity is constant through the stage and equal to 250 m/s. Mean blade speed is 350 m/s.
Mass flow rate of gas is 15 kg/s and assume equal inlet and outlet velocities.
Nozzle efflux angle is 63°, stage exit swirl angle equal to 9°. Determine the rotor-blade gas angles, degree of reaction, and power output.
Refer to Fig. 7.7.
Ca_{1} = Ca_{2} = Ca_{3} = Ca = 250 m/s
From velocity triangle (b)
C_2 = \frac{Ca_{2}} {\cosα_2} = \frac{250} {\cos 63°} = 550.67 m/s
From figure (c)
C_3 = \frac{Ca_{3}} {\cos α_3} = \frac{250} {\cos 9°} = 253 m/s
C_{w3} = Ca_{3} \tanα_3 = 250 tan 9° = 39.596 m/s
\tanβ_3 = \frac{U + C_{w3}} {Ca_{3}} = \frac{350 + 39.596}{250} = 1.5584
i.e., β_3 = 57.31°
From figure (b)
C_{w2} = Ca_{2} \tanα_2 = 250 tan 63° = 490.65 m/s
and
\tanβ_2 = \frac{C_{w2} – U} {Ca_{2}} = \frac{490.65 – 350} {250} = 0.5626
∴ β_2 = 29°21′
Power output
W = mUCa (\tanβ_2 + \tanβ_3)
= (15)(350)(250)(0.5626 + 1.5584)/1000
= 2784 kW
The degree of reaction is given by
Λ = \frac{Ca} {2U} (\tanβ_3 – \tanβ_2)
= \frac{250} {2 × 350} (1.5584 – 0.5626)
= 35.56%
