Question 7.DE.11: A small inward radial flow gas turbine operates at its desig......
A small inward radial flow gas turbine operates at its design point with a total-to-total efficiency of 0.90. The stagnation pressure and temperature of the gas at nozzle inlet are 310 kPa and 1145 K respectively. The flow leaving the turbine is diffused to a pressure of 100 kPa and the velocity of flow is negligible at that point. Given that the Mach number at exit from the nozzles is 0.9, find the impeller tip speed and the flow angle at the nozzle exit.
Assume that the gas enters the impeller radially and there is no whirl at the impeller exit. Take
C_p = 1.147 kJ/kgK, γ = 1.333.
The overall efficiency of turbine from nozzle inlet to diffuser outlet is given by
η_{tt} = \frac{T_{01} – T_{03}} {T_{01} – T_{03 ss}}
Turbine work per unit mass flow
W = U^2_2 = C_p(T_{01} – T_{03}), (C_{w3} = 0)
Now using isentropic p–T relation
T_{01} \left(1 – \frac{T_{03ss}} {T_{01}}\right) = T_{01} \left[1 – \left(\frac{p_{03}} {p_{01}}\right)^{γ – 1/γ}\right]
Therefore
U_2^2 = η_{tt}C_pT_{01}\left[1 – \left(\frac{p_{03}} {p_{01}}\right)^{γ – 1/γ}\right]
= 0.9 × 1147 × 1145 \left[1 – \left(\frac{100} {310}\right)^{0.2498}\right]
∴ Impeller tip speed, U_2 = 539.45 m/s
The Mach number of the absolute flow velocity at nozzle exit is given by
M = \frac{C_1} {a_1} = \frac{U_1} {a_1 \sinα_1}
Since the flow is adiabatic across the nozzle, we have
T_{01} = T_{02} = T_2 + \frac{C_2^2}{2C_p} = T_2 + \frac{U_2^2 }{2C_p \sin^2α_2}
or \frac{T_2} {T_{01}} = 1 – \frac{U^2_2} {2C_pT_{01} \sin^2α_2} , \text{but} C_p = \frac{γR} {γ – 1}
∴ \frac{T_2} {T_{01}} = 1 – \frac{U^2_2 (γ – 1)} {2γRT_{01} \sin^2α_2} = 1 – \frac{U^2_2 (γ – 1)} {2a^2_{01} \sin^2α_2}
But \left(\frac{T_2} {T_{01}}\right)^2 = \frac{a_2}{a_{01}} = \frac{a_2}{a_{02}} since T_{01} = T_{02}
and \frac{a_2} {a_{02}} = \frac{U_2} {M_2a_{02} \sinα_2}
∴ \left(\frac{U_2} {M_2a_{02} \sinα_2}\right)^2 = 1 – \frac{U^2_2 (γ – 1)} {2a^2_{02} \sin^2α_2}
and 1 = \left(\frac{U_2} {a_{02} \sinα_2}\right)^2\left(\frac{(γ – 1)}{2} + \frac{1}{M^2_2}\right)
or \sin^2α_2 = \left(\frac{U_2} {a_{02}}\right)^2\left(\frac{(γ – 1)}{2} + \frac{1}{M^2_2}\right)
But a^2_{02} = γRT_{02} = (1.333)(287)(1145) = 438043 m²/ s²
∴ \sin^2α_2 = \frac{539.45^2} {438043} \left(\frac{0.333} {2} + \frac{1} {0.9^2}\right) = 0.9311
Therefore nozzle angle α_2 = 75°