A single stage steam turbine is supplied with steam at 5 bar, 200°Cattheratec 50kg/min. It expands into a condenser at a pressure of O.2 bar. The blade speed is 400 m/s. The nozzles are inclined at an angle of 20° to the plane of the wheel and the outlet blade angle is 30°. Neglecting friction losses, Determine power developed, blade efficiency and stage efficiency.

Mumbai University > MECH > Sem 6 > Thermal and Fluid Power Engineering

Marks: 10 M

Year: May 2015

$\text{Steam at 5 bar} 200^0C, Ms = 50 kg/min \\ \text{Condenser pressure = 0.2 bar} \\ C_0 = 400 m/s \ \ α = 200 ø = 300 \\ Solution: \\ \text{From steam tables} \\ @ 5 bar, 20^0C, h_1 = 2855.4 KJ/kg, S_1 = 7.0592 KJ/kg \\ @ 0.2 bar \space h_2 = 251.5 KJ/kg , hf_{f2} = 2358.4 KJ/kg \\ \hspace{0.5cm} S_{f2} = 0.832 KJ/kg, s_{fg} = 7.07773 KJ/kg \\ S_1 = S_2 = \gt 7.0592 = 0.8321 + x_2 * 7.0773 \\ \hspace{0.5cm} X_2 = 0.88 \\ \text{Enthalpy of steam at 2 barh} \\ h_2 = hf_2 + xhf_{g2} = 251.5 + 0.88 * 2358.4 \\ \hspace{2cm} = 2326.9 KJ/kgs \\ \text{Enthalpy drop} = h_1 – h_2 = 2855.4 – 2326.9 = 528.5 KJ/kg \\ \text{Velocity oleam enteiny blades} = G = 44.7 \sqrt{h_1-h_2} = 1028 m/s$

From velocity diag., $C_{w1} = 1028 \cos 20^0 \\ = 966 m/s \\ C_f = 1028 \sin 20 \\ 351.6 m/s$

$\tan \theta=\dfrac{c_{f1}}{c_1 \cos 20-G_01}=0.6212 \Rightarrow \theta=31.85^0 \\ c_{r1}=\dfrac{c_{f1}}{\sin \theta}=\dfrac{351.6}{\sin 31.85}=666 m/s \\ C_{w1}= C_{r2} \cos \emptyset-G_0=666 \cos 30-400=177 m/s$

Fower developed P $= Ms (C_{w1} + C_{w2}) C_{b1} \\ = 50/60 (966+177) 400 x 10^{-3} kw \\ = 381 kw$

Blade efficiency, $n_{bl}$ $=\dfrac{2c_{bi}(C_{w1}+C_{w2})}{C_1^2} \\ =\dfrac{2 \times 400 \times (966+177)}{(1028)^2} \\ = 86.5 \%$

Since there are no losses.

Blade efficiency = Stage efficiency = 86.5%