written 3.8 years ago by
Sayali Bagwe • 4.1k

•
modified 3.8 years ago

$\text{Steam at 5 bar} 200^0C, Ms = 50 kg/min \\
\text{Condenser pressure = 0.2 bar} \\
C_0 = 400 m/s \ \ α = 200 ø = 300 \\
Solution: \\
\text{From steam tables} \\
@ 5 bar, 20^0C, h_1 = 2855.4 KJ/kg, S_1 = 7.0592 KJ/kg \\
@ 0.2 bar \space h_2 = 251.5 KJ/kg , hf_{f2} = 2358.4 KJ/kg \\
\hspace{0.5cm} S_{f2} = 0.832 KJ/kg, s_{fg} = 7.07773 KJ/kg \\
S_1 = S_2 = \gt 7.0592 = 0.8321 + x_2 * 7.0773 \\
\hspace{0.5cm} X_2 = 0.88 \\
\text{Enthalpy of steam at 2 barh} \\
h_2 = hf_2 + xhf_{g2} = 251.5 + 0.88 * 2358.4 \\
\hspace{2cm} = 2326.9 KJ/kgs \\
\text{Enthalpy drop} = h_1 – h_2 = 2855.4 – 2326.9 = 528.5 KJ/kg \\
\text{Velocity oleam enteiny blades} = G = 44.7 \sqrt{h_1h_2} = 1028 m/s$
From velocity diag., $C_{w1} = 1028 \cos 20^0 \\
= 966 m/s \\
C_f = 1028 \sin 20 \\
351.6 m/s$
$\tan \theta=\dfrac{c_{f1}}{c_1 \cos 20G_01}=0.6212 \Rightarrow \theta=31.85^0 \\
c_{r1}=\dfrac{c_{f1}}{\sin \theta}=\dfrac{351.6}{\sin 31.85}=666 m/s \\
C_{w1}= C_{r2} \cos \emptysetG_0=666 \cos 30400=177 m/s$
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%