Mumbai University > Mechanical Enginerring > SEM 6 > Thermal and Fluid Power Engineering

Marks: 8M

The following particulars refer to a single row impulse turbine: Mean diameter of blade ring = 2.5 m ; speed = 3000 rpm ; nozzle angle = 20°; ratio of blade velocity to steam velocity = 0.4; blade friction factor = 0.8 ; blade angle at exit = 3° less than that at inlet; steam flow =36,000 kg/hr.

Draw velocity diagram for moving blade and estimate (a) power developed ; (b) blade efficiency; and (c) steam consumption in kg/kW hr.

$$D_m = 2.5 m, \hspace{1cm} N = 3000 rpm\ d = 20^0 \hspace{1cm} \frac{C_b}{C_1} = 0.4, \hspace{1cm} k = 0.8, \hspace{1cm}V_{r2}=0.8 V_1\ \phi = \theta - 3 , m_s =36000kh/k_u=10kg/s$$

To find 1) Power developed 2) Blade efficiency 3) Steam consumption

$C_{w1}=C_1 cos \alpha\\ c_b = \frac{\pi D_mN}{60} = 392.7 m/s\\ c_1 = \frac{C_b}{0.4} = 981.748\\ \therefore C_{w1}=c_1 sin \alpha\\ \hspace{1.1cm} = 981.748 \times cos 20\\ \hspace{1.1cm} = 922.54 m/s ---- (1)\\ cf_1 = C_1 sin \alpha = 981.748 \times sin \alpha\\ \hspace{2.3cm} = 335.78 m/s ----- (2)\\ \theta = tan^{-1}\Big(\frac{cf_1}{C_{w1}-C_b}\Big) = tan^{-1}\Big(\frac{335.78}{922.54-392.7}\Big)=32.36^0\\ C_{r1} = \frac{Cf_1}{sin \theta}=\frac{335.78}{sin 32.36}=627.348m/s\\ C_{w2}=0.8 \times C_{r1} = 501.878 \text{and} \phi = \theta -3 = 29.36^0\\ C_{w2}=C_{r2}cos \phi -C_b=501.878 cos 29.36 - 392.7 = 44.7149mm/s$

Power developed $\hspace{1cm}= m_s (c_{w1} + c_{w2}). c_b\\ \hspace{1cm}= 10 (922.54 + 44.7149) x 392.7\\ \hspace{1cm}= 3.7984 Mw ---------------- (1)$

Blade efficiency $\hspace{1cm} = \frac{Ms(c_{w1}+c_{w2})c_b}{Ms_2^{C^2_1}}\\ \hspace{1cm} = \frac{3.7984 \times 10^2 \times 2 }{10\times (981.748)^2}\\ \hspace{1cm} = 78.8% ------------- (2)$

Steam consumption $hspace{1cm} = \frac{36000kg/hr}{3798.4kw}\\ \hspace{1cm} = 9.4776kh/kwh ------------- (3)$