Answer to Question #128906 in Electrical Engineering for Amiru Hardi Sulemana

Question

An outward radial - flow impulse turbine has nozzles with a total effective area of 970mm² and the guide vanes make an angel of 20 degree to the tangent at exit. The inner and outer diameter of the runner are 508mm and 712mm respectively, and the moving vanes have an outlet angle making 20 degree to the tangent. When running at 620rev/min under a head of 61m the discharge is 1.8m cube/min and the power outlet is 11.9kw . The water is observed to leave the runner in a forward direction and inclined at 15 degree to the radius. Assuming the water leaves tangentially to the moving vanes, calculate the head lost in

a) the nozzles

b) the moving vanes

c) bearing friction and windage .

Answer 1

To solve this question, we need to understand the construction of the turbine.

As per the given question,

The area of the Nozzle $(A_n)= 970mm²$

Angle made by guide veins $\theta_v=20^\circ$

The inner diameter of the runner $(d_{in})=508mm$

Outer diameter of the runner $(d_{outer})= 712mm$

Angle made by outlet $(\theta_{outlet})=20^\circ$

angular speed of the runner $(N)=620 rev/min =\frac{31}{3} rev/sec$

Head length (d)= 61m

discharge volume $(\frac{dV}{dt})=\frac{1.8m^3}{60 sec}=0.03 m^3/sec$

Power outlet $=11.9kW$

The angle made by runner $=15^\circ$

i) Tangential velocity at inlet and outlet

$u_1=\pi D_1N=3.14\times0.508\times \frac{31}{3}=16.48m/sec$

$u_2=\pi D_2 N=3.14\times 0.712\times \frac{31}{3} =23.10 m/sec$

$V_f=16.48\times \tan (20) =5.99 m/sec$

Head lost $F=PA=11.9\times 1000\times 0.00097 =11.54cN$