Answer to Question #271373 in Mechanical Engineering for Bagath

Question

An inward flow reaction turbine is required to produce a power of 280 kW at 200

rpm. The effective head on the turbine is 20 m. The inlet diameter is twice as the

outlet diameter. Assume hydraulic efficiency as 80%. The radial

velocity is 3.5 m/s and is constant. The ratio of breadth to wheel diameter is 0.1

and 5% of the flow area is blocked by vane thickness. Determine the inlet and

outlet diameters, inlet and exit vane angle and guide blade angle at the inlet.

Assume radial discharge.

Answer 1

External Dia =D1 =0.9m =D1=0.9m

Internal Dia =D 2 =0.45 =D2=0.45

Speed N=200rpm N=200rpm

Width at inlet B 1 =200mm=0.2m B1=200mm=0.2m

Velocity of flow vf 1 =vf 2 =1.8m/s

vf1=vf2=1.8m/s Guide blade angle α=10∘

α=10∘

Discharge at outlet = Radial

 B=90∘

∴    B=90∘ and Vw 2 =0

The tangential velocity of the wheel at inlet and outlet are:-

$U1 = \pi*D1 N/60 = \pi*.9*200/60$

=9.424m/s

$U2 = \pi*D2 N/60 = \pi*0.45*200/60$

=4.712m/s

ii. Absolute velocity of water at the inlet of the runner i.e. V1

from the

$V1 = sin\alpha = Vf1$ inlet velocity triangle.

$v1 = Vf1/sin\alpha = 1.8/sin 10\degree = 10.365m/s$

ii. the velocity of whirl at inlet i.e. Vw1

$vw1 =V1 cos\alpha =10.365*cos 10 \degree$

=10.207m/s

iv. the runner blade angle means the angle $\theta$ and $\phi$

$tan\theta = Vfi/(Vwi -U1)= 1.8/(10.2017- 9.424)= 2.298$

$\theta =tan^-1 2.298 = 66.48\degree 0r 66\degree 29\min$

from outlet velocity triangle we have

$tan\theta = Vf2/U2 =1.8 /4.712 = tan 20.9\degree$

$\phi =20.9\degree$

v. width or runner at outlet i.e. B2

$\pi D1B1Vf1 = \pi D2B2 Vf2; D1B1 =D2B2$

$B2 = D1B1/D2= 0.9*.20/.45= 0.40m$

$Q =\pi D1B1 Vf1 =\pi *0.8*0.2*1.8$

= $1.0178m^3/s$