Question #271373

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.



1
Expert's answer
2021-11-30T02:49:01-0500

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=πD1N/60=π.9200/60U1 = \pi*D1 N/60 = \pi*.9*200/60

=9.424m/s

U2=πD2N/60=π0.45200/60U2 = \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=sinα=Vf1V1 = sin\alpha = Vf1 inlet velocity triangle.

v1=Vf1/sinα=1.8/sin10°=10.365m/sv1 = Vf1/sin\alpha = 1.8/sin 10\degree = 10.365m/s

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

vw1=V1cosα=10.365cos10°vw1 =V1 cos\alpha =10.365*cos 10 \degree

=10.207m/s

iv. the runner blade angle means the angle θ\theta and ϕ\phi

tanθ=Vfi/(VwiU1)=1.8/(10.20179.424)=2.298tan\theta = Vfi/(Vwi -U1)= 1.8/(10.2017- 9.424)= 2.298

θ=tan12.298=66.48°0r66°29min\theta =tan^-1 2.298 = 66.48\degree 0r 66\degree 29\min

from outlet velocity triangle we have

tanθ=Vf2/U2=1.8/4.712=tan20.9°tan\theta = Vf2/U2 =1.8 /4.712 = tan 20.9\degree

ϕ=20.9°\phi =20.9\degree

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

πD1B1Vf1=πD2B2Vf2;D1B1=D2B2\pi D1B1Vf1 = \pi D2B2 Vf2; D1B1 =D2B2

B2=D1B1/D2=0.9.20/.45=0.40mB2 = D1B1/D2= 0.9*.20/.45= 0.40m

Q=πD1B1Vf1=π0.80.21.8Q =\pi D1B1 Vf1 =\pi *0.8*0.2*1.8

= 1.0178m3/s1.0178m^3/s


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