Question

A 0.1 m diameter water jet with velocity of 35 m/s strikes on a vane which is moving with a velocity of 20 m/s . The water jet enters the vane at an angle of 35° with the direction of motion of the vane . The jet leaves the vane at an angle of 125 ° . The vane angle at outlet is 20° Determine
a) Force of water jet exerted on the vane
b) Work done by the jet on the vane

Answer 1

Answer:

Given

Jet veloicty v2=35m/s

vane velocity, 4=20m/s

α=30∘,β=180−120=60∘

1] Vane angle at inlet Θ and exit , ϕ

from inlet velocity △ACD

Vw=Ac=v1cosα=30.31 m/s

vs1=CD=V1sinalpha

35 sin 30=17.50m/s

BC=Vw1−μ=30.31-20=10.31m/s

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vr1BD=(BC)2+(D)2−−−−−−−−−−−√

(10.31)2+(17.8)2−−−−−−−−−−−−−−√

20.31 m/s

Θ=tan−1(CDBC)

tan−1(17.510.31)=59.5

Consider outlet velocity △EGH and neglecting blade friction i,e

Vr2=vr1=20.31 m/s

Applying sine rule to △EFG

Vr2sin(180−β)=4sin(B−ϕ)

20.31sin(180−60)=20sin(60−Φ)

sin(60−ϕ)=0.8528

(60−ϕ)=58.52∘

ϕ=1.48∘

FH==Vw2=EH-EF

vr2cosϕ-4

20.31cos1.48-20

=0.303 m/s

ii) work done /N weight of water,W

w=12(Vw1+Vw2)u=19.81×(30.31+0.303)

=62.41 Nm/N weight of water

iii) Hydraullic efficiency

nh=Workdone/kg.wk.Esupplied(w2/2)

2(Vw1+Vw2)v2

2×(30.31+0.303)×20352

=0.9996

=99.96