A single jet Pelton wheel runs at 300rev/min under a head of 510m. The jet diameter is 200mm, its deflection inside the bucket is 16.5° and its relative velocity is reduced by 15% due to friction Determine 10) Water power. (ii) Resultant force on the bucket (ii) Overall efficiency. (Mechanical losses = 3%, coefficient of velocity = 0.98 and speed ration=0.46)

Question-Answer › Category: Fluid Mechanics › A single jet Pelton wheel runs at 300rev/min under a head of 510m. The jet diameter is 200mm, its deflection inside the bucket is 16.5° and its relative velocity is reduced by 15% due to friction Determine 10) Water power. (ii) Resultant force on the bucket (ii) Overall efficiency. (Mechanical losses = 3%, coefficient of velocity = 0.98 and speed ration=0.46)

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Kristin asked 1 year ago

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MG answered 1 year ago

Given:
Speed of Pelton Wheel, N = 300 RPM
Head, Hnet = 510 m
Jet diameter, d = 200 mm = 0.2 m
Deflection inside bucket, $\phi$ = 16.5 $\degree$
Relative velocity is reduced by 15%, Vr2 = ( 1 – 0.15 ) Vr1 = 0.85 Vr1
Mechanical Loss = 3%, then Mechanical Efficiency $Пm$ = 97% = 0.97
Co efficient of velocity, Cv = 0.98
Speed ratio, u / V1 = 0.46 where u is velocity of bucket and V1 is velocity of jet
For Pelton Wheen, u1 = u2 = u
The Velocity triangles for inlet and outlet are as below:

Where Vr1 and Vr2 are relative velocities, V1 and V2 are Absolute Jet velocities, Vw1 and Vw2 are whirl velocities, u1 and u2 are velocity of buckets which are equal in Pelton wheel
We have,
$Vi = C, 29 H net = 0.98 * 2 * 9.81 * 510$
V1 = 98.0304 m/s => Vw1 = V1 = 98.0304 m/s
We have Speed ratio, u / V1 = 0.46 => u = 0.46 V1
Then, u = 0.46 * 98.0304 => u = 45.0939 m/s
From Inlet Velocity triangle, Vr1 = V1 – u = 98.0304 – 45.0939 => Vr1 = 52.9365 m/s
We have,
Vr2 = 0.85 * Vr1 = 0.85 * 52.9365 => Vr2 = 44.9960 m/s
We have, Vr2 cos( $\phi$ ) = 44.9960 * cos(16.5) = 43.1431 < u2 = 45.0939
Then, $B<$ and the actual Velocity triangle for output is as below:

(Note: If Vr2 cos( $\phi$ ) > u2, then $\beta > \phi$ and the output velocity triangle considered initially would be correct and Vw2 = Vr2 cos( $\phi$ ) – u2 )
From Outlet Velocity triangle, Vr2 cos( $\phi$ ) + u2 = Vw2 [ where u2 = u1 = u ]
Then, Vw2 = 44.9960 * cos(16.5) + 45.0939 => Vw2 = 88.237 m/s
i) We have,
Water Power, $w = p*9*A*V1 * Hnet$
where $\rho$ = density of water, $A=* d$ is the area of jet => $A=*0.22 = 0.0314m$
Substituting the values,
Pw = 1000 * 9.81 * 0.0314 * 98.0304 * 510 = 15400315.08 W
Pw = 15400.31508 kW
ii) Resultant Force on Bucket,
   $F, =P* A* V1 * (Vw1 =Vw2$
   Here, if $\beta > \phi$ , consider positive sign and if $B<$ , consider negative sign. In this case, $B<$ then
   $F, =P* A* V1 * (Vw1 – Vw2$
Substituting values,
Fr = 1000 * 0.0314 * 98.0304 ( 98.0304 – 88.2369 ) = 30145.9067 N
Fr = 30145.9067 N
iii) Runner Power, Pr = Fr * u = 30145.9067 * 45.0939 => Pr = 1359396.502 W = 1359.396502 kW
Hydraulic Efficiency, $nh$ = Runner Power / Water Power
$nh$ = 1359.396502 / 15400.31508 = 0.0883 = 8.83%
Overall efficiency,
$= nm *nh$
$no$ = 0.97 * 0.0883 = 0.0856
$no$ = 8.56 %
Conclusion:
i) Water Power, Pw = 15400.31508 kW
ii) Resultant Force on Bucket, Fr = 30145.9067 N
iii) Overall efficiency, $no$ = 8.56 %
The values obtained here are based on correct calculations and seem not viable. There may be an issue with the data given in the question.