Question

Pelton Wheel Problem

Conditions at the inlet to the nozzle of a Pelton wheel are:

• Pressure, p=700 psig
• Velocity, V=15 mph

The jet diameter is d=7.5 in. and the nozzle loss coefficient is Knozzle=0.04. The wheel diameter is D=8 ft. At this operating condition, η=0.86. Calculate:

1. The power output
2. The normal operating speed
3. The approximate runaway speed
4. The torque at normal operating speed
5. The approximate torque at zero speed

Answer 1

Given:

• Pressure: P=700 psig 
• Jet velocity: V=15mph
• Jet diameter: d=7.5 in 
• Nozzle Loss Coefficient: Knozzle=0.04
• Peltom Wheel Diameter: dp=8ft
• Turbine Efficiency: η=0.86
• The density of water is ρ=1.94 slug /ft3

Calculate:

1. The power output

   We have, the power produced by the Turbine:

   P=η×ρ×Q×g×H

   Effective head available sum of pressure head and the velocity head

   H=p1ρg+V122g
   H=700psi1.94slug/ft3×32.2+(15mph)22×32.2
   H=700psi1.94slug/ft3×32.2+(22ft/s)22×32.2
   H=1621.13ft

   Velocity at Turbine Exit

   Applying Bernoulli equation at Turbine input and output

   (p1ρg+V122g+z1)=(p2ρg+V222g+z2)+hL
   (7001.94×32.2+(15)22×32.2)=(0ρg+V222g)+kV222g
   (1621.13ft)=(1+0.04)V222×32.2
   V22=100385.35
   V2=316.83ft/s

   Flow Rate in Turbine

   Area of the turbine is A=π4d2=π4(7.512ft)2=0.3068ft2

   Thus, flow rate is Q=VA=316.83ft/s×0.3068ft2=97.20ft3/s

   Output Power

   P=η×ρ×Q×g×H
   =0.86×1.94×97.20×32.2×1621.13ft550hp
   =15391.37hp

2. Normal Operating Speed

   Turbine speed is 0.5 times the jet velocity for maximum efficiency, hence

   U=0.5V2=0.5×316.83ft/s=158.415ft/s

   The corresponding angular velocity of turbine is

   ω=Ur=2UD=2×158.415ft/s8ft=39.603rad/s

   The operating speed is

   ω=2πN60⇒N=60×ω2π
   =60×39.603rad/s2π
   =378.18rpm

3. Approximate runaway speed

   The runaway speed is equal to velocity of jet at exit: U=V2=316.83ft/s

4. Torque at Normal operating speed

   The expression for Torque is Ti=ρQR(V2−U)(1−cos⁡θ)

   Taking maximum feasible value of angle: θ=165∘

   The Torque generated is

   Ti=ρQR(V2−U)(1−cos⁡θ)
   =1.94×97.2×82×(316.83−158.415)×(1−cos⁡165∘)
   =2.349×105lb×ft

5. Torque at U=0

   T0=ρQR(V2)(1−cos⁡θ)
   =1.94×97.2×82×(316.83)×(1−cos⁡165∘)
   =4.6981×105lb×ft

Therefore there is all the answer.