Pelton Wheel: Solve 200kW Diameter Problem

[lennonconnor]

Think I posted in the wrong section previously.

1. Homework Statement

The buckets of a Pelton wheel deflect a jet, having a velocity of 60m/s, through an angle of 160o. Assuming that the velocity of the jet relative to the buckets is reduced by 15% as the jet water moves over them, find the efficiency of the wheel if the speed of the buckets is 27m/s and, using this efficiency, calculate the diameter of the jet so that the wheel may develop 200kW.


2. Homework Equations

I believe the following are relevant:

0.5(1+cos θ )

or 0.5 Rp (1+ cos θ) To be honest I can't find anything in my classroom notes to suggest there should be a Rp in this formula.

V(bucket) = 0.5 x V(jet)


3. The Attempt at a Solution

Using the equation: 0.5 Rp (1+ cosθ)

The first problem would give me an efficiency of 0.5 x 60 x (1 + cos 20 degress)

= 28.691

V(jet) at 85% = 27 / 0.5 = 54 m/s

At this speed using again the above formula:

0.5 x 54 x (1 + cos 20 degrees) = 25.871

25.871 / 28.691 = approx 90%

The answer has been given as 89% which is pretty close. However, I don't feel given all the different numerical values given that I have quite grasped the problem and I have just been lucky!

Any help would be appreciated.

 

[haruspex]

0.5(1+cos θ )

That's an expression, not an equation. What does it represent?

or 0.5 Rp (1+ cos θ) To be honest I can't find anything in my classroom notes to suggest there should be a Rp in this formula.

Rp being?

V(bucket) = 0.5 x V(jet)

Always? Why?

The first problem would give me an efficiency of 0.5 x 60 x (1 + cos 20 degress) = 28.691

That's a speed, not an efficiency, right? Is this the speed the buckets would move at if the wheel were 100% efficient? And you mean the first equation?

V(jet) at 85% = 27 / 0.5 = 54 m/s

Where did the 27 come from? 85% of 60 = 51

At this speed using again the above formula:

0.5 x 54 x (1 + cos 20 degrees) = 25.871

25.871 / 28.691 = approx 90%

I don't understand any basis for calculating the efficiency that way. Efficiency should be a matter of energy, which varies according to squares of velocities typically.