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Power Output of a water wheel

EDIT : Please take a look at the post below this if you have time !

I'm having an issue with (a)(ii).

Taking the diameter to be 5 metres and hence the radius to be 2.5 metres :

2*pi*2.5*7 = distance moved by water wheel = 109.9557 metres/minute

109.9557/60 = 1.8326 metres/second = speed of water wheel.

Finding the force acting on the edge of wheel :

7400/2.5 = 2960 N

Since Power = Force*velocity

Power = 2960*1.8326 = 5424 W

Is this the correct method ? Do you have some other method of doing this question ?

NOTE : The stated answer is 5.5 KW.

(edited 13 years ago)

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Reply 1

Ari Ben Canaan

This is the (b) part of the same question attached in the above question.

For part (b)(i)

How would you do it ?

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Reply 2

Joinedup

I'd have gone

power = (2*pi*torque*rpm)/60
= (6.283*7.4*7)/60
5.424

same answer tho. :frown:

Reply 3

Ari Ben Canaan

Original post by Joinedup

I'd have gone

power = (2*pi*torque*rpm)/60
= (6.283*7400*7)/60
5424

same answer tho. :frown:

Yeah, exact same answer. You made a small error.

Could you check my second post ?

Reply 4

Joinedup

Meh kilonewtons in = kilowatts out.

2nd

Should just be able to calc power input from the mass of water per sec and the distance, multiply by the conversion efficiency and sub into the formula for power to get torque (rpm still = 7)

don't really see how that answers the question introduced in the text though - maybe it's too early in the morning for me.

Reply 5

Ari Ben Canaan

Original post by Joinedup

Meh kilonewtons in = kilowatts out.
Should just be able to calc power input from the mass of water per sec and the distance, multiply by the conversion efficiency and sub into the formula for power to get torque (rpm still = 7)

I did all of that but I dont understand why we multiply by the efficiency ratio !

Doing so would give us the USEFUL energy output, no ?

Shouldnt we multuply by the conversion efficiency and then subtract that answer from the input power to get the output power that is wasted in friction ?

Reply 6

Joinedup

The question said the wheel was uncoupled from all machinery so I thought all the useful energy was being used up keeping it the wheel turning against the friction of it's bearing? That wheel can't be turning for free unless we've invented a perpetual motion machine.