Windmill Problem

1. Mar 12, 2006

Jacob87411

Air moving at 14.5 m/s in a steady wind encounters a windmill of diameter 2.30 m and having an efficiency of 28.0%. The energy generated by the windmill is used to pump water from a well 36.5 m deep into a tank 2.30 m above the ground. At what rate in liters per minute can water be pumped into the tank?

Confused on where to start. Do you take the energy created by the wind needs to equal the energy required to move the water from 36.5 m deep to 2.3 m above the ground? Any help is appreciated

2. Mar 12, 2006

vaishakh

Calculate the rotational Kinetic energy transferred to the windmill. 28% will be converted into power. Calculate the amount of enrgy needed or work to be done to lift the given amount of water - the change in potential energy of the water lifted. But I am sensing some missing informations here.

3. Mar 12, 2006

Jacob87411

Isnt rotational energy given by E=Iw^2? What is I or do you need to find it

4. Mar 12, 2006

lightgrav

No, you can ignore the rotational KE of the windmill,
since this whole scenario is STEADY.

Wind has Kinetic Energy Density ... 1/2 rho v^2 ...
some wind goes thru (pierces) the windmill "disk" Area each second.
How much KE does that carry? (what's the mass flow rate?) per minute?

what mass water would 28% of that Energy lift 38.8 meters in Earth gravity?

5. Mar 14, 2006

Jacob87411

Ok so the kinetic energy of the wind is .5 * 1.12 * 14.5^2 (is that the right rho?)

This equals 117.74 J. Every second this hits the total area of the windmill which is 4.155 m^2. So the total energy on the windmill is (.28)(117.74)(4.155) = 137 J every second on the windmill. We need the amount every minute so 137*60 = 8220 J.

So we have 8220 J to move the water. Is that right so far?

6. Mar 15, 2006

vaishakh

When will you relate density with mass? You have the area of contact.
So the mass of air hitting it must be? Think a bit. I don't like giving you everything readymade.