Efficient Water Pumping with Wind: Calculating Rate in Liters per Minute

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In summary, the conversation discusses using a windmill with an efficiency of 28.0% to pump water from a well 36.5 m deep into a tank 2.30 m above the ground. The wind energy is calculated as 117.74 J per second and 8220 J per minute, which is used to lift the water. The mass of air hitting the windmill is needed to determine the amount of energy available.
  • #1
Jacob87411
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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
 
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  • #2
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
Isnt rotational energy given by E=Iw^2? What is I or do you need to find it
 
  • #4
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
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
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.
 

FAQ: Efficient Water Pumping with Wind: Calculating Rate in Liters per Minute

1. How does a wind-powered water pump work?

A wind-powered water pump uses the energy from wind to rotate a turbine, which powers a pump that pulls water from a source, such as a well or reservoir, and pushes it through a system of pipes to a desired location.

2. What factors affect the efficiency of a wind-powered water pump?

The efficiency of a wind-powered water pump can be affected by various factors such as wind speed, turbine size and design, pump efficiency, pipe diameter, and the height and distance the water needs to be pumped.

3. How do you calculate the rate of water pumping in liters per minute?

The rate of water pumping in liters per minute can be calculated by dividing the volume of water pumped by the time it takes to pump that volume. For example, if 500 liters of water is pumped in 10 minutes, the rate would be 500/10 = 50 liters per minute.

4. Can a wind-powered water pump be used for large-scale water supply?

Yes, wind-powered water pumps can be used for large-scale water supply. However, the efficiency and output may vary depending on the factors mentioned above. It is important to carefully consider the design and location of the pump to ensure optimal performance.

5. Are there any maintenance requirements for a wind-powered water pump?

Like any mechanical device, a wind-powered water pump may require regular maintenance to ensure optimal functioning. This may include monitoring and cleaning the turbine, pump, and pipes, as well as checking for any wear and tear and replacing parts when necessary.

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