Help : isolated system-conservation of mechnaical energy

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The discussion centers on solving a physics problem involving a windmill generating energy from wind to pump water. The wind moves at 11.0 m/s and the windmill has a diameter of 2.3 m with an efficiency of 27.5%. The water is pumped from a well 32.5 m deep to a tank 2.30 m above ground. Key calculations involve kinetic energy (Ekin) using the formula Ekin = 1/2 mv^2, where air density is 1.29 kg/m^3. The area of the windmill is calculated using the formula A = π(diameter/2)^2. The energy generated by the windmill is determined by multiplying Ekin by its efficiency. The potential energy required to lift the water is expressed as mgh, where g is the acceleration due to gravity (9.
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can canyone here help me with this physics problem:
Air moving at 11.0 m/s in a steady winds encounters a windmill of diameter 2.3m and having an efficiency of 27.5 %. The energy generated by the windmill is used to pump water from a well 32.5m deep into a tank 2.30m above the ground. At what rate in liters per minute can water be pumped into the tank?

This is what I have so far:
Ekin = 1/2mv^2
Density for air is: 1.29 kg/m^3
The time for this energy to form is 1 sec.
m = ( a * v * A ) ... where A is air density and v is velocity, a in this case is area of the windmill, which is pi(diameter/2)^2

Thus
Ekin = 1/2 (a * v * A)

So :
Ekin * 25% = energy generated by the windmill

After that, I'm lost. But I think it has to do something with potential energy.
 
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Originally posted by cgt32

This is what I have so far:
Ekin = 1/2mv^2
Density for air is: 1.29 kg/m^3
The time for this energy to form is 1 sec.
m = ( a * v * A ) ... where A is air density and v is velocity, a in this case is area of the windmill, which is pi(diameter/2)^2

Thus
Ekin = 1/2 (a * v * A)
That should be v^3
Originally posted by cgt32

So :
Ekin * 25% = energy generated by the windmill

After that, I'm lost. But I think it has to do something with potential energy.
The energy needed to raise mass m a height h is mgh where g is acceleration of gravity (9.8 m/sec/sec). This should give you enough info to do the problem.
 
For simple comparison, I think the same thought process can be followed as a block slides down a hill, - for block down hill, simple starting PE of mgh to final max KE 0.5mv^2 - comparing PE1 to max KE2 would result in finding the work friction did through the process. efficiency is just 100*KE2/PE1. If a mousetrap car travels along a flat surface, a starting PE of 0.5 k th^2 can be measured and maximum velocity of the car can also be measured. If energy efficiency is defined by...

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