Determine the largest current the generator can produce

[Vladi]

Homework Statement

A 120-V generator is run by a windmill that has blades 2.0 m long. The wind, moving at 12 m/ s, is slowed to 7.0 m/ s after passing the windmill. The density of air is 1.29 kg/ m3. If the system has no losses, what is the largest current the generator can produce? [Hint: How much energy does the wind lose per second?]

Homework Equations

P=(1/8)(density)(pi)(diameter)^2(v)^3
P=(V)(I)
Speed=Distance/Time

The Attempt at a Solution

I need to figure out the speed of the windmill. Can I determine the speed of the windmill by figuring out how much time it takes to change the speed of the wind? I want to make sure my logic is correct.

 

[NFuller]

I need to figure out the speed of the windmill

Actually, you don't. The problem says the system has no losses so the change in the kinetic energy of the wind is the energy that the generator will produce.

 

[Vladi]

Ahh.

Actually, you don't. The problem says the system has no losses so the change in the kinetic energy of the wind is the energy that the generator will produce.

You're saying that the wind turbine converted the wind's kinetic energy into electrical energy, which makes perfect sense. The wind is spinning the blades after all. Does this mean that I must include the change in velocity within the equation?
(1/8)(1.29 kg/m^3)(pi)(2 m)^2(12 m/s-7 m/s)^3=120 volts*(I)
---->I=2.1 Amps
The largest current the generator can produce is 2.1 amps.

 

[gneill]

You're saying that the wind turbine converted the wind's kinetic energy into electrical energy, which makes perfect sense. The wind is spinning the blades after all. Does this mean that I must include the change in velocity within the equation?
(1/8)(1.29 kg/m^3)(pi)(2 m)^2(12 m/s-7 m/s)^3=120 volts*(I)
---->I=2.1 Amps

Almost. Consider the power of the wind before and after it encounters the blades separately; don't just take the velocity difference. That is, $(V_1^3 - V_2^3) \ne (V_1 - V_2)^3$.

Also, your relevant equation for the power associated with the moving air uses the diameter of the cross section. You're given the individual blade length. Is the blade length the same as the diameter of the area swept out by the blades?

 

[Vladi]

Almost. Consider the power of the wind before and after it encounters the blades separately; don't just take the velocity difference. That is, $(V_1^3 - V_2^3) \ne (V_1 - V_2)^3$.

Also, your relevant equation for the power associated with the moving air uses the diameter of the cross section. You're given the individual blade length. Is the blade length the same as the diameter of the area swept out by the blades?

You're saying the diameter should be 4 meters. I shouldn't be using the radius.
I got the equation from the following derivation:
P = 1/2 ρ A v^3 = (1/8)(p)(pi)(d)^2(v)^3

 

[NFuller]

Yes, this looks right.

 

[Vladi]

Yes, this looks right.

Thank you for all your help.

 

[Olalekan]

I still not

Yes, this looks right.

I still not get it wright sir, I will be glad if you can elaborate it more

 

[Olalekan]

I still not

Yes, this looks right.

I still not get it wright sir, I will be glad if you can elaborate it mo