# Electrical power generated by wind turbine

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1. Jul 6, 2015

### toforfiltum

1. The problem statement, all variables and given/known data

2. Relevant equations
P=FV

3. The attempt at a solution
To calculate the mass of wind swept by turbine in a second, I take 2000m2 ×10ms-1×1.3kgm-3.Then, to calculate the force I multiply the answer with 10ms-1. Since they want the power and the yield is 50%. I multiply the answer again with 10ms-1 and divide it by two. This gives me C as the answer. However, the answer is B. So where did I go wrong?

2. Jul 6, 2015

### Orodruin

Staff Emeritus
It is unclear to me why you would do this. Instead, try to figure out the kinetic energy contained in the air which passes the turbine per time unit.

3. Jul 6, 2015

### toforfiltum

I did that because I was using the equation power = force x velocity.
If I use the KE approach, is it 1/2 (2000x1.3x10) x 102? I am still getting the wrong answer.

4. Jul 6, 2015

### Orodruin

Staff Emeritus
Because now you forgot about the 50% efficiency factor.

5. Jul 6, 2015

### toforfiltum

I'm confused. Isn't m*v/t =ma? Anyway, I was using the units to guide what I was doing. By multiplying 2000m2 with 10ms-1 and 1.3kgm-3, I will end up with 26000kgs-1. Since force is kgms-2, I multiplied my answer with 10ms-1 to get 26000kgms-2. Since power = force x velocity, I multiplied the answer again with 10 ms-1, and divide it by 2 to get effective power.

I could get the answer by using KE now that you have told me, I just want to know what's wrong with my approach, since it seems logical to me.

6. Jul 6, 2015

### Orodruin

Staff Emeritus
While this is often a good approach to get at order of magnitudes, you will always miss factors of order unity (as in this case a factor of two). By the same argumentation, the kinetic energy of a mass m moving with velocity v would be mv2. Dimensional analysis will only give you the exact answer if you are lucky and the prefactor of order one is actually one.

Edit: Similarly, the area of a circle of radius 1 m would be given by 1 m2 with the same reasoning. It gives you the correct order of magnitude byt is off by a factor of $\pi$, which is roughly of order one.

7. Jul 6, 2015

### toforfiltum

Oh I see. So my approach is wrong. Then, I need to use the KE approach in solving problems like this one next time. So, this is the only right approach, is it?

8. Jul 6, 2015

### Orodruin

Staff Emeritus
Since you are told that 50% of the wind energy is transferred to electrical power, it does seem reasonable to compute the total wind energy I would say. I do not see any other way of solving this the way it is written.

9. Jul 6, 2015

### toforfiltum

Ok, thanks for your help. This mistake is a good one, I will not solve problems by looking at the units again.

10. Jul 6, 2015

### Orodruin

Staff Emeritus
Do not get me wrong, looking at units is a very valuable tool in physics and it will very often tell you how your output depends on your input parameters. However, in order to use it properly, you need to be aware of the limitations.

11. Jul 6, 2015

### toforfiltum

So it's just important in ensuring homogeneity on both sides, but not much use in ensuring the right answer. Thanks again. I will keep it in mind.

12. Jul 6, 2015

### haruspex

The reason the force method gave you double the answer is that if you think of a 100% efficient turbine as slowing the air to a halt then the average speed of the air through the turbine is only 5m/s. P=Fv only works where F and v are constant.