Electrical power generated by wind turbine

In summary, the conversation discusses using different methods to calculate the mass of wind swept by a turbine in a second. The first method involves using the equation P=FV, while the second method uses the kinetic energy approach. The expert advises against using the first method as it often leads to missing factors of order unity. Instead, they suggest using the kinetic energy approach and taking into account the 50% efficiency factor. The conversation concludes with a reminder to use dimensional analysis as a tool, but to be aware of its limitations.
  • #1
toforfiltum
341
4

Homework Statement


upload_2015-7-6_21-57-11.png


Homework Equations


P=FV

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?
 
Physics news on Phys.org
  • #2
toforfiltum said:
Then, to calculate the force I multiply the answer with 10ms-1.

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
Orodruin said:
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.
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
toforfiltum said:
If I use the KE approach, is it 1/2 (2000x1.3x10) x 102? I am still getting the wrong answer.

Because now you forgot about the 50% efficiency factor.

toforfiltum said:
I did that because I was using the equation power = force x velocity.

I am not asking you about that equation, I am asking you to justify your expression for the force as m*v/t.
 
  • #5
Orodruin said:
Because now you forgot about the 50% efficiency factor.
I am not asking you about that equation, I am asking you to justify your expression for the force as m*v/t.
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
toforfiltum said:
Anyway, I was using the units to guide what I was doing.

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
Orodruin said:
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.
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
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
Orodruin said:
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.
Ok, thanks for your help. This mistake is a good one, I will not solve problems by looking at the units again.
 
  • #10
toforfiltum said:
This mistake is a good one, I will not solve problems by looking at the units again.

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
Orodruin said:
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.
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
toforfiltum said:
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?
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.
 

Related to Electrical power generated by wind turbine

1. How does a wind turbine generate electricity?

A wind turbine works by harnessing the power of the wind to turn its blades, which are connected to a rotor. The rotor is connected to a main shaft, which spins a generator to produce electricity. The wind's energy is converted into mechanical energy, which is then converted into electrical energy.

2. What is the average lifespan of a wind turbine?

The average lifespan of a wind turbine is around 20-25 years. However, with proper maintenance and upgrades, some turbines can last up to 30 years or more.

3. How much electricity can a single wind turbine produce?

The amount of electricity a wind turbine can produce depends on its size, location, and wind conditions. On average, a single wind turbine can produce enough electricity to power around 600 homes.

4. What are the environmental benefits of wind power?

Wind power is a renewable energy source, meaning it does not rely on finite resources and does not produce harmful emissions. It also has a small carbon footprint and helps reduce our dependence on fossil fuels, which contribute to climate change.

5. Can wind turbines be used in all locations?

Wind turbines are most effective in areas with consistent and strong winds. However, they can still be used in areas with lower wind speeds, as long as there is enough wind to turn the blades and generate electricity. Factors such as topography and surrounding structures can also affect the efficiency of a wind turbine.

Similar threads

  • Introductory Physics Homework Help
Replies
10
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
985
  • Introductory Physics Homework Help
Replies
3
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
16
Views
2K
Replies
2
Views
3K
  • Introductory Physics Homework Help
Replies
4
Views
3K
  • Introductory Physics Homework Help
Replies
16
Views
2K
  • General Engineering
Replies
33
Views
5K
  • Introductory Physics Homework Help
Replies
1
Views
2K
Back
Top