## Wind Power B/C

andrewwski Posts: 953
Joined: January 12th, 2007, 7:36 pm

### Re: Wind Power B/C

My team and I are having a very hard time understanding how to solve for calculating power loss along a line. I think the terms / equations are similar to Ohms Law but not quite the same and that is mixing us up. Here is an example question for an older test:

A power plant in Chicago is generating 900 MW on 700 KV. Assuming the line is 100 kilometers long, with a resistance along the line of 0.2 ohms
A. What is the current flowing along the line?
Using Ohms Law, current is I=P/V 900 MW / 700 KV = 1286 amps

B. How much power is lost in the lines?
This is where we get confused. The answer key says find the voltage drop = current (amps) * resistance (ohms) 1286 * 0.2 = 257.2 kV. Then find power lost = amps * kV 1286 amps * 257.2 kV = 330.76 MW
We don't understand these equations and how does the line distance factor in?

C. What is the percentage of the power lost?
This should be easy taking the answer to B / 900 MW.

Can someone please help explain how to solve these kinds of problems or point us to resources to learn from? Thank you for the help.
For part B, you are not looking at the units. You should do 1286 * 0.2 = 257.2 kV (like you did). But then, 1286 amps * 257.2 kV = 330,759 kW. The reason that the correct answer is in MW, is because the values that started the problem were given in MW. To convert your answer, know that 1 MW is 1000 kW. So it is 330.76 MW.
For part C, you are correct, it is 330.76MW / 900MW = 36.75%.

I once had a teacher that took off points if we did not carry the units throughout the problem. We could only cross off the unit if they matched. This habit stuck with me, and it has really helped me, so try that.

I hope this helps
Thanks but this does not answer my question. I understand the units of measure changes. It is the overall question itself. Intuitively it seems that power loss should factor in to distance but we don't see that included in any of the above answers - even though it is included in the question. What if the line was 50 KM long - how would that change things? The easy way is just to memorize the following equations, but I want my team to understand the concepts as well and I can't explain it to them.
Voltage Drop = Current * Resistance
Power Loss = Current * Voltage Loss

Thanks for the help
(your poor suffering Industrial Engineer who should have apparently taken more Electrical Engineering classes)
This question has been posted here before, and the posted answer is wrong.

You're correct that distance is a factor in power loss. For a constant gauge wire, it will have a constant resistivity - or resistance per unit length. So as the length increases, so does the distance.

But here, they've given the resistance - so that's the total resistance of the entire length. The given distance is irrelevant (although you could be asked to compute the resistivity).

For the above question itself, the answers should be:

A. I = P/V = 900 MW / 700 kV = 900e6 W / 700e3 V = 1286 A

B. The methodology in the solution is correct. Voltage Drop = Current * Resistance (Ohm's Law). The resistance is given, not the resistivity, so the distance is irrelevant here. So you just put in the 0.2 ohms.

So Vdrop = 1286 A * 0.2 ohm = 257.2 V (not kV!).

Then power lost = Vdrop * I = 257.2 V * 1286 A = 330.8 kW (not MW!).

Alternatively if you plug in symbolically, you get Vdrop = I * R, and P = Vdrop * I, so P = I * (I * R) or P = I^2 * R. Then P = (1286 A)^2 * 0.2 ohms.

C. Percent loss = 330.8 kW/900 MW = 330.8e3/900e6 = 0.0367% loss. The posted solution of 36.7% is wrong, and a line loss of over 1/3 would be absurd.

To answer your question about what if the line was 50 km - if it was 50 km but the resistance was still given as 0.2 ohms, the solution would not change.

But, if the same gauge line was used, and the resistance was not given, we could determine the resistance. The resistivity of the line is 0.2 ohms/100 km = 0.002 ohm/km. So if the distance was 50 km of the same line, the resistance would be 0.002 ohms/km * 50 km = 0.1 ohm. Then proceed as above.

NeilM
Member Posts: 2
Joined: March 24th, 2017, 8:58 pm

### Re: Wind Power B/C

Does anybody have an estimate of the 1st place voltage for at the Northern California competition last year? Our balsa turbine is attaining about 1.2V, and I do know that varying setups will yield varying voltages, but it would be nice to have a ballpark estimate. Also, any idea what material and general design was used for those blades? Thank you!

ashmmohan
Member Posts: 47
Joined: March 18th, 2016, 6:36 am
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State: FL

### Re: Wind Power B/C

Does anybody have an estimate of the 1st place voltage for at the Northern California competition last year? Our balsa turbine is attaining about 1.2V, and I do know that varying setups will yield varying voltages, but it would be nice to have a ballpark estimate. Also, any idea what material and general design was used for those blades? Thank you!
Last year's rules had a resistance of 5 ohms; now it can be anywhere between 5 and 25...scores also vary due to different voltmeters/fans. I'm not in North Cali. but at our state champhionship in FL, we had the highest high and low speed voltages with around 331 and 251 mV. On our setup, our high speed with the same turbine was 625 mV...this just shows how much scores fluctuate between competition and our own testing setups.
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BasuSiddha23
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### Re: Wind Power B/C

What angle do you put your blades to the wind(angle of attack)?
Just to see the general idea
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NeilM
Member Posts: 2
Joined: March 24th, 2017, 8:58 pm

### Re: Wind Power B/C

For us, having an angle of about 10-15 degrees worked the best.

drcubbin
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### Re: Wind Power B/C

Question: for the past few years, I have seen lots of discussion regarding the electrical output from the fan's rotation. Since there is such variation in electrical recording devices, why don't we then just concern ourselves with # of rotations the fan makes? Any thoughts?

Ionizer
Member Posts: 32
Joined: March 20th, 2014, 6:03 pm
State: PA
Location: Pitt

### Re: Wind Power B/C

Question: for the past few years, I have seen lots of discussion regarding the electrical output from the fan's rotation. Since there is such variation in electrical recording devices, why don't we then just concern ourselves with # of rotations the fan makes? Any thoughts?
There will still be variations depending on fan strength, resistance, friction, etc. Also, how would you suggest we measure the frequency of the fan blade? Some blades go very fast and I feel as though measuring frequency would be more time consuming and not as accurate as voltage.
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Anomaly
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### Re: Wind Power B/C

I'm trying to think of different formulas they ask for wind power... can anyone help? I mean, i've been doing wind power for a bit and i have a fairly decent binder but the formulas for calculations always confuse me. Can anyone help with this?

PS: i don't know how the heck we got 2nd for wind power at regionals
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freed2003
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### Re: Wind Power B/C

This question has been posted here before, and the posted answer is wrong.

You're correct that distance is a factor in power loss. For a constant gauge wire, it will have a constant resistivity - or resistance per unit length. So as the length increases, so does the distance.

But here, they've given the resistance - so that's the total resistance of the entire length. The given distance is irrelevant (although you could be asked to compute the resistivity).

For the above question itself, the answers should be:

A. I = P/V = 900 MW / 700 kV = 900e6 W / 700e3 V = 1286 A

B. The methodology in the solution is correct. Voltage Drop = Current * Resistance (Ohm's Law). The resistance is given, not the resistivity, so the distance is irrelevant here. So you just put in the 0.2 ohms.

So Vdrop = 1286 A * 0.2 ohm = 257.2 V (not kV!).

Then power lost = Vdrop * I = 257.2 V * 1286 A = 330.8 kW (not MW!).

Alternatively if you plug in symbolically, you get Vdrop = I * R, and P = Vdrop * I, so P = I * (I * R) or P = I^2 * R. Then P = (1286 A)^2 * 0.2 ohms.

C. Percent loss = 330.8 kW/900 MW = 330.8e3/900e6 = 0.0367% loss. The posted solution of 36.7% is wrong, and a line loss of over 1/3 would be absurd.

To answer your question about what if the line was 50 km - if it was 50 km but the resistance was still given as 0.2 ohms, the solution would not change.

But, if the same gauge line was used, and the resistance was not given, we could determine the resistance. The resistivity of the line is 0.2 ohms/100 km = 0.002 ohm/km. So if the distance was 50 km of the same line, the resistance would be 0.002 ohms/km * 50 km = 0.1 ohm. Then proceed as above.
But to get the resistance from the resistivity won't you also have to know things like the cross section area?
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Ionizer
Member Posts: 32
Joined: March 20th, 2014, 6:03 pm
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### Re: Wind Power B/C

You're correct that distance is a factor in power loss. For a constant gauge wire, it will have a constant resistivity - or resistance per unit length. So as the length increases, so does the distance.

...

But, if the same gauge line was used, and the resistance was not given, we could determine the resistance. The resistivity of the line is 0.2 ohms/100 km = 0.002 ohm/km. So if the distance was 50 km of the same line, the resistance would be 0.002 ohms/km * 50 km = 0.1 ohm. Then proceed as above.
But to get the resistance from the resistivity won't you also have to know things like the cross section area?
I think there is a misunderstanding here. Resistivity is not resistance per length. Resistivity has units of ohm meters, not ohms per meter. The technique andrewwski describes works, but it's not using resistivity, it's using some other quantity. You can use the resistivity equation to get the same answer through proportions. The cross sectional area will cancel out.
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