## Wind Power B/C

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?

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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?
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
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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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heiber
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### Re: Wind Power B/C

[Resistance changes with length yeah but I think you don't need length in this problem. Resistivity I believe measures resistance per length, but the resistance is given to you so I think the length can safely be ignored.
OK. That helps. We've been trying to find equations and information that include distance. So the distance is included in the resistance calculation. Here is another example from the same test.

21. If a power plant is generating 100 MW, running on a 20 KV line (20 kilometers long), loses 10 MW of power after running along the line, how much resistance does the line have? (2 pts) Their answer:
1) Find Amps: P/V 100MW / 20KV= 5000A
2) Find Ohms: V/I= 20KV / 5MA= 4 ohms

My questions:
From Ohms Law, Resistance = Power / (Current * Current) or 100 MW / (20 KV * 20 KV) = 250 which is different. Why can't you use this equation and why is the answer different from above. All of the equations appear to be coming from the electrical formula wheel so would think it would work out either way.
Again - have information such as power loss of 10 MW and length of 20 kilometers that does not appear to even be used.

Thanks again.[/quote]

Bumping to see if I can get help understanding the solution to this problem. Thanks.

freed2003
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### Re: Wind Power B/C

Ummm volts don't measure current
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.
I'm still confused, you said they weren't using resistivity but then said it works? What would the cross section cancel with
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Ionizer
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### Re: Wind Power B/C

I'm still confused, you said they weren't using resistivity but then said it works? What would the cross section cancel with
$\frac {R_{1}}{R_{2}}=\frac {p_{1}\frac {l_{1}}{A_{1}}}{p_{2}\frac {l_{2}}{A_{2}}}$

In this case ${p_{1} = p_{2}}$ and ${A_{1} = A_{2}}$, so ${p_{1}, p_{2}, A_{1}, A_{2}}$ all cancel out, so it turns out you don't need resistivity or cross sectional area to find ${R_{2}}$. Once you plug in all your numbers you can just solve for ${R_{2}}$.
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### Re: Wind Power B/C

Check this out if you need ideas for wind turbines: https://www.youtube.com/watch?v=603F7vPUbrg

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freed2003
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### Re: Wind Power B/C

Nice, what resistor are you on?

Since I didn't want to double post I'll edit this in
33) The power generated by a turbine is 18,000 Watts. If the wind speed, which is initially 30 m/s decreases by 25%, what must the lift force be in order to generate the same amount of power? A) 600 N B) 2,400 N C) 60 N D) 240 N E) 800 N
Can someone help? How would you know the initial lift force?
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