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.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?
But to get the resistance from the resistivity won't you also have to know things like the cross section area?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.
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.But to get the resistance from the resistivity won't you also have to know things like the cross section area?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.
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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.
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.[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.
I'm still confused, you said they weren't using resistivity but then said it works? What would the cross section cancel withI 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.But to get the resistance from the resistivity won't you also have to know things like the cross section area?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.
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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