To reduce cost and weight, power transmission lines are made

AI Thread Summary
The discussion focuses on calculating the resistance of a 30 km aluminum wire, the current through it when transferring 1000 MW at 500 kV, and the power loss during transmission. The initial resistance calculation yielded an incorrect value of 4.23 x 10^14 ohms, prompting requests for clarification on the calculations and the wire's specifications. Participants emphasize the importance of using consistent units for resistivity, length, and cross-sectional area in the resistance formula R = ρ(l/A). There is confusion regarding whether to use square meters or square millimeters for area, with a consensus that consistent units are crucial for accurate results. The thread highlights the need for accurate resistance values to determine power loss and its fraction of the transmitted power effectively.
g98
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Homework Statement


Determine the resistance of a 30 km long wire. b. Assume that the line transfers 1000 MW of power at a potential difference of 500 kV. What is the current through the wires? c. How much power is lost during transmission? What fraction of the transmitted power is lost?

Homework Equations


R=p*l/A P=I*deltaV P=I^2*R

The Attempt at a Solution


so, for the first one i determined the resistance to be 4,23*10^14 ohm, the current through the wire is 2000A . I am not quite sure for the above values and because of that, i get weird values for the lost power in c) I would very much appreciate a bit of help with the exercise
 
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The wire resistance can't be 10^14 ohms. You have done something wrong. Show us how you did the calculations of the wire resistance.
 
I think there must be some important information we're missing, probably given in a "part a" portion of the problem?
 
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g98 said:
How much power is lost during transmission? What fraction of the transmitted power is lost?
That can't be determined unless the resistance of the line is specified.
 
Last edited:
gneill said:
I think there must be some important information we're missing, probably given in a "part a" portion of the problem?
The thing i didn't mention is that the wire is aluminium and the diameter is 4cm
 
phyzguy said:
The wire resistance can't be 10^14 ohms. You have done something wrong. Show us how you did the calculations of the wire resistance.
I was wondering for the formula R=p(l/A) (p being the specificity) what should be the units of the area? i though it should be m^2 but i checked in internet and in many places i saw mm^2 and i got slightly confused
 
g98 said:
I was wondering for the formula R=p(l/A) (p being the specificity) what should be the units of the area? i though it should be m^2 but i checked in internet and in many places i saw mm^2 and i got slightly confused

OK, You're right that the resistance is given by R=ρ(l/A). You can use any units you want as long as you are consistent. If your value for resistivity is in Ohm-m, then you would want to use m for the length and m^2 for the area. Why don't you set up this calculation and show us the values you are using and the results.
 

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