To reduce cost and weight, power transmission lines are made

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SUMMARY

The discussion centers on calculating the resistance of a 30 km aluminum wire with a diameter of 4 cm, transferring 1000 MW of power at a potential difference of 500 kV. The resistance was incorrectly calculated as 4.23 x 1014 ohms, leading to confusion regarding the current, which was determined to be 2000 A. The correct approach involves using the formula R = ρ(l/A) and ensuring consistent units, specifically using meters for length and square meters for area.

PREREQUISITES
  • Understanding of electrical resistance and Ohm's Law
  • Familiarity with the formula R = ρ(l/A) for calculating resistance
  • Knowledge of power transmission concepts, including current and voltage
  • Basic proficiency in unit conversion, particularly between meters and millimeters
NEXT STEPS
  • Calculate the correct resistance of the aluminum wire using R = ρ(l/A)
  • Determine power loss during transmission using P_loss = I²R
  • Explore the impact of wire diameter on resistance and power loss
  • Research the properties of aluminum as a conductor, including resistivity values
USEFUL FOR

Electrical engineers, physics students, and professionals involved in power transmission and electrical system design will benefit from this discussion.

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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