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Power loss in transmission lines

by IniquiTrance
Tags: lines, loss, power, transmission
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IniquiTrance
#1
Feb25-09, 11:30 PM
P: 183
1. The problem statement, all variables and given/known data

A power station delivers 890 kW of power at 12 kV to a factory through wires with total resistance 5.0 [tex]\Omega[/tex].

How much less power is wasted if the electricity is delivered at 50 kV rather than 12 kV?


2. Relevant equations

eq-1) P = IV
eq-2) P = I2R

Ohm's Law: V = IR

3. The attempt at a solution

I know the problem can be solved by solving for the current value in each case from eq-1, and plugging it in to eq-2 to calculate power loss.

My real question is why can't we apply Ohm's law to calculate I, and then plug it into eq-2? Why must we use eq-1 to get the current? Doesn't Ohm's law apply for all ohmic conductors, and since we have both the resistance and voltage of and across the conductor, can't we get the current that way?

Thanks!
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Delphi51
#2
Feb25-09, 11:51 PM
HW Helper
P: 3,394
If you apply Ohm's Law to the line resistance, you would need to know the voltage drop in the line.

If you apply it to the load plus the line, you would have to use the combined resistance of line and load. You could use the power formula to find the load resistance, then use Ohm's Law to find the current.
IniquiTrance
#3
Feb26-09, 12:13 AM
P: 183
Quote Quote by Delphi51 View Post
If you apply Ohm's Law to the line resistance, you would need to know the voltage drop in the line.

If you apply it to the load plus the line, you would have to use the combined resistance of line and load. You could use the power formula to find the load resistance, then use Ohm's Law to find the current.
Thanks for the reply!

So when you have a 12kV transmission line, what is the 12 kV drop across? The negative terminal is obviously at the powerplant, the positive, after passing through the load, back at the powerplant?

Delphi51
#4
Feb26-09, 03:21 PM
HW Helper
P: 3,394
Power loss in transmission lines

A power station delivers 890 kW of power at 12 kV to a factory
The wording is careful, as it should be, to make it clear that the potential at the load is 12 Kv. It will have to be larger at the source to make up for the loss in the line resistance.

Virtually all transmission lines are AC, so there is no negative or positive terminal.
mgb_phys
#5
Feb26-09, 03:26 PM
Sci Advisor
HW Helper
P: 8,954
Quote Quote by IniquiTrance View Post
so when you have a 12kV transmission line, what is the 12 kV drop across?
It's between the output of the power line and ground - and in an ideal world is the same as the difference between the input to the powerline and ground.

This confuses everybody who first sees this question - do a search here and you will see dozens of threads on it!
IniquiTrance
#6
Feb26-09, 03:37 PM
P: 183
Thank you!

I have a much better understanding of the concept now.
Andrew Mason
#7
Feb26-09, 04:10 PM
Sci Advisor
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P: 6,653
Quote Quote by IniquiTrance View Post
Thanks for the reply!

So when you have a 12kV transmission line, what is the 12 kV drop across? The negative terminal is obviously at the powerplant, the positive, after passing through the load, back at the powerplant?
The voltage is the potential difference with no load. It does not matter what the load is. The question does not ask how much power is lost. It just asks for the relative power loss at these two different line voltages. All you have to know is that for a given load, if the 12Kv line has current I the 50Kv line the current will be 12I/50. The power loss across the line will be I^2R where R is 5 ohms so you can easily calculate the relative line losses.

AM
Rang
#8
Nov12-11, 01:03 AM
P: 2
KW KV R I=KW/KV Loss=I*I*R Loss reduction
890 12 5 74.17 27503.47
890 50 5 17.8 1584.2 25919.27
PF=1
Rang
#9
Nov12-11, 01:11 AM
P: 2
KW KV R I=KW/KV Loss=I*I*R Loss reduction
890 12 5 74.17 27503.47
890 50 5 17.8 1584.2 25919.27
PF=1


890 kw 12kv current= 890/12=74.17 Amp assuption pf=1, loss= I*I*R=27503 W
890kw, 50kv, current=890/50=17.8 amp, loss= 1584.2 w,
loss reduction= 25919.27 w,
As the voltage increases, current decreases for same power. current decreases loss also decreases. current has a squred relation with loss.


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