Energy Loss vs Energy Delivered and Voltage Drop - Confused!


by mathological
Tags: energy loss, power lossess, power transmission, voltage drop
mathological
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#1
Feb11-11, 05:15 AM
P: 10
Hi,

ok so everytime I think that I have understood the concept of Energy transmission, losses and voltage drop, I get even more confused about things.

I have searched several threads on this forum and the physics forum but failed to find anything that directly answers my query. So I kindly request your help on the fundamental concept.

I will try to give an example and I would like you guys to correct me where I am wrong -- Thanks in advance!!!
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If a 1 MW load is to be supplied by a 10kV feeder, and the cable resistance is 1 ohms, then the current drawn by the load (or the current travelling on the cable) is:

I = P/V = 100 A

The line losses are I^2*R = 10 kW and the Voltage Drop = I*R = 100 V

Is it right to say that:

a) Voltage at supply point is 10kV
b) Voltage at consumer end is 9.9kV (10kV - 100V)
c) The current through the cable is 100A
d) The power required is 1 MW but the actual power being delivered is 0.99 MW [ (P - losses) OR (V*I = 9.9kV * 100A) ]
e) hence the efficiency of this system is 0.99MW/1MW * 100 = 99% efficient?
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Please correct me where I am wrong.

Thank you.
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Fish4Fun
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#2
Feb11-11, 07:33 AM
P: 247
All looks right to me.

Fish
lorenb
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#3
Feb11-11, 01:07 PM
P: 28
nevermind... looks good

sophiecentaur
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#4
Feb11-11, 03:38 PM
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Energy Loss vs Energy Delivered and Voltage Drop - Confused!


@mathological
It's near enough OK but, to be accurate, you need to know the actual resistance of the load (assume it's V2/Nominal Power where V is the nominal operating voltage) - that would imply 100Ω load resistance.
You have assumed a cable resistance of 1Ω which would be in series with a load of 100Ω.
The voltage drop across the feeder will be 10kV*1/101.
This gives a voltage across the load of 10kV/1.01 which gives about 98% of the power (proportional to the square of the volts). So it's less efficient than you would initially have thought. There's a double whammy in there! Weird huh.

Edit - Tidied up the bit about the volts on the load - no change in info- just removed a duplication
lorenb
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#5
Feb11-11, 03:50 PM
P: 28
what if the load was hooked up in parallel?
sophiecentaur
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#6
Feb11-11, 04:38 PM
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In parallel with what??
You wouldn't connect the cables straight across the mains!!!!
lorenb
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#7
Feb11-11, 05:22 PM
P: 28
true
mathological
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#8
Feb12-11, 06:17 AM
P: 10
Quote Quote by sophiecentaur View Post
@mathological
It's near enough OK but, to be accurate, you need to know the actual resistance of the load (assume it's V2/Nominal Power where V is the nominal operating voltage) - that would imply 100Ω load resistance.
You have assumed a cable resistance of 1Ω which would be in series with a load of 100Ω.
The voltage drop across the feeder will be 10kV*1/101.
This gives a voltage across the load of 10kV/1.01 which gives about 98% of the power (proportional to the square of the volts). So it's less efficient than you would initially have thought. There's a double whammy in there! Weird huh.

Edit - Tidied up the bit about the volts on the load - no change in info- just removed a duplication
You have assumed a cable resistance of 1Ω which would be in series with a load of 100Ω.
The voltage drop across the feeder will be 10kV*1/101. Are you using voltage division here?

This gives a voltage across the load of 10kV/1.01 which gives about 98% of the power (proportional to the square of the volts). Sorry I dont get how you calculated this...can you please elaborate? Thanks!
Phrak
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#9
Feb12-11, 06:33 AM
P: 4,513
Look, mathlogical, you have stipulated that your load dissipates 1MW. Now. you have to be clear whether this 10KV pair of wires you call a feeder has 10KV at the source or 10KV at the load. The voltage at the source is not the same as the voltage at the load.

Pick one or the other and the answer is determinate.
sophiecentaur
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#10
Feb12-11, 11:34 AM
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@mathological
Yes, I am using voltage division (it's the same old potential divider circuit that we come across everywhere).

I was assuming that the original "1MW load" was one which is designed to dissipate 1MW when presented with 10Kv. Few loads adjust themselves to take their specified power; they are mostly 'dumb' resistors, or equivalent.
The power dissipated in a resistor is V2/R so the power you get at the end of the cable will be (1/1.01)2 (=0.98) as much as without the cable.
Actually, to be fair, the supply (generator) would be delivering less current (also 1/1.01 as much) so, defining efficiency as:
power out/power supplied
the actual efficiency will not be as low as 98% because less actual power will be put into the system by the generator. But you still get only 98% of the power you wanted.
mathological
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#11
Feb12-11, 05:50 PM
P: 10
@sophiecentaur

Thanks for the clarification! :)


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