High voltage power transmission?

Click For Summary

Discussion Overview

The discussion revolves around the principles and reasoning behind high voltage power transmission, particularly focusing on the role of transformers and the implications of voltage and current on power loss in transmission lines. Participants explore various aspects including Ohm's law, conductor and insulator losses, and economic considerations related to wire sizing.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Conceptual clarification

Main Points Raised

  • One participant questions the rationale behind high voltage transmission, noting that while reducing current (I) decreases power loss (P=I^2 R), it seems counterintuitive that stepping up voltage (V) would not similarly affect power loss.
  • Another participant clarifies that Ohm's law applies to loads, and the primary reason for high voltage transmission is indeed the I^2 R losses in conductors.
  • A different perspective introduces the idea that there are two types of losses: conductor loss (I^2 R) and insulator loss (V^2 G), suggesting that insulator losses are significantly lower at high voltages compared to conductor losses.
  • One participant provides a numerical example illustrating the vast difference in losses at high voltage versus low voltage, emphasizing that insulators are less lossy than conductors.
  • Economic factors are discussed, with a participant noting that higher voltage allows for smaller diameter wires, which reduces costs for power transmission.
  • Another participant mentions that AC transmission requires matching line and load impedance to minimize power loss, hinting at the complexities involved in practical applications.
  • A later reply acknowledges a misunderstanding regarding voltage and resistance, indicating a learning moment for the original poster.

Areas of Agreement / Disagreement

Participants express varying viewpoints on the principles of high voltage transmission, with some agreeing on the significance of I^2 R losses while others introduce additional factors like insulator losses and economic considerations. The discussion does not reach a consensus, as multiple competing views remain.

Contextual Notes

Participants highlight the distinction between different types of losses in transmission lines and the implications of AC transmission, but the discussion does not resolve the complexities or assumptions involved in these concepts.

Wallace
Science Advisor
Messages
1,256
Reaction score
0
Okay I have a question for you engineer types. I teach a little high school physics and something has struck me as slightly odd in what they get taught. They learn about transformers and are told that a major application of them is in electrical power transmission. They are told that since the power dissipated in a resistor is P=I^2 R then by reducing I (by stepping up V) the power loss is reduced in high voltage lines.

My obvious question then is why the reverse is not true, i.e. using Ohm's law it's clear that P=V^2 / R and the most obvious form P=IV shows that it doesn't make any difference to the power loss by stepping up or down the voltage/current.

So, am I wrong at this simple level?

Alternatively, what is the real reason for high voltage transmission? One guess I would have is that the wires are non-ohmic and high currents increase the resistance of the wire more than high voltage. This could well be bollocks.

Alternatively, is the AC nature of the transmission the key. I.E. does it cost more energy to correct the power factor due to the phase shift of the inductive transmission lines if the current is high but it easier if the voltage is high? I have no idea why this should be the case, I'm just guessing here.

Can anyone clear this up for me?
 
Engineering news on Phys.org
Ohm's law referrs to a load and a transmission line is not a load. V is the voltage drop across a resistor (for example).

The IR^2 loss really is the reason for transmission lines being high voltage.
 
The "R" in your 2 equations are not the same resistance.

There are 2 loss elements in transmitting power. The conductor loss is I^2*R, and the insulator loss is V^2*G. R is the line series resistance, and G is the line shunt conductance. As an example let's just consider a power level of 150 MW, or 750 kV with 200 A (one phase for illustration). Why, one may ask, 750kV/200A, and why not 200V/750kA?

The V^2*G insulation loss is MUCH MUCH LOWER than the I^2*R conduction loss at voltages up to somewhere in the MV range. As an example, if the shunt conductance of a cable is 1/(10 Gohm) or 0.1 nS, and the series resistance is 10 milliohm, at 200V/750kA, the insulator loss is (200V)^2*1e-10S = 4.0 uW. The conductor loss is (750kA)^2*1e-2 ohm = 5.625 GW!

At 750kV/200A, the insulator loss is (750kV)^2*1e-10S = 56.25 W. The conductor loss is (200A)^2*1e-2 ohm = 400 W. Quite a difference!

As the voltage is raised, the V^2*G insulator loss increases, but the I^2*R conductor loss decreases. Since the conductor losses are enormous in comparison, it pays greatly to transmit at high voltages. In a nutshell, insulators are less lossy than conductors. Insulators are "superinsulating" at normal temperatures, whereas conductors are not superconducting unless cooled down to very low temperatures. In the future when and if high temperature superconductors are available, things could change. BR.
 
From the economic standpoint, transmission line wires can be much smaller in diameter when the line voltage is high. To get the same power transfer at a low voltage, the conductors would cost a fortune.

About loss; AC transmission means one must match the line and load impedance to get the best transfer. If this is not considered, then power would be lost by radiation. (ever hear the hum on your radio when driving under a transmission line?)
 
Wallace said:
Okay I have a question for you engineer types. I teach a little high school physics and something has struck me as slightly odd in what they get taught. They learn about transformers and are told that a major application of them is in electrical power transmission. They are told that since the power dissipated in a resistor is P=I^2 R then by reducing I (by stepping up V) the power loss is reduced in high voltage lines.

My obvious question then is why the reverse is not true, i.e. using Ohm's law it's clear that P=V^2 / R and the most obvious form P=IV shows that it doesn't make any difference to the power loss by stepping up or down the voltage/current.

So, am I wrong at this simple level?

Yes. The problem with this line of reasoning is that P = IV is the power delivered by an electrical energy source not the power dissipated by it.

Hope that helps.

CS
 
Thanks all, yep I was mixing up the supplied nominal V of the power lines with the V lost by the resistance. Makes perfect sense now!

Cheers
 

Similar threads

Electric power distribution from powerplant to homes
  • · Replies 38 ·
2
Replies
38
Views
5K
How transmission lines deliver (established?) power from generator
  • · Replies 8 ·
Replies
8
Views
3K
Propagation speed and LC model of transmission line
  • · Replies 21 ·
Replies
21
Views
3K
Can Superconducting Cables Overcome AC Transmission Distance Limits?
  • · Replies 2 ·
Replies
2
Views
2K
Would high voltage DC power grids work the same?
  • · Replies 4 ·
Replies
4
Views
2K
Power, current, voltage transfer principles
  • · Replies 10 ·
Replies
10
Views
2K
A single wire electrical energy transmission....
  • · Replies 5 ·
Replies
5
Views
3K
Charge distribution on a power line
  • · Replies 7 ·
Replies
7
Views
3K
Power Dissipation along power lines and Resistance value
  • · Replies 9 ·
Replies
9
Views
5K
Alternator torque when connected to a phase-shifted AC source
  • · Replies 3 ·
Replies
3
Views
2K