Why is high voltage better for power transmission than high current?

[berlinpose2]

I've done some research and the following is what I understand of the topic. Please correct me if I'm wrong.

When power is transmitted, electricity is sent to a transformer, which increases the voltage and decreases the current according to the relationship S = IV. The reason for doing this is to minimize power losses along the transmission line, which is equal to RI2 . However, isn't power loss also equal to V2 /R, so having a large voltage would also cause a large power drop? Clearly there is a fault in my logic here because the power calculated with current and the power calculated with voltage would not be equal, so I am looking for an explanation of this. Thanks!

 

[crshears]

I do not know where you're getting your formulas, but it is high current that causes losses, not high voltage. Losses through either resistance or impedance, viz., inductive or capacitive reactance, will cause the voltage drops that cause the losses, therefore the higher the operating voltage and the lower the current, the better.

CR

"As iron sharpens iron, so one person sharpens another." [Proverbs 27:17, NIV]

 

[zeusfaber]

You can't just plug numbers into a formula without first making sure they represent the quantity the formula is talking about.

In the case of[tt] [/tt]W = V[sup]2[/sup]/R,[tt] [/tt]V is the voltage drop along the length of the conductor whose power loss you're calculating, not the voltage between that conductor and the others in the circuit - and this V goes down when you increase the system voltage.

Many years ago, I spotted an (extremely clever) boss blindly plugging a totally inappropriate height into a complex formula to be used as evidence for certifying something as safe for flight. Ever since then, "Read the bit that says 'Where V is...'" has been a bit of a hobby horse for me.

 

[crshears]

Nice one, Zeus! LPS for you! But how do you manage to write in superscript like that? I'm somewhat eng-tips-typing challenged . . .

CR

"As iron sharpens iron, so one person sharpens another." [Proverbs 27:17, NIV]

 

[zeusfaber]

The fifth button from the left in the toolbar at the top of the box that you type your reply into inserts a pair of [ignore][sup][/sup][/ignore] tags and you just need to ensure that the text you want to superscript lands between those.

A.

 

[cranky108]

For the same resistance, if you double the voltage the loss decreases by a factor of four.

Losses = I squared Z. The lower the current, the lower the losses.

Electrical Engineering is a 4 or 5 year degree. It might takt me that long to explain everything.

 

[FacEngrPE]

Or if you want to lower the voltage with the same loss the conductor cross sections get really large, fault currents get so large circuit protection becomes ineffective. and you get some of this

https://youtu.be/tJ1yEndX6ng

 

[LionelHutz]

Actual losses, as in power lost as heat, are only due to resistances in the circuit. Voltage drop across an inductive element aren't lost as heat.

 

[sushilksk]

In our country it is I[sup]2[/sup]R

 

[crshears]

Actual losses, as in power lost as heat, are only due to resistances in the circuit. Voltage drop across an inductive element aren't lost as heat.

Ah! That's something I never quite wrapped my head around, but now that you say that, it makes sense. I've read up about an older technology called the magnetic amplifier and how it controls current flow but without overheating; based on voltage drop without heat buildup, that computes.

Thanks for that! LPS to you!

Addition via edit: and that's why even though VARs don't "travel" well, at least they don't incur additional losses . . . then again, I was always taught that current is current, in phase or not, and that if a system element reaches its current limit, whether due to reactive or real power flow, something's gotta give, because there's no such thing as a free lunch. The corollary was that if you reduced the amount of reactive power flowing in a circuit it releases more of that circuit's capacity to transfer real power.

What am I missing?

 

[xnuke]

I don't think you're missing anything. There are two components to total current, the real part that does reals work or is turned to heat in the resistive portion of a circuit, and the reactive part that flows to/from the reactive energy elements in the circuit, I = I[sub]R[/sub] + jI[sub]X[/sub]. Losses still equal I[sup]2[/sup]R, so if the reactive component of the current grows, it limits the real portion of current available for a given total current. This is why having more vars limit the kW flow on a line; since the total current on the line is limited by the line's ampacity, total maximum current I[sub]max[/sub] must be constant, so if I[sub]X[/sub] goes up, I[sub]R[/sub] must go down. Reduce the reactive component of current, and more real current can flow without destroying the component through which it is flowing.

xnuke
"Live and act within the limit of your knowledge and keep expanding it to the limit of your life." Ayn Rand, Atlas Shrugged.
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[LionelHutz]

The current in a wire always flows through the resistance of the wire, regardless of the inductance or capacitance that wire also has. So, yes if you try to push too much current through that wire it will overheat and fail. The wire can be overhead, underground or inside a transformer, capacitor or MCC. It could be shaped like a piece of bus bar. The same too much current = overheating applies.

 

[davidbeach]

Voltage drop across the impedance of the circuit, not operating voltage. With the correct V the two equations give exactly the same result for any circuit.

I’ll see your silver lining and raise you two black clouds. - Protection Operations

 

[bacon4life]

The V^2 portion is difference in voltage between ends of wire rather than the line voltage. For example:
A nominal 120V circuit would have 120 V at one end and 110 V at the other end while carrying 10 amp on a cable with 1 ohms of resistance. This gives (120 V-110 V)^2/ 1 ohm = 100 watts of losses

A nominal 240 V circuit would have 240 V at one end and 235 V at the other end while carrying 5 amp on a cable with 1 ohms of resistance. This gives (240 V-235 V)^2/ 1 ohm = 25 watts of losses

 

[LittleInch]

This appears to be an excuse to insert a very dodgy link into a post and isn't a proper question.

The offending response by the OP has been deleted, but it does look like this wasn't a proper question. The answers were pretty good though!

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[Sn00ze]

Because of the relationship that has been mentioned. These usually translate to a lot of cost when it comes to construction. Upsizing conductors/connectors, Clearances can become a nightmare when it comes to maintenance as well as lattices, etc etc. Keeping all of these down would be the main driver in deciding the best method to transmit power.

 

[LionelHutz]

Voltage drop across a high voltage transmission line is voltage dropped across both the resistance and impedance of the line, so it's not really this voltage drop that causes the loss. Yes, it is the voltage drop, but only the voltage drop across the resistance part of the line. How these pesky AC circuits work always screw up any simplified circuit analysis.

 

[waross]

You said better than I Lionel.
In practice, the line voltage drop is the current times the impedance. The losses are I[sup]2[/sup]R.
But, the voltage drop is IZ (Impedance). So the On-Load Tap Changer in the substation will ramp up to compensate for the lower incoming voltage. That drives the real current up and results in added indirect losses from inductive voltage drop.

How these pesky AC circuits work always screw up any simplified circuit analysis.

--------------------
Ohm's law
Not just a good idea;
It's the LAW!

 

[LionelHutz]

Bill, you'll fit the line to voltage drop is IZ, not I^2Z?

 

[waross]

Thanks Lionel. Correction made.

--------------------
Ohm's law
Not just a good idea;
It's the LAW!