Why transmission lines are high voltage?

JMW

Joined Nov 21, 2011
137
Yeah... we recently discussed that recently in this forum... it's due to the skin-effect
Skin effect deals with current flowing on the skin of the conductor, it increases with frequency. This is irrelevant on power lines and even more so on the DC filament voltage of a magnetron
 

cmartinez

Joined Jan 17, 2007
8,218
Skin effect deals with current flowing on the skin of the conductor, it increases with frequency. This is irrelevant on power lines and even more so on the DC filament voltage of a magnetron
I'm well aware that the skin effect is dependent on frequency, and that at 60Hz it's neglegible...

My comment was meant in the context of his message:
"One other point, look up the "ampacity" of a cable. Say you want 1 KW (1 volt at 1000 amps). I recall a 60's era air search RADAR with a 1 volt magnetron supply, the current was over 100 amps, at any rate the "cables" were copper pipe."

I'm not sure at what frequency the magnetron that he talks about worked. But I'm sure there must be a reason for them to have used copper pipe... if not for skin effect, then for running a coolant through it?
EDIT: Or maybe because of rigidity issues?
 

WBahn

Joined Mar 31, 2012
29,976
Skin effect deals with current flowing on the skin of the conductor, it increases with frequency. This is irrelevant on power lines and even more so on the DC filament voltage of a magnetron
It's not irrelevant on power lines as the size of many transmission line cables are large enough that skin-effect plays a significant role. It's one of the reasons that they use bundled conductors.
 

RamaD

Joined Dec 4, 2009
328
P=I^I*R, V*V*R, right as you said.
Here we are concerned about power loss on the wire during transmission.
The voltage is the voltage drop, and not the high voltage itslef!
Since I is reduced with higher transmission voltage V, IR drop or Vdrop is also reduced!
 

JMW

Joined Nov 21, 2011
137
I'm well aware that the skin effect is dependent on frequency, and that at 60Hz it's neglegible...

My comment was meant in the context of his message:
"One other point, look up the "ampacity" of a cable. Say you want 1 KW (1 volt at 1000 amps). I recall a 60's era air search RADAR with a 1 volt magnetron supply, the current was over 100 amps, at any rate the "cables" were copper pipe."

I'm not sure at what frequency the magnetron that he talks about worked. But I'm sure there must be a reason for them to have used copper pipe... if not for skin effect, then for running a coolant through it?
EDIT: Or maybe because of rigidity issues?
Magnetron filament voltage is DC, the plate is typically several thousand volts minus WRT ground. And yes it was pipe to contain the dionized water coolant. Several hundred amps even through copper pipe heats things up.
 

JMW

Joined Nov 21, 2011
137
It's not irrelevant on power lines as the size of many transmission line cables are large enough that skin-effect plays a significant role. It's one of the reasons that they use bundled conductors.
IIRC the bundled conductors are for strength and flexibility. Copper pipe or waveguide don't survive well in a swaying environment. Even Cmartinez agrees that skin effect at 50/60 Hz is negligible. My statement was to the OP who asked why the high voltage. Ampacity wasn't mentioned on this thread, it has been at other times. Yes, lower voltages at higher amperages can be used, but you need Godzilla to run the lines I don't recall the reasoning behind a 1 volt filament voltage.
 
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Thread Starter

Vorador

Joined Oct 5, 2012
87
You can use either expression and you will get the same result. The key is that to use P=(I^2)/R to find the power loss IN THE LINE you need to use the current THROUGH the line. Similarly, to use P=(V^2)/R to find the power loss IN THE LINE you need to use the voltage ACROSS the line. You are trying to use the voltage across B (the load or the generator) to find the power loss in A (the line).

Simple example. You have a line that has 0.1Ω of resistance and you need to deliver 100kW to a load. You can do this either with a 1kV line pulling 100A or you can do it at 100kV pulling 1A. How much power is lost in the line in both cases?

Using P=(I^2)/R:
1kV -> P = (100A)^2 * 0.1Ω = 1000W
100kV -> P = (1A)^2 * 0.1Ω = 0.1W

Using P=(V^2)/R:
1kV -> V = (100A)(0.1Ω) = 10V; P = (10V)^2 / 0.1Ω = 1000W
100kV -> V = (1A)(0.1Ω) = 0.1V; P = (0.1V)^2 / 0.1Ω = 0.1W
Wonderful! That makes it absolutely clear. I should have tried doing the math myself.

Thank you, WBahn!

The V^2/R you refer to is load power, whereas I^2*R is power lost in line. Big difference. The reason is straight forward as to why high voltage low current is employed.
Insulators are very good compared to conductors with regard to losses. Insulators incur a loss in the form of shunt conductance, denoted by G. G is reciprocal of R, but this R is the insulation resistance, different from load R, or cable R. Power lost in insulation is V^2*G. For any good insulator over a wide span of temperatures in common use, this loss is tiny. But it increases with square of voltage.
In conductors, power lost is I^2*R, where R is the cable resistance and I is load current. Load current and cable current are equal as the cable is in series with the load. As current increases, loss increases with square of current.
Increasing V results in lower I for a constant power load. High V incurs greater insulation loss and lower cable loss. But insulation loss is so many orders of magnitude below cable loss, that it is very advantageous to transmit power at high V values.
If cables could be designed for superconductivity at normal temperatures, then things might be different. THer limitation with SC cables is the critical current value where SC is lost. Depending on the values, the transmission voltages could be lowered, but only to a point. I doubt we will ever see 125V with 1 million amps. Even SC cables are limited in their current capacity. Instead of 750kV at 200 amp, we may see 75kV at 2,000 amp, depending on the SC cable.
Anyway, as long as insulators are ultra-low loss, and cable losses are much greater, which at this time is the case, higher voltage with lower current will incur much lower losses. So that is why it is done this way. Did I help.

Claude
Very nice explanation. You certainly helped a lot, Claude! :)

Thanks all for all your informative answers! You guys are the best! :)
 

cmartinez

Joined Jan 17, 2007
8,218
IIRC the bundled conductors are for strength and flexibility. Copper pipe or waveguide don't survive well in a swaying environment. Even Cmartinez agrees that skin effect at 50/60 Hz is negligible. My statement was to the OP who asked why the high voltage. Ampacity wasn't mentioned on this thread, it has been at other times. Yes, lower voltages at higher amperages can be used, but you need Godzilla to run the lines I don't recall the reasoning behind a 1 volt filament voltage.
To clarify my comment further (btw, thanks for clearing that the magnetron runs on DC... I didn't know that) I meant that skin effect is negligible at 60 Hz in ordinary loads... When thousands of amps are required then it does become important:
At 60 Hz in copper, the skin depth is about 8.5 mm.

This means that the maximum practical wire (solid wire, that is... one could use Litz wire to increase efficiency to a further point) diameter at our normal household frequency is 17mm (0.670") ... haven't done the calculations for the max amps of that wire at 60 Hz ... a correction factor involving temperature would also be involved
 

WBahn

Joined Mar 31, 2012
29,976
IIRC the bundled conductors are for strength and flexibility. Copper pipe or waveguide don't survive well in a swaying environment. Even Cmartinez agrees that skin effect at 50/60 Hz is negligible. My statement was to the OP who asked why the high voltage. Ampacity wasn't mentioned on this thread, it has been at other times. Yes, lower voltages at higher amperages can be used, but you need Godzilla to run the lines I don't recall the reasoning behind a 1 volt filament voltage.
http://en.wikipedia.org/wiki/Overhead_power_line#Bundled_conductors
 

ian field

Joined Oct 27, 2012
6,536
Skin effect deals with current flowing on the skin of the conductor, it increases with frequency. This is irrelevant on power lines and even more so on the DC filament voltage of a magnetron
That would be a bit puzzling!

The magnetrons used in microwave ovens have a little screened box housing a pair of chokes and feed-through filter capacitors to *STOP* any microwaves getting out that way.
 
A couple of things you don't hear alot, is Galloping Power lines and a Deer on the line.

Max.
50 years ago when I was a young idiot kid I use to have fun attaching a 25 foot long wire #30 copper wire to a helium balloon then release it under the megawatt power lines then take off on my bicycle to get away. It worked best to have the wire almost too heavy for the balloon to lift so the balloon would go up very slow and the wind would blow it into the HV wires. Release the balloon 200 ft from the power lines the wind did the rest. When the #30 wire shorted out the HV wires it was amazing. LOL. Crazy.
 

MaxHeadRoom

Joined Jul 18, 2013
28,617
50 years ago when I was a young idiot kid I use to have fun attaching a 25 foot long wire #30 copper wire to a helium balloon then release it under the megawatt power lines then take off on my bicycle to get away. It worked best to have the wire almost too heavy for the balloon to lift so the balloon would go up very slow and the wind would blow it into the HV wires. Release the balloon 200 ft from the power lines the wind did the rest. When the #30 wire shorted out the HV wires it was amazing. LOL. Crazy.
.....So that was YOU!:(
Max.
 

JMW

Joined Nov 21, 2011
137
To clarify my comment further (btw, thanks for clearing that the magnetron runs on DC... I didn't know that) I meant that skin effect is negligible at 60 Hz in ordinary loads... When thousands of amps are required then it does become important:
At 60 Hz in copper, the skin depth is about 8.5 mm.

This means that the maximum practical wire (solid wire, that is... one could use Litz wire to increase efficiency to a further point) diameter at our normal household frequency is 17mm (0.670") ... haven't done the calculations for the max amps of that wire at 60 Hz ... a correction factor involving temperature would also be involved
Thank you sir!!!
All vacuum tubes and transistors, run on DC. Configured appropriately, they modify, amplify, or generate AC. Magnetrons are unique, the physical design actually creates a resonant chamber. The British developed the "Maggie" prior to WWII, Churchill hand carried it to the US after we entered the war. He was afraid it would be handed over to the Axis or the plant would be bombed. IIRC the RADAR at Pearl Harbor operated at at 6 meters. 2 meters was considered UHF in those days. The maggie introduced true UHF, in fact they are available into the SHF range. I think the Air Search I worked on was in 220 MHz range. Don't even remember the nomenclature. I never went to school on, I was a helper to the actual tech. The Senior Chief was working on our SPS 10, and got bit, 8KV as I recall. fortunately the rubber mats, shoes etc were in place.
 

cmartinez

Joined Jan 17, 2007
8,218
Thank you sir!!!
All vacuum tubes and transistors, run on DC. Configured appropriately, they modify, amplify, or generate AC. Magnetrons are unique, the physical design actually creates a resonant chamber. The British developed the "Maggie" prior to WWII, Churchill hand carried it to the US after we entered the war. He was afraid it would be handed over to the Axis or the plant would be bombed. IIRC the RADAR at Pearl Harbor operated at at 6 meters. 2 meters was considered UHF in those days. The maggie introduced true UHF, in fact they are available into the SHF range. I think the Air Search I worked on was in 220 MHz range. Don't even remember the nomenclature. I never went to school on, I was a helper to the actual tech. The Senior Chief was working on our SPS 10, and got bit, 8KV as I recall. fortunately the rubber mats, shoes etc were in place.
The history of radar technology development by the brits and their handling over to the americans in WWII is another fascinating story... I had forgotten that the magnetron was one of it's key elements...
 

JMW

Joined Nov 21, 2011
137
I've read on a number of places that, since P=I^2*R, the lower the current the lower the power losses will be, so transmission lines carry low current, high voltage. But Power can also be written as P=V^2/R, so if we want to avoid losses, voltage also needs to be down, not just current. The way I understand it, it doesn't matter which quantity you step up and which you step down, because power is compensated by the reciprocal effect current and voltage have upon each other.

I've always been confused by this. I'll be really grateful for your illuminating response.

Thank you!:)
Did you get a satisfactory answer to your quest?
 

SoftwareGuy

Joined Oct 29, 2013
29
Think of it like this. Power losses are given off as heat. Heat is created by current. More current in a wire equals more power losses to heat. For efficient power transmission, the goal is to reduce current as much as possible. P=I^2 * R, so for any given wire with a fixed R, the power dissipated by the wire goes up as the square of the current.
 
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