# Power Line design specs

How does the amount of power a transmission line can handle vary with its voltage and the distance? In other words, how much power could a 10 mile
long 69kV line carry during normal usage (excluding unusual conditions that may produce temporary overloads)? Or, put another way, if I had to transmit 100 megawatts 20 miles, what voltage would be necessary?
How does the answer vary if there are two circuits used rather than one? Is it just double?
I know the size of the conductors affects the answer, I'm curious about the largest commonly used size for a given voltage.
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I'm not a power transmission engineer, but to send that kind of juice that far, would require six figure voltages or you have to use expensive, heavy cables. At 69KV you're around 1000 amps on each leg.

Half the current, so it is possible to use lower transmission voltages.

Yes, it would be interesting if a power transmission engineer would jump in here. John
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I'm not a power transmission engineer, but I did stay at a Holiday Inn last night... :>
My old college reference book gives these capacity examples for a single-circuit, 3 phase transmission line.
138 kV ---> 80,000 kW 345 kV ---> 500,000 kW 765 kV ---> 2,500,000 kW
There are other specific values for factors such as sag, conductor size, overhead vs. underground, etc. The above figures are ballpark.
The important relationship is that the theoretical power capacity increases as the 'square' of increased voltage, thus the higher voltages are much more efficient at transmitting large quantities of power, with all else being equal. The practical upper limit for voltage, however, is determined by the point at which corona breakdown takes place, and, of course, the size-expense of the insulators, bushings, conductors that require spacers, extra surface area, etc.
Double circuit towers essentially give you redundancy and double the capacity of one three-phase line, but with the limitations of limited work clearances for maintenance and the potential dangers and undesirable effects of inductive coupling when a de-energized line is situated next to an energized line.
Also, theoretically, high voltage DC is more efficient than AC for long distance and large loads since there is only resistance and no reactance, opposing the current flow. In practice, DC transmission is more complex and expensive since AC-DC conversion must be done at either end of the line.
Beachcomber
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Also, during the early years of the 20th Century, it was determined that efficient AC power transmission could be done in the range of about 1000 volts/mile, as a minimum.
Beachcomber
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-------------- P =[(Vs*Vr)sin(theta)]/Zo =(SIL)sin(theta) where Vs and Vr are sending end and receiving end voltages in KV (line to line ) P is the power transmitted in MW (total) Zo is nearly root (L/C) where L and C are the inductance and capacitance per mile (Resistance can be neglected in most cases) and theta is the angle between the sending and receiving end voltages(line to neutral) . Outside limit on theta is 30 degrees (approx 0.12 degrees/mile) to maintain a stability margin and 15-20 degrees is better. A rough estimate for Vs =Vr is that Vs^2=Mw*miles for a 30 degree phase shift which puts the levels above at about 230-240 miles at the 30 degree limit and 150-200 miles as a practical limit. Certainly longer and shorter lines exist and power capacity increases or decreases accordingly.
Lots of factors involved but resistance is usually (except for short lines) the least of these. For HV lines, bundled conductors are used to reduce surface fields as well as inductance and generally the mechanical requirements mean that the conductor current capacity is not a problem. This also means that Zo for high voltage lines is about 250-275 ohms vs 350-400 ohms for single conductor, lower voltage lines. (so for a 20 degree phase shift at 765KV and a loading of 2350MW the distance would be about 170 miles. Whether this load could be transmitted a longer distance would depend on a host of other factors.
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Don Kelly snipped-for-privacy@shawcross.ca
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Thanks for the replies.
Another question: How do they normally transfer power from point A to point B when there are multiple lines of different voltages? I'm thinking as an example of a specific case where A and B are substations, there's a large power source adjacent to A (so power normally would flow from A to B), and A and B are connected with multiple lines: 2 345kV, 2 138kV, and 2 69kV. (The 69kV lines go a different route and feed 3 smaller substations along the way, and in normal use may be broken into 2 parts so no power goes from A to B along them, I don't know. The other lines have no taps between A and B)
Would they normally try to run most of the loads at substation B off the lower voltages while passing the 345kV through, would they load B so power is drawn proportional to each line's capacity, or can they somehow "parallel" the 138kV lines with the 345kV with 345-138 transformers at each end and something (what?) to balance the power according to capacity? 345kV and 138kV lines continue past B, also off from A.
Just curious.
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There are many different schemes in the substations The lower voltage lines would be the oldest and would come from nominal 69KV busses to 69KV busses at B .The same thing would be the case for the 138 and 345KV lines. As the system grew, there would be a need for more transmission capacity but that doesn't mean throwing out existing infrastructure- it gets demoted to a "lower" level. The 345KV line is now the backbone of the system while the 69KV mainly serves secondary loads. Depending on needs the three systems may be independent or can be interconnected at each end by transformers in the substations. There would be little or no need for control of individual lines as the power handling capacity of each is related to the square of the voltage level and, generally there is a natural balance that takes place. The lower voltage lines would be feeding the smaller loads at the in between substations while the 345KV line would bypass these (uneconomic to feed these loads from the 345KV lines) and only feed the major substation at B from which there may be a feed back to intermediate points at the lower voltage. Think of a net made with strong ropes, interliked with lighter ropes and cords.
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Don Kelly snipped-for-privacy@shawcross.ca
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Thanks for the info.

That certainly appears to be the case. The 69kV are on wooden poles, and while one of the 345kV lines is older than the 138kV towers, it's because they replaced two single circuit set of pylons with 2 circuit tower, to make room for the other 345kV line.

Are you saying that if I have a 345kV line, a parallel 138kV line connected to the 345 line with 345-138 transformers at each end, and a parallel 69kV line connected with appropiate transformers at each end, the power will naturally balance itself over the lines in a reasonable fashion without additional controls?
What happens when the routes are different? Consider an equilateral triangle A B C where the legs are long enough that phase shift is significant, and you wish to move power from A to B using both A-B and A-C-B paths (all paths same voltage), and a smaller load at C.
What about phase shift in the transformers in the first question?
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writes:

------ Basically yes. To analyse a particular situatio would require sufficient line data and load data to allow a load flow study which is a common procedure even for situations involving hundreds to thousands of lines where thousands of simultaneous non-linear equations have to be solved- quickly.

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The flow would depend on all the loads but note that the voltages at A and B
are the same for both lines. Is there a chance of overloading one line or
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It will "balance out" but the balance point may not be where you like it.
One would want to send the most current over the link with the least losses. One would model it as there independent voltage sources feeding three resistors connected to ground at the receiving end. The total current is what the load demands. The resistances are scaled with the square of the voltage. (A 1 ohm resistance on a 50 K line scales as 4 ohms on the 100 K line.) The currents are scales with the inverse of the voltage. You throw in some linear restraints (the maximum current on each line) and you have a problem that a bright high school student could solve but that your typical EE profession would not attempt untill you came up with his fee or a charge number for his time sheet.

Again, simple to model. High School stuff.

OK. Now you are generating some problems. If the voltage sources aren't in phase you have to make the currents complex.
But all you care about is the magnitude of the vector/phasor sum of the currents at the receiving end.
It looks like a problem that would be fun to solve if someone were paying for my time. On my own time, it just looks like homework and since I already have completed my education ....
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----------- Unfortunately that would lead to the wrong answer. A resistance model is not applicable as the current distribution is mainly controlled by reactance which is typically far greater than the resistance. Power flow on a given line depends on the voltage , reactance and phase angle with resistance having relatively little effect - The easy approximation simply ignores line losses. There is no way to avoid the use of complex numbers unless you want to go back to differential equations. Scaling is easy- it's called the per unit system-done all the time by utility engineers.
Some idea of what exists is important in trying to set up a model.
In this case, you have 3 busses, a known voltage magnitude at one of them and known power and reactive loads at the others- Only 5 non-linear simultaneous equations are needed to get a correct answer (or an approximation if line/source resistance is ignored). This is not a difficult problem but somewhat beyond high school in general. Oh, yes, the negligable (with respect to that in the lines) phase shift in the transformers can be included.
--

Don Kelly snipped-for-privacy@shawcross.ca
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