Electric Transmission Losses

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Opening Thoughts

I think everyone guesses that transferring electricity from point A to point B with the lowest possible loss is not a small task as it may seem at the first glance. Let's start the story - only in broad lines - at the very beginning and take a closer look at the industrial-sized items and in the second half of this topic let’s look at some calculations in household scales as well. Of course, if you do not have an appropriate qualification in the field of electronics, always call a specialist!
So, every power plant, that produces electricity has an efficiency factor, which indicates to us that certain losses must be taken into account in the very first step too. Let’s take a simple example very quickly to make this thing clear. The wind turbine generator converts mechanical energy into electrical power with very good efficiency of 95-97%, but that means already a loss of 3-5%, which is caused mainly by the internal resistance of the generator windings and the friction of the bearings. I think the point is clear, so the losses of an electrical system - even if it seems small in this case - cannot be forgotten.

The power lines, which cover down the countries for many-many thousands of kilometers, also have internal - what's more temperature-dependent - resistance, which refers to the efficiency of the electric transport in terms of the economy as well. I don’t think many people know that these wires, which deliver the electricity to the consumers are heating the sky at the same time due to their internal electrical resistance. This is the reason very briefly why the voltage has to be transformed up to the highest possible value (e.g.: 400 kV) because as a result of this the current will be inversely smaller, and this is perhaps the greatest opportunity to save losses. I think this is easy to understand because the infrastructure is built only once, as the material, the internal mechanical structure, the cross section and the length of the wires are already fixed as economically as possible, but the line voltage and the current values can be adjusted relative to each other as desired.

Anyway, there are well-known electric and magnetic fields around the high-voltage power lines, so always keep a safe distance from them in everyday life to avoid the biological consequences. The appropriate biological limits have always been and will always be a highly disputed topic. In my personal opinion, that would be more acceptable, as for example in Sweden, where it is not allowed - to the best of my knowledge and if I remember well - to give a building permit for an area within 300 m of the high voltage transmission lines, but I emphasize it once again that I am far from being an expert in this very sensitive subject and I only deal with this topic as a hobby. Let me note here that, we don't have to be so scared of this transmission lines so much, since according to my simple measurements we are talking about only 50 meters in general, but I like to overprotect things multiple times. In addition, I hope we will see in another article that we can measure surprising values in our households as well.

Let's continue our train of thought with Corona discharge. Perhaps one of the main undesirable losses around the high voltage lines is due to the just mentioned Corona discharge, especially in the area of sharper conductive surfaces. The main reason for this phenomenon is that the air can be easily ionized in the very close environment of the high-voltage lines, because the highly inhomogeneous air mass contains different amount of dust and humid particles, thus this ionized mass is able to conduct the electricity to a very variable degree.
Whereas the internal resistance of the here and there flowing air mass is quite high, therefore the corona discharge current is pretty low, so its light is not really visible, however its voltage level is very high. This explains briefly the fact that only a soft crackling sound is present near the high voltage pylons and wires.
Consequently, this loss strongly depends on the humidity and the amount of dust in the air and we can easily talk about - if I remember well - thousands of W-s only between two pylons of the high voltage transmission line. So, in case of bad weather - such as rain or fog - this loss could be quite significant, especially if we try to estimate quickly this energy loss for hundreds of kilometers, then in my opinion many-many thousands of kW will easily come out.

If we go further in this line of thought, so if the electric charge exceeds the breakdown voltage level of the above mentioned air mass, an electric discharge arc would be appeared between the two isolated conductors, and it could already be well seen and heard with a very bright light phenomenon as well, as in the case of lightning, or just like in my 12,000 V simulation, which gave out only 6 Joules of energy, but still created a fairly thick spark channel.

If I have already mentioned the high voltage, let's not forget the creeping current either, as a well-known physical phenomenon. It doesn't belong so much to the main topic, but I find it very interesting if anyone didn't know this yet.

So, the creeping current could flow - unexpectedly - on the wet and dirty surfaces of the high voltage insulator materials. This could mean an occasional energy loss as well, especially in case of wet weather. As a brief explanation of this let me point out that strictly speaking, there is no perfect insulator at all. Therefore, all materials are able to conduct electricity to a certain degree, even the air, especially if it is dusty. I think it is easy to imagine.

Anyway, the characteristic corrugated porcelain insulators of electrical wiring pylons - see the pictures below - must prevent the above-mentioned creeping current in all circumstances. Thanks to their knurled shape, the dusty raindrops have to travel much more on these ceramic insulator materials to create an unexpected electrical discharge. In other words, the chance of an electric arc discharge is very small over the entire length. In addition to this, the creeping current can escape more easily into the air at the knurled peaks - since the electric field strength of the peaks is much higher - so that the insulating surface can not melt either, as the discharge is distributed in the atmosphere.

The picture above shows a smaller - 132 kV - ribbed ceramic insulation components, but this is already a more serious - 400 kV - double ceramic insulation in the picture below, where the wire is almost as thick as a beer can. I remember when it was on the ground before the installation I couldn't even lift it a little. So no wonder that there are several tons of wires between two such transmission line pylons because it is more advantageous if the cross-section is as large as possible, but then the weight will be also greater.

Returning to the loss of infrastructure components, let’s not forget either that large transformer stations and smaller local transformer units and load distribution switching stations have also an own goodness factor too. Therefore, without the need for completeness, taking note of all these, it could be clear to everyone, that the electricity transmission must be a very precisely designed infrastructure, where a significant portion of calculations is only about the losses. Thus, the electrical network design requires a lot of optimization to minimize mainly these inescapable and unnecessary costs, which I have touched here only very roughly on such a hobby culinary level.

Losses in the cables at home?

So, it was not my goal to introduce the large-scale industrial network design - I wouldn't even know - but I just wanted to show a comprehensive picture of current transfer in general and to show how very important is this topic in more serious sizes. So, in the second half of this article, we can get to know a bit more about the electrical wirings and their losses in much much shorter and smaller household sizes - the quality of the connections will be another issue - based mainly on my individual interest.
Of course, if you do not have an appropriate qualification in the field of electronics, always call a specialist! Before I move on, I would like to emphasize that it is strictly forbidden to carry out any electrical installation or reparation without confident knowledge and proper qualification!

So, I would like to continue this article with my local example, because this topic was inspired partially by my high-performance semi-industrial welding machine as well, which could have an enormous and continuous current consumption. Unfortunately, we can use this machine most often only far away from the proper power source. Therefore, I had to design a suitable extension cable for its safe operation, and during this process, I had to calculate the appropriate cross-section for the desired wire length too.

So, in the next few lines, I would like to give a brief overview of these calculations. For better comprehensibility, I illustrated this story with nice pictures as well.

Before I move on, I would like to emphasize that it is strictly forbidden to carry out any electrical installation or reparation without confident knowledge and proper qualification.

Furthermore, I would NOT start the electrical works even if I am tired, because I have already got an electric shock under these conditions, so I learned that it is not worth forcing things. This was the well-known small print section, which must be taken care of in all circumstances.

Sizing of Wires - Basics

In this section, I try to summarize the simple and completely ordinary basics of the sizing of electrical wires. The story is much more complicated than that, but it can already be used at least for some estimation. I think it is worth paying attention to the below-mentioned factors, especially at the beginning of planning.

When we are designing a circuit - even only an extension cable - it is always worth considering that some voltage drop occurs on the resistance (more exactly on the impedance) of the DC and AC wires. Moreover, keep in mind that the phase and null wires have the same cross-section and the consumer current causes on both wires the same voltage drop too.

In general, it can be said, that care must be taken to ensure that the voltage drop (ΔU%) should not exceed the 2-3% threshold. In addition, the temperature of the cable must be checked as well to avoid the possible damages, even for completely ordinary consumers - such as an electric stove - which could have a continuously high current consumption too.

Perhaps even professionals don’t know by heart that how many twisted or solid cables with completely different cores exist in the market. The image below shows just a few of the wire types and cables I use more often.

I would say that the entertaining part has been done so far, and now let’s start with the more complicated calculations. :)

To whom the notations are still new in the formulas, I say that the current is denoted by a large I, the voltage by a large U, and the power by a large P. Moreover I will write the lower indexes completely to make them even more understandable.

I have just forgotten to say that there is no need to be afraid of cos φ either, because all you need to know is that it is the power factor of the given AC consumer, which is simply a number between 0 and 1.

So, in the first step, I read out the maximum power consumption from the consumer nameplate, and after that from the absorbed power, the current load can be easily calculated, or better said only estimated.

So, the calculation of the current load on the wire for DC consumers looks as follows:

I = Pconsumer / Uconsumer.

Furthermore, this formula is a little more complicated for a single-phase AC system

(for me during calculation the Uconsumer = Unominal = 230 V, because that's enough for me):

I = Pconsumer / (Uconsumer * cosφconsumer)

In the next step can we calculate the exact value of the allowed voltage drop per wire:
In the case of direct current and single-phase line, the formula looks like as follows:

Uallowed = (Ɛ * Uconsumer) / 100 => per wire / 2

where Ɛ is the allowable voltage drop expressed as a percentage, which is usually 2-3%.

Uallowed = I * R = I * ρ * l / A

Finally, the required cross-section of the wires can be calculated by knowing the current load and the allowable voltage drop:

A = (ρ * I * l) / Uallowed

where the ρ = resistivity, l = wire length and A = wire cross-sectional area.

Last but not least, don't forget that due to the ohmic resistance of the cable it also generates some heat, so checking the heating of the cable is also an important step. Furthermore, in the case of temperature check always the absolute value of the current - the total current - must be taken into account:

I = Iconsumer / cosφ

I think it is obvious to everyone, that the allowable maximum current load depends on the cross-section of the given wire. So, in the case of the usual copper wires, I used to calculate with the following values, which can apply to the simple wires as well - of course in proper PVC tubes - in the standard single-phase households.

A = 1,5 mm² => Imax. = 16 A
A = 2,5 mm² => Imax. = 21 A
A = 4,0 mm² => Imax. = 27 A
A = 6,0 mm² => Imax. = 35 A

A = 10 mm² => Imax. = 78 A

In the case of those high-current current sources - such as batteries - where there are freely mountable (i.e. not under plaster) simple wires, I think the table below is right for me.

A = 16 mm² => Imax. = 104 A

A = 25 mm² => Imax. = 137 A

A = 35 mm² => Imax. = 168 A

A = 50 mm² => Imax. = 210 A

A = 70 mm² => Imax. = 260 A

A = 95 mm² => Imax. = 310 A

These are strictly just my views only, and of course no one has to design or install anything based on these values. In addition, it is possible and worthwhile to use cables or wires with a much higher load capacity than the calculated current values.

I think so, it is particularly important - for example - for those high-performance battery packs which are driving solar island inverters in continuous operation, and it is worth saving every single W from the dissipation.

So, I think we're fine with these formulas so far, but let's go further into the deeper waters.

Calculation of Current Loss

I would like to present just a simplified and clear line of thought through the below-mentioned example, instead of a complex calculation. In the practice, I have only time for this simplified and reasonable model in most cases, to move forward with my work as well, because there is not really time to calculate the current density and similarly complicated things on the wire.

Thus, for the sizing of low voltage twisted copper wires I consider only the real part of the impedance. So in other words, I consider that the voltage drop is only due to the ohmic resistance of the wire. Moreover, I used to calculate only the effective component of the load current for sizing the voltage drop. Last but not least the changes in resistance due to the changes in temperature are ignored from my side. I hope everything is clear, and let's take a brief look at the task I brought with me.

Let's see an average single-phase 230 V electrical network, where the short-circuit current does not exceed the usual 16 A. Let's suppose, that after connecting the above-mentioned welding machine to the socket next to the electricity meter, the welding machine consumes 2300 W power after switching on, and a voltage of U = 230 V can be measured at the connection point of this electrical consumer.

This welding machine is located at the end of an electric extension cord, which has a usually twisted copper wire with a length of 15 m and a cross-section of q = 2.5 mm², measured from the supply point.

Let's calculate the electrical loss and the voltage drop across the wires based on the provided data when the power factor of the welding machine is: cos φ = 0.7.

By the way, I calculate this because what is most important for the inverter is the voltage drop across the 15 m long wire resistance. According to the manufacturer's datasheet of our device, the network voltage has to be at least 230 V AC ± 15% to avoid unexpected inconveniences.

Furthermore, I know from my experience - thanks mainly to the MOSFET-s - that inverters are very sensitive to voltage fluctuations. Anyway, the MOSFET abbreviation means the Metal–Oxide–Semiconductor Field-Effect Transistor, which is essentially a voltage-controlled transistor. The normal bipolar transistor is current-controlled and that is a big difference, but let’s get back to the task.

I made a nice sketch as well to make the thing more understandable.

So, we are talking about the P = 2300 W single-phase power consumption, furthermore, the phase and null lines can be modeled in this case quite calmly with the RL and RN resistors, which is derived from the specific resistance of the copper wires.

For easier understanding, I illustrated this separately too.

Anyway, in most electrical network calculations, the internal resistance of the wires can be negligible, but now I take them into account because that is what the task is about.

Therefore I'm going to use the following formula for this:

Rwire = ρ * l / A

where the ρ = resistivity, l = wire length and A = wire cross-sectional area.

Let's illustrate the task in the next step with formulas as well:

Ploss = Psupply - Pconsumer

ΔU = Usupply - Uconsumer

We are talking about alternating current, so I'm going to use the following formulas to calculate the performance:

Pconsumer = Uconsumer * I * cosφconsumer

Qconsumer = Uconsumer * I * sinφconsumer

So, according to the above-mentioned formula, the internal resistance of phase and null wires to the 15 m cable length can be calculated, if the copper resistivity is: ρ = 0,0175 Ω*mm²/m.

Rwire = ρ * l / A = 0,0175 Ωmm²/m 15m / 2.5 mm² = 0.105 Ω

BUT I emphasized that I am talking about the phase and null wires because the copper wire runs back and forth in the circuit:

R = 2 * Rwire = 0.21 Ω

Anyway, if Q = 0 and cosφ = 1 then the welding machine can be modeled with simple resistance. So the internal resistance of the welding machine as an electrical consumer:

Rconsumer = U² / P = (230V)² / 2300W = 23 Ω

So, the current drawn by the welding machine for the P = 2300W power:

I = Pconsumer / (Uconsumer * cosφconsumer) =

= 2300 W / (230 V * 0.7) = 14.29 A

The effective power loss due to the internal resistance of the wires:

Ploss = I² * (RL + RN) = 42.88 W

So there is a power loss of 1.43 W per meter on this twisted copper wire.

Thus, the percentage value of the loss in relation to the absorbed power:

P % = 100 * 42.88 / 2300 = 1.86 %

Finally, the voltage drop will be as follows:

ΔU = R * I * cosφ = 0.21 Ω * 14.29 A * 0.7 = 2.1 V

The percentage value of the voltage drop: 100 * 2.1 / 230 = 0.91 %

I would like to mention that in the practice, the relative short wire sections (e.g.: 25 m) are only sized for their current carrying capacity range. So in other words, I can select the wire-specific value from the table, but the much longer wires are sized for the voltage drop.

Finally, based on the calculations above, I made my dedicated extension cable as well that can be used for the welding machine most likely more calmly. In the learning phase I will use this machine only from a 16 A network, as I don't want to weld plates that are too thick yet.
Furthermore, only a maximum power of 3500 W can be drawn from the usual single-phase network - if the circuit breaker allows it - in the best case, because do not forget that the connectors are also sized most often up to the 3500 W power transmission.

Closing Thoughts

I think everyone experiences a rapid increase in electricity demand because more and more powerful electrical machines are appearing in our direct environment. For example, it is not difficult to guess, that with the proliferation of electric cars, the demand for electricity continues to grow enormously.
Of course, I do not see through the whole energy sector, but in this question, I have a feeling sometimes that we are slowly pushing the physical boundaries. In any case, it is quite clear to me that we need to support this seemingly exponential development with appropriate - as green as possible - high-performance infrastructure solutions as well.

So, I just think about what it means in a household the 32 A load at 230 V on a single-phase wire, and it results only a slightly more than 7 kW of power when an average smaller car has already an engine over 100 kW. Of course, I know that the efficiency of internal combustion engines is around 30% and the electric motors run at 90% efficiency. Furthermore, I am also familiar with the different torque characteristics of the two technologies, but at the end of these less interesting comparisons, I think it’s easy to see that, serious Kilowatts are required for fast traveling and long distance trips as well. So either way I look at this issue, the battery charger puts a tremendously extra load on the electrical network to store the necessary energy, and that is exactly the point.

Of course, solution can be found to almost everything and that is why I think it is very important to manage the instantaneous current consumption wisely in order to make things a lot easier.

As I remember, the instantaneous electricity demand can move on a quite wide range almost everywhere because the instantaneous energy consumption of around 10 million people in Europe is just 3.5 MW in the morning hours, but this value can easily jump to above 8 MW in the afternoon. It follows to me that the increased power demand can even be handled smartly too.

ELECTRICAL