Electrical Engineering

This is done in order to minimize power losses in the power distribution network due to the resistance of the transmitting cables. It should be noted that for a given cable resistance, voltage drop, and thus power dissipated in the cable and not available to use, is directly related to the current flow through the conductor.

According to the Power Law: P = I2 ×R, that is power (in this case, power lost) is equal to current squared times resistance. To deliver power, it takes amps and volts. If you raise the volts, you can reduce the amps and still get the same power. If you reduce the amps, you lower the losses. Did you notice the squared term in the formula? That means if you reduce the current to one-tenth of the original value, your losses go down to one one-hundredth of what they were.

This is a huge issue for the utilities. Every kW lost is one they cannot collect money for, yet they still have to pay for fuel to generate it, they have to size the generator bigger to supply it, and they have to size the transmission system to carry it. There are other good reasons too (see below), but minimizing line loss is the $main$ one. A few transmission systems have been designed at 1.2 million volts. The utilities would have billion-volt systems if they could figure out how to do it.

2nd Answer

A major reason is that, to carry the same amount of power, if the transmission voltage is made higher, then, even though a thinner cable has a higher resistance for a given length, the cables can be made thinner and lighter in weight.

Use of a higher transmission voltage saves a tremendous amount of money in many ways. For example for the expensive material used for the cables (often a steel multi-strand core wound with an outer skin of copper, aluminum, or similar good conducting wires) and for the weight and costs of construction and erection of the towers that carry the cables across the countryside.

To carry 400 kV (= 400 kilovolts = 400 thousand Volts) the steel towers have to be taller and the porcelain insulators have to be longer than they would have to be for cables carrying lower voltages but the cost of making the towers taller and the insulators longer is far less than the cost of the extra weight of the much thicker cables that would be needed to carry the same power at a lower voltage. *** (See Note below for more explanation)

There are many other costs which have to be reckoned when deciding what voltage to use for long-distance power distribution. For example the high cost of the massive power Transformers and big switching stations that have to be included in the power distribution network; the power that is lost from the cables - radiated to the surrounding air as heat - because of the electrical resistance of the materials from which the cables are made.

The above answer just gives a very simplified overview of the kinds of things a skilled power transmission engineer has to work with and calculate when designing a new power transmission network.

3rd Answer

Transmission and quantum of electricity can be considered analogous to hydraulics. Reckon voltage as pressure, the longer the distances, the higher the pressure required to pump. That is why for long distance transmission high pressure (voltage here) is required, failing which, the power will not reach the destined end. It will dissipate on the way. Reckon current as quantity which will be drawn from the pipeline (cables here) at the pressure/voltage required.

*** Note:

If we use the Electric Power Equation we can get an idea of what the descriptions given in the answers above really mean:

P = V × I or, in words:

P (power) = potential difference V (voltage) times current I (amperage)

So, using simple mathematics we can say that:

I = P / V or, in words:

Current I (amperage) = P (power) divided by potential difference V (voltage)

Now, as an example, if a small town needs to have a supply of power W of say 1 MW (= 1 megawatt = 1 million watts) to be delivered over cables from a power generation station:

Calculation A:

If the voltage V used for transmission through the cables is 1000 Volts then the current I in the cables would have to be:

P / V = 1,000,000 / 1,000 = 1,000 amps

which would require a very thick and heavy, and therefore very expensive, cable and associated support towers.

Calculation B:

If the voltage V used for transmission through the cables is 400 kV (= 400 kilovolts = 400,000 volts) then the current I in the cables would have to be:

P / V = 1,000,000 / 400,000 = 2.5 amps

which can be carried safely in a very much thinner, lighter and less expensive cable and support tower.

Line supports are mainly two types:

1. poles

Poles are classified as wood poles, concrete poles and steel/aluminum poles.

2. towers

Towers are classified as self supporting towers and stayed/guyed towers.

Self supporting towers are in two types: wide base and narrow base.

Stayed towers are classified as portal type and V-type.

First of all, it's energy that's being transmitted, not power! Power merely tells us the rate at which energy is being transmitted.

Next, it's a complete misconception to think that energy is delivered by electrons which, then, return to the source to collect more energy -rather like a conveyor belt! Electrons move far too slowly for that to happen (an individual electron is unlikely to travel the length of a flashlight's filament before the battery runs down!). It's far more likely that energy is delivered in the space around (rather than through) the supply wires due to a combination of the electric and magnetic fields, called the 'Poynting Field' (check this out on the internet).

Finally, if we think of 'positive' power as the rate at which energy is being transmitted in the 'forward' direction (i.e. from the source to the user) then, during the negative half-cycles of a.c. current and voltage, the product (P =EI) is still positive. So it matters not whether it is being 'delivered' by d.c. or a.c.