Transmitting Electricity

Printer-friendly versionPrinter-friendly version Share this

If high voltage is so dangerous, why do we transmit electricity that way?

Big Ideas: 
  • When determining the optimum voltage to transmit electricity from power plants, it is necessary to examine resistance power consumption in transmission lines and the transmission distance
  • Transformers can be used to increase or decrease voltage
Electricity in our houses runs in the walls, but you may have noticed that in the countryside you occasionally see huge towers carrying electrical lines. If you are brave enough to get close to one of them you’ll see it’s marked “DANGER: HIGH VOLTAGE”

If high voltage is so dangerous, why do we need it in the first place?

 

 Electrical Transmission Model
To answer this question, let’s think about a simple model of electricity generation and consumption.

Here’s a basic picture of a power plant connected by transmission lines to your town. In reality, the power plant would be connected to many more houses than shown in this diagram, but let’s just keep our diagram simple. (We will be focusing on those wires anyways, so the details of the houses aren’t important)

Here’s a circuit diagram that shows the basic elements of this generator – wire – house system:

The power plant acts as a voltage source that sends current through the wires to your town. The town is a very complicated network of various devices that convert the electricity into any number of useful things, but for the purposes of this analysis we can treat them as just one big resistor. We don’t care about the exact details of what is happening in the town; we only care only that it needs to get power from the power plant through those wires.

The key thing that we want to accomplish in electrical transmission is to get as much of the energy to your town as possible. We can’t actually get 100% of the energy to the you, because the transmission wires have some resistance so there is always some resistive heating of those wires (which costs us energy). But we want to try to keep that to a minimum.

Power Consumption in Transmission Lines
In order to show why we need high voltage for transmission, let’s figure out what would happen if we just used household voltage of 120 V. We’ll figure out the resistance of a typical transmission wire, and then see how far we can transmit with that wire without losing too much energy.

The power loss in the wire is given by:

$ {P}_{\textnormal{wire}}={I}_{\textnormal{wire}}\times\Delta{V}_{\textnormal{wire}} $

We don’t want to calculate the ∆V of the line, but we can substitute using Ohm’s Law:

$ \Delta{V}_{\textnormal{wire}}={I}_{\textnormal{wire}}\times{R}_{\textnormal{wire}} $

This gives us

$ {P}_{\textnormal{wire}}={I}_{\textnormal{wire}}^2\times{R}_{\textnormal{wire}} ( *\*) $

Now, the current will be the same throughout the circuit. This means we can calculate it from the total power output of the plant.

$ {P}_{\textnormal{plant}}={I}_{\textnormal{plant}}\times{V}_{\textnormal{plant}} $

$ {I}_{\textnormal{plant}}=\dfrac{{P}_{\textnormal{plant}}}{{V}_{\textnormal{plant}}} $

Now we substitute back into the (*) equation to get the following result for the power loss in the wire.

$ {P}_{\textnormal{wire}}={I}_{\textnormal{wire}}^2\times{R}_{\textnormal{wire}} $

$ {P}_{\textnormal{wire}}=(\dfrac{{P}_{\textnormal{plant}}}{{V}_{\textnormal{plant}}})^2\times{R}_{\textnormal{wire}}  (*\**\*) $

To make use of this equation we need to choose a reasonable value for Pplant and figure out the typical resistance of the transmission wires. This will let us see how much power is wasted for a given voltage and distance.

Resistance of Transmission Lines
Transmission wires are made of layers of braided wire. Aluminum is usually used as a conductor because it has very low resistance and is relatively cheap. There are also a few strands of steel braided in to make the wire stronger.

While copper is a better conductor, it’s much more expensive and much heavier. This would mean we would need to put more steel in a copper cable to hold it up, making it even more expensive.

A typical wire used for long-distance transmission is called 4/0 AWG, and it has an effective diameter of 11.7 mm. We can look up the resistivity of Aluminum to figure out the resistance of a particular wire.

$ \textnormal{Resistivity of Aluminum}{ (\rho)}=28.2\times{10}^{-9}\Omega\textnormal{m} $

Resistance is given by:

$ \textnormal{R}={\rho}\times\dfrac{\textnormal{L}}{\textnormal{A}} $

Now we want to figure out the resistance of a transmission line per kilometer. This will let us use equation ** to figure out how far we can transmit power.

The resistance for one kilometer of wire will be given by:

$ \textnormal{R}=\dfrac{28.2\times{10}^{-9}\Omega\textnormal{m}\times1\textnormal{km}}{\pi\times(\dfrac{11.7\textnormal{mm}}{2})^2} $

$ \textnormal{R}=0.263 \Omega $

Transmission Distance

Now we can figure out how far we can transmit using the standard voltage of 120V. Let’s choose that we want less than 10% of our power to be used up in the transmission line, so we choose

$ {P}_{\textnormal{wire}}=10\%{P}_{\textnormal{plant} $

For the plant power we can use 25 MW, which is the power of a hydroelectric dam recently constructed near Squamish, BC1.

And using:

$ {R}_{\textnormal{wire}}= 0.263 \Omega\textnormal{/km} $

our expression ** becomes

$ {P}_{\textnormal{wire}}=(\dfrac{{P}_{\textnormal{plant}}}{{V}_{\textnormal{plant}}})^2\times{R}_{\textnormal{wire}} $

$ 25\textnormal{MW}=(\dfrac{2.5\textnormal{MW}}{120\textnormal{V}})^2\times0.263\Omega\textnormal{/km}\times\textnormal{distance}( *\* *\*) $

Solving for distance we get 0.0002 km, or only 20 cm!! This would mean that in order to get 90% of the power to your house, using the biggest wire commercially available, the power plant would have to be only 20 cm away!

This is an extreme example because we are using modern power plant capacities with very inappropriate voltages, but it shows a real problem: if you transmit at lower voltage, you need to have a power plant very nearby.

The solution is to use a higher transmission voltage. You can see that in equation ** the power dissipated in the wire decreases as the voltage squared, so when you multiply the voltage by 10 you decrease the power (or increase the distance) by 100. This relationship is shown below.


 

Modern transmission systems have a line loss of around 6.5%2, so we can see that by going up to 300 kV we can transmit around 1000 km. This is a much more sensible transmission distance for a province as big as BC.

(Note: you can do your own calculations for any % loss, wire resistance, or plant power in the following excel document: TRANSMISSION_LOSSES_CALCULATOR.xls )

Transforming Voltage

In order to transmit electrical energy at high voltages, we need to be able to convert from one voltage to another. This is essential, as we wouldn’t want to have 300 kV power in our homes! A simple reliable devise allowing to change alternating current (AC) voltage is called a transformer.

A transformer consists of two coils of wire wound around an ferromagnetic (most often iron) core. By winding a different number of turns of wire on each side, the transformer can increase the voltage at the cost of decreased current or vice versa. We discuss how a transformer works in this lecture.
 

This technology is easy to build, but it ONLY works for AC power. This is one of the main reasons that we use AC power today for transmission. When electricity was first being introduced to North America there were two different systems being proposed: AC power with transformers to enable long distance transmission, and DC power with many regional power plants3. The competition between the two sides was fierce and acrimonious, but in the end the ability to transform AC power to higher voltages played a key part in its widespread adoption.

In modern times, there are some technologies that allow transformation of DC voltages, and because they avoid some problems associated with AC transmission High-Voltage DC (HVDC) transmission is used in some parts of the world4. Because of problem associated with transmitting AC through saltwater (which is a conductive fluid) HVDC is especially suited to undersea lines that connect major power grids together.

 

 

Comments

Post new comment

Please note that these comments are moderated and reviewed before publishing.

The content of this field is kept private and will not be shown publicly.
By submitting this form, you accept the Mollom privacy policy.

a place of mind, The University of British Columbia

C21: Physics Teaching for the 21st Century
UBC Department of Physics & Astronomy
6224 Agricultural Road
Vancouver, BC V6T 1Z1
Tel 604.822.3675
Fax 604.822.5324
Email:

Emergency Procedures | Accessibility | Contact UBC | © Copyright The University of British Columbia