Power Transmission and Usage
Supplying Cities with Electrical Energy from Remote Power Plants
Jul 29, 2008
All modern countries are crisscrossed with high-voltage transmission lines, which transport electrical energy that can only be supplied from generators located at remote, large power plants. Substations on the electricity network, transform the voltage generated so it can be distributed to homes, schools and businesses.
High Current is Not Desirable
The transmission of electrical energy over large distances is a very important consideration for power engineers, particularly in countries with widely separated population centers. The power P carried by a transmission line is determined by the product of the current I carried and the voltage V:
Power: P = I V
If a large city would use, say over 5000 MW of power at peak times, then at 250 V operating voltage, this would require a current of 20 million Amperes! Way too large, no practical conductor could carry this current over long distances. So how can the power be transmitted from the generating station to the users?
The solution is to use much higher voltages. According to the above formula, for a given power, the higher the voltage, the lower the current needed. For example, at a voltage of 500 kV, the 5000 MW power could be carried by a much lower current of 10,000 A. This is much more practically feasible.
Power Losses
Any practical power line has significant resistance R, which causes a power loss Po, according to the formula:
Power Loss: Po = I² R
To highlight how much resistance change would be needed to compensate for any increased current in the line, consider a doubling of current in the above formula. In this argument, zero power loss is to be maintained.
- A doubling of current would cause power to increase by a factor of four, therefore
- Resistance must be reduced by a factor of four to compensate.
- For such reduction of line resistance, the weight of conductors would then be four times greater
Clearly, as power depends on the square of current, it is important to keep line current as low as possible. The percentage power loss is often of interest. This is given by power loss divided by rated power multiplied by 100.
Measurement of Electric Power
Power companies measure the electrical energy we use at home by installing a kilowatt hour meter (or joule meter). It has a small rotor spinning around and is connected to a series of dials, which register the energy in terms of the number of kilowatt hours (kWh). That is, energy E is the product of power P and time t.
Power companies charge at a rate R' in cents per kWh. To obtain cost, one must multiply this rate by the energy consumed, as summarised by the following cost formula:
Cost: C = R' x P x t
Worked Examples
A generator produces 5kW of electrical power at 500V, which is transmitted to a distant house via twin cables of total resistance of 4.0 ohm. What voltage is available at the house?
- Current in the cables would be I = P/V = 5000/500 = 10 A
- Power loss = I² R = 10 x 10 x 4 = 400 W.
- % power loss = Power loss / Rated Power x 100 = (400/5000) x 100 = 8%
- Voltage available at the house is V = P/I = (5000 - 400)/10 = 460 V
What is the cost of running a 80W light light globle for 30minutes if the rate is 15 cents per kWh? Using the cost formula, C = 15 x 0.080 x 1/2 = 45 cents
Conclusion
Most people in the developed world consume electric power. This power comes from generator plants located remotely where there are natural resources to provide this energy. Transmission lines transport power to where people are located, so that they may tap into. As a result of great distances usually involved, power losses along the lines occur and this presents engineers with design challenges, such as the types of cables used. Simple formulas exist to calculate power, line current and voltage, and cost of power.
The reader may be interested in more details on this topic, or perhaps learn about electromagnetic induction.
