9.4 Alternating current versus direct current (new)  (Page 11/7)

Power losses are less for high-voltage transmission

(a) What current is needed to transmit 100 MW of power at 200 kV? (b) What is the power dissipated by the transmission lines if they have a resistance of $1\text{.}\text{00}\Omega$ ? (c) What percentage of the power is lost in the transmission lines?

Strategy

We are given $P=\text{100 MW}$ , $V=\text{200 kV}$ , and the resistance of the lines is $R=1\text{.}\text{00}\Omega$ . Using these givens, we can find the current flowing (from $P=\text{IV}$ ) and then the power dissipated in the lines ( $P=I^{2}R$ ), and we take the ratio to the total power transmitted.

Solution

To find the current, we rearrange the relationship $P=IV$ and substitute known values. This gives

Knowing the current and given the resistance of the lines, the power dissipated in them is found from $P=I^{2}R$ . Substituting the known values gives

The percent loss is the ratio of this lost power to the total or input power, multiplied by 100:

Discussion

One-fourth of a percent is an acceptable loss. Note that if 100 MW of power had been transmitted at 25 kV, then a current of 4000 A would have been needed. This would result in a power loss in the lines of 16.0 MW, or 16.0% rather than 0.250%. The lower the voltage, the more current is needed, and the greater the power loss in the fixed-resistance transmission lines. Of course, lower-resistance lines can be built, but this requires larger and more expensive wires. If superconducting lines could be economically produced, there would be no loss in the transmission lines at all. But, as we shall see in a later chapter, there is a limit to current in superconductors, too. In short, high voltages are more economical for transmitting power, and AC voltage is much easier to raise and lower, so that AC is used in most large-scale power distribution systems.