## A 40 MW Run Off River, ROR, system.

AC electricity is generated in power stations at a fairly low voltage (

Substitution [

P = I X I X R or P =( VXV)/R

P = I X IX R(cable) or P = (Delta V)2/R(cable) - For resistive heat loss in a wire.

P = (I X I) R(load) or P = (V X V)/R(load) - For power dissipated as heat by load (city).

P = (110 volt)(110 volt)/R

If power delivered to city was only 1,000,000 watts [1.0 megawatt].

R = (110)(110)/1000000 = 12100/1000000 = .0121 ohms.

If wanted the heat loss in cable to be less than 10% of total power, then cable or cables in parallel would have to have a resistance less than .00121 ohms. Of course the use of transformers make using all that copper wire unnecessary.

*i.e.*, about 440V), and is consumed by the domestic user at a voltage of 110V (in the U.S.). However, AC electricity is transmitted from the power station to the location where it is consumed at a very high peak voltage (typically 50kV). In fact, as soon as an AC voltage comes out of a generator in a power station it is fed into a step-up transformer which boosts its peak voltage from a few hundred volts to many kilovolts. The output from the step-up transformer is fed into a high tension transmission line, which typically transports the electricity over many tens of kilometers. Once the electricity has reached its point of consumption, it is fed through a series of step-down transformers until, by the time it emerges from a domestic power socket, its voltage is only 110V. AC electricity is both generated and consumed at comparatively low voltages. Why than go to the trouble of stepping up the voltage to a very high value at the power station, and then stepping down the voltage again once the electricity has reached its point of consumption? Why not generate, transmit, and distribute the electricity at a voltage of 110V? Well, consider an electric power line which transmits a peak electric power between a power station and a city. We can think of the city as a resistance, which depends on the number of consumers in the city, and the nature of the electrical devices which they operate, as an essentially a fixed resistance. We can think V and I as being variables, since we can change them using a transformer. . The rate at which electrical energy is lost due to resistive heating in the line is**P =****I X IX R**. This equation shows that keeping**the current "I"**small is all that is required to reduce the energy lost as heat in a transmission line.Substitution [

**V = I X R**or**I = V/R ]**law.] in Power Formulae**P = I X V result in:**P = I X I X R or P =( VXV)/R

P = I X IX R(cable) or P = (Delta V)2/R(cable) - For resistive heat loss in a wire.

P = (I X I) R(load) or P = (V X V)/R(load) - For power dissipated as heat by load (city).

P = (110 volt)(110 volt)/R

If power delivered to city was only 1,000,000 watts [1.0 megawatt].

R = (110)(110)/1000000 = 12100/1000000 = .0121 ohms.

If wanted the heat loss in cable to be less than 10% of total power, then cable or cables in parallel would have to have a resistance less than .00121 ohms. Of course the use of transformers make using all that copper wire unnecessary.

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