Question

Question #ea004

Answer 1

No. AC is much more efficient for transmission over long distances than DC, because AC allows you to use transformers step up the voltage for transmission and step down the voltage at the load.

Explanation:

Let's consider a simple case that I will deliberately make a "best case" scenario.

An industrial customer was a facility 10 km away from your power plant that requires `1 xx 10^6" W"` at `440" V"` and does not care whether you supply it do him using AC or DC.

Your power lines have the incredibly low resistance of `0.1 Omega"/km"`

If you choose DC, the current in the power lines must be the same as the current in the load, we can use `P=VI` to compute the current in the load:

`1 xx 10^6" W" = (440" V")I`

`I ~~ 2200" A"`

Compute the resistance of the power lines

`R = 2(10" km")(0.1Omega"/km")`

`R = 2Omega`

NOTE: The factor 2 is required, because you need a 2 wires to make a complete circuit, one out and one back.

We can use `P =I^2R` to compute the power lost in the power lines:

`P_"lost" = (2200" A")^2(2Omega)`

`P_"lost"~~ 9.6xx10^6" W"`

Using DC, you loose 9.6 times the power that you deliver.

Lets consider the AC solution where you step of the voltage to `250000" VAC"` for transmission.

It is easy to obtain a transformer with an efficiency of `95%` so we shall use that number; this makes the power in the transmission lines at the primary of the step down transformer `1.1xx10^6" W"`

At `250000" VAC"`, the current in the transmission lines is:

`I = (1.1xx10^6" W")/(250000" VAC")`

`I ~~ 4.5" A"`

Compute the power lost in the same two transmission lines:

`P_"lost"= (4.5" A")^2(2Omega)`

`P_"lost" = 40.5" W" larr` this is insignificant

Compute the power lost in the step up transformer:

`P_"generated" = (1.1xx10^6" W")/0.95`

`P_"generated" ~~ 1.2xx10^6" W"`

Using the step up / step down transformer, you loose approximately `20%` of the power generated; this is far better than `960%` for DC.