Comparison of 2-wire and 3-wire Distribution Systems

We will now compare the 2-wire and 3-wire systems from the point of view of economy of conductor material. 
For this comparison, it will be assumed that—

  1. the amount of power transmitted is the same in both cases;
  2. the distance of transmission is the same;
  3. the efficiency of transmission (and hence losses) is the same;
  4. voltage at consumer’s terminals is the same;
  5. the 3-wire system is balanced; and
  6. in the 3-wire system, the mid-wire is of half the cross-section of each outer.

Let W be the transmitted power in watts and V the voltage at the consumer's terminals. 

Also, let
R2 = resistance in ohms of each wire of 2-wire system.
R3 = resistance in ohms of each outer in 3-wire system.

The current in 2-wire system is W/V and the losses are 2 *(W/V)sqr*R2.

In the case of 3-wire system, voltage between outers is 2 V, so that current through outers is
(W/2V), because there is no current in the neutral according to our assumption (5) above. Total losses
in the two outers are 2*(W/2V)sqr*R3.

Since efficiencies are the same, it means the losses are also the same.

∴ 2(W/V)sqr*R2 = 2(W/2V)sqr*R3 or

Since the cross-section and hence the volume of a conductor of given length, is inversely propor-
tional to its resistance,

∴ (volume of each 3-wire conductor)/(volume of each 2-wire conductor) = 1/4

Let us represent the volume of copper in the 2-wire system by 100 so that volume of each
conductor is 50.

Then, volume of each outer in 3-wire system = 50/4 = 12.5
volume of neutral wire ,, ,, = 12.5/2 = 6.25

∴ total volume of copper in 3-wire system = 12.5 + 6.25 + 12.5 = 31.25
∴ (total coper vol. in 3-wire feeder)/(total copper vol. in 2-wire feeder) = 31.25/100 6=5/6

Hence, a 3-wire system requires only 5/16th (or 31.25%) as much copper as a 2-wire system.

1 comment:

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