Generator efficiency

A star connected balanced three phase load of 30 ohm per phase is supplied by a 400 V, three phase generator of efficiency 90%. Calculate the power input to the generator.

Thanks

> Somehow the answer given for this problem is 6.4 kW.
> Can anyone tell me how?

Yes! The 400 V in the question is a misprint. It should be 240 V.

Okay, first let me first refresh my memory: (It's been 35 years.)

- "Star Connected" means that each load is connected between one of the phases and neutral.
- "Balanced Load" means simply that each load has the same resistance/impedance.
- "Voltage" specified is measured between each phase and neutral. (As opposed to between two phases.)
- Efficiency is the ratio Pout/Pin at the stated load. Pin is the level of power driving the generator, and Pout is the total power delivered to the load.

What I don't know is whether the 240 VAC specified is with or without the loads connected. I'll assume it is with the loads connected. (Since that will give the correct answer. ;o)

First calculate the power consumed by each load:

P = V^2 / R (variation of Ohm's Law)
P = (240)^2 / 30 = 1920 W

This is the output power, Pout. Now calculate the input power as follows:

E = Pout / Pin
Pin = Pout / E
Pin = 1920 / 0.90 = 2133.33 W

Multiply that by three to get the total power input, 6400 W.

ANSWER: 6.4 kW

Data
-Load
R=30ohm
V=400v

-Gen
N=90%

P(load)= V(squared)/R = (400*400)/30 = 5333.333 W
P(load 3phase) = P(load) * 3 = 16 KW (16 000 W)
N(efficiency) = (Pout/Pin)*100
Pin = (Pout/N)*100 = (16 KW/90)*100 = 17.777 KW