A 400hp SCIM and a 400hp synchronous motor both operating at 90% efficiency. The SCIM (Squirrel Cage Induction Motor) power factor is 0.8% lag. The synchronous motor power factor is unity. How many amperes of incandescent lighting may be added to the plant at the same kVA load if the synchronous motor is operated at a leading power factor of 0.8%? The supply is 600V, 3 phase. The answer to the question is 43.78A.

I am trying to work out the calculations to get the answer. So far this is what I have........

P motor output = hp x watts/horsepower......400 x 746/hp = 298kw.

P motor input = output power/% efficiency x 100 = 298.4kw/90 = 3315.56
3315.56 x 100 = 331.6 kw
Qt = Pt x tan(theta)
331.6kw x tan 36.87degrees
= 248.7 kvars

Thanks!

 

I = P / √3 x V x cos x eff

 

Still having difficulty.

331.6kw/1.732x 600V x 0.8 x .90 = 0.44318279?

 

For the initial configuration you know

Total Reactive Power=248.7 kvars
Synchronous Motor Real Power= 331.6 kw
Induction motor Real Power= 331.6 kw

The total real power = 663.2 kw

This will enable you to find the initial apparent power.

In the second configuration the reactive powers from both machines will cancel - the leading synchronous motor reactive power will balance the lagging induction motor reactive power.
Total apparent power (reconfigured case) = Total real power = 663.2 kw

The difference in apparent power = VA_diff

This difference is then attributed to the proposed additional incandescent (unity pf) load.

I_incansdescent (per phase) =VA_diff/(√3*600)

 

Do I use the 663.2kw and the 248.7 kvars and plug them in the formula (attached file) to get my apparent power? I'm still confused.

 

Do I use the 663.2kw and the 248.7 kvars and plug them in the formula (attached file) to get my apparent power? I'm still confused.

Yes - that will give you the apparent power value for the initial configuration.

Then subtract the 663.2 kw from this initial value to give you the notional difference in apparent power.

Remember that the final configuration (with the synchronous motor having 0.8pf leading) has zero reactive power draw from the supply so the apparent and real power draw in that situation are the same.

BTW power factor is not usually expressed as a percentage, since it corresponds to the cosine of the phase difference between the current & voltage phasors. A pf of 0.8 would be effectively 80%.

 

Thank you for your help, t_n_k!