# Synchronous machine problem (not homework)

#### cnh1995

Homework Helper
Gold Member
< Mentor Note -- thread moved to HH from the technical forums, so no HH Template is shown >

Here's a question from a national-level engineering exam in India called GATE (2017).
It is for 2 marks, numerical answer type.

Two synchronous motors A and B in parallel with identical O.C. characteristics are acting as synchronous condensers and supply a total reactive power of 50MVAR to the grid. The grid is operating at 11kV. If the synchronous reactances of A and B are 1 ohm and 3 ohm respectively and current through both the machines is the same, find the ratio of excitations of A and B i.e. excitation of A/excitation of B.

I got the answer as 0.744 and it is correct. (The answer key says anything in the range 0.72 to 0.78 is acceptable).

Here is my solution:
Since the machines are acting as synchronous condensers, active power is zero, which means power angle for both the machines is zero.

So, we can write the equation for reactive power as,
EA*11-112+EB*11/3-112/3=50.
This simplifies to,

33EA+11EB=634.....1)

Since both the currents are same,
EA-11=(EB-11)/3
which simplifies to

3EA-EB=22..........2)

Solving 1) and 2) gives EA=13.27 kV and Eb=17.82kV.
So, their ratio is 0.744.
(I did not convert line quantities into phase quantities as the equations are scaled down by a factor of 3, which gets cancelled out and we again get the equation in terms of line quantities).
Is there a simpler and faster way of doing it?

Because these calculations take time, while the expected time is roughly 2.7 minutes/problem. There are 65 questions and you get 3 hrs.
Of course no one attempts all the questions but still, the maximum time you can afford to spend on a question is 4-5 minutes and they asked this question for only 2 marks. Is there any short-cut or some conceptual trick? This exam is all about using conceptual tricks and short cuts and the way I solved it, it is worth at least 6 marks.

Any help is appreciated.

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#### scottdave

Homework Helper
You may want to do more research on this particular test. You should definitely consider employing some test taking strategies, which ensure that you do not get bogged down taking too much time on some questions (especially in relation to their point values).

#### cnh1995

Homework Helper
Gold Member
Thanks @berkeman for moving the thread here. And I'm glad that after drawing a couple of phasor diagrams, I found a much easier way to do it. It does save a lot of time! We know Q=50MVAR.
So,
Q=S=VI
∴I=Q/V i.e I=50/11 kA.
Since I=IA+IB and IA=IB,
IA=IB=50/22 kA.

Since the network is reactive and active power is zero, EA, EB and grid voltage 11kV are all in phase.
So,
EA/EB=(11+50*1/11)/(11+50*3/11)=292/392=0.744

This looks like the most obvious way of doing it . Don't know why I complicated it earlier.. I'll mark it solved now.

• jim hardy and berkeman

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