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# Calculating neutral currents with varying power factors

Discussion in 'Homework Help' started by smithbrian, Sep 3, 2012.

1. ### smithbrian Thread Starter New Member

Sep 3, 2012
8
0
Hello I am having trouble calculating the neutral currents for the following system
Phase A = 10A @ PF of 1
Phase B = 9A @ PF of 0.9
Phase C = 8A @ PF of 0.8
The answer is meant to be 4.55A but I get 1.73

I used the equation from the following website.

Here they seem to not take power factor into account there can someone please point out to me how to calculate this taking power factor into account.

2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
If all pf's are lagging then

Phase A current

$I_A=10\angle{0^o} \ Amps$

Phase B current

$I_B=9\angle(-120^o-{\arccos(0.9)}^o) \ Amps$

Phase C current

$I_C=8\angle(-240^o-{\arccos(0.8)}^o) \ Amps$

Then

$I_N=I_A+I_B+I_C \ Amps$

3. ### smithbrian Thread Starter New Member

Sep 3, 2012
8
0

But I still don't get the right answer. I don't understand some of your symbols I am guessing you meant.

Ia = 10* cos (0)
Ib = 9* cos (-120 - 25.84)
Ic = 8* cos (-240 - 36.87)

If I do the above I get Ia=10, Ib=-7.45 and Ic= 0.957

In=3.507A (not 4.55A which is the correct answer)

4. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
Hi there - remember you have to think in terms of complex quantities.

So

$I_A=10+j0 \ Amps$

$I_B=9 \angle (-145.84^o)=-7.447-j5.054 \ Amps$

$I_C=8 \angle (-276.87^o)=0.957+j7.942 \ Amps$

And finally

$I_N=I_A+I_B+I_C=3.51+j2.89 \ Amps=I_x \angle(\theta^o) \ Amps$

5. ### mlog Member

Feb 11, 2012
276
36
You're only finding the real components. You can't ignore the imaginary components. The answer you want is the vector magnitude of the real and imaginary parts. In other words, take the square root of the sum of the squares.

6. ### smithbrian Thread Starter New Member

Sep 3, 2012
8
0
Yes I understand it now. I realised that I was making that mistake.Thank you guys for your help.

7. ### smithbrian Thread Starter New Member

Sep 3, 2012
8
0
I repeated the same question but this assuming the sequence to be 0,120,-120. But I get a different answer.
Ia = 10* cos(0)
= 10 + J0 A
Ib= 9* cos(120-25.84)
= -0.65 + J8.98 A
Ic= 8*(-120-36.87)
= -7.36 -J3.14 A
In = Ia + Ib +Ic
= 1.99 +J5.84
=6.17A
I don't get it why is the answer different?
Shouldn't it be the same answer?

8. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
Once again it boils down to correctly tracking phase rotation and power factor angles.

If you make A phase the reference [0°], B at 120° and C at -120° I would assume a clockwise rotation sequence A-B-C-A-B ....etc.

With lagging power factors for the non-unity cases [B & C phases] then the three phasors would be ..

$I_A=10\angle{0^o} \ Amps \\ I_B=9\angle{145.84^o} \ Amps \\ I_C=8\angle{-83.13^o} \ Amps$

9. ### smithbrian Thread Starter New Member

Sep 3, 2012
8
0
Well why did you do the following can you please explain.

Ib= 9* cos(120+25.84)

and

Ic= 8* cos(-120 + 36.87)

when previously you said

Ib = 9* cos (-120 - 25.84)
Ic = 8* cos (-240 - 36.87)

my question is why in one instance we are subtracting and in one instance we are adding?

10. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
In that earlier case you quote I was using the phase rotation convention I normally adopt - a counter-clockwise convention. The A phase is reference at 0°. The B phase lags the A phase by 120° which I denote as -120°. The C phase lags the A phase by 240° which I denote as -240°. The current must therefore either lead or lag those values depending upon whether the power factor is leading or lagging. For a counter-clockwise convention a lagging current of θ degrees on the B phase would be interpreted as a phasor angle of -120°-θ°.

Your proposed convention differs to mine so the angles would differ.

If you were taught how to draw phasor diagrams then this all becomes second nature.