Discussion in 'Homework Help' started by matt_t, Sep 9, 2012.

1. ### matt_t Thread Starter Member

Aug 18, 2012
30
0
Hi I have the following question:

The following are connected in delta to a 600 Volt three-phase supply.
• A-B has a 20 Amp load with a power factor of .9 lag.
• B-C has a 30 Amp load with a power factor of 1
• C-A has a 40 Amp load with a power factor of .9 lag.
Calculate the current in line A:
Calculate the current in line B:
Calculate the current in line C:

I have drawn out the circuit in delta form to try and understand where the current is going but without the neutral of a star system I am confused when using kirchoffs law.

I know that two of the currents are not in phase so I have:

A-B 20A /_ 25.85 deg
B-C 30A /_ 0 deg
C-A 40A /_ 25.85 deg

I can't use the formula I line = I phase X √3 as the loads are not balanced. So I am unsure how the current is distributed between the phases.

2. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Any of the line currents would be the vector summation of the two associated delta currents. So Ia=Iab-Ica and so forth.

3. ### matt_t Thread Starter Member

Aug 18, 2012
30
0
So to find the current in line B I would do the following?

Ib= Iba-Ibc

20/_ 35.85 deg - 30/_ 0 deg
18+j8.717 - 30+j0 = -12-j8.717 = 14.83/_ -144 which isnt the correct answer.

Think im missing something here.

Mar 6, 2009
5,448
783
But

Ib=Ibc-Iab

etc....

5. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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Also how could a lagging pf current give a +ve imaginary part.

6. ### matt_t Thread Starter Member

Aug 18, 2012
30
0
I don't understand, the correct answers I have are B = 37.2A and C = 67A.

Would you be able to show me the vector sums so I can see clearly how i've gone wrong?

7. ### matt_t Thread Starter Member

Aug 18, 2012
30
0
Am I missing the fact that all of the phases are 120 deg out of phase with eachother? Do I need to take this into account?

8. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Yes - absolutely you must remember the individual line voltage phase displacements are an important consideration.

9. ### matt_t Thread Starter Member

Aug 18, 2012
30
0
But how do I add currents which are normally 120 deg out of phase with eachother when two of them are out of phase with themselves?

10. ### mlog Member

Feb 11, 2012
276
36
I'm giving my solution since we've hashed it out enough. Don't you think?

Here's the way I see it. Given Iab, Ibc, & Ica, find Ia, Ib, & Ic.

Given:

Iab = 20A @ pf 0.9 lag
Ibc = 30A @ pf 0.0
Ica = 40A @ pf 0.9 lag

Since all currents will be in Amperes, I will drop the "A."

Iab = 20.0 ang(-25.85 deg)
Ibc = 30.0 ang(0.00 deg)
Ica = 40.0 ang(-25.85 deg)

I shall define Ia as the current entering node A, etc. As t_n_k said (at least for Ib):

Ia = Iab+Iac = Iab-Ica
Ib = Ibc+Iba = Ibc-Iab
Ic = Ica+Icb = Ica-Ibc

Ia = Iab-Ica = -18.00 + j8.72 = 20.0 ang(154.2 deg)
Ib = Ibc-Iab = +12.00 + j8.72 = 14.83 ang(36.0 deg)
Ic = Ica-Ibc = +6.00 - j14.44 = 18.00 ang(-71.0 deg)

I think you'll find if you sum Ia, Ib, & Ic, you will get zero.

If you draw a phasor diagram, you'll see the 3 currents, Ia, Ib, & Ic are almost 120 degrees apart. I say "almost," because A & B are ~118 deg apart, B & C are ~107 deg apart, and C & A are ~135 deg apart.

cork_ie likes this.
11. ### matt_t Thread Starter Member

Aug 18, 2012
30
0
Thanks for the reply this is what I understand according to hat tnk has said. I do not understand why the textbook answers are line currents B=37.2A and C=67A?

12. ### mlog Member

Feb 11, 2012
276
36
Is that the entire problem? I noticed the supply voltage was given as 600 volts. Is there something else given? What does it say for the A current?

13. ### matt_t Thread Starter Member

Aug 18, 2012
30
0
That is the entire problem in the first post the question only asked to work out lines A and B so I've been spending alot of time trying to figure out the answers given.

14. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Unfortunately this solution is incorrect - the Op's answers are good.

15. ### mlog Member

Feb 11, 2012
276
36
What did I do wrong?

16. ### matt_t Thread Starter Member

Aug 18, 2012
30
0
I have been on this same question for 2 days now and still cannot come up with the answers. This is as close as I can get:

B = 20/_ -25.84 - 30/_ 120 = 18+j8.717 - -15+j25.98 = 33+j34.697 = 47.88 /_46.435

17. ### matt_t Thread Starter Member

Aug 18, 2012
30
0
OK so i've finally realised what I did wrong.

B = 20/_ -25.84 - 30/_ 120 = 18+j8.717 - -15+j25.98 = 33-j17.26 = 37.24/_-27.61A

FINALLY.

18. ### matt_t Thread Starter Member

Aug 18, 2012
30
0
C= 30/_ 0 - 40/_ 145.84 = 30+j0 - -33.09+j22.46 = 63.09-j22.46 = 66.96/_19.59A

19. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783

$I_a=52.92 \angle{-66.74^o} \ A$

$I_b=37.2 \angle{ -152.5^o} \ A$

$I_c=66.9 \angle{ 79.58^o} \ A$

Last edited: Sep 12, 2012
20. ### matt_t Thread Starter Member

Aug 18, 2012
30
0
how did you get those different angles?