# current draw

Discussion in 'Homework Help' started by bwd111, Jul 28, 2013.

1. ### bwd111 Thread Starter Member

Jul 24, 2013
117
1
25kW operating across a balanced three-phase 460v what is the current draw?

I came up with 54.35A

2. ### odinhg Active Member

Jul 22, 2009
65
15
I think you need to include √3 in your calculation since this is 3-phase.

3. ### bwd111 Thread Starter Member

Jul 24, 2013
117
1
Still comes up to 54.34A

What did you come up with

4. ### odinhg Active Member

Jul 22, 2009
65
15
This is what I get:

$\frac{25000W}{\sqrt{3}\ast460V}\approx31,34A$

5. ### bwd111 Thread Starter Member

Jul 24, 2013
117
1
only had 31.41,54.35or 163.04 to chose from not 31.35 to choose from

Apr 27, 2013
920
161
7. ### odinhg Active Member

Jul 22, 2009
65
15
You will get 31.41A if your round √3 down to 1.73

8. ### WBahn Moderator

Mar 31, 2012
20,214
5,743
Those two answers only differ by 0.2%. What more can you expect, especially when odinhg indicated that the answer was approximately 31.34A.

9. ### bwd111 Thread Starter Member

Jul 24, 2013
117
1
Why wont this formula work P = I*V, so 25,000W = I * 460v? The answer comes up to 54.35A

10. ### WBahn Moderator

Mar 31, 2012
20,214
5,743
Because that formula applies to a device in which you have two terminals across which V appears and the I going in one terminal and out the other. It also assumes that the V and the I are in phase.

With a three-phase netwrok, you have three terminals. Across any two of them you have V (but the pairs have phase shifts of 0, +120°, and -120°). This means that you have three supplies dumping power into it. But the reason why the total power isn't 3VI is because the voltage and the current aren't in phase on any pair of legs.

Work through the analysis on either a delta or a wye connected load and you should see how it works out. In particular, focus on the relationship between the line-to-line voltages compared to the line-to-neutral voltages.

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11. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
784
Keep in mind the 25kW is perhaps distributed over 3 phases.

If the line-to-line voltage is 460 volts then the line current would be

I_line=25000/(Vll*√3)=31.38 Amps.

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12. ### donpetru Senior Member

Nov 14, 2008
186
26
Current draw depends on the power factor, power load and voltage.
Mathematical relations for the calculation are:

Option 1: - use voltage between phases:

I = P / (1.73 * V_line * cos_Fi)

where: I - current draw, P = 25kW, V_line - voltage between phases; cos_Fi - power factor (<=1).

Option 2: - use voltage between phase and neutral:

I = P / (3 * V_phase * cos_Fi)

where: I - current draw, P = 25kW, V_phase - voltage between phase and neutral (V_phase = V_line / 1.73); cos_Fi - power factor (<=1).

If you apply any of the two above mathematical relationships, the result is the same.

You should take into account a power factor of 0.5, to cover all cases.

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Mar 6, 2009
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14. ### donpetru Senior Member

Nov 14, 2008
186
26
My statement above, about power factor of 0.5, was made in the event that it is not known type of load.

LATER EDIT: If the load is resistive then the power factor is equal to 1.

Last edited: Jul 29, 2013
15. ### bwd111 Thread Starter Member

Jul 24, 2013
117
1
this was the formula we were shown and is different that what you all have shown me. kw*100/1.73*e*PF

16. ### WBahn Moderator

Mar 31, 2012
20,214
5,743
Perhaps you should consider changing schools and going to one that is actually going to teach you correct information.

17. ### bwd111 Thread Starter Member

Jul 24, 2013
117
1
The answer came out the same so both equations worked. Schools not the answer its who you know. Thanks again for your help and I learned soething as well

18. ### WBahn Moderator

Mar 31, 2012
20,214
5,743
Didn't one giver you 54A and the other 31A?

I don't follow. If you believe this, then why are you in school at all? Or do you figure that all you need to do is get a piece of paper and then rely on getting someone to do a favor for you and give you a job?

Your welcome, and I hope so.

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