25kW operating across a balanced three-phase 460v what is the current draw?

I came up with 54.35A

 

I think you need to include √3 in your calculation since this is 3-phase.

 

I think you need to include √3 in your calculation since this is 3-phase.

Still comes up to 54.34A

What did you come up with

 

This is what I get:

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

 

This is what I get:

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

only had 31.41,54.35or 163.04 to chose from not 31.35 to choose from

 

I found 2 pages that will help.
http://myelectrical.com/notes/entryid/8/three-phase-power-simple-calculations
http://ncalculators.com/electrical/kva-calculator.htm

 

You will get 31.41A if your round √3 down to 1.73

 

only had 31.41,54.35or 163.04 to chose from not 31.35 to choose from

Those two answers only differ by 0.2%. What more can you expect, especially when odinhg indicated that the answer was approximately 31.34A.

 

Those two answers only differ by 0.2%. What more can you expect, especially when odinhg indicated that the answer was approximately 31.34A.

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

 

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.

 

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.

 

25kW operating across a balanced three-phase 460v what is the current draw?

I came up with 54.35A

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.

 

In the original post here

http://forum.allaboutcircuits.com/showthread.php?t=87691

the OP stated this was a 25kW 3 phase heater. A power factor of 0.5 seems rather extreme.

 

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.

 

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.

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

 

Perhaps you should consider changing schools and going to one that is actually going to teach you correct information.

 

Perhaps you should consider changing schools and going to one that is actually going to teach you correct information.

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

 

The answer came out the same so both equations worked.

Didn't one giver you 54A and the other 31A?

Schools not the answer its who you know.

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?

Thanks again for your help and I learned soething as well

Your welcome, and I hope so.