Resistance Calculation 3Phase Delta Load

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fifty60

fifty60

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I have a 20,000W 460V 3Phase Delta heater load. I am trying to calculate the resistance.

I-Line = 20000W/460V/Sqrt3 = 25.1A
I-Phase = 25.1/Sqrt3 = 14.5A

R-Phase = 460/14.5 = 31.7 Ohms

What I am actually measuring with my Fluke is approximately 19 Ohms.

If I take 460VAC divided by the line current (25.1A) I do get 18.3 Ohms. Is my original calculation wrong? I definitely though I had to divide a second time by the square root of three to get the phase current, and then calculate phase resistance from there. Is that wrong, or is there something wrong with my readings. The readings are all the same from each leg to each leg as well, so I am thinking the reading is correct.
 
It is possible (probably likely) that the cold resistance and hot resistance of your elements is not the same.
 
fifty60 said:
I have a 20,000W 460V 3Phase Delta heater load. I am trying to calculate the resistance.

I-Line = 20000W/460V/Sqrt3 = 25.1A
I-Phase = 25.1/Sqrt3 = 14.5A

R-Phase = 460/14.5 = 31.7 Ohms

What I am actually measuring with my Fluke is approximately 19 Ohms.

If I take 460VAC divided by the line current (25.1A) I do get 18.3 Ohms. Is my original calculation wrong? I definitely though I had to divide a second time by the square root of three to get the phase current, and then calculate phase resistance from there. Is that wrong, or is there something wrong with my readings. The readings are all the same from each leg to each leg as well, so I am thinking the reading is correct.
Click to expand...
I-Line and I-Phase are the same in a Delta configuration, you used one too many sq. rt. of 3s...

Also, 460V is a utilization voltage, line voltage should be 480V. So 24.06A into 480V = 19.95ohms
 
140923-1051 EDT

fifty60:

I think the answer is simple.

You have a delta load that consists of three equal resistors connected in delta.

From your information the total power is 20000 W. Divide this by 3, the power in one of the three resistors. One resistor has 460 V applied so its resistance is 460^2*3/20,000 = 460*460/ 6666.7 = 211,600/6666.7 = 31.74 ohms.

Next you have a 31.74 ohms resistor in parallel with two 31.74 ohm resistors in series in your delta load when viewed from any two terminals of the delta.

Thus, the measured resistance should be 31.74 in parallel with 63.48 . You do the calculation.

.
 
Jraef said:
I-Line and I-Phase are the same in a Delta configuration, you used one too many sq. rt. of 3s...

Also, 460V is a utilization voltage, line voltage should be 480V. So 24.06A into 480V = 19.95ohms
Click to expand...
I-line and I-phase are the same for a single line to line load whether the source is wye or delta. But when you have three balanced line to line loads I-line equals I-phase(sqrt(3)).
You also have to be careful for a single line to line load not to try to calculate power by multiplying I-line times V(line-to-neutral) because kW is not equal to kVA in this situation.
I will not go into what this means in terms of PF, since that discussion was never settled last time. :)
 
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