Ohms = E squared / Power

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tryinghard

tryinghard

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California
I have a heater that is labeled 3PH 440V 3000W, it was manufactured in 1964.

We have been rewiring a unit that actually includes 4 of these and just for kisses and giggles I checked the resistance. My results are 118 ohms from A-B, B-C, & A-C on each heater.

I think these elements are losing resistance through use and age. The branch is currently a 480V 30A circuit, if the 118 ohms holds true after warm up I should read 28 amperes, I know this will be the final test.

BUT if Ohms = E squared / P shouldn?t I read 194 ohms? I understand reading resistance cold may be lower than reading resistance heated up, but would it be this much? Any advice/opinions?
 
tryinghard said:
I have a heater that is labeled 3PH 440V 3000W, it was manufactured in 1964.

We have been rewiring a unit that actually includes 4 of these and just for kisses and giggles I checked the resistance. My results are 118 ohms from A-B, B-C, & A-C on each heater.

I think these elements are losing resistance through use and age. The branch is currently a 480V 30A circuit, if the 118 ohms holds true after warm up I should read 28 amperes, I know this will be the final test.

BUT if Ohms = E squared / P shouldn’t I read 194 ohms? I understand reading resistance cold may be lower than reading resistance heated up, but would it be this much? Any advice/opinions?
Click to expand...
Normally the resistance of the elements is somewhat higher at elevated temperatures.

If these are in a delta configuration you would be reading 194 ohms in parallel with 388 ohms. About 129 ohms.
 
Last edited:
tryinghard said:
I have a heater that is labeled 3PH 440V 3000W, it was manufactured in 1964.

We have been rewiring a unit that actually includes 4 of these and just for kisses and giggles I checked the resistance. My results are 118 ohms from A-B, B-C, & A-C on each heater.

I think these elements are losing resistance through use and age. The branch is currently a 480V 30A circuit, if the 118 ohms holds true after warm up I should read 28 amperes, I know this will be the final test.

BUT if Ohms = E squared / P shouldn?t I read 194 ohms? I understand reading resistance cold may be lower than reading resistance heated up, but would it be this much? Any advice/opinions?
Click to expand...
If the elements are connected in delta then yes, it should be 194 ohms each. But, if you measure resistance between two input lines you will measure across that arm of the delta with the other two arms in series with each other but in parallel with the 194 ohms. So you have 193.6 in parallel with 387.2, or about 129 ohms. Not so far off your 118 ohms, and as the iwire guy says, resistance will increase with temperature.
I can see where you got your 28A at 480V.
It's about 23kW from 12kW of heaters. The 118 ohms isn't the real resistance.

That said, if you operate a 440V heater at 480V, you will subject it to about a 20% overload.
I'm not sure that it is good or safe.
 
This is a particularly tricky problem to set up mathematically. It doesn?t follow all the usual formulas. This is how I would go about it; I?d take it one step at a time.

Step 1: If the rated voltage is 440 volts, three phase, then the phase-to-neutral voltage is 440/1.732, or 254 volts. That is the value of ?E? I would use in the OP?s formula.

Step 2: If the rated power for the entire heater is 3000 watts, then each heater element is rated for 1000 watts. That is the value of ?P? I would use in the OP?s formula.

Step 3: Using R = E^2/P, I get R = 254 x 254 / 1000, or 64.5 ohms. That would be the resistance of the heater when new, and when running at full power (normal operating temperature). The resistance measurement taken when cold would be lower.

Step 4: Look at your method of measuring resistance. You measured phase-to-phase, but with no power applied, and no current flowing. Also, when you measured any two phases, nothing was connected to the third phase. Thus, you were not measuring a three-phase configuration. You were measuring two resistances that were, at the time of your measurement, and under the configuration of your measurement, in series with each other.

Step 5: Thus, to predict the measurement of resistance phase-to-phase, I would start with the 64.5 ohms, and double it, since you measured two elements in series. That predicts a measurement of 129 ohms. You measured 118 ohms, a value that is about 91% of the prediction. I can account for the difference by noting that the heater was not new, and the measurements were taken cold.

Step 6: If you wish to check my analysis, and if you can do this both quickly and (more importantly) safely, try to take resistance measurements when the elements are hot. You do that by letting the heater run for a while, then turn it off, and take the measurements before it has much of a chance to cool. Did I emphasize the ?safely? part strongly enough?
 
charlie b said:
Step 6: If you wish to check my analysis, and if you can do this both quickly and (more importantly) safely, try to take resistance measurements when the elements are hot. You do that by letting the heater run for a while, then turn it off, and take the measurements before it has much of a chance to cool. Did I emphasize the ?safely? part strongly enough?
Click to expand...
I would instead calculate the hot-n-running resistance by measuring voltage and current while operating.
 
Besoeker said:
If the elements are connected in delta then yes, it should be 194 ohms each. But, if you measure resistance between two input lines you will measure across that arm of the delta with the other two arms in series with each other but in parallel with the 194 ohms. So you have 193.6 in parallel with 387.2, or about 129 ohms. Not so far off your 118 ohms...
Click to expand...

I can see by drawing how my reading was one arm of the delta and the other two in series but I still don't understand how to get the 129 ohms you come up with.
 
charlie b said:
...Step 1: If the rated voltage is 440 volts, three phase, then the phase-to-neutral voltage is 440/1.732, or 254 volts. That is the value of ?E? I would use in the OP?s formula...
Click to expand...

Is this how I should calculate even though I don't have a neutral?
 
tryinghard said:
Is this how I should calculate even though I don't have a neutral?
Click to expand...
If you know the heater is connected as a Delta (come to think of it, that is more likely than a WYE), then look at Besoeker?s post #5.
 
081009-0908 EST

tryinghard:

Assuming balanced source voltages, equal phase angles, and a balanced load, then there are two equivalent circuits you can draw for your resistive load. One is a delta configuration and the other a Y. The load can be built either way and either configuration can be fed from either a delta or Y source and the same heat will be generated. Obviously the values of the individual resistors are different between the delta and Y loads. But from the terminals alone you can not tell the difference assuming the Y connection does not have its neutral point brought out to a fourth terminal.

charlie b gave you a very good description.

Neutral is not an issue. Do not blindly follow the use of some numbers like 1.732 without an understanding of its derivation.

The delta is probably the easiest to work with since you know the delta voltage. What I am saying has already been said above. The power dissipated in a resistor is Pr = Vr^2/R. A delta has a resistor between each pair of legs.

Since everything is equal based on the above assumptions, then 1/3 of the total power is dissipated in one resistor. This is 1000 W. The voltage across the resistor is 440 V. The current thru the resistor is 1000/440 = 2.27 A. This is not the current in the supply leg, but just the current in the one resistor.

Also the value of this one resistor is R1 = Vll^2/P1 = 440^2/1000 = 193,600/1000 = 193.6 ohms.

When you measure between two terminals of the load you have 2*193.6 = 387.2 ohms in parallel with 193.6 ohms. This combination is Rll = 1/(1/193.6 + 1/387.2) = 1/(5.1653*10^-3 + 2.583*10^-3) = 1/(7.748*10^-3) = 10^3/7.748 = 129.1 ohms. All very easy to calculate on an HP RPN calculator.

The other way to look at the problem is to consider the load to be in the Y configuration. Now you need to know the voltage from a supply leg to the neutral line. This is where 1.732 comes into play. Do you know how to derive this value?

Line to neutral voltage is 440/1.732 = 254.04 . Again knowing the power in one resistor is 1000 W we get R1y = 254.04^2/1000 = 64537.1/1000 = 64.54 ohms. Now as said before the terminal to terminal resistance is 2*64.54 = 129.1 ohms. The same value we saw above.

.
 
tryinghard said:
I can see by drawing how my reading was one arm of the delta and the other two in series but I still don't understand how to get the 129 ohms you come up with.
Click to expand...
OK.
Each element is 194 ohms.
The two in series make 388 ohms (194*2)
So, you have 194 in parallel with 388.
This gives 194*388/(194+388)
129 ohms
 
charlie b said:

If you know the heater is connected as a Delta (come to think of it, that is more likely than a WYE), then look at Besoeker?s post #5.
Click to expand...
Thank you for that.
Yes, if it just a three terminal heater, I agree that it is more likely to be delta - as it happens, I chose to take it as as a delta because tryinghard had already come up with the 194 ohms.

As a Y connection a short circuit on one leg woud subject the other two to line voltage with the possible danger of serious overheating.
That said, it would make no difference to the measured resistance between terminals whether the connection is Y or delta.
 
Besoeker said:
OK.
Each element is 194 ohms.
The two in series make 388 ohms (194*2)
So, you have 194 in parallel with 388.
This gives 194*388/(194+388)
129 ohms
Click to expand...

Besoeker, I knew there would be a math answer to this issue and the last part of the equation is not something I would stumble to. Without really understanding the math I suppose someone like me could just read the resistance of each individual element and arrive at 194 ohms (193,600 / 1000) rather than reading the phases. I want to use the resistance to check an element but it?s not real accurate unless the element is at its operating temperature.
 
gar said:
?

Neutral is not an issue. Do not blindly follow the use of some numbers like 1.732 without an understanding of its derivation.

?Do you know how to derive this value?

Line to neutral voltage is 440/1.732 = 254.04 . Again knowing the power in one resistor is 1000 W we get R1y = 254.04^2/1000 = 64537.1/1000 = 64.54 ohms. Now as said before the terminal to terminal resistance is 2*64.54 = 129.1 ohms. The same value we saw above.
Click to expand...

Sounds like I can calculate this heater either way as delta or wye. The ?wye? way does seem simpler to me though.

I do understand 1.732 is the square root of 3 but I do not understand why the value of 1.732 applies?
 
tryinghard said:
Besoeker, I knew there would be a math answer to this issue and the last part of the equation is not something I would stumble to. Without really understanding the math I suppose someone like me could just read the resistance of each individual element and arrive at 194 ohms (193,600 / 1000) rather than reading the phases.
Click to expand...
If you measure from just the three terminals you can't measure the elements individually. The mathematics isn't so difficult.
For two resistors in parallel the total resistance, Rt, can be determined from:
1/Rt = 1/R1 + 1/R2
A little bit of manipulation gives you:
Rt = R1*R2/(R1+R2)
 
Besoeker said:
If you measure from just the three terminals you can't measure the elements individually. The mathematics isn't so difficult.
For two resistors in parallel the total resistance, Rt, can be determined from:
1/Rt = 1/R1 + 1/R2
A little bit of manipulation gives you:
Rt = R1*R2/(R1+R2)
Click to expand...

True, thanks
 
081010-1805 EST

tryinghard:

I thought the derivation of the 1.732 factor might be unknown to many users of the number. Later I will try to explain this, or maybe someone else will before I have the time.

.
 
tryinghard said:
Sounds like I can calculate this heater either way as delta or wye. The ?wye? way does seem simpler to me though.

I do understand 1.732 is the square root of 3 but I do not understand why the value of 1.732 applies?
Click to expand...

If you know a little trig, you can see that the line-to-line voltage is 1.732 x the line to neutral voltage. For example,

208/120 = 1.73 = 2cos(30)

If you don't know trig, draw out the wye to scale. Then carefully measure the line to line voltage and compute the ratio, L-L/L-N.
 
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