# 3-Phase Heater Question



## frorge (Jul 18, 2012)

Hey guys, I'm new to the forum, but thanks in advance for any help!

I've got a question concerning an electrical element resistive heater. 

It runs off of 480 3Phase Voltage
The heater is a delta configuration, 60 kW

We've had a new heater made to replicate one that is in service currently. The new heater is measuring 7.6 ohms between each of the phases.

My calculations came up with 11.5 ohms.

I was using the formula R (ohms) = (3 * Volts^2) / Watts

If I plug in the measured 7.6 ohms into my formula I get ~90.0 kW

So my question is, am I using the wrong equation, or have we got an improperly sized heater?

Thanks!


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## Big John (May 23, 2010)

The resistance of the element increases with temperature. If you measure the resistance of a cold light-bulb filament it also appears to defy ohm's law.

-John


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## frorge (Jul 18, 2012)

I definitely understand that. But I am measuring both circuits at the same room temperature in the same conditions.

I guess my question is: is my formula correct for calculating the resistance of the heater? If so, can I be fairly certain the heater measuring 7.6 ohms is NOT a 60 kW heater? And if my equation is not right....then what is the right formula! haha


Thanks


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## Dennis Alwon (May 9, 2009)

I don't understand the ohms measuring 7.6 between the phases. I assume 3 elements and each element is 7.6 ohms???? What is the wattage of the element? Is it marked? 

If you get 7.6 ohms on one element then at 480V this is a 30kw element if you have 3 elements then you have 60 kw load on each phase

AB BC CA


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## frorge (Jul 18, 2012)

Yes, sorry. The resistance measured across one element is 7.6 ohms.

Ie. if you imagine the circuit as a triangle with vertices A,B,C and resistors between AB, BC, CA. Then AB=BC=CA= 7.6


The total wattage of the system is suppose to be 60 kW


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## Big John (May 23, 2010)

I do not believe you can calculate the true wattage of a heater when all you know is the room temperature resistance.

What I would do is find an element of known wattage, and see if it also has a 7.6Ω resistance. If it does, the wattages of the two elements are probably the same.

-John


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## RePhase277 (Feb 5, 2008)

You are reading through the other elements in the delta. You would have to disconnect one element end entirely to get an accurate resitance reading.


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## elmecinchennai (Jul 31, 2012)

am looking for some info regarding manifold heaters? do anyone heard about it, what will be its resistance and wattage?


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## hardworkingstiff (Jan 22, 2007)

InPhase277 said:


> You are reading through the other elements in the delta. You would have to disconnect one element end entirely to get an accurate resitance reading.


R total = (R1*R2)/(R1+R2)

So, if all of the elements are connected in the delta, and you put your meter across one element, you have 2 paths for current, one through the element you are testing across the other is through the other 2 elements in series, so ...

R total = (11.5*23)/(11.5+23) = 264.5/34.5 = 7.67

I think this proves what Inphase said, and also matches (pretty close) what the OP posted.

Edit for the OP:

480V / 11.5 ohms = 41.74A
41.74A * 480V = 20,035 VA (since PF is not a factor, it's 20kW)

20kW * 3 elements = a 60kW heater.


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