frorge

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!

Big John

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

frorge

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

Dennis Alwon

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

__________________

frorge

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

Big John

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

InPhase277

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.

elmecinchennai

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

hardworkingstiff

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.

__________________
asomatous
Lou