Calculating resistance in a 3 phase heater circuit

[PaulKraemer]

Hi,

I have a three phase resistance heater as shown in the attached diagram.

I know the formula for power is P = I * E * 1.732. Based on the information shown, this comes to P = 22.6 * 480 * 1.732 = 17,789 watts. This is pretty close to the 18,650 watts shown on the diagram, so this makes sense to me.

In the location where I am hoping to use this heater, I don't have 480 VAC. I only have 208 VAC. I am hoping to calculate what my current and power would be at this lower voltage. To do this, it seems like I might have to calculate what the resistance of each heater element is using the data provided for 480 VAC, and use this resistance to calculate what the current would be at 208 VAC. After I know what the current would be at 208 VAC, it would be easy to then calculate the power using the formula I've mentioned above.

I don't know how to calculate the resistance of the heater elements in a three phase circuit like this. If anyone can point me in the right direction, I would greatly appreciate it.

Thanks and best regards,
Paul

 

[FacEngrPE]

V/I=R --- so 480 V / 22.6 A = 21.23 ohms (each resistor)

V/R=I --- so 208 V / 21.23 ohms = 9.35 A

Now if we assume the load on the circuit is resistive only, power factor is unity (or one) which reduces the formula to P = V x I x square root of three.

208 V x 9.35 A x SQRT 3 = 3368 watts

Resistance heater capacity is a strong function of voltage, as voltage is needed to push the current through the wire.

If this is not enough watts for your application, it is likely less cost to order the correct heater vs ordering boost transformer.

 

[che12345]

Mr. PaulKraemer (Electrical)(OP)18 Aug 23 20:01
"...I have a three phase resistance heater as shown in the attached diagram..... this comes to P = 22.6 * 480 * 1.732 = 17,789 watts. This is pretty close to the 18,650 watts shown on the diagram, I don't have 480 VAC. I only have 208 VAC...."
My suggestion as following for your consideration:
1. P = I[sup]2[/sup]x R (W) basic
I = V/R, therefore I[sup]2[/sup]= (V/R)[sup]2[/sup]
That is, P is proportional to the V[sup]2[/sup]
Therefore, P2 = P1 x (208V/480V)[sup]2[/sup]
P2 = P1 x 0.1874
Based on P1 = 17789 W, P2 = 3340.77 W.
2. You can calculate the current according to your formula , which is in order.
Che Kuan Yau (Singapore)

 

[waross]

18650W/3phases = 6217 Watts per heater.
6217W/480V = 12.95 Amps per heater.
480V/12.95A = 37 Ohms.
(12.95A √3 = 22.4 Amps (Close enough for rounding error, both mine and theirs))

Or;
208V/480V = .433
.433[sup]2[/sup] = .188
18650W x .188 = 3502W @ 208V

--------------------
Ohm's law
Not just a good idea;
It's the LAW!

 

[che12345]

Mr. PaulKraemer (Electrical)(OP)18 Aug 23 20:01
"....I have a three phase resistance heater as shown in the attached diagram".
site revisited.
1. All the learned advice are close enough.
2. There is some discrepancies, however.
Based on your given 1.732 x 480V x 22.6A = 17789W is incorrect. Should be 18788.736W.
3. P2 = P1 (V2/V1)[sup]2[/sup] ...W .......(1)
= 18788.736W x (208V/480V)[sup]2[/sup]= 3526.645W
4. I2 = (P1/P2) x (V1/V2) x (I1) ...A......(2)
= (3526.645/18788.736)W x (480V/208V) x (22.6)A = 9.7836A
5. Check: 1.732 x 208V x 9.7836A = 3524.60W. The difference is due to rounding of decimal point.
6. See above 3 and 4, it is evident that there is no requirement to calculate the value of R .
Che Kuan Yau (Singapore)

 

[FacEngrPE]

There are actually many other ways to arrange this calculation, some more cumbersome than others.
I prefer simple, and easy to explain.

I agree about the rounding error. Too often we calculate insignificant digits that don't matter. Here we calculate 3 or 4 digits on an electric circuit where the voltage can be +/- 10%. Only 2 digits provide useful information.

 

[7anoter4]

Srat=sqrt(3)*480*22.6=18789.3 VA, I think.
However, if the heater impedance [I think it is a slight reactance] it is:
Ztotal=480^2/18789.3=12.26 ohm
Snew=208^2/12.26=3528.9 VA

 

[waross]

Remember folk;
We have not only our own rounding errors but rounding errors in the original spec sheet.

--------------------
Ohm's law
Not just a good idea;
It's the LAW!

 

[LionelHutz]

sqrt(3) x 480 x 22.6 = 18789W is what I get.

As for at 208V, 2 calculations can get the current and wattage you want.

208V/480V x 22.6A = 9.79A
answer (9.79) x sqrt(3) x 208V = 3528W

I got told years ago to never type 1.732 or 1.414 into a calculator, always type in sqrt(3) or sqrt(2). After the current calculation, I use the answer feature for the current in the next calculation. Stops accumulating rounding errors.

 

[waross]

I got told years ago to never type 1.732 or 1.414 into a calculator, always type in sqrt(3) or sqrt(2). After the current calculation, I use the answer feature for the current in the next calculation. Stops accumulating rounding errors.

--------------------
Ohm's law
Not just a good idea;
It's the LAW!

 

[zeusfaber]

Anyone got a view on how much lower the resistance will be at the lower voltage?

A.

 

[IRstuff]

The resistance is proportional to temperature, and temperature is roughly proportional to power

TTFN (ta ta for now)
I can do absolutely anything. I'm an expert!

https://www.youtube.com/watch?v=BKorP55Aqvg

faq731-376 forum1529 Entire Forum list

http://www.eng-tips.com/forumlist.cfm

 

[waross]

At the lower voltage, the current will be lower.
With lower current the heater will run cooler.
At a lower temperature, the resistance will be slightly lower.
Unless the heater is running at incandescent temperatures, we tend to ignore the difference.
At one half voltage, the heater is running at 1/4 the wattage, plus a little bit.

Exception: The resistance of a tungsten incandescent lamp increases by a factor of about 10:1 from cold to rated voltage.
If you are using incandescent lamps as heaters, the resistance calculation at lower voltages becomes challenging.

--------------------
Ohm's law
Not just a good idea;
It's the LAW!