# Heater wiring in Star or Delta

Discussion in 'General Electronics Chat' started by kallileo, Jul 23, 2014.

1. ### kallileo Thread Starter New Member

Dec 25, 2013
3
0
I have a 12kW heater which has 3x 4kW 230V elements.
I wire it into Star connection and apply ~3, 400V in order to have 230V phase voltage and the line current is around 18A.
1. How much will the power increase if the heater is wired in Delta at 400V (if it doesn't burn)?
2. What will be the line current if I wire it in Delta and apply ~3, 240V (I suppose the power will remain the same)?

2. ### inwo Well-Known Member

Nov 7, 2013
2,441
315
If you almost double the voltage, elements will burn out.
Ohms law. Resistance stays the same,

Each element drawing 29 amps. 11.6 KW

Line current 50 amps. total of 35KW

Line current in delta is greater than element current by the square root of 3. ~1.732

Not considering increased resistance at incandescence.

Last edited: Jul 23, 2014
3. ### Dodgydave AAC Fanatic!

Jun 22, 2012
6,416
1,017
Approx 12KW per element @ 30A each phase , thats nearly 36KW total power. Like bang!! man

4. ### kallileo Thread Starter New Member

Dec 25, 2013
3
0
At the moment I wire it in Star and I get 18A line current and 12kW at 400V, which is normal.
If the voltage is lowered to 230v and I wire it in Delta shouldn't the line current remain the same since I lowered the voltage?

5. ### Dodgydave AAC Fanatic!

Jun 22, 2012
6,416
1,017
If your going to end up with the same wattage and current, why change it?

6. ### inwo Well-Known Member

Nov 7, 2013
2,441
315
Which is it?

The current will follow the voltage across each element.

edit: amplitude

Last edited: Jul 24, 2014
7. ### kallileo Thread Starter New Member

Dec 25, 2013
3
0
I need it be installed on ~3, 230V and I need to calculate the line current to choose proper circuit breaker.
The resistance of every heating element is 13Ω.
According to the calculation I did the power remains the same but the line current will increase to 30A.

Delta:
I(phase) = 230V / 13Ω = 17,5A
I(line) = 1,73*I(phase) = 30A

I(line) = 30A
P=1,73*30*230 = 12kW

I believe that the calculations are correct.