# 3 Phase heating elements

Discussion in 'General Electronics Chat' started by Jsw123, Jun 20, 2009.

1. ### Jsw123 Thread Starter Active Member

Jun 20, 2009
43
0
I am trying to design a set of 3 phase heating elements for a 120/208 volt system. The purpose of these is to maintain a constant load on a generator when no load is present. The load will be 30 amps.
My question is:
- Do I find the necessary resistance of individual elements by E/I (120 volts/30 amps) or is it different since the three phases are connected together?

- Will an individual heating element consume more power on 3 phase Y than it would on single phase 120volt?

- If I choose to use elements connected in delta what what would the calculation for resistance be?

- What would be the increase in power consumtion for a given set of elements switched from wye to delta?
Thank you

2. ### mik3 Senior Member

Feb 4, 2008
4,846
63
This looks like a school assignment.

Where is your work until now?

When you say 30 Amps, you mean per phase?

3. ### GetDeviceInfo Senior Member

Jun 7, 2009
1,571
230
On the purely resistive balanced load side, power = Vline * Iline * sqrt(3). This being the case for both Y or Delta.
If connected in Y, phase voltages are line voltages/sqrt(3). In your case of 120/208, the lines are 208v and the phases 120. Each leg is a single phase 120.
In Delta, your line current is sqrt(3) times phase current.
In Delta, your phase voltage is sqrt(3) higher than in Y. As well, your line current is sqrt(3) times phase. This results in 3 times the power.

4. ### Jsw123 Thread Starter Active Member

Jun 20, 2009
43
0
Ok so if I use elements rtated at 240 v0lts, 2400 watt in delta will they consume more wattage causing the elements to burn up? Also for a given set of elements in delta to find amps do I devide 208 by resistance or is it something more?

5. ### Jsw123 Thread Starter Active Member

Jun 20, 2009
43
0
This is a home generator project.

6. ### GetDeviceInfo Senior Member

Jun 7, 2009
1,571
230
if power in a balanced resistive 3 phase = Vline * Iline * sqrt(3),
because it's balanced, then phase power = 1/3 total power.

7. ### Jsw123 Thread Starter Active Member

Jun 20, 2009
43
0
The power source is wye connected but the heating loads will be delta connected to provide more power. My question is if I use three sets of elements rated at 2400 watt 240 volt single phase, since there is a amperage gain in delta (I line x 1.73) will the individual elements draw more than 2400 watt thus causing them to burn up? do i need tho use elements rated at a higher voltage?

Also you said your phase current is Vline / Rphase. How do you find R Phase?

8. ### Jsw123 Thread Starter Active Member

Jun 20, 2009
43
0
If I go with wye connected I don't think there is a difference.But if I go with delta is there?

9. ### GetDeviceInfo Senior Member

Jun 7, 2009
1,571
230
For your resistive elements at 2400w/240v;
P=V*I
P=V*(V/R)
P=Vsq/R
R=Vsq/P
so, 240V sq / 2400w = 24 ohm

In your Y configuration, total power = 3 * (120Vphase)sq / 24 ohm = 1800w. As line current = phase current, Iline = Iphase = 5amp.

In your Delta config, total power = 3 * (208Vphase)sq / 24 ohm = 5408w
Iphase = 8.67amp. Iline = Iphase * sqrt(3) = 15amp.

Notice that power dissipated is 3 times greater in Delta compared to Y. Because you are using the same line voltage, the difference is in the current consumption.

Again, total power in a balanced resistive circuit can be calculated as Vline * Iline * sqrt(3).

10. ### Jsw123 Thread Starter Active Member

Jun 20, 2009
43
0
Thank you the line system is 120/208v wye but I hope to connect heating elements in delta. is the formula for the elements above then
3* (120v phase)sq * 24ohm?