#### bill@usps06492

##### Member

- Location
- Connecticut

[ January 23, 2004, 09:27 AM: Message edited by: bill@usps06492 ]

- Thread starter bill@usps06492
- Start date

- Status
- Not open for further replies.

- Location
- Connecticut

[ January 23, 2004, 09:27 AM: Message edited by: bill@usps06492 ]

- Location
- Seattle, WA

- Occupation
- Electrical Engineer

I think you have a typo, and surmise that you meant three heaters at 3000 watts each.Originally posted by bill@usps06492:I am installing 3-2000w single phase quartz heaters @ 208volt@14.4amps each.

No. The formula you started with (I = 3kw x 1000/208 = 14.4a) tells you the current in a single heater. Multiplying by the 125% (to get 18a) gives you the required sizes for the branch circuit conductors and the overcurrent protection, but it applies only to a single phase circuit that serves a single heater. When you connect the three heaters in a three phase balanced configuration, you do not add 18 + 18 + 18 to get 54 amps. Rather, you multiply the current in a single phase by the square root of three. 18 times 1.732 is 31.2 amps. The next higher standard size OCPD is 35 amps.Should I be sizing my branch circuit, controller and overload protection on the 54 amp?

- Location
- Seattle, WA

- Occupation
- Electrical Engineer

But it?s not a single phase system. Bill said he planned to balance the loads on a three phase feeder. That tells me that one heater is connected from A to B, another from B to C, and the third from C to A. Anytime you get involved with a three phase system, the factor of "the square root of 3" is going to find its way into the calculations. I will stick with my answer.

Charlie,

You said, "Bill said he planned to balance the loads on a three phase feeder. That tells me that one heater is connected from A to B, another from B to C, and the third from C to A."

I agree that thats what Bill is doing. If we use Phase A as an example, it would have the first and third heater on it. The current from the first and third heater would be 14.4 amps each, or 28.8 amps, correct? Seems to me this would be the same for B and C as well. Shouldn't each phase be sized for that? Current per leg is not affected by 1.73 in a wye connection. I line = I phase for wye. Only voltage. E line = E phase x 1.73 ie 208V = 120 x 1.73.

I appreciate your input as always. Thank you.

- Location
- Seattle, WA

- Occupation
- Electrical Engineer

If a single load touches two phases (and the other two loads do the same), then you are describing a DELTA configuration, not a WYE.Originally posted by spsnyder:If we use Phase A as an example, it would have the first and third heater on it.

Sorry, not correct. If you are adding up apples, oranges, ducks, telemarketers, or single-phase amps, then 14.4 + 14.4 will equal 28.8. Not so for three phase amps. That is because just calling it ?14.4 amps? is not a complete description. What you are adding is the value ?14.4 amps at an angle of zero degrees with respect to the Phase A Voltage? to the value ?14.4 amps at an angle of 120 degrees with respect to the Phase A Voltage.? The result is ?25.0 amps at an angle of 60 degrees with respect to the Phase A Voltage.? When you factor in the 125% for the heater load, this becomes ?31.2 amps at an angle of 60 degrees with respect to the Phase A Voltage.?The current from the first and third heater would be 14.4 amps each, or 28.8 amps, correct?

If Bill did choose to construct his heater bank in a WYE configuration, then you would be right in saying that line current and phase current would be the same. However, we started with a 3000 watt heater. If you connect it in a WYE configuration, then the voltage it would see is not 208, but rather 120. The difference between the 208 and the 120 is, as you have already pointed out, tat factor of 1.732 (i.e., the square root of 3). The current in each heater would be 25 amps (i.e., 3000/120 = 25). Multiplying this by the 125% factor for the heater load brings you back to 31.2 amps. Bottom line, when you deal with three phase, there will be no escape from that strange ?square root of three? factor.Current per leg is not affected by 1.73 in a wye connection. I line = I phase for wye. Only voltage. E line = E phase x 1.73 ie 208V = 120 x 1.73.

- Location
- Connecticut

Charlie and spsnyder,I was concerned about oversizing the ocpd. I know that the 1.73 had to factor into the load calculation some where, but because the heaters are single phase loads I assumed that the total (i) would some how have to be additive.I did contact the company who makes the heaters(Fostoria) and their engineer told me that the total load would be based on the 14.4per heaterx125%=18ax1.73=31.14.

thanks for the help guys

i

Re: elec heater calc

How do you calculate an electric heater load in single phase?

How do you calculate an electric heater load in single phase?

- Location
- Seattle, WA

- Occupation
- Electrical Engineer

A new thread has been opened to address this question.Originally posted by hde_electric:How do you calculate an electric heater load in single phase?

- Status
- Not open for further replies.