Running a 240V heater on 208V power

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deb4523

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Alaska, USA
So... two heaters were purchased (without anyone checking with our team) that require 36A each on a three phase 240V system. We have three phase 208V power available. :roll:

Is it going to produce less heat or draw more current?

Do I need to install buck/boost transformers to boost the voltage or will it run as is?

Will running on a lower voltage shorten the lifespan or potentially damage elements within the heater?

Thanks and Happy New Year!
 
deb4523 said:
So... two heaters were purchased (without anyone checking with our team) that require 36A each on a three phase 240V system. We have three phase 208V power available. :roll:

Is it going to produce less heat or draw more current?

Do I need to install buck/boost transformers to boost the voltage or will it run as is?

Will running on a lower voltage shorten the lifespan or potentially damage elements within the heater?

Thanks and Happy New Year!
Click to expand...

Assuming there are not some form of electronics in the circuit, the heaters will draw less current and produce less heat. The element life, if anything has the potential of being longer, not shorter. You need to know how to do the math so that you can size circuits properly in the future. This same phenomenon happens with water heater. If it were a single phase heater here is the math. First how much work does the element do? P=IxE 36x240=8640 watts Second find the element resistance R=E/I 240/36=6.66666 ohms. third find the 208 volt amps I=E/R 208/6.6 equals 31.5 amps Power is P=IxE 31.5x208=6555 watts 8640 watts divided in to 6555 watts equal %76 of the heat output. See if you can extend that to three phase.
 
Strathead said:
the heaters will draw less current and produce less heat.
Click to expand...


Yes and the only time I have had this become a problem is if a certain recovery rate is needed/required to furnish a certain number of gallons of hot water pr hr.

I have never had an electrical inspector even mention it but have been caught by the health inspector ( not enough hot water furnished as required by heath department ).

If this is a restaurant or kitchen ( or anything else the health dept would be interested in) then check to see how much hot water is required.
 
It is easiest for me to understand when I remember that the heater wire has a constant resistance. Since the resistance is a fixed value, all of the other units will vary.
 
Electronic controls will almost certainly work fine, as they'll almost certainly have a wide-input-voltage-range switching power supply.
If the controller uses 240-volt relays, they might inconsistently pull in on 208 volts. If this is the case, they'll probably work just fine when new, then become problematic later as wear & tear & crud accumulate.
 
drcampbell said:
Electronic controls will almost certainly work fine, as they'll almost certainly have a wide-input-voltage-range switching power supply.
If the controller uses 240-volt relays, they might inconsistently pull in on 208 volts. If this is the case, they'll probably work just fine when new, then become problematic later as wear & tear & crud accumulate.
Click to expand...

For the record, I knew this, but my math would account for electronics. So the caveat.
 
Strathead said:
Assuming there are not some form of electronics in the circuit, the heaters will draw less current and produce less heat. The element life, if anything has the potential of being longer, not shorter. You need to know how to do the math so that you can size circuits properly in the future. This same phenomenon happens with water heater. If it were a single phase heater here is the math. First how much work does the element do? P=IxE 36x240=8640 watts Second find the element resistance R=E/I 240/36=6.66666 ohms. third find the 208 volt amps I=E/R 208/6.6 equals 31.5 amps Power is P=IxE 31.5x208=6555 watts 8640 watts divided in to 6555 watts equal %76 of the heat output. See if you can extend that to three phase.
Click to expand...


Power is about 15kW
[P=E*I*sqrt(3) = 240*36*sqrt(3) = 14964.92]
(plus the spec sheet confirms it)​


What I'm having a problem with is the resistance. I end up getting
Element resistance is about 3.85 ohms
(R = E^2/P = 240^2/15000 = 3.849)
Or R = E/I = 240/36A = 6.667 ohms
or using the three phase ohms law calculators I'm getting 7.68 ohms.

Current at 208V is 31.2A
Or I=E/R = 208V/6.667 ohms = 31.2A

Power =31.2*208*sqrt(3) = 11.2KW

Does this look right?
​
 
deb4523 said:
Power is about 15kW
[P=E*I*sqrt(3) = 240*36*sqrt(3) = 14964.92]
(plus the spec sheet confirms it)​


What I'm having a problem with is the resistance. I end up getting
Element resistance is about 3.85 ohms
(R = E^2/P = 240^2/15000 = 3.849)
Or R = E/I = 240/36A = 6.667 ohms
or using the three phase ohms law calculators I'm getting 7.68 ohms.

Current at 208V is 31.2A
Or I=E/R = 208V/6.667 ohms = 31.2A

Power =31.2*208*sqrt(3) = 11.2KW

Does this look right?
​
Click to expand...

Yes. Easiest way is to multiply the KW by the ratio of voltages squared.
(208/240)^2 x 15 = 11.2.
 
deb4523 said:
You're right. That is easier... Thanks!
Click to expand...

Just something to think about. My brain and I assume others, only have so much room. retirede's formula is definitely easier. But is it is derived from the combining and breaking down the formulas you and I use. If your job requires you to do these calculations on a regular basis, the shorter versions are well worth memorizing or remembering. If it is just on occasion, I just remember the base formulas and do it the long way. In my opinion you HAVE to understand the long way either way. So, for example, I teach and remember E=I*R and P=I*E I don't automatically remember P=E2/I or even I=E/R I do that with algebra first and then sole. Of course that is if I don't have access to the little yellow wheel.
 
Strathead said:
. . . I teach and remember E=I*R and P=I*E I don't automatically remember P=E2/I or even I=E/R I do that with algebra first and then sole. Of course that is if I don't have access to the little yellow wheel.
Click to expand...
Just write down the two formulae and cover the unknown value with your fingertip. To wit:

ohm-law-chart.png
 
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