gdub0000
Member
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- Concord, Ca., USA
10.1 kw 3 ph 208 what is the draw on a 1 ph 240 volt line.
single ph to 3 ph amps in unequal voltage
- Thread starter gdub0000
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- Placerville, CA, USA
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- Retired PV System Designer
Well the stupidly simple answer is 10,100 divided by 240.
Now to get real, do you mean resistive heating elements wired line to line on a 208Y/120 supply? Then maybe the answer is eventually zero since they will burn out.
What voltage are the elements rated for?
If they can handle 240, then 10,100/(3*208) times 240*3.
And you will get 10.1kW* (240/208)^2 for the power.
If the 10.1 was really the rating of the elements when run off 240, then everything will be different.
Tapatalk!
Now to get real, do you mean resistive heating elements wired line to line on a 208Y/120 supply? Then maybe the answer is eventually zero since they will burn out.
What voltage are the elements rated for?
If they can handle 240, then 10,100/(3*208) times 240*3.
And you will get 10.1kW* (240/208)^2 for the power.
If the 10.1 was really the rating of the elements when run off 240, then everything will be different.
Tapatalk!
Last edited:
kwired
Electron manager
- Location
- NE Nebraska
Don't you have basically the same topic in another thread?
Like GD mentions the simple conversion is assuming you still have same kW in the end, but that means voltage transformation and phase conversion to achieve that. In the other thread it was mentioned you may (or may not) be able to reconfigure equipment to connect directly to the other phase and/or voltage - but you never gave us any other details to help with that possibility. It was also mentioned that voltage transformation and phase conversion may possibly not be cost effective over getting a different oven rated for the voltage you do have.
JoeStillman
Senior Member
- Location
- West Chester, PA
If the 10.1kW 3? heater is delta connected, and you connect 240V single phase to two of the terminals, the power is 6.7kW.
10.1 kW 3? = 3 @ 33.67kW/?. R/? = 208?/3,367 = 12.8ohms
The total 1? resistance is one phase leg in parallel with two phase legs in series.
R1? = 1/((1/12.8)+(1/25.6)) = 8.55ohms
P1? = 240?/8.55 = 6,738 W or 6.73 kW
10.1 kW 3? = 3 @ 33.67kW/?. R/? = 208?/3,367 = 12.8ohms
The total 1? resistance is one phase leg in parallel with two phase legs in series.
R1? = 1/((1/12.8)+(1/25.6)) = 8.55ohms
P1? = 240?/8.55 = 6,738 W or 6.73 kW
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kwired
Electron manager
- Location
- NE Nebraska
And when connected to 240 volt single phase the element pictured on the bottom is operating at 240 volts but is possibly only designed for 208 volts which may shorten its life. The other two however only have a drop of 120 volts across them and will have extended life, but at a reduction in output from what they were designed for.
JoeStillman
Senior Member
- Location
- West Chester, PA
Interestingly, one phase leg at 12.8ohms is exactly 4500W at 240V. This tells me they are probably 240V elements derated for 208V. I've seen unit heaters with ratings like this.
If the oven was connected to 240V 3?, it would draw 13.5 kW which is (4500x3) or (10.1 x 2402/2082).
If the oven was connected to 240V 3?, it would draw 13.5 kW which is (4500x3) or (10.1 x 2402/2082).
- Location
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Yes. And because of that fact, I am closing this thread. The original discussion is here:
http://forums.mikeholt.com/showthread.php?t=160717
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