Hi, I have been using my electric 5500w herms rig at home for a year now. It has been working flawlessly. My boss said I could brew at work but they have three phase power there. I guess when you split the power the outlet only gets 208 volts. I know my elements will have less power, but will it slow the pumps or have any other issues? Will the contactors with a 120v coil still latch with 104 volts? I kinda want to know if it will still function the same or will there be problems? Thanks
Ok to run my 240v brew rig on 208v?
- Thread starter JRems
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With 3 phase power delivered in a Y configuration (208V) the power for 120 volt devices is 120V. Here is an illlustration.
I don't think it will be a problem. It may draw slightly higher current, but it will still be less than 30 amps. Wish I could brew at work, that sounds like you'll be sharing a lot of brew.
JRems
Well-Known Member
With 3 phase power delivered in a Y configuration (208V) the power for 120 volt devices is 120V. Here is an illlustration.
Thanks p-j. I built my panel with your diagram. 3 phase still confuses me. Your diagram looks like a flux capaitor. 120+120=240. I still don't understand where the 208 comes from. As long as it work I don't care.
240V*sin(360/3)
Stauffbier
Well-Known Member
I don't use 3 phase very often, but I do work with 240V all the time at work. We have a 30 amp, 240V floor sanding machine. I can tell you that every house I work in is different. Voltage ranges from as low as 205V up to 245V and everything in between. My machinery always runs the same in any of those scenarios. I can only assume that your brewing equipment would react the same way. Just my 2 cents....
No, the current draw will be less. The heating element is a purely resistive device. With a resistive circuit, the current draw will be directly proportional to the voltage, and the power consumption will be directly proportional to the square of the voltage. A 240V, 5500W element will output 5500 * (208^2 / 240^2) = 4131 watts with 208 volts input.
Yep, your right, when working with resistive loads compare it to a light bulb, lowering the voltage will result in a dimmer bulb, which means lower efficiency for the heater at 208. Too many years working with motors.
Thinking about the pumps though, I thought current draw for an inductive load goes up with decreased voltage? The pumps may run hotter but I wouldn't think a small decrease in voltage would be a big issue.
I'll wait to be corrected now
Thinking about the pumps though, I thought current draw for an inductive load goes up with decreased voltage? The pumps may run hotter but I wouldn't think a small decrease in voltage would be a big issue.
I'll wait to be corrected now
The pumps won't see a decreased voltage. They will still be on 120V. From any leg of the three phase service to neutral is 120V.
What about if they were 240V pumps? (just for those who do have 240V pumps)
Your pump will draw more current and depending on what type of current protection it has, it may trip out. If it has a thermal switch inside the motor that could also trip as the motor will run hotter. A test run would be necessary to monitor current and temperature.
Disagreeing with fbold1:
Heating efficiency is the heat energy output divided by the electric energy input. The efficiency for electric resistance heating is always 100%, whether it is a heater, or a light bulb used as a heater, regardless of the voltage.
The capacity of the heater is lower when voltage is lower. But the efficiency remains at 100%.
The efficiency for incandescent lighting, you are correct, it is way below 100% and is worse when it is less hot as occurs with lower voltage.
Agreeing and adding to what danb and P-J said:
The reason you can not add the three phases (120+120=240) and you actually get only 208 volt, is because the phases are not peaking at the same time. When one leg is peaking, the other leg is at a much lower volt. Then a fraction of a second later the 2nd leg peaks but the first leg has gone down in volts.
In three phase, none of the legs are in phase with each other. They are spread equally apart by 1/3 cycle.
Thanks for clearing that up, I feel so much smarter now!
Light bulbs give off radiant heat as part of their energy conversion. Light is not heat until it gets absorbed by material.
So, light bulbs are definitely not 100% heat efficient. This might be a semantic argument, so I apologize in advance for that.
hahahaha a few hours ago I thought of posting something simliar... but thought better of it
But if you put that light bulb in a sealed coffee can would it not be 100% effcient
hahahaha a few hours ago I thought of posting something simliar... but thought better of it
But if you put that light bulb in a sealed coffee can would it not be 100% effcient
Yep. Everything's relative, isn't it?
raleighwood
Member
I wouldn't worry unless it was pushing higher than 240v
But is the light bulb really making light if no one can see it... ... ... I need to go drink some homebrews so I can blame my ramblings on that
Thinking about the pumps though, I thought current draw for an inductive load goes up with decreased voltage? The pumps may run hotter but I wouldn't think a small decrease in voltage would be a big issue.
I'll wait to be corrected now
this is the difference between an inductive and a resistive load.
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