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Old 11-16-2009, 11:28 PM   #1
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Default Is 41.666 amps too much on a 50 amp breaker? 80% rule?

I am going to be running two 240v Camco Ripp elements off of my 50 amp breaker. The rest of my equipment will be running off a seperate plug for all the 120v stuff. One of the Ripp elements is 5500 Watts and the other is 4500 Watts. Together they will draw 5500+4500/240 = 41.6666 Amps. I know you are suppose to follow the 80% rule but do you think 83% is really that big a deal? One is for my BK the other is for my HLT/Herms Kettle. There is a good chance that for the most part they will not be running at the same time but I started thinking that if I needed more hot water I could use my BK to heat up the initial mash in water, that way I wouldn't have to worry about running out of water for big bill brews. I'm really trying to be able to brew some high gravity 15 gallon batches out of my keggles. Does anyone else pull this off or do i need to deal with 10-12 gallon batches and live with it?

As a side note when I checked the voltage off the 50 amp plug I was actually getting 250v. That actually brings me down to 40 amps on the nose but is that the way is works? how is the heater set up? does it max out on amps or watts? Like since im running 250v am i really putting out 5500/240 = 22.92 -> 22.92 * 250v = 5729 Watts? Or am I drawing less amps?

Theres a lot of questions in there. Any help is much appreciated.

Thanks,
Craig

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Old 11-16-2009, 11:45 PM   #2
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That should work as long as you DO NOT turn both elements on at the same time.

the 80% rule is to help with the high draw from the initial powering up.

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Old 11-16-2009, 11:56 PM   #3
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That should work as long as you DO NOT turn both elements on at the same time.

the 80% rule is to help with the high draw from the initial powering up.
Ahhh. That's interesting. I guess I'll have to work some sort of delay into my programming. Good to know! Thanks Cheeto!

Anyone have any thoughts on the 15 gallon batch size (or any other comments on the 240 vs. 250 power consumption question)?
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Old 11-17-2009, 12:05 AM   #4
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Measure it if you have an ammeter. Most likely the 4500 watt element is only really about 4200 and the 5500 watt about 5200. They may pull slightly more current when cold but we're talking fractions of an amp difference here.

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Old 11-17-2009, 12:27 AM   #5
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Originally Posted by cruelkix View Post
As a side note when I checked the voltage off the 50 amp plug I was actually getting 250v. That actually brings me down to 40 amps on the nose but is that the way is works? how is the heater set up? does it max out on amps or watts? Like since im running 250v am i really putting out 5500/240 = 22.92 -> 22.92 * 250v = 5729 Watts? Or am I drawing less amps?

Thanks,
Craig

5500/240 = 22.92 Amps (and 10.5 ohms, that never changes)

but if you have 250 volts, you current increases the same
22.92*250/240= 23.87 amps

22.87*250 = 5968 Watts. The current increases also.

I still think you're OK that voltage will drop some when you load it.
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Old 11-17-2009, 12:31 AM   #6
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Measure it if you have an ammeter. Most likely the 4500 watt element is only really about 4200 and the 5500 watt about 5200. They may pull slightly more current when cold but we're talking fractions of an amp difference here.
As Ohio stated use a Amprobe on it besides a voltage meter while under load, I bet there's some voltage drop also as your reading under no load conditions. Sounds like your power company has decided to run high on the pig vs the next lower tap. Size and length of the feeders to your 50 amp twistloc receptacle plus the cord in gauge and length all add up to your VD under a load. (Not that other kind of VD). Remember to add your pump as well controller and control panel lighting load with your circulation pump if your using a HERMS system to the elements load.

If you like electric heating with dual elements in eack keggle heat each keggle to near target temp in each keggle then switch over to one element in each keggle to reach your final target temp as well maintain these set temps in both keggles at the same time. You'll still be within your maximum current limits as well maintaining two keggles temps. Same with the boil keggle, start with two elements until near the boil then switch over to one element to reach and maintain your amount of boil. This is my plan with using the BCS 460 controller to control the elements in my brewing system.
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Old 11-17-2009, 12:41 AM   #7
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It will work just fine although you really should have a 60A breaker.

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Old 11-17-2009, 12:45 AM   #8
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Originally Posted by cruelkix View Post

As a side note when I checked the voltage off the 50 amp plug I was actually getting 250v. That actually brings me down to 40 amps on the nose but is that the way is works? how is the heater set up? does it max out on amps or watts? Like since im running 250v am i really putting out 5500/240 = 22.92 -> 22.92 * 250v = 5729 Watts? Or am I drawing less amps?

Theres a lot of questions in there. Any help is much appreciated.

Thanks,
Craig
Actually elements are not "smart" and do not regulate current or power. They have a fixed resistance that can varry a bit with temperature. There are a series of formulas invoving current, voltage and resistance to find power and current usage. Here are the some relevant ones. They achieve the same goal, they just use different electrical properties to get there.
Where P = power in watts
Where V = voltage in volts
Where I = Current in amps
Where R = resistance of the load

P = I(squared) x R
P = V(squared) / R
P = E x I
I = E / R

So as your voltage goes up and your elements resistance remains fixed, your current goes up as well. With voltage higher, and current higher, power draw will be higher.

Youe 5500 W element probably has a resistance between 10 and 11 ohms. Should be right around 10.5 ohms depending on if it rated for 220 or 240VAC. (10.5 ohms assumes 240VAC) Your 4500W element probably has a resistance around 13 ohms.

Rather than doing all the math, once you understand that a rise in voltage across a fixed resistance creates a direct rise in current, the you could just as easily divde 250 by 240 and come up with 1.04. Basically you can expect 104% of the current that you would have at 240 volts to occur at 250 volts. So if you expect a draw of 41.7 amps at 240, you should expect a draw of approximately 43.4 amps at 250. If the elements are rated to deliver their wattage at 220 or 230 V, your increase in current at 250 would be higher.
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Last edited by Zen_Brew; 11-17-2009 at 12:48 AM.
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Old 11-17-2009, 03:38 AM   #9
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Quote:
Originally Posted by Zen_Brew View Post
Actually elements are not "smart" and do not regulate current or power. They have a fixed resistance that can varry a bit with temperature. There are a series of formulas invoving current, voltage and resistance to find power and current usage. Here are the some relevant ones. They achieve the same goal, they just use different electrical properties to get there.
Where P = power in watts
Where V = voltage in volts
Where I = Current in amps
Where R = resistance of the load

P = I(squared) x R
P = V(squared) / R
P = E x I
I = E / R

So as your voltage goes up and your elements resistance remains fixed, your current goes up as well. With voltage higher, and current higher, power draw will be higher.

Youe 5500 W element probably has a resistance between 10 and 11 ohms. Should be right around 10.5 ohms depending on if it rated for 220 or 240VAC. (10.5 ohms assumes 240VAC) Your 4500W element probably has a resistance around 13 ohms.

Rather than doing all the math, once you understand that a rise in voltage across a fixed resistance creates a direct rise in current, the you could just as easily divde 250 by 240 and come up with 1.04. Basically you can expect 104% of the current that you would have at 240 volts to occur at 250 volts. So if you expect a draw of 41.7 amps at 240, you should expect a draw of approximately 43.4 amps at 250. If the elements are rated to deliver their wattage at 220 or 230 V, your increase in current at 250 would be higher.
Very clearly explained. Thank you. Thanks to the previous posts as well. It makes a lot of sense. I think I may just down size the HLT to a 3500 Watt element to be on the safe side.

@BrewB everything besides the elements will be run off a different breaker box. My 240 box will run the elements and then the standard house box will run the pumps and motors.

All very good info and thanks for the quick replies.

Thanks,
Craig
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Old 11-17-2009, 02:54 PM   #10
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Craig
To your other question: Really 12 gallon batches is all you can plan to get out of a 15.5g keg. By cutting off the lid you've lost maybe 1/3 gallon, and to do a 60 minute boil you're going to have boiled off a couple gallons at least. Even there you will have to address boilovers. There are products like to prevent boilover (fermcap?). You can do "no sparge" to make a big beer with less water to boil off, but that takes a bigger grain bill per gallon.
I suppose you could use your HLT to take on some of the boil volume to make bigger batches but then your MLT size may become your limitation. See Bobby's chart to give you an idea:
http://www.homebrewtalk.com/f11/how-...-needs-123585/

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