Author Topic: Kw's and quantity  (Read 1038 times)

Offline weazletoe

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Kw's and quantity
« on: March 13, 2017, 04:38:48 PM »
  The prpoerty manager of the condo I live in now will not allow me to add a 50amp 220v circuit. So when I built my control panel I designed it for use with 120v 30amp that will be easily convertable to 220v when I either a) buy a place, or b) just go ahead and add the circuit where I am now.
 I planned on doing it in the kitchen with my 4 gallon kettle. I would mash and boil a high og 3.5 gallons of wort then top off to five gallons.
 This weekend, I set about installing the elements in my kettle and ran into two issues I had not taken into account. One is that my IC will sitting on top of the element, thus a large portion will be above the actual wort. And second, with the smaler diameter kettle, I cannot get a good seal where the element goes into the kettle. I have two kegs I am planning on making keggles out of once I have 220v. Due to the 120 v limitations I'm limited on what I can boil. I should be able to comfortable boil 4 four gallons, if my math is correct. (8.3#x4gal=33.2 x40*=1328kw.
 Do you feel that boiling four gallons in a 15.5 gallon keggle will present any problems or should I be all right? Thanks! 
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Offline a10t2

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Re: Kw's and quantity
« Reply #1 on: March 13, 2017, 06:35:28 PM »
I'm not sure what the math is supposed to represent, but you can boil an arbitrary amount of liquid with relatively little power provided the kettle is well insulated. The only questions are how quickly you need/want to come up to a boil, and the boiloff rate once there. For example, if you have a 1500 W element and 80% heat transfer efficiency (pretty conservative), you can heat 7 gal from room temperature to 212°F in about two hours:

7 gal * 3.8 L/gal * 4.2 kJ/L-°C * 80°C / 1.2 kJ/s = 7448 s

And boil off about half a gallon per hour:

1.2 kJ/s * 3600 s/hr / (2257 kJ/L * 3.8 L/gal) = 0.50 gal/hr

Given that it's a 3600 W circuit, I'd assume you'll have more power than that going into the kettle. It should be totally reasonable for full-volume 5 gal boils.
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Offline weazletoe

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Re: Kw's and quantity
« Reply #2 on: March 13, 2017, 07:04:58 PM »
Sorry, should have laid my math out a little better.

Weight of water (8.30#) times the desired temp rise. In my case, I'm figuring 60*. My hot tap is about 125*, so 40* is plenty for strike water. So, 40* times my weight of the water, in this case 33.2 = 1328kw will heat that amount of water in one hour.
 I do have the option of putting in two 1650kw elements. That would speed it up, and give me the ability to do larger volumes. 

 Also, with a submersed element not give me 100% efficiency? Why would you figure 80%?
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Offline a10t2

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Re: Kw's and quantity
« Reply #3 on: March 13, 2017, 07:40:19 PM »
Weight of water (8.30#) times the desired temp rise. In my case, I'm figuring 60*. My hot tap is about 125*, so 40* is plenty for strike water. So, 40* times my weight of the water, in this case 33.2 = 1328kw will heat that amount of water in one hour.

That would give you BTU (I think - I have very little experience with non-metric units). Google says 1328 BTU/hr is just under 400 W, which sounds more reasonable.

Also, with a submersed element not give me 100% efficiency? Why would you figure 80%?

You will get nearly 100% heat transfer to the liquid, but you need to account for the losses to the environment. FWIW my electric HLT averages about 88% efficiency, but it isn't particularly well insulated.
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Offline mabrungard

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Re: Kw's and quantity
« Reply #4 on: March 13, 2017, 07:46:04 PM »
I have a 5500w element in my system and recently started covering my kettle more completely during the boil to reduce evaporative losses. My element controller was formerly operating at about 45% (~2500w) with the kettle partially open. I had to drop that power setting to about 20% (~1100w) when I significantly reduced the kettle opening. I'd say that your 1300w of power should be fine for producing and maintaining a boil. The big deficiency is the amount of time it will take to bring the kettle to a boil. That's when the excess wattage is very handy (I do use the full 5500w during that time).
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Offline The Beerery

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Kw's and quantity
« Reply #5 on: March 13, 2017, 10:50:32 PM »
I run a 3750watt boil coil in my BK. I run that at 47%. I am considering dropping down to 120v and run that wide open.  This is for 6.5g post boil.  I could pick up vegetal hop flavors from using the 5500 way ULWD elements. Going to a boil coil solved the issue, much lower w/sqin.  Remember a 5500watt element ran with a PID is not really half power at 50%. It's still on 100% just for a lower amount on time. Phase angle is the only way to drop wattage. 


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« Last Edit: March 13, 2017, 10:53:32 PM by The Beerery »
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Offline Stevie

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Re: Kw's and quantity
« Reply #6 on: March 13, 2017, 11:04:17 PM »
I run a 3750watt boil coil in my BK. I run that at 47%. I am considering dropping down to 120v and run that wide open.  This is for 6.5g post boil.  I could pick up vegetal hop flavors from using the 5500 way ULWD elements. Going to a boil coil solved the issue, much lower w/sqin.  Remember a 5500watt element ran with a PID is not really half power at 50%. It's still on 100% just for a lower amount on time. Phase angle is the only way to drop wattage. 


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That's why it's important to adjust the amount of power as well. Some PIDs allow this, but a dial works as well.

Offline natebrews

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Re: Kw's and quantity
« Reply #7 on: March 14, 2017, 12:31:00 AM »
I also have a 3750W boil coil in my 8.5g kettle.  I find that I run it at about 70% with the top off and 55-60% with the top largely covered to get a gentle boil.  I think if I had an insulated kettle (wrap it with something heat resistant) then I could easily get away with a 120W element running close to full out.  It would take a bit longer to get up to temp for my strike water, but not a big amount on top of the whole brew day.

To his point of the element being on full blast or off, I think the only way to mitigate that (without using a phase angle controlled dimmer type thing) is to change the pulsing rate.  I set mine to 10Hz, which I think is pretty fast compared to most.  That way the temperature ripple at the coil is lower.
« Last Edit: March 14, 2017, 12:42:50 AM by natebrews »
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Offline mabrungard

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Re: Kw's and quantity
« Reply #8 on: March 14, 2017, 01:14:24 AM »
That way the temperature ripple at the coil is lower.

Um, you guys are overthinking this...or underthinking this. Don't forget that there is a significant thermal mass between the actual resistive element and the interface with the wort. The on-pulses aren't sufficient to significantly alter the interface temperature beyond the average value.
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Offline natebrews

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Re: Kw's and quantity
« Reply #9 on: March 14, 2017, 01:18:50 AM »
Martin, do you have any numbers for what the ripple is that you see at the element?  After I wrote that message I started thinking about how I could measure it...because who doesn't love telemetry?
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Offline satchman

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Re: Kw's and quantity
« Reply #10 on: March 14, 2017, 01:34:29 AM »
Back to the OPs question, it seems like the there is plenty of energy available to make a full 5 gallon batch. I use a keggle to make 5 gallon batches, and it works quite well. The only down side is that the valve is not well located, so I use a racking cane to get the last couple of gallons out.

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Offline weazletoe

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Re: Kw's and quantity
« Reply #11 on: March 14, 2017, 02:40:48 AM »
Yeah,  the more math I do on this I think I'll be fine for five gallon batches. Maybe even a hair larger. When I used to do five gallon batches, I actually did 5.5 to have a little extra for sampling along the way. Seems like 1650kw should boil just fine. It will just take a little longer to get there. But avoiding to the numbers I've run, I should be able to get five gallons from mash out temp to boil in about 27 minutes. And I'm okay with that. With that being said, I got stayed on keggle #1 tonight.  8)
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