What power % do you use?

Wondering what others are using for their boil. I use a Camco 5500W element on a 5 gallon batch (about 7 gallons preboil) at 55% power. After one hour, I get 22% evaporation rate which seems high to me, but I don't think my boil is too crazy.

What do others use?

 

I boiled 13.5 gallons with the same element this weekend. I use 90% for the first half hour, then cut it down to 70-80%. I have to use Fermcap-S or I get boilovers.

This is a boil in an uninsulated keggle.

 

The shape of the kettle also effect boil off. Larger, diameter, larger surface area has more boil off. I can maintain a boil and adjust boil off rate between 1-1.5gal per hour using the element power.

 

Shape of kettle should only affect boil off using gas burners since what changes is energy input.

 

Increased surface area affects increases rate of boil-off regardless of the heating process. That is, for the same volume of water, and the same net energy input, the vessel with the larger surface area will experience more evaporation.

 

Evaporation is different than vaporization due the heat from the element. So, while you are technically right, I don't think it's germane here.

Almost all of the vapors leaving the boil pot during the boil were vaporized either at the bottom of the pot (gas flame) or in the middle of the pot (element). They would have left the pot regardless of the surface area.

 

Thanks. In thinking about it further, I agree regarding surface area not being germane to vaporization in an electric heating setting. Evaporation does come into play though, in pre-boil, during boil, and post-boil, and those losses are measurable.

In the gas flame setting, the larger surface area of the bottom in a cylindrical vessel generally increases the amount of energy that can be transferred to the water, so the larger surface area of the bottom leads to higher vaporization rates for the same amount of gross energy applied, and the larger surface area of the top leads to higher evaporation rates.

I'm a long way out of physics class, but it's good to exercise that part of the brain every once in a while.

 

Not that it particularly matters to the overall discussion (I use 60-65% for 7gal boils in an unisulated keggle, btw), but I would say the shape does play into things. A tall, skinny vessle will have more liquid above the element. The rising steam bubbles that originate on / near the element will have more time to cool and be reabsorbed into the liquid as they make their way to the top. Given a theoretically infinitely tall keggle, you would eventually see no bubbles reaching the surface and no vaporization due to the heating from the element.

That being said - I'm not sure that the relatively small differences in the range of pots / kettles we use would have much observable effect. What might cause a larger effect to percieved boil off rate is the size of the opening of your keggle. If you have a relatively small opening at the top, there is a large surface area for the vapors to condense on and roll back into the keggle, thus reducing your overall boil-off rate. If you were to cut the keggle top all the way at the shoulder, there'd be significantly less area for the steam to condense on, and you would find yourself with higher recorded boil-off rates.

-Kevin

 

I'm using a 5500w element and once boiling can keep a boil with as little as 55-60% with 14 gallons. I usually run it around 70% to keep a very vigorous boil. I'll usually boil off 15-17%

 

Boil-off should be at constant rate proportional to energy input and independent of volumes, I'm not sure who decided to start using percentages but it is inappropriate.

 

I dial my PID back to 83% for my 12gal batches, after I hit a boil. Using a 5500W camco. I'm brewing in my garage, so I expect there are a number of weather related factors that could be influencing my number.

 

I typically start off with 13.9 or 14.9 gallons and boil off about 1.9 gallons/hour, running the 5500W element at 85%.



It's a pretty vigorous boil that results in a great hot break and clear wort:









Kal

 

I'm in a similar boat to Kal, but I opted not to throttle my camco 5500w at all. I'm usually at 14g boil or so for 10-11g batches and near 19g for half bbl ones. I boil off right at 2g in either case. Boil is vigorous, and I have to watch for boil overs with first wort hopping or bigger batches, but I switch contactor til I hit boil.

 

i've only done two batches, but i believe i had mine at about 60% and it was a good boil. doing 10G batches in 15G pots.

using 5500W camco's.

 

i run 2 x 2000w 120v elements.

100% on either the 6.5 preboil or 11.5 preboil... obviously one is more vigerous then the other, but both get about a gallon of boil off in an insulated keggle

 

In the BK ive got 1 5.5kw element, to keep a nice rolling boil with 10gal batch in a keggle I normally have the PID set for 60% duty cycle
for 5gal its around 40%

this maintains a nice rolling boil

 

Atmospheric pressure and humidity seem to come into play for exact volume loss, but typically 55-66% duty cycle on my rig (5500watt kettle element) will result in my desired 1.5 gallons of boil off.

As someone pointed out earlier, this boil off should be fairly linear based upon applied wattage. I get the same 1.5 gallon boil off whether it's 13.0 down to 11.5 or 7.5 down to 6.0. Some other thread once listed a formula that ignored a few key variables (like delta temp for instance) and the math confirmed that which I physically proved: turning my throttle to the 60%ish area gets me what I'm looking for...

Okay, I couldn't leave well enough alone... And I also couldn't find the thread I was referring to... So without further ado, here is math:
(Math and physics trolls, be on flame stand-by)

965 BTUs are required to convert 1 lb of water (at 212F) to 1lb of steam (at 212F)
Water at 212F weighs approximately 7.99 lbs per gallon

My target boil off is 1.5 gallons per hour

965 BTU x 1.5 gal x 7.99 lbs = 11565.53 BTUs required (ignoring ambient temperature, altitude above sea level, etc, etc)

1 BTU/hr = 0.29307107 Watts; so (assuming a 1 hour boil):

11565.53 x .29 = 3354 watts

Assuming a perfect world where your 5500 watt element puts out exactly 5500 watts when set to 100% duty cycle:

3354 / 5500 = .61

In other words: In a perfect world setting the duty cycle of your E-kettle to 61% would result in 1.5 gallons of boil off per hour

Of course, there are other factors to consider, such as actual voltage potential available at your wall outlet, relative humidity, so on and so forth. That is why I advise using the above math as a guideline to find your initial boil setting but then make adjustments as required while monitoring your volume and gravity compared to time.

Cheers

 

With the same heat input, a taller narrower pot would appear to have a more vigorous boil than a shorter wider pot, with the same boil off rate.

 

It maybe appear that way, but the same amount of vapors leave both pots.

 

Quote:
Originally Posted by DaleHair View Post
With the same heat input, a taller narrower pot would appear to have a more vigorous boil than a shorter wider pot, with the same boil off rate.


It maybe appear that way, but the same amount of vapors leave both pots.

 

5500W camco, 6-7 gallon boils, 60%.
Haven't dialed in my boil-off yet.

 

5500W HD camco in keggle. 80-85% for 10 gal batch, 65-70% for 5 gal batch, both at 1.5 gallon boil off per hour. In South Florida so at sea level and high humidity.

 

My apologies - you're right. I misread what you wrote.