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Heat Losses while boiling a pot of water

  1. Aug 20, 2014 #1
    Hi all,
    I'm a homebrewer and i'm trying to find the boil off rate in gal/hour when using an electric heating element in my kettle.

    If i'm using a 5500Watt heating element at 100%, i'm inputting about 312 BTU/min...Using this site and the table for the losses due to evaporation and radiation of the water surface and the losses from the vessel, I can calc the losses in BTU/min for temperatures LESS than boiling, and use the total BTU's from the element less the losses to determine the BTU's available for heating the water and the associated ΔT(°F)/min...but i'm sure these losses change while boiling and i'm not properly accounting for them and i cant find any examples of this problem online.

    Now what i am looking to understand is the losses when i'm boiling the water/wort so i can determine the net BTU's available for converting water to steam (at 970 BTU/lb, i want to boil off 1 gal per hour, 8.34 lbs/gal, so i would need 8092 BTUs to boil one gallong in one hour)

    Any information on how i can determine these losses while boiling would be greatly appreciated!!

    (PS, i understand the losses are not static and dependent upon how vigorous the boil is as well, I'm guessing the element will need to be fired around 65% to overcome losses and input enough BTU's to boil at that rate/hour)

    Thanks!!
     
  2. jcsd
  3. Aug 20, 2014 #2

    anorlunda

    Staff: Mentor

    Detailed calculations of the type you ask for are very difficult. They also depend greatly on the size/shape/properties of the containing vessel. That's probably why you haven't gotten more answers.

    From an engineering point of view, I would estimate the losses while boiling as the same as losses at 1 degree below boilint temperature. If you want to be conservative, double or triple those losses.

    Even tripled, it should be a small fraction of the heat necessary to boil the fluid at 970 BTUl/LB. If the losses are a large fraction of the total requirement, your equipment design is poor.

    For small jobs, engineers use generous safety margins and estimates partially to avoid the need to do detailed calculations. If they did do extensive calculations, they might need to validate them with experiment to guard against error.

    In other words, I am suggesting that you may not need such an accurate calculation.
     
  4. Aug 20, 2014 #3
    Of course i dont NEED it :) I just like solving problems and getting as close as I can...and i totally understand this is a very compicated problem, i'm just looking for some guidelines to be as close as i can.

    Like I said, i'm brewing beer, people did it for thousands of years before understanding what yeast did. My goal is to be consistent, and if i can dial in a boil off rate that i can use each time, i'm hoping to be repeatable in my process.

    Using the losses i referenced in that link above i feel overestimates the losses in that system, as it seems that boiling would negate some of the radiation and evaporation losses from the surface of the vessel (i'm just guessing that the temp differential btw the surface and ambient that causes losses when at temps less than boiling would not be there, as steam would be insulating the surface area.)

    When I only consider the losses from the sides and bottom of the vessel, and not evaporation/radiation from the surface of the water, i feel as though my model is more representative of reality...

    Thanks for the reply!!

    Edit: I should have mentioned, i'm using a stainless steel keg as the boiling vessel...the losses for a SS vessel are on that link in the first post.
     
  5. Aug 20, 2014 #4
    Maybe put a thermometer in the pot and take temp every minute till it boils. Use that to calculate the heat loss. Or, measure level when it begins to boil, then measure level vs time later; heat input vs boil off rate gives you losses under boiling conditions...
     
  6. Aug 20, 2014 #5

    That is exactly my next step (actually monitor losses while boiling), I just know that I live in a very dry climate that also effects the rate of boil off...my goal was to find a generic formula that might work for other ambient temps and humidity as a basis.

    Thanks all for your input!!
     
  7. Aug 20, 2014 #6

    russ_watters

    User Avatar

    Staff: Mentor

    The boiling loss is the evaporation loss. So you need only the surface radiation and tank wall loss, since you already calculated the boiling loss. You are right that the boiling will reduce the radiation loss somewhat (some condenses into a fog above the tank, reducing radiation).

    So what are the dimensions of the tank/pot? That's what you need to solve the problem, since the table gave values per square foot.
     
  8. Aug 20, 2014 #7
    The keg is 23.375 inches high and 16.125 inches in diameter...thanks for your input, i kind of thought that was the case (that the boiling loss is or was close to the evaporation loss).

    edit: I also should mention that i scale the side surface area of the keg to the amount of water I'm heating inside of it. Meaning that when i have 10 gallons in the 15.5 gallon vessel, i scale the losses from the sides to (10/15.5), but leave the losses from the surface area of the bottom the same.

    I'm not sure if that is the right approach, but it makes sense as the top of the keg is definitely not hot when only partially full.
     
  9. Aug 21, 2014 #8

    russ_watters

    User Avatar

    Staff: Mentor

    Your approach makes sense -- so what answer did you get?
     
  10. Aug 21, 2014 #9
    Generally people shoot for between 1gal and 1.25gal per hour boil off...so my assumptions are:

    10 gallons heating volume
    15.5 gallons heating vessel capacity
    5500 watts from electric element
    3.41 BTU/watt
    23.375 inches keg height
    16.125 inches keg diameter

    This gives me:
    1.418 sqft for surface of water in keg
    4.071 sqft for outside surface area radiating heat (sides and bottom)

    This tells me i am losing about 109.51 BTU/min through the sides and water surface at 210 degrees, and that i need to fire the element approx 35% just to maintain 210F.

    The exact interchange between just under boiling and boiling i'm not sure how to figure, but i'm not interested in having the a very slow boil anyway, as i'm shooting for 1-1.5gal per hour. More or less, i dont want to boil off too much, or worse, boil too little and end up with weak beer!! :)

    After the boil has started, i use only the losses associated with the sides and walls of the vessel (assuming the losses from the surface of the water are negated by the boil), and any "extra" BTU's generated are used to convert water to steam at (970 BTU/lb*8.34lb/gal) = 8,092 BTU to boil 1 gal of water.

    In conclusion:
    -I have to fire the element approx 40% just to maintain 210 degrees.
    -I have to fire the element approx 56% to boil off 1 gal/hr
    -I have to fire the element approx 67% to boil off 1.25 gal/hr
    -I have to fire the element approx 78% to boil off 1.5 gal/hr
    -The max i can boil off would be 2.02 gal/hr @100%

    I would have to say these figures would change for any other ambient temps and probably for any wind blowing steam away from the top of the kettle, but they provide a baseline to use to calculate how much water will be converted and how much to expect as an end result.

    Thanks for your help!!

    Edit: I had an error in my calcs for the losses on the sides of the vessel
     
    Last edited: Aug 21, 2014
  11. Aug 21, 2014 #10

    russ_watters

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    Staff: Mentor

    Did you answer the right question? Why did you include the evaporation loss...?
     
  12. Aug 21, 2014 #11
    I included it at 210 degrees because the water is not boiling but the evaporation loss is still there, this more or less told me the minimum i need to fire the element just to maintain 210F. (this is higher than the percent needed to boil when not including that radiation from the surface, i considered this a kind of "grey" area, since i dont care about boiling slowly, and just moved on to using the element at percentages greater than this minimum.

    While at boiling, i used only the radiation from the sides/bottom of the vessel to determine how much excess energy (net of losses) i was using to boil.

    This is why i said that the "interaction" at the boiling point is hard to figure, because i switch from vessel sides and top losses (for temps 210 and lower), to vessel sides only while boiling (assuming the boil negates the surface water losses).

    I hope that explanation makes sense!! Let me know if you have any other concerns or questions!!

    Thanks again!!
     
  13. Aug 21, 2014 #12

    russ_watters

    User Avatar

    Staff: Mentor

    Apologies, you did go all the way through, I just apparently didn't read closely enough and you organized it differently from how I would. I do have a couple of potential issues though:

    In the beginning you say you need 35% to maintain 210F, but later say you need 40%. I agree with 35%.

    What makes it "grey" as you point out, is that with that you get 4.7 lb/hr of evaporation. So the additional energy for boiling a total of 1 gal/hr is 3531 BTU/hr or 19%, for a total of 54%.
    Or, the other way:
    Losses not including evaporation at 210F is 360*(1.42+4.07)/3413/5.5=10.5%
    Boiling 1gal = 8.34*970/3413/5.5 = 43%
    Total: 54.5%. (close enough to your 56%)

    And yes, they can change for ambient temps and wind, but the faster the boil, the less influence there is from those other factors (percentagewise).
     
  14. Aug 21, 2014 #13
    I'm sorry, i've edited some portions of previous posts but not others. I should have just left it alone and added another post.

    For surface areas i have:
    16.125 inch diameter is a 1.344 ft diameter, .67 ft radius
    23.375 inch height is a 1.95 ft height

    1.42 sqft = ∏*(.67)^2 <<<<Liquid surface area
    6.72 sqft = 2*∏*(.67)*(1.95) <<<<Vessel surface area (sides and bottom, with the sides scaled for percentage of volume in the keg (10g/15.5g)

    At 210F for losses i have:
    85.09 BTU/min from the liquid surface (evaporation and radiation) (1.42 sqft @ 60 BTU/sqft/min)
    40.34 BTU/min from the tank surface (6.72 sqft @ 6 BTU/sqft/min)

    for a total of 125.43 BTU/min. This gives me 40.13% to maintain 210F. (The 109 BTU/min from before was because i was not correctly calcing the losses on the vessel sides, it was an error in my spreadsheet)

    (125.43 BTU/min)/(312.58 BTU/min from element) = 40.13% Did you independently calc the 35%? If you used some of my figures from before, please accept my apology, I had incorrectly divided the height of the vessel to get it in ft.

    I'm realizing this is sort of hard to follow also because i'm using a spreadsheet, and not rounding (whereas it makes some of these calcs slightly off), but i'm sure you get the picture.

    Can you please explain to me this line in more detail? (it looks like you are using my dated figure of 4.07 sqft for the sides, where the actual sides and bottom are 6.72sqft, again, an error in my previous post)

    Then this line:
    Losses not including evaporation at 210F is 360*(1.42+4.07)/3413/5.5=10.5%
    Should be:
    Losses not including evaporation at 210F is 360*(1.42+6.72)/3413/5.5=15.6%

    The second line takes into account the Radiation losses of the water surface, but not the Evaporation.

    My model ignores both the radiation and evaporation losses while boiling, do you think the radiation portion should be included? Its a few percent different, i would get

    40.13% to maintain 210F (this is calcd using all losses)
    59% for 1 gal/hour (calc'd using the radiation from vessel AND radiation from liquid surface area)
    70% for 1.25 gal/hour
    80% for 1.5 gal/hour

    Would you say that i need to include the radiation losses of the liquid surface temp as above?

    Sorry again for the confusion of editing posts, and thanks for the help!!
     
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