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Physics and Beer, the perfect union

  1. Feb 11, 2006 #1
    I brew beer as a hobby and in order to avoid the cold i want to take my boiling inside. Because propane inside my apartment is dangerous I was looking to heat my boil with a modified electric heating element from a hot water heater.

    I need to figure out an equation that will tell me how long it will take me to boil a 19 Liter solution that ranges from zero (water for my mash) to 25% (pre-beer) sugar saturation. It is usually around 22 degrees C in my apartment if that helps for the heat disappation factor. I boil with an open pot.

    For practical purposes i have access to 120V outlets which limits me to a 1500W heating element. I want to find out if 3500W would do the trick and how long it would take to take it from about 70 degrees celcius to a rolling boil, or whether i have to get multiple 1500 W systems and put them all into the pot.

    Many thanks in advance!

    I'm an idiot when it comes to physics. Could you help me?
  2. jcsd
  3. Feb 11, 2006 #2


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    Why reinvent the wheel? Get an appropriate-size stainless steel stock pot and heat the solutions on your kitchen stove. For around $40 you can get a 30 quart stock pot with a stainless strainer basket, so when you're not making beer, you can make a killer batch of gumbo.
    Last edited: Feb 11, 2006
  4. Feb 11, 2006 #3
    haha, great idea but the problem is that i can't fit the pot on my stove. When i did that at my last place i scorched teh stove and counter and my girlfriend had my head. I want to be able to do it all in the living room on a ceramic plate. Thanks for the tip though.
  5. Feb 11, 2006 #4
    The amount of time it takes is going to depend on the amount of Amps your heating element uses. So unless you have a way to know how much electrical energy per unit time you are putting into your solution, you wont be given a good answer.
  6. Feb 11, 2006 #5


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    Let's say you get about 65% of that 1.5kW as usable heat. That's about 1kW. Let's assume further, that no heat is lost radiatively (which is a good assumption), convectively (which is not) or through mass transfer.

    You want to heat 19 L (about 20 kg) of "nearly" water through about 30C. That takes about 2500kJ. It will take you at least 2500 seconds or about 45 minutes.

    If you include say, 25% convective heat loss, that brings the time to about an hour. Can you live with that ?
  7. Feb 11, 2006 #6
    hmm, probably not quick enough for my purposes. What are the assumptions you made to get that answer so I can plug it in for different scenarios? Any simple equations would help. Do I really only get 65% efficiency on something like that? What else is it being translated into if it's not heat? Thanks
  8. Feb 11, 2006 #7
    How long did you have in mind?
  9. Feb 11, 2006 #8
    i guess i was looking more for some sort of explanation as to how those numbers above were derived so i can plug in different factors and see what's acceptable. I'd obviously like the least amount of time possible but it's going to have to be weighed against more practical things like the amount of outlets i have in my kitchen, the cost of each one of these heating elements, ect.

    Here's what i think i have decuded thus far by reading these posts and wikipedia, and please tell me where i'm wrong.

    a joule is one pound (0.4536 kilogram) of water raised by one degree F (9/5 degree C)
    This comes out to 1 kilogram raising temp by .816 Celcius and pushing it one step further by dividing both by .816......
    1.225 joule should raise 1 kg of water by 1 degree C

    Now i'm going to be dealing with sugar water which could weigh up to 5% more then regular water because the gravity often gets up around 1.050 (kg/L) so lets calculate using that.

    19 Liters shoudl weight about 20 kilograms like you said.

    Therefore, 20 kilograms multiplied by 30 degrees celcius multiplied by a factor of 1.225 equals 735 kJ needed to do that work. To me that sounds like 735 seconds at the rate of 1kJ/s (1.0 kW). Where am I going wrong. I want to really understand what I'm doing going forward. Thanks to all again.
  10. Feb 11, 2006 #9
    wait! I just figured out what I did wrong.

    0.4536 kilogram by 5/9 a degree C (NOT 9/5 A DEGREE)
    or rather
    1 kg by .252 degrees C
    or even
    3.968 joule should raise 1 kg of water by 1 degree C

    That should give me 2380 seconds or 39.6 minutes. Am I doing it right now? How do I get a greater efficiency then 65% on my heating element?
  11. Feb 11, 2006 #10
    additionally, how much further should i get to get a violent boil rather then a simmering boil. Is there any way to quantify that? To effectively get ride of certain undesirables in beer (the reason why budweiser smells like cooked corn), one needs to give a vigorous boil.
  12. Feb 12, 2006 #11


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    That's actually 4.17 kJ for 1 kg of water by 1deg C. Did you forget a factor of 1000 somewhere ?


    Either you've made another error involving a factor of 1/1000 or you've corrected the old error, so you're good either way.

    Wait, are you using some kind of an immersion heater (I missed the sentence where you said it's an element from a water heater) ? In that case, you'll get a much higher efficiency (pretty close to 100%), especially if the heater is preheated before immersing it.

    The hard part of the calculation is determining convective heat loss. That depends so much on the shape and type of vessel and the air-flow characteristics that it's virtually impossible to calculate the amount of loss here. One just has to go by experience.

    Alternatively, if you have a scaled down version of the pot you intend to use, then an experiment on your kitchen stove (followed by a tiny calculation) will tell you how much convective heat loss there is (assuming your living room is not a whole lot more/less windy than the kitchen).

    Out of curiosity, what makes your pre-beer brew be at 70 deg C ?

    PS : If the total heat loss is as low as 10% (which I think is unlikely), then the time to boil is not much more than 30 minutes (assuming an immersion heater).
    Last edited: Feb 12, 2006
  13. Feb 12, 2006 #12
    Thanks, that works great. The reason it comes out to be 70 degress C is because that's what tempt it comes out of the mash at. For starches to be converted to sugars the grains must be heated in a sticky mash between 150F and 160F for about an hour. The grain husks then becomes a filtering bed as the hot sticky "wort" is drained out from the bottom. From there it goes straight into the boiling pot.

    Just to confirm, the way that I would figure it out would be the following:

    (Change in tempt [C]) x (weight of substance [kg]) x 4.17 = Joules required

    I then take this number and divide it by the number of watts I have and this gives me seconds. In the case of a 1500 Watt element, this number would be 1.5

    30 x 20 x 4.17 = 2502

    2502/1.5 = 1668 seconds or ~28 minutes

    As this boil is going, the intention is for a lot of steam to be lost through the top (in order to insure that the DMS, or cooked corn, is lost through the vapor). I won't be able to cover the top for that reason, but is there enough heat lost through the sides of the pot, which is a pretty standard large cooking pot, to be worth it? Also, how much energy is required to get a really violent boil? I've had simmering boils on my stove and I've had explosive boils using propane. I want the latter, so is there a way I can calculate for this mathmatically rather then empirically? I have to figure out whether or not I have to build another one.


    I haven't done this much math since I graduated college! Thanks.
  14. Feb 12, 2006 #13


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    That question is impossible to answer without a whole lot more info (pot dimensions, shape, material, airflow conditions)...and with the info, it's just an extremely hard calculation that is typically done using Finite Element Analysis on a really fast computer.

    The greater the power supplied, the more violent will be the boil (assuming the liquid being boiled is the same). Again, being able to calculate buble dynamics in turbulent fluids is the kind of stuff you can get a PhD thesis for.

    I hope all this "heavy weightlifting" is making you feel good ! :smile:

    The best way to determine convective heat loss is either (i) experimentally, or (ii) with additional info.

    For instance, do you know how long it would take to boil off all (or some known fraction of) the 19L of brew ? With that number, you can come up with a reasonable estimate of convective losses.
  15. Feb 12, 2006 #14
    Thanks guys, you've been a lot of help.
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