The Maths Behind Heating Water

[Steve1262]

Hi all.

Can anyone tell me if they know the formulae for calculating the size of a heating element to use in a HLT and Kettle.
I'm sure there will be such a thing and I'd like to take some of the guesswork out of my build, I'm currently fitting a 3kw element to a 50 litre pot but I'm not sure it will be big enough to maintain a rolling boil and need to consider if a second element is needed.

I'm thinking of: element size to raise 50 litres of water to 100c in say 30 mins. . . or something of that nature.

Regards, Steve (G6JEF)

[Steve1262]

Having just asked the question I dropped just about straight away on this:

http://processheatingservices.com/water ... alculator/

Might be of interest to others with the same question.

[boingy]

I can't help you with the formula but I can tell you that the formula would be useless to you because one of the variables is very difficult to measure. I'm talking about heat loss. It depends on your pot construction, brewing location (outdoors?), ambient temperature, whether you cover the pot etc.

I do know that 3kW is more than adequate for a rolling boil at your brew length as long as you are somewhere sheltered and perhaps with a bit of insulation on the pot. Many folk use two elements to get to the boil quickly and then just one to maintain it.

Incidentally, we have a calculator on this very website: http://www.jimsbeerkit.co.uk/calc.html but it also does not take losses into account.

[Eric]

Yes, losses are likely to be quite significant, they certainly are in my setup.

1 kilowatt hour is equivalent to about 860 kilocalories and 1 kilocalorie will raise 1 litre of water 1 degree centigrade.
We might assume when brewing there are 2 stages of heating liquor, each by about 50C, from cold for mash and sparge and a similar order of rise in the copper to boil runnings. Based on that, heating 50 litres by 50 degrees without losses will take 2500 kcals or 2.9 kwh, meaning about an hour with a single 3 kw element.
Heat losses are greater at higher temperature differential, and while, as said, 3kw will be fine to keep the boil going, in a strong cold wind with no shelter or insulation it might not get there.

[Andy]

Hi all.

Can anyone tell me if they know the formulae for calculating the size of a heating element to use in a HLT and Kettle.
I'm sure there will be such a thing and I'd like to take some of the guesswork out of my build, I'm currently fitting a 3kw element to a 50 litre pot but I'm not sure it will be big enough to maintain a rolling boil and need to consider if a second element is needed.

I'm thinking of: element size to raise 50 litres of water to 100c in say 30 mins. . . or something of that nature.

Regards, Steve (G6JEF)

A mate of mine has a 3kw element in a 50L pot. He has insulated the pot with camping mat foam and foil bubble wrap. With the element on full chat the boil is not very vigorous, it serves him well but he could do with a little more ooomph. His electricity supply restriction is holding him back from fitting a second element but he would if he didn't have that power restriction.

[lord groan]

If it would be relatively easy to put 2 elements in the pot then I'd definitely do it. As Andy says you may need to sort out the electricity supply if you want both on at the same time, 3kw being the upper limit for a 13A socket and you cannot safely plug 2 into a double socket. My temporary solution was 1 plugged into the downstairs ring main, the other plugged into a separately fused 16a supply.
Even if you can only use one to start with, the 2nd one is a backup in case of problems with the first, if you're in mid boil and the element cuts out you wouldn't believe how grateful you are to have another fitted that you can just switch on - this has happened to me.

I've now sorted the supply to my brewshed and my boiler now has a 3kw element which is either off or on, and a 2nd 2.8kw element which is powered via a simple variable power controller so I can control how vigorous the boil gets. Both are full on until a boil is achieved,then 1 gets reduced until I'm happy with a steady boil.

[themadhippy]

Can anyone tell me if they know the formulae for calculating the size of a heating element to use in a HLT and Kettle

E = m × c × θ and P = E/ t should do it but easier to use http://www.jimsbeerkit.co.uk/calc.html