How do you calculate the heating requirements for a brewery or anything else?
First up there is the heat capacity, or specific heat of water. It takes 4.2 kilojoules of energy to raise the temperature of 1 kg of water by 1 degree C. Power is energy over time. One kilowatt is one kilojoule per second, or 3,600 kJ an hour. So to raise the temperature of 1 kg of water in one second would require one kilowatt of power.
The brewery has a capacity of 15 US barrels, about 1,750 litres. The amount of mash water will be about 1,000 litres. I want to heat the mash water from 10 degrees to 72 degrees over an hour:
(1000 kilograms x 62 degrees x 4.2 kJ) / 1 hour = 260,400 kJ/hour
Dividing 260,400 kJ per hour by 3,600 gives 73 kilowatts of power required to heat the mash water.
To heat 1300 litres of sparge water from 10 degrees to 80 degrees over the course of an hour would require
(1300 x 70 x 4.2) / (1 x 3,600) = 106 kilowatts
How much power is needed to lift the temperature of the mash 10 degrees over 15 mins? The specific heat of malt is about 40% of that of water. Let’s say the mash is 1000 litres of water and 325 kg of malt.
((1000 kgs of water + 325 kgs of malt x 0.4) x 10 x 4.2) / (0.25 hour x 3,600) = (1130 x 10 x 4.2) / 900 = 53 kilowatts
Once the mash and sparge are finished the next step is to boil the wort. To heat the collected wort of say 1900 litres from 60 degrees to 100 degrees in 45 minutes would require
(1900 x 40 x 4.2) / (0.75 x 3,600) = 319,200/2,700 = 118 kilowatts
This is heating from a standing start, of course as the wort is run into the kettle from the mash and sparge heat would be applied so the actual power required would be somewhat less.
Now heating a liquid is one thing, getting it to boil and change state is another. While it only takes 4.2 kilojoules to lift the temperature of 1 kg of water by one degree, it takes 2260 kilojoules to change the state of the 1 kg water from a liquid at 100 degrees to steam at 100 degrees. Explains why a watched pot never boils. This is the latent heat of vaporisation.
So to boil this 1900 litres of wort with 7% evaporation an hour would require
(1900 x 0.07 x 2260) / 1 x 3600 = 84 kilowatts
So you can see that a 100 kilowatt (10 boiler horsepower) boiler would be the barest minimum. It would allow only one heating task at a time. For example, to raise temperature of the mash for a multi-temperature mash would mean cutting back the heating of sparge water. Lack of boiler power could limit the type of mash regimes and would make for a longer brew day.
A 150 kilowatt boiler would be a lot easier but brewing twice in a day would be quite tricky. Ideally you would want to start the second brew as early as possible, rather than waiting until the first brew is completely finished. Once the run-off is finished you would try to empty the mash tun as quickly as possible – remembering the grain will be hot. During this time you would be heating both the first wort number and mash water for brew number two. Next you would be simultaneously boiling the first wort and heating the second sparge water. If you were not doing a step mash, and with some careful planning and juggling then it might be possible to squeeze two brews out of a long day with a 150 kilowatt boiler. So, in the end it really needs to be a 200 kilowatt boiler.