We are building a new kettle and want to heat it with steam. Our boiler will provide plenty of steam (1200lb boiler) and the batch size 25HL.
We are trying to find the surface area required to bring the kettle to the boil in 30min.
The kettle will have an agitator to move the wort in the kettle.
Can you point me in the right direction to find this info please?
I had to make a few assumptions to answer your question, and pardon me for using imperial units. Your boiler at 1200 lbs of steam per hour looks adequate, I calculate a need for 510 lbs/hour of dry steam to pre-heat to boiling and approximately 375 lbs of steam per hour to maintain the kettle boil.
25 hL cold wort = 21.3 US bbls and assuming you are evaporating 5%/hr with a one hour boil time your total wort collection should be 22.4 bbls
Assuming a 12P wort the SG is 1.048
Heat load= 5/100 x 22.4 bbls x 258.5 lbs water/bbl x 1.048 x 970.6 BTU/Hr/ lb water = 294,500 BTU/hour
Surface area of heating required assuming 45 psig dry steam (292F from steam table):
Heat Load= UA(T1-T2)
Where U for boiling wort is 220 BTUs/Hr/SqFt/DegF
A is the square feet of surface area needed
T1 = temperature of 45 psig steam
T2= boiling temperature
So with the givens we can solve for A:
A= 294,500/220-(292-212)= 16.7 SF
Taking into consideration a 5% factor for radiant heat loss, make the SF equal 17.6 SF (1.64 Square Meters)
You should check these assumed figures against your actual values ie wort deg P, desired evaporation rate, steam pressure, etc and with your fabricator as well to determine an adequate surface area for your kettle. Depending on the heating format you are using you can turn on bottom dimple jackets while collecting wort to achieve boil by the time the wort collection is completed.