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Master Brewers Association of the Americas > BREWING RESOURCES > Ask the Brewmaster > Posts > Kettle steam load calculations
August 09
             Kettle steam load calculations

​Q:  How do you calculate the steam load needed to boil a 100 bbl brew?

A:  Here is a steam calculation for a 105 bbl (90 hL) wort heat up and boil based on the following assumptions:
We will ignore the 4% contraction factor
5% kettle evaporation per hour
Heat exchange inefficiency and thermal losses of 10%
Steam quality of 100%

Heat up from 165F to 212F with steam at 45 psig  @ 292F and latent heat of 915.9 BTU/lb steam (from steam table):
BTU's= M x Cp x (T1- T2)

Cp (heat capacity) of wort =  0.92 BTU/LbF

M = Mass of wort expressed in lbs.

T1 temperature of the wort at boiling

T2 starting temperature of the wort in the brew kettle

 In this example, heat up starting with 105 bbls and ignoring the 4% thermal expansion of hot liquids the equation is:

= (105 bbls)(258.9 lb/bbl)(0.92 BTU/LbF)(212F-165F) = 1,175,457 BTU/Hr

 At 90% heat exchange inefficiencies and losses =

(1,175,457 BTU/hr)/ 0.90 = 1,306,063 BTU/hr

1,303,063 BTU/Hr / 915.9 BTU/lb steam = 1,426 lbs steam per hour (0.64 metric ton/hr)


Once the wort is heated up the kettle is now maintaining the boil and evaporating water from the wort. It takes 970 BTU to evaporate one pound of water.  The calculation for heat load for the kettle boil assuming 5% evaporation over one hour:

(Volume in bbls) x (% Evaporation/hour) x (Weight of water per bbl) x (BTU's per lb evaporation)

 (105 bbls)(5%/Hr)(258.9 lb water/bbl)(970 BTU/lb water evaporated) = 1,318,448 BTUs/Hr

 1,318,448 BTUs/Hr  / 915.9 BTU/lb steam= 1,439 lbs of steam per hour (0.65 metric ton/Hr)


Since 1 lb of steam condenses when evaportating about one lb of water we can check the calculation based on the percentage of lbs water in and lbs water out compared with the desired evaportation:

 1,439 lbs condensed steam/ 258.9 lbs/bbl = 5.55 bbls evaporated out

 5.55 bbls out/ 105 bbls in = 5.3% evaporation


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