Boiler Makeup from Condensate Return

M=S(1−%CR100)M = S\left(1 - \frac{\%CR}{100}\right)

Worked example: 10,000 kg/h steam, 7716.18 lb/h makeup → 65% return — press Try an example to run it live, then adjust anything.

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Boiler Makeup from Condensate Return explained

S%CRM

Whatever steam does not come back has to be replaced with cold, hard city water, and that makeup is the entire reason a boiler house has softeners, a dealkaliser and a chemical room. A 20,000 lb/h plant returning 70% of its condensate must treat 6000 lb/h of makeup — about 12 gpm — and if the return drops to 50% that jumps to 10,000 lb/h and the softener that was comfortably sized is now regenerating twice as often.

Cost the difference and the argument for repairing traps makes itself: makeup enters at maybe 55 °F while condensate returns near 190 °F, so every pound of lost condensate costs roughly 135 BTU of extra fuel on top of the water, salt, chemical and sewer charges. Strictly, feedwater is steam plus blowdown, so the true makeup is slightly higher than this equation gives — add the blowdown term when you are sizing equipment rather than benchmarking. And do not forget deaerator vent losses and steam-driven pumps, which quietly consume a few percent of production and never return.

Boiler Makeup from Condensate Return formula

M=S(1−%CR100)M = S\left(1 - \frac{\%CR}{100}\right)
Where
  • MM= Makeup water rate (kg/h)
  • SS= Steam production rate (kg/h)
  • %CR\%CR= Condensate return percentage (%)

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