Assume that the feed water consists of 60% condensate and 40% make-up. By recovering additional condensate, the feed water quality is improved, resulting in a lower blow down rate. The blow down rate reduction and corresponding fuel savings can be calculated. Thus if the additional recovery results in a feed water of 67% condensate rather than 60%, the total alkalinity will be reduced from 70 ppm to 58 ppm and we can increase the feed water concentration from 10 to 12. The blow down rate can then be reduced from 10% to 8-1/3%. The actual blow down and feed water requirement in pounds can be calculated as follows: Assume a steam production of 1,000.000 lbs / day.
Where; F = feed water requirement (lbs) S = steam generated (lbs) % B D = percent blow down, expressed as a decimal.
1,111,110 lbs. - 1,090,870 lbs. = 20,240 lbs. reduction Apply the following equation to determine the fuel costs savings:
Where: Br = blow down reductions (lbs/day) H = heat content of blow down (from Table II) C = cost of fuel ($/unit) V = heating value of fuel (Btu/unit) % E = boiler efficiency Using our former example and burning No. 6 fuel oil with a heating value of 142,440 Btu/gallon at a cost of $0.32 per gallon (adjust to market cost as desired), we can calculate the following daily savings:
In this example, by returning only an additional 7% of condensate, a significant savings has been realized. Also, the heating value of the returned condensate would yield additional savings. These calculations are based on the assumption that blow down heat is not being recovered. A blow down heat recovery system would, reduce the potential savings.
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