26 - Question 20 - Boiler Deaerator energy balance - Post by Kambiz Ehsani - 5/12/2018

A boiler generates 50,000 lb/hr of saturated steam at 300 psia, out of which 10,000 lb/hr is taken for process and is returned to the deaerator as condensate at 180 degree F, the remainder is consumed. Makeup water enters the DA tank at 70 degree F and steam is available at 300 psia for deaeration. The DA tank operates at 25 psia. The blow down has TDS of 1,500 ppm by weight and make up has a TDS of 100 ppm. Evaluate the blow down and deaeration steam quantities.

Solution:

If boiler concentration is 0.5 ppm solid and water hardness concentration is 2,500 ppm: then the percent moisture in steam = Steam Purity, ppm / Boiler water concentration, ppm or

percent moisture in steam = (0.5 / 2,500) * 100 = 0.02 so the steam quality = 100 – 0.02 = 99.98 %

DA Energy balance:

The mass balance around two control volumes gives us:

 If M is the make up water to DA, D is the steam for deaeration to DA, F the BFW and B is the Blowdown

For the boiler 50,000 + B = F and for DA: 10,000 + M + D = F                         Eq (1)

by equating this two we will have: 50,000 + B = 10,000 + M + D  

 Energy Balance around the two control volumes gives:

10,000 * 148 + 1,202.8 * D + M * 38 = 209 * F = 209 * (50,000 + B)           Eq (2)

Balance of the solids in the makeup and blowdown gives 100 * M = 1,500 B   Eq (3)