Industrial Steam Systems, Boiler Water Chemistry & Energy Recovery
Steam Boiler TDS Blowdown Rate & Flash Heat Loss Formula
Calculate Steam Boiler TDS Cycles of Concentration, Continuous Drum Blowdown Mass Rate (kg/hr), Unrecovered Sensible Blowdown Heat Loss (kW), and Flash-Economizer Heat Recovery (kW).
Do not confuse blowdown % of feedwater with blowdown % of steam—Steam / (COC - 1) sets drum mass balance!
In a steam boiler, pure dry steam exits with ~0 ppm Total Dissolved Solids (TDS), leaving all incoming feedwater minerals inside the steam drum. Because Feedwater = Steam + Blowdown and Feedwater * TDS_fw = Blowdown * TDS_max, Boiler Cycles of Concentration is COC = TDS_max / TDS_fw, Blowdown as a fraction of FEEDWATER is 1 / COC, and Blowdown Mass Rate relative to STEAM generation is Blowdown = Steam_Rate * TDS_fw / (TDS_max - TDS_fw) = Steam_Rate / (COC - 1)! Applying 1 / COC directly to Steam_Rate under-blows the boiler drum, risking carryover foaming and superheater salt scaling!
Verify on a 15-bar Packaged Watertube Boiler—Steam_Generation = 20,000 kg/hr (B2), Blended_Feedwater_TDS = 250 ppm (B3), Max_ASME_Drum_TDS = 2,750 ppm (B4 -> Boiler COC = 2,750 / 250 = 11.00 in E2), Saturated_Drum_Liquid_Enthalpy = 850 kJ/kg (B5), Cold_Makeup_Enthalpy = 130 kJ/kg (B6), and Flash_Tank_Economizer_Recovery = 0.70 (70.0%, B7): Continuous Blowdown Rate equals 20,000 * 250 / (2,750 - 250) = 2,000.00 kg/hr (E3, with 22,000 kg/hr total feedwater), Unrecovered Blowdown Heat Loss is 2,000 * (850 - 130) / 3,600 = 400.00 kW (E4), and Flash-Economizer Heat Recovered is 280.00 kW (E5)!
| Steam Boiler Blowdown KPI (20,000 kg/hr Steam, 250 -> 2,750 ppm TDS) | Formula | Verified Output |
|---|---|---|
| Boiler Cycles of Concentration | Max_Drum_TDS / Feedwater_TDS | 11.00 (5.50 @ 500 ppm FW) |
| Continuous Blowdown Rate (kg/hr) | Steam_kg_hr * TDS_fw / (TDS_max - TDS_fw) | 2,000.00 kg/hr (4,444.44 @ 500) |
| Unrecovered Blowdown Loss (kW) | Blowdown_kg_hr * (h_drum - h_makeup) / 3600 | 400.00 kW (888.89 kW @ 500) |
| Flash + HX Heat Recovered (kW) | Unrecovered_Loss_kW * Recovery_Eff | 280.00 kW |
Steam-Basis TDS Blowdown Mass & Heat-Loss Formula
Validate Steam_Rate > 0, 0 < Feedwater_TDS < Max_Drum_TDS, Drum_Enthalpy > Makeup_Enthalpy >= 0, and Recovery_Eff in [0, 0.95]!
=LET(st_m, B2, td_f, B3, td_d, B4, h_dr, B5, h_mk, B6, rc_e, B7, IF(OR(st_m <= 0, td_f <= 0, td_d <= td_f, h_dr <= h_mk, h_mk < 0, rc_e < 0, rc_e > 0.95), "Invalid Boiler TDS Input", LET(bd_m, st_m * td_f / (td_d - td_f), ROUND(bd_m * (h_dr - h_mk) / 3600, 2))))Hardness Leakage Feedwater Spike Check (Set Feedwater TDS = 500 ppm)
Double Feedwater_TDS B3 from 250 ppm to 500 ppm (a softener regeneration breakthrough): Boiler COC E2 falls to 5.50, Continuous Blowdown E3 more than doubles from 2,000.00 kg/hr to 4,444.44 kg/hr (because TDS_max - TDS_fw also shrinks!), and Unrecovered Heat Loss E4 surges to 888.89 kW!
=IF(ISTEXT(E4), E4, ROUND(E4 * B7, 2))Try the example
Use the original values or change them to test your understanding.