caneng2002
Civil/Environmental
- Aug 15, 2002
- 35
Greetings
Based on the following equation taken from Water Treatment - Principles and Design (2nd Edition)
4Fe(HCO3)2 + O2 + H2O >>> 4Fe(OH)3 + 8CO2
I'm trying to figure out how Alkalinity used (mg/ mg Fe) of 1.8 was calculated.
I can see how oxidant needed of 0.14 mg/mg Fe was calculated (32/223.2) and that sludge production is 1.9 kg/kg Fe (233 +204)/233.2.
The closest I can get is bicarbonate on lefts side = 488 subtract hydroxide on right side of 51 divided by Fe. which is more a guess.
Any advice appreciated.
Based on the following equation taken from Water Treatment - Principles and Design (2nd Edition)
4Fe(HCO3)2 + O2 + H2O >>> 4Fe(OH)3 + 8CO2
I'm trying to figure out how Alkalinity used (mg/ mg Fe) of 1.8 was calculated.
I can see how oxidant needed of 0.14 mg/mg Fe was calculated (32/223.2) and that sludge production is 1.9 kg/kg Fe (233 +204)/233.2.
The closest I can get is bicarbonate on lefts side = 488 subtract hydroxide on right side of 51 divided by Fe. which is more a guess.
Any advice appreciated.