Acid Feed to Reduce Alkalinity

m˙=ΔAlkQ×EW50.04p/100%\dot m = \frac{\Delta \mathrm{Alk} \, Q \times \frac{\mathrm{EW}}{50.04}}{p/100\%}

Enter your known values, leave one input blank, and solves for the missing one. Try different units for next level excitement!

Learning zone

Acid feed is how a cooling tower runs at high cycles without cementing itself shut. Alkalinity is what drives the LSI upward, so knocking it down from, say, 200 to 150 mg/L as CaCO₃ buys back most of a scaling index unit. The arithmetic is equivalent-for-equivalent: every equivalent of alkalinity needs one equivalent of acid. Because CaCO₃ has an equivalent weight of 50.04 and sulphuric acid 49.04, pure H₂SO₄ neutralizes alkalinity at almost exactly 0.98 pounds per pound — then divide by the commercial strength. Removing 50 mg/L from 1 MGD with 93% sulphuric acid takes 50 × 1 × 8.34 × 0.98 / 0.93 ≈ 440 lb/day, and the same job with 100% pure acid would take 409.

Two hard-won cautions. Acid feed and pH control are not the same knob: you are removing buffer capacity, so as alkalinity falls the water's pH becomes progressively twitchier and a pump that was fine yesterday can overshoot into the low 6s and start dissolving copper. Never run a tower below about 50 mg/L residual alkalinity for that reason. And plumb it properly — feed into a turbulent, well-mixed return line, never into the tower basin or a dead leg, use a corrosion-resistant quill that discharges into the middle of the pipe, and interlock the pump to the recirculation flow switch. A concentrated acid slug sitting in a stagnant carbon-steel header will eat a hole through schedule 40 pipe in a single afternoon.

Acid Feed to Reduce Alkalinity
m˙=ΔAlkQ×EW50.04p/100%\dot m = \frac{\Delta \mathrm{Alk} \, Q \times \frac{\mathrm{EW}}{50.04}}{p/100\%}
Where
  • m˙\dot m= Acid feed rate
  • ΔAlk\Delta \mathrm{Alk}= Alkalinity reduction as CaCO₃
  • QQ= Water flow rate
  • EW\mathrm{EW}= Equivalent weight of the acid
  • pp= Acid strength