The same 750 gpm tower is going out to tender on a full-service treatment contract, and the price has to cover the worst case the site can legitimately produce — so the sizing is done off the spec sheet, not off the best month in the logbook. Read the contract the way the water balance will: the tower online 24 hours a day for 365 days, the conductivity controller allowed to fall to its contractual floor of 4 cycles, the design 10 °F range, the eliminators at their rated 0.005 % drift, and a maintained inhibitor residual of 100 ppm as product written into the scope. Every gallon that leaves the system as liquid leaves carrying that residual, and the vapour going over the fan carries none of it, so what reads like a chemical question is a water balance run at the ugly end of the permitted range. The drums delivered to site have to cover all of it before the price goes on the tender. Find the evaporation, the blowdown at four cycles, the drift, the flow that actually carries chemical out of the system, the volume of water leaving as liquid in a year, and the mass of inhibitor that has to be delivered to the site to cover it.
Every number in this problem is editable — change any value below and the whole chain recalculates.
Given
R = 750 gpm — Recirculation rate
ΔT = 10 °F — Range
COC = 4 cycles — Contractual floor on cycles
d = 0.005 % — Drift
C = 100 ppm — Maintained inhibitor residual, as product
t = 365 d — Tower online, the full year
Determine
(a)the evaporation
(b)the blowdown at four cycles
(c)the drift
(d)the flow that actually carries chemical out
(e)the water leaving as liquid in a year
(f)the inhibitor that has to be delivered to cover it
Step 1 of 6(a) · solve for Evaporation rate
Start with evaporation even though it consumes no chemical at all, because it is what sets the blowdown. 7.5 gpm of pure vapour leaves every mineral and every molecule of inhibitor behind in the basin.
Rearranged for E
E=0.001RΔT
Your values, in your units
E=0.001(750gpm)(10F∘)
Converted to base units
E=0.001(2,839.06L/min)(5.55556C∘)
Answer
E=28.391L/min
Carried onward at full precision, not this rounded figure.
Design maximum means the lowest cycles the contract permits, not the best the program can hold — four cycles, not six. Lower cycles means more bleed, and more bleed is the whole cost of the chemistry.
Rearranged for B
B=COC−1E
28.391 L/mincarried from step 1
Your values, in your units
B=(4)−1(0.000473176m3/s)
Converted to base units
B=(4)−1(28.3906L/min)
Answer
B=9.4635L/min
Carried onward at full precision, not this rounded figure.
Drift is only 0.0375 gpm, but it leaves as liquid at full basin concentration, so every drop of it is carrying inhibitor over the fence. It counts here even though it barely registers on the water meter.
Rearranged for D
D=100dR
Your values, in your units
D=100(0.005%)(750gpm)
Converted to base units
D=100(0.005%)(2,839.06L/min)
Answer
D=141.95cm3/min
Carried onward at full precision, not this rounded figure.
Step 4 of 6(d) · solve for Chemical-consuming loss
L = B + D, and evaporation is simply not in the equation — not zeroed out of a makeup sum, but absent, because vapour carries no inhibitor. What comes out is the total flow of treated water leaving as liquid, 2.5375 gpm, and that is the only flow that takes chemical with it.
Rearranged for L
L=B+D
9.4635 L/mincarried from step 2
141.95 cm³/mincarried from step 3
Your values, in your units
L=(0.000157725m3/s)+(0.00000236588m3/s)
Converted to base units
L=(9.46353L/min)+(0.141953L/min)
Answer
L=9.6055L/min
Carried onward at full precision, not this rounded figure.
Run that loss for a full contract year — V = L·t, throughput and not tank holdup: 5,048.6 m³, exactly 1,333,710 US gallons of treated water sent to drain or over the roof.
Rearranged for V
V=Qt
9.6055 L/mincarried from step 4
Your values, in your units
V=(0.000160091m3/s)(365d)
Converted to base units
V=(9.60548L/min)(31,536,000s)
Answer
V=4.093ac⋅ft
Carried onward at full precision, not this rounded figure.
Every one of those gallons leaves at the maintained 100 ppm residual, so the annual consumption is simply that dose applied to that volume: about 505 kg, or 1,112 lb — a little over two 55-gallon drums a year at a typical 9.5 lb/gal. That is the upper limit the contract price is built on.
Rearranged for m
m=100CVρw
4.093 ac·ftcarried from step 5
Your values, in your units
m=100(100ppm)(5,048.64m3)ρw
Converted to base units
m=100(0.01%)(5,048,640L)ρw
Answer
m=504.54kg
Carried onward at full precision, not this rounded figure.
Therefore the design-maximum tower evaporates 7.5 gpm and bleeds 2.5 gpm at the four-cycle floor against 0.0375 gpm of drift; chemical rides out on 2.5375 gpm of liquid, a full year of which is 1,333,710 US gallons — and holding 100 ppm across all of it takes 504.5 kg of inhibitor, about 1,112 lb, the ceiling the contract price is built on.
Why this order
The order of this chain encodes one physical fact: evaporation leaves the inhibitor behind. Pure water vapour goes up the stack carrying no phosphonate, no azole, no polymer and no dissolved solids whatsoever — that is precisely why the basin concentrates and why cycles of concentration exists as a concept. Chemical can only leave the system dissolved in liquid water, which means it leaves through the bleed valve and over the drift eliminators, and nowhere else. So evaporation is computed first only because blowdown depends on it through B = E/(COC − 1); the moment blowdown and drift are known, evaporation is dropped out of the accounting entirely, which is what step 4 does — the chemical-consuming loss L = B + D, a relation that names what it means instead of being a makeup sum with one term quietly knocked out. From there the arithmetic is one idea repeated: the flow that carries chemical out, times a year, times the residual you have promised to maintain.
Size that contract on makeup instead and you will be wrong by a factor of four. Makeup here is 10.0375 gpm against 2.5375 gpm of chemical-consuming loss, so the makeup basis predicts about 1,996 kg of inhibitor a year where the true design maximum is 505 kg — 3.956 times the real consumption, because evaporation is the largest single loss and it consumes nothing. Quote that number and you lose the job to anyone who did the balance properly; buy that much product and three quarters of it sits in a bunded store going out of date. The error runs the other way too: a program fed proportional to makeup rather than to bleed will overfeed by exactly the same factor and the residual will climb until someone throttles the pump and stops trusting the numbers. Two constants to keep honest while you do it. The 8.34 lb/gal inside the dose-to-mass step is water at about 60 °F, so it is 8.29 in a 100 °F basin and it is simply the wrong number for a brine or a glycol loop — use the real density there. And the ppm has to be labelled: 100 ppm of product on a 20 % active blend is 20 ppm of active inhibitor, and a test kit reads the active while the drum log records the product. Write which one you mean at the top of the log sheet, in the contract, and on the invoice.
Carried values move at full precision, not the rounded figure shown — chaining rounded numbers compounds error.
