Chiller tonnage to daily makeup water

Cooling tower · water balance and chemical treatment

A 250 ton electric centrifugal chiller serves a downtown office building and rejects to a single induced-draft cooling tower designed for a 10 °F range — 95 °F back from the condenser, 85 °F out of the basin. The conductivity controller is set to hold 5 cycles of concentration, and the tower's cellular eliminators are rated at 0.005 % drift. Find the heat the tower actually has to reject, the condenser water flow the spray pump has to deliver to carry it, the evaporation, the bleed needed to hold five cycles, the drift, the total makeup rate, and the gallons of makeup water the building is billed for in a day.

Step 1 of 7 · solve for Heat rejection rate

Tonnage is the only number the customer can hand you, and it is the wrong number for the tower. The heat rejection factor of 1.25 adds the compressor's own work to the building load, so the 250 ton nameplate becomes 312.5 tons of rejection — exactly 3.75 million BTU/h. Everything downstream is sized on this, never on the nameplate.

Rearranged for Q_r
Qr=QeHRFQ_r = Q_e \, \mathrm{HRF}
Your values, in your units
Qr=(250 ton)(1.25)Q_r = \left( 250\ \text{ton} \right) \, \left( 1.25 \right)
Converted to base units
Qr=(879,213 W)(1.25)Q_r = \left( 879{,}213\ \text{W} \right) \, \left( 1.25 \right)
Answer
Qr=1.1Q_r = 1.1

Carried onward at full precision, not this rounded figure.

Open the Chiller Heat Rejection solver →

Step 2 of 7 · solve for Recirculation rate

Now the heat buys the flow, which is the direction the trade actually reasons in: Q = 500·R·ΔT solved for R. 3.75 MBTU/h across a 10 °F range is exactly 750 gpm — 3.0 gpm per ton of nameplate, which is where the rule of thumb comes from rather than the other way round. Evaporation, drift and system volume are all fractions of this circulating rate, so nothing further can start until it exists.

Rearranged for R
R=Q500ΔTR = \frac{Q}{500 \, \Delta T}
1.1 MWcarried from step 1
Your values, in your units
R=(1,099,020 W)500(10 F)R = \frac{\left( 1{,}099{,}020\ \text{W} \right)}{500 \, \left( 10\ \text{F}^{\circ} \right)}
Converted to base units
R=(1,099,020 W)500(5.55556 C)R = \frac{\left( 1{,}099{,}020\ \text{W} \right)}{500 \, \left( 5.55556\ \text{C}^{\circ} \right)}
Answer
R=47.3R = 47.3

Carried onward at full precision, not this rounded figure.

Open the Cooling Tower Heat Rejection solver →

Step 3 of 7 · solve for Evaporation rate

Evaporation is the first loss and almost always the largest: 0.1 % of the circulating flow per Fahrenheit degree of range, so 750 gpm on a 10 °F range boils off exactly 7.5 gpm. It is also the only loss that leaves the dissolved minerals — and the inhibitor — behind in the basin.

Rearranged for E
E=0.001RΔTE = 0.001 \, R \, \Delta T
47.3 L/scarried from step 2
Your values, in your units
E=0.001(0.0473176 m3/s)(10 F)E = 0.001 \, \left( 0.0473176\ \text{m}^{3}\text{/s} \right) \, \left( 10\ \text{F}^{\circ} \right)
Converted to base units
E=0.001(2,839.06 L/min)(5.55556 C)E = 0.001 \, \left( 2{,}839.06\ \text{L/min} \right) \, \left( 5.55556\ \text{C}^{\circ} \right)
Answer
E=28.4E = 28.4

Carried onward at full precision, not this rounded figure.

Open the Cooling Tower Evaporation Rate solver →

Step 4 of 7 · solve for Blowdown rate

Blowdown is not a loss you suffer, it is a loss you choose. Holding five cycles against that evaporation costs a bleed of E/(5 − 1), 1.875 gpm; ask for three cycles instead and it doubles.

Rearranged for B
B=ECOC1B = \frac{E}{\text{COC} - 1}
28.4 L/mincarried from step 3
Your values, in your units
B=(0.000473176 m3/s)(5)1B = \frac{\left( 0.000473176\ \text{m}^{3}\text{/s} \right)}{\left( 5 \right) - 1}
Converted to base units
B=(28.3906 L/min)(5)1B = \frac{\left( 28.3906\ \text{L/min} \right)}{\left( 5 \right) - 1}
Answer
B=7.1B = 7.1

Carried onward at full precision, not this rounded figure.

Open the Blowdown Rate from Cycles solver →

Step 5 of 7 · solve for Drift loss

Drift is basin water flung out of the stack as droplets — 0.005 % of recirculation on modern eliminators, 0.0375 gpm here. Small enough to ignore in the water bill, never small enough to ignore in the chemistry, because it leaves at full basin strength.

Rearranged for D
D=d100RD = \frac{d}{100} \, R
47.3 L/scarried from step 2
Your values, in your units
D=(0.005 %)100(0.0473176 m3/s)D = \frac{\left( 0.005\ \text{\%} \right)}{100} \, \left( 0.0473176\ \text{m}^{3}\text{/s} \right)
Converted to base units
D=(0.005 %)100(2,839.06 L/min)D = \frac{\left( 0.005\ \text{\%} \right)}{100} \, \left( 2{,}839.06\ \text{L/min} \right)
Answer
D=54D = 54

Carried onward at full precision, not this rounded figure.

Open the Cooling Tower Drift Loss solver →

Step 6 of 7 · solve for Makeup water rate

Water leaves through exactly three doors and the makeup valve replaces all three. 9.4125 gpm — 1.255 % of the recirculation rate, which is the sanity check to carry in your head.

Rearranged for M
M=E+B+DM = E + B + D
28.4 L/mincarried from step 3
7.1 L/mincarried from step 4
54 gpdcarried from step 5
Your values, in your units
M=(0.000473176 m3/s)+(0.000118294 m3/s)+(0.00000236588 m3/s)M = \left( 0.000473176\ \text{m}^{3}\text{/s} \right) + \left( 0.000118294\ \text{m}^{3}\text{/s} \right) + \left( 0.00000236588\ \text{m}^{3}\text{/s} \right)
Converted to base units
M=(28.3906 L/min)+(7.09765 L/min)+(0.141953 L/min)M = \left( 28.3906\ \text{L/min} \right) + \left( 7.09765\ \text{L/min} \right) + \left( 0.141953\ \text{L/min} \right)
Answer
M=35.6M = 35.6

Carried onward at full precision, not this rounded figure.

Open the Cooling Tower Makeup Water Rate solver →

Step 7 of 7 · solve for Volume over the period

Finally turn the rate into the number on the invoice, with the relation that means throughput rather than holdup: V = Q·t. A full day of 9.4125 gpm is 13,554 US gallons, about 51.3 m³ — the line the building pays for, not the water standing in the system.

Rearranged for V
V=QtV = Q \, t
35.6 L/mincarried from step 6
Your values, in your units
V=(0.000593836 m3/s)(1 d)V = \left( 0.000593836\ \text{m}^{3}\text{/s} \right) \, \left( 1\ \text{d} \right)
Converted to base units
V=(35.6302 L/min)(86,400 s)V = \left( 35.6302\ \text{L/min} \right) \, \left( 86{,}400\ \text{s} \right)
Answer
V=51.3V = 51.3

Carried onward at full precision, not this rounded figure.

Open the Volume of Water Over a Period solver →

Why this order

The order is forced by what you can actually know on site. Nobody meters the makeup on a first visit, but everybody can read a chiller nameplate, so tonnage is the entry point — and the first honest move from tonnage is not flow, it is heat. A tower is bought, selected and rated on heat rejection, so step 1 turns the 250 tons of cooling into the 312.5 tons the tower has to carry, at a heat rejection factor of 1.25: 3.75 million BTU/h exactly. Only then does flow follow, because Q = 500·R·ΔT solved for R is the calculation a tower engineer actually performs — 3.75 MBTU/h across a 10 °F range is 750 gpm, or 3.0 gpm per ton. Evaporation, drift and system volume are all fractions of that circulating rate and none of them can be computed until it is known. Blowdown then has to wait for evaporation, because B = E/(COC − 1) — the bleed is set by what the tower concentrates, not by the size of the plant — and makeup can only be summed once all three losses exist. Running heat and flow through the same relation also keeps one set of constants underneath the chain: the 500 is 8.34 lb/gal × 60 min/h × 1 BTU/(lb·°F). Solve the identical duty through the ρc_p physics instead, at 998.3 kg/m³ and 8.33 lb/gal, and the flow comes out 750.19 gpm — 0.025 % away, a quarter of a tenth of a percent, and far inside the error of any field measurement.

The expensive mistake here is buying tower on the wrong number. Order a 250 ton tower for a 250 ton chiller and you are 25 % short of the duty on the hottest afternoon of the year; the approach walks out, condensing pressure climbs, and the chiller loses capacity and burns compressor power for the rest of its life — a far bigger bill than the tower ever saved. The second is trusting the rules of thumb past their range. The 3 gpm per ton comes from 15,000 BTU/h per ton of rejection divided by 500 × 10 °F, so it dies the moment the range changes (2 gpm/ton at 15 °F), the moment the loop carries glycol (specific heat near 0.9 rather than 1.0), or the moment the machine is an absorption chiller, whose heat rejection factor is 1.7–1.8 and whose tower is correspondingly enormous. The 0.001 evaporation rule has the same character: it comes from a pound of water absorbing roughly 1000 BTU to boil against 1 BTU per degree to cool, so one pound in a hundred leaves per 10 °F of range — honest as a design-day figure, but it runs nearer 0.00085 on a cold dry night, nearer 0.0011 on a humid afternoon, and it is flatly zero on a closed-circuit cooler or a dry cooler, which lose no water at all. Bill a customer for makeup on a dry cooler once and you will not be asked back.

Carried values move at full precision, not the rounded figure shown — chaining rounded numbers compounds error.