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Cooling Towers & Heat Rejection

5% of exam

Tower range, approach, and effectiveness, evaporation and makeup water, cycles of concentration, and condenser-water heat balance.

3 concepts

A. Cooling Towers and Fluid Coolers

Cooling-Tower Range, Approach, Effectiveness and Water Balance

Range, approach, and effectiveness define what a cooling tower delivers; the wet-bulb sets the thermodynamic floor and the 500-coefficient ties heat rejection to water flow.

A cooling tower does one job: it throws building heat away to the atmosphere by evaporating a little of the water that circulates through it. Two temperature differences describe how well it does that job, and almost every tower question on the PE Mechanical exam turns on keeping them straight. The first is the

\textbf{range}

— how far the water temperature drops as it falls through the tower. The second is the

\textbf{approach}

— how close the cooled water gets to the outdoor air wet-bulb, the thermodynamic floor a wet tower can never quite reach. Confuse the two, or measure the approach against dry-bulb instead of wet-bulb, and the rest of the calculation collapses. The NCEES PE Mechanical Reference Handbook fixes these definitions in §9.3.9 Cooling Towers and Fluid Coolers, and they recur in §8.3 Condensers when the tower serves a chiller.

Range: the heat the water gives up

Range is simply the entering-hot minus the leaving-cold water temperature,

R = T_{hw}-T_{cw}

. It is set by the load and the flow, not by the weather: if you reject more heat into the same gallons, the water arrives hotter and the range grows. For a condenser-water loop the range equals the temperature rise the water picked up across the chiller condenser, because the loop is closed — what the condenser adds, the tower must remove. A typical comfort-cooling tower runs a

10\,^\circ\text{F}

range, which is why the

3\ \text{gpm/ton}

rule of thumb falls out so cleanly.

R = T_{hw} - T_{cw}

Approach: distance from the wet-bulb floor

Approach is the leaving cold-water temperature minus the entering-air

\textbf{wet-bulb}

temperature,

A = T_{cw}-T_{wb}

. The wet-bulb — not the dry-bulb — is the limit because the tower cools by evaporation; the coldest water a wet surface can theoretically reach is the wet-bulb of the air blowing past it. Approach can never be zero (that would demand an infinitely large tower) and is the single most expensive parameter in a selection: tightening the approach from

7\,^\circ\text{F}

to

5\,^\circ\text{F}

can add half again as much tower. Unlike range, approach is fixed by the tower size and the design wet-bulb, so it is the number a manufacturer guarantees.

A = T_{cw} - T_{wb}, \qquad T_{cw} = T_{wb} + A

Tower effectiveness

Thinking of the tower as a heat exchanger whose cold-side limit is the wet-bulb gives a clean effectiveness: the actual cooling achieved (the range) divided by the maximum possible cooling (range plus approach, i.e. hot water all the way down to the wet-bulb). Effectiveness rises as the approach shrinks, and a real tower lands in the

0.5

–

0.8

band. It is dimensionless, so it travels between SI and USCS untouched — handy when a problem mixes unit systems.

\varepsilon = \frac{R}{R + A} = \frac{T_{hw}-T_{cw}}{T_{hw}-T_{wb}}

The heat-rejection balance and the 500 coefficient

The water side carries away sensible heat at a rate set by flow and range. With standard water (

\rho \approx 8.34\ \text{lb/gal}

,

c_p \approx 1\ \text{Btu/lb·}^\circ\text{F}

), the familiar coefficient is

8.34\ \tfrac{\text{lb}}{\text{gal}} \times 60\ \tfrac{\text{min}}{\text{hr}} \times 1\ \tfrac{\text{Btu}}{\text{lb·}^\circ\text{F}} \approx 500\ \tfrac{\text{Btu·min}}{\text{gal·hr·}^\circ\text{F}}

. Memorize

500

the way you memorize the

1.08

for air-side sensible heat; it bakes in standard water properties and only shifts if a glycol mix or very hot water changes

\rho c_p

. Solving for flow turns a heat-rejection duty into a required gpm in one line.

\dot q_o = 500\,(\text{gpm})\,R \quad\Longrightarrow\quad \text{gpm} = \frac{\dot q_o}{500\,R}

Cooling-tower tons and selection at a design wet-bulb

Beware the units of a tower rating. A refrigeration ton is

12{,}000\ \text{Btu/hr}

at the evaporator, but a

\textbf{cooling-tower ton}

is

15{,}000\ \text{Btu/hr}

— the handbook's industry figure that adds the compressor heat to the chiller load at a nominal

0.88\ \text{kW/ton}

efficiency. A tower is selected at a design wet-bulb (a local statistic, e.g. the

1\%

ASHRAE wet-bulb), a design range, and a design approach. Because approach is a property of the iron, a tower picked for a

78\,^\circ\text{F}

wet-bulb will, on a milder day, deliver colder water at the same approach; on a hotter, more humid day its leaving water rises one-for-one with the wet-bulb. Selecting for the wrong wet-bulb is the classic field failure — the tower is fine in spring and short in August.

Exam strategy

Write the three temperatures down the moment you read the problem —

T_{hw}

,

T_{cw}

,

T_{wb}

— and label range

(T_{hw}-T_{cw})

and approach

(T_{cw}-T_{wb})

before touching a formula; swapping them is the number-one trap. Confirm whether 'ton' means

12{,}000

(load) or

15{,}000

(cooling-tower) Btu/hr. When a duty is given, get flow from

\text{gpm}=\dot q_o/(500R)

and remember

500

assumes standard water. If the wet-bulb shifts, the new cold-water temperature is

T_{wb}+A

at constant approach — no re-rating needed for a quick estimate.

Key equations

Range

R = T_{hw} - T_{cw}

Tower range (°F): entering hot minus leaving cold water. Equals the condenser-water rise across the chiller in a closed loop; set by load and flow.

Approach

A = T_{cw} - T_{wb}

Approach (°F): leaving cold water minus entering-air WET-BULB. Fixed by tower size and design wet-bulb; never zero.

Cold-water temperature

T_{cw} = T_{wb} + A

Leaving water at a given approach tracks the wet-bulb one-for-one; the quick way to re-estimate at a new design wet-bulb.

Tower effectiveness

\varepsilon = \dfrac{R}{R+A} = \dfrac{T_{hw}-T_{cw}}{T_{hw}-T_{wb}}

Water-side heat rejection

\dot q_o = 500\,(\text{gpm})\,R

Sensible heat carried by the circulating water (Btu/hr). The 500 = 8.34 lb/gal × 60 min/hr × 1 Btu/lb·°F assumes STANDARD water.

Required water flow

\text{gpm} = \dfrac{\dot q_o}{500\,R}

Cooling-tower ton

1\ \text{CT ton} = 15{,}000\ \tfrac{\text{Btu}}{\text{hr}}

Industry rating adding compressor heat to the 12,000 Btu/hr refrigeration ton at ≈0.88 kW/ton. Do not confuse with the evaporator ton.

Heat-rejection from refrigeration tons

\dot q_o = \dot q_{evap}\left(1 + \dfrac{3412\,(\text{kW/ton})}{12{,}000}\right)

Worked examples

Range, approach, and effectiveness

Problem. A cooling tower receives condenser water at $95\,^\circ\text{F}$ and returns it at $85\,^\circ\text{F}$ . The design entering-air wet-bulb is $78\,^\circ\text{F}$ . Find the range, the approach, and the tower effectiveness.

Solution. Range:

R = T_{hw}-T_{cw} = 95-85 = 10\,^\circ\text{F}

\varepsilon = \frac{95-85}{95-78} = \frac{10}{17} = 0.588

Heat rejection to condenser-water flow

Problem. A tower circulates $1{,}200\ \text{gpm}$ of standard condenser water at a $10\,^\circ\text{F}$ range. Find the heat-rejection rate in Btu/hr and express it in both refrigeration tons and cooling-tower tons.

Solution. Heat rejected:

\dot q_o = 500(\text{gpm})R = 500(1{,}200)(10) = 6.00\times10^{6}\ \text{Btu/hr}

Cold-water temperature at a hotter design wet-bulb

Problem. The tower above (approach $7\,^\circ\text{F}$ ) was selected for a $78\,^\circ\text{F}$ wet-bulb. A revised weather analysis raises the design wet-bulb to $80\,^\circ\text{F}$

Common pitfalls

•Swapping range and approach. Range = hot − cold water; approach = cold water − wet-bulb. Write all three temperatures and label each difference before using any formula.
•Measuring approach against the DRY-bulb. A wet tower's floor is the wet-bulb temperature; using dry-bulb makes the approach look impossibly small. Always use the entering-air wet-bulb.
•Treating a cooling-tower ton as 12,000 Btu/hr. A CT ton is 15,000 Btu/hr (it includes compressor heat at ≈0.88 kW/ton); a refrigeration ton is 12,000 Btu/hr. Confirm which the problem means.
•Applying the 500 coefficient to glycol or non-standard water. 500 = 8.34 × 60 × 1 bakes in standard water ρ and c_p; a glycol mix or very hot water changes ρc_p and the coefficient.
•Expecting zero approach. A finite tower can never cool water to the wet-bulb; a guaranteed approach is typically 4–7°F, and tightening it sharply enlarges the tower.
•Selecting at the wrong design wet-bulb. Pick the local design (e.g. ASHRAE 1% or 0.4%) wet-bulb; a tower sized for a mild wet-bulb runs short of capacity on the hottest, most humid afternoons.

References

NCEES PE Mechanical Reference Handbook — §9.3.9 Cooling Towers and Fluid Coolers
NCEES PE Mechanical Reference Handbook — §8.3 Condensers
ASHRAE Handbook — HVAC Systems and Equipment, Cooling Towers
Cooling Technology Institute (CTI) — Cooling Tower Performance terminology

Evaporation, Drift, Blowdown and Cycles of Concentration

A cooling tower's makeup water is the sum of evaporation, drift, and blowdown; cycles of concentration set the blowdown that keeps the loop from scaling.

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Condenser-Water Heat Balance and Tower Selection

A water-cooled chiller's condenser must reject the cooling load plus the compressor heat; the 3 gpm/ton rule and a heat-rejection factor size the loop and the tower.

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Dimensionless thermal effectiveness; cold-side limit is the wet-bulb. Real towers ≈ 0.5–0.8.

Inverts the rejection equation; with a 10°F range and 15,000 Btu/hr per cooling-tower ton this gives the 3 gpm/ton rule.

Adds heat of compression to the chiller load; gives Btu/hr to reject for an electric chiller (3412 Btu/hr per kW).

. Approach:

A = T_{cw}-T_{wb} = 85-78 = 7\,^\circ\text{F}

(measured to wet-bulb, not dry-bulb). Effectiveness:

\varepsilon = \dfrac{R}{R+A} = \dfrac{10}{10+7} = 0.588

. Sanity check:

\varepsilon

is dimensionless and lands in the realistic

0.5

–

0.8

band, so the selection is plausible. Answer:

R = 10.0\,^\circ\text{F}

,

A = 7.00\,^\circ\text{F}

,

\varepsilon = 0.588

.

. Refrigeration tons:

6.00\times10^{6}/12{,}000 = 500

tons. Cooling-tower tons:

6.00\times10^{6}/15{,}000 = 400

CT tons. Sanity check: dimensions are

\tfrac{\text{Btu·min}}{\text{gal·hr·}^\circ\text{F}}\cdot\text{gpm}\cdot{}^\circ\text{F} = \text{Btu/hr}

; the CT-ton count is smaller because each CT ton bundles more Btu/hr. Answer:

\dot q_o = 6.00\times10^{6}\ \text{Btu/hr}

(

500

ref. tons,

400

CT tons).

\dot q_o = 500(1200)(10) = 6.00\times10^{6}\ \text{Btu/hr}

. Holding the approach constant, what leaving cold-water temperature should the designer now expect, and what does that do to the chiller?

Solution. At constant approach,

T_{cw} = T_{wb} + A = 80 + 7 = 87\,^\circ\text{F}

. The cold water leaves

2\,^\circ\text{F}

warmer than before (

85\to 87\,^\circ\text{F}

), so the chiller condenses against warmer water, its saturated condensing temperature rises, and its kW/ton increases (capacity falls). Sanity check: leaving water moves one-for-one with the wet-bulb, the defining behavior of a fixed-approach tower. Answer:

T_{cw} = 87.0\,^\circ\text{F}

, degrading chiller performance.