Cooling towers are among the most critical utility systems in any industrial plant—chemical, pharma, power, refinery, HVAC, steel, and more. They ensure reliable process cooling by rejecting heat to the atmosphere through evaporation. A deep understanding of cooling tower working principles, design basics, and performance calculations is essential for any engineer involved in utilities, operations, maintenance, process, or project engineering.

This guide is written for beginners but structured with the depth of an industry expert. By the end, you will fully understand cooling tower fundamentals, performance parameters, calculations, and selection guidelines.

1. What is a Cooling Tower?

A cooling tower is a heat-rejection device that cools down hot water by bringing it into contact with ambient air. A small portion of water evaporates during this contact, removing heat from the remaining water.

Key principle:

Evaporative cooling — when water evaporates, it absorbs heat (latent heat of vaporization), thus cooling the water.

Cooling towers are used in:

Power plants (condenser cooling)
Chemical & pharmaceutical industries
HVAC systems (chiller cooling)
Steel & cement plants
Oil refineries & petrochemical complexes

2. Cooling Tower Working Principle (Simplified)

Hot water from the process enters the tower and is sprayed over fill media.
Fans draw or push air through the tower.
Air picks up heat from water through:
- Evaporation (major contribution)
- Convection
A small fraction of water evaporates, causing the remaining water to cool.
Cold water collects in the basin and returns to the process.

3. Types of Cooling Towers

A. Based on Air Flow

Induced Draft (most common)
- Fan at the top pulls air upward.
- Energy efficient and widely used.
Forced Draft
- Fan at air inlet pushes air inside.
Natural Draft
- Uses chimney effect; no electrical fan.
- Used in large power plants.

B. Based on Construction

Counterflow Towers
- Air flows upward, water flows downward.
Crossflow Towers
- Air flows horizontally across falling water.

4. Key Cooling Tower Components

Hot Water Distribution System – nozzles, headers
Fill Media – splash or film fill
Drift Eliminators
Louvers
Fan & Motor
Casing & Structure
Cold Water Basin
Make-up Water System
Bleed-Off/Blowdown Line

5. Psychrometric Basics for Cooling Towers

Dry Bulb Temperature (DBT)

Normal air temperature measured by a thermometer exposed to air.

Wet Bulb Temperature (WBT)

Lowest temperature air can be cooled to by evaporation of water at constant pressure.

Critical point:

Cooling towers cannot cool water below WBT.

Relative Humidity (RH)

Amount of moisture in air relative to its moisture-holding capacity.

Why WBT matters?

It defines the theoretical limit of cooling.

6. Cooling Tower Key Performance Parameters

1. Range

Range=Thot−Tcold\text{Range} = T_{\text{hot}} – T_{\text{cold}}Range=Thot−Tcold

Example:
Hot water = 40°C
Cold water = 32°C
Range = 8°C

2. Approach

Approach=Tcold−TWBT\text{Approach} = T_{\text{cold}} – T_{\text{WBT}}Approach=Tcold−TWBT

Example:
Cold water = 32°C
WBT = 27°C
Approach = 5°C

Smaller approach = better tower performance.

3. Cooling Capacity (kcal/h or kW)

Q=m×Cp×ΔTQ = m \times Cp \times \Delta TQ=m×Cp×ΔT

Where:

mmm = water flow (m³/hr × 1000 = kg/hr)
CpCpCp = 1 kcal/kg°C

Example:
Water flow = 100 m³/hr
Range = 8°C

Q=100,000×1×8=800,000 kcal/hrQ = 100,000 \times 1 \times 8 = 800,000 \text{ kcal/hr}Q=100,000×1×8=800,000 kcal/hr

Convert to kW:

kW=kcal/hr860=800,000860=930 kW\text{kW} = \frac{\text{kcal/hr}}{860} = \frac{800,000}{860} = 930 \text{ kW}kW=860kcal/hr=860800,000=930 kW

4. Evaporation Loss

Evaporation Loss (m³/hr)=Range100×Circulation flow\text{Evaporation Loss (m³/hr)} = \frac{\text{Range}}{100} \times \text{Circulation flow}Evaporation Loss (m³/hr)=100Range×Circulation flow

Rule of thumb:

Evaporation=0.68% of circulation per 5.56°C drop\text{Evaporation} = 0.68\% \text{ of circulation per } 5.56°C \text{ drop}Evaporation=0.68% of circulation per 5.56°C drop

Example:
Range = 8°C
Flow = 100 m³/hr
Evaporation ≈ 1.2 m³/hr

5. Drift Loss

Very small:

Drift=0.02% of circulation\text{Drift} = 0.02\% \text{ of circulation}Drift=0.02% of circulation

(if drift eliminators are good)

6. Blowdown (Bleed Off)

Controls cycles of concentration (COC).

Blowdown=EvaporationCOC−1\text{Blowdown} = \frac{\text{Evaporation}}{\text{COC}-1}Blowdown=COC−1Evaporation

Example:
Evaporation = 1.2 m³/hr
COC = 4
Blowdown = 1.2 / 3 = 0.4 m³/hr

7. Make-up Water Requirement

Make-up=Evaporation+Drift+Blowdown\text{Make-up} = \text{Evaporation} + \text{Drift} + \text{Blowdown}Make-up=Evaporation+Drift+Blowdown

Example:
1.2 + 0.02 + 0.4 = 1.62 m³/hr

7. Cooling Tower Fan Power Calculation (Basic)

P=Q×ΔPη×102P = \frac{Q \times \Delta P}{\eta \times 102}P=η×102Q×ΔP

Where:

QQQ = air volume flow (m³/s)
ΔP\Delta PΔP = static pressure (mmWC)
η\etaη = overall efficiency

8. Heat Transfer Mechanism Inside Cooling Tower

Evaporation

85–90% of heat removed
(latent heat: 540 kcal/kg at 32°C)

Sensible Heat Transfer

10–15%

9. Wet Bulb Temperature – Explained with Example

Statement:
“WBT is the lowest temperature air can be cooled to by evaporation at constant pressure.”

Meaning:
If you wrap a thermometer bulb in wet cloth and blow air on it, it cools due to evaporation. It stops cooling at the WBT – this is the limit for evaporative cooling.

Example:
DBT = 35°C
RH = 50%
WBT ≈ 26°C

Even the best cooling tower cannot cool water below 26°C in this condition.

10. Psychrometric Chart Use for Cooling Tower

On the chart:

Plot DBT.
Move to the RH line to find the air state.
Move along constant enthalpy line to saturation curve → WBT.
Cooling tower performance depends on WBT and humidity ratio.

11. Heat Load Calculation – Practical Example

Given:

Flow = 150 m³/hr
Hot water = 42°C
Cold water = 33°C
WBT = 27°C

Step 1 – Range

42 – 33 = 9°C

Step 2 – Approach

33 – 27 = 6°C

Step 3 – Cooling capacity

Q=150,000×1×9=1,350,000 kcal/hrQ = 150,000 \times 1 \times 9 = 1,350,000 \text{ kcal/hr}Q=150,000×1×9=1,350,000 kcal/hr

Convert to kW:

1,350,000/860=1,569.8 kW1,350,000 / 860 = 1,569.8 \text{ kW}1,350,000/860=1,569.8 kW

12. Cooling Tower Selection – What to Consider

Required cooling capacity (kcal/hr or kW)
Flow rate (m³/hr)
Expected range & approach
Site wet bulb temperature
Type: induced or forced draft
Material: FRP, HDG steel, RCC
Fill type: film or splash
Fan type: axial or centrifugal
Cycles of concentration (water quality)
Space availability

13. Best Practices for Cooling Tower Efficiency

✔ Maintain Approach

Approach < 5°C gives excellent performance (if WBT allows).

✔ Ensure Clean Fill

Scaling reduces surface area → lowers cooling.

✔ Maintain Fan Belt Tension

Loose belts reduce air flow.

✔ Maintain Correct COC

Higher COC saves water but increases scaling.

✔ Ensure Proper Air Flow

No air short-circuiting around tower.

✔ Keep Drift Eliminators Clean

Reduces drift loss and PM emission.

14. Troubleshooting Cooling Tower Problems

Problem: Cold water temperature rising

Causes:

High wet bulb due to climatic conditions
Scaling in fill
Insufficient airflow
Damaged nozzles
Incorrect fan rotation

Problem: Excess water consumption