In this post I want to share to you how to do preliminary sizing and horse power estimation for counterflow induced-draft cooling tower. Calculation in this post ** do not apply to counterflow** cooling tower. Also they

**.**

__do not apply when the approach to cold water temperature is less than 5__^{o}F (2.8^{o}C)Approach temperature is the temperature of the water leaving the cooling tower minus ambient wet bulb temperature.

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Assessment of Cooling Tower Performance]

To do this, we need this data:

- Hot water temperature
- Cold water temperature
- Wet-bulb temperature
- Water rate

Let’s start sizing the estimation. Let say we have cooling tower with the following process data:

- Hot water temperature = 102
^{o}F - Cold water temperature = 78
^{o}F - Wet-bulb temperature = 70
^{o}F - Water rate = 1000 gallon per minute

**Step 1: Find water concentration by using chart**

Chart below is used to find water concentration, which is expressed in gallon per minute per square feet (gal/(min.ft^{2})).

When hot water temperature is 102^{o}F and cold water temperature is 78^{o}F, we get water concentration at around ** 2.8 gal/(min.ft^{2})**.

**Step 2: Calculate required area of cooling tower**

Required area of cooling tower is quantity of water circulated divided by water concentration = 1000/2.8 = ** 357.14 ft^{2}**.

**Step 3: Find horsepower per area of cooling tower**

Horsepower per area of cooling tower is estimated by using the following chart. Connecting the point representing 100 percent of standard tower performance with the turning point and extending this straight line to the horsepower scale show that it will give around 0.04 hp/ft^{2} of actual effective tower area. For a tower 357.14 ft^{2} (see Step 2), ** 14.6 hp** is required to perform the necessary cooling.

**Step 4: Check if commercial tower size is less than required area**

Suppose the actual commercial tower size has an area of only 300 ft^{2}. Within reasonable limit, the shortage of actual area can be compensated by increasing air velocity through the tower. This requires boosting fan horsepower to achieve 110 percent of standard tower performance. From chart in Step 3, we found the fan horse power is 0.057 hp/ft2 of actual area, or 0.057 x 300 = __17.1 hp.__

**Step 5: Check if commercial tower size is more than required area**

On the other hand, if the actual commercial tower size is 370 ft^{2}, the cooling equivalent to 357.14 ft^{2} of standard tower area can be accomplished with less air and less fan horsepower. From figure in Step 3, the fan horsepower for a tower operating at 90 percent of standard performance is 0.028 hp/ft^{2} of actual tower area, or 370 x 0.028 = ** 10.4 hp**.

Reference:

Perry’s Chemical Engineers’ Handbook.