# How to calculate the cooling coil capacity

In this article guide, I will use an example problem to calculate the cooling coil capacity.

## Example 1:

Air at 27 deg dry bulb and at a flow of 1 m3/sec and 60% relative humidity is cooled and dehumidified. Accordingly, the temperature of the exiting air is 16 deg and at 80% relative humidity.

You need to use the psychrometric chart shown below in order to calculate the total cooling coil capacity.

### Solution:

You first add **point 1** on the chart at 60% relative humidity line and at 27 degrees as shown in the x-axis (dry bulb temperature). Additionally, you extract the specific enthalpy for **point 1** as shown in the chart below. Next, you add **point 2** on the chart above at 80% relative humidity and at 16 degrees Celsius. In the same way, you extract the specific enthalpy for **point 2**.

To calculate the total load of the air handling unit you will use the following thermodynamic rule:

**Q = Air mass flow rate x (Enthalpy point 1 – Enthalpy point 2)**

**Q= Cooling Load in KW**

**Enthalpy point 1 = 67 **Kj**/kg**

**Enthalpy point 2 = 38 **Kj**/kg**

Accordingly, you need to remember that you still don’t have the air mass flow rate, in order to calculate it you need the specific volume of air at **point 1**. Therefore, you need to use the psychrometric chart again to extract that information:

As shown in the figure above, the green line is between 0.85 and 0.9, it is actually **0.878**. Consequently, now you can calculate the air mass flow:

**Air Mass Flow Rate= (1 / 0.878) = 0.878 kg/s**

Finally, using the total load equation ‘**Q**‘, you can insert all the values extracted from the psychrometric chart:

**Q= 0.878 x (67 – 38) = 25.4 Kw**

In the next article, we will learn how to calculate the sensible and latent heat loads using the psychrometric chart.

This article is part of a series of documents discussing HVAC Design, stay tuned for more design guides.