Evaporative Cooling System’s Water Requirement

Objective

To determine mass flow rate of water required per TR of cooling in an evaporative cooling system. 

Outcome

Students will realize that irrespective of the ambient conditions or the type of climate, water requirement in an evaporative cooler per TR remains the same. 

Requirements

None

Prerequisites

1. Conceptual knowledge of psychrometry.
2. Understanding of psychrometric chart.

Given Data:

Design Conditions:

1. Cooling Load : 1 TR
2. Latent heat of vaporization of water : 540 kcal/kg

Conversion Factors:

1. 1 kW = 1 kJ/s
2. 1 kcal = 4.18 kJ
3. 1 hr = 3,600 s
4. 1 TR = 3.517 kW of cooling

Procedure

Step 1: Convert the Cooling Load in TR to kcal/s.

Step 2: From Cooling Load and Latent Heat of Vaporization, determine the mass flow rate of water in kg/s, convert into kg/hr.

Step 3: Determine the flow rate of water per TR (kg/hr/TR). 

Calculations:

Step 1: Cooling Load (kcals) = (cooling load (TR) 3.517) (kJs)4.18 (kJkcal)  

= 100 3.5174.18    

= 84.1kcals

Step 2: Mass flow rate of water (m_W) (kghr) =cooling load (kcals) 3600 (shr)latent heat of vaporization of water (kcalkg) 

              = 84.1 3600540   

              = 560 kghr

Step 3: Flow rate of water per TR (kghr-TR) = mW (kg H2Ohr)cooling load (TR) 

= 560100  

            = 5.6 kghr-TR 

References

None

Source of Idea