Evaporative Cooling System’s Air Requirement

Objective

To determine volume flow rate of air required per TR of cooling in an evaporative cooling system.

Outcome

Students will realize that unlike water requirement, air requirement in an evaporative cooler per TR varies. It depends on the ambient and desired air conditions.

Requirements

Psychrometric Chart

Prerequisites

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

Given Data: 

Design Conditions (Peak Summer)

1. Ambient : DB = 38°C, RH = 30%
2. Effectiveness of Evaporative Cooler : 90%
3. Density of air : 1.12 kg/m³
4. Mass flow rate of water (m_W) : 5.6 kg/hr/TR (refer last activity)

Conversion Factors

1. 1 kW = 1 kJ/s
2. 1 hr = 3,600 s
3. 1 TR = 3.517 kW of cooling
4. 1 m³ = 35.32 ft³

Procedure

Step 1: Estimate the Desired Air Condition from effectiveness of the Evaporative Cooling System.

Step 2: Estimate Specific Humidity Ratios (⍵) of Ambient and Desired Air Conditions (g H₂O/kg air) and their difference (∆⍵).

Step 3: Estimate the mass flow rate of air (kg/hr).

Step 4: Estimate the required volumetric air flow rate (m³/hr), convert into cfm.

Calculations: 

Step 1: Desired condition: DB = 26 °C, RH = 87%

Step 2: Humidity ratio of

1. Ambient condition (⍵_a) = 13 g H2Okg air
2. Desired condition (⍵_d) = 18 g H2Okg air

Δw = ⍵_d – ⍵_a = 5 g H2Okg air

Step 3: Flow rate of air per TR (kghr-TR) =_ma (kg H2OTR) 1000 (g H2Okg H2O)Δ⍵ (g H2Okg air)

= 5.6 10005         

= 1,120 kghr-TR

Step 4: Volumetric air flow rate per TR (cfmTR) = flow rate of air (kghr-TR) 35.32 (ft³m³) 60 (minhr) density of air (kgm³) 

   = 1,120 x 35.321.12 x 60  

   = 588.6 cfm/TR

Image:

Humidity Ratios for different air conditions

References

None

Source of Idea