Cooling Load Estimation of a Space

Objective

To estimate the AC Tonnage (TR) required for combating the typical fresh-air cooling load in a bedroom.

Outcome

Students will understand the procedure to be followed in calculating heat load requirements of a built space.

Requirements

Psychrometric chart

Prerequisites

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

Given Data:

Design Conditions (peak summer)

1. Ambient : DB = 40°C, RH = 60%
2. Desired : DB = 24°C, RH = 50%
3. Density of air : 1.12 kg/m³
4. 4 air changes per hour
5. 120 sq. ft. room, 10 ft. ceiling height

Conversion Factors:

1. 1 kW = 1 kJ/s
2. 1 hr = 3,600 s
3. 1 m³ = 35.32 ft³

1 TR = 3.517 kW of cooling

Procedure

Step 1: Estimate Enthalpy (H) of ambient condition and desired air conditions (kJ/kg) and find the difference (∆H).

Step 2: Estimate required volumetric air flow rate (m³/hr).

Step 3: Convert to mass flow rate (kg/hr).

Step 4: Estimate cooling required (kJ/s).

Step 5: Convert kJ/s to TR.

Calculations:

Step 1: Enthalpy of

1. Ambient condition (H_a) = 114kJkg
2. Desired condition (H_d) = 48.5kJkg

ΔH= H_a – H_d = 114-48.5 = 65.5kJkg 

Step 2: Volumetric air flow rate (v_a) (m3hr) = air changes per hour (1hr) volume of the room (ft3)35.32 (ft3m3)

= 4 120 1035.32  

= 135.84 m3hr

Step 3: Mass flow rate (m_a) (kghr) = v_a (m3hr) x density of air (kgm3)

= 135.8 x 1.12

= 152.1kghr

Step 4: Cooling required (kJs) =_ma (kghr) ΔH (kJkg)3600 (shr) 

=152.14 65.53600 

= 2.768kJs

Step 5: Cooling required (TR) =cooling required (kJs)3.517 (kJs-TR) 

= 2.7683.517 

≈ 0.8 TR 

Image:

Enthalpies of air at different conditions

References

None

Source of Idea