Fair Conditioning Energy Saving from Higher Temperature Setting – Fair Conditioning

Energy Saving from Higher Temperature Setting

Subject

Building Services

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Climatology

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Academic Year

2'nd Year

,

3'rd Year

,
Complexity Level

No Complexity Level available

Content tags

Air Conditioning

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Climate Justice

,

Energy Consumption

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Psychrometry

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Activity Type

Numerical Exercise

,
Activity duration

15 mins

,
Objective

To estimate the energy savings from an AC by maintaining the thermostat at a higher temperature.

Outcome

Students will realise that considerable energy savings can be achieved just through setting higher temperature in cooled spaces.  

Requirements

Psychrometric chart

Prerequisites

Conceptual knowledge of psychrometry.

Given Data:

Design Conditions (peak summer)

  1. Ambient condition : DB = 40°C, RH = 60%
  2. Desired condition for scenario 1 : DB = 22°C, RH = 50%
  3. Desired condition for scenario 2 : DB = 25°C, RH = 50%
  4. Density of air : 1.12 kg/m3
  5. 4 air changes per hour
  6. 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
  4. 1 TR = 3.517 kW of cooling
Procedure

Step 1: On the psychrometric chart, estimate Enthalpy (H) of ambient air condition and desired conditions for both scenarios (kJ/kg) and find the differences (∆H1 & ∆H2).

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

Step 3: Convert volumetric air flow rate (m3/hr) to mass flow rate (kg/hr).

Step 4: Estimate cooling required for scenario 1 (kJ/s).

Step 5: Estimate cooling required for scenario 2 (kJ/s).

Step 6: Estimate the percentage energy savings.

Calculations:

Step 1: Enthalpy of

  1. Ambient condition (Ha) = 114kJkg
  2. Desired condition 1 (Hd1) = 42kJkg
  3. Desired condition 2 (Hd2) = 50kJkg

ΔH1 = Ha – Hd1 = 114-42 = 72kJkg 

ΔH2 = Ha – Hd2 = 114-50 = 64kJkg

Step 2: volumetric air flow rate (va) (m3hr) = air changes per hour (1hr) volume of the room (ft3)35.32 (ft3m3)
= 4 120 1035.32  
= 135.84m3hr 

Step 3: mass flow rate (ma) (kghr) = va (m3hr) x density of air (kgm3)

= 135.8 x 1.12

= 152.1kghr

Step 4: Cooling required for case 1 (L1) (kJs) =ma (kghr) ΔH1(kJkg)3600 (shr) 

=152.14 723600 

= 3.043kJs

Step 5: Cooling required for case 2 (L2) (kJs) =ma (kghr) ΔH2(kJkg)3600 (shr) 

=152.14 643600 

= 2.705kJs

Step 6: Percentage energy savings = L1 – L2L1x 100

= 3.043 – 2.7053.043

=  11.1% 

Image:

References

None

Source of Idea

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