Energy Saving from Higher Temperature Setting
Subject
Building Services
,Climatology
,Academic Year
2'nd Year
,3'rd Year
,Complexity Level
No Complexity Level available
Content tags
Air Conditioning
,Climate Justice
,Energy Consumption
,Psychrometry
,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)
- Ambient condition : DB = 40°C, RH = 60%
- Desired condition for scenario 1 : DB = 22°C, RH = 50%
- Desired condition for scenario 2 : DB = 25°C, RH = 50%
- Density of air : 1.12 kg/m3
- 4 air changes per hour
- 120 sq. ft. room, 10 ft. ceiling height
Conversion Factors:
- 1 kW = 1 kJ/s
- 1 hr = 3,600 s
- 1 m3 = 35.32 ft3
- 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
- Ambient condition (Ha) = 114kJkg
- Desired condition 1 (Hd1) = 42kJkg
- 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