Fair Conditioning Air Conditioner’s Water Generation – Fair Conditioning

Air Conditioner’s Water Generation

Subject

Building Services

,

Climatology

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Academic Year

2'nd Year

,
Complexity Level

Advanced

,

Intermediate

,
Content tags

Air Conditioning

,

Cooling Load

,

Energy Efficiency

,

Psychrometry

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Activity Type

Numerical Exercise

,
Activity duration

< 30 mins

,
Objective

To estimate the water generated from the dehumidification of fresh-air cooling load in a bedroom.

Outcome

Students will realize that water generated from dehumidification of air while cooling a room by Airconditioning, can be used for handy purposes (like gardening)

Requirements

Psychrometric chart

Prerequisites
  1. Conceptual knowledge of psychrometry and psychrometric chart
  2. Understanding of the phenomenon of Enthalpy and Humidity ratio.

Given Data:

Design Conditions (peak summer)

  1. Ambient : DB = 40°C, RH = 60%
  2. Desired : DB = 24°C, RH = 50%
  3. Cooling Load : 0.8 TR (calculated in the last activity)
  4. Density of air : 1.12 kg/m3
  5. AC Operation hours (daily average) : 2 hours

DB – Dry Bulb temperature, RH – Relative Humidity

Conversion Factors:

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

Step 1: Estimate the Enthalpies and Specific Humidity Ratios (⍵) of Ambient and Desired Air Conditions (g H2O/kg air) and ∆⍵.

Step 2: Estimate the required volumetric air flow rate (m3 air/hr); convert to mass flow rate (kg air/hr).

Step 3: Estimate cooling required (kg air/hr x g H2O/kg air). 

Step 4: Determine annual water generation by multiplying with hours of AC operation per year.

Step 5: Reflect on ways to capture the water generated from Air conditioners.

Calculations:

Step 1: Enthalpy and Humidity Ratio values

UnitAmbientDesiredDifference
Enthalpy (H)kJkg48.511465.5
Humidity Ratio (w)g H2Okg air29920

Step 2: Mass flow rate of air (ma) (kghr) = (Heat Load (TR) 3.517) (kJs) 3600 (shr)ΔH (kghr)

= 0.8 3.517 360065.5 

= 154.64 kghr

Step 3: Water generated (kghr) (Wg) = mass flow rate of air (kg airhr) x Δ⍵ (g H20kg air)1000 g H2Okg H2O 

= 154.64 22.51000

= 3.48 kghr

Step 4: Annual water generation (kgyr) = Wg (kghr) x average daily operation x 365 (no. 

of days in a year)

= 3.48 x 2 x 365

= 2,540 kgyr

Reference image:

Enthalpies & Humidity Ratios for different air conditions

References

None

Source of Idea

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