Structure Cooling: System Design

Objective

Estimating Structure Cooling Water Flow Rate Requirements for Concrete Slab

Outcome

Students will have hands-on experience of determining the water required for structure cooling and enhance their learning about sustainable cooling systems and technologies which can help them integrate it into their designs and reduce the use of Air Conditioners.

Prerequisites

Knowledge required about Structure Cooling system and intermediate level knowledge about calculations.

Given Data

Design Conditions:

Rooftop Temperature = 45°C

Solar Radiation (peak) = 500 W/m2

10,000 sq. ft. slab

Wet Bulb Temperature  = 20°C

Pipe Material = Polypropylene

Pipe Wall Thickness = 3 mm

Pipe Inside Diameter = 25 mm

Total Pipe Length = 6200 m

Conversion Factors & Constants:

1 m² = 10.76 ft²

Specific Heat Capacity (Water) =4.186 kJ/kg ᵒC

Thermal Conductivity (Polypropylene) = 0.44 W/m ᵒC

Calculations

Step 1: Calculate Rooftop Heat Load (Q, Solar) = 500 wm² x 10,000 ft² x 1 m²10.76 ft2 = 4,64,684 W

Step 2: Calculate Pipe Area: A = 2πrl= 2*3.14*25m*6200m1000*2 = 486.9 m²

Step 3: Calculate R = d (thickness)𝝺(thermal conductivity) = 0.003m0.44 Wm°C = 0.0068 m²CW

Step 4: Calculate U = 1R = 10.0068= 146.67 Wm²°C

Step 5: Calculate LMTD based on known Heat Transfer to Water (Q, Solar) = U x A x LMTD

4,64,684 W= 146.67 Wm²°C  x 486.9 m² x LMTD 

LMTD =6.5 °C

Step 6: Calculate ∆T1 = 45-20 = 25°C

Step 7: Calculate ∆T2 from LTMD =∆T1 – ∆T2ln ∆T1∆T2

6.5= (25-∆T2)ln 25∆T2

Using Spreadsheet: ∆T2 = 0.6

Step 8: Calculate water exit temperature T_COLD-OUT, assuming constant slab temperature:

∆T2 = 45-T_{COLD OUT} 

T_{COLD OUT} = 44.4

References

Fairconditioning Structure cooling ppt and animation video.