1. Lecture Objectives Learn about Cooling Load Calculation
- Solve an example

2. Solar Gain
Affects conductive heat gains because outside surfaces get hot
Use Q = U·A·ΔT
Replace ΔT with TETD – total equivalent temperature differential
Q = U·A· TETD
Tables 2-12 – 2-14 in Tao and Janis
Replace ΔT with CLTD (Tables 1 and 2 Chapter 29 of ASHRAE Fundamentals)

3. Solar Gain TETD depends on: orientation, time of day, wall properties
surface color
thermal capacity

4. Glazing Q = U·A·ΔT+A×SC×SHGF Calculate conduction normally Q = U·A·ΔT
Use U-values from NFRC National Fenestration Rating Council
ALREADY INCLUDES AIRFILMS
Use the U-value for the actual window that you are going to use
Only use default values if absolutely necessary
Tao and Janis - no data
Tables 4 and 15, Chapter 31 ASHRAE Fundamentals

5. Shading Coefficient - SC
Ratio of how much sunlight passes through relative to a clean 1/8” thick piece of glass
Depends on
Window coatings
Actually a spectral property
Frame shading, dirt, etc.
Use the SHGC value from NFRC for a particular window
SC=SHGC/0.87
Lower it further for blinds, awnings, shading, dirt

6. More about Windows Spectral coatings (low-e) Tints Polyester films
Allows visible energy to pass, but limits infrared radiation
Particularly short wave
Tints
Polyester films
Gas fills
All improve (lower) the U-value

7. Low- coatings

8. Internal gains What contributes to internal gains? How much?
What about latent internal gains?

9. Internal gains Tao and Janis - People only - Table 2.17
ASHRAE Fundamentals ch. 29 or handouts
Table 1 – people
Table 2 – lighting, Table 3 – motors
Table 5 – cooking appliances
Table Medical, laboratory, office

10. Summary: Heating and cooling loads
Heating - Everything gets converted to a UA, UF, mcp
Sum and multiply it by the design temperature difference
Cooling loads have additional components
Internal gains
Solar gain
Increased gain through opaque surfaces
Also need to calculate latent cooling load

11. Heating and Cooling Load Procedures
Handout
Calculate heating load
Calculate cooling load
Need to also calculate latent cooling load

12. Conclusions
Conduction and convection principles can be used to calculate heat loss for individual components
Air transport principles used to account for infiltration and ventilation
Radiation for solar gain and increased conduction
Include sensible and internal gains

13. Example problem
Calculate the cooling load for the building in Pittsburgh PA with the geometry shown on figure. On east north and west sides are buildings which create shade on the whole wall.
Windows: Champion Window CDP number CHW-A-8
Walls: 4” face brick + 2” insulation + 4” concrete block, Uvalue = 0.1, Dark color
Roof: 2” internal insulation + 4” concrete , Uvalue = , Dark color
Below the building is basement wit temperature of 75 F.
Internal design parameters:
air temperature 75 F
Relative humidity 50%
Find the amount of fresh air
that needs to be supplied by
ventilation system.

14. Example problem (continuing)
Internal loads:
10 occupants, who are there from 8:00 A.M. to 5:00 P.M. doing moderately active office work
1 W/ft2 heat gain from computers and other office equipment from 8:00 A.M. to 5:00 P.M.
0.2 W/ft2 heat gain from computers and other office equipment from 5:00 P.M. to 8:00 A.M.
0.5 W/ft2 heat gain from suspended fluorescent lights from 8:00 A.M. to 5:00 P.M.
0.1 W/ft2 heat gain from suspended fluorescent lights from 5:00 P.M. to 8:00 A.M.
Infiltration:
0.5 ACH per hour

15. Example solution SOLUTION steps (see handouts):
1. Calculate cooling load from conduction through opaque surfaces using TETD.
2. Calculate conduction and solar transmission through windows.
3. Add sensible internal gains and infiltration.
4. The result is your raw sensible cooling load.
5. Calculate latent internal gains.
6. Calculate latent gains due to infiltration.
7. The sum of 5 and 6 is your raw latent cooling load.

16. Example solution SOLUTION: For which hour to do the calculation ?
With computer calculation for all and select the largest.

17. Example solution
For which hour to do the calculation when you do manual calculation?
Identify the major single contributor to the cooling load and do the calculation for the hour when the maximum cooling load for this contributor appear.
For example problem major heat gains are
through the roof or solar through windows!
Roof: maximum TETD=61F at 6 pm (Table 2.12)
South windows: max. SHGF=109 Btu/hft2 at 12 am (July 21st Table 2.15 A)
If you are not sure, do the calculation for both hours:
at 6 pm
Roof gains = A x U x TETD = 900 ft2 x 0.12 Btu/hFft2 x 61 F = 6.6 kBtu/h
Window solar gains = A x SC x SHGF =80 ft2 x 0.71 x 10 Btu/hft2 = 0.6 kBtu/h total = 7.2 kBtu/h
at 12 am
Roof gains = A x U x TETD = 900 ft2 x 0.12 Btu/hFft2 x 30 F = 3.2 kBtu/h
Window solar gains = A x SC x SHGF =80 ft2 x 0.71 x 109 Btu/hft2 = 6.2 kBtu/h total= 9.4 kBtu/h
For the example critical hour is July 12 AM.

18. Solution
On the board