1. Why we need to calculate heating load ?
To find how much energy we are going to use?
To size the HVAC equipment?
To estimate the cost of the building?
A and C
B and C
A, B and C

2. Objectives Finish the heating Load Calculation
Learn and Practice the calculation of cooling load of buildings

3. Heat transfer in the building Not only conduction and convection !

4. Infiltration Q = m × cp × ΔT [BTU/hr, W]
Air transport Sensible energy
Previously defined
Q = m × cp × ΔT [BTU/hr, W]
ΔT= T indoor – T outdoor
or Q = 1.1 BTU/(hr CFM °F) × V × ΔT [BTU/hr]

5. Latent Infiltration and Ventilation
Can either track enthalpy and temperature and separate latent and sensible later:
Q total = m × Δh [BTU/hr, W]
Q latent = Q total - Q sensible = m × Δh - m × cp × ΔT
Or, track humidity ratio:
Q latent = m × Δw × hfg

6. Ventilation Example Supply 500 CFM of outside air to our classroom
Outside 90 °F 61% RH
Inside 75 °F 40% RH
What is the latent load from ventilation?
Q latent = m × hfg × Δw
Q = ρ × V × hfg × Δw
Q = lbair/ft3 × 500 ft3/min × 1076 BTU/lb × ( lbH2O/lbair lbH2O/lbair) × 60 min/hr
Q = kBTU/hr

7. What is the difference between ventilation and infiltration?
Ventilation refers to the total amount of air entering a space, and infiltration refers only to air that unintentionally enters.
Ventilation is intended air entry into a space. Infiltration is unintended air entry.
Infiltration is uncontrolled ventilation.

8. Ground Contact Receives less attention:
3-D conduction problem
Ground temperature is often much closer to indoor air temperature
Use F- value for slab floor [BTU/(hr °F ft)]
Note different units from U-value
Multiply by slab edge length
Add to ΣUA
Still need to include basement wall area
Tao and Janis Tables 2.10 and 2.11
More details in ASHRAE handbook -Chapter 29

9. Ground Contact 3-D conduction problem
Ground temperature is often much closer to indoor air temperature
Use F- value for slab floor
Multiply by slab edge length
and Add to ΣUA

10. Where do you get information about amount of ventilation required?
ASHRAE Standard 62
Table 2
Tao and Janis Table 2.9A

11. Weather Data Table 2-2A (Tao and Janis) or
Chapter 28 of ASHRAE Fundamentals
For heating use the 99% design DB value
99% of hours during the winter it will be warmer than this Design Temperature
Elevation, latitude, longitude

12. Weather Data For cooling use the 1% DB and
coincident WB for load calculations
1% of hours during the summer will be warmer than this Design Temperature
Use the 1% design WB for specification of equipment

13. Cooling Load Calculation

14. 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)

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

16. 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

17. 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

18. 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

19. Low- coatings

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

21. 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

22. The latent load of a building is needed to calculate __________.
Heating only
Cooling only
Heating and cooling

23. 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

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

25. 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

26. Reading Assignment
Readings: Tao and Janis
Chapter 2

27. 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: Horizontal slider, Manufacturer:  American Window Alliance, Inc,
CDP number AMW-K ttp://cpd.nfrc.org/pubsearch/psMain.asp
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.

28. Example problem Internal loads: Infiltration:
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.
1.5 W/ft2 heat gain from suspended fluorescent lights from 8:00 A.M. to 5:00 P.M.
0.3 W/ft2 heat gain from suspended fluorescent lights from 5:00 P.M. to 8:00 A.M.
Infiltration:
0.5 ACH per hour

29. 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.

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

31. 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.

32. Solution
On the board

33. Example 2 How to calculate Cooling Load for HVAC design
If the room with no outdoor influence has 4 lighting fixtures with 100 W each and 10 students,
what is the needed relative humidity and temperature of supply air if only required amount of fresh air is supplied
and room temperature is 75 F and RH 50%?