1. Quiz
First 10 minutes of the class

2. Lecture Objectives Define Heating and Cooling Loads
Learn how to Calculate
Heating Loads
Cooling Loads

3. Why do we calculate heating and cooling loads?
10/21/2003
Heating and Cooling Loads
Why do we calculate heating and cooling loads?
To estimate amount of energy used for heating and cooling by a building Or
To size heating and cooling equipment for a building
ARE 346N

4. Introduction to Heat Transfer
Conduction
Components
Convection
Air flows (sensible and latent)
Radiation
Solar gains (cooling only)
Increased conduction (cooling only)
Phase change
Water vapor/steam
Internal gains (cooling only)
Sensible and latent

5. 1-D Conduction U = k/l Q = UAΔT l k U U-Value [W/(m2 °C)] A
90 °F
70 °F l k A
U U-Value [W/(m2 °C)]
U = k/l
k conductivity [W/(m °C)]
l length [m]
Q heat transfer rate [W]
ΔT temperature difference [°C]
A surface area [m2]
Q = UAΔT

6. Material k Values Material k [W/(m K)] Steel 64 - 41 Soil 0.52 Wood
Fiberglass
Polystyrene
0.029
At 300 K
Table 2-3 Tao and Janis (k=λ) values in [Btu in/(h ft2 F)]

7. Wall assembly l1 l2 R = l/k Q = (A/Rtotal)ΔT Add resistances in series
Add U-values in parallel k1 k2
90 °F
70 °F R1 R2
Tout
Tmid
Tin

8. Rsurface= 1/h Rtotal= ΣRi Surface Air Film Table 2-5 Tao and Janis
h - convection coefficient
- surface conductance
[W/m2, Btu/(h ft2)]
Direction/orientation
Air speed
Table 2-5 Tao and Janis
Tout
Tin
Rsurface= 1/h Ro R1 R2 Ri
Rtotal= ΣRi
Tout
Tin

9. What if more than one surface?l1 l2
k1, A1
k2, A2
Qtotal = Q1,2 + Q3
Q1,2
A2 = A1
U1,2 = 1/R 1,2=1/(R1+R2)
k3, A3
Q1,2 = A1U1,2ΔT Q3
Q3 = A3U3ΔT l3

10. Tin Tout Qtotal= Σ(UiAi)·ΔT
Relationship between temperature and heat loss
U1A1
U2A2
U3(A3+A5)
U4A4
U5A5 A2 A3 A1 A4
Tin
Tout A5 A6
Qtotal= Σ(UiAi)·ΔT

11. Example Consider a 1 ft × 1 ft × 1 ft box
Two of the sides are 2” thick extruded expanded polystyrene foam
The other four sides are 2” thick plywood
The inside of the box needs to be maintained at 120 °F
The air around the box is still and at 80 °F
How much heating do you need?

12. The Moral of the Story Calculate R-values for each series path
Convert them to U-values
Find the appropriate area for each U-value
Multiply U-valuei by Areai
Sum UAi
Calculate Q = Σ(UAi)ΔT

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

14. 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]

15. 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

16. 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

17. Where do you get information about amount of ventilation required?
ASHRAE Standard 62
Table 2
Hotly debated – many addenda and changes
Tao and Janis Table 2.9A

18. 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

19. 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

20. 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

21. Summary of Heating Loads
Conduction and convection principles can be used to calculate heat loss for individual components
Convection principles used to account for infiltration and ventilation

22. Cooling loads

23. 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)

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

25. 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

26. 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

27. 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

28. Low- coatings

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

30. 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

31. 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

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

33. 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

34. Reading Assignment
Readings: Tao and Janis
Chapter 2

35. 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.

36. 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

37. 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.

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

39. 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.

40. Solution
On the board

41. 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%?