1. Objectives Review heating and cooling load calculation
Practice the calculation of cooling load
Learn about heating systems

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

3. 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

4. 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

5. 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

6. Low- coatings

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

8. 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

9. 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

10. 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
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.

11. 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

12. 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.

13. Heating systems

14. Choosing a Heating System
What is it going to burn?
What is it going to heat?
How much is it going to heat it?
What type of equipment?
Where are you going to put it?
What else do you need to make it work?

16. Choosing a Fuel Type Availability Storage Cost Code restrictions
Emergencies, back-up power, peak demand
Storage
Space requirements, aesthetic impacts, safety
Cost
Capital, operating, maintenance
Code restrictions
Safety, emissions

17. Selecting a Heat Transfer Medium
Air
Not very effective (will see later)
Steam
Necessary for steam loads, little/no pumping
But: lower heat transfer, condensate return, bigger pipes
Water
Better heat transfer, smaller pipes, simpler
But: requires pumps, lower velocities, can require complex systems

18. Choosing Water Temperature
Low temperature water (180 °F – 240 °F)
single buildings, simple
Medium and high temperature (over 350 °F)
Campuses where steam isn’t viable/needed
Requires high temperature and pressure equipment
Nitrogen system to prevent steam formation

19. Choosing Steam Pressure
Low pressure (<15 psig)
No pumping for steam
Requires pumping/gravity for condensate
Medium and high-pressure systems
Often used for steam loads

20. Steam Systems Steam needs bigger pipes for same heat transfer
Water is more dense and has better heat transfer properties
You can use steam tables and water properties to calculate heat transfer
Vary design parameters

21. What About Air? Really bad heat transfer medium But !
Very low density and specific heat
Requires electricity for fans to move air
Excessive space requirements for ducts
But !
Can be combined with cooling
Lowest maintenance
Very simple equipment
Still need a heat exchanger

23. Furnace Load demand, load profile Efficiency Combustion air supply
Amount and type of heat
Response time
Efficiency
80 – 85 % is typical
Electricity is ~100 %
Combustion air supply
Flue gas discharge (stack height)

24. Choosing a Boiler Fuel source Transfer medium
Operating temperatures/pressures
Equipment
Type
Space requirements
Auxiliary systems

25. Water Boilers Types Water Tube Boiler Fire Tube Boiler
Water in tubes, hot combustion gasses in shell
Quickly respond to changes in loads
Fire Tube Boiler
Hot combustion gasses in tubes, water in shell
Slower to respond to changes in loads

27. Electric Types Resistance Electrode
Resistor gets hot
Typically slow response time (demand issues)
Electrode
Use water as heat conducting medium
Bigger systems
Cheap to buy, very expensive to run
Clean, no local emissions

28. Auxiliary Burner type (atmospheric or power vented) Feedwater systems
Returns steam condensate (including accumulator)
Adds water to account for blowdown and leaks
Preheats the water
Removes dissolved gasses
Blowdown system
Periodically drain and cool water

29. Auxiliary Water treatment Treatment options
Dissolved minerals and gasses cause:
Reduced heat transfer
Reduced flow (increased pressure drop)
Corrosion
Treatment options
Chemical (add bases, add ions, add inhibitor)
Temperature (heat to remove oxygen)

30. Location Depends on type Aesthetics Stack height
Integration with cooling systems

31. Reading Assignment
Tao and Janis Chapter 5