Quiz

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? 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

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 l k A 90 °F 70 °F 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 At 300 K Table 2-3Tao 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. Surface Air Film 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 Ri Ro R1 R2 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. 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 • Air transport Sensible energy Previously defined • Q = m× cp × ΔT [BTU/hr, W] • ΔT= T indoor – T outdoor • or Q = 1.1BTU/(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 = 0.076 lbair/ft3 × 500 ft3/min × 1076 BTU/lb × (0.01867 lbH2O/lbair - .00759 lbH2O/lbair) × 60 min/hr • Q = 26.3 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 • http://cpd.nfrc.org/pubsearch/psMain.asp • 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 • http://cpd.nfrc.org/pubsearch/psMain.asp

27. More about Windows • Spectral coatings (low-e) • 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 • ASHRAE Fundamentals ch. 29 or handouts • Table 1 – people • Table 2 – lighting, Table 3 – motors • Table 5 – cooking appliances • Table 6 -10 Medical, laboratory, office • Tao and Janis • - People only - Table 2.17

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-3-00028 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 = 0.120 , 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: • 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 theroof 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 61F = 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 30F = 3.2 kBtu/h Window solar gains = A x SC x SHGF =80 ft2 x 0.71 x 109Btu/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%?