Lecture Objectives Learn about Cooling Load Calculation - Solve an example

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)

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

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

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

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

Low- coatings

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

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 6 -10 Medical, laboratory, office

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

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

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

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 https://www.energystar.gov/products/most_efficient/horizontal_slider_windows 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.

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

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.

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

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.

Solution On the board