1) Why do we calculate heating and cooling loads? A)To estimate amount of energy used for heating and cooling by a building B)To size heating and cooling equipment for a building C)Because Dr. Siegel tells us to

A note about units Quiz #6 question asked for the humidity ratio No one included units Every quantity that has dimensions needs to have those units included Failure to included units will be counted as a wrong answer in the future

Objectives Review 1-D conduction Use knowledge of heat transfer to calculate heating and cooling loads Conduction Ventilation Ground contact Solar gains Internal gains

1-D Conduction 90 °F 70 °F q heat transfer rate [W, BTU/hr] q heat flux [W/m2, BTU/(hr ft2)] k conductivity [W/(m °C), BTU/(hr ft °F)] l length [m, cm, ft, in] ΔTtemperature difference [°C, K, R, °F] A surface area [m 2, ft 2 ] l k A U = k/l = 1 / R U-Value [W/(m 2 °C), BTU/(hr ft 2 °F)] R = l/k = 1 / U R-Value [m 2 °C/W, hr ft 2 °F/BTU]

Material k Values Materialk W/(m K) 1 Steel Soil0.52 Wood Fiberglass Polystyrene At 300 K Incropera and DeWitt (2002) Appendix A

1-D Conduction qheat transfer rate [W] k conductivity [W/(m °C)] l length [m] 90 °F 70 °F l k A U = k/l ΔTtemperature difference [°C] Asurface area [m 2 ] U U-Value [W/(m 2 °C)] q = UAΔT

90 °F 70 °F l1l1 k1k1 k2k2 l2l2 R = l/k q = (A/R total )ΔT Add resistances in series Add U-values in parallel R1/AR1/AR2/AR2/A T out T in T mid

2) To find a total U value for a wall, which of the following formulas is not acceptable? A)U total = 1/R total B)U total = U 1 +U 2 +U 3 C)U total = 1/R 1 +1/R 2 +1/R 3 D)U total = 1/(1/U 2 +1/U 2 +1/U 3 ) E)B and C

T out T in R1/AR1/A R2/AR2/A Ro/ARo/A T out Ri/ARi/A T in Air film (Table 25-1) Surface conductance Convection heat transfer coefficient Use ε = 0.9 Direction/orientation Air speed Air space (Table 25-3) Orientation Thickness

The Drape Defense hs/ hs/ Do drapes limit heat loss? Do drapes limit heat gain?

l1l1 k 1, A 1 k 2, A 2 l2l2 l3l3 k 3, A 3 A 2 = A 1 (l 1 /k 1 )/A 1 R 1 /A 1 T out T in (l 2 /k 2 )/A 2 R 2 /A 2 (l 3 /k 3 )/A 3 R 3 /A 3 1.Add resistances for series 2.Add U-Values for parallel

R1/A1R1/A1 T out T in R2/A2R2/A2 R3/A3R3/A3 1.R 1 /A 1 + R 2 /A 2 = (R 1 + R 2 ) /A 1 = R 12 /A 1 =1/(U 12 A 1 ) 2.R 3 /A 3 =1/(U 3 A 3 ) 3.U 3 A 3 + U 12 A 1 4.q = (U 3 A 3 + U 12 A 1 )ΔT A 1 =A 2

U 1 A 1 U 2 (A 2 +A 4 ) U 3 A 3 U 5 A 5 Relationship between temperature and heat loss

3) Which of the following statements about a material is true? A)A high U-value is a good insulator, and a high R-value is a good conductor. B)A high U-value is a good conductor, and a high R-value is a good insulator. C)A high U-value is a good insulator, and a high R-value is a good insulator. D)A high U-value is a good conductor, and a high R-value is a good conductor.

Example Consider a 1 ft × 1 ft × 1 ft box Two of the sides are 1” thick extruded expanded polystyrene foam The other four sides are 1” thick plywood The inside of the box needs to be maintained at 40 °F The air around the box is still and at 80 °F How much cooling do you need?

4)What is the R-value of 1” of plywood? A.0.62 BTU/(hr∙°F∙ft 2 ) B.1.24 BTU/(hr∙°F∙ft 2 ) C.0.81 hr∙°F∙ft 2 /BTU D.0.62 hr∙°F∙ft 2 /BTU E.1.24 hr∙°F∙ft 2 /BTU

5)What is the U-value of the plywood walls? A.0.81 BTU/(hr∙°F∙ft 2 ) B.0.38 BTU/(hr∙°F∙ft 2 ) C.1.24 BTU/(hr∙°F∙ft 2 ) D.0.48 BTU/(hr∙°F∙ft 2 )

The Moral of the Story 1.Calculate R-values for each series path 2.Convert them to U-values 3.Find the appropriate area for each U-value 4.Multiply U-value i by Area i 5.Sum UA i 6.Calculate q = UA total ΔT

Infiltration (Convection) Air carries sensible energy q = M × C × ΔT [BTU/hr, W] M mass flow rate = ρ × Q [lb/hr, kg/s] ρ air density (0.076 lb/ft 3, 1.2 kg/m STP) Q volumetric flow rate [CFM, m 3 /s] C specific heat of air 0.24 BTU/(lb °F), 1007 kJ/(kg STP For similar indoor and outdoor conditions ρ and C are often combined q = 1.08 BTU min/(ft 3 °F hr ) × Q × ΔT

Latent Infiltration and Ventilation Can either track enthalpy and temperature and separate latent and sensible later q = M × ΔH[BTU/hr, W] Or, track humidity ratio q = M × h fg × ΔW h fg = ~1076 BTU/lb, 2.5 kJ/kg M = ρ × Q [lb/hr, kg/s]

Ventilation Example Supply 500 CFM of outside air to our classroom Outside 90 °F61% RH Inside 75 °F40% RH What is the latent load from ventilation? q = M × h fg × ΔW q = ρ × Q × h fg × ΔW q = lb air /ft 3 × 500 ft 3 /min × 1076 BTU/lb × ( lb H2O /lb air lb H2O /lb air ) × 60 min/hr q = 26.3 kBTU/hr

6) What is the difference between ventilation and infiltration? A)Ventilation refers to the total amount of air entering a space, and infiltration refers only to air that unintentionally enters. B)Ventilation is intended air entry into a space. Infiltration is unintended air entry. C)The terms can be used interchangeably.

Where do you get information about amount of ventilation required? ASHRAE Standard 62 Table 2 Available on website from library Hotly debated – many addenda and changes

Ground Contact Receives less attention: 3-D conduction problem Ground temperature is often much closer to indoor air temperature Use F- value (from simulations) [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 WA State Energy Code heat loss tables

Weather Data Chapter 27 of ASHRAE Fundamentals For heating use the 99% DB value 99% of hours during the winter it will be warmer than this Design Temperature Elevation, latitude, longitude Heating dry-bulb –99.6% and 99% values

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 Facing page 0.4%, 1%, 2% cooling DB and MWB 0.4%, 1%, 2% cooling WB and MDB

Solar Gain Increased conduction because outside surfaces got hot Use q = UAΔT 1.Replace ΔT with TETD Tables 2-11 – 2-13 in Tao and Janis (2001) 4 pm for a dark colored surface 2.Replace ΔT with CLTD (Tables 1 and 2 Chapter 28 of ASHRAE Fundamentals) 3.Sol-air temperature Table (example)

Glazing q = UAΔT+A×SC×SHGF Calculate conduction normally q = UAΔT Use U-values from NFRC Certified Products Directory ALREADY INCLUDES AIRFILMS Use the U-value for the actual window that you are going to use Only use default values if absolutely necessary (Tables 4 and 15, Chapter 30 ASHRAE Fundamentals)

Solar Gain Through Windows Add to conduction A× SHGF × SC SHGF = solar heat gain factor Measure of how much energy comes through an average “perfect” window Depends on –Latitude –Orientation –Time of Day –Time of Year Tabulated in ASHRAE Fundamentals 1997 Chapter 29 Table 15 Tao and Janis Table 2-15 for 40° latitude (July 8 am)

Shading Coefficient 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 Lower it further for dirt, blinds, awnings, shading

More about Windows Spectral coatings (low-e) Allows visible energy to pass, but limits infrared radiation Particularly short wave Can see it with a match/lighter in older windows 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?

Conclusions Conduction and convection principles can be used to calculate heat loss for individual components Convection principles used to account for infiltration and ventilation Radiation for solar gain and increased conduction Include sensible and internal gains