Objectives Lean about energy transport by air Calculate Cooling and Heating loads Solve 1-D conduction Design whether condition Use knowledge of heat transfer.

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Presentation transcript:

Objectives Lean about energy transport by air Calculate Cooling and Heating loads Solve 1-D conduction Design whether condition Use knowledge of heat transfer to calculate Solar gains Internal gains

Equations for sensible energy transport by air Energy per unit of mass Δh sensible = c p × ΔT [Btu/lb] c p - specific heat for air (for air 0.24 Btu/lb°F) Heat transfer (rate) Q s = m × c p × ΔT [Btu/h] m - mass flow rate [lb/min, lb/h], m = V ×  V – volume flow rate [ft 3 /min or CFM]  – air  density (0.076lb/ft 3 ) Q s = 1.1 × CFM × ΔT (only for IP unit system)

Equations for latent energy transport by air Energy per unit of mass Δh latent = Δw × h fg [Btu/lb da ] h fg - specific energy of water phase change (1000 Btu/lb w ) Heat transfer (rate) Q l = m × Δw × h fg [Btu/h] Q l = 1000 × WaterFloowRate (only for IP units)

Total energy transport calculation using enthalpies from chat Energy per unit of mass Δh=h 1 -h 2 [Btu/lb da ] Heat transfer (rate) Q total = m × Δh [Btu/h] Q total = Q sensible + Q latent

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 my supervisor request that Heating and Cooling Loads

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

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

Material k Values Materialk [W/(m K)] 1 Steel Soil0.52 Wood Fiberglass Polystyrene At 300 K Table 2-3 Tao and Janis (k=λ) values in [Btu in/(h ft 2 F)]

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 R1R1 R2R2 T out T in T mid Wall assembly

T out T in R1R1 R2R2 RoRo T out RiRi T in Surface Air Film h - convection coefficient - surface conductance [W/m 2, Btu/(h ft 2 )] Direction/orientation Air speed Table 2-5 Tao and Janis R total = ΣR i R surface = 1/h

l1l1 k 1, A 1 k 2, A 2 l2l2 l3l3 k 3, A 3 A 2 = A 1 What if more than one surface? Q3Q3 Q 1,2 Q total = Q 1,2 + Q 3 U 1,2 = 1/R 1,2 =1/(R 1 +R 2 ) Q 3 = A 3 U 3 ΔT Q 1,2 = A 1 U 1,2 ΔT

U 1 A 1 U 2 A 2 U 3 (A 3 +A 5 ) U 4 A 4 U 5 A 5 Q total = Σ(U i A i )·ΔT Relationship between temperature and heat loss A1A1 A5A5 A4A4 A3A3 A2A2 A6A6 T in T out

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 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?

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 = Σ(UA i )ΔT

Heat transfer in the building Not only conduction and convection !

Infiltration Air transport Sensible energy Previously defined Q = m × c p × ΔT [BTU/hr, W] ΔT= T indoor – T outdoor or Q = 1.1 BTU/(hr CFM °F) × V × ΔT [BTU/hr]

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 × c p × ΔT Or, track humidity ratio: Q latent = m × Δw × h fg

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 latent = m × h fg × Δw Q = ρ × V × 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

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)Infiltration is uncontrolled ventilation.

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

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

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

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

Where do you get information about amount of ventilation required? ASHRAE Standard 62 Table 2 Tao and Janis Table 2.9A

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

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 Weather Data

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 _default.aspx?type=Whttp://cpd.nfrc.org/search/cpd/cpd_search _default.aspx?type=W

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

Low-  coatings

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

Internal gains ASHRAE Fundamentals ch. 29 or handouts Table 1 – people Table 2 – lighting, Table 3 – motors Table 5 – cooking appliances Table Medical, laboratory, office Tao and Janis - People only - Table 2.17

Readings: Tao and Janis