objectives lean about energy transport by air calculate cooling and heating loads solve 1-d...
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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 Δhsensible = cp × ΔT [Btu/lb]
cp - specific heat for air (for air 0.24 Btu/lb°F)
• Heat transfer (rate) Qs = m × cp × ΔT [Btu/h]
m - mass flow rate [lb/min, lb/h], m = V × V – volume flow rate [ft3/min or CFM]
– airdensity (0.076lb/ft3)
Qs = 1.1 × CFM × ΔT (only for IP unit system)
Equations for latent energy transport by air
• Energy per unit of mass Δhlatent = Δw × hfg [Btu/lbda]
hfg - specific energy of water phase change (1000 Btu/lbw)
• Heat transfer (rate) Ql = m × Δw × hfg [Btu/h]
Ql = 1000 × WaterFloowRate (only for IP units)
Total energy transport calculation using enthalpies from chat
• Energy per unit of mass Δh=h1-h2 [Btu/lbda]
• Heat transfer (rate) Qtotal = m × Δh [Btu/h]
Qtotal = Qsensible + Qlatent
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
Q heat transfer rate [W]
TR
AQ
RU ,
1
k conductivity [W/(m °C)]l length [m]
90 °F
70 °F
lk
AU = k/l
ΔT temperature difference [°C]A surface area [m2]
U U-Value [W/(m2 °C)]
Q = UAΔT
Material k Values
Material k [W/(m K)]1
Steel 64 - 41
Soil 0.52
Wood 0.16 - 0.12
Fiberglass 0.046 - 0.035
Polystyrene 0.0291At 300 K
Table 2-3 Tao and Janis (k=λ) values in [Btu in/(h ft2 F)]
90 °F
70 °F
l1k1 k2
l2• R = l/k
• Q = (A/Rtotal)ΔT
• Add resistances in series• Add U-values in parallel
2
2
1
1
2121
111
k
l
k
lR
UUURRR
total
totaltotal
R1 R2
Tout TinTmid
Wall assembly
Tout
Tin
R1 R2Ro
Tout
Ri
Tin
•Surface Air Film h - convection coefficient - surface conductance [W/m2, Btu/(h ft2)]
•Direction/orientation•Air speed
• Table 2-5 Tao and Janis
Rtotal= ΣRi
Rsurface= 1/h
l1k1, A1 k2, A2
l2
l3
k3, A3
A2 = A1
What if more than one surface?
Q3
Q1,2
Qtotal = Q1,2 + Q3
U1,2 = 1/R 1,2=1/(R1+R2)
Q3 = A3U3ΔT
Q1,2 = A1U1,2ΔT
U1A1
U2A2
U3(A3+A5)
U4A4
U5A5
Qtotal= Σ(UiAi)·ΔT
Relationship between temperature and heat loss
A1
A5
A4
A3A2
A6
TinTout
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-valuei by Areai
5. Sum UAi
6. Calculate Q = Σ(UAi)ΔT
Heat transfer in the building Not only conduction and convection !
Infiltration
• Air transport Sensible energy
Previously defined
• Q = m × cp × 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 × cp × ΔT
• Or, track humidity ratio:• Q latent = m × Δw × hfg
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
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 floorMultiply 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• 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
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/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 6 -10 Medical, laboratory, office
• Tao and Janis - People only - Table 2.17
Readings:
• Tao and Janis 2.4-2.8.10