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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 TUATl
kAq
q heat transfer rate [W, BTU/hr]q heat flux [W/m2, BTU/(hr ft2)]
TR
Aq
k conductivity [W/(m °C), BTU/(hr ft °F)]l length [m, cm, ft, in]ΔT temperature difference [°C, K, R, °F]A surface area [m2, ft2]
lk
A
U = k/l = 1/R U-Value [W/(m2 °C), BTU/(hr ft2 °F)]R = l/k = 1/U R-Value [m2 °C/W, hr ft2 °F/BTU]
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 0.029
1At 300 KIncropera and DeWitt (2002) Appendix A
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
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/A R2/A
Tout TinTmid
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 = U1+U2+U3
C) U total = 1/R1+1/R2+1/R3
D) U total = 1/(1/U2+1/U2+1/U3)
E) B and C
Tout
Tin
R1/A R2/ARo/A
Tout
Ri/A
Tin
•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
• http://utwired.engr.utexas.edu/conservationmyths/
• Do drapes limit heat loss?
• Do drapes limit heat gain?
l1k1, A1 k2, A2
l2
l3
k3, A3
A2 = A1
(l1/k1)/A1
R1/A1
ToutTin
(l2/k2)/A2
R2/A2
(l3/k3)/A3
R3/A3
1. Add resistances for series
2. Add U-Values for parallel
R1/A1
ToutTin
R2/A2
R3/A3
1. R1/A1 + R2/A2 = (R1 + R2) /A1 = R12 /A1=1/(U12A1)
2. R3 /A3=1/(U3A3)
3. U3A3 + U12A1
4. q = (U3A3 + U12A1)ΔT
A1=A2
U1A1
U2(A2+A4)
U3A3
U5A5
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∙ft2)
B. 1.24 BTU/(hr∙°F∙ft2)
C. 0.81 hr∙°F∙ft2/BTU
D. 0.62 hr∙°F∙ft2/BTU
E. 1.24 hr∙°F∙ft2/BTU
5)What is the U-value of the plywood walls?
A. 0.81 BTU/(hr∙°F∙ft2)
B. 0.38 BTU/(hr∙°F∙ft2)
C. 1.24 BTU/(hr∙°F∙ft2)
D. 0.48 BTU/(hr∙°F∙ft2)
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 = UAtotalΔT
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/ft3, 1.2 kg/m3 @ STP)
• Q volumetric flow rate [CFM, m3/s]
• C specific heat of air• 0.24 BTU/(lb °F), 1007 kJ/(kg K) @ STP
• For similar indoor and outdoor conditions• ρ and C are often combined
• q = 1.08 BTU min/(ft3 °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 × hfg × ΔW
• hfg = ~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 °F 61% RH
• Inside 75 °F 40% RH
• What is the latent load from ventilation?• q = M × hfg × ΔW
• q = ρ × Q × 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
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 = UAΔT1. 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 29-15 (example)
Glazing
• q = UAΔT+A×SC×SHGF• Calculate conduction normally q = UAΔT
• Use U-values from NFRC Certified Products Directory• ALREADY INCLUDES AIRFILMS
• http://www.nfrc.org/nfrcpd.html
• 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 21 @ 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
•http://www.nfrc.org/nfrcpd.html
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