1) Why do we calculate heating and cooling loads? A)To estimate amount of energy used for heating...
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Transcript of 1) Why do we calculate heating and cooling loads? A)To estimate amount of energy used for heating...
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