Technology in Architecture
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Technology in ArchitectureTechnology in ArchitectureTechnology in ArchitectureTechnology in Architecture
Lecture 7Degree Days Heating Loads
Annual Fuel ConsumptionSimple Payback Analysis
Lecture 7Degree Days Heating Loads
Annual Fuel ConsumptionSimple Payback Analysis
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Heating Degree DaysHeating Degree Days
Balance Point Temperature (BPT): temperature above which heating is not needed
DDBPT= BPT-TA
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Sample CalculationSample Calculation
January TA=28ºF
DD65=65-28= 37 Degree-days/day
x 31 days = 1,147 degree-days
S: p. 1562, T.C.19
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Heating LoadsHeating LoadsHeating LoadsHeating Loads
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Heating LoadsHeating LoadsComputed for worst case scenario: Pre-dawn at outdoor
design dry bulb temperature
Do not include: Insolation from sun Heat gain from people,
lights, and equipment Infiltration in
nonresidential buildings
Ventilation in residential buildings
SR-3
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Outdoor Dry Bulb Outdoor Dry Bulb TemperatureTemperature
Use Winter Conditions
S(10th): T.B.1 p. 1496
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Determine Temperature Determine Temperature DifferenceDifference
Indoor Dry Bulb Temperature (IDBT): 68ºF
Outdoor Dry Bulb Temperature (ODBT): 8ºF
ΔT=IDBT-ODBT=68ºF - 8ºF = 60ºF
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Determine Envelope U-Determine Envelope U-valuesvalues
Calculate ΣR and then find U for walls, roofs, floors.
Obtain U values for glazing from manufacturer or other reference
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Determine Area QuantitiesDetermine Area Quantities
Perform area takeoffs for all building envelope surfaces on each facade:
gross wall areawindow areadoor areanet wall area
4’
Elevation
4’
12’
100’
8’
1200 sf
64 sf
368 sf
768 sf
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Floor SlabsFloor Slabs
For floor slabs at grade, there are two heat loss components: slab to soil losses edge losses
S: p. 1624, F.E.1
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Slab to Soil LossesSlab to Soil Losses
Q=Uslab x 0.5 x Aslab x (TI-TGW)
TI=Indoor Air Temperature
TGW=Ground Water Temperature
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Edge LossesEdge Losses
Method I
Determine F2 based on heating degree days
S: p. 1624, T.E.11/F. E.1
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Slab Edge LossesSlab Edge Losses
Method II
Select F2 based on insulation configuration
S: 1625, T.E.12
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Slab Edge LossesSlab Edge Losses
Q=F2 x Slab Perimeter Length x (TI-TO)
where,TI= Indoor air temperature
TO=Outdoor air temperature
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Heating Load Example Heating Load Example ProblemProblem
Building: Office BuildingLocation: Salt Lake CityΔT=IDBT-ODBT=68-8=60ºF
Building: 200’ x 100’ (2 stories, 12’-6” each)
Uwall= 0.054 Btuh/sf-ºF
Uroof= 0.025 Btuh/sf-ºF
Uwindow= 0.31 Btuh/sf-ºF
Uslab= 0.16 Btuh/sf-ºF
Udoor= 0.20 Btuh/sf-ºF
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Heating Load Example Heating Load Example ProblemProblem
Determine Building Envelope Areas (SF)
Building: 200’ x 100’ (2 stories, 12’-6” each)
N E S WGross Wall 5,000 2,500 5,000 2,500Windows 1,000 500 2,000 500Doors 20 20 50 20Net Wall 3,980 1,980 2,950 1,980
Roof/Floor Slab 20,000
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Heating LoadsHeating LoadsInsert roof values
Insert wall values
Insert glass values
Insert door values
Insert floor values
SR-3
0.025 20,000 60 30,000 30,000
N 0.054 3,980 60 12,895E 0.054 1,980 60 6,415S 0.054 2,950 60 9,558W 0.054 1,980 60 6,415 38,555
N 0.31 1,000 60 18,600E 0.31 500 60 9,300S 0.31 2,000 60 37,200W 0.31 500 60 9,300 74,400
0.20 110 60 1,320 1,320
N/A N/A N/A N/A
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Slab to Soil LossesSlab to Soil Losses
Q=Uslab x 0.5 x Aslab x (TI-TGW)
TI=Indoor Air Temperature
TGW=Ground Water Temperature
Ground Water= 53ºFΔT=68ºF-53ºF=15ºF
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Heating LoadsHeating LoadsInsert floor values
SR-3
0.025 20,000 60 30,000 30,000
N 0.054 3,980 60 12,895E 0.054 1,980 60 6,415S 0.054 2,950 60 9,558W 0.054 1,980 60 6,415 38,555
N 0.31 1,000 60 18,600E 0.31 500 60 9,300S 0.31 2,000 60 37,200W 0.31 500 60 9,300 74,400
0.20 110 60 1,320 1,320
N/A N/A N/A N/A
0.16 20,000 15 24,000
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Edge LossesEdge Losses
Method I
Determine F2 based on heating degree days
S: p. 1624, T.E.11/F.E.1
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Heating Degree DaysHeating Degree Days
Salt Lake CityHDD65=5983
S: p. 1562, T.C.10
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Edge LossesEdge Losses
Method I
Interpolate to find F2
at 5983 DD
5350 5983 74330.50 F2? 0.56
S: p. 1624, T.E. 11/F.E.1
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Interpolate to Find FInterpolate to Find F22
Find difference in Degree Days: 5983-5350=633 7433-
5350=2083
Find difference in F2: F2?-0.50=x
0.56-0.50=0.06
Set up proportion, solve for x: 633/2083=x/0.06
x=0.018 F2?-0.50=0.018
F2?=0.518
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Edge LossesEdge Losses
Method I
Interpolate to find F2
at 5983 DD
5350 5983 74330.50 F2= 0.56
0.518
S: p. 1624, T.E. 11/F.E.1
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Heating LoadsHeating LoadsInsert floor values
SR-3
0.025 20,000 60 30,000 30,000
N 0.054 3,980 60 12,895E 0.054 1,980 60 6,415S 0.054 2,950 60 9,558W 0.054 1,980 60 6,415 38,555
N 0.31 1,000 60 18,600E 0.31 500 60 9,300S 0.31 2,000 60 37,200W 0.31 500 60 9,300 74,400
0.20 110 60 1,320 1,320
N/A N/A N/A N/A
0.16 20,000 15 24,000
0.518 600 60 18,648 42,648
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InfiltrationInfiltrationResidential buildings use infiltration to provide fresh air
“Air change/hour (ACH) method” (see S: p.1601, T. E.27)
or
“Crack length method” (see S: p. 1603, T. E.28)
Prone to subjective interpretationVulnerable to construction defects
Provides a relatively approximate result
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Ventilation AnalysisVentilation Analysis
Non-residential buildings use ventilation to provide fresh air and to offset infiltration effects.
ASHRAE Standard 62-2001 (S: p. 1597-99, T.E.25)
Estimates the number of people/1000 sf of usage typePrescribes minimum ventilation/person for usage type
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ASHRAE 62-2001ASHRAE 62-2001
Defines space occupancy and ventilation loads
S: p. 1639, T.E.25
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ASHRAE 62-2001ASHRAE 62-2001
Defines space occupancy and ventilation loads
S: p. 1639, T.E.25
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Ventilation Load — Sensible Ventilation Load — Sensible
40,000 sf x 5people/1,000sf = 200 people
200 people x 17 cfm/person = 3,400 cfm
3,400 cfm x 60min/hr = 204,000cfh
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Heating LoadsHeating LoadsInput Ventilation Load—Sensible
SR-3
0.025 20,000 60 30,000 30,000
N 0.054 3,980 60 12,895E 0.054 1,980 60 6,415S 0.054 2,950 60 9,558W 0.054 1,980 60 6,415 38,555
N 0.31 1,000 60 18,600E 0.31 500 60 9,300S 0.31 2,000 60 37,200W 0.31 500 60 9,300 74,400
0.20 110 60 1,320 1,320
N/A N/A N/A N/A
0.16 20,000 15 24,000
0.518 600 60 18,648 42,648
204,000 60 220,320
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Ventilation Load — Latent Ventilation Load — Latent
Determine ΔW
WI=0.0066 #H2O/#dry air
-WO=0.0006 #H2O/#dry air
ΔW= 0.0060 #H2O/#dry air
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Heating LoadsHeating LoadsInput Ventilation Load — Latent
SR-3
0.025 20,000 60 30,000 30,000
N 0.054 3,980 60 12,895E 0.054 1,980 60 6,415S 0.054 2,950 60 9,558W 0.054 1,980 60 6,415 38,555
N 0.31 1,000 60 18,600E 0.31 500 60 9,300S 0.31 2,000 60 37,200W 0.31 500 60 9,300 74,400
0.20 110 60 1,320 1,320
N/A N/A N/A N/A
0.16 20,000 15 24,000
0.518 600 60 18,648 42,648
204,000 60 220,320
204,000 0.0060 97308 317628
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Heating LoadHeating LoadTotal Load
504551 Btuhor
505 MBH
SR-3
0.025 20,000 60 30,000 30,000
N 0.054 3,980 60 12,895E 0.054 1,980 60 6,415S 0.054 2,950 60 9,558W 0.054 1,980 60 6,415 38,555
N 0.31 1,000 60 18,600E 0.31 500 60 9,300S 0.31 2,000 60 37,200W 0.31 500 60 9,300 74,400
0.20 110 60 1,320 1,320
N/A N/A N/A N/A
0.16 20,000 15 24,000
0.518 600 60 18,648 42,648
204,000 60 220320
204,000 0.0060 97,308 317628
504551
5.9
7.6
14.7
0.3
8.4
63.1
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Annual Fuel Annual Fuel ConsumptionConsumptionAnnual Fuel Annual Fuel ConsumptionConsumption
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Annual Fuel Usage (E)Annual Fuel Usage (E)
E= UA x DDBPT x 24
AFUE x V
where:UA: heating load/ºFDDBPT: degree days for given balance pointAFUE: annual fuel utilization efficiencyV: fuel heating value
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Calculating UACalculating UA
QTotal= UA x ΔT
UA= QTotal/ΔT
From earlier example:QTotal=504,551 Btuh ΔT= 60ºF
UA=504,551/60=8,409 Btuh/ºF
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Determine AFUEDetermine AFUEAnnual Fuel Utilization Efficiency of an electric heating system is 100%
S: p. 262, T.8.7
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Determine Heat Content (V)Determine Heat Content (V)
Heat content is the quantity of Btu/unit
Note: Natural Gas is sold in therms (100 cf)
S: p. 259, T.8.5
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Annual Fuel Usage ExampleAnnual Fuel Usage Example
What is the expected annual fuel usage for a house in Salt Lake City if its peak heating load is 39,000 Btuh?
UA=Q/ΔTUA=39,000/60= 650 Btuh/ºF
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Determine AFUEDetermine AFUEAnnual Fuel Utilization Efficiency of an electric heating system is 100%
S: p.262, T.8.7
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Determine Heat Content (V)Determine Heat Content (V)
Heat content is the quantity of Btu/unit
S: p. 259, T.8.5
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Annual Fuel Usage — Annual Fuel Usage — Electricity Electricity
E= UA x DDBPT x 24
AFUE x V
EELEC =(650)(5,983)(24)/(1.0)(3,413)
=27,347 kwh/yr
If electricity is $0.0735/kwh, thenannual cost = $2,010
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Annual Fuel Usage — Gas Annual Fuel Usage — Gas
E= UA x DDBPT x 24
AFUE x V
EGas =(650)(5,983)(24)/(0.8)(105,000) =1,111 therms/yr
If gas is $0.41/therm, thenannual cost = $456
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Simple Simple Payback Payback AnalysisAnalysis
Simple Simple Payback Payback AnalysisAnalysis
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Simple PaybackSimple Payback
Heating System Cost ComparisonFirst
Cost ($)
Electricity 6,000Oil 8,000
Gas 8,900
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Simple PaybackSimple Payback
Heating System Cost ComparisonFirst AnnualIncremental Incremental Simple
Cost Fuel Cost First Cost Annual Savings Payback ($) ($/yr) ($) ($/yr) (yrs)
Electricity 6,000 2,010 --- --- ---Oil 8,000 1,152 2,000 858 2.3
Gas 8,900 456 2,900 1,554 1.9
If money is available, select gas furnace system
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