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Lecture 7Degree Days Heating Loads
Annual Fuel ConsumptionSimple Payback Analysis
Lecture 7Degree Days Heating Loads
Annual Fuel ConsumptionSimple Payback Analysis
Heating Degree DaysHeating Degree Days
Balance Point Temperature (BPT): temperature above which heating is not needed
DDBPT= BPT-TA
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
Heating LoadsHeating LoadsHeating LoadsHeating Loads
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
Outdoor Dry Bulb Outdoor Dry Bulb TemperatureTemperature
Use Winter Conditions
S(10th): T.B.1 p. 1496
Determine Temperature Determine Temperature DifferenceDifference
Indoor Dry Bulb Temperature (IDBT): 68ºF
Outdoor Dry Bulb Temperature (ODBT): 8ºF
ΔT=IDBT-ODBT=68ºF - 8ºF = 60ºF
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
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
-
-
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
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
Edge LossesEdge Losses
Method I
Determine F2 based on heating degree days
S: p. 1624, T.E.11/F. E.1
Slab Edge LossesSlab Edge Losses
Method II
Select F2 based on insulation configuration
S: 1625, T.E.12
Slab Edge LossesSlab Edge Losses
Q=F2 x Slab Perimeter Length x (TI-TO)
where,TI= Indoor air temperature
TO=Outdoor air temperature
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
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
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
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
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
Edge LossesEdge Losses
Method I
Determine F2 based on heating degree days
S: p. 1624, T.E.11/F.E.1
Heating Degree DaysHeating Degree Days
Salt Lake CityHDD65=5983
S: p. 1562, T.C.10
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
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
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
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
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
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
ASHRAE 62-2001ASHRAE 62-2001
Defines space occupancy and ventilation loads
S: p. 1639, T.E.25
ASHRAE 62-2001ASHRAE 62-2001
Defines space occupancy and ventilation loads
S: p. 1639, T.E.25
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
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
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
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
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
Annual Fuel Annual Fuel ConsumptionConsumptionAnnual Fuel Annual Fuel ConsumptionConsumption
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
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
Determine AFUEDetermine AFUEAnnual Fuel Utilization Efficiency of an electric heating system is 100%
S: p. 262, T.8.7
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
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
Determine AFUEDetermine AFUEAnnual Fuel Utilization Efficiency of an electric heating system is 100%
S: p.262, T.8.7
Determine Heat Content (V)Determine Heat Content (V)
Heat content is the quantity of Btu/unit
S: p. 259, T.8.5
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
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 = $456
Simple Simple Payback Payback AnalysisAnalysis
Simple Simple Payback Payback AnalysisAnalysis
Simple PaybackSimple Payback
Heating System Cost ComparisonFirst
Cost ($)
Electricity 6,000Oil 8,000
Gas 8,900
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