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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