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Technology in Architecture. Lecture 7 Degree Days Heating Loads Annual Fuel Consumption Simple Payback Analysis. Heating Degree Days. Balance Point Temperature (BPT): temperature above which heating is not needed DD BPT = BPT-TA. Sample Calculation. January TA=28ºF

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Technology in Architecture

Lecture 7

Degree Days

Annual Fuel Consumption

Simple Payback Analysis

Heating Degree Days

Balance Point Temperature (BPT): temperature above which heating is not needed

DDBPT= BPT-TA

Sample Calculation

January TA=28ºF

DD65=65-28= 37 Degree-days/day

x 31 days

= 1,147 degree-days

S: p. 1562, T.C.19

Computed 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 Temperature

Use Winter Conditions

S(10th): T.B.1

p. 1496

Determine Temperature Difference

Indoor Dry Bulb Temperature (IDBT): 68ºF

Outdoor Dry Bulb Temperature (ODBT): 8ºF

ΔT=IDBT-ODBT=68ºF - 8ºF = 60ºF

Determine Envelope U-values

Calculate ΣR and then find U for walls, roofs, floors.

Obtain U values for glazing from manufacturer or other reference

Determine Area Quantities

Perform area takeoffs for all building envelope surfaces on each facade:

gross wall area

window area

door area

net wall area

1200 sf

100’

-

368 sf

-

64 sf

768 sf

4’

12’

4’

8’

Elevation

Floor 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 Losses

Q=Uslabx 0.5 x Aslabx (TI-TGW)

TI=Indoor Air Temperature

TGW=Ground Water Temperature

Edge Losses

Method I

Determine F2 based on heating degree days

S: p. 1624, T.E.11/F. E.1

Slab Edge Losses

Method II

Select F2 based on insulation configuration

S: 1625, T.E.12

Slab Edge Losses

Q=F2x Slab Perimeter Length x (TI-TO)

where,

TI= Indoor air temperature

TO=Outdoor air temperature

Building: Office Building

Location: Salt Lake City

ΔT=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

Determine Building Envelope Areas (SF)

Building: 200’ x 100’ (2 stories, 12’-6” each)

N E S W

Gross Wall 5,000 2,500 5,000 2,500

Windows 1,000 500 2,000 500

Doors 20 20 50 20

Net Wall 3,980 1,980 2,950 1,980

Roof/Floor Slab 20,000

0.025 20,000 60 30,000 30,000

N 0.054 3,980 60 12,895

E 0.054 1,980 60 6,415

S 0.054 2,950 60 9,558

W 0.054 1,980 60 6,415 38,555

Insert roof values

Insert wall values

Insert glass values

Insert door values

Insert floor values

N 0.31 1,000 60 18,600

E 0.31 500 60 9,300

S 0.31 2,000 60 37,200

W 0.31 500 60 9,300 74,400

0.20 110 60 1,320 1,320

N/A N/A N/A N/A

SR-3

Slab to Soil Losses

Q=Uslabx 0.5 x Aslabx (TI-TGW)

TI=Indoor Air Temperature

TGW=Ground Water Temperature

Ground Water= 53ºF

ΔT=68ºF-53ºF=15ºF

0.025 20,000 60 30,000 30,000

N 0.054 3,980 60 12,895

E 0.054 1,980 60 6,415

S 0.054 2,950 60 9,558

W 0.054 1,980 60 6,415 38,555

Insert floor values

N 0.31 1,000 60 18,600

E 0.31 500 60 9,300

S 0.31 2,000 60 37,200

W 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

SR-3

Edge Losses

Method I

Determine F2 based on heating degree days

S: p. 1624, T.E.11/F.E.1

Heating Degree Days

Salt Lake City

HDD65=5983

S: p. 1562, T.C.10

Edge Losses

Method I

Interpolate to find F2

at 5983 DD

5350 5983 7433

0.50 F2? 0.56

S: p. 1624, T.E. 11/F.E.1

Interpolate to Find F2

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 Losses

Method I

Interpolate to find F2

at 5983 DD

5350 5983 7433

0.50 F2= 0.56

0.518

S: p. 1624, T.E. 11/F.E.1

0.025 20,000 60 30,000 30,000

N 0.054 3,980 60 12,895

E 0.054 1,980 60 6,415

S 0.054 2,950 60 9,558

W 0.054 1,980 60 6,415 38,555

Insert floor values

N 0.31 1,000 60 18,600

E 0.31 500 60 9,300

S 0.31 2,000 60 37,200

W 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

SR-3

Infiltration

Residential 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 interpretation

Vulnerable to construction defects

Provides a relatively approximate result

Ventilation 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 type

Prescribes minimum ventilation/person for usage type

ASHRAE 62-2001

Defines space occupancy and ventilation loads

S: p. 1639, T.E.25

ASHRAE 62-2001

Defines space occupancy and ventilation loads

S: p. 1639, T.E.25

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

0.025 20,000 60 30,000 30,000

N 0.054 3,980 60 12,895

E 0.054 1,980 60 6,415

S 0.054 2,950 60 9,558

W 0.054 1,980 60 6,415 38,555

N 0.31 1,000 60 18,600

E 0.31 500 60 9,300

S 0.31 2,000 60 37,200

W 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

SR-3

Determine ΔW

WI= 0.0066 #H2O/#dry air

-WO= 0.0006 #H2O/#dry air

ΔW= 0.0060 #H2O/#dry air

0.025 20,000 60 30,000 30,000

N 0.054 3,980 60 12,895

E 0.054 1,980 60 6,415

S 0.054 2,950 60 9,558

W 0.054 1,980 60 6,415 38,555

N 0.31 1,000 60 18,600

E 0.31 500 60 9,300

S 0.31 2,000 60 37,200

W 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

SR-3

5.9

0.025 20,000 60 30,000 30,000

N 0.054 3,980 60 12,895

E 0.054 1,980 60 6,415

S 0.054 2,950 60 9,558

W 0.054 1,980 60 6,415 38,555

7.6

504551 Btuh

or

505 MBH

N 0.31 1,000 60 18,600

E 0.31 500 60 9,300

S 0.31 2,000 60 37,200

W 0.31 500 60 9,300 74,400

14.7

0.20 110 60 1,320 1,320

0.3

N/A N/A N/A N/A

0.16 20,000 15 24,000

8.4

0.518 600 60 18,648 42,648

204,000 60 220320

63.1

204,000 0.0060 97,308 317628

SR-3

504551

Annual Fuel Usage (E)

E= UA x DDBPT x 24

AFUE x V

where:

DDBPT: degree days for given balance point

AFUE: annual fuel utilization efficiency

V: fuel heating value

Calculating UA

QTotal= UA xΔT

UA= QTotal/ΔT

From earlier example:

QTotal=504,551 Btuh

ΔT= 60ºF

UA=504,551/60=8,409 Btuh/ºF

Determine AFUE

Annual Fuel Utilization Efficiency of an electric heating system is 100%

S: p. 262, T.8.7

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 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/ΔT

UA=39,000/60= 650 Btuh/ºF

Determine AFUE

Annual Fuel Utilization Efficiency of an electric heating system is 100%

S: p.262, T.8.7

Determine Heat Content (V)

Heat content is the quantity of Btu/unit

S: p. 259, T.8.5

Annual Fuel Usage — 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, then

annual cost = \$2,010

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, then

annual cost = \$456

Simple Payback

Heating SystemCost Comparison

First

Cost

(\$)

Electricity 6,000

Oil 8,000

Gas 8,900

Simple Payback

Heating SystemCost Comparison

First Annual Incremental Incremental Simple

Cost Fuel Cost First Cost Annual Savings Payback

(\$) (\$/yr) (\$) (\$/yr) (yrs)

Electricity 6,000 2,010 --- --- ---

Oil 8,0001,152 2,0008582.3

Gas 8,900 456 2,900 1,554 1.9

If money is available, select gas furnace system