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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

Technology in Architecture

Lecture 7

Degree Days

Heating Loads

Annual Fuel Consumption

Simple Payback Analysis


Heating degree days

Heating Degree Days

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

DDBPT= BPT-TA


Sample calculation

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


Heating loads

Heating Loads


Heating loads1

Heating Loads

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

Outdoor Dry Bulb Temperature

Use Winter Conditions

S(10th): T.B.1

p. 1496


Determine temperature difference

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

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

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

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

Slab to Soil Losses

Q=Uslabx 0.5 x Aslabx (TI-TGW)

TI=Indoor Air Temperature

TGW=Ground Water Temperature


Edge losses

Edge Losses

Method I

Determine F2 based on heating degree days

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


Slab edge losses

Slab Edge Losses

Method II

Select F2 based on insulation configuration

S: 1625, T.E.12


Slab edge losses1

Slab Edge Losses

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

where,

TI= Indoor air temperature

TO=Outdoor air temperature


Heating load example problem

Heating Load Example Problem

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


Heating load example problem1

Heating Load Example Problem

Determine Building Envelope Areas (SF)

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

NESW

Gross Wall5,0002,5005,0002,500

Windows1,0005002,000500

Doors20205020

Net Wall3,9801,9802,9501,980

Roof/Floor Slab20,000


Heating loads2

Heating Loads

0.025 20,000 60 30,000 30,000

N0.054 3,9806012,895

E 0.054 1,980606,415

S0.0542,950609,558

W0.0541,980606,415 38,555

Insert roof values

Insert wall values

Insert glass values

Insert door values

Insert floor values

N0.31 1,0006018,600

E 0.31 500609,300

S0.312,0006037,200

W0.31500609,300 74,400

0.20 110 60 1,3201,320

N/A N/AN/AN/A

SR-3


Slab to soil losses1

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


Heating loads3

Heating Loads

0.025 20,000 60 30,000 30,000

N0.054 3,9806012,895

E 0.054 1,980606,415

S0.0542,950609,558

W0.0541,980606,415 38,555

Insert floor values

N0.31 1,0006018,600

E 0.31 500609,300

S0.312,0006037,200

W0.31500609,300 74,400

0.20 110 60 1,3201,320

N/A N/AN/AN/A

0.16 20,0001524,000

SR-3


Edge losses1

Edge Losses

Method I

Determine F2 based on heating degree days

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


Heating degree days1

Heating Degree Days

Salt Lake City

HDD65=5983

S: p. 1562, T.C.10


Edge losses2

Edge Losses

Method I

Interpolate to find F2

at 5983 DD

535059837433

0.50F2?0.56

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


Interpolate to find f 2

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 losses3

Edge Losses

Method I

Interpolate to find F2

at 5983 DD

535059837433

0.50F2=0.56

0.518

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


Heating loads4

Heating Loads

0.025 20,000 60 30,000 30,000

N0.054 3,9806012,895

E 0.054 1,980606,415

S0.0542,950609,558

W0.0541,980606,415 38,555

Insert floor values

N0.31 1,0006018,600

E 0.31 500609,300

S0.312,0006037,200

W0.31500609,300 74,400

0.20 110 60 1,3201,320

N/A N/AN/AN/A

0.16 20,0001524,000

0.518 6006018,64842,648

SR-3


Infiltration

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

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

ASHRAE 62-2001

Defines space occupancy and ventilation loads

S: p. 1639, T.E.25


Ashrae 62 20011

ASHRAE 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 loads5

Heating Loads

0.025 20,000 60 30,000 30,000

N0.054 3,9806012,895

E 0.054 1,980606,415

S0.0542,950609,558

W0.0541,980606,415 38,555

Input Ventilation Load—Sensible

N0.31 1,0006018,600

E 0.31 500609,300

S0.312,0006037,200

W0.31500609,300 74,400

0.20 110 60 1,3201,320

N/A N/AN/AN/A

0.16 20,0001524,000

0.518 6006018,64842,648

204,000 60220,320

SR-3


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 loads6

Heating Loads

0.025 20,000 60 30,000 30,000

N0.054 3,9806012,895

E 0.054 1,980606,415

S0.0542,950609,558

W0.0541,980606,415 38,555

Input Ventilation Load — Latent

N0.31 1,0006018,600

E 0.31 500609,300

S0.312,0006037,200

W0.31500609,300 74,400

0.20 110 60 1,3201,320

N/A N/AN/AN/A

0.16 20,0001524,000

0.518 6006018,64842,648

204,000 60220,320

204,000 0.006097308317628

SR-3


Heating load

Heating Load

5.9

0.025 20,000 60 30,000 30,000

N0.054 3,9806012,895

E 0.054 1,980606,415

S0.0542,950609,558

W0.0541,980606,415 38,555

7.6

Total Load

504551 Btuh

or

505 MBH

N0.31 1,0006018,600

E 0.31 500609,300

S0.312,0006037,200

W0.31500609,300 74,400

14.7

0.20 110 60 1,3201,320

0.3

N/A N/AN/AN/A

0.16 20,0001524,000

8.4

0.518 6006018,64842,648

204,000 60220320

63.1

204,000 0.006097,308317628

SR-3

504551


Annual fuel consumption

Annual Fuel Consumption


Annual fuel usage e

Annual Fuel Usage (E)

E= UA x DDBPT x 24

AFUE x V

where:

UA: heating load/ºF

DDBPT: degree days for given balance point

AFUE: annual fuel utilization efficiency

V: fuel heating value


Calculating ua

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

Determine AFUE

Annual 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 example

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 afue1

Determine AFUE

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

S: p.262, T.8.7


Determine heat content v1

Determine Heat Content (V)

Heat content is the quantity of Btu/unit

S: p. 259, T.8.5


Annual fuel usage electricity

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

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 analysis

Simple Payback Analysis


Simple payback

Simple Payback

Heating SystemCost Comparison

First

Cost

($)

Electricity6,000

Oil8,000

Gas8,900


Simple payback1

Simple Payback

Heating SystemCost Comparison

FirstAnnualIncremental Incremental Simple

Cost Fuel CostFirst CostAnnual SavingsPayback

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

Electricity6,0002,010--- ------

Oil8,0001,1522,0008582.3

Gas8,9004562,9001,5541.9

If money is available, select gas furnace system


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