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Technology in Architecture Lecture 7 Degree Days Heating Loads Annual Fuel Consumption Simple Payback 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 Heating Degree Days Balance Point Temperature (BPT): temperature above which heating is not needed DDBPT= BPT-TA

Sample Calculation January TA=28ºF DD 65=65 -28= 37 Degree-days/day x 31 days = 1, Sample Calculation January TA=28ºF DD 65=65 -28= 37 Degree-days/day x 31 days = 1, 147 degree-days S: p. 1562, T. C. 19

Heating Loads Heating Loads

Heating Loads Computed for worst case scenario: n Pre-dawn at outdoor design dry bulb Heating Loads Computed for worst case scenario: n Pre-dawn at outdoor design dry bulb temperature Do not include: n n Insolation from sun Heat gain from people, lights, and equipment Infiltration in nonresidential buildings Ventilation in residential buildings SR-3

S: T. B. 1 p. 1513 Outdoor Dry Bulb Temperature Use Winter Conditions S: T. B. 1 p. 1513 Outdoor Dry Bulb Temperature Use Winter Conditions

Determine Temperature Difference Indoor Dry Bulb Temperature (IDBT): 68ºF Outdoor Dry Bulb Temperature (ODBT): Determine Temperature Difference Indoor Dry Bulb Temperature (IDBT): 68ºF Outdoor Dry Bulb Temperature (ODBT): 13ºF ΔT=IDBT-ODBT=68ºF - 13ºF = 55ºF

Determine Envelope U-values Calculate ΣR and then find U for walls, roofs, floors. Obtain 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: 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’ Elevation 8’

Floor Slabs For floor slabs at grade, there are two heat loss components: n Floor Slabs For floor slabs at grade, there are two heat loss components: n slab to soil losses n edge losses S: p. 1624, F. E. 1

Slab to Soil Losses Q=Uslab x 0. 5 x Aslab x (TI-TGW) TI=Indoor Air Slab to Soil Losses Q=Uslab x 0. 5 x Aslab x (TI-TGW) TI=Indoor Air Temperature TGW=Ground Water Temperature

Edge Losses Method I Determine F 2 based on heating degree days S: p. Edge Losses Method I Determine F 2 based on heating degree days S: p. 1624, T. E. 11/F. E. 1

Slab Edge Losses Method II Select F 2 based on insulation configuration S: 1625, Slab Edge Losses Method II Select F 2 based on insulation configuration S: 1625, T. E. 12

Slab Edge Losses Q=F 2 x Slab Perimeter Length x (TI-TO) where, TI= Indoor Slab Edge Losses Q=F 2 x Slab Perimeter Length x (TI-TO) where, TI= Indoor air temperature TO=Outdoor air temperature

Heating Load Example Problem Building: Office Building Location: Salt Lake City ΔT=IDBT-ODBT=68 -13=55ºF Building: Heating Load Example Problem Building: Office Building Location: Salt Lake City ΔT=IDBT-ODBT=68 -13=55ºF 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 Problem Determine Building Envelope Areas (SF) Building: 200’ x 100’ (2 Heating Load Example Problem Determine Building Envelope Areas (SF) Building: 200’ x 100’ (2 stories, 12’-6” each) Gross Wall Windows Doors Net Wall N 5, 000 1, 000 20 3, 980 Roof/Floor Slab E S 2, 500 5, 000 500 2, 000 20 50 1, 980 2, 950 20, 000 W 2, 500 20 1, 980

Insert glass values Insert door values Insert floor values SR-3 N E S W Insert glass values Insert door values Insert floor values SR-3 N E S W 0. 054 3, 980 1, 980 2, 950 1, 980 55 55 0. 31 1, 000 500 2, 000 55 55 110 55 N/A Insert wall values 55 0. 20 Insert roof values 20, 000 N E S W Heating Loads 0. 025 N/A N/A

Slab to Soil Losses Q=Uslab x 0. 5 x Aslab x (TI-TGW) TI=Indoor Air Slab to Soil Losses Q=Uslab x 0. 5 x Aslab x (TI-TGW) TI=Indoor Air Temperature TGW=Ground Water Temperature Ground Water= 53ºF ΔT=68ºF-53ºF=15ºF

55 N E S W 0. 054 3, 980 1, 980 2, 950 1, 55 N E S W 0. 054 3, 980 1, 980 2, 950 1, 980 55 55 0. 31 1, 000 500 2, 000 55 55 0. 20 110 55 N/A Insert floor values 20, 000 N E S W Heating Loads 0. 025 N/A 0. 16 SR-3 20, 000 15 N/A

Edge Losses Method I Determine F 2 based on heating degree days S: p. Edge Losses Method I Determine F 2 based on heating degree days S: p. 1624, T. E. 11/F. E. 1

Heating Degree Days Salt Lake City HDD 65=5983 S: p. 1562, T. C. 10 Heating Degree Days Salt Lake City HDD 65=5983 S: p. 1562, T. C. 10

Edge Losses Method I Interpolate to find F 2 at 5983 DD 5350 5983 Edge Losses Method I Interpolate to find F 2 at 5983 DD 5350 5983 7433 0. 50 F 2? 0. 56 S: p. 1624, T. E. 11/F. E. 1

Interpolate to Find F 2 Find difference in Degree Days: 7433 -5350=2083 5983 -5350=633 Interpolate to Find F 2 Find difference in Degree Days: 7433 -5350=2083 5983 -5350=633 Find difference in F 2: 0. 56 -0. 50=0. 06 F 2? -0. 50=x Set up proportion, solve for x: 633/2083=x/0. 06 x=0. 018 F 2? -0. 50=0. 018 F 2? =0. 518

Edge Losses Method I Interpolate to find F 2 at 5983 DD 5350 5983 Edge Losses Method I Interpolate to find F 2 at 5983 DD 5350 5983 7433 0. 50 F 2= 0. 56 0. 518 S: p. 1624, T. E. 11/F. E. 1

55 N E S W 0. 054 3, 980 1, 980 2, 950 1, 55 N E S W 0. 054 3, 980 1, 980 2, 950 1, 980 55 55 0. 31 1, 000 500 2, 000 55 55 0. 20 110 55 N/A Insert floor values 20, 000 N E S W Heating Loads 0. 025 N/A 0. 16 0. 518 SR-3 20, 000 600 15 55 N/A

Infiltration Residential buildings use infiltration to provide fresh air “Air change/hour (ACH) method” (see 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 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. 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. ASHRAE 62 -2001 Defines space occupancy and ventilation loads S: p. 1639, T. E. 25

Ventilation Load — Sensible 40, 000 sf x 5 people/1, 000 sf = 200 Ventilation Load — Sensible 40, 000 sf x 5 people/1, 000 sf = 200 people x 17 cfm/person = 3, 400 cfm x 60 min/hr = 204, 000 cfh

55 N E S W 0. 054 3, 980 1, 980 2, 950 1, 55 N E S W 0. 054 3, 980 1, 980 2, 950 1, 980 55 55 0. 31 1, 000 500 2, 000 55 55 0. 20 110 55 N/A Input Ventilation Load—Sensible 20, 000 N E S W Heating Loads 0. 025 N/A 0. 16 0. 518 20, 000 600 204, 000 SR-3 N/A 15 55 55

Ventilation Load — Latent Determine ΔW W I= -WO= ΔW= 0. 0066 #H 2 Ventilation Load — Latent Determine ΔW W I= -WO= ΔW= 0. 0066 #H 2 O/#dry air 0. 0006 #H 2 O/#dry air 0. 0060 #H 2 O/#dry air

55 N E S W 0. 054 3, 980 1, 980 2, 950 1, 55 N E S W 0. 054 3, 980 1, 980 2, 950 1, 980 55 55 0. 31 1, 000 500 2, 000 55 55 0. 20 110 55 N/A Input Ventilation Load — Latent 20, 000 N E S W Heating Loads 0. 025 N/A 0. 16 0. 518 20, 000 600 N/A 15 24, 000 55 18, 648 204, 000 SR-3 55 204, 000 0. 0060 220, 320 97308

N E S W 0. 054 3, 980 1, 980 2, 950 1, 980 N E S W 0. 054 3, 980 1, 980 2, 950 1, 980 55 55 11, 583 5, 880 8, 762 5, 880 32105 6. 9 0. 31 1, 000 500 2, 000 55 55 17, 050 8, 525 34, 100 8, 525 68, 200 14. 6 110 55 1, 210 N/A 468, 377 Btuh 55 0. 20 Total Load 20, 000 N E S W Heating Load N/A N/A 0. 16 0. 518 20, 000 600 27, 500 15 1, 210 0. 3 24, 000 55 17, 094 204, 000 SR-3 27, 500 5. 9 0. 025 55 204, 000 0. 0060 41, 094 8. 8 201, 960 97, 308 299, 268 468, 377 63. 9

Annual Fuel Consumption Annual Fuel Consumption

Annual Fuel Usage (E) E= UA x DDBPT x 24 AFUE x V where: 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 QTotal= UA x ΔT UA= QTotal/ΔT From earlier example: QTotal=468, 377 Btuh Calculating UA QTotal= UA x ΔT UA= QTotal/ΔT From earlier example: QTotal=468, 377 Btuh ΔT= 55ºF UA=468, 377/55=8, 516 Btuh/ºF

Determine AFUE Annual Fuel Utilization Efficiency of an electric heating system is 100% S: 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 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 Annual Fuel Usage Example What is the expected annual fuel usage for a house in Salt Lake City if its peak heating load is 35, 750 Btuh? UA=Q/ΔT UA=35, 750/55= 650 Btuh/ºF

Determine AFUE Annual Fuel Utilization Efficiency of an electric heating system is 100% S: 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, 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 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 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 Heating System Cost Comparison First Cost ($) Electricity Oil Gas 6, 000 Simple Payback Heating System Cost Comparison First Cost ($) Electricity Oil Gas 6, 000 8, 900

Simple Payback Heating System Cost Comparison First Annual Incremental Simple Annual Savings ($/yr) Payback Simple Payback Heating System Cost Comparison First Annual Incremental Simple Annual Savings ($/yr) Payback (yrs) --- Cost ($) Electricity Oil Gas First Cost ($) Incremental Fuel Cost ($/yr) 6, 000 2, 010 8, 000 1, 152 2, 000 858 2. 3 8, 900 456 2, 900 1, 554 1. 9 --- If money is available, select gas furnace system