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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. 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

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  1. Technology in Architecture Lecture 7 Degree Days Heating Loads Annual Fuel Consumption Simple Payback Analysis

  2. Heating Degree Days Balance Point Temperature (BPT): temperature above which heating is not needed DDBPT= BPT-TA

  3. 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

  4. Heating Loads

  5. 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

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

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

  8. Determine Envelope U-values Calculate ΣR and then find U for walls, roofs, floors. Obtain U values for glazing from manufacturer or other reference

  9. 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

  10. 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

  11. Slab to Soil Losses Q=Uslabx 0.5 x Aslabx (TI-TGW) TI=Indoor Air Temperature TGW=Ground Water Temperature

  12. Edge Losses Method I Determine F2 based on heating degree days S: p. 1624, T.E.11/F. E.1

  13. Slab Edge Losses Method II Select F2 based on insulation configuration S: 1625, T.E.12

  14. Slab Edge Losses Q=F2x Slab Perimeter Length x (TI-TO) where, TI= Indoor air temperature TO=Outdoor air temperature

  15. 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

  16. Heating Load Example Problem 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

  17. Heating Loads 0.025 20,000 55 N 0.054 3,980 55 E 0.054 1,980 55 S 0.054 2,950 55 W 0.054 1,980 55 Insert roof values Insert wall values Insert glass values Insert door values Insert floor values N 0.31 1,000 55 E 0.31 500 55 S 0.31 2,000 55 W 0.31 500 55 0.20 110 55 N/A N/A N/A N/A SR-3

  18. 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

  19. Heating Loads 0.025 20,000 55 N 0.054 3,980 55 E 0.054 1,980 55 S 0.054 2,950 55 W 0.054 1,980 55 Insert floor values N 0.31 1,000 55 E 0.31 500 55 S 0.31 2,000 55 W 0.31 500 55 0.20 110 55 N/A N/A N/A N/A 0.16 20,000 15 SR-3

  20. Edge Losses Method I Determine F2 based on heating degree days S: p. 1624, T.E.11/F.E.1

  21. Heating Degree Days Salt Lake City HDD65=5983 S: p. 1562, T.C.10

  22. 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

  23. 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

  24. 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

  25. Heating Loads 0.025 20,000 55 N 0.054 3,980 55 E 0.054 1,980 55 S 0.054 2,950 55 W 0.054 1,980 55 Insert floor values N 0.31 1,000 55 E 0.31 500 55 S 0.31 2,000 55 W 0.31 500 55 0.20 110 55 N/A N/A N/A N/A 0.16 20,000 15 0.518 600 55 SR-3

  26. 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

  27. 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

  28. ASHRAE 62-2001 Defines space occupancy and ventilation loads S: p. 1639, T.E.25

  29. ASHRAE 62-2001 Defines space occupancy and ventilation loads S: p. 1639, T.E.25

  30. 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

  31. Heating Loads 0.025 20,000 55 N 0.054 3,980 55 E 0.054 1,980 55 S 0.054 2,950 55 W 0.054 1,980 55 Input Ventilation Load—Sensible N 0.31 1,000 55 E 0.31 500 55 S 0.31 2,000 55 W 0.31 500 55 0.20 110 55 N/A N/A N/A N/A 0.16 20,000 15 0.518 600 55 204,000 55 SR-3

  32. Ventilation Load — Latent Determine ΔW WI= 0.0066 #H2O/#dry air -WO= 0.0006 #H2O/#dry air ΔW= 0.0060 #H2O/#dry air

  33. Heating Loads 0.025 20,000 55 N 0.054 3,980 55 E 0.054 1,980 55 S 0.054 2,950 55 W 0.054 1,980 55 Input Ventilation Load — Latent N 0.31 1,000 55 E 0.31 500 55 S 0.31 2,000 55 W 0.31 500 55 0.20 110 55 N/A N/A N/A N/A 0.16 20,000 15 24,000 0.518 600 55 18,648 204,000 55 220,320 204,000 0.0060 97308 SR-3

  34. Heating Load 5.9 0.025 20,000 55 27,500 27,500 N 0.054 3,980 55 11,583 E 0.054 1,980 55 5,880 S 0.054 2,950 55 8,762 W 0.054 1,980 55 5,880 32105 6.9 Total Load 468,377 Btuh N 0.31 1,000 55 17,050 E 0.31 500 55 8,525 S 0.31 2,000 55 34,100 W 0.31 500 55 8,525 68,200 14.6 0.20 110 55 1,210 1,210 0.3 N/A N/A N/A N/A 0.16 20,000 15 24,000 8.8 0.518 600 55 17,094 41,094 204,000 55 201,960 63.9 204,000 0.0060 97,308 299,268 SR-3 468,377

  35. Annual Fuel Consumption

  36. 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

  37. Calculating UA QTotal= UA xΔT UA= QTotal/ΔT From earlier example: QTotal=468,377 Btuh ΔT= 55ºF UA=468,377/55=8,516 Btuh/ºF

  38. Determine AFUE Annual Fuel Utilization Efficiency of an electric heating system is 100% S: p. 262, T.8.7

  39. 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

  40. 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

  41. Determine AFUE Annual Fuel Utilization Efficiency of an electric heating system is 100% S: p.262, T.8.7

  42. Determine Heat Content (V) Heat content is the quantity of Btu/unit S: p. 259, T.8.5

  43. 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

  44. 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

  45. Simple Payback Analysis

  46. Simple Payback Heating SystemCost Comparison First Cost ($) Electricity 6,000 Oil 8,000 Gas 8,900

  47. 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

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