1 / 48

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

olympia-daugherty
Télécharger la présentation

Technology in Architecture

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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. Outdoor Dry Bulb Temperature Use Winter Conditions S(10th): T.B.1 p. 1496

  7. Determine Temperature Difference Indoor Dry Bulb Temperature (IDBT): 68ºF Outdoor Dry Bulb Temperature (ODBT): 8ºF ΔT=IDBT-ODBT=68ºF - 8ºF = 60ºF

  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=68-8=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

  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 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

  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 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

  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 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

  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 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 Input Ventilation Load—Sensible 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

  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 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 Input Ventilation Load — Latent 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

  34. Heating Load 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 Total Load 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

  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=504,551 Btuh ΔT= 60ºF UA=504,551/60=8,409 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 39,000 Btuh? UA=Q/ΔT UA=39,000/60= 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

More Related