Environmental Controls I/IG Lecture 11 Degree Days Heating Loads 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. 1524, T.C.15

Heating Loads

Heating Loads Computed for worst case scenario: Do not include: 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: p. 1496, T.B.1

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 12’ 4’ 4’ 8’ Elevation

Floor Slabs For floor slabs at grade, there are two heat loss components: slab to soil losses edge losses S: p. 1583, T.E.11

TI=Indoor Air Temperature TGW=Ground Water Temperature 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 F2 based on heating degree days S: p. 1582, T.E.11

Select F2 based on insulation configuration Slab Edge Losses Method II Select F2 based on insulation configuration S: 1583, T.E.12

TI= Indoor air temperature TO=Outdoor air temperature Slab Edge Losses Q=F2 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=68-8=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 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

Heating Loads Insert roof values Insert wall values 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

TI=Indoor Air Temperature TGW=Ground Water Temperature 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

Heating Loads Insert floor values SR-3 0.025 20,000 60 30,000 30,000 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 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. 1582, T.E.11

Heating Degree Days Salt Lake City HDD65=5983 S: p. 1524, T.C.15

Edge Losses Method I Interpolate to find F2 at 5983 DD 5350 5983 7433 0.50 F2? 0.56 S: p. 1582, T.E. 11

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. 1582, T.E.11

Heating Loads Insert floor values SR-3 0.025 20,000 60 30,000 30,000 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 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

ASHRAE Standard 62-2001 (S: p. 1597-99, T.E.25) 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. 1598, T.E.25

ASHRAE 62-2001 Defines space occupancy and ventilation loads S: p. 1598, T.E.25

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 Loads Input Ventilation Load—Sensible SR-3 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

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

Heating Loads Input Ventilation Load — Latent SR-3 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

Heating Load Total Load 504551 Btuh or 505 MBH 5.9 7.6 8.4 SR-3 14.7 0.025 20,000 60 30,000 30,000 5.9 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 Total Load 504551 Btuh or 505 MBH 7.6 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 0.518 600 60 18,648 42,648 8.4 204,000 60 220320 204,000 0.0060 97,308 317628 63.1 SR-3 504551

Annual Fuel Consumption

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=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. 258, 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. 255, 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.258, T.8.7

Determine Heat Content (V) Heat content is the quantity of Btu/unit S: p. 255, 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 =1,111 therms/yr If gas is $0.41/therm, then annual cost = $456

Simple Payback Analysis

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

Simple Payback Heating System Cost Comparison First Annual Incremental Incremental Simple Cost Fuel Cost First Cost Annual Savings Payback ($) ($/yr) ($) ($/yr) (yrs) Electricity 6,000 2,010 --- --- --- Oil 8,000 1,152 2,000 858 2.3 Gas 8,900 456 2,900 1,554 1.9 If money is available, select gas furnace system