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Environmental Controls I/IG Lecture 11 Degree Days Heating Loads Annual Fuel Consumption Simple Payback Analysis Lecture 11 Degree Days Heating Loads Annual Fuel Consumption Simple Payback Analysis

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

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Sample Calculation January TA=28ºF DD 65 =65-28= 37 Degree-days/day x 31 days = 1,147 degree-days S: p. 1524, T.C.15

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Heating Loads

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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

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Outdoor Dry Bulb Temperature Use Winter Conditions S: p. 1496, T.B.1

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

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

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Determine Area Quantities Perform area takeoffs for all building envelope surfaces on each facade: gross wall area window area door area net wall area 4’ Elevation 4’ 12’ 100’ 8’ 1200 sf 64 sf 368 sf 768 sf - -

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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

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Slab to Soil Losses Q=U slab x 0.5 x A slab x (T I -T GW ) T I =Indoor Air Temperature T GW =Ground Water Temperature

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Edge Losses Method I Determine F 2 based on heating degree days S: p. 1582, T.E.11

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Slab Edge Losses Method II Select F 2 based on insulation configuration S: 1583, T.E.12

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Slab Edge Losses Q=F 2 x Slab Perimeter Length x (T I -T O ) where, T I = Indoor air temperature T O =Outdoor air temperature

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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) U wall = Btuh/sf-ºF U roof = Btuh/sf-ºF U window = 0.31 Btuh/sf-ºF U slab = 0.16 Btuh/sf-ºF U door = 0.20 Btuh/sf-ºF

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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, , Doors Net Wall3,9801,9802,9501,980 Roof/Floor Slab20,000

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Heating Loads Insert roof values Insert wall values Insert glass values Insert door values Insert floor values SR , ,000 30,000 N , ,895 E ,980606,415 S0.0542,950609,558 W0.0541,980606,415 38,555 N0.31 1, ,600 E ,300 S0.312, ,200 W ,300 74, ,3201,320 N/A N/AN/AN/A

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Slab to Soil Losses Q=U slab x 0.5 x A slab x (T I -T GW ) T I =Indoor Air Temperature T GW =Ground Water Temperature Ground Water= 53ºF ΔT=68ºF-53ºF=15ºF

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Heating Loads Insert floor values SR , ,000 30,000 N , ,895 E ,980606,415 S0.0542,950609,558 W0.0541,980606,415 38,555 N0.31 1, ,600 E ,300 S0.312, ,200 W ,300 74, ,3201,320 N/A N/AN/AN/A , ,000

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Edge Losses Method I Determine F 2 based on heating degree days S: p. 1582, T.E.11

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Heating Degree Days Salt Lake City HDD 65 =5983 S: p. 1524, T.C.15

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Edge Losses Method I Interpolate to find F 2 at 5983 DD F 2 ?0.56 S: p. 1582, T.E. 11

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Interpolate to Find F 2 Find difference in Degree Days: = =2083 Find difference in F 2 :F 2 ?-0.50=x =0.06 Set up proportion, solve for x:633/2083=x/0.06 x=0.018 F 2 ?-0.50=0.018 F 2 ?=0.518

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Edge Losses Method I Interpolate to find F 2 at 5983 DD F 2 = S: p. 1582, T.E.11

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Heating Loads Insert floor values SR , ,000 30,000 N , ,895 E ,980606,415 S0.0542,950609,558 W0.0541,980606,415 38,555 N0.31 1, ,600 E ,300 S0.312, ,200 W ,300 74, ,3201,320 N/A N/AN/AN/A , , ,64842,648

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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

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Ventilation Analysis Non-residential buildings use ventilation to provide fresh air and to offset infiltration effects. ASHRAE Standard (S: p , T.E.25) Estimates the number of people/1000 sf of usage type Prescribes minimum ventilation/person for usage type

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

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

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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

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Heating Loads Input Ventilation Load—Sensible SR , ,000 30,000 N , ,895 E ,980606,415 S0.0542,950609,558 W0.0541,980606,415 38,555 N0.31 1, ,600 E ,300 S0.312, ,200 W ,300 74, ,3201,320 N/A N/AN/AN/A , , ,64842, , ,320

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Ventilation Load — Latent Determine ΔW W I = #H 2 O/#dry air -W O = #H 2 O/#dry air ΔW= #H 2 O/#dry air

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Heating Loads Input Ventilation Load — Latent SR , ,000 30,000 N , ,895 E ,980606,415 S0.0542,950609,558 W0.0541,980606,415 38,555 N0.31 1, ,600 E ,300 S0.312, ,200 W ,300 74, ,3201,320 N/A N/AN/AN/A , , ,64842, , , ,

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Heating Load Total Load Btuh or 505 MBH SR , ,000 30,000 N , ,895 E ,980606,415 S0.0542,950609,558 W0.0541,980606,415 38,555 N0.31 1, ,600 E ,300 S0.312, ,200 W ,300 74, ,3201,320 N/A N/AN/AN/A , , ,64842, , , ,

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Annual Fuel Consumption

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Annual Fuel Usage (E) E= UA x DD BPT x 24 AFUE x V where: UA: heating load/ºF DD BPT : degree days for given balance point AFUE: annual fuel utilization efficiency V: fuel heating value

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Calculating UA Q Total = UA x ΔT UA= Q Total /ΔT From earlier example: Q Total =504,551 Btuh ΔT= 60ºF UA=504,551/60=8,409 Btuh/ºF

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Determine AFUE Annual Fuel Utilization Efficiency of an electric heating system is 100% S: p. 258, T.8.7

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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

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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

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Determine AFUE Annual Fuel Utilization Efficiency of an electric heating system is 100% S: p.258, T.8.7

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Determine Heat Content (V) Heat content is the quantity of Btu/unit S: p. 255, T.8.5

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Annual Fuel Usage — Electricity E= UA x DD BPT x 24 AFUE x V E ELEC =(650)(5,983)(24)/(1.0)(3,413) =27,347 kwh/yr If electricity is $0.0735/kwh, then annual cost = $2,010

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Annual Fuel Usage — Gas E= UA x DD BPT x 24 AFUE x V E Gas =(650)(5,983)(24)/(0.8)(105,000) =1,111 therms/yr If gas is $0.41/therm, then annual cost = $456

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Simple Payback Analysis

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Simple Payback Heating System Cost Comparison First Cost ($) Electricity6,000 Oil8,000 Gas8,900

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Simple Payback Heating System Cost Comparison FirstAnnualIncremental Incremental Simple Cost Fuel CostFirst CostAnnual SavingsPayback ($)($/yr)($)($/yr)(yrs) Electricity6,0002, Oil8,0001,1522, Gas8, ,9001, If money is available, select gas furnace system

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