Math in HVAC

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Math in HVAC

• Presented by Mike Veeder
• HVAC instructor
• Columbia Greene Questar

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Whole House Heat Lossquestion 1

• Determine design temp difference
• Find total square foot of windows and doors
• Find gross walls
• Subtract windows and doors from gross walls to
  find net walls
• Determine exposed ceiling area
• Find floor area

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• Multiply each by U factor
• Add all btus/degree Fahrenheit
• Multiply by temp difference
• Multiply by 15 for duct losses
• Arrive at heating btu loss for house

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Calculate annual heat lossQuestion 2

• BTU annual BTUH x HDD x 24hr/day x .75

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• BTU annual 875 x 6860 x 24 x .75
• Annual heat loss is 108,045,000 btus

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Propane

• Determine how many gallons of propane burned per
  year
• Gal/yr annual btu / 92,000btu / gal
• Gal/yr 108,045,000/92,000
• Gal/yr 1174 gal
• Cost/yr 1174 gal x 2.60/gal
• Cost/yr 3052.40 x stack loss of furnace
• Cost/yr 3662.88

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Electricity

• Determine the amount of electricity used on a
  annual basis
• Watt/yr btu annual / 3.41 btu/watt
• Watt/yr 108,045,000/3.41
• Watt/yr 31,684,750 / 1000
• KW/yr 31,684 x .14/kw
• Yearly cost of electric is 4435.87

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Calculate BTU output of a Heat PumpQuestion 5

• BTU1.08 x cfm x delta T
• CFM FPM x .66 x sq. ft of duct area

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Determine cfm of a Geothermal Heat pumpQuestion
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• CFM fpm x .66 x sq. ft
• CFM 606 fpm x .66 x 2
• CFM 800 cfm
• Btu 1.08 x cfm x delta t
• Btu 1.08x 800 x (92 65)
• Btu 23,328 btu

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Geothermal loop performanceQuestion 6

• Loop filled with 20 Methanol
• Btu 485 x gpm x Delta t
• Btu 485 x 6 x (40 32)
• Btu 23,280 btu

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BPI Building Airflow Standard(BAS)

• Living Space Area 1500 sq. ft.
• Basement area 700 sq. ft.
• Ceiling height 8 ft.
• of Occupants 4
• of stories above grade 2
• Location Albany, NY

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Calculate ventilation required for building

• Airflow(b) .35 x volume/60
• Volume 8 x (1500 700) 17,600 cu. ft.
• Airflow(b) .35 x 17,600 / 60
• Airflow(b) 102 cfm

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Calculate ventilation required for people

• Airflow(p) 15 x of occupants
• Airflow(p) 15 x 4 60 cfm

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Using the higher airflow requirement

• Convert cfm to cfm50
• Minimum cfm50 airflow x N
• N LBL conversion factor
• Minimium cfm50 102 cfm x 15.4 1570 cfm50

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• When we read the cfm50 from the blower door, 1570
  is what is recommended for infiltration for this
  house

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Determine course of action for home

• Cfm50 x .7 minimum ventilation
• 1570 cfm50 x .7 1099 cfm50

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• If the blower door reads
• Above 1570 cfm then the house is too loose and
  you would recommend air sealing

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• If the blower door reads
• Between 1570 cfm50 and 1099 cfm50, then the house
  is within the normal range for infiltration and
  ventilation

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• If the blower door reads
• Below 1099 cfm50 then you need to mechanically
  ventilate the house using a hrv or equivalent
  device. The house is too tight at this point.

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My class after a math lesson
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Thanks for your attention !!

• Any questions??????????