Transcript Math in HVAC

Math in HVAC
Presented by Mike Veeder
HVAC instructor
Columbia Greene Questar
Whole House Heat Loss
question #1
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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 btu’s/degree Fahrenheit
Multiply by temp difference
Multiply by 15% for duct losses
Arrive at heating btu loss for house
Calculate annual heat loss
Question #2
 BTU annual = BTUH x HDD x 24hr/day x
.75
• BTU annual = 875 x 6860 x 24 x .75
• Annual heat loss is 108,045,000 btus
Propane
• Determine how many gallons of propane burned
per year
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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
Electricity
• Determine the amount of electricity used on a
annual basis
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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
Calculate BTU output of a Heat
Pump
Question 5
• BTU=1.08 x cfm x delta “T”
• CFM= FPM x .66 x sq. ft of duct area
Determine cfm of a Geothermal
Heat pump
Question #4
• 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
Geothermal loop performance
Question #6
Loop filled with 20% Methanol
Btu = 485 x gpm x Delta “t”
Btu = 485 x 6 x (40 – 32)
Btu = 23,280 btu
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
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
Calculate ventilation required for
people
Airflow(p) = 15 x # of occupants
Airflow(p) = 15 x 4 = 60 cfm
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
When we read the cfm50 from the
blower door, 1570 is what is
recommended for infiltration for this
house
Determine course of action for home
 Cfm50 x .7 = minimum ventilation
 1570 cfm50 x .7 = 1099 cfm50
 If the blower door reads:
– Above 1570 cfm then the house is too loose
and you would recommend air sealing
 If the blower door reads:
– Between 1570 cfm50 and 1099 cfm50, then the
house is within the normal range for infiltration
and ventilation
 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.
My class after a math lesson
Thanks for your attention !!
 Any questions??????????