Department of Natural Sciences
Clayton State University
Morrow, Georgia

SCI 1901E
Class Example
Home Heating Audit
Supporting Calculations

This document provides calculations associated with the heating audit for a very simple example home, consisting of only two rooms, and is used to introduce SCI 1901E students to the types of calculations that are required for the Home Heating Audit during a class period.

Information about the simple home is provided for:

Calculations provided are for:

Degree days:

The degree day calculation requires:

You also need to know how many days are in each month of the heating season.

Examine the table in Appendix D of your text, and select information only for those months for which the average thermostat setting is greater than the outdoor average temperature - these are the months when you use heat. In our example, for a location near Atlanta, Georgia with an average thermostat setting of 70°F, the following months require heat:

January, February, March, April, October, November and December.

Note: the actual months used in the calculation for your own home heating audit will depend on your thermostat setting. For example,if your average thermostat setting is 64°F, the months of April and October would not be included in the degree day calculations because the average monthly temperature in Atlanta for those months is greater than 64°F.

The degree days for each month are calculated by multiplying the difference between the thermostat temperature and the outdoor temperature by the number of days in the month.

For example, the calculation for January would be:

Ttherm = 70°F

Tavg = 47.2°F
(from the table in Appendix D of your text)

DT = 70°F - 47.2°F = 22.8°F

The number of days in January is 31.

Therefore,

DD(Jan) = (22.8°F) (31 days) = 706.8 °F·days

To determine the number of degree days in the entire heating season, make the degree day calculation for each month and then sum. This is most easily done by constructing a table:

month Tavg
(°F)		 Ttherm - Tavg
(°F)		 days/month
(days)		 DD/month
(°F·days)
Jan 47.2 22.8 31 706.8
Feb 49.6 20.4 28 571.2
Mar 55.9 14.1 31 437.1
Apr 65.0 5.0 30 150.0
Oct 66.5 3.5 31 108.5
Nov 54.8 15.2 30 456.0
Dec 47.7 22.3 31 691.3
Total for entire heating season 3120.9

There is no space on the form for the number of degree days.  You will need this number later, so write it on page 1.

The total number of degree days in the heating season is entered on page 1 of home heating audit forms.

Floor area:

The floor area is the product of the interior length and width of the home. This home (see dimensions) has a length of 40 ft and a width of 30 ft.

Therefore,

floor area = (40 ft) (30 ft) = 1200 ft2

The floor area is entered on page 1 of the heating audit forms.

Volume:

The volume is the product of the interior length, width and height of the home (see description, #14). This is also the floor area times the interior height.

Therefore,

volume = (40 ft) (30 ft) (10 ft) = 12,000 ft3

The volume is entered on page 1 of the heating audit forms.

Energy source cost per MBtu:

To make this calculation you must know

The two sources that are most common are electricity or natural gas.

Electricity is sold by the kWh; natural gas, by the therm. Your instructor will provide a cost per kWh for electricity or a cost per therm for natural gas.

We will assume a conversion efficiency for natural gas furnace to be 0.80, unless better information is available.

We will assume a conversion efficiency for direct electric heating (without a heat pump) to be 0.90 unless better information is available.

We will use the seasonal performance factor (SPF) for you heat pump for the conversion efficiency of that device.

In our example, the source and cost are specified. To complete the calculation:

($1.38/therm)(10 therm/MBtu)/(0.80) = $17.25/MBtu

The energy source cost is entered on page 1 of the home heating audit forms.

Heat loss rates by component for Rooms 1 and 2:

Calculating areas first (see dimensions):

Room 1:

Windows: Room 1 has two 3 ft x 4 ft windows and one 8 ft x 4 ft window.

Awindow = (2)[(3 ft)(4 ft)] + (8 ft)(4 ft) = 56 ft2

Doors: Room 1 has one 3 ft x 7 ft door.

Adoor = (3 ft)(7 ft) = 21 ft2

Walls: To find the wall area, calculate the interior area of the sides of the house that are adjacent to the outdoors, and then subtract the areas of the windows and the door. There are two sides of 25 ft x 10 ft, and one side of 30 ft x 10 ft.

Asides = (2)[(25 ft)(10 ft)] + (30 ft)(10 ft) = 800 ft2

Awall = Asides - Awindow - Adoor

Awall = 800 ft2 - 56 ft2 - 21 ft2= 723 ft2

Ceiling and floor: The area of the ceiling and floor are the same in this example. Both the ceiling and floor are 30 ft x 25 ft in Room 1.

Aceling = Afloor = (30 ft)(25 ft) = 750 ft2

Room 2:

Windows: Room 2 has three 3 ft x 4 ft windows.

Awindow = (3)[(3 ft)(4 ft)] = 36 ft2

Doors: Room 2 has one 3 ft x 7 ft door.

Adoor = (3 ft)(7 ft) = 21 ft2

Walls: To find the wall area, calculate the interior area of the sides of the house that are adjacent to the outdoors, and then subtract the areas of the windows and the door. There are two sides of 15 ft x 10 ft, and one side of 30 ft x 10 ft.

Asides = (2)[(15 ft)(10 ft)] + (30 ft)(10 ft) = 600 ft2

Awall = Asides - Awindow - Adoor

Awall = 600 ft2 - 36 ft2 - 21 ft2= 543 ft2

Ceiling and floor: The area of the ceiling and floor are the same in this example. Both the ceiling and floor are 30 ft x 15 ft in Room 2.

Aceling = Afloor = (30 ft)(15 ft) = 450 ft2

Determining surface factors:

A table of surface factors is provided in Appendix C, starting on page C.8., of your text. In this example the windows, doors, ceiling and floor are constructed of materials that have tabulated surface factors that may be determined by examining the table. Results of this lookup are:

component surface type surface factor
(Btu/hr/ft2)
windows single pane 71.4
doors 1 2/3" wood solid core 21
ceiling 12" fiberglass insulation 1.5
floor unheated crawl space
no insulation		 15

The wall construction is not listed in the table and its surface factor must be calculated. To obtain a surface factor that is not tabulated, use the R-values provided in Table 5-2 on page 134 of your text and the following relationships:

Rtotal = R1 + R2 + R3 + ...

surface factor = (65°F) / Rtotal

From that table, R-values of the materials that make up the walls are

material thickness R-value
(ft2·hr·°F / Btu)
common brick 1" 0.20
fiberglass insulation 3 1/2" 10.9
sheetrock 1/2" 0.45

Therefore,

Rtotal = [(4)(0.20) + (10.9) + (0.45)] ft2·hr·°F / Btu

Rtotal = 12.15 ft2·hr·°F / Btu

surface factor = (65°F) / (12.15 ft2·hr·°F / Btu) = 5.35 Btu / (ft2·hr)

Finally, to calculate the conductive heat loss rate for each of the surface types, multiply the surface factor and the area for each type together. For example, for the windows in Room 1:

Qc / t (windows) = SFwindow x Awindow

Qc / t (windows) = [71.4 Btu / (ft2·hr)] (56 ft2) = 3998 Btu / hr

This calculation is then made for each surface in each room.

Information from the calculations associated with the conductive heat loss rates through the exterior surfaces of the examples home has been entered into the second page of the forms for the Home Heating Audit.

Summary page.

       Section III.A.

Conduction loss rate:

Calculate the conductive heat loss rate for each surface type in the entire house by summing heat loss rates for each surface type from each room. Examine the results of calculations tabulated on the second page of the forms for the home heating audit to obtain the required data.

For example, for the windows in the example home.

Qc / t (windows,Room 1) = 3998 Btu / hr

Qc / t (windows,Room 2) = 2570 Btu / hr

Qc / t (windows, all) = 3998 Btu / hr + 2570 Btu / hr = 6568 Btu / hr

Perform these calculations for each surface type and enter into the "Sum of Rows" column of the summary table on the 3rd page of the heating audit forms.

 

    Air infiltration heat loss rates

To make this calculation you must estimate which description of the tightness of your home is best from those provided on the 3rd page of the home heat audit forms.

In this example, the home is described as an "average insulated house, well maintained." This results in a selection of K = 1 turnover of air per hour.

Use the infiltration gain equation on the 3rd page of the home heat audit forms with the volume previously calculated and listed on the 1st page of the home heat audit forms.

Qinfil / t = [0.018 Btu / (ft3·°F)] K V (65°F)

Qinfil / t = [0.018 Btu / (ft3·°F)] (1/hr) (12000 ft3) (65°F)

Qinfil / t = 14040 Btu / hr

 

    Total peak hourly loss rate:

 Complete this column by summing all the heat loss rates to provide the total peak hourly loss rate.

In this example the total is 48,063 Btu / hr.

Complete the summary table by calculating the percentage of the heat loss rate from each of the surfaces and infiltration.

For example, the percentage of the total heat loss rate due to windows is:

% total loss (windows) = (6568 Btu / hr ) / (48063 Btu / hr) (100%)

% total loss (windows) = 13.67%

These percentages for the example home are entered into the "Percent Total Loss" column of the summary table in on the 3rd page of the audit forms.

The remaining calculations on page 3 of the form are described on the form itself. The descriptions of the calculations below refer to the letter of the lines on the 3rd page of the heating audit form.

     Losses as a result of heat transfer from central heating systems (B):

Because some heat energy is lost in the process of transfer from the central heating system to the rooms of the house, the net heat loss rate must be adjusted to determine the peak heating load that the heating system must deliver. The size of the adjustment depends upon the location and insulation of the home's air ducts.

In this example, with insulated heating ducts:

Q / t = (1.1 ) Q / t (net loss)

Q / t = (1.1)(48063 Btu / hr) = 52,869 Btu / hr

Internal heat gains (C):

To complete this portion of the calculations, determine:

In this example, the water heater is not located in a heated area; two persons occupy the home; and there are two South facing windows, having dimensions of 8 ft x 4 ft and 3 ft x 4 ft. With this information the internal heat gains can be determined.

Q / t (appl & lights) = 2000 Btu / hr

Q / t (water heater) = 0 Btu / hr

Q / t (people) = (2 people)(400 Btu / hr / person) = 800 Btu / hr

A (South windows) = (8 ft)(4 ft) + (3 ft)(4 ft) = 44 ft2

Q / t (South windows) = (35 Btu / hr / ft2)(44 ft2) = 1540 Btu / hr

Q / t (internal, total) = 2000 Btu / hr + 0 Btu / hr + 800 Btu / hr + 1540 Btu / hr

Q / t (internal, total) = 4340 Btu / hr

Peak net heating loss (D):

The net heat loss rate is total heat loss rate by conduction and infiltration (A) minus the internal gain rate (B).

In this example:

Q / t (net loss) = 52,869 Btu / hr - 4340 Btu / hr = 48,529 Btu / hr

Seasonal heating needs (E):

Use the equation provided on the form to calculate the amount of heat energy required for the entire heating season. You will need the number of degree days (DD) calculated previously and entered on page 1 of the heating audit forms.

Q = (Q / t)(DD)(24 hr / day) / (65°F) / (106 Btu / MBtu)

In this example,

Q = (48,095 Btu / hr)(3120.9 °F·days)(24 hr / day) / (65°F) / (106 Btu / MBtu)

Q = 55.92 MBtu

Total heating season costs (F):

The predicted cost for a heating season is then given by the product of the seasonal energy heating needs (E) and the cost per MBtu of the energy source calculated previously and entered on page 1 of the heating audit forms.

In this example,

Cost = (55.92 MBtu) ( $17.25 / MBtu) = $964.62

Results of the calculations in parts A through F are entered on the 3rd page of the home heating audit forms.

Last update: February 11, 2009