1. HEATING LOAD ANALYSIS
Ethan Ehrman and John Lamb
Professor Ware
CSM 2310
April 19, 2013

2. The work for the heat load analysis of 374 East 13 Avenue apartment E was
divided equally among team members. Mr. Lamb was in charge of and completed r-value
calculations for exterior walls. Mr. Ehrman was in charge of and completed dimensioning
through Auto CAD. Both members worked on excel spreadsheets and charts as well as
preparing for the in lab presentation.
The dimensions of the apartment are 25’ long by 32’ wide with 8’ wall height.
There are two exterior walls, east and west, and flat roof overhead. Two windows on each
exterior wall that are double pane insulated with low-e and argon that measure 3’ by 4’ ,
and single steel door 1 ¾” thick with a rigid foam core standard height with a width of 3’
on the west wall are the only areas where it is not solid wall. Based upon these
dimensions it is calculated that the volume of the living space is approximately 6400
square feet with an 800 square foot roof and that the square footage of east and west wall
respectively are 176 and 156 square feet with the removal of windows and doors taken
into account.
The construction of exterior walls determined from removing a receptacle cover
and depth of window sill consists of 5/8” gypsum, 2x6 framing, ½” OSB, and 3 ½” face
brick. Due to the fact that we were unable to precisely determine the construction of the
roof it is assumed that because it is a flat roof with a single ply membrane, ¾” OSB
decking, 2x12 framing, and 5/8” interior gypsum. Based on this construction we
determined that the r-value for exterior walls during the winter is 20.695 and that the rvalue of the roof for winter is 42.42. Also because this is considered campus housing and
there are places where air infiltration around windows and doors can be felt an air
exchange rate of .87 is assumed and used for calculations. The r-value for windows and
doors respectively is 3.1 and 5.3. The inside design temp is 72 degrees Fahrenheit and 1%
criteria for Columbus Ohio is 6 degrees Fahrenheit.
Based on these calculated values we were able to use formulas provided in our
text and in class to calculate heat lost through each exterior medium and through
infiltration. Total transmission through the west mediums was 1257.53 BTU/HR, east
wall was 1072.2627 BTU/HR, and through roof materials was 1244.7 BTU/HR. This
equates to 3574.4945 BTU/HR through transmission above grade. (Refer to attached
spreadsheets for complete breakdown) Calculated heat loss through infiltration based on
assumed ACH was 6614.7840 BTU/HR. Combined heat losses through infiltration and
transmission is 10189.2785 BTU/HR.
Upon completion of these calculations the cover from furnace was removed and
determined from manufacturers tag that the existing furnace is a 50000 BTU/HR natural
gas unit. Compared to calculated heat load the unit is oversized, but unfortunately a
smaller unit is hard to obtain. Based on heat load calculated and fuel cost of $0.60/ccf
annual heating costs are determined to be $145.45 which is very close to actual costs.
To save 10% on seasonal heating costs it was determined that we needed to
reduce heat load by a minimum of 1018.18 BTU/HR. To do so it was determined that
interior design temperature would be changed from 72 degrees Fahrenheit to 68 degrees

3. Fahrenheit, new triple pane windows with ½” insulation gap between panes would be
installed, and that the ACH rate would need to be reduced to 0.80 by effectively caulking
areas where infiltration could be felt.
Based on labor rate of $20.00/ HR , window cost of $300.00/window (4
windows), 2 hours to remove 1 old window and install 1 new window, $5.50/tube
caulking (3 tubes) and 1 hour to perform caulking work, it was calculated to cost
$1396.50 for alterations. These alterations were then calculated to reduce heat load by
1294.3476 BTU/HR. This would result in yearly savings of $18.48 on heating costs. It
was then determined that payback for the repairs based on yearly savings would take
75.58 years. This improvement would not be worth the payback because likely the costs
could be put toward other improvements to the facility.
Through calculation of r-values and techniques used in class we were able to
determine heat loss for the residence. Based on those calculations we were able to
calculate heating costs for the residence as well as size appropriate equipment. Based on
these findings we were then able to devise a plan to improve efficiency and reduce costs
by a minimum of ten percent.