2345 project
- 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.