INTRODUCTION:PROBLEM STATEMENT:ASSUMTIONS:CALCULATIONS:ANALYSIS:Using the data from Table 2, we can calculate the cost of heating the building for one day (45ºF temperature difference) using various fuel sources. Table 3 shows the results of this analysis:CONCLUSION:EGR 468 PROJECT:KELLER HEAT LOSS CALCULATIONDr. David BlekhmanDr. Shirley FleischmannFebruary 16, 2003Brad Vander VeenINTRODUCTION:The Keller Engineering Lab at 301 West Fulton is a 30,000 square foot facility that accommodates the Grand Valley engineering students 24 hours a day, 7 days a week. During the winter months, heating the building to a comfortable level is important to the students. The lab consists of two levels of rooms for different experiments, machining, aswell as workstations for student to do their homework at. One unique feature that makes Keller difficult to heat is the amount of windows that are present. Most of the front and sides of the building are covered with windows. This makes the Keller Labs a nice place to work, but a very expensive building to heat! The Keller Lab building is heated with two large boilers on the lower floor of the building. The air is preheated on the roof and then vented into the lab. Hot water from the boilers is then pumped through finned, copper pipes around the building, and free convection heats the air in Keller to a comfortable temperature. In this document, the total heat loss for Keller will be calculated for a given temperature difference.PROBLEM STATEMENT:We need to determine the heat loss for the entire Keller Engineering Lab when the inside temperature is 65ºF and the temperature outside is -10ºF.ASSUMTIONS:There are some assumptions that will be made for this analysis. These assumptions are listed below:a) steady state conditionsb) convection on inside and outside walls wills be neglectedc) the lab will be treated as a “big box”d) outside surface area is treated as either a brick wall or a windowe) the lab undergoes 1.5 air exchanges per hourf) assume a constant temperature throughout the buildingCALCULATIONS:Below is a listing of all the measurements taken of the Keller Engineering Lab, as well asall other values used in the analysis:total_window_area 7.47 103 ft2total_wall_area 1.224 104 ft2total_window_area window_south window_east window_north window_westtotal_wall_area wall_south wall_east wall_north wall_westtotal_roof_area 15572 ft2total_floor_area 15572 ft2Total Area:wall_west 2492 ft2window_west 1076 ft2total_west 3568 ft2West Side:wall_north 5063 ft2window_north 845 ft2total_north 5908 ft2North Side:wall_east 1892 ft2window_east 436 ft2total_east 2328 ft2East Side:wall_south 2789 ft2window_south 5113 ft2total_south 7902 ft2South Side:Area Declaration:Height:average_height 32 ftVolume:total_volume total_floor_area average_heighttotal_volume 4.983 105 ft3Perimeter:perimeter 500 ftTemperature Difference:delta_T 75 RProperties of Air: .00238slugft3 c 0.240BTUslug RAir Exchanges per Hour:occurrence_per_hour 1.51hrNext we will find the resistance to heat loss for each of the boundaries: Resistance and Heat Loss Coefficents:Windows:R_window 1.538hr ft2 RBTUWalls:R_cinder_block 1.72hr ft2 RBTUR_insulation 9hr ft2 RBTUR_air_gap .97hr ft2 RBTUR_face_brick .44hr ft2 RBTUR_wall_total R_cinder_block R_insulation R_air_gap R_face_brickR_wall_total 12.13hr ft2 RBTURoof:R_roof 13.5hr ft2 RBTUFloor:floor_loss 65BTUhr1ftNow using the building properties and the heat loss coefficients, we will now calculate the overall heat loss:Heat Loss Calculations:heat_thru_wallstotal_wall_area delta_TR_wall_totalheat_thru_walls 7.566 104BTUhrheat_thru_windowstotal_window_area delta_TR_windowheat_thru_windows 3.643 105BTUhrheat_thru_rooftotal_roof_area delta_TR_roofheat_thru_roof 8.651 104BTUhrheat_thru_floor floor_loss perimeterdelta_T70 Rheat_thru_floor 3.482 104BTUhrheat_from_infiltration  total_volume c delta_T occurrence_per_hourheat_from_infiltration 3.202 104BTUhrtotal heat_thru_walls heat_thru_windows heat_thru_roof heat_thru_floor heat_from_infiltrationtotal 5.933 105BTUhrANALYSIS:In Table 1 below, a summary of the heat loss through each boundary is shown:Table 1 – Heat Loss by Category for 75º Temperature Difference heat loss [BTU/hr]wall 75660window 364300roof 86510floor 34820infiltration 32020TOTAL 593310In Figure 1 below, the heat loss by category can be seen graphically in a pie graph:Figure 1 – Pie Graph of Heat Loss by Category for 75º Temperature DifferenceUsing the same process as outlined in the calculations section, we can also find the heat loss from Keller with a more reasonable temperature difference. For this analysis, we will use an inside temperature is 65ºF and an outside temperature of 20ºF. This will give results typical for the winter months.In Table 2 below, a summary of the heat loss through each boundary:Table 2 – Heat Loss by Category for 45º Temperature Difference heat loss [BTU/hr]wall 45390window 218600roof 51910floor 20890infiltration 19210TOTAL 356000In Figure 2 below, the heat loss by category can be seen graphically in a pie graph:Figure 2 – Pie Graph of Heat Loss by Category for 45º Temperature DifferenceNote that although the values change for a different temperature difference, the percentage of heat through each boundary does not change. This is because the heat transfer rate is directly proportional to the temperature difference across the resistive boundary.Using the data from Table 2, we can calculate the cost of heating the building for one day (45ºF temperature difference) using various fuel sources. Table 3 shows the results of this analysis:Table 3 – Comparison of Fuel SourcesFuel Source BTU/W*hr W*hr/dollar BTU/dollar BTU needed/day dollars/dayElectricity 3.4121 12704.000 43347 8544000 $197.11Fuel Source BTU/ft^3 ft^3/dollar BTU/dollar BTU needed/day dollars/dayGas 1,000 25.445 25445 8544000 $335.78Fuel Source BTU/gal gal/dollar BTU/dollar BTU needed/day dollars/dayOil 130,000 0.735 95588