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NORA Advanced Hydronics

ue. Our example has two outside walls—one ten feet long and the other 15 feet long. To make it easier, simply add the lengths together and make it one big wall. (NOTE—with heating, north, south, east or west facing walls are not a factor as the room will be heated after sundown when no solar gain is applied). In this case, there are 25 linear feet of wall times 9 feet high for a total area of 225 square feet. Subtract the window area from that total—225 minus 60 equals 165 square feet of net wall area. This is the number to use in calculations. The insulation (Figure 5 on preceeding page) in the wall will determine its U value—in this case, it’s six-inches of fiberglass bat at an R-19. R-values for the exterior sheathing, house wrap and finish, as well as for the inside sheetrock may be available and lead to a more accurate calculation. For simplicity and to provide a safety margin, simply use the insulation Rvalue for the calculations. In this case, divide 1 by 19 to find the wall U-value of .05. 165 x 70 x .05 equals 578 Btuh lost through the walls. Outside doors, if there were any, would be calculated in the same way and their area would also be deducted from the net wall area. The ceilings and floors also need to be evaluated, but only if they are below or above unheated space. If the area above the ceiling is heated, there would be no heat loss through the ceiling, and if the floor is above a heated area, there would be no heat loss through the floor. For the ceiling: Multiply the length times the width times the DTD times the U-value. For simplicity, this example will use insulation values. With R-38 in the ceiling, U-value would be 0.02. 10 x 15 x 70 x .02 and there is a loss of 210 Btuh. 24 National Oilheat Research Alliance The floors have R-19 insulation, so calculate a U-value of 0.05. Then multiply 10 x 15 x 70 x .05 for a floor loss of 525 Btuh. Finally, add them all together: Infiltration 1,701 Windows 1,512 Walls 578 Ceiling 210 Floors 525 Total 4,256 The total heat loss for the room is 4,526 Btuh This procedure would be repeated for each room of the building to determine the terminal unit capacity needed. The total of heat loss for all the rooms determines the heat loss for the entire building. Assume the complete heat loss for the building is 70,500 Btuh. A boiler of suitable output must be selected. Depending on the manufacturer’s terminology, choose a boiler based on DOE capacity, the NET IBR RATING or the output Btu/hr. In this case, the proper boiler is the EK-1 model.


NORA Advanced Hydronics
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