2. EXAMPLE EXERCISE CALCULATING HEAT LOSS & HEAT GAIN  Several exhibits in the class packet are necessary to understand the entries into the Heat Loss / Heat Gain calculation sheet.  The example floor plan will be used to make calculations required for the selection of mechanical equipment necessary to maintain comfort heating and cooling.  Certain criteria are given on the lower left section of the floor plan to be used in the calculation. These are criteria established in the design of the building envelope, for the type building for which it is intended.

4.  Heat flow by conduction in Btu/h, through low mass or thin surfaces, such as doors and glass, is calculated by multiplying the area, times the “U” factor, times the difference in temperature from one side of the material to the other.  The temperature differential is the difference between the recognized high (for summer) or low (for winter), from climatilogical data, and the temperature desired to be maintained within the space.  The quantity of heat by conduction that passes through surfaces of greater mass, such as walls and roofs is calculated by multiplying the area, times the “U” factor, times the Equivalent Temperature Difference.

5. RADIANT HEAT FROM DIRECT SUN IN SUMMER Radiant heat that enters a space by shining through a transparent or semi-transparent surface, such as glass is calculated by multiplying the area times the amount of Solar Gain, reduced only by the effectiveness of a shading coefficient AS AN ILLUSTRATION, calculate the total heat loss (winter) and heat gain (summer) for the plan of the example building.The construction is medium weight masonry, such as brick veneer over concrete masonry units. In this area, the peak summer daytime temperature occurs around 4:00 P.M. Consider that the color of masonry is light, such as would be the shade of the brick on the architecture building.

7.  “U” values were selected from reasonable allowance of materials. ETD and Solar Gain values were ETD and Solar Gain values were selected from the charts. selected from the charts. Shading coefficient was selected Shading coefficient was selected from a reflective glass. from a reflective glass.

8. 21 26 31 36 48 ETD values

9. Another factor that relates to direct sunlight is solar gain through glass. Direct sunlight will pass through clear glass without an appreciable effect on the glass itself, since glass is light in mass compared to the building envelope. Solar gain, referenced by S g on the chart, is the amount of heat gain in Btu per square foot, as the result of sun radiation that penetrates glass. The angle of the sun is taken into account as the result of the tilt of the earth, as well as the time of the day. In this area, 4:00 P.M. is the peak summer temperature. June 21, the beginning of summer produces the worst sun angle on the north and east side of a building, while September 21, the beginning of fall produces the worst sun angle for south and north.

10. 66 196 SOLAR GAIN 23

11.  In order to determine a reasonable temperature difference between outside and inside, consult with climatilogical data for the area a building is located. The chart that follows is taken from the appendix of the text, and gives the outside design temperatures for winter and summer for Lubbock, Texas, in terms of dry-bulb temperatures. The wet-bulb temperature given in the summer column is a measure of the relative humidity. Inside temperature is set by the designer as the maintained desired interior temperature.

12. CLIMATILOGICAL DATA PAGE 1630 OF TEXT

13.  In the chart, note that winter outside design temperature is 15 degrees, while summer design temperature is 96 degrees.  For the purpose of the example problem, say it is desired to maintain a temperature inside of 75 degrees, for both summer and winter. So, temperature difference for summer conditions would be 96 – 75 = 21 degrees. And for winter conditions the temperature difference is 75 – 15 = 60 degrees.

16. 339.075 21534 60 1,526

17. 339 534.075 21 26 534 1,041 60 1,526 2,403

18. 339 534 216.075 21 26 31 534 1,041 502 60 1,526 2,403 972

19. 339 534 216 484.075 21 26 31 36 534 1,041 502 1,307 60 1,526 2,403 972 2,178

20. 339 534 216 484.075 21 26 31 36 534 1,041 502 1,307 60 1,526 2,403 972 2,178 21.55 21 243 693 60

21. 339 534 216 484.075 21 26 31 36 534 1,041 502 1,307 60 1,526 2,403 972 2,178 21.55 21 243 693 2,200.05048 60 5,280 6,600

22. 339 534 216 484.075 21 26 31 36 534 1,041 502 1,307 60 1,526 2,403 972 2,178 21.55 21 243 693 2,200.05048 60 5,280 6,600 0 21 60

23. 339 534 216 484.075 21 26 31 36 534 1,041 502 1,307 60 1,526 2,403 972 2,178 21.55 21 243 693 2,200.05048 60 5,280 6,600 0 96.60 21 60 1,2103,456

24. 339 534 216 484.075 21 26 31 36 534 1,041 502 1,307 60 1,526 2,403 972 2,178 21.55 21 243 693 2,200.05048 60 5,280 6,600 0 96 144.60 21 60 1,210 1,814 3,456 5,184

25. 339 534 216 484.075 21 26 31 36 534 1,041 502 1,307 60 1,526 2,403 972 2,178 21.55 21 243 693 2,200.05048 60 5,280 6,600 0 96 144 128.60 21 60 1,210 1,814 1,613 3,456 5,184 4,608

26. 339 534 216 484.075 21 26 31 36 534 1,041 502 1,307 60 1,526 2,403 972 2,178 21.55 21 243 693 2,200.05048 60 5,280 6,600 0 96 144 128 0.60 21 60 1,210 1,814 1,613 3,456 5,184 4,608

27. 339 534 216 484.075 21 26 31 36 534 1,041 502 1,307 60 1,526 2,403 972 2,178 21.55 21 243 693 2,200.05048 60 5,280 6,600 0 96 144 128 0 96.60 21 60 1,210 1,814 1,613 3,456 5,184 4,608 23.75 1,656

28. 339 534 216 484.075 21 26 31 36 534 1,041 502 1,307 60 1,526 2,403 972 2,178 21.55 21 243 693 2,200.05048 60 5,280 6,600 0 96 144 128 0 144 96.60 21 60 1,210 1,814 1,613 3,456 5,184 4,608 23 66.75 1,656 7,128

29. 339 534 216 484.075 21 26 31 36 534 1,041 502 1,307 60 1,526 2,403 972 2,178 21.55 21 243 693 2,200.05048 60 5,280 6,600 0 96 144 128 0 144 96 128.60 21 60 1,210 1,814 1,613 3,456 5,184 4,608 23 66 196.75 1,656 7,128 18,816 41,144 Sub Totals27,620

31. The heat gain and heat loss calculated in the top part of the chart involved the integrity of the building envelope. An examination of the components would reveal that the foremost consideration of this building design should be an analysis of what can be done about the windows. Realize the AREA of the windows is not the main concern, but rather the orientation of the building with regard to window placement. Glass area on the south side (next to worse solar gain) is 144 sq.ft. compared to 128 on the west side, yet the solar heat gain on the west side is more than 2 ½ times that on the south. A more judicious concern for plan arrangement would result in better conservation of energy. The bottom half of the chart is concerned mainly with internal conditions of the building and how they affect the overall heat loss/gain.

32.  PERIMETER: Heat loss only. Refers to the perimeter of the building; specifically at floor level near the ground outside. During winter the ground will remain cold, since soil does not readily convert electromagnetic energy to heat because of it’s relatively light density. The ground temperature at the building will probably be cold at least 12” deep. So, the temperature difference between the soil and inside the building at the floor level will be sufficient to cause a significant heat loss around the perimeter of the building. On the page you have in the packet labeled “ETD”, find the written material on the left column of the page under the words PERIMETER HEAT LOSS reads: Perimeter Insulation:... Use a value of.81 btu/h for each UNINSULATED foot of building perimeter. Perimeter Insulation:... Use a value of.81 btu/h for each UNINSULATED foot of building perimeter. Use a value of.55 btu/h for each INSULATED foot of building perimeter. Use a value of.55 btu/h for each INSULATED foot of building perimeter. Assume the example problem is INSULATED.

34. Building perimeter = 68+40+68+40 = 216’

35. 216 x 0.55 x 60 = 7,128 21 60 41,14427,620 Sub-totals from above... 7,128

36. VENTILATION is the removal of unwanted air from a space. Ventilation cannot happen unless stale air is replaced by new air – and the only source for new air is outside the space. VENTILATION is the removal of unwanted air from a space. Ventilation cannot happen unless stale air is replaced by new air – and the only source for new air is outside the space. If outside air is brought into the space, it must be thermally treated in order to blend with comfort air. So, in summer, heat must be removed from the air, and in winter, heat must be added. The measurement for handling air quantity is CUBIC FEET PER MINUTE (cfm), so a transition must be made to convert cubic feet per minute to btu per hour. The ventilation requirement for the building is 400 cubic feet per minute – 200 for toilets and 200 for the room on the southwest side of the building.

38. So, heat gain, or heat loss from ventilation air is calculated simply ; CFM ventilation x 60 x.018 x temp. diff. = btu/h But remember to use the temperature difference that applies to summertime for heat gain, and the temperature difference that applies to wintertime for heat loss.

39. 216 x 0.55 x 60 = 7128 400 x 60 x.018 x 21 = 9072 400 x 60.018 x 60 = 25,920 9,072 7,128 25,920 21 60 41,14427,620 Sub-totals from above...

40.  INFILTRATION is unwanted air that gets into the space by infiltrating through cracks around doors and windows and by other means by which un-conditioned outside air can get inside. To calculate the amount of air, use a factor of.50 for each linear foot of crack around doors and windows. The number,.50 represents ½ CFM per foot of crack. The amount of crack is the measurement of the perimeter of each window sash that is movable, and doors. So the heat loss or gain from infiltration equals total length of crack x.5 x 60 x.018 x temperature difference (summer or winter)

42. 216 x 0.55 x 60 = 7128 400 x 60 x.018 x 21 = 9072 400 x 60.018 x 60 = 25,920 9,072 7,128 25,920 21 60 1,2473,564 110 x.5 x 60 x.018 x 21 = 1247 110 x.5 x 60 x.018 x 60 = 3564 41,14427,620 Sub-totals from above...

43.  PEOPLE means the number of people that occupy the space and contribute heat. The amount varies with the size of individuals and their level of activity. For the purpose of this calculation, assume that approximately ten people will occupy the space, and their physical activity is rather tranquil. So use 250 btu/h for each person = 250 x 10 = 2500 Btu/h (the number of people that occupy public buildings is also given in the building code)

44. 216 x 0.55 x 60 = 7128 400 x 60 x.018 x 21 = 9072 400 x 60.018 x 60 = 25,920 9,072 7,128 25,920 21 60 1,2473,564 110 x.5 x 60 x.018 x 21 = 1247 110 x.5 x 60 x.018 x 60 = 3564 10 x 250 = 2,5002,500 41,14427,620 Sub-totals from above...

45.  ELECTRIC WATTS involves the amount of heat generated by the consumption of electricity within the space, such as the operation of mechanical equipment and electric lights. Since all the answers as to the electrical design is generally not known at the time these calculations are done, use an allowance of two watts per square foot of space. 2200 square feet x 2 = 4400 watts Since one electrical watt produces heat of 3.4 btu / h, the total number of btu equals, 4400 x 3.4 = 14,960 Btu/h

46. 216 x 0.55 x 60 = 7128 400 x 60 x.018 x 21 = 9072 400 x 60.018 x 60 = 25,920 9,072 7,128 25,920 21 60 1,2473,564 110 x.5 x 60 x.018 x 21 = 1247 110 x.5 x 60 x.018 x 60 = 3564 10 x 250 = 25002,500 2200 x 2 x 3.4 = 14,96014,960 41,14427,620 Sub-totals from above...

47.  Note on the chart that PERIMETER is heat loss only, since there is no place for values in the heat gain column.  VENTILATION and INFILTRATION both contribute to heat gain and heat loss.  PEOPLE and ELECTRIC WATTS produce heat gain only.  At this point the chart asks for a total of the amount of heat gain, called SUB TOTAL 1. This amount is the sum of SENSIBLE HEAT from heat gain within the space.  Here we also total the column under heat loss.

48. 216 x 0.55 x 60 = 7128 400 x 60 x.018 x 21 = 9072 400 x 60.018 x 60 = 25,920 9,072 7,128 25,920 21 60 1,2473,564 110 x.5 x 60 x.018 x 21 = 1247 110 x.5 x 60 x.018 x 60 = 3564 10 x 250 = 25002,500 2200 x 2 x 3.4 = 14,96014,960 41,14427,620 Sub-totals from above... 68,923 64,232 Add heat gain / loss columns from top of chart down through “elec. Watts”

49.  The next line of the chart is “Latent Heat Load.”  Realize that moisture is present within the space, that will condense to water when the dew point temperature is reached. And remember that an amount of heat is required to change the state of a substance. (water vapor to water)  Where moist air passes over the air conditioner’s cooling coil, the air will be at room temperature, but the cooling coil will be slightly above freezing, so the dew point temperature will occur somewhere in between – resulting in condensed water forming on the coil, collected in a drain pan.  So, latent heat is a COOLING LOAD because some of the capacity of the air conditioning unit is required to condense moisture into water.

50.  But how much energy is wasted on condensation of moisture... ?  Because of the mean wet bulb temperature in Lubbock, Texas, (from the climatilogical chart, pg.1630 of text) the ratio of Sensible Heat to Latent Heat within a space such as an office building is approximately a 70 / 30 ratio. So, multiply the sensible heat load (sub total 1) times 0.30 to get the amount of latent heat in btu/h. Realize that latent heat is heat gain and involves cooling load only. So, multiply the sensible heat load (sub total 1) times 0.30 to get the amount of latent heat in btu/h. Realize that latent heat is heat gain and involves cooling load only. Add this amount to sub total 1 to get SUB TOTAL 2.

51. CLIMATILOGICAL DATA PAGE 1630 OF TEXT

52. 216 x 0.55 x 60 = 7128 400 x 60 x.018 x 21 = 9072 400 x 60.018 x 60 = 25,920 9,072 7,128 25,920 21 60 1,2473,564 110 x.5 x 60 x.018 x 21 = 1247 110 x.5 x 60 x.018 x 60 = 3564 10 x 250 = 25002,500 2200 x 2 x 3.4 = 14,96014,960 41,14427,620 Sub-totals from above... 68,923 64,232 30% of sensible load = 68,275 x.30 = 20,697 Add heat gain / loss columns from top of chart down through “elec. Watts” Add Sub Total 1 & Latent Heat Load = 89,620

53.  The last entry in the chart is a calculation that the designer must make based on the location of ductwork used to move conditioning air from the mechanical equipment to the spaces to be conditioned. If ductwork is installed in interstitial space such as inside chase space or attic space between floors of a multi-story building where there is no temperature difference, there will be no duct heat loss and gain.

54. If ductwork is installed in a ventilated, un- insulated attic space, the walls of the duct will be subject to higher temperatures in summer and lower temperatures in winter, so one must compensate for heat flow in the form of heat gain in summer and heat loss in winter. For purposes of this calculation, take ten percent of sub total 2 as duct loss and add the sum to sub total 2 for a grand total of heat gain and heat loss.

55. 216 x 0.55 x 60 = 7128 400 x 60 x.018 x 21 = 9072 400 x 60.018 x 60 = 25,920 9,072 7,128 25,920 21 60 1,2473,564 110 x.5 x 60 x.018 x 21 = 1247 110 x.5 x 60 x.018 x 60 = 3564 10 x 250 = 25002,500 2200 x 2 x 3.4 = 14,96014,960 41,14427,620 Sub-totals from above... 68,923 64,232 30% of sensible load = 68,275 x.30 = 20,697 68,275 + 20,482 =89,620 6,423 88,757 x.10 = 64,232 x.10 = Only if duct is in Un-conditioned space 8,962 Add heat gain / loss columns from top of chart down through “elec. Watts” 98,58270,655

56.  COMPARISON:  Total Heat Gain = 98,582 btu/h Which equals 98,582/ 2200 = 44.81 btu / h per sq.ft. Which equals 98,582/ 2200 = 44.81 btu / h per sq.ft.  Total Heat Loss = 70,655 btu/h Which equals 70,655 / 2200 = 32.12 but / h per sq.ft. Which equals 70,655 / 2200 = 32.12 but / h per sq.ft.

57. Which indicates that orientation and consideration of sun affects on a building envelope, coupled with the availability of daylight as an illumination source is of major importance to architectural design in the use and conservation of energy.  Total Heat Gain = 98,582 Heat gain based on The building envelope 41,144 = 41.74 % Heat gain based on The use of the building*57,438 = 58.26 %

58. end