EXAMPLE EXERCISE CALCULATING HEAT LOSS & HEAT GAIN

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 selected from the charts. Shading coefficient was selected from a reflective glass.

8. ETD values 21 26 31 31 36 48

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 Sg 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. SOLAR GAIN 23 66 196

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 21 534 1,526 60

17. 339 .075 21 534 1,526 534 .075 26 1,041 2,403 60

18. 339 .075 21 534 1,526 534 .075 26 1,041 2,403 60 972 216 .075 31 502

19. 339 .075 21 534 1,526 534 .075 26 1,041 2,403 60 972 216 .075 31 502 484 2,178 .075 36 1,307

20. 339 .075 21 534 1,526 534 .075 26 1,041 2,403 60 972 216 .075 31 502 484 2,178 .075 36 1,307 21 243 693 21 .55 60

21. 339 .075 21 534 1,526 534 .075 26 1,041 2,403 60 972 216 .075 31 502 484 2,178 .075 36 1,307 21 243 693 21 .55 60 2,200 .050 48 5,280 6,600

22. 339 .075 21 534 1,526 534 .075 26 1,041 2,403 60 972 216 .075 31 502 484 2,178 .075 36 1,307 21 243 693 21 .55 60 2,200 .050 48 5,280 6,600 0 60 21

23. 339 .075 21 534 1,526 534 .075 26 1,041 2,403 60 972 216 .075 31 502 484 2,178 .075 36 1,307 21 243 693 21 .55 60 2,200 .050 48 5,280 6,600 0 96 .60 1,210 3,456 60 21

24. 339 .075 21 534 1,526 534 .075 26 1,041 2,403 60 972 216 .075 31 502 484 2,178 .075 36 1,307 21 243 693 21 .55 60 2,200 .050 48 5,280 6,600 0 .60 96 1,210 3,456 60 21 1,814 144 .60 5,184

25. 339 .075 21 534 1,526 534 .075 26 1,041 2,403 60 972 216 .075 31 502 484 2,178 .075 36 1,307 21 243 693 21 .55 60 2,200 .050 48 5,280 6,600 0 .60 96 1,210 3,456 60 21 .60 1,814 144 5,184 128 .60 4,608 1,613

26. 339 .075 21 534 1,526 534 .075 26 1,041 2,403 60 972 216 .075 31 502 484 2,178 .075 36 1,307 21 243 693 21 .55 60 2,200 .050 48 5,280 6,600 0 .60 96 1,210 3,456 60 21 .60 1,814 144 5,184 128 .60 4,608 1,613 0

27. 339 .075 21 534 1,526 534 .075 26 1,041 2,403 60 972 216 .075 31 502 484 2,178 .075 36 1,307 21 243 693 21 .55 60 2,200 .050 48 5,280 6,600 0 .60 96 1,210 3,456 60 21 .60 1,814 144 5,184 128 .60 4,608 1,613 0 96 23 .75 1,656

28. 339 .075 21 534 1,526 534 .075 26 1,041 2,403 60 972 216 .075 31 502 484 2,178 .075 36 1,307 21 243 693 21 .55 60 2,200 .050 48 5,280 6,600 0 .60 96 1,210 3,456 60 21 .60 1,814 144 5,184 128 .60 4,608 1,613 0 96 23 .75 1,656 144 66 .75 7,128

29. 339 .075 21 534 1,526 534 .075 26 1,041 2,403 60 972 216 .075 31 502 484 2,178 .075 36 1,307 21 243 693 21 .55 60 2,200 .050 48 5,280 6,600 0 .60 96 1,210 3,456 60 21 .60 1,814 144 5,184 128 .60 4,608 1,613 0 96 23 .75 1,656 144 66 .75 7,128 128 196 .75 18,816 41,144 Sub Totals 27,620

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. 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. Sub-totals from above . . . 41,144 27,620 216 x 0.55 x 60 = 7,128 7,128 60 21

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. 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. Sub-totals from above . . . 41,144 27,620 216 x 0.55 x 60 = 7128 7,128 400 x 60 x .018 x 21 = 9072 9,072 25,920 400 x 60 .018 x 60 = 25,920 60 21

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. Sub-totals from above . . . 41,144 27,620 216 x 0.55 x 60 = 7128 7,128 400 x 60 x .018 x 21 = 9072 9,072 25,920 400 x 60 .018 x 60 = 25,920 110 x .5 x 60 x .018 x 21 = 1247 1,247 60 3,564 21 110 x .5 x 60 x .018 x 60 = 3564

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. Sub-totals from above . . . 41,144 27,620 216 x 0.55 x 60 = 7128 7,128 400 x 60 x .018 x 21 = 9072 9,072 25,920 400 x 60 .018 x 60 = 25,920 110 x .5 x 60 x .018 x 21 = 1247 1,247 60 3,564 21 110 x .5 x 60 x .018 x 60 = 3564 10 x 250 = 2,500 2,500

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. Sub-totals from above . . . 41,144 27,620 216 x 0.55 x 60 = 7128 7,128 400 x 60 x .018 x 21 = 9072 9,072 25,920 400 x 60 .018 x 60 = 25,920 110 x .5 x 60 x .018 x 21 = 1247 1,247 60 3,564 21 110 x .5 x 60 x .018 x 60 = 3564 10 x 250 = 2500 2,500 2200 x 2 x 3.4 = 14,960 14,960

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 gainonly. • 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. Sub-totals from above . . . 41,144 27,620 216 x 0.55 x 60 = 7128 7,128 400 x 60 x .018 x 21 = 9072 9,072 25,920 400 x 60 .018 x 60 = 25,920 110 x .5 x 60 x .018 x 21 = 1247 1,247 60 3,564 21 110 x .5 x 60 x .018 x 60 = 3564 10 x 250 = 2500 2,500 2200 x 2 x 3.4 = 14,960 14,960 Add heat gain / loss columns from top of chart down through “elec. Watts” 64,232 68,923

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. Add this amount to sub total 1 to get SUB TOTAL 2.