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EXAMPLE EXERCISE CALCULATING HEAT LOSS

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Title: EXAMPLE EXERCISE CALCULATING HEAT LOSS


1
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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.

3
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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 400 P.M. Consider
    that the color of masonry is light, such as would
    be the shade of the brick on the architecture
    building.

6
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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, 400 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.

14
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15
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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
30
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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 its 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.

33
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34
Building perimeter 68406840 216
35
41,144
27,620
Sub-totals from above . . .

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.

37
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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
41,144
27,620
Sub-totals from above . . .

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)

41
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42
41,144
27,620
Sub-totals from above . . .

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
60
1,247
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
41,144
27,620
Sub-totals from above . . .

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
60
1,247
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
41,144
27,620
Sub-totals from above . . .

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
60
1,247
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 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
41,144
27,620
Sub-totals from above . . .

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
60
1,247
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 conditioners
    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.

51
CLIMATILOGICAL DATA PAGE 1630 OF TEXT
52
41,144
27,620
Sub-totals from above . . .

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
60
1,247
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
30 of sensible load 68,275 x .30
20,697
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
41,144
27,620
Sub-totals from above . . .

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
60
1,247
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
30 of sensible load 68,275 x .30
20,697
68,275 20,482
89,620
88,757 x .10 64,232 x .10
Only if duct is in Un-conditioned space
8,962
6,423
98,582
70,655
56
  • COMPARISON
  • Total Heat Gain 98,582 btu/h
  • 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.

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
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