EXAMPLE EXERCISE CALCULATING HEAT LOSS & HEAT GAIN Several exhibits in the class packet are necessary to understand the entries into the Heat Loss

EXAMPLE EXERCISE CALCULATING HEAT EXAMPLE EXERCISE CALCULATING HEAT LOSS & HEAT GAINLOSS & HEAT GAIN

Several exhibits in the class packet are Several exhibits in the class packet are necessary to understand the entries into necessary to understand the entries into the Heat Loss / Heat Gain calculation the Heat Loss / Heat Gain calculation sheet.sheet.

The example floor plan will be used to The example floor plan will be used to make calculations required for the make calculations required for the selection of mechanical equipment selection of mechanical equipment necessary to maintain comfort heating necessary to maintain comfort heating and cooling.and cooling.

Certain criteria are given on the lower left Certain criteria are given on the lower left section of the floor plan to be used in the section of the floor plan to be used in the calculation. These are criteria established calculation. These are criteria established in the design of the building envelope, for in the design of the building envelope, for the type building for which it is intended.the type building for which it is intended.

12'-4" 8'-0"

68'-0"

8'-0" 7'-8"32'-0"

40

'-0"

20

'-0"

20

'-0"

6'-0

"1

2"

12

"1

2'-0

"

20'-0"8'-0"

8'-0"4'-4" 7'-8"

3'x7'opaquedoor

window area= 64 sq.ft.


window areaincluding entry

door = 144 sq.ft.

windowarea =32 sq.ft.

14

'-0"

26

'-0"

OFFICE OFFICE OFFICE OFFICE

OFFICE

WORK ROOM

MECHANICAL

TOILET

TOILET

STORAGE

SECRETARY

RECEPTION

WAITING ROOM

"U" VALUES

WALLS

ROOF

GLASS

OPAQUEDOORS

0.075

0.050

0.60

0.55

FLOOR PLANSCALE 1/8"=1'-0"

20 lineal ft.of crack

20 lineal ft.= 64 sq.ft.

window area

of crack


total crack =20 lineal ft.

10 linealft. of crack

20 lin.ft. ofcrack

no

rth

wa

ll n

et

are

a =

33

9 s

q.ft.

sou

th w

all

ne

t a

rea

= 2

16

sq

.ft.

east wall net area = 534 sq.ft.

west wall net area = 484 sq.ft.

ETD VALUES

no. wall = 21ea. wall = 26so. wall = 31we. wall = 36roof = 48

Solar Gain forglass

north = 26east = 23south = 66west = 196

Shading Coefficient

for glass = .75

ventilate 100 CFM

ventilate 100 CFM

ventilate 200 CFM

400 CFMoutside airto mechanicalunit

40'-0"

6

Heat flow Heat flow by conductionby conduction in Btu/h, in Btu/h, through through low mass or thin surfaceslow mass or thin surfaces, such as , such as doors and glass, is calculated by multiplying doors and glass, is calculated by multiplying the area, times the “U” factor, times the the area, times the “U” factor, times the difference in temperaturedifference in temperature from one side of from one side of the material to the other. the material to the other.

The The temperature differentialtemperature differential is the is the difference between the recognized high (for difference between the recognized high (for summer) or low (for winter), from summer) or low (for winter), from climatilogical data, and the temperature climatilogical data, and the temperature desired to be maintained within the space. desired to be maintained within the space.

The quantity of heat The quantity of heat by conductionby conduction that that passes through passes through surfaces of greater masssurfaces of greater mass, , such as walls and roofs is calculated by such as walls and roofs is calculated by multiplying the area, times the “U” factor, multiplying the area, times the “U” factor, times the Equivalent Temperature times the Equivalent Temperature Difference.Difference.

RADIANT HEAT FROM DIRECT SUN IN SUMMERRADIANT HEAT FROM DIRECT SUN IN SUMMER

Radiant heat that enters a space by shining Radiant heat that enters a space by shining through a transparent or semi-transparent surface, through a transparent or semi-transparent surface, such as glass is calculated by multiplying the area such as glass is calculated by multiplying the area times the amount of Solar Gain, reduced only by times the amount of Solar Gain, reduced only by the effectiveness of a shading coefficient the effectiveness of a shading coefficient

AS AN ILLUSTRATION, calculate the total heat loss AS AN ILLUSTRATION, calculate the total heat loss (winter) and heat gain (summer) for the plan of the (winter) and heat gain (summer) for the plan of the example building.example building. The construction is medium The construction is medium weight masonry, such as brick veneer over weight masonry, such as brick veneer over concrete masonry units. concrete masonry units.

In this area, the peak summer daytime In this area, the peak summer daytime temperature occurs around 4:00 P.M. Consider temperature occurs around 4:00 P.M. Consider that the color of masonry is light, such as would be that the color of masonry is light, such as would be the shade of the brick on the architecture building. the shade of the brick on the architecture building.

12'-4" 8'-0"

68'-0"

8'-0" 7'-8"32'-0"

40

'-0"

20

'-0"

20

'-0"

6'-0

"1

2"

12

"1

2'-0

"

20'-0"8'-0"

8'-0"4'-4" 7'-8"

3'x7'opaquedoor




door = 144 sq.ft.


14

'-0"

26

'-0"


OFFICE

WORK ROOM

MECHANICAL

TOILET

TOILET

STORAGE

SECRETARY

RECEPTION

WAITING ROOM

"U" VALUES

WALLS

ROOF

GLASS

OPAQUEDOORS

0.075

0.050

0.60

0.55




window area

of crack




20 lin.ft. ofcrack

no

rth

wa

ll n

et

are

a =

33

9 s

q.ft.

sou

th w

all

ne

t a

rea

= 2

16

sq

.ft.



ETD VALUES


Solar Gain forglass


Shading Coefficient

for glass = .75

ventilate 100 CFM

ventilate 100 CFM

ventilate 200 CFM


40'-0"

6

“ “U” values were selected from U” values were selected from reasonable allowance of reasonable allowance of materials.materials.

ETD and Solar Gain values wereETD and Solar Gain values were

selected from the charts.selected from the charts.

Shading coefficient was Shading coefficient was selectedselected

from a reflective glass.from a reflective glass.

21

26

3131

36

48

ETD values

Another factor that relates to direct Another factor that relates to direct sunlight is solar gain through glass. Direct sunlight is solar gain through glass. Direct sunlight will pass through clear glass without sunlight will pass through clear glass without an appreciable effect on the glass itself, since an appreciable effect on the glass itself, since glass is light in mass compared to the building glass is light in mass compared to the building envelope.envelope.

Solar gain, referenced by Solar gain, referenced by SSgg on the chart, on the chart, is the amount of heat gain in Btu per square is the amount of heat gain in Btu per square foot, as the result of sun radiation that foot, as the result of sun radiation that penetrates glass.penetrates glass.

The angle of the sun is taken into account The angle of the sun is taken into account as the result of the tilt of the earth, as well as 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 time of the day. In this area, 4:00 P.M. is the peak summer temperature.the peak summer temperature.

June 21, the beginning of summer June 21, the beginning of summer produces the worst sun angle on the north and produces the worst sun angle on the north and east side of a building, while September 21, east side of a building, while September 21, the beginning of fall produces the worst sun the beginning of fall produces the worst sun angle for south and north.angle for south and north.

66

196

SOLAR

GAIN

23

In order to determine a reasonable In order to determine a reasonable temperature difference between outside temperature difference between outside and inside, consult with climatilogical data and inside, consult with climatilogical data for the area a building is located.for the area a building is located.

The chart that follows is taken from The chart that follows is taken from the appendix of the text, and gives the the appendix of the text, and gives the outside design temperatures for winter and outside design temperatures for winter and summer for Lubbock, Texas, in terms of summer for Lubbock, Texas, in terms of dry-bulb temperatures. The wet-bulb dry-bulb temperatures. The wet-bulb temperature given in the summer column is temperature given in the summer column is a measure of the relative humidity. a measure of the relative humidity.

Inside temperature is set by the Inside temperature is set by the designer as the maintained desired interior designer as the maintained desired interior temperature. temperature.

CLIMATILOGICAL DATA

PAGE 1630 OF TEXT

In the chart, note that winter outside In the chart, note that winter outside design temperature is 15 degrees, while design temperature is 15 degrees, while summer design temperature is 96 summer design temperature is 96 degrees.degrees.

For the purpose of the example For the purpose of the example problem, problem, say it is desired to maintain a say it is desired to maintain a temperature inside of 75 degrees, for both temperature inside of 75 degrees, for both summer and winter.summer and winter.

So, temperature difference for So, temperature difference for summer conditions would be 96 – 75 = summer conditions would be 96 – 75 = 21 21 degreesdegrees. And for winter conditions the . And for winter conditions the temperature difference is 75 – 15 = temperature difference is 75 – 15 = 60 60 degreesdegrees..

12'-4" 8'-0"

68'-0"

8'-0" 7'-8"32'-0"

40

'-0"

20

'-0"

20

'-0"

6'-0

"1

2"

12

"1

2'-0

"

20'-0"8'-0"

8'-0"4'-4" 7'-8"

3'x7'opaquedoor




door = 144 sq.ft.


14

'-0"

26

'-0"


OFFICE

WORK ROOM

MECHANICAL

TOILET

TOILET

STORAGE

SECRETARY

RECEPTION

WAITING ROOM

"U" VALUES

WALLS

ROOF

GLASS

OPAQUEDOORS

0.075

0.050

0.60

0.55




window area

of crack




20 lin.ft. ofcrack

no

rth

wa

ll n

et

are

a =

33

9 s

q.ft.

sou

th w

all

ne

t a

rea

= 2

16

sq

.ft.



ETD VALUES


Solar Gain forglass


Shading Coefficient

for glass = .75

ventilate 100 CFM

ventilate 100 CFM

ventilate 200 CFM


40'-0"

6

339 .075 21 534

60

1,526

339

534

.075

.075

21

26

534

1,04160

1,526

2,403

339

534

216

.075

.075

.075

21

26

31

534

1,041

50260

1,526

2,403 972

339

534

216484

.075

.075

.075

.075

21

26

31

36

534

1,041

502

1,307

60

1,526

2,403 972

2,178

339

534

216484

.075

.075

.075

.075

21

26

31

36

534

1,041

502

1,307

60

1,526

2,403 972

2,178

21 .55 21 243

69360

339

534

216484

.075

.075

.075

.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

.050 48

60

5,280 6,600

339

534

216484

.075

.075

.075

.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

.050 48

60

5,280 6,600

0

21 60

339

534

216484

.075

.075

.075

.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

.050 48

60

5,280 6,600

0

96 .6021 60

1,210 3,456

339

534

216484

.075

.075

.075

.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

.050 48

60

5,280 6,600

0

96

144

.60

.6021 60

1,210 1,814

3,456

5,184

339

534

216484

.075

.075

.075

.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

.050 48

60

5,280 6,600

0

96

144128

.60

.60

.60

21 60 1,210 1,814

1,613

3,456

5,1844,608

339

534

216484

.075

.075

.075

.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

.050 48

60

5,280 6,600

0

96

144128

0

.60

.60

.60

21 60 1,210 1,814

1,613

3,456

5,1844,608

339

534

216484

.075

.075

.075

.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

.050 48

60

5,280 6,600

0

96

144128

0

96

.60

.60

.60

21 60 1,210 1,814

1,613

3,456

5,1844,608

23 .75 1,656

339

534

216484

.075

.075

.075

.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

.050 48

60

5,280 6,600

0

96

144128

0

144

96

.60

.60

.60

21 60 1,210 1,814

1,613

3,456

5,1844,608

23

66

.75

.75

1,656

7,128

339

534

216484

.075

.075

.075

.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

.050 48

60

5,280 6,600

0

96

144128

0

144

96

128

.60

.60

.60

21 60 1,210 1,814

1,613

3,456

5,1844,608

23

66

196

.75

.75

.75

1,656

7,128

18,816

41,144 Sub Totals

27,620

The heat gain and heat loss calculated in the The heat gain and heat loss calculated in the top part of the chart involved the top part of the chart involved the integrity of the integrity of the building envelopebuilding envelope. .

An examination of the components would An examination of the components would reveal that the foremost consideration of this reveal that the foremost consideration of this building design should be an analysis of what can be building design should be an analysis of what can be done about the windows. Realize the AREA of the done about the windows. Realize the AREA of the windows is not the main concern, but rather the windows is not the main concern, but rather the orientation of the building with regard to window orientation of the building with regard to window placement.placement.

Glass area on the south side (next to worse Glass area on the south side (next to worse solar gain) is 144 sq.ft. compared to 128 on the west solar gain) is 144 sq.ft. compared to 128 on the west side, yet the solar heat gain on the west side is side, yet the solar heat gain on the west side is more than 2 ½ times that on the south. more than 2 ½ times that on the south.

A more judicious concern for plan arrangement A more judicious concern for plan arrangement would result in better conservation of energy. would result in better conservation of energy.

The bottom half of the chart is concerned The bottom half of the chart is concerned mainly with internal conditions of the building and mainly with internal conditions of the building and how they affect the overall heat loss/gain.how they affect the overall heat loss/gain.

PERIMETERPERIMETER: : Heat loss onlyHeat loss only. Refers to the . Refers to the perimeter of the building; specifically at floor level perimeter of the building; specifically at floor level near the ground outside. During winter the ground near the ground outside. During winter the ground will remain cold, since soil does not readily convert will remain cold, since soil does not readily convert electromagnetic energy to heat because of it’s electromagnetic energy to heat because of it’s relatively light density.relatively light density.

The ground temperature at the building will The ground temperature at the building will probably be cold at least 12” deep. So, the probably be cold at least 12” deep. So, the temperature difference between the soil and inside temperature difference between the soil and inside the building at the floor level will be sufficient to the building at the floor level will be sufficient to cause a significant heat loss around the perimeter cause a significant heat loss around the perimeter of the building. of the building. On the page you have in the packet labeled “ETD”, On the page you have in the packet labeled “ETD”, find the written material on the left column of the find the written material on the left column of the page under the words page under the words PERIMETER HEAT LOSSPERIMETER HEAT LOSS reads: reads:

Perimeter InsulationPerimeter Insulation: . . . Use a value of : . . . Use a value of .81 .81 btu/hbtu/h for each for each UNINSULATEDUNINSULATED foot of building foot of building perimeter.perimeter. Use a value of Use a value of .55 btu/h.55 btu/h for each for each INSULATEDINSULATED foot foot of building perimeter.of building perimeter.

Assume the example problem is Assume the example problem is INSULATEDINSULATED..

12'-4" 8'-0"

68'-0"

8'-0" 7'-8"32'-0"

40

'-0"

20

'-0"

20

'-0"

6'-0

"1

2"

12

"1

2'-0

"

20'-0"8'-0"

8'-0"4'-4" 7'-8"

3'x7'opaquedoor




door = 144 sq.ft.


14

'-0"

26

'-0"


OFFICE

WORK ROOM

MECHANICAL

TOILET

TOILET

STORAGE

SECRETARY

RECEPTION

WAITING ROOM

"U" VALUES

WALLS

ROOF

GLASS

OPAQUEDOORS

0.075

0.050

0.60

0.55




window area

of crack




20 lin.ft. ofcrack

no

rth

wa

ll n

et

are

a =

33

9 s

q.ft.

sou

th w

all

ne

t a

rea

= 2

16

sq

.ft.



ETD VALUES


Solar Gain forglass


Shading Coefficient

for glass = .75

ventilate 100 CFM

ventilate 100 CFM

ventilate 200 CFM


40'-0"

6Building perimeter = 68+40+68+40 = 216’

216 x 0.55 x 60 = 7,128

21 60

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

7,128

VENTILATIONVENTILATION is the removal of unwanted is the removal of unwanted air from a space. Ventilation cannot happen air from a space. Ventilation cannot happen unless stale air is replaced by new air – and unless stale air is replaced by new air – and the only source for new air is outside the the only source for new air is outside the space. space.

If outside air is brought into the space, it If outside air is brought into the space, it must be thermally treated in order to blend must be thermally treated in order to blend with comfort air. So, in summer, heat must be with comfort air. So, in summer, heat must be removed from the air, and in winter, heat must removed from the air, and in winter, heat must be added. be added.

The measurement for handling air The measurement for handling air quantity is CUBIC FEET PER MINUTE (cfm), so quantity is CUBIC FEET PER MINUTE (cfm), so a transition must be made to convert cubic a transition must be made to convert cubic feet per minute to btu per hour.feet per minute to btu per hour.

The ventilation requirement for the The ventilation requirement for the building is 400 cubic feet per minute – 200 for building is 400 cubic feet per minute – 200 for toilets and 200 for the room on the southwest toilets and 200 for the room on the southwest side of the building.side of the building.

for glass = .75

Shading Coefficient

we. wall = 36so. wall = 31ea. wall = 26no. wall = 21

ETD VALUES"U" VALUES

OPAQUEDOORS

WALLS

GLASS

ROOF

0.55 roof

0.075

0.050

0.60

west = 196south = 66east = 23north = 26

Solar Gain for

= 48

glass

SCALE 1/8"=1'-0"

FLOOR PLAN 6

to mechanicaloutside air400 CFM

ventilate 100 CFM

ventilate 100 CFM


TOILET

TOILET

MECHANICAL

32'-0"

sou

th w

all

net

are

a =

216

sq.

ft.

WAITING ROOM

window area

20'

-0"

door = 144 sq.ft.including entry

8'-0"

12"

12'

-0"

20 lineal ft.total crack =

ft. of crack10 lineal32 sq.ft.area =window

12"

6'-0

"

40'

-0"

40'-0"


STORAGE

RECEPTIONunit

SECRETARY

window area

OFFICE

20'

-0"

12'-4"

of crack

ventilate 200 CFM

WORK ROOM OFFICE

8'-0"


68'-0"

nor

th w

all

net

are

a =

339

sq.

ft.

20 lin. of crack

4'-4" 8'-0"

20'-0"

crackft. of

dooropaque3'x7'

20 lineal ft.

= 64 sq.ft.window area

7'-8"

14'-0

"

OFFICE

OFFICE

of crack

OFFICE

8'-0"

window area= 64 sq.ft.20 lineal ft.

26'-0

"

7'-8"

So, So, heat gainheat gain, or , or heat lossheat loss from from ventilation air is calculated simply ;ventilation air is calculated simply ;

CFM ventilation x 60 x .018 x temp. CFM ventilation x 60 x .018 x temp. diff. = btu/hdiff. = btu/h

But remember to use the temperature But remember to use the temperature difference that applies to difference that applies to summertimesummertime for for heat gainheat gain, and the temperature difference , and the temperature difference that applies to that applies to wintertimewintertime for for heat lossheat loss..

216 x 0.55 x 60 = 7128

400 x 60 x .018 x 21 = 9072400 x 60 .018 x 60 = 25,920

9,072

7,128

25,920

21 60


INFILTRATION INFILTRATION is unwanted air that gets into is unwanted air that gets into the space by infiltrating through cracks around the space by infiltrating through cracks around doors and windows and by other means by which doors and windows and by other means by which un-conditioned outside air can get inside. un-conditioned outside air can get inside.

To calculate the amount of air, To calculate the amount of air, use a factor use a factor of .50 for each linear foot of crackof .50 for each linear foot of crack around doors around doors and windows. The number, .50 represents ½ and windows. The number, .50 represents ½ CFM per foot of crack.CFM per foot of crack.

The amount of crack is the measurement of The amount of crack is the measurement of the perimeter of each window sash that is the perimeter of each window sash that is movable, and doors. movable, and doors.

So the heat loss or gain from infiltration So the heat loss or gain from infiltration equalsequalstotal length of crack x .5 x 60 x .018 x total length of crack x .5 x 60 x .018 x temperature difference (summer or winter)temperature difference (summer or winter)

for glass = .75

Shading Coefficient

we. wall = 36so. wall = 31ea. wall = 26no. wall = 21

ETD VALUES"U" VALUES

OPAQUEDOORS

WALLS

GLASS

ROOF

0.55 roof

0.075

0.050

0.60

west = 196south = 66east = 23north = 26

Solar Gain for

= 48

glass

SCALE 1/8"=1'-0"

FLOOR PLAN 6

to mechanicaloutside air400 CFM

ventilate 100 CFM

ventilate 100 CFM


TOILET

TOILET

MECHANICAL

32'-0"

sou

th w

all

net

are

a =

216

sq.

ft.

WAITING ROOM

window area

20'

-0"

door = 144 sq.ft.including entry

8'-0"

12"

12'

-0"

20 lineal ft.total crack =

ft. of crack10 lineal32 sq.ft.area =window

12"

6'-0

"

40'

-0"

40'-0"


STORAGE

RECEPTIONunit

SECRETARY

window area

OFFICE

20'

-0"

12'-4"

of crack

ventilate 200 CFM

WORK ROOM OFFICE

8'-0"


68'-0"

nor

th w

all

net

are

a =

339

sq.

ft.

20 lin. of crack

4'-4" 8'-0"

20'-0"

crackft. of

dooropaque3'x7'

20 lineal ft.

= 64 sq.ft.window area

7'-8"

14'-0

"

OFFICE

OFFICE

of crack

OFFICE

8'-0"

window area= 64 sq.ft.20 lineal ft.

26'-0

"

7'-8"

216 x 0.55 x 60 = 7128

400 x 60 x .018 x 21 = 9072400 x 60 .018 x 60 = 25,920

9,072

7,128

25,920

21 601,247 3,564110 x .5 x 60 x .018 x 21 = 1247110 x .5 x 60 x .018 x 60 = 3564


PEOPLE PEOPLE means the number of people means the number of people that occupy the space and contribute that occupy the space and contribute heat. The amount varies with the size of heat. The amount varies with the size of individuals and their level of activity. For individuals and their level of activity. For the purpose of this calculation, assume the purpose of this calculation, assume that approximately ten people will occupy that approximately ten people will occupy the space, and their physical activity is the space, and their physical activity is rather tranquil. rather tranquil.

So So use 250 btu/h for each personuse 250 btu/h for each person = =

250 x 10 = 2500 Btu/h250 x 10 = 2500 Btu/h

(the number of people that occupy public (the number of people that occupy public buildings is also given in the building buildings is also given in the building code)code)

216 x 0.55 x 60 = 7128

400 x 60 x .018 x 21 = 9072400 x 60 .018 x 60 = 25,920

9,072

7,128

25,920

21 601,247 3,564110 x .5 x 60 x .018 x 21 = 1247110 x .5 x 60 x .018 x 60 = 3564

10 x 250 = 2,500 2,500


ELECTRIC WATTSELECTRIC WATTS involves the amount of involves the amount of heat generated by the consumption of heat generated by the consumption of electricity within the space, such as the electricity within the space, such as the operation of mechanical equipment and operation of mechanical equipment and electric lights. electric lights.

Since all the answers as to the Since all the answers as to the electrical design is generally not known at electrical design is generally not known at the time these calculations are done, the time these calculations are done, use an use an allowance of two watts per square foot of allowance of two watts per square foot of spacespace..

2200 square feet x 2 = 4400 watts2200 square feet x 2 = 4400 watts

Since one electrical watt produces heat Since one electrical watt produces heat of 3.4 btu / h, the total number of btu of 3.4 btu / h, the total number of btu equals,equals,

4400 x 3.4 = 14,960 Btu/h4400 x 3.4 = 14,960 Btu/h

216 x 0.55 x 60 = 7128

400 x 60 x .018 x 21 = 9072400 x 60 .018 x 60 = 25,920

9,072

7,128

25,920

21 601,247 3,564110 x .5 x 60 x .018 x 21 = 1247110 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


Note on the chart that Note on the chart that PERIMETER is PERIMETER is heat loss onlyheat loss only, since there is no place for , since there is no place for values in the heat gain column. values in the heat gain column.

VENTILATION and INFILTRATION both VENTILATION and INFILTRATION both contribute to contribute to heat gainheat gain and and heat lossheat loss. .

PEOPLE and ELECTRIC WATTS produce PEOPLE and ELECTRIC WATTS produce heat gainheat gain onlyonly..

At this point the chart asks for a total At this point the chart asks for a total of the amount of heat gain, called of the amount of heat gain, called SUB SUB TOTAL 1TOTAL 1. . This amount is the sum of This amount is the sum of SENSIBLE HEAT from heat gain within the SENSIBLE HEAT from heat gain within the space. space.

Here we also total the column under heat Here we also total the column under heat loss.loss.

216 x 0.55 x 60 = 7128

400 x 60 x .018 x 21 = 9072400 x 60 .018 x 60 = 25,920

9,072

7,128

25,920

21 601,247 3,564110 x .5 x 60 x .018 x 21 = 1247110 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


68,923 64,232Add heat gain / loss columns from top of chart down through “elec. Watts”

The next line of the chart is “Latent Heat The next line of the chart is “Latent Heat Load.”Load.”

Realize that moisture is present within the Realize that moisture is present within the space, that will condense to water when the space, that will condense to water when the dew point temperature is reached. And dew point temperature is reached. And remember that an amount of heat is required remember that an amount of heat is required to change the state of a substance. (water to change the state of a substance. (water vapor to water)vapor to water)

Where moist air passes over the air Where moist air passes over the air conditioner’s cooling coil, the air will be at conditioner’s cooling coil, the air will be at room temperature, but the cooling coil will be room temperature, but the cooling coil will be slightly above freezing, so the dew point slightly above freezing, so the dew point temperature will occur somewhere in temperature will occur somewhere in between – resulting in condensed water between – resulting in condensed water forming on the coil, collected in a drain pan.forming on the coil, collected in a drain pan.

So, latent heat is a COOLING LOAD because So, latent heat is a COOLING LOAD because some of the capacity of the air conditioning some of the capacity of the air conditioning unit is required to condense moisture into unit is required to condense moisture into water.water.

But how much energy is wasted on But how much energy is wasted on condensation of moisture . . . ?condensation of moisture . . . ?

Because of the Because of the mean wet bulb mean wet bulb temperature in Lubbock, Texas,temperature in Lubbock, Texas, (from the (from the climatilogical chart, pg.1630 of text)climatilogical chart, pg.1630 of text) the ratio the ratio of Sensible Heat to Latent Heat within a of Sensible Heat to Latent Heat within a space such as an office building is space such as an office building is approximately a 70 / 30 ratio.approximately a 70 / 30 ratio.

So, multiply the sensible heat load (sub So, multiply the sensible heat load (sub total 1) times 0.30 to get the amount of total 1) times 0.30 to get the amount of latent heat in btu/h. Realize that latent latent heat in btu/h. Realize that latent heat is heat gain and involves cooling load heat is heat gain and involves cooling load only.only.

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

CLIMATILOGICAL DATA

PAGE 1630 OF TEXT

216 x 0.55 x 60 = 7128

400 x 60 x .018 x 21 = 9072400 x 60 .018 x 60 = 25,920

9,072

7,128

25,920

21 601,247 3,564110 x .5 x 60 x .018 x 21 = 1247110 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


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

The last entry in the chart is a The last entry in the chart is a calculation that the designer must make calculation that the designer must make based on the location of ductwork used to based on the location of ductwork used to move conditioning air from the mechanical move conditioning air from the mechanical equipment to the spaces to be equipment to the spaces to be conditioned.conditioned.

If ductwork is installed in If ductwork is installed in interstitialinterstitial space such as inside chase space or attic space such as inside chase space or attic space between floors of a multi-story space between floors of a multi-story building where there is no temperature building where there is no temperature difference, there will be no duct heat loss difference, there will be no duct heat loss and gain.and gain.

If ductwork is installed in a If ductwork is installed in a ventilated, un-insulated attic space, the ventilated, un-insulated attic space, the walls of the duct will be subject to higher walls of the duct will be subject to higher temperatures in summer and lower temperatures in summer and lower temperatures in winter, so one must temperatures in winter, so one must compensate for heat flow in the form of compensate for heat flow in the form of heat gain in summer and heat loss in heat gain in summer and heat loss in winter.winter.

For purposes of this calculation, For purposes of this calculation, take take ten percent of sub total 2 as duct lossten percent of sub total 2 as duct loss and and add the sum to sub total 2 for a grand add the sum to sub total 2 for a grand total of heat gain and heat loss. total of heat gain and heat loss.

216 x 0.55 x 60 = 7128

400 x 60 x .018 x 21 = 9072400 x 60 .018 x 60 = 25,920

9,072

7,128

25,920

21 601,247 3,564110 x .5 x 60 x .018 x 21 = 1247110 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


68,923 64,232

30% of sensible load = 68,275 x .30 = 20,697

68,275 + 20,482 = 89,620

6,42388,757 x .10 =64,232 x .10 =

Only if duct is inUn-conditioned space

8,962

Add heat gain / loss columns from top of chart down through “elec. Watts”

98,582 70,655

COMPARISON:COMPARISON:

Total Heat Gain = 98,582 btu/h Total Heat Gain = 98,582 btu/h Which equals 98,582/ 2200 = Which equals 98,582/ 2200 = 44.81 btu / 44.81 btu /

h per sq.ft.h per sq.ft.

Total Heat Loss = 70,655 btu/hTotal Heat Loss = 70,655 btu/h Which equals 70,655 / 2200 = Which equals 70,655 / 2200 = 32.12 but / 32.12 but /

h per sq.ft.h per sq.ft.

Which indicates that orientation and Which indicates that orientation and consideration of sun affects on a building consideration of sun affects on a building envelope, coupled with the availability of envelope, coupled with the availability of daylight as an illumination source is of daylight as an illumination source is of major importance to architectural design in major importance to architectural design in the use and conservation of energy.the use and conservation of energy.

Total Heat Gain = 98,582Total Heat Gain = 98,582

Heat gain based on Heat gain based on The building envelopeThe building envelope 41,144 = 41,144 = 41.7441.74 % %

Heat gain based onHeat gain based onThe use of the building*The use of the building* 57,438 = 57,438 = 58.26 %58.26 %

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Documents

EXAMPLE EXERCISE CALCULATING HEAT LOSS & HEAT GAIN Several exhibits in the class packet are necessary to understand the entries into the Heat Loss