CHAPTER 5 SOLUTIONS
Heating Degree-days
1. Step 1: Identify the equation to calculate heating degree days, in which HDD is
heating degree days, Tbase is the base temperature or inside temperature, usually
65°F, and Ta is the average outside temperature.
HDD  Tbase  Ta
Step 2: Substitute the known variables and solve for heating degree days.
HDD  65F  13F
HDD = 52 Degree-days
2. Step 1: Convert the average outside temperature to degrees Fahrenheit. Recall the
following equation:
5
C  ( F  32) 
9
2C  ( F  32) 
5
9
F  35.6
Step 2: Substitute the known variables and solve for heating degree days.
HDD  65F  35.6F
HDD = 29.4 Degree-days
3. Step 1: Calculate the heating degree days for each day of the week.
Day
Average Temperature
Sunday
49°F
Monday
47°F
Tuesday
51°F
Wednesday
60°F
Thursday
65°F
Friday
67°F
Saturday
58°F
HDD
16
18
14
5
0
0
7
Step 2: To calculate the HDD for the week add the HDD for each day of the week.
HDDweek = 60 Degree-days
4. Step 1: Calculate the daily HDD. Then, multiply by the number days in the month
of January to calculate the HDD for the month.
HDD  65F  25F
HDD = 40 Degree-days
40 Degree-days  31days  1,240 Degree-days
5. Step 1: Calculate the heating degree days for each month to determine the heating
season. Then, multiply the number days in each month by its HDD to calculate
the HDD for the month.
Month
January
February
March
April
May
June
July
August
September
October
November
December
Average Temperature
25°F
28°F
37°F
48°F
59°F
67°F
71°F
70°F
62°F
51°F
41°F
31°F
HDD HDDmonth
40
1240
37
1036
28
868
17
510
6
186
0
0
0
0
0
0
3
90
14
434
24
720
34
1054
Step 2: Add the HDD for each month to calculate the HDD for the heating season.
HDDheating season = 6,138 Degree-days
6. Step 1: Find the increase in HDD by subtracting Birmingham, AL HDD from
State College, PA HDD. Then, divide the difference by the HDD for the location
she is moving from and multiply by 100% to calculate the percent increase.
(6,000 HDD  2,800 HDD )
 100%  114%
2,800 HDD
Composite R-values
7. Step 1: Add the R-values of the materials the wall consists of to calculate the
composite R-value of the wall.
Composite R-value = 15.2 ft2 °F hr/BTU
8. Step 1: Add the R-values of the materials the wall consists of to calculate the
composite R-value of the wall. Remember to multiply the R-value of the brick by
its total thickness.
Composite R-value = 2.94 ft2 °F hr/BTU
9. Step 1: Add the R-values of the materials the wall consists of to calculate the
composite R-value of the wall. Remember to multiply the R-value of the
fiberglass by its total thickness.
Composite R-value = 19.76 ft2 °F hr/BTU
10. Step 1: Add the R-values of the materials the wall consists of to calculate the
composite R-value of the wall.
.94  (2in  5)  (3in  3.7)  22.04 ft2 °F hr/BTU
11. Step 1: Add the R-values of the materials the wall consists of to calculate the
composite R-value of the wall.
(8in  .08)  (3in  6.25)  .94  20.33 ft2 °F hr/BTU
Heat Loss
12. Step 1: Recall that Area  Length  Height .
Area  20 ft  8 ft  160 ft 2
13. Step 1: Find the wall area of the room by adding the area for each wall.
Area  16 ft  8 ft  128 ft 2  2walls  256 ft 2
Area  12 ft  8 ft  96 ft 2  2walls  192 ft 2
Total Area  256 ft 2  128 ft 2  448 ft 2
14. Step 1: Find the wall area of the room by adding the area for each wall.
Area  20 ft  10 ft  200 ft 2  2walls  400 ft 2
Area  12 ft  10 ft  120 ft 2  2walls  240 ft 2
Total Area  400 ft 2  240 ft 2  640 ft 2
Step 2: Find the area of openings in the room by adding the area of both windows
and the door.
Area window1  3 ft  4 ft  12 ft 2
Area window2  6 ft  4 ft  24 ft 2
Area door  4 ft  7 ft  28 ft 2
Total Area  12 ft 2  24 ft 2  28 ft 2  64 ft 2
Step 3: Then, subtract the area of openings from the total area to calculate the
total area of the wall space.
640 ft 2  64 ft 2  576 ft 2
15. Step 1: Identify the equation to calculate heat loss.
Heat Loss (BTU/h) 
Area  (Tinside  Toutside)
R  value
Step 2: Calculate the heat loss by substituting the known variables.
Heat Loss 
(10 ft  8 ft )  (70 F  43 F )
22
Heat Loss = 98.18BTU/h
16. Step 1: Calculate the heat loss by substituting the known variables.
Heat Loss 
(12 ft  9 ft )  (72 F  27 F )
19
Heat Loss = 255.79BTU/h
17. Step 1: Calculate the heat loss per hour by substituting the known variables.
(360 ft 2 )  (65 F  37 F )
Heat Loss 
22
Heat Loss = 458.18BTU/h
Step 2: Calculate the heat lost through the room in 12 hours.
458.18BTU
 12h  5,498.18BTU
h
18. Step 1: Calculate the heat loss through windows, the walls, and the roof using
their respective areas and R-values.
Heat Losswindows
(120 ft 2 )  (68F  33F )

 4,200 BTU h
1
Heat Losswalls 
(776 ft 2 )  (68F  33F )
 1,429.47 BTU h
19
Heat Lossroof 
(900 ft 2 )  (68 F  33 F )
 1,431.82 BTU h
22
Step 2: Calculate the heat lost through the house by totaling the heat loss values.
Heat Losshouse = 7,061.29BTU/h
19. Step 1: Calculate the heat loss per hour by substituting the known variables.
(246 ft 2 )  [70 F  (4 F )]
Heat Loss 
19
Heat Loss = 958.11BTU/h
Step 2: Calculate the heat lost through the wall during a 36 hour period.
958.11BTU
 36h  34,491.79 BTU
h
20. Step 1: Identify the equation to calculate seasonal heat loss.
Seasonal Heat Loss (BTU) 
Area  HDD  24h day
R  value
Step 2: Calculate the seasonal heat loss by substituting the known variables.
184 ft 2  6,000 F day  24h day
Seasonal Heat Loss 
19
Seasonal Heat Loss = 1,394,526.32BTU
21. Step 1: Calculate the seasonal heat loss by substituting the known variables.
Seasonal Heat Loss 
(2 ft  3 ft )  11,000 F day  24h day
1
Seasonal Heat Loss = 1,584,000BTU
22. Step 1: Calculate the HDD for the 150 day heating season in Roanoke, VA.
HDD  65F  47F  18 Degree-days
18 Degree-days  150days  2,700 Degree-days
Step 2: Calculate the seasonal heat loss in Roanoke, VA.
(176 ft 2 )  2,700 F day  24h day
 712,800 BTU
Seasonal Heat Loss 
16
23. Step 1: Calculate the HDD for the 220 day heating season in Fargo, ND.
HDD  65F  27F  38 Degree-days
38 Degree-days  220days  8,360 Degree-days
Step 2: Calculate the seasonal heat loss through the window.
Seasonal Heat Losswindow 
(8 ft  6 ft )  8,360 F day  24h day
1
Seasonal Heat Losswindow  9,630,720 BTU
24. Step 1: Add the R-values of the materials the wall consists of to calculate the
composite R-value of the wall.
(3in  .2)  1.89  (2in  3.7)  .45  10.34 ft2 °F hr/BTU
Step 2: Calculate the heat lost through the wall during the heating season in State
College, PA.
Seasonal Heat Losswall 
(160 ft 2 )  6,000F day  24h day
10.34
Seasonal Heat Losswall  2,228,239.85BTU
25. Step 1: Add the R-values of the materials the wall consists of to calculate the
composite R-value of the wall.
.81  .94  (3in  3.7)  (1.5  6.25)  45  22.68 ft2 °F hr/BTU
Step 2: Calculate the seasonal heat loss through windows, the walls, and the roof
using their respective areas and R-values.
Seasonal Heat Losswindows 
(580 ft 2 )  6,000 F day  24h day
1
Seasonal Heat Losswalls 
(1,920 ft 2 )  6,000 F day  24h day
22.68
Seasonal Heat Lossroof 
(2,750 ft 2 )  6,000 F day  24h day
22
Step 3: Calculate the seasonal heat lost through the house in State College, PA by
totaling the seasonal heat loss values.
Seasonal Heat Losshouse = 113,710,476.2BTU
26. Step 1: Find the wall area for the room by adding the area for each wall.
Area  50 ft  8 ft  400 ft 2  2walls  800 ft 2
Area  70 ft  8 ft  560 ft 2  2walls  1,120 ft 2
Total Area  800 ft 2  1,120 ft 2  1,920 ft 2
Step 2: Find the area of openings in the single-story house by totaling the area of
the six windows.
Area windows  4 ft  6 ft  24 ft 2  6windows  144 ft 2
Step 3: Then, subtract the area of openings from the total area to calculate the
total area of the wall space.
Area walls  1,920 ft 2  144 ft 2  1,776 ft 2
Step 4: Add the R-values of the materials the wall consists of to calculate the
composite R-value of the wall.
.81  (2in  6.25)  (4in  3.7)  .45  28.56 ft2 °F hr/BTU
Step 5: Calculate the seasonal heat loss through windows, the walls, and the roof
using their respective areas and R-values.
Seasonal Heat Losswindows 
(144 ft 2 )  11,000 F day  24h day
1
Seasonal Heat Losswalls 
(1,776 ft 2 )  11,000 F day  24h day
28.56
Seasonal Heat Lossroof 
(1,920 ft 2  1.1)  11,000 F day  24h day
30
Step 6: Calculate the seasonal heat lost through the house by totaling the seasonal
heat loss values.
Seasonal Heat Losshouse = 73,018,406.72BTU
Heating and Fuels
26. Step 1: Convert CCF to BTUs of natural gas. Recall that 1 CCF is equal to
100,000BTU.
34CCF 
100,000 BTU
 3.4MMBTU
1CCF
27. Step 1: Convert gallons to BTUs of natural gas. Recall that 1 gallon is equal to
130,000BTU.
83gal 
130,000 BTU
 10.79MMBTU
1gal
28. Step 1: Convert the seasonal heating requirement from BTUs to CCF.
236MMBTU 
1CCF
 2,360CCF
100,000 BTU
29. Step 1: Convert the heating requirement from BTUs to gallons of fuel oil.
173MMBTU 
1gal
 1,330.77 gal
130,000 BTU
30. Step 1: Convert CCF to BTUs of natural gas.
106CCF 
100,000 BTU
 10.6MMBTU
1CCF
Step 2: Convert BTUs to Kilowatt-hours of electricity.
10.6MMBTU 
1kWh
 3,106.68kWh
3,412 BTU
31. Step 1: Calculate the seasonal heat loss through windows, the walls, and the roof
using their respective areas and R-values.
Seasonal Heat Losswindows 
Seasonal Heat Losswalls
(580 ft 2 )  6,000 F day  24h day
1
(1,920 ft 2 )  6,000 F day  24h day

19
Seasonal Heat Lossroof 
(2,750 ft 2 )  6,000 F day  24h day
22
Step 6: Calculate the seasonal heat lost through the house by totaling the seasonal
heat loss values.
Seasonal Heat Losshouse = 116,071,579BTU