BPI BUILDING SCIENCE PRINCIPLES QUESTIONS AND
ANSWERS A+ GRADED WITH EXPERT SOLUTIONS - 90
Questions with Answers
Page 1
,Q1. A building assembly consists of (from inside to outside): 1/2" gypsum board
(R=0.45), 6" fiberglass batt (R=19), 1" extruded polystyrene (R=5), and 1/2" OSB
sheathing (R=0.62). The interior temperature is 70°F, exterior is 30°F. What is the
temperature at the interface between the fiberglass batt and the extruded polystyrene,
assuming steady-state heat flow?
A. 45.5°F
B. 38.2°F
C. 52.1°F
D. 41.8°F
Correct Answer: B. 38.2°F
Rationale: The temperature drop across each layer is proportional to its R-value. Total R
= 0.45+19+5+0.62 = 25.07. Temperature drop from interior to the interface = (R_inside
+ R_batt)/R_total * (70-30) = (19.45/25.07)*40 = 31.0°F. Thus interface temp = 70 - 31.0
= 39°F, closest to 38.2°F. Other options misapply the ratio.
Why Wrong:
A - This would result from using only the batt R-value, ignoring the interior air film
and gypsum.
C - This would result from incorrectly adding the exterior R-values to the numerator.
D - This would result from using the exterior temperature drop rather than the interior.
Reference: BPI Building Science Principles, Chapter 4: Heat Transfer
Q2. In a cold climate, a wall assembly has an interior vapor retarder (Class I) and an
exterior insulating sheathing. Which condition is most likely to cause moisture
accumulation within the wall cavity during winter?
A. Exterior sheathing has a high permeance (more than 10 perms).
B. The vapor retarder is installed on the interior side of the insulation.
C. The wall cavity is vented to the exterior through weep holes.
D. The insulation is unfaced fiberglass batt with a low R-value.
Correct Answer: A. Exterior sheathing has a high permeance (more than 10 perms).
Rationale: In a cold climate, the interior vapor retarder keeps interior moisture out of the
cavity. If the exterior sheathing is highly permeable, in winter the cavity may become
colder than the dew point of the interior air, but more importantly, the lack of an exterior
vapor barrier allows moisture from outside (or from exfiltration) to enter and condense on
the cold interior side. However, the classic problem is that a Class I vapor retarder on the
interior prevents drying to the inside, and if exterior sheathing is permeable, moisture can
accumulate if the cavity is wetted. The correct answer is A because high permeance
exterior sheathing allows exterior moisture to enter and, combined with the interior vapor
retarder, creates a moisture trap.
Why Wrong:
B - This is the standard practice for cold climates, not a cause of accumulation.
Page 2
, C - Venting the cavity to the exterior would help dry the cavity, not cause
accumulation.
D - Low R-value insulation would increase heat loss but not directly cause moisture
accumulation.
Reference: BPI Building Science Principles, Chapter 5: Moisture Management
Q3. A technician performs a blower door test on a 2,500 sq ft house. The measured
CFM50 is 3,200. The house has a volume of 20,000 cubic feet. What is the
approximate natural air changes per hour (ACH_natural) using the default divisor of
20?
A. 0.35 ACH
B. 0.48 ACH
C. 0.20 ACH
D. 0.60 ACH
Correct Answer: B. 0.48 ACH
Rationale: ACH50 = CFM50 * 60 / Volume = 3200*60/20000 = 9.6 ACH50. Then
ACH_natural = ACH = 9.6/20 = 0.48 ACH. Other options misapply the formula or
use incorrect volume conversions.
Why Wrong:
A - This would result from dividing by 25 instead of 20.
C - This would result from using a divisor of 50.
D - This would result from using a divisor of 16 or incorrect volume.
Reference: BPI Building Science Principles, Chapter 8: Air Leakage Testing
Q4. Which of the following is the primary reason that a naturally aspirated gas water
heater in a tightly sealed mechanical room may backdraft when an exhaust fan is
operating?
A. The exhaust fan creates a negative pressure that overcomes the draft hood's
buoyancy.
B. The water heater's flue is too short to generate sufficient draft.
C. The combustion air supply is too large, causing excessive cooling of the flue gases.
D. The water heater is located too close to the exhaust fan, causing direct interference.
Correct Answer: A. The exhaust fan creates a negative pressure that overcomes the
draft hood's buoyancy.
Rationale: Exhaust fans depressurize the space, reducing the pressure available to push
combustion gases up the flue. If the negative pressure exceeds the draft hood's ability, flue
gases spill or reverse direction. The other options are not primary causes of backdrafting.
Why Wrong:
B - Flue length affects draft but is not the primary cause when an exhaust fan operates.
Page 3
, C - Excess combustion air would not cause backdraft; it might cool flue gases but not
reverse flow.
D - Proximity to the fan is not the main mechanism; pressure difference is.
Reference: BPI Building Science Principles, Chapter 10: Combustion Safety
Q5. A building has a measured duct leakage to outside of 400 CFM25. The total duct
leakage is 800 CFM25. What is the approximate percentage of duct leakage that is to
the outside?
A. 50%
B. 33%
C. 25%
D. 75%
Correct Answer: A. 50%
Rationale: The percentage to outside is simply (leakage to outside / total leakage) * 100 =
(400/800)*100 = 50%. The other options are incorrect because they misapply the ratio or
confuse with other metrics.
Why Wrong:
B - This would be the result if total leakage were 1200 CFM25.
C - This would be the result if leakage to outside were 200 CFM25.
D - This would be the result if leakage to outside were 600 CFM25.
Reference: BPI Building Science Principles, Chapter 9: Duct Leakage Testing
Q6. In a hot-humid climate, which of the following is the most effective strategy to
prevent condensation on a cold interior surface of a wall assembly during summer?
A. Increase the interior air temperature.
B. Add a vapor barrier on the exterior side of the wall.
C. Vent the wall cavity to the exterior.
D. Use a smart vapor retarder that changes permeability with humidity.
Correct Answer: D. Use a smart vapor retarder that changes permeability with
humidity.
Rationale: In hot-humid climates, the dominant moisture drive is from outside to inside. A
smart vapor retarder allows drying to the interior during summer when needed but blocks
moisture in winter. Adding an exterior vapor barrier would trap moisture. Venting the
cavity to the exterior would bring in humid air. Increasing interior temperature would help
but is not a structural strategy.
Why Wrong:
A - Raising the interior temperature might reduce condensation but is not a building
assembly solution.
B - An exterior vapor barrier would trap moisture and cause problems.
Page 4
ANSWERS A+ GRADED WITH EXPERT SOLUTIONS - 90
Questions with Answers
Page 1
,Q1. A building assembly consists of (from inside to outside): 1/2" gypsum board
(R=0.45), 6" fiberglass batt (R=19), 1" extruded polystyrene (R=5), and 1/2" OSB
sheathing (R=0.62). The interior temperature is 70°F, exterior is 30°F. What is the
temperature at the interface between the fiberglass batt and the extruded polystyrene,
assuming steady-state heat flow?
A. 45.5°F
B. 38.2°F
C. 52.1°F
D. 41.8°F
Correct Answer: B. 38.2°F
Rationale: The temperature drop across each layer is proportional to its R-value. Total R
= 0.45+19+5+0.62 = 25.07. Temperature drop from interior to the interface = (R_inside
+ R_batt)/R_total * (70-30) = (19.45/25.07)*40 = 31.0°F. Thus interface temp = 70 - 31.0
= 39°F, closest to 38.2°F. Other options misapply the ratio.
Why Wrong:
A - This would result from using only the batt R-value, ignoring the interior air film
and gypsum.
C - This would result from incorrectly adding the exterior R-values to the numerator.
D - This would result from using the exterior temperature drop rather than the interior.
Reference: BPI Building Science Principles, Chapter 4: Heat Transfer
Q2. In a cold climate, a wall assembly has an interior vapor retarder (Class I) and an
exterior insulating sheathing. Which condition is most likely to cause moisture
accumulation within the wall cavity during winter?
A. Exterior sheathing has a high permeance (more than 10 perms).
B. The vapor retarder is installed on the interior side of the insulation.
C. The wall cavity is vented to the exterior through weep holes.
D. The insulation is unfaced fiberglass batt with a low R-value.
Correct Answer: A. Exterior sheathing has a high permeance (more than 10 perms).
Rationale: In a cold climate, the interior vapor retarder keeps interior moisture out of the
cavity. If the exterior sheathing is highly permeable, in winter the cavity may become
colder than the dew point of the interior air, but more importantly, the lack of an exterior
vapor barrier allows moisture from outside (or from exfiltration) to enter and condense on
the cold interior side. However, the classic problem is that a Class I vapor retarder on the
interior prevents drying to the inside, and if exterior sheathing is permeable, moisture can
accumulate if the cavity is wetted. The correct answer is A because high permeance
exterior sheathing allows exterior moisture to enter and, combined with the interior vapor
retarder, creates a moisture trap.
Why Wrong:
B - This is the standard practice for cold climates, not a cause of accumulation.
Page 2
, C - Venting the cavity to the exterior would help dry the cavity, not cause
accumulation.
D - Low R-value insulation would increase heat loss but not directly cause moisture
accumulation.
Reference: BPI Building Science Principles, Chapter 5: Moisture Management
Q3. A technician performs a blower door test on a 2,500 sq ft house. The measured
CFM50 is 3,200. The house has a volume of 20,000 cubic feet. What is the
approximate natural air changes per hour (ACH_natural) using the default divisor of
20?
A. 0.35 ACH
B. 0.48 ACH
C. 0.20 ACH
D. 0.60 ACH
Correct Answer: B. 0.48 ACH
Rationale: ACH50 = CFM50 * 60 / Volume = 3200*60/20000 = 9.6 ACH50. Then
ACH_natural = ACH = 9.6/20 = 0.48 ACH. Other options misapply the formula or
use incorrect volume conversions.
Why Wrong:
A - This would result from dividing by 25 instead of 20.
C - This would result from using a divisor of 50.
D - This would result from using a divisor of 16 or incorrect volume.
Reference: BPI Building Science Principles, Chapter 8: Air Leakage Testing
Q4. Which of the following is the primary reason that a naturally aspirated gas water
heater in a tightly sealed mechanical room may backdraft when an exhaust fan is
operating?
A. The exhaust fan creates a negative pressure that overcomes the draft hood's
buoyancy.
B. The water heater's flue is too short to generate sufficient draft.
C. The combustion air supply is too large, causing excessive cooling of the flue gases.
D. The water heater is located too close to the exhaust fan, causing direct interference.
Correct Answer: A. The exhaust fan creates a negative pressure that overcomes the
draft hood's buoyancy.
Rationale: Exhaust fans depressurize the space, reducing the pressure available to push
combustion gases up the flue. If the negative pressure exceeds the draft hood's ability, flue
gases spill or reverse direction. The other options are not primary causes of backdrafting.
Why Wrong:
B - Flue length affects draft but is not the primary cause when an exhaust fan operates.
Page 3
, C - Excess combustion air would not cause backdraft; it might cool flue gases but not
reverse flow.
D - Proximity to the fan is not the main mechanism; pressure difference is.
Reference: BPI Building Science Principles, Chapter 10: Combustion Safety
Q5. A building has a measured duct leakage to outside of 400 CFM25. The total duct
leakage is 800 CFM25. What is the approximate percentage of duct leakage that is to
the outside?
A. 50%
B. 33%
C. 25%
D. 75%
Correct Answer: A. 50%
Rationale: The percentage to outside is simply (leakage to outside / total leakage) * 100 =
(400/800)*100 = 50%. The other options are incorrect because they misapply the ratio or
confuse with other metrics.
Why Wrong:
B - This would be the result if total leakage were 1200 CFM25.
C - This would be the result if leakage to outside were 200 CFM25.
D - This would be the result if leakage to outside were 600 CFM25.
Reference: BPI Building Science Principles, Chapter 9: Duct Leakage Testing
Q6. In a hot-humid climate, which of the following is the most effective strategy to
prevent condensation on a cold interior surface of a wall assembly during summer?
A. Increase the interior air temperature.
B. Add a vapor barrier on the exterior side of the wall.
C. Vent the wall cavity to the exterior.
D. Use a smart vapor retarder that changes permeability with humidity.
Correct Answer: D. Use a smart vapor retarder that changes permeability with
humidity.
Rationale: In hot-humid climates, the dominant moisture drive is from outside to inside. A
smart vapor retarder allows drying to the interior during summer when needed but blocks
moisture in winter. Adding an exterior vapor barrier would trap moisture. Venting the
cavity to the exterior would bring in humid air. Increasing interior temperature would help
but is not a structural strategy.
Why Wrong:
A - Raising the interior temperature might reduce condensation but is not a building
assembly solution.
B - An exterior vapor barrier would trap moisture and cause problems.
Page 4
