AABC CERTIFIED TEST
AND BALANCE
TECHNICIAN TBT
PRACTICE EXAM
COMPLETE 107
QUESTIONS
Question 1: A pitot-tube traverse gives an average total pressure of 1.05 in. w.g. and an average static pressure of 0.69 in. w.g. What is the average velocity pressure?
Choices:
1) 0.36 in. w.g.2) 0.34 in. w.g.3) 0.69 in. w.g.4) 1.74 in. w.g.
Correct Answer: 0.36 in. w.g.
Explanation: Velocity pressure equals total pressure minus static pressure: 1.05 - 0.69 = 0.36 in.w.g.Page 1
Question 2: At standard air density, a measured velocity pressure is 0.25 in. w.g.Using V = 4005 × √VP, what is the air velocity?
Choices:
1) About 1,001 fpm 2) About 2,003 fpm 3) About 4,005 fpm 4) About 8,010 fpm
Correct Answer: About 2,003 fpm
Explanation: V = 4005 × √0.25 = 4005 × 0.5 = 2,002.5 fpm, or about 2,003 fpm.Question 3: A rectangular duct is 24 in. by 18 in. and the average measured velocity is 4,500 fpm. What is the air�ow?
Choices:
1) 9,000 cfm 2) 18,000 cfm 3) 13,500 cfm 4) 27,000 cfm
Correct Answer: 13,500 cfm
Explanation: The duct area is () × () = 3.0 ft². Air�ow = area × velocity = 3.0 × 4,500 = 13,500 cfm.Page 2
Question 4: A fan delivers 10,000 cfm at 900 rpm. If fan speed is increased to 1,080 rpm and the system is otherwise unchanged, what air�ow is predicted by the �rst fan law?
Choices:
1) 10,800 cfm 2) 14,400 cfm 3) 17,280 cfm 4) 12,000 cfm
Correct Answer: 12,000 cfm
Explanation: Air�ow varies directly with fan speed. 10,000 × (1,) = 12,000 cfm.Question 5: A fan develops 2.0 in. w.g. static pressure at 1,000 rpm. If speed rises by 20% with no other change, what static pressure is predicted by the fan laws?
Choices:
1) 2.88 in. w.g.2) 2.20 in. w.g.3) 2.40 in. w.g.4) 3.46 in. w.g.
Correct Answer: 2.88 in. w.g.
Explanation: Fan pressure varies with the square of speed. 2.0 × 1.20² = 2.88 in. w.g.Page 3