NICET LEVEL 1 WATER BASED LAYOUT EXAM
QUESTIONS AND ANSWERS | 100% VERIFIED
ANSWERS | LATEST UPDATE 2026/2027 | GRADED
A+.
Toggle hangers shall be permitted only for the support of pipe
__________ inches or smaller in size under ceilings of hollow tile or
metal lath and plaster. - CORRECT ANSWER - 1 2/2 in. (ref. NFPA
13 2016 Sec. 9.2.1.1.2.
While conducting a flow test, the chosen hydrant should be ______ -
CORRECT ANSWER - As close to the property where the sprinkler
system is being installed. (Ref, NFPA 13 Sec. A.24.2.2. or NFPA 291)
Factors that define Occupancy Group Hazard 2 include: - CORRECT
ANSWER - Quantity and combustibility of contents are moderate to
high.
Stockpiles of contents with moderate rates of heat release do not
exceed 12 ft and stockpiles with high heat release do not exceed 8 ft
(Ref NFPA 13 Sec. 5.3.2)
100 GPM is flowing through 100' of 1 1/4 in. Sch. 40 pipe. How
many feet of 1
,1/2 in. pipe will produce the same friction loss? - CORRECT
ANSWER - Assume this is Schedule 40 pipe, C = 120. Use the
Hazen-Williams formula to calculate the friction loss for the 1 ¼ in.
pipe and the 1 ½ in. pipe.
For 1 ¼ in. pipe (ID = 1.38):
P friction loss = 4.52 x Q 1.85
c 1.85 x d 4.87
P friction loss = 4.52 x 100 1.85
120 1.85 x 1.38 4.87
P friction loss = 0.672 psi/ft
For 1 ½ in. pipe (ID = 1.61):
P friction loss = 4.52 x Q 1.85
c 1.85 x d 4.87
, P friction loss = 4.52 x 100 1.85
120 1.85 x 1.61 4.87
P friction loss = 0.317 psi/ft
Step #2: Determine the total friction loss for 100 gpm flowing through
100 ft. of
1 ¼ in. Schedule 40 black steel pipe: 100 ft. x 0.672 psi/ft = 67.2 psi
Step #3: Determine the pipe length to produce the same friction loss
in the 1 ½ in. pipe:
y ft. x 0.317 psi/ft = 67.2 psi
y = 67.2 ÷ 0.317
y = 212 ft.
QUESTIONS AND ANSWERS | 100% VERIFIED
ANSWERS | LATEST UPDATE 2026/2027 | GRADED
A+.
Toggle hangers shall be permitted only for the support of pipe
__________ inches or smaller in size under ceilings of hollow tile or
metal lath and plaster. - CORRECT ANSWER - 1 2/2 in. (ref. NFPA
13 2016 Sec. 9.2.1.1.2.
While conducting a flow test, the chosen hydrant should be ______ -
CORRECT ANSWER - As close to the property where the sprinkler
system is being installed. (Ref, NFPA 13 Sec. A.24.2.2. or NFPA 291)
Factors that define Occupancy Group Hazard 2 include: - CORRECT
ANSWER - Quantity and combustibility of contents are moderate to
high.
Stockpiles of contents with moderate rates of heat release do not
exceed 12 ft and stockpiles with high heat release do not exceed 8 ft
(Ref NFPA 13 Sec. 5.3.2)
100 GPM is flowing through 100' of 1 1/4 in. Sch. 40 pipe. How
many feet of 1
,1/2 in. pipe will produce the same friction loss? - CORRECT
ANSWER - Assume this is Schedule 40 pipe, C = 120. Use the
Hazen-Williams formula to calculate the friction loss for the 1 ¼ in.
pipe and the 1 ½ in. pipe.
For 1 ¼ in. pipe (ID = 1.38):
P friction loss = 4.52 x Q 1.85
c 1.85 x d 4.87
P friction loss = 4.52 x 100 1.85
120 1.85 x 1.38 4.87
P friction loss = 0.672 psi/ft
For 1 ½ in. pipe (ID = 1.61):
P friction loss = 4.52 x Q 1.85
c 1.85 x d 4.87
, P friction loss = 4.52 x 100 1.85
120 1.85 x 1.61 4.87
P friction loss = 0.317 psi/ft
Step #2: Determine the total friction loss for 100 gpm flowing through
100 ft. of
1 ¼ in. Schedule 40 black steel pipe: 100 ft. x 0.672 psi/ft = 67.2 psi
Step #3: Determine the pipe length to produce the same friction loss
in the 1 ½ in. pipe:
y ft. x 0.317 psi/ft = 67.2 psi
y = 67.2 ÷ 0.317
y = 212 ft.
