NICET LEVEL 1 - WATER BASED
LAYOUT QUESTIONS & ANSWERS
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. - ANSWER1 2/2 in.
(ref. NFPA 13 2016 Sec. 9.2.1.1.2.
While conducting a flow test, the chosen hydrant should be ______ - ANSWERAs close
to the property where the sprinkler system is being installed. (Ref, NFPA 13 Sec.
A.24.2.2. or NFPA 291)
What formula is used to determine the discharge coefficiant relating pressure and flow
rate? - ANSWERK = Q / (Square root of P)
What best describes a tree system? - ANSWERCross mains and the branch lines are
not tied together, providing only one path of water flow.
Factors that define Occupancy Group Hazard 2 include: - ANSWERQuantity 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)
Which of the following standards contain reqs. pertaining to the installation of
underground piping? - ANSWERNFPA 13 & 24 (Both in Chapter 10)
Which of the following should you inspect on an extension cord before each use? -
ANSWERThe grounding prong & the insulation for nicks, cuts, or damage
________ indicates a mandatory req. according to NFPA definitions. - ANSWERShall.
(Ref. NFPA 13 2016, Sec. 3.2.4.)
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? - ANSWERAssume 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.
The owners info cert. indicates the property to be protected will contain nitrate film.
Which of the following is also req. to be provided? - ANSWERType, arrangement,
location, and intended maximum quantities.
Reference NFPA 13 2016, Figure A.23.1(b).
LAYOUT QUESTIONS & ANSWERS
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. - ANSWER1 2/2 in.
(ref. NFPA 13 2016 Sec. 9.2.1.1.2.
While conducting a flow test, the chosen hydrant should be ______ - ANSWERAs close
to the property where the sprinkler system is being installed. (Ref, NFPA 13 Sec.
A.24.2.2. or NFPA 291)
What formula is used to determine the discharge coefficiant relating pressure and flow
rate? - ANSWERK = Q / (Square root of P)
What best describes a tree system? - ANSWERCross mains and the branch lines are
not tied together, providing only one path of water flow.
Factors that define Occupancy Group Hazard 2 include: - ANSWERQuantity 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)
Which of the following standards contain reqs. pertaining to the installation of
underground piping? - ANSWERNFPA 13 & 24 (Both in Chapter 10)
Which of the following should you inspect on an extension cord before each use? -
ANSWERThe grounding prong & the insulation for nicks, cuts, or damage
________ indicates a mandatory req. according to NFPA definitions. - ANSWERShall.
(Ref. NFPA 13 2016, Sec. 3.2.4.)
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? - ANSWERAssume 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.
The owners info cert. indicates the property to be protected will contain nitrate film.
Which of the following is also req. to be provided? - ANSWERType, arrangement,
location, and intended maximum quantities.
Reference NFPA 13 2016, Figure A.23.1(b).
