Nicet Level 1 - Water Based Layout Practice Exam Questions And Answers
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Nicet Level 1 - Water Based Layout
Institution
Nicet Level 1 - Water Based Layout
Nicet Level 1 - Water Based Layout Practice
Exam Questions And Answers
Which of the following standards contain reqs. pertaining to the installation of
underground piping? - answer -NFPA 13 & 24 (Both in Chapter 10)
Which of the following should you inspect on an extension cord before each use...
Nicet Level 1 - Water Based Layout Practice
Exam Questions And Answers
Which of the following standards contain reqs. pertaining to the installation of
underground piping? - answer -NFPA 13 & 24 (Both in Chapter 10)
Which of the following should you inspect on an extension cord before each use? -
answer -The grounding prong & the insulation for nicks, cuts, or damage
________ indicates a mandatory req. according to NFPA definitions. - answer -Shall.
(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? - 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.
A basic contract: (choose all that apply) - answer -Contains the names of the parties
involved, References the contract documents, Discusses the work that is to be done,
and Discusses start and completion dates.
Define both the constant, y, and the exponent ,z, for q in the Hazen-Williams friction
loss formula shown Below:
P = L x (y x Q^z)
(c^1.85 x d^4.87) - answer -4.52, 1.85.
Reference NFPA 13 2016 Section 23.4.2.1.1.
A water flow test was conducted using 1 hydrant, flowing with two 2 1/2 inch outlets.
Which is the approximate water flow rate, if the pitot reading was 15 psi. The hydrant
outlet coefficiant is 0.8. - answer -The correct answer is 1156 gpm.
Solution #12:
Q 1 = 29.84 cd 2 √ P
Q 1 = 29.84 x 0.8 x (2.5) 2 x √15
Q 1 = 577.8 gpm
Both outlets are 2 ½ in., with the same outlet coefficient. Therefore, Q 1 = Q 2
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