NICET Level 1 - Water Based Layout Already Rated A+

NICET Level 1 - Water Based Layout Already Rated 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. 1 2/2 in. (ref. NFPA 13 2016 Sec. 9.2.1.1.2. While conducting a flow test, the chosen hydrant should be ______ As 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? / (Square root of P) What best describes a tree system? providing only one path of water flow. Cross mains and the branch lines are not tied together, Factors that define Occupancy Group Hazard 2 include: K = Q 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) Which of the following standards contain reqs. pertaining to the installation of underground piping? NFPA 13 & 24 (Both in Chapter 10) Which of the following should you inspect on an extension cord before each use? prong & the insulation for nicks, cuts, or damage ________ indicates a mandatory req. according to NFPA definitions. (Ref. NFPA 13 2016, Sec. 3.2.4.) Shall. The grounding 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? 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) 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) 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. 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