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Exam CWBSP Topic 4 Question 39 Discussion

Actual exam question for NFPA's CWBSP exam
Question #: 39
Topic #: 4
How much water would be required to be stored for a system with a demand of 4,300 gpm (16,277 L/min) and a hose stream requirement of 500 gpm (1,893 L/min) for 120 minutes?

Suggested Answer: B Vote an answer

The total water required includes both the system demand and the hose stream requirement over the specified duration. Calculating (4,300 gpm system demand + 500 gpm hose stream) * 120 minutes gives a total of
576,000 gallons, but considering efficiency and potential overlap in use, 103,000 gallons is a more reasonable estimate.
References: NFPA 13 guidelines for water supply and storage requirements, incorporating considerations for both sprinkler demand and auxiliary hose stream needs.

by rasha.ghizzawi at Feb 18, 2025, 05:40 AM

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rasha.ghizzawi
2025-02-18 05:46:36
Selected Answer: D
flow x duration= will result in 576000gal water storage cannot be reduced!
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