Civil FE/PE Exam - Water Resources - Solve for Required Pipe Size given Flow Demand

[@AndrewMcGinley-te7ym]

There is a simpler equation - D = [C0*Q*n/Sqrt(S)]^(3/8) that should be used, in my opinion. Also, as others have noted, the manning's value is 0.013 for plastic.

[@Imlivinlife27]

Hello on PDF pg 346 for the PE plastic is listed as " n = 0.013 "

[@jagc1998]

With corrected n = 0.013 value I got d = 11.064 in

[@andrewlecheminant1079]

n=0.010 if pipe was PVC but problem statement states that pipe is plastic. Per the table, n=0.013. Please revise video.

[@oleopathic]

ncees pe hb 1.1, p. 3, 328-329. used hazen-williams to get a close-enough answer. anyone thinks this was dead wrong ? here are my numbers: . q = 1.81215 cfs . c = 150 . s = 0.004 solve for d. d = 0.7966' = 9.56" = 10.0". always round up pipes hence the 10". (B)

[@abradablincoln3052]

I would not say it is dead wrong as the results from using the hazen-williams equation versus manning's equation are around the same. Although, in industry, we typically don't use the hazen-williams equation for sizing sewer pipes, therefore I would use the manning's equation. The PE exam is the "profession & practice of engineering", so you would need to use the engineering theory mixed with what is typically done by most engineers in the practice of the profession.