Question

Homework 4 Problem 1: Solver Hw#3 using Klentrance)= 0.5, Kelbow)= 0.25, Kvalve)=1.5, Klexit)= 1 Problem 2: As engineer, you are required to design a single pipe that connected manhole A to manhole B. The design discharge for the system is 5 cfs and ground elevations at A and B are 950 ft and 945 ft, respectively. Your job to come up with design system (Commercial Diameter) that satisfied all requirements for cheapest costs. Keep in mind, you are dealing with storm water system. 950 ft L= 25 ft 945 ft

Answer 1

Solution of Problem 2:

Assume the dia of Storm water sewer = D

The design discharge of sewer can be computed from Manning’s Equation as below:

1.486AR2/351/2 Q=

Where A = Cross Section area of pipe

R = Hydraulic radius = A/P

P = Wetted perimeter

S = Slope

n = Manning’s roughness coefficient

Given,

Q = 5 cfs

TD2

P = TD (for full pipe)

RE=480 = ATD2D

SE25 (950 - 945) = 0.20

Assume, n = 0.012 for concrete pipes

Keeping the above values in manning’s equation,

5 - 1.486 TEDx @*** X0.201/2

D2 x D2/3 = 5 x 0.012 X 4 X 42/3 1.486 x 3.1416 x 0.201/2

D^8/3 = 0.2897

D= 0.2897^8/3 = 0.628 feet = 7.54 inch say 8 inch

Hence it would be sufficient to install an 8 inch dia circular concrete pipe between manholes to facilitate the design discharge of 5cfs.

Problem 1 is not clear. Please post it separately with references.

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