Calculating pipe diameter for half-flow condition/mannings equation

jeudi 29 mai 2014

Hello,



My knowledge on fluid mechanics is on a introductory level. When explaining, it would be great if you could explain in simple English and also like teaching to a student who isn't very good at the subject area!

I've been reading a lecture slide and have been trying to follow the example to solve my given problem. I don't really understand how the lecturer got certain parts of the solution though (equation hasn't been explained/defined on any of the slides). I don't understand how to find the diameter or what the suitable velocity is for half-flow.



I have attached the slide that I'm using as an example below.



I used A = pi*D^2/8 and P = pi*D/2 as my problem is for half flow condition.



What is wrong? How do I do this correctly?



Thank you for reading. Any help would be appreciated, thank you.





1. The problem statement, all variables and given/known data



If water flows from Point D at Kinglake to Point E at Sugarloaf in a lined circular channel laid on a slope of 0.001. If Manning’s n = 0.013 and if the design discharge is 0.07 m^3/s, what diameter channel should be designed for a half-flow condition? What will be the mean velocity in the channel for this condition?



Manning’s n = 0.013

Slope = 0.001

Q, Discharge = 0.03m^3/s



Find diameter of channel for half flow condition.

Find mean velocity of channel in half flow condition.



2. Relevant equations



Mannings equation:


















Velocity equation:














3. The attempt at a solution









A = pi*D^2/8 and P = pi*D/2





How do I find the diameter required for the channel from this?



------





V = 0.07m^3/s/((pi*(0.057)^2)/8) = 3.127 m/s



Is 3.127 m/s suitable for half-flow?

What is the velocity for a half-flow pipe? How do I find whether the velocity I calculated is suitable?




Attached Images





File Type: png manningseq.png (19.9 KB)







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