Runoff from a subdivision is collected in a basin then released at 43 CFS through a circular corrugated metal pipe. ( n=0.025) The pipe slope is 0.001ft/ft
Find the diamter (ft) of the pipe that should be used to deliver the flow.
A. 3
B. 5
C. 6
D. 18
I ran out of time yesterday and skipped this. However I just did it and while it took a few minutes I thought I had it correct. However I was wrong.
I used the same method and equation as the solution, expect the solution peformed an odd mathematical step. ( or I'm wrong)
They used the manning equation. Because the diamter is unknow they used (3.14*d^2)/4 for area and (d/4)2/3 for the R value
But when they plugged in the numbers they get 43 = (1.49/.025) * (0.785d^2)* (0.397d^2/3)* (0.001^1/2)
Where is the 0.397 coming from? I had 0.25d^2/3
Using my numbers i get a 6' pipe. There is no errata for this. WTF , am I missing something very obivous here? I do feel like I'm starting to burnout a little.
Thanks
Find the diamter (ft) of the pipe that should be used to deliver the flow.
A. 3
B. 5
C. 6
D. 18
I ran out of time yesterday and skipped this. However I just did it and while it took a few minutes I thought I had it correct. However I was wrong.
I used the same method and equation as the solution, expect the solution peformed an odd mathematical step. ( or I'm wrong)
They used the manning equation. Because the diamter is unknow they used (3.14*d^2)/4 for area and (d/4)2/3 for the R value
But when they plugged in the numbers they get 43 = (1.49/.025) * (0.785d^2)* (0.397d^2/3)* (0.001^1/2)
Where is the 0.397 coming from? I had 0.25d^2/3
Using my numbers i get a 6' pipe. There is no errata for this. WTF , am I missing something very obivous here? I do feel like I'm starting to burnout a little.
Thanks
The 0.397 is simply (1/4)(2/3)
