# Appendix 19C (Circular channel Ratio) quaetion



## Calixico (Aug 17, 2009)

So I am working problem #6 in open channel flow in the practice problem of Cerm and have gotton stuck on one concept.

given diameter, depth of flow, slope of a channel, mannings coefficient and roughness coefficent, where n varies with depth.

The problem asked to find (a) velocity of water in the storm sewer. I found that with mannings no problem by doing the following calculation:

V=16.21 ft/sec

Q=203.7ft^3/sec

so d/D= 0.375

Using Circular channel Ratio table, v/v(full)=0.68

so v=0.68*16.21ft/sec =11ft/sec

*This is where I get left behind:* The next question is:

(b)what is the maximum velocity that can be achieved

After numerous attempt and finally looking at the soluton I cannot figure out how it was done. Kinda like does the chicken comes before the egg question.

The solution states: go to Using Circular channel Ratio table. Maximum value of v/v(full)= 1.04 (at d/D= aproximately 0.96)

I cannot see this. do you just find the maximum value of v(varied) line (almost at the top of the page) by just looking at the maximum value of that curve and simply move vertical down to the horizontal axis to find v/v(full)= 1.04? Or d/D= aproximately 0.96, is found first, if yes, then how.....

c) next problem asked for maximum capacity of the storm drain and gives the solution in the smae manner.. Stated that Qmax/Qfull= 1.025 (at d/D = 0.96)

and Q(max)= 1.02 * 203.7ft^3/sec = 207.8 ft^3/sec

Thank for all your help, in advance!


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