NCEES WR/ENV 2011 sample question 509

[Jacob_PE]

For an open channel flow question where you have a known Q and are solving for an unknown rectangular width, using Manning's you'll have the unknown width in the numerator and denominator of the left side of the equation. I have an HP33s and Casio fx115es and have briefly reviewed their equation solving abilities, but I'm still not clear on how to solve the problem without trial and error. Is it even possible?

 

[civilized_naah]

For an open channel flow question where you have a known Q and are solving for an unknown rectangular width, using Manning's you'll have the unknown width in the numerator and denominator of the left side of the equation. I have an HP33s and Casio fx115es and have briefly reviewed their equation solving abilities, but I'm still not clear on how to solve the problem without trial and error. Is it even possible?

Using tables for these unknown depth (or breadth) problems is best. The CERM and the All In One have these tables.

 

[Jacob_PE]

For an open channel flow question where you have a known Q and are solving for an unknown rectangular width, using Manning's you'll have the unknown width in the numerator and denominator of the left side of the equation. I have an HP33s and Casio fx115es and have briefly reviewed their equation solving abilities, but I'm still not clear on how to solve the problem without trial and error. Is it even possible?

Using tables for these unknown depth (or breadth) problems is best. The CERM and the All In One have these tables.

Right on.

Design Q=3000cfs

Q=KS^(1/2)

Q=(K1+K2)S^(1/2)

K2S^(1/2) = 1149.82cfs [10' deep end]

K1S^(1/2) = 1850.18cfs [6' deep end]

All-in-one K' = 28.818542

use table on of 507, x=.035938, b=166.95'

CERM K' = 42.676

use table on pg A-40, x=.034, b=176.45'

CERM is far more accurate since table gives x value every hundredth instead if every 2 hundredth.

Available answers are spread very far apart so using the All-in-one still golden.

 

[civilized_naah]

Design Q=3000cfsK2S^(1/2) = 1149.82cfs [10' deep end]

K1S^(1/2) = 1850.18cfs [6' deep end]

Not sure what you mean by 10 ft deep end vs 6 ft deep end. Can you post the entire problem?

 

[Jacob_PE]

Design Q=3000cfsK2S^(1/2) = 1149.82cfs [10' deep end]

K1S^(1/2) = 1850.18cfs [6' deep end]

Not sure what you mean by 10 ft deep end vs 6 ft deep end. Can you post the entire problem?

It's a rectangular channel with a 'step in the middle' right side is 50' wide and 10' deep, left side is unknown' wide and 6' deep. n=.14 left side and n=.05 right side, s=.001 ft per ft. Need 3000 cfs, find unknown width.

 

[Environmental_Guy]

Which NCEES sample exam book is this question from? Is it the 2008 book with the color photo cover?

It sounds like a good problem and I'd like to work it but I'm not quite understanding the problem or solution. Thanks!

 

[Jacob_PE]

Which NCEES sample exam book is this question from? Is it the 2008 book with the color photo cover?
It sounds like a good problem and I'd like to work it but I'm not quite understanding the problem or solution. Thanks!

It's from the 2011 sample questions, there's one similar in the All-in-one book but it only has you solve for Q and it's trapezoidal, but you use the d/b tables just like when solving for K and K'. I might be able to do a write up of the problem, scan and post it some time.