Open channel flow, mannings equation and specific energy

[ShawnCohen]

Homework Statement

A 4m wide channel has a bed slope of 1 in 1500 and a Manning's roughness coef n=0.01. A broad crested measuring weir in the form of a streamlined hump 0.4m high spans across the full width of the channel. the discharge is 10.5m^3/s. Determine
i)the uniform depth of flow
iv)the depth of water on the weir, is this critial

Homework Equations

Mannings equation: Q=A^(5/3)*(S)^(1/2)/(n*P^(2/3)) where A is cross sectional area, S is bed slope, P is wetted perimeter

Bernoulli equation elevation+(v^2)/(2*g)=constant (i ignore pressure head here as I am working on surface so use atmospheric pressure as zero)

Specific energy - E=d+(Q/w)^2/(2*g*d^2) where d is the depth and w is the width

The Attempt at a Solution

i use mannings equation to determine the depth and i get 1.22m, but then when i try and find the depth on top of weir i get nowhere, in face if i work out the minimum possible specific energy i get 1.333 and on top of the weir i get 1.05, so I am clearly going wrong somewhere. any help anyone??

 

[gneill]

I admit up front that I don't know much about channel flow rates... but I took your Manning's equation and assumed that if ww = 4m is the channel width, and wd is the unknown water depth, then A = ww*wd and P = ww + 2wd, and take the other constants as given. I inserted a constant ##k = 1m^{1/3}s^{-1}## to make the units balance for flow rate.

Then, using a numerical solver I find that wd = 1.891m.

Of course I may be missing some procedural details regarding taking the weir into account...EDIT: I discovered later that I had used S = 1/5000 instead of 1/1500 for this calculation. See below for further groveling.

 

[ShawnCohen]

thanks for the reply, but I am pretty sure that's not right, if you put in 1.22m you get Q=10.47 as required, but if you put in 1.891 you get Q=19.16 don't you?

 

[gneill]

thanks for the reply, but I am pretty sure that's not right, if you put in 1.22m you get Q=10.47 as required, but if you put in 1.891 you get Q=19.16 don't you?

Hmm. If I plug wd = 1.891 m into the formula I find Q = 10.5 m³/s. That formula being:

$$Q = k \frac{(ww\;wd)^{5/3}\sqrt{S}}{n (ww + 2wd)^{2/3}}~~~~~~~k = m^{1/3}s^{-1}$$

 

[ShawnCohen]

really?? i keep trying it again and again and i get 1.22 as the answer. what value are you using for S? 1/1500=6.67x10^-4 right?

 

[gneill]

really?? i keep trying it again and again and i get 1.22 as the answer. what value are you using for S? 1/1500=6.67x10^-4 right?

D'OH! I've been wasting your time For some reason when I copied down the constants I "saw" the slope as 1 in 5000. Much apologies.

If I use the correct value for S then I too get 1.22 m for the depth.

 

[ShawnCohen]

aha i thought i was going insane for a minute. thanks for attempting anyway