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Forum » civil eng brahs, how the fukk do i answer this question?
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Thread: civil eng brahs, how the fukk do i answer this question?

post 1620269461 10-23-2020, 08:36 PM
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civil eng brahs, how the fukk do i answer this question?

calculate depth of flow in open rectangular channel of width 2m, slope 2%, n=0.013 (manning coeff.), and peak flow 1.19m/s.

srs.

i can't find the formula that connects all of these.
post 1620269491 10-23-2020, 08:37 PM
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Bout 350
post 1620270421 10-23-2020, 08:49 PM
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PhD in M.E. checking in

Spoiler!
3.50
post 1620270541 10-23-2020, 08:51 PM
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It's quite simple, 5x70= (your answer)
post 1620270701 10-23-2020, 08:52 PM
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700/2 = X

X is the answer
There is no they…
post 1620271051 10-23-2020, 08:59 PM
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Originally Posted By 3basic5me⏩
calculate depth of flow in open rectangular channel of width 2m, slope 2%, n=0.013 (manning coeff.), and peak flow 1.19m/s.

srs.

i can't find the formula that connects all of these.
just multiply 2 x 0.02 x 0.013 x 1.19 x 5.656x10^3
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post 1620271191 10-23-2020, 09:01 PM
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Originally Posted By frezKo⏩
Simple

Use manning's equation:


But how do I get the velocity and area? All the question provides is the width of the channel.
post 1620271271 10-23-2020, 09:01 PM
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Originally Posted By Better
just multiply 2 x 0.02 x 0.013 x 1.19 x 5.656x10^3
I knew it
post 1620271411 10-23-2020, 09:03 PM
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Fd = ((2m x 2%) + (0.013 + 1.19)) x 281.577


Fukn wizarded by Better Unborn
post 1620271671 10-23-2020, 09:07 PM
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Originally Posted By 3basic5me⏩
But how do I get the velocity and area? All the question provides is the width of the channel.
did they give you peak flow or peak velocity?
post 1620271761 10-23-2020, 09:08 PM
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Originally Posted By elterrible987⏩
did they give you peak flow or peak velocity?
Peak flow, yes
post 1620272251 10-23-2020, 09:14 PM
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Originally Posted By frezKo⏩
OP if you want to be an engineer one day you need to improve your critical thinking skills




You have your flow (Q), coefficient (n), and slope (s).

The only variables you have left is your A = area and hydraulic radius


You can substitte A = W * H

where w = width = 2'
and H = height


R is the only variable left.


Rectangular Cross Section
From the hydraulic radius definition: RH = A/P, where A is the cross sectional area of flow and P is its wetted perimeter.

Both A and P can be simplified as a Height * Width & W+ H + H where we already have the width.


Plug in the hydraulic radius in the complex equation and the only variable left to solve for is the flow height.

Hope that makes sense
Yeah that makes sense. Fukk I shud have thought of that. Thanks
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