Storm of the day

[Suns Den]

A Storm drain with the following given:

Diameter = 36 inch

Q = 40 cfs

S0 = 0.010

n = 0.015

Find:

a) Normal depth

B) Normal velocity

c) Critical depth; and

d) Flow regime

 

[owillis28]

Can you post the answers?

Thanks

owillis

 

[Suns Den]

Can you post the answers?
Thanks

owillis

Here we go... Owillis

a) Normal depth

1.85'

b) Normal velocity

8.82 fps

c) Critical depth; and

2.06'

d) Flow regime

Yc > Yn

SuperCritical

 

[owillis28]

Here we go... Owillis
a) Normal depth

1.85'

B) Normal velocity

8.82 fps

c) Critical depth; and

2.06'

d) Flow regime

Yc > Yn

SuperCritical

Can you post your solutions for these problems? I came up with most of the answers but have a question on one of them. Not sure if it is just rounding or just fuzzy math.

Thanks,

owillis

 

[roadwreck]

I'm not a water resources guy, so maybe I am missing something...

...how do you calculate the normal depth of a round pipe?

Just curious, all the problems I have worked usually give you a different channel geometry to try and solve for (rectangular, trapezoidal, etc.) or give you the normal depth for a circular pipe.

The CERM makes it very clear that it isn't easy to calculate the normal depth of a round pipe directly unless the pipe is running gull or half full.

 

[WaterGirl]

I'm not a water resources guy, so maybe I am missing something...
...how do you calculate the normal depth of a round pipe?

Just curious, all the problems I have worked usually give you a different channel geometry to try and solve for (rectangular, trapezoidal, etc.) or give you the normal depth for a circular pipe.

The CERM makes it very clear that it isn't easy to calculate the normal depth of a round pipe directly unless the pipe is running gull or half full.

There is an add-in function in Excel called Goal Seek. It does the iterative procedure necessary to solve for normal depth. You would put in the diameter, the n-value, and the slope with a flow value and it will iterate until it finds a depth of water that satisfies both sides of Manning's equation.

 

[RIP - VTEnviro]

:multiplespotting: Welcome watergirl.

I hadn't thought of using Goal-Seek for something like that but it's a great idea.

 

[roadwreck]

There is an add-in function in Excel called Goal Seek. It does the iterative procedure necessary to solve for normal depth. You would put in the diameter, the n-value, and the slope with a flow value and it will iterate until it finds a depth of water that satisfies both sides of Manning's equation.

Thanks, I just wanted to know if there was a way other then finding it iteratively. My NCEES approved calculator doesn't seem to have excel on it.

 

[jfusilloPE]

Thanks, I just wanted to know if there was a way other then finding it iteratively. My NCEES approved calculator doesn't seem to have excel on it.

I'm looking through this problem right now, and I agree, the CERM does say that "normal depth in circular channels can be calculated directly only under limited conditions..."

I'm going to work on it this morning/afternoon, and I will attempt to post my solutions today.

 

[roadwreck]

I'm looking through this problem right now, and I agree, the CERM does say that "normal depth in circular channels can be calculated directly only under limited conditions..."
I'm going to work on it this morning/afternoon, and I will attempt to post my solutions today.

Thanks for looking into it. Any example problem of this type I have come across so far has either been a shape where the normal depth can be computed directly or is a round pipe where you are told to assume the pipe is flowing full or half full. I'm not sure if that is meant to indicate that calculating normal depth in a round pipe isn't expected on the exam (b/c it seems that it can only be achieved iteratively) or if their is truly an alternative (less time consuming) method that the CERM doesn't want you to know .

 

[Guest]

I have attached a nomograph that provides a way to determing normal depth for either circular or rectangular/trapezoidal cross-sections. It is pretty straight-forward, no iterations necessary. I am also adding a nomograph for computing critical depth for good measure. Maybe this might help some

JR

 

[owillis28]

Nice work, JR

 

[Suns Den]

Kings tables may be one way to go baby!

 

[Guest]

^^^ I will start working that angle as well ... I haven't forgotten I have just been a little side tracked lately :laugh:

JR

 

[IlPadrino]

I have attached a nomograph that provides a way to determing normal depth for either circular or rectangular/trapezoidal cross-sections. It is pretty straight-forward, no iterations necessary. I am also adding a nomograph for computing critical depth for good measure. Maybe this might help some




JR

JR,

Good stuff! But... I don't think they're called nomographs. I thought nomographs were those things that have three vertical axis and you pick points on any two, connect them with a straight line and read where the line crosses the third axis. Or am I confused? I only ask because I usually hate them - I always have a hard time connecting the points with a piece of paper!

 

[Guest]

^^^ A nomograph is considered to be a graph consisting of three coplanar curves, each graduated for a different variable so that a straight line cutting all three curves intersects the related values of each variable (according to American Heritage Dictionary).

Your most popular nomographs are the straight line connectors for mannings equation, hazen-williams formula, or the various culvert geometrical sections, so in a sense you are correct. However, the main object is that you are plotting two variable types and each axis and representing a third variable by means of the curve(s). In the case of the two charts I posted you are plotting a variable (expression) on each axis and the curves plotted represent geometry (circle, trapezoid, or rectangle) as well as different values of z (side slope of channel). So technically they are nomographs

Another statement about nomographs taken from McGraw-Hill:

Nomograph A graphical relationship between a set of variables that are related by a mathematical equation or law. The fundamental principle involved in the construction of a nomographic or alignment chart consists of representing an equation containing three variables, f(u,v,w) = 0, by means of three scales in such a manner that a straight line cuts the three scales in values of u, v, and w, satisfying the equation. The cutting line is called the isopleth or index line. Numbers may be quickly and easily read from the scales of such a chart even by one unfamiliar with the construction of the chart and the equation involved. The illustration shows such an example. Assume that it is desired to find the value of E when D = 2 and Q = 50. Lay a straightedge through 50 on the Q scale and through 2 on the D scale and read 11.8 at its intersection with the E scale. As another example, it might be desired to know what value or values of D should be used if E and Q are required to be 10 and 60, respectively. A straightedge through E = 10 and Q = 60 cuts the D scale in two points, D = 2.8 and D = 9.4. This is equivalent to finding two positive roots of the cubic equation D3 − 10D2 + 56.25 = 0. It is assumed that g = 32 ft/s2 in this equation.

JR

 

[IlPadrino]

^^^ A nomograph is considered to be a graph consisting of three coplanar curves, each graduated for a different variable so that a straight line cutting all three curves ..

.

I'm going to have to start paying for these lessons!

 

[cantaloup]

I'm going to have to start paying for these lessons!

This is the nomograph for trapezoid channel.

 

[IlPadrino]

This is the nomograph for trapezoid channel.

Now that's a NOMOGRAPH!