Flow in Open Channels

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Do curso por Georgia Institute of Technology

Fundamentos da Engenharia - Revisão

215 classificações

Georgia Institute of Technology

Fundamentos da Engenharia - Revisão

215 classificações

The purpose of this course is to review the material covered in the Fundamentals of Engineering (FE) exam to enable the student to pass it. It will be presented in modules corresponding to the FE topics, particularly those in Civil and Mechanical Engineering. Each module will review main concepts, illustrate them with examples, and provide extensive practice problems.

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Hydraulics and Hydrologic Systems

This module applies basic principles of fluid mechanics to practical problems in hydraulics, hydrology, and groundwater flow. We first discuss the generalized and one-dimensional momentum theorem and apply it to various typical problems. Flow in pipes and non-circular conduits is discussed beginning with the Bernoulli equation accounting for energy losses and gains. Calculation of head loss due to friction and minor losses due to valves and other accoutrements are presented. Friction losses are calculated for laminar Poiseuille flow and turbulent flow using the Moody chart; examples include computation of pressure drop in laminar pipe flow and turbulent water flow. Methods to calculate flow in pipe networks consisting of multiple connecting pipes and other fittings is then discussed with examples for parallel pipes. Pipes and turbines are then discussed along with their basic equations and definitions. Characteristic curves, especially of centrifugal pumps, are presented and it is shown how to match a pump to a system head.Flow in open channels are discussed including classification of flow types and prediction of uniform flow by the Manning equation. The use of specific energy concepts to solve gradually varying flows, and the importance of the Froude number and sub and supercritical flows are presented. Predictions of hydraulic jumps and flow over weirs are given.Hydrological principles include predictions of surface runoff by the curve number method and peak runoff by the rational formula. Groundwater principles include Darcy’s law for flow through porous media and prediction of drawdown by wells in confined and unconfined aquifers by the Dupuit and Thiem equations.In all cases, basic ideas and equations are presented along with sample problems that illustrate the major ideas and provide practice on expected exam questions.Time: Approximately 3 hours | Difficulty Level: Medium

Momentum Theorem9:26

Momentum Theorem Continued15:26

Flowin Pipesand Conducts12:44

Flow in Pipes10:56

Flow in Pipes Continued9:57

Pumps and Turbines6:51

Pumps and Turbines Continued11:38

Flow in Open Channels10:04

Flow in Open Channels Continued13:56

Groundwater5:56

Groundwater Continued7:20

Conheça os instrutores

Dr. Philip Roberts
Professor
School of Civil and Environmental Engineering

The next topic we'll cover in hydraulics is flow in open channels.

First of all, we'll look at different flow types, and then the equations for

predicting uniform flow, the Manning equation, and then do some examples.

Flow in open channels refers to any flow with a free surface.

And in hydraulics, of course, we're mostly looking at, flow of water.

There are different types of flow types that can occur.

For example, if I consider this flow from this, reservoir here under a slew skate,

we can have initially a region close to the slew skate here.

Where the depth and velocity are varying quite quickly, and

the stream lines are not straight and parallel.

This type of flow we call a Rapidly Varying Flow.

Further down-stream we get to a region where the friction and

gravity forces are in balance.

And at that point the velocity and depth no longer change with distance.

And that type of a flow we call a Uniform Flow.

We may have another type of rapidly varying flow called a Hydraulic Jump.

Wherein the water surface suddenly raises and flows down-stream at a lower velocity.

This can be followed by another region of uniform flow and

then possibly flow over a sluice gate.

Which is again a type of rapidly varying flow.

We may have flow along a horizontal boundary,

in which case the flow will always be gradually varying.

Following a change in slope, we can have another period of rapidly varying flow

until eventually again, far down-stream, we get to uniform flow again.

For this segment, the main interest will be in predicting uniform flow.

But we will also look somewhat at hydraulic jumps and

flow over slew skates or weirs.

The first thing we want to arrive at is the equation for uniform flow.

So, let's suppose we have a flow down a slope like this, and,

because it's uniform flow, the depth and the velocity are constant down the slope.

In cross section we might have some arbitrary shape.

And around here we have a perimeter, P, called the wetted

perimeter which is in contact with the water, and a free surface.

And I'll suppose that the cross sectional area here is A.

Now in principle we could

the Darcy-Weisbach equation that we used for round pipes.

And the Darcy-Weisbach equation was Hf,

the head loss due to friction is F L over D V squared over 2 g.

To apply this to non-circular conduits, for example an open-channel flow,

we replace the diameter by the hydraulic diameter, defined as 4A over P.

So, in principle, we could solve our problems in this way.

However, we don't usually do it that way for open-channel flows.

More commonly, we use the Manning equation given here, which says that the velocity

is equal to K over N, R H to the two thirds, S to the one half power.

And this is the corresponding section from the reference handbook.

Where in this equation N is a constant,

known as the Manning coefficient or Manning's N.

Rh is the hydraulic radius, divided by the cross section area,

divided by the wetted perimeter.

In other words, a quarter of the hydraulic diameter.

S is the channel slope, and

K is a factor which is different depending on what units we're working in.

The Mannings equation is unfortunate in fluid mechanics,

in that it's not dimensionally homogeneous.

And it's a different equation, depending on what system of units you're,

you're working in.

In metric units or SI units, the value of K is 1.

But, in US customary system of units,

the value of the constant is one point four eight six.

Here is the equation again in terms of velocity.

And because the volume flow rate Q is equal to velocity times area,

we can also write the Manning equation in terms of volume flow rate as shown there.

And N in this equation, Manning's N, is a type of roughness parameter.

And the reason that Manning's equation is useful often

is that there is a lot of empirical evidence built up over the years

on what is a suitable value of N to use for different situations.

For example, for natural channels here a typical value of Manning's N for

clean and straight channels is 0.03.

For sluggish or major rivers with a lot of vegetation and

meandering, it increases to about 0.035.

For very, large roughness elements, for example heavy brush or

trees such as you would encounter in a flood plane,

Manning's N increases substantially to 0.075 or as much as 0.15.

On the other hand, for very smooth channels, for example,

artificial channels with glass lining, it's 0.01.

For steel, etc., it increases to 0.015.

So, N is a type of a roughness coefficient.

To illustrate that, let's do some examples.

So in this question we have a rectangular cross section channel four meters wide,

carrying a discharge of two cubic meters per second,

at a uniform depth of 1.5 meters.

Manning's N is 0.012.

And the question is the slope in percent of the channel bed

is most nearly which of these alternatives?

So here's the solution.

We begin with our Manning's equation in terms of volume flow rate, and

in this case, because we're working in SI or metric units, the value of K is one.

So the equation becomes as shown here.

In this case, we're trying to calculate the slope, so recasting that in-,

equation in terms of the slope, we have that expression.

And to calculate the hydraulic, radius, first of all,

we see that the area is the width times the depth, which is six square meters.

The hydraulic radius, which is A over P, is equal to area, W times D,

and the wetted perimeter here is the width of the bottom plus two channel depths.

W plus 2D.

So substituting in the numbers we get the hydraulic radius is 0.86 meters.

And now we're ready to compute the bed slope is equal to, this expression,

which is equal to 1.96 times ten to the minus five.

Or, expressing that in terms of a percentage,

0.00196%, which we note is a very gradual slope.

It's a drop of only one meter in 51 kilometers.

The closest answer is B, 0.002%.

The next question, we have a circular sewer, which is two meters in diameter.

When the pipe flows full, the velocity is 0.8 meters per second.

Assuming Manning's N to be constant,

the flow velocity when it's flowing half full is most nearly which of these?

So again we start with our basic Manning's equation in terms of velocity this time.

And again K is equal to one, because we're working in metric units.

So we get that.

Now when the pipe is flowing full,

in other words the pipe is completely filled with water here,

the hydraulic radius is the area, which is pi by four times the diameter squared,

divided by the wetted perimeter, which is the length of the circumference here,

which is pi D, which is equal to D over four.

When it's flowing half full, again, the hydraulic radius is A over P,

but now the area on top is 1/2 pi by 4 D squared.

And the wetted perimeter is 1/2 pi D.

So the factor one half cancels.

So the hydraulic radius is exactly the same.

It's also D over four.

Looking at this equation, then, we see that N is the same,

the hydraulic radius is the same, the slope is the same.

In other words, the velocity is the same.

So the answer is C, 0.8 meters per second.

A similar question, but in, USCS units.

In this case, the diameter of the sewer is two feet, and for

full flow, the flow velocity is one foot per second.

If Manning's N is constant, the flow rate when the sewer is flowing half full

is most nearly which of these three, four alternatives?

So again, we start off with Manning's equation.

But in this case, and this is the most common mistake to make here,

because we're working in USCS units, we have the factor 1.49 here.

So the pers-, first procedure is exactly the same as for the last problem.

The hydraulic radius when it's flowing full is D over 4.

And when it's flowing half full it's also D over 4.

But the area now is reduced to pi by eight D squared.

So, for the same reasoning in, in the last case,

the velocity when it's flowing half full is the same as when it's full.

It's one feet per second.

However, the volume of flow rate is not the same because the area is different.

The volume flow rate is velocity times area.

Velocity is one foot per second.

The area is pi by eight, times the diameter squared,

which gives me 1.57 cubic feet per second.

The closest answer is C.

This finishes our discussion of the Manning equation.

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