# Uniform Flow Computations

Good morning to everyone, welcome back to

our lecture series on advanced hydraulics. So, for last few weeks, we are dealing with

the module on uniform flows. And in the last class, we have started the portion on uniform

flow computations. We had elaborated various methods on that we were discussing on the

various methodologies of for computing uniform flow and all. First let us start with the

quiz for the last class. In the last class, whatever we have taught based on that we will

take a brief quiz. After that we will just go through that briefly again those portions,

and today we will start the next portion of the module. So, please begin your quiz, you

will be given five minutes to answer the quiz. First question for the quiz is what is Manning’s

formula? So, will you please write down what is manning’s formula, it is a very easy

one. The second question for you is enumerate five factors that may affect the Manning’s

roughness coefficient. You know the Manning’s formula in that there is a coefficient called

Manning’s roughness coefficient. So, what are the five factors some just enumerate some

five factors; there are many factors that affect roughness manning’s roughness coefficient.

You enumerate any five factors that may affect the Manning’s roughness coefficient. I hope

you have answered that. The next one is, what is the depth of water

in uniform flow called? What is the depth of water in uniform flow called? It is just

a one word answer; it should these answer should be given in one or two words. Now the

last question for you is, what is conveyance factor for uniform flow? What is conveyance

factor for uniform flow? I hope you have answered that. So, this quiz was a just a brief quiz,

just to test your updation of the course, how you are going in proper time and proper

methodology. So, we hope that you have answered all the questions. The solutions for this quiz is Manning’s

formula, I will just write it. So, the solutions first one, the average velocity in any cross

section of the channel for uniform flow, it is nothing but 1 by n R to the power of 2

by 3 S naught to the power of half; where n is called manning’s coefficient – roughness

coefficient, r is your hydraulic radius, s naught is your bed slope. The solution for your second question, your

second question was enumerate five factors that may affect the manning’s roughness

coefficient. So, you can enumerate any one of them. You can tell that say surface roughness,

for example, you can tell channel irregularity. You can also tell

channel alignment, silting, scouring, off

course one important point, vegetation. Vegetation also considerably affects your roughness coefficient.

If any obstruction is there in the channel – seasonal changes. So, you can enumerate

any one of any five of this or even if you know more than these things that can also

be that is also most welcome from your side. So, what was your third question? Your third

question is what is the depth of water in uniform flow called? I will just mention it

here itself; it is called normal depth. No need to further elaborate that expression.

Then the next question is what is conveyance factor for uniform flow? If you recall the

discharge the uniform flow discharge, it was given as a function of normal depth y n and

manning’s roughness this thing into s naught to the power of half. This function of normal

depth y n and manning’s roughness coefficient n was subsequently given as conveyance factor

k into s naught to the power of half. So, your conveyance factor k is a function of,

your conveyance factor k is a function of normal depth y n and roughness coefficient

n. It is given as 1 by n a r to the power of 2 by 3. Then let us start today’s lecture now. So,

today we will be continuing with uniform flow computations. So, in the last class, we had

studied on manning’s equation. We have studied normal depth conveyance factors, section factor;

we have also briefly discussed that what are the procedures

to compute normal depth. So, in that we mentioned

that you can use design charts, you can imply trial and error method, you can use numerical

methods. So, how to develop your design charts, this was explained in the last class. We had,

in fact, developed some design charts for trapezoidal channel or how the table how to

compute design charts and all that we have done it, we have derived. In fact, derived

it for trapezoidal channel for any sides slope one is to v, you recall them. Any slope one

is to v, and v is equal to one, v when we substitute that the corresponding table was

also developed. You can see how the trial and error method

is also employed to compute normal depth, and how to use numerical methods also to compute

normal depth. We will just see it in the following portion now. So, the trial and error method.

So, you recall your manning situation for discharge – v 1 by n r to the power of 2 by

3 s naught to the power of half or your discharge Q – this is one by n AR to the power of 2

by 3 s naught to the power of half. In this relationship, if you remember then you had

obtained the following relationship, AR to the power of 2 by 3, this was clubbed together

this is nothing but, n times the discharge Q by s naught to the power of half. You had

obtained such a relationship in the last class, subsequently we suggested this quantity as

a section factor and all if you recall them. Now, the point here is to use your trial and

error method, in the left hand side, this is your left hand side, this is your right

hand side, for a given channel usually n is a given quantity; q is also specified for

you, bed slope is also given, or it is a channel property. You are now requested to find the

normal depth y n, that is the objective here, that is you are using trial and error method

to compute your normal depth. So, how will you compute normal depth using trial and error

procedure? Again just as an illustration, let us go with

the trapezoidal channel; you are following the same norms as we are doing in our earlier

classes. The depth of uniform flow is called normal depth, the bottom width is B 0, side

slope 1 is to b. If you recall the area of cross section for such a trapezoidal channel

is y n into B 0 plus small b into y n, your wetted perimeter p is equal to B 0 plus 2

y n into root of 1 plus b square. Therefore, your hydraulics radius R is equal to a by

p

y n y B 0 plus twice y n into root of 1 plus b square.

Therefore, in your relationship A R to the power of 2 by 3 is equal to n q by s naught

to the power of half; substitute the quantities whatever relationship here with respect to

y n, you substitute it here. Now what about the r h s term your r h s terms here all the

quantities are known, all the quantities here are known to you. So, you can just specify

it as a known quantity J fine. So, you we you were now going to do your equating A R

to the power of 2 by 3 is equal to a known quantity J. Further proceed, substitute the value of A

and R, you will see that you are getting the following relationship y n into B 0 plus b

y n square b 0 plus 2 y n into root of 1 plus b square whole to the power of 2 by 3. This

is equal to a known quantity J. You can do one thing you can just bring all the terms

here. So, B 0 plus b into y n whole to the power of 5 by 3 into y n whole to the power

of 5 by 3 by B 0 plus twice y n root of 1 plus b square whole to the power of 2 by 3.

This is equal to a known quantity j. So, once you have such a relationship, now you are

having such a relationship. Now substitute for various value of y, you

try with various values of y n, you see which value of, which value of y n satisfy your

equation one, satisfies equation one. You have to find that which value of y n satisfies

that equation one; you have to select that this is called the trial and error method.

So, you have to try it for several values of y, may be some of you, if you are lucky

you may get it in two or three trials itself. Some of you may have to go for some nearly

hundred trials then only they may arrive at the solution, depends on luck also. And also

through intuition based on your problem and all, what could be the initial depth means

or guess value, how you try initial guess and all how you do that depends on your intuition

as well. So, you can try it many problems for various this is only for a trapezoidal

cross section. You can try it for various. So, similar to, similar to equation one for

other cross sections, you can develop corresponding equations. Now it is off to you, say whether

it is for rectangular, triangular, even circular, it is up to you to do work it out and just

find it. What are those corresponding equation try the trial, I mean use the trial and error

method to find the normal depth. Now let us go with the next method. So, now,

we next method, we are going to do is the how to apply numerical methods, to obtain

the normal depth for uniform flow. Here, if you recall the equation one previously develop

that is B 0 plus b y n whole to the power of five by three y n whole to the power of

five by three by B 0 plus two y n one plus b square. This equation, it was of course

developed for the trapezoidal channel, you can obtain similar equations for other cross

sectional channels also. In this equation, this is a non-linear

in y n.

So to solve this equation, you have to use non-linear solutions techniques. This in the

numerical methods, there are methods or in the methods or techniques to solve non-linear

equations. So, you have to employ them here to solve and get the normal depth. Some of

them, I will just briefly mentioned the you might of heard about Newton-Raphson method,

conjugate method – by conjugate method, sequence method, there are various method to solve

non-linear equations in the numerical methods portion. So I am not going to describe all

of them, just briefly we will see through the Newton-Raphson method. So you have equation, I can now obtain the

equation as from this B 0 b y n, again I am iterating here, iterating that this equation

was developed for the trapezoidal channel for other cross section it is quite different.

So in this equation, L H S is unknown; R H S is the known quantity. So the Newton Raphson

iteration or Newton Raphson method, it suggest that if you can develop a function of y n

and if that function is differentiable, so you can obtain the first differentiation f

dash y n. You can iterate and obtain the value of y as y n new is equal to y n old minus

f of y n by f dash y n. This is the standard Newton- Raphson iteration or Newton Raphson

method to find non-linear solution. So here what does y n new and y n old mean? This is

the previous value or guessed value; y n new is it is the modified value or improved value.

So the steps of that if you have any non-linear equation in y n, you can give an initial guess

of y n some value. And start using this relationship you first obtain the function of y n which

is also possible to be differentiated then using your initial guess or old guess, you

can get a new value of y n or improved value of y n using the given equation mentioned

here, so that we can employ here. In our case for open channel for a computation

of uniform flow, we can suggest that your f of y n is equal to B 0 plus b y n whole

to the power of five by three y n to the power of five by three minus J. Let us consider

that your function of y n is equal to the is in the following form. In this following

form now, if you obtain this function, this is also quite possible to be differentiated.

This was for trapezoidal for a general case, you can suggest f of y n for any cross sectional

method you can suggest f of y n is equal to A R to the power of two by three, minus J.

And for exact value of y n, this relationship A R two to the power of three minus J should

be equal to zero. Now your initial guess, y n initial, it may not equal to be your y

n actual. In that case, then f of y n initial not equal

to zero. So, this is the principle behind that. Suppose if you are guessing initially

and if you are getting a function f of y n then that suggest that for the exact value

of y n, if your initial guess was an exact value of y n then f of y n would have directly

yield you zero. Then there is no need of further solving the equation. If your initial guess

is not the actual normal depth y n, what you can do is that you are now going to evaluate

the function f of y n, you are now going to evaluate this function f of y n, subsequently

you are going to evaluate f dash y n also. You are going to evaluate f dash y n. Now what is f dash y n? This is d f by d y

n, so as taken from Anil Chaudhry, two thousand eight on his book in flow through open channels.

So, d f by d y n, you can give it as d by d y of A to the power of five by three P to

the power of two by three minus J. So this is nothing but five by three A to the power

of two by three P to the power of two by three d A by d y n minus two by three P to the power

of minus five by three A five by three d P by d y n. So you rearrange the terms, recall

that d A by d y n in any cross section of the channel, any channel we had suggested

that the top width T is nothing but equal to d A by d y. So therefore, the uniform flow

d A by d y n will give you the top width of that uniform flow in the channel. If you recall

them, we had discussed these things in the earlier classes.

So, just substitute those quantities here, if you substitute them appropriately you will

get the corresponding relationship for the trapezoidal channel. Your f dash y n is nothing

but equal to one by three R to the power of two by three into five times the top width

minus twice R d p by d y n. And you also know that f of y n is equal to A R to the power

of two by three minus J. So what you can do is the f of y n is equal to A R to the power

of two by three minus J. So you have now two function f of y n and its corresponding differentiated

function f dash y n. Start with initial guess value for y n that

is y n 0 is equal to some value, some value you can specify, we start with the initial

guess value. Using your, so using your y n value what you have to do is evaluate f of

y n, using your initial guess; evaluate f dash y n using your initial guess. Now your

improved value for your normal depth y n, so I am just giving it as super fixed one

the improved value which is nothing but a old value minus f of y n that has been obtained

using your initial guess and f dash y n that has been obtained using your initial guess.

So this will give you an improved value for your normal depth y n, so this is given as

y n one. Once you obtain y n one, you can check whether f of y n one whether it is equal

to zero. You can check that. If not again go for the

next iteration, that is in the next iteration y n two is equal y n one minus f of y n. So

again check if your y n two satisfies that condition f of y n or using this y value,

if it satisfies that if we get equated to 0 then find the that is the solution. If not,

then go again for the next iteration y n three. Like this any general iteration can be given

as i minus one minus f of y n using the i minus one by f dash, so this you go on. So

you can stop your iterations once y n converges. What do you mean by converging? Convergence

means, you can put some convergence criteria, such that between two iterations y n in the

ith iteration minus y n in the i minus one eth iteration by y n in the i minus one eth

iteration mod of this thing if this difference is some less than some tolerance value.

Tolerance, you can specify according to your the some of them some people may desired one

into ten to the power minus three, one into ten to the power of minus four, one into ten

to the power of minus five, what are we according to the requirement, you can specify any convergence

criteria. Use them, and see whether your y n is getting converge. Once it converges,

you can use that value of y as the normal depth, so that is the method of using numerical

methods to compute normal depth. If you recall in the critical flow, in critical

flow computation we had we had use the concept called

hydraulic exponent, so the corresponding section

factor for critical flow, please note that this section factor is for critical flow.

This was given as zeta square is equal to some coefficient C into y to the power of

M, where y is your critical depth; M was given as hydraulic exponent for critical flow. As

similar analogous is there for uniform flow computations also. In uniform flow, you have already seen section

factor, conveyance factor. So in uniform flow also, the conveyance factor now can be given

as K square is equal to some coefficient C into the normal depth raise to and exponent

capital N. So this capital N, this is called the hydraulic exponent, it is called the hydraulic

exponent for uniform flow. So these you can use, why this is been used, it gives some

certain characteristic of the channel. You can, it will be of a if one know the hydraulic

exponent of the particular channel section , it will be quite useful to compute the normal

depth. There is no need to further go and measure

the depth or no need to go and measure the various other features; using simple discharge

and roughness coefficient, you will be able to evaluate the normal depth. So now the hydraulic

exponent is also use, therefore hydraulic exponent is also used in computation of uniform

flow. How we are arrival that, so in this thing I hope you every one know the K is your

conveyance factor, N is your hydraulic exponent for uniform flow. In this case, I can just

derive the following quantity. K square is equal to C into y to the power

of capital N; taking logarithm both sides, this is twice log k is equal to log c plus

N log of normal depth. If you differentiate the quantity

with respect to your normal depth, this is

nothing but N by twice y n. Just keep this as equation number two.

Manning’s equation, from the Manning’s equation, you remember the conveyance factor

K is equal to one by n A R to the power of two by three. Use the logarithm here. Differentiate

that

differentiating this thing, you will get d by d y of log K is equal to this quantity,

differentiating this quantity in zero. So you will see that this is nothing but one

by A d A by d y n plus twice by three R d R by d y n. So you know the quantity d R by

d y n, because R is equal to A by P, this is we are quite aware. So therefore, your

d R by d y n term this is nothing but equal to one by P d A by d y n minus A by P square

d p by d y n. Again, if you recall that d A by d y n is equal to top width T, use the

following relationship. Go ahead, your d by d y n of log K is now

equal to T by A plus two by three P by A into T by P minus A by P square d P by d y n. Or

this is equal to one by thrice area into 5 times that the top width minus 2 R d P by

d y n, from this becomes your equation three. Comparing equations two and three, you will

get

the hydraulic exponent for uniform flow N is equal to twice y n by 3 A into 5 times

T minus twice R d P by d y n. Based on say, based on cross section, whether it is trapezoidal,

whether it is rectangular, triangular, parabolic, circular what are the you can identified as

you can find, you can find N for each type of cross section. Now for example, a trapezoidal channel with

bottom width B zero, side slope one is to be normal depth of flow y n. What can you

expect now? In this thing, again your area of cross section, I can write it in the following

form one plus b into y n by B naught. R is equal to one plus b into y n by B naught whole

thing into y n by one plus two y n by B naught root of one plus b square. Top width T is

equal to B naught into one plus two b y n by B naught, is not it? This quantities are

already aware, you can evaluate it on your own also, using these things on substituting

them in the equation for N hydraulic exponent N, I am not going to do it for here, you can

do it as homework. You will see that N becomes ten by three into

one plus twice b y n by B naught by one plus b into y n by B naught minus eight by three

into root of one plus b square into y n by B naught one plus twice root of one plus b

square into y n by B naught. Like this you will get expression for N for trapezoidal,

channel. So why I wrote N in such a form is that, you can have a relationship of N versus

non-dimensional depth y n by B naught. You can plot them, your plots may be, you know

how to plot by this time now. So these figures, you see it may range from

two to five point five and your N value may be something it is going like this for various

cross sections. So if you have any particular value, say if you happened

to observe N value for any particular channel, say it is here then you can just check the

corresponding depth of flow, you can interpret that and from that you can get your normal

depth of flow. Similarly, if it is here corresponding thing, you can easily interpret them. So it

is up to you to determine that, so that is you can use the normal, you can use your hydraulic

exponent for uniform flow, you can compare it with the non dimensional normal depth of

flow. Use those graphs and interpret the normal depth of flow. So next we are going to deal with is what

happens to channel flow or uniform flow if composite properties are there. If composite

properties are there, what happens to your channel flow. First one, if your channel section

is having composite roughness, if it is having composite roughness what happens to your uniform

flow. See it can be, say for example, in the trapezoidal channel, the sidewalls may be

of one particular material, the bottom bed may be of some material, this may be of another

material, still the uniform flow is there in the channel. How will you compute uniform

flow? It is quite difficult or it is deviates, if you then take the corresponding areas allotted

to this perimeter and try to evaluate it independently to be quite different fondly to get your normal

flow, normal or normal depth. So how will you evaluate the uniform flow

in such situation? This is different, this is different, this is different, so you can

just suggest now

that the wetted perimeter is of different material; or for example, another simple case

is in a rectangular channel the sidewalls are made up of glass and the bottom is made

up of wood, what happens to your uniform flow. How will you compute them? So in these situation,

you require the concept called equivalent roughness coefficient. So, we can use the

same equations to compute the uniform flow, but using your equivalent roughness coefficient.

So how will you evaluate equivalent roughness coefficient? What is the procedure? You may or may not divide the areas wherever

the properties are changing, corresponding perimeters say cross sectional area divided into sub areas,

into N sub areas, each having wetted perimeter P i, its roughness coefficient n i. Now, we

are assuming that here the channel cross section it is not that much, the roughness things

are the roughness is the thing that is getting differed here. As so we are assuming that

the velocity the average velocity in each section, each sub area is same as that of

average velocity for entire flow, if you assume that. So that means that say v one bar is

equal to v two bar is equal to v three bar equal to v i bar equal to v n bar. And these

are all equal to the entire average velocity of the cross section of the channel. Then

you can evaluate the equivalent roughness coefficient in the following form. This is one particular formula, i is equal

to one to N p i n i power of three by two by i equal to one to N p I whole to the power

of two by three. So students please note that the capital N used here is not your hydraulic

exponent, it is just to show the summation. The total number of sub areas divided in the

channel cross section, this is just to denote that. We can also evaluate equivalent roughness

coefficient, I assuming that the total force resisting the flow in the channel cross section,

it is equal to the summation of the forces, individual forces that are opposing the flowing

individual sub areas that can also be used. So you can use that method, you can give the

equivalent roughness coefficient n e is equal to i is equal to one to N p i n i square p

i whole to the power of half. This is one method. So was in some other literature you may see,

some scientist they might have used that the total discharge in the channel is equal to

the summation of the discharges in individual areas. If that concept is used, then your

roughness equivalent coefficient n e can be computed as P R to the power of five by three,

where P is the total wetted perimeter of the cross section; R is the hydraulic radius for

the entire cross section of the channel then i is equal to one to N p i R i to the power

of five by three by n i. Like this you can evaluate it. So you can use this particular

formula, once get your equivalent roughness coefficient, you can use it to compute your

average velocity R to the power of two by three S naught to the power of half. You can

also use it to compute your discharge, A R to the power of two by three S naught to the

power of half. So we will conclude here, so we have discussed

on the how to compute the composite of effective roughness coefficient or you can tell it as

the equivalent coefficient of roughness. To compute the roughness coefficient, if there

are different surfaces or there are different surfaces in the channel cross section. You

can use them to evaluate the velocity and discharge. Next time, conclude the portion

here, next time we will be dealing with how to evaluate uniform flow, if there is difference

in cross sectional areas, if shape of the area and all if it is quite differ how to

evaluate the uniform flow. Thank you.

## One Reply to “Uniform Flow Computations”

hello sir,….. i appreciated on you