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Lecture – 22: River Equilibrium-I
Very good morning all of you today will have very interesting lectures on river equilibriums and which talk about how the river reaches the equilibrium positions that is what we today will discuss. And also I will show a case studies of predicting the river bank erosions in the Brahmaputra river so that the part that we will cover today. And if you look at that mostly we are following this book P.Y. Julian books and it has chapters on river equilibriums. So we are following that book with some of the case studies from Indian rivers conditions.
Now if you look at the next part see if you look at that many of the times it was challenging task to design stable alluvial rivers. What could be the cross section, what could be the slope of the channels so that river remains at stable conditions that means there is not significant order of scouring or the siltation process. So there was a challenging task and that is what we solved it by thorough field observations of the rivers and modifying the river such a way that river can come to an equilibrium stage where there is no much change of the channel is a stables.
Where the particles along these wetted perimeter are not moving it which we talk about in
terms of the stable channels but if you look at this the stable channel part which is is a
concrete line canal but can you make a stable channel of alluvial rivers and the
sedimentations if you look at these rivers can you make a stable channels can you make a the
river at the equilibrium stage.
What do we mean by that the equilibriums of alluvial channels is a balance between the
incoming and outgoing discharge and the sediment loads what is talking about that you say
no scouring and no depositions of sediment passing through the particular river reach. The
concept is now it's a quite challenging and quite interesting to make a river cross section such
a way that the channel bed slope such a way that there will be no scouring and there will be
That means that whatever the sediment in flux is coming from a particular reach upstream
that should be equal to the sediment outflux coming from that. So if you look at that at thereach levels equilibriums examples will come it when the rate of erosions of the outside of
river bank is a rate of the sedimentations on the point bar. So that means we are talking about
the reach scale reach scale river equilibriums where the rate of erosions of the outside of a
river bank that should be equal to rate of sedimentations on the point bar.
Or we can talk about the amount of the sediments is coming from a upstream a particular
reach that should be equal to the amount of sediment out flux from that the downstream rates.
So, that means we can say that reach is at the equilibrium stage but not only that it is not easy
to achieve that or we can make a flow such a way that the all the particles along the wetted
perimeters that means in a bed bank they are not at moving it that is the conditions we do not
look it for alluvial channel cases.
But this is the conditions we look at the for the stone riprap or protection system such a way
that the particles which is on the bed or the bank in a wetted zones are not moving it that is
the channel stability. But besides that we also talk about that if we plot the time versus let be
channel depth h if I locate that there will be erosions and depositions and all but if I talk
about long term average that is there is constant.
So if I take a long term average if I plot the depth versus time and take a long term for
average at a particular stations if that long term average becomes a steady there no trend the
increasing trend and decreasing trend. The significant trend then also I can say the river at
equilibriums. So this is the concept if you look at that is quite challenging that is the reasons
there is 2 way 1 way to try to understand the river mechanics point of view the how how river
can be at the equilibrium stage.
Other ways that go to the field collect the river cross section data collect the velocity the
discharge area of the flow and then try to establish the reaches where the equilibrium stage is
achieved in the last 5-10 years what is that geometric relationship between flow
characteristics and degree of freedoms like in terms of width which in terms of the depth in
terms of the bed slopes.
(Refer Slide Time: 06:49)So that what we will be discussing more and more details. Let us go for next ones which is
the particle stability. The basically we look it like stone riprap or we try to design a channel
such a alluvial channel such a way that there will be no motions of the particles on the bank
as well as on the bed so that is what we are looking at. That whether we can design a channel
such a way that the particle which is on the bed bank that is not in motions not in a motions
not in incipient conditions.
So that means we are trying to look it at what conditions what the configurations of the drag
force the lift force and the uplift forces those force how the equilibrium and how they are
initiating the bed materials to roll it. So now if you look at that let be a bed materials are here
and you have the channels with a Q discharge is coming it it has a side slope ϴ1, ϴo is the
downstream slope of the channels ok.
The channels also have a slope and if you look at that part now if you look at this component
we can see particle path line we can see the stream line components the particles once
detached from the surface which is the path is follows it is that what is the particle path line at
that point because we are talking about the drag forces and the lift forces. That is what it
happens it and we look it what is the stream line components and if I result in the different
angle like ϴ, β, α and λ.
So λ is a deviation of the angle from this slope to the stream lines. So the stream lines having
the angle of deviations from this slope surface. So, if you look at that part and if you look at
this downstream part so you can have the stream lines and you have the the force componentsin terms of F s force. The F s force is a submerged weight of the particles. So let you look at
that if you have a stones you can visualize that stones will have the packing like that so we
are trying to look it that is a stone is a particle for us.
But if you look at the sand; so if you microscopically if you look at the sand compositions
sand can be located like a stacking of the particles. So if you look at that particles either a
stone in a gravel river or if you talk about the sandy river the sand particles this is very finer
particle but they are the compositions could be like that.
Because of that we need to find out submerged weight of the particles which you know it
very well is a difference between the weight and the buoyancy force that is the differentiate
and the small water surface slope in the downstream directions. So you have a channel
parameter like ϴ1 side slope angles ϴo the downstream bed slopes the F L is stands for the
lift force. So you can imagine it the particles you are talking about is just starting the incipient
motion it is a detaching from that and its moving along the particle path lines.
So you have a drag force we have a buoyancy force and the weight of the particles so thats
what again I need to repeat it if you look at that sand particles these are all cohesion less
particles we are not talking about the clay we are not talking about compositions of the silt.
We are talking about big boulders or we are talking about the sand particles. So if you look at
that and we are looking at these particles where it will be stability and these are the force
component if you can result it geometrically if you look at that part.
(Refer Slide Time: 10:58)Now if you look at the next part which is more geometric point of view that you can find out
the ϴ E J functions of ϴo and ϴ1, tan ϴ we can confine it when the angles are the smalls in
most of the times of the alluvial rivers when you design it the downstream slope these slopes
are quite gentle similarly aϴ can be approximated as sin2ϴ1 – sin2ϴo
Basically what I am looking at you want to look at the projections on the plane on the slope
the all this component we are looking at the projections on the bank slope. If you look at that
and you have the lambda is a deviations of the stream line from the downstream directions
positive downward that means how much of angle deviations are there of the stream lines the
The beta is angle of the particles from the direction of steepest descent the deviations of angle
between the particles direction and streamline in case of the rotation the particles will be
rotated. So if you are approximated most of the ϴ, ϴo is very close to the 0 and aϴ can be as
close to the cosϴ1 because ϴo is very very close to 0 as we discussed is the like Brahmaputra
rivers we have the slope gradient 1 is to 10000 scale.
So you can find out what will be the ϴo value for that so you can easily cut bar but no doubt
in case of the hilly regions it can go up to 1 is to 100. So you can also come the ϴo value so it
is generally ϴo as close to the 0 and we can approach means aϴ you will be cos ϴ1 so that is
the approximations we can do it and its quite valid for the river systems when you do not
have a steep slope of the river but that though conditions you should consider all other
(Refer Slide Time: 13:22)Now if you look at particles levels I am looking at how it is stables that means what are the
forces acting on these particles 1 is lift force because the water is flowing through that there
is a drag force and there is a submerged weight of the particles. So if you look at that what
are the force is a lift force you have a drag force FD is a drag force FS is submerged in weight
of the particles.
If is that depending on particles size you will have these force acting like a lift force and drag
force acting components also you will have the components due to the submerged weight in 2
components will have and they will have a if I to take a O is a point the pivotal point where
the particles will be rotate out rotate from that that means particle will be detaching from that.
If that is the conditions, if you look at that and you can find they are the distance of l 1, l 2, l
3 and l 4.
So basically I am representing that you microscopically if you look at these sand
compositions are the stones packing’s and you are looking at a particular stones at what flow
conditions it will be lift out from that it will be retained out from that. So if these are the
stone 1 2 3 and our target stone is the particle p if is that it will rotate from this because these
are the force component will act on these particles.
The lift force the drag force and the submerged weight and they are depending upon particles
dimensions they will have a l 1, l 2, l 3, l 4 distance we are looking microscopically we will
talk about a sand composites. If it is that what I try to look at how much a movement isworking on to rotate this ones 1 is a restoring moments that is what it happens it that the
distance into the force component.
And another is the overturning moments that because of F X F D and F L components as you
can see these force components. So you can find out the movement of rotations 1 is a
restoring another is overturning moment that what we can compute it and you can equate it
that is a positions if this force component is larger than this one then it will be overturned that
is the reasons the stabilizing movement due to the particle weight and there is a lift moment
to destabilize the particles.
You have a lift moment to destabilize. So basically we try to locate factor of safety for
overturning will it have a resist ratio between the resisting moments and moments generating
the motions or overturning that that is what if I substitute I will get it this ones. The basically
I am looking it as f naught should be a much, much lesser than the value of 1. If its value
equal to the 1 there is a chance to have the particle will be detached from that rotate from
When you do design a alluvial channels we try to look at this SFo the that is a factor of safety
for overturning this particle should be lesser than 1 value should be lesser than 1 value.
(Refer Slide Time: 17:22)
If you look it that way and if I somewhat simplified it that for example when fluid at the rest
conditions. So in that case we will not have a drag and lift forces you can see the channels the
flow is at the right conditions that means the the factors of t equal to 1 and your the ϴo, ϴ1will come you will be the equal to angle of repose and if I have a tanϕ in terms of angle of
repose is l by 2 by l 1 then if I just change that equations I will get a factor of safety of
overturning is mathematically we are just manipulating it.
To get η functions of a naught 10 of 5 phi is the angle of repose the β part and we have a η1
part. So if you look at this η1 is called stability number of the particles on the embankment on
the side. So that is what will be equal to this value m plus delta and m and can be defined like
this just we are rising looking it the ratio of the lift to drag moments of the force okay this is
just the simplifications of the previous equations.
To look at that the factor of safety is a functions of phi angle of repose it depends upon your
stability numbers size stability numbers which is a functions of M and N, M and N ratio
representing us lift to drag moments of the force that is the concept we brought it and we try
to look it is how these functions relationship is there.
(Refer Slide Time: 19:26)
If you look at that further we can simplify that as we go for a plane horizontal surface where
you have a ϴo, ϴ1, δ becomes 0 and this is a sub assumption of angle is a 90 degree and you
can simplify the η1 is a function of ηo in the function of M and N α, β and ϴ and ϴo can
divide is a τo is applied shear stress that is what is τo stands for applied shear stress and τ*
stands for the critical shield shear stress.
And basically when you have a fully turbulent flow and the hydraulically rough surface at the
incipient motions the ηo =1 and the critical shear stress the shield critical shear stress is equalto 0.047 that is what we discussed earlier. So we can find out the ηo part you have a
relationship between applied shear stress and the critical shear stress and that is what we
define the critical shear stress in terms of shield critical shear stress. And that is what we can
have very simple equations to find out what will be the η1.
Now if you look it next 1 when you talk about its not the it will be rotate on this the plane
surface on the bank plane surface. It can rotate to perpendicular to that that means we are
looking at particle b which can rotate along these directions perpendicular to inclined surface.
You put that you can again come get it the components here there are the 2 components are
there 1 is the drag force components and another is submerged weight components.
And if you look at that what is the moment at this point again we can put it to the same
conditions are the restoring moment is equal to the yours the overturning moments and that
what if you simplify it will get it tan β or beta = tan-1 will be a function of this. So more
detailed derivations you can get it from reference like Julian book or respective publications
who are not going more details.
But analytically if I consider the particles which are there on the river or the it is bank they
are the particles and we are looking at what force components are there because of the flow
the drag force the lift force and some modulate and what conditions of the layouts are there
that is what is talking about in terms of angles in terms of the path line particle path line in
terms of stream, path lines stream lines.
All we can result it to find out the factor of safety of this particle. If I know the factor of
safety of the particles that means I can know it at what conditions if a factor of safety is more
than 1 no doubt the particles will remove from that the erosions will start it on the bank or the
bed. If a factor of safety is lesser than 1 I can say it it is a stable but significantly if it is a
much, much lesser than 1 then I can say it the bed or bank cases it remains a stable
So we try to look it at the particles levels which you know it the river bed does not have a
uniform bed materials or the bank materials and the particle level of concept to implement at
the rivers is always a questions mark. We can derive a good analytical equation to know itwhat could be a factor of safety for a particle to initiate the motion so in spanned motions that
what we can do it. Looking these safety factors of the particles ok.
Whether it is a greater than 1 its unstable the particles will remove from the bed or bank in
that case it will initiate the river is not equilibrium stage or particle number is less than 1 this
we can achieve the stable but if your factor of safety is much lesser than 1 then we can say
the channels are the stable. But the limitations here that as you know it river banks are
heterogeneous mixed bank similar way the bed is a mix is not uniform sand or the uniform
the gravels are there.
So those conditions we should look it not only that we should talk about more detail about
river bend and the geometric conditions.
(Refer Slide Time: 25:05)
So another interesting concept will let us talk about not the particle level stability talk about
channel stability that means I have a river channels ok. So most of the times it will be
parabolic shape and I have the flow here I this is the flow is moving with the velocity v and
there are the variations of the shear stress the variations of the shear stress happening along
the perimeter of the channels. How can I say the channel is stable?
If I consider a straight channel it is and all the particles on the bank or the bed they are at the
inspired motion they are about to move out from that bank or the bank materials along these
wetted parameters having a weight part of the particles is F S. So that means I consider a wet
particles which is on the bank or the wet particles here on the back on the bed I try to look itwhat conditions it should prevail it that all these things and what could be the shape of these
channels such a way that these 2 particles are remains on the same locations they are not we
just starting the incipient motions.
So that means we are tracking it if it is ϴ1 angle is there and looking it critical shear stress is
tau as c along these things use analogous to critical shear stress corresponding angle of repose
and if I consider the lanes 1953 the critical c stress is equal to F S then phi, phi stands for
angle of repose. And the resultant component on the side slope is tan phi R by F S cos ϴ1 I
can have a ratio between shear stress critical shear stress on the bank on the side and the
relations will be come to a functions of ϴ1 and the phi.
So ϴ1 is angle, so if I look it even if I have a flow depth is h naught the fluid depth is h
naught then I have a the critical shear stress acting on this and applied shear stress on this side
flow. And if I just equate all these equations and do isolating this omega s per omega S value
S stands for here is the bed slope.
(Refer Slide Time: 28:01)
If you look at that parameters and if I make it the simplifications of this.
(Refer Slide Time: 28:13)We are going to get it a differential equations of ideal cross sections geometry that is what
will be tanϴ1 is dh/dy so this is h naught if you look at that we are looking at τo is equal to
applied stress acting on this bank or the bed that is what you equal to the τc that is the
incipient conditions. If that is the conditions you tau c is a functions of omega hs cosϴ1 and
ϴ1can re write it in terms of geometrically tan ϴ1 =-dh/dy.
And if you cancel it you will get these equations and if I solve these equations h 1 is flowed
up center line of the channel the hydraulic radius comes like this. So now if you look it so no
doubt the tau sc on the side surface which is much lesser than the critical shear stress on the
bed. So if that is the conditions if you look at that the here we have the ratio between tau sc
the critical shear stress on the bank divide by the critical shear stress on the bed.
If I consider that the conditions definitely this value should of course this value should be less
than 1 and this is the angles okay the ϴ1 angles and this is the different slope 20 degree 25
degree 30 degree and 40 degree of the phi value angle of repose. So we can get this
relationship with the solutions of these equations. So this is what is indicating it the shape.
The same way for this equilibrium surface will get it this once.
The shape will come it to like this so we can define the shape of the river which is the
equilibrium shape. The shape are all these bed material and the bank materials they are for a
uniform bed size bed and sediment glue conditions we will have a the relationship between
these ones as well as the graphically you can see it how these ratios are varying with angle of
repose and the ϴ1, ϴ1 is a downstream slope.So it is a quite interesting the analytical derivations of the relationship with a side critical
shear stress and the critical shear stress as well as getting a equilibrium shape the channel
shape for uniform bed and the bank materials. If you know it angle of repose and all we can
compute it what could be the channel shapes so most of the times we show the sum of the
channel cross sections the data from Brahmaputra rivers and you can see it, it does not follow
But that means it does not follow these equilibrium constants concept as we are looking at
more details and we will do discuss more details in about Brahmaputra rivers and why do not
get a equilibrium cross sections. But smaller rivers we generally get a single channels close to
this resembled to this equilibrium ideal cross section geometry of a river.
(Refer Slide Time: 31:56)
Next very interesting thing is regime relationship. Way back we used to have a constructions
of great Ganga canal or Indus canals in Pakistan it was a quite challenging task for the
engineers that how to design a slope of the channels and the cross section such a way that
unlined canals in alluvial plain of Ganga and also alluvial from Indus such a way that we
should get a stable channel cross sections.
That means they are not a significant amount of erosions or depositions are happening it they
are not significant amount of changing of the bed width or the change of for a particular
constant discharge. So what they try to do it that is that is what is the contributions of Indianengineers by the Lacies, Kennedy Way back in 1929's before our country independence we
used to have a to be a leader on designing this alluvial channels.
In terms of getting the field data try to establish the relationship between for a river or stable
rivers stable canal reach what is the relationship between geometry of the river with the flow
discharge. Because most of the canals we have the constant discharge we do not make a
transient flow in that conditions what is the geometry it shapes the rivers. If we follow that
geometry of the rivers that means we try to understand from the river perspectives or the
channel perspective that how the river shaping it.
And if we try to find out the same geometry of the rivers in geometry and the slope and that
empirical relationship will give a design relations. The Lacey equations in 1929 and Blench
in 1969's which is very widely used Lacey’s equations still now also we have been use it for
scouring computations we have been using for designing the bridge piers and we are also we
have been using steel.
This Lacey’s equations is similar to the meanings equations its empirical natures since it is a
field data collected data based a plus a data mining or data science today we talk about. The
combinations of both after extensive data collections and the data mining they try to do it
establish the relationship which hold goatee for a river or the reach at the equilibrium stage.
The relationship what they got it is a functions of controlling part and the dependent part.
So if you look at the river as you say that river is author of its geometry the author of its cross
sections and author of its slope. If I talk about that river shapes it is the path, river shapes it is
cross sections. If I look at that concept and try to look at that there are the dependent
variables like the velocity area if I talk about the velocity and if you talk about the area and if
you talk about the relationship with the slopes these are empirical equations are derived for
wetted parameters and all what they do it because very easy to know the discharge.
And also its depends upon another particles is a particle dimension DMN is the particle
diameters we can consider D50 values can have a silt factors. So the empirical they introduce
a factor which is called Lacey silt factors and they establish the relationship of velocity area
and hydraulic radius and the parameters wetted perimeters and the slope these are allempirical data collecting the field data do a data mining develop a correlations between the
And here is independent variable is the discharge and the D50 value that is what they
introduce the Lacey shield factors this is very well known the Lacey’s equations and the
velocity area radius and speed and still we use for designing a online canal scouring depth of
plot plane width all while we do it based on this Lacey’s equations which is way back in 1929
still it has a lot of relevance to a to hydraulic engineers or the sediment engineers to know it
how they have divide these empirical equations which is still has a lot of utility even if you
have a lot of understanding more.
So this is called regime relationship and that is you have to try to understand it and try to
make it how it is very good equations has come up in terms of Q. And so this is now if you
look it that is what is representing you can make it as equivalent to rectangular cross sections
with a plot plane and river and you can know it whether the river is equilibrium conditions if
it follows these regime relationship.
(Refer Slide Time: 38:12)
Now if you look it very interesting part if you look at this large gorges in Brahmaputra rivers
still it is a inaccessible to these river bends which is there in part of China and India border
and 1 of the largest gorge is there and huge energy is dissipated in these 3 gorge still we have
a lot of questions mark on these gorges what it happens to that. So let me come back to the
river bend a simple river bends if you make it.We can understand it that will have a surface stream lines will have the surface stream lines
like this and you will have a the near bed stream lines very interesting that you will have a
suppose surface stream lines which are attacking this outer bank you have a near bed stream
line which scours this bed materials and bring it to the inner bend so that is the reasons point
bar formations happens it. So the you have this outer bank this surface you can look it and
there is angle there is a angle of deviations of nearby stream lines.
So how much of angle of deviations are happening it that was also play the major roles. Now
if you look it if I have a river bend ok and if I consider river bank at the equilibrium stage
what do you mean by that? That the force components are some of the net force acting on this
control volume should be equal to the 0 that is what the equilibrium conditions. That means
we look at that centrifugal force per unit mass that is the accelerations is equal to balance by
this unbalance the pressure force components.
And also the unbalance the force due to the shear stress components. So ready so So stand for
here radial water surface slope r is a radius of curvature z is in vertical directions. See if you
look at that what we are looking at centrifugal force per unit mass that is the accelerations
this is the unbalanced pressure force because there will be super elevations per unit mass that
is what also the shear force per unit mass that is what we have done it.
For equilibrium condition that is the net force acting on this control volumes is equal to 0
here as the water is rotated and you will have a centrifugal forces per unit mass you look at
the pressure force difference what is comes it and the shear force per unit mass that is what
we equate here.
(Refer Slide Time: 41:28)Now if you look at the figure levels that you have the river like this and you have the slope of
the free-board this is the inner bank this is the outer bank you have a W is a width r is a radius
and you have a z and y directions tau r the subscript r is a transversal shear stress. So if you
look at that g S r can have a this directions if you look it and v square by r which is a
centrifugal forces components that is what will be net will come like this you can see that
there will be negative and the positive part net force acting that.
So transverse boundary shear stress we can we can have a radial sales force part and we can
average flow and all so we can make a non dimensionally try to look this part but I try to
convince you that you take a control volumes and try to look it what are the force
components are there and that the force components we are looking at how it varies along the
depth that is what is the showing it the net components of the shear stress component how its
acting it that is what is showing it.