Course on Transport Phenomena

Professor Sunando DasGupta

Department of Chemical Engineering
Indian Institute of Technology Kharagpur
Module No 2
Lecture 10
Equations for Isothermal Systems (Contd.)

This class we are going to talk about equation of motion. So what is equation of motion and how
it can be derived? The derivation itself is not that important because ultimately we are going to
use the various forms of equation of motion, the one which is relevant to the system, to the
geometry. We are trying to find out what’s the velocity distribution.

But for any system, if one wants to write the equation of motion essentially what we are doing
for a fluidic system is we are writing the equation of Newton’s 2nd law of motion for an open
system in which the fluid is allowed to enter and leave the system. So there is some momentum
which is being added to the system due to the motion of the fluid and there are two ways by
which due to the motion of the fluid, momentum can come into the system which we have seen
before. One is the convective momentum which is due to the flow of the fluid. So some amount
of mass of fluid per unit time is coming, crossing the surface area. The amount of mass which
crosses the surface area has a velocity at that point. So mass flow rate, mass × velocity would
give you amount of momentum which comes in due to the actual motion, crossing the interface
which is nothing but the convective flow of momentum into the control volume.

There are certain cases in which let’s say the fluid is moving in upward direction (y direction)
along a vertical plane which is the control surface and there is a gradient in velocity between two
vertical planes. So y momentum gets transported in the x direction (horizontal direction) due to
viscosity which is going to manifest itself as shear stress on the surface. So the y momentum
getting transported in the x direction must also be taken into account as a source of momentum
coming in to the control volume. Since its taking place in a direction perpendicular to that of the
motion and since the principle reason by which this kind of momentum transfer takes place is
molecular in nature, it’s also known as the molecular transport of momentum or conductive
transport of momentum. So we have two different momentum due to the flow of the fluid, one is
the convective transport of momentum which is due to velocity and the second is conductive
transport of momentum which is due to the velocity gradient. You remember Newton’s law of
viscosity in which the viscous transport is expressed by velocity gradient and not by velocity
itself. So convective transport is due to velocity, conductive is due to velocity gradient. So these
are different ways by which net rate of momentum coming into the control volume.

This control volume can also be acted upon by different forces. So, all the forces which are
acting on the fluid inside the control volume should also be taken into account in the difference
equation. As a result of all these, there may be an unbalanced force on the control volume and
whenever there is an unbalanced force on the control volume, its momentum may change. So

Rate of momentum Rate of momentum Rate of momentum Forces acting on

= - +∑ .
accumulation coming in going out the system

This is the most general form of equation of motion for a fluid which is nothing but again,
Newton’s 2nd law for an open system. (Refer Slide Time: 5:17)

vx

That’s what I have written over here. You can see that if this is defined as the control volume of
side Δx, Δy and Δz, So you understand that the equation that I have just described
Rate of momentum Rate of momentum Rate of momentum Forces acting on
= - +∑ would be
accumulation coming in going out the system
applicable on the system. So we would see how the convective momentum comes into the
system.
If I only look at the face ∆y∆z , and let’s assume that the velocity here is vx, so the amount of
mass which comes in through the ∆y∆z face must be equal to ∆y∆z ρ vx kg/s. Now in order to

obtain the momentum multiply it with another vx. So the x component of momentum must be
equal to ∆y∆z ρ vx vx . So, rate of momentum coming into the control volume is ∆y∆z ρ vx vx x . And

when we talk about the momentum that is going out, so this would be ∆y∆z ρ vx vx x +∆x
. Similarly

when we talk about the ∆x∆z face, the amount of mass which is coming in through the face must
be equal to ∆y∆z ρ v y . But this amount of mass has an x component of velocity. So in order to

obtain the x component contribution of this much of mass, this must be multiplied with vx. So
mass coming in multiplied by the component of velocity in the x direction at that point would
give me the amount of momentum that comes in to the control volume through the face ∆x∆z at
y. So, ∆y∆z ρ v y vx is the total amount of momentum coming in through the y face. And the x
y

momentum that goes out of the control volume through the face at y would be ∆y∆z ρ v y vx .
y +∆y

Similarly I can write the z component, the amount of mass coming in through the z face is the
area, density and velocity of z component, ∆y∆x ρ vz . Now this amount of mass has some x

component of momentum associated with it. In order to obtain that, I simply multiply it with vx.
So momentum in through the z face is ∆y∆z ρ vz vx z
and out is ∆y∆z ρ vz vx z +∆z
. So all are x

momentum. So these six terms tells us the x component contribution of momentum to the control
volume.

(Refer Slide Time: 11:54)

Next we go into the contribution of molecular momentum or conductive momentum or viscous
momentum. Let’s start with the y face. The area of the y face is ∆x∆z . So the x component of
momentum getting transported in the y direction must act on an area which is ∆x∆z . So y
direction has this much of surface area associated with it. So the x component contribution in the
y direction must be equal to ∆x∆z × the shear stress acting on the y face. The one that goes out
would also be same thing but evaluated at y + Δy.

Let us try to again work out this. x component of momentum getting transported in the z
direction. So if x component of momentum gets transported in the z direction, in order to obtain
the total amount of viscous momentum which is coming in through the z face, I must multiply
this τzx with the area of the z face which is ∆x∆y . So x component getting transported in the z
direction multiplied by the area of the z direction, this gives me the rate of viscous momentum in
through the z face. So, the rate of x momentum in z dirction is ∆x∆y τ zx z , and rate of x

momentum getting out of the z face is ∆x∆y τ zx z +∆z

. So what is τxx? τxx is the x component of

momentum getting transported in the x direction. This is slightly unusual. the packets of fluid
and this is the.

Let us say three packets of fluids are coming towards the x face which is ∆x∆y and there is a
variation on velocity between the 3 packets. So since there is a variation in vx in the x direction,
dvx
that means ≠ 0 . Then by Newton’s law, over our understanding of viscosity, there must be a
dx
stress between these 3 packets of fluids which will also be transmitted on this face. So this kind
of stress where the principle direction of motion and the direction in which the momentum gets
transported are identical, they are commonly called as the normal stress. So τxx is nothing but the
normal stress exerted by the fluid on the x face due to a variation in velocity of the x component.
So ∆y∆z τ xx x is going to be the in term and ∆y∆z τ xx x +∆x
is going to be the out terms. So these 6

terms in total would give you the amount of momentum which is coming in to the control
volume as a result of viscous transport of momentum.
So we have correctly identified in our equation of motion, the rate of momentum in and the rate
of momentum out, both for convective motion and the conductive motion.

(Refer Slide Time: 16:21)

So what is left right now is to identify what is going to be the pressure forces and what is going
to be the body forces. So let us assume that the pressure at x is p x and the pressure at x + ∆x is

p x +∆x . So, the difference in pressure force acting on the x face is ∆y∆z ( p x − p x +∆x ) , the other

pressure differences do not contribute to the x component of momentum. Only the pressure over
on acting on the x face and on the x + ∆x face, they have contribution in the x direction.

Please remember I am pointing out once again that we are writing this equation for the x
component of the forces, be it pressure or be it other body forces. All the other components, y
and z components, can be written in a similar fashion and you do not have to write each one of
those components separately. So we have identified the momentum, we have identified the
pressure forces, the only thing that is remaining is the body force which is acting on it.

So what is the body force? Body force must be equal to the M, the total mass of the system
multiplied by the component of gravity, if gravity is the only body force. Component of gravity
in the x direction is gx, since we are writing this equation for the x component of equation of
motion. So the mass of the control volume would be ∆x∆y∆z ρ which makes it kg multiplied by
gx would be the x component of the body force. And the rate of accumulation of momentum
 ∂
inside the control volume would simply be ∆x∆y∆z ( ρ vx ) kg/m2s. So if I take the convective
∂t
momentum, conductive momentum, the x component of the pressure force, the x component of
the body force and the rate of accumulation of the x momentum, then according to the equation,
which we have written,

Rate of momentum Rate of momentum Rate of momentum Forces acting on

= - +∑
accumulation coming in going out the system

all these terms together can be written and I am not going to write the full form of the
expression.

(Refer Slide Time: 21:00)

What I am going to do is I am going to simply give you the compact form of this equation after
Dv
some simplification which is ρ = −∇p − [∇.τ ] + ρ g . This equation in tensor notation is
Dt
Dv
known as the equation of motion, where the ρ is essentially
Dt
 Dv 
mass per unit volumeρ)×acceleration
(   . ∇p is pressure, Newton, force per unit volume.
 Dt 
∂
Then you have ∇.τ which is (τ ) , shear force per unit volume and ρ g is gravitational force
∂x
per unit volume. So how did we get to this equation from all the previous equations. Remember
what we have done for the case of x component of equation of motion. I have identified all the
terms those can be put into the equation which is

Rate of momentum Rate of momentum Rate of momentum Forces acting on

= - +∑
accumulation coming in going out the system

So both sides can then be divided by ∆x∆y∆z and in the limit ∆x∆y∆z → 0 , then one can get a
differential equation which is the x component of equation of motion.

Hence in a similar fashion, one can write the y component of equation of motion and the z
component of equation of motion. All these three equations can be added to obtain the compact
equation of motion once you express them in a vector-tensor notation. So no new concepts are
involved beyond what I have taught you in this part. So you can see the text and you can yourself
see the simplifications that are made which are only algebraic in nature without the involvement
of any additional concepts.

So I did not derive the entire equation in this. I have given you enough pointers for the
fundamental development of equation, fundamental development of the equation and then I have
told you how to combine these 3 equations in vector tensor notation and what you get is the
equation of motion considering all the 3 directions. I would like to draw your attention to this
equation once again, because each term of this equation is essentially force per unit volume and
everything possible has been taken into account in here. So if in this equation you introduce the
restriction of constant ρ and constant μ and if you add the equation of continuity, the previous
Dv
equation simply boils down to ρ = −∇p + µ∇ 2 v + ρ g . This is known as the famous Navier
Dt
Stoke’s equation.

(Refer Slide Time: 26:02)

So the equation of motion for the special case when the row and mue of the fluid are constant,
then what we get is Navier Stoke’s equation and this is the one which we are going to use in all
our subsequent studies of fluid motion. There is one more simplification that can be thought of,
is if viscous effects are absent, that means we are dealing with an invicid fluid, in that case,
Dv
ρ = −∇p + ρ g . This special form of Navier Stoke’s equation is known as the well-known
Dt
Euler equation.

So what we have done in this class and the previous class is introduced the concept of different
derivatives, introduced the concept of equation of continuity, equation of motion, the equation of
motion for the special case where the density and viscosity of the fluid are constant, would revert
to an equation which is more commonly known as the Navier Stoke’s equation. But all these
equations, the equation of motion or the Navier Stoke’s equation are nothing but the statement of
the Newton’s 2nd law for an open system. Another fundamental relation in fluid mechanics can
be the special case of Navier Stoke’s equation where there is no effect of viscosity. So for a fluid
of very low viscosity or in other words truly for an invicid fluid under idealized condition, what
you get is from Navier Stoke’s equation is known as the Euler’s equation. Euler’s equation has
so many fundamental uses in fluid mechanics. It is the starting point from where you can start to
obtain the the Bernoulli’s equation. So Euler equation is for invicid fluid, Navier Stoke’s
equation is more common for viscous fluids in which case ρ and μ are assumed to be constant. If
they are not, then one has to go back to the fundamental equation which is valid both for steady
and unsteady, constant ρ or variable ρ, so constant μ and variable μ, Newtonian or non-
Newtonian.

So equation of motion itself is a complete expression that gives you the the entire physics behind
the motion of fluids and the momentum transfer. Now all these equations, the Navier Stoke’s
equation as well as equation of motion are available in different coordinate systems. What would
be the equation for x component in cartesian, cylindrical and spherical coordinates or similarly
for other components, in all 3 coordinate systems, are available in your textbook.

So what we would do is in the next class we would see these components and we will try to
figure out how to use those for all the problems that we have dealt with so far using a shell
momentum balance. And I am sure you will see with me that the use of the right components of
Navier Stoke’s equation for the problems that we have dealt before would essentially simplify
our life a lot. We would be able to much more conveniently handle problems of momentum
transfer if we start with the right component of the Navier Stoke’s equation. And from our
understanding we would cancel the terms which are not relevant and what would be left with is a
compact governing equation that we should be able to solve using appropriate boundary
conditions. That’s what we are going to do in the next class.