Video summary
The lecture begins by reviewing the fundamental equations derived from the conservation of mass and momentum, setting the stage for the derivation of the conservation of energy. The instructor recaps the continuity equation and the Navier-Stokes equations, which describe fluid motion in terms of inertia, pressure, viscous, and body forces. These momentum equations are identified as non-linear second-order differential equations that are mathematically well-posed but difficult to solve generally without simplifying assumptions. The discussion highlights how removing the viscous terms from these equations leads to Euler's equation for inviscid flow, which can further be integrated along a streamline to yield Bernoulli's principle, illustrating the logical progression from general fluid dynamics to specific energy relationships.
Transitioning to the main topic, the lecture introduces the conservation of energy based on the first law of thermodynamics, stating that the rate of change of total energy within a system equals the sum of heat transfer and work done on that system. The instructor applies the Reynolds Transport Theorem to a control volume to express the rate of increase of energy as the difference between the energy entering and leaving through the control surfaces. By analyzing a small parallelepiped element, the derivation expands the convective transport terms for kinetic and thermal energy. Through careful expansion and utilizing the continuity equation, the instructor simplifies the left-hand side of the energy balance, effectively grouping terms to represent the material derivative of total energy per unit mass.
The right-hand side of the energy equation is then addressed by expanding the heat transfer and work terms in differential form. For heat transfer, Fourier's law of conduction is applied to show that the net heat flux into the control volume is proportional to the Laplacian of temperature, assuming constant thermal conductivity. The derivation also begins to address the work done by pressure forces, noting that pressure acts inward on the surfaces of the fluid element and contributes to the energy balance through the product of pressure and velocity. Although the full integration of the work terms is deferred to the next class, the current session successfully establishes the structural framework for the energy equation, linking the temporal change in internal and kinetic energy to conductive heat flux and mechanical work within a Cartesian coordinate system.
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>> So, we had just completed conservation
of mass and momentum. Just let us
quickly take a recap of what we had
derived. So, you can see that we had
derived the conservation of the mass
either as this form or this form. We did
use this form in the conservation of
momentum. We took the conservation of
momentum, applied RTT, that is the
Reynolds transport theorem, and we
derived the surface forces and the body
forces in the differential form
and got the Cauchy's equation and
substituted the stresses in terms of the
velocity gradients and the viscosity,
and we got this equation. So, if you
write the
complete equations, so what is that you
have gotten here is that
the conservation of mass, that is
continuity equation, is [clears throat]
d rho by dt plus rho of del dot v equal
to zero. This is continuity equation.
Now, the x momentum equation is that rho
du by dt is equal to minus of do p by do
x plus
mu into del squared u by del x squared
plus del squared u by del y squared plus
del squared u by del z squared
plus del by del x of
mu by three into del dot v
plus f of x.
Similar This is x momentum equation.
Similarly, Y momentum equation is rho DV
by DT equal to minus do P by do Y plus
mu into do squared V by do X squared
plus do squared V by do Y squared plus
do squared V by do Z squared plus do by
do Y of mu by 3
del dot V plus FY. This is Y momentum
equation. Similarly, X momentum equation
is [clears throat]
rho DW by DT is equal to minus do P by
do Z plus mu into do squared W by do X
squared plus do squared W by do Y
squared plus do squared W by do Z
squared plus do by do Z of mu by 3
do del dot V
plus
FZ.
So, you have four equations.
You have four equations.
And the four unknowns The four unknowns
are
which you are going to solve is
P
U V W. So, you get the complete pressure
distribution and the velocity
distribution if you solve this
equations. So, this is That means it is
mathematically
well posed.
That means number of equations are equal
to the number of unknowns.
Mathematically well posed. It is a
non-linear. It is Unfortunately, it is
not possible to solve except few
simplifying cases, simple cases,
non-linear second-order
differential equation this is.
And this was first derived by Navier
and Stokes
parallelly. Navier and Stokes parallelly
derived these equations. So, that is why
this is called Navier-Stokes equation.
Navier is French mathematician
and Stokes is a English He's from
England English mechanician, that is
fluid mechanician.
Okay, they independently derived them.
So, now if you just see this
each of these terms actually, the first
term, this is what is called as This is
what is called as the inertia term.
This is mass into acceleration. So,
that's why this is inertia term. This is
because of the pressure force.
And this are because of the viscous
forces.
This is because of the viscous forces
because by virtue of viscosity.
Okay? And this is body force. So, this
is the physical significance of the each
of these terms. So, you should remember
that this is inertia force. This is mass
into acceleration. This is the pressure
force term and this is viscous force and
this is the body force term. Okay? So,
having understood this, actually if you
if you remove the viscous term, this
equation should reduce to Bernoulli's
equation. Does that happen? Yes, it does
happen. So, if you see that
if you just see, so this is you can
write this in the you can write this
actually in the vectorial form in the
vectorial form that is here. I'm not
going to write that here. Rho DV by DT
you can combine all of this you can
write that Rho DV by DT equal to minus
del P mu
del squared V plus mu by 3 del of del
dot V plus F.
So, if viscous term is zero, so you have
constant viscosity, you can write this
as
mu of del of del dot V and if viscosity
is zero, if viscosity is zero, then what
will happen? If it is for incompressible
fluids, del dot V is zero, so this term
gets
canceled out and that's what we have
written here.
Now, okay, we'll write this. So, if we
write this, what is that I get here is
that I get in the vectorial form as rho
DV by DT
is equal to minus del P plus mu del
squared V
plus mu by 3 del of del dot V.
This is for
viscous
compressible
constant viscosity fluids.
But, if it is incompressible viscosity
if it is incompressible, we know that
for incompressible fluids
we know that del dot V is equal to zero.
So, you get rho DV by DT is equal to
minus del P plus mu del squared V.
This is for incompressible fluids with
constant viscosity. Now, if you take
inviscid
inviscid means ideal fluid where there
is no viscosity. Viscosity is zero. So,
that is rho DV by DT equal to minus of
del P.
This is nothing but my Euler's equation,
which can be reduced to Bernoulli's
equation, which is P by rho plus U
squared by 2 plus GZ equal to constant
for along a streamline.
Along a streamline. So, that I'm not
going to do, but I will just show you
the steps which are involved in that.
So, you just take the Euler's equation,
take a streamline, okay, which is of
length delta S. Consider steady
incompressible inviscid fluid. So, then
that means you have whole of this total
derivative reduces to rho dou U by dou S
because it is steady, so dou U by dou T
is zero. And you have dou P by dou S
along a streamline. F is the body force
because of the gravity that is acting
along the streamline, that is minus rho
G sin theta. So, if you write this
sin theta as dou Z by dou S. So, if you
integrate this equation, so you get P by
rho plus U squared by 2 plus GZ equal to
constant, which is nothing but my
Bernoulli's equation, which we have
studied in uh
fluid mechanics. So, that's the That's
how we have completed the momentum
equation derivation. So, the next
logical thing after conservation of mass
and momentum is conservation of energy.
So, we shall do the conservation of the
energy in Cartesian coordinates. So, for
that we need to take the recourse of
first law of thermodynamics.
So, let us get started with first law of
conservation
of
energy.
So, for that but from first law of
thermodynamics
first law of
thermodynamics
First law of thermodynamics, you have d
e equal to d q plus d w. If you take the
total derivative, you get d e by d t
equal to d q by d t plus d w by d t.
Or
>> [snorts]
>> what is d e? D e is increment in
increment
in
kinetic energy plus
thermal energy of the system. Kinetic
energy plus thermal energy of the
system.
Okay? And d q is heat transferred
to the system. Heat transferred
to the system.
Now, d w is work done on the system.
This we have studied in thermodynamics.
That's why I am stating it directly.
Now, kinetic energy kinetic energy is
u squared plus v squared by two. I'm
doing 2D because 3D becomes very clumsy,
but we can extend this to 3D little
later. And the internal energy that is
the thermal energy is not thermal
internal energy. Thermal internal
energy.
That is represented by E.
That is represented by E. So now, let us
try to apply the Reynolds transport
theorem for conservation of energy. RTT,
Reynolds transport theorem. That is DB
system
divided by DT
is equal to do by do T of row B DV
for control volume plus row BV dot N DA
across the control surface CS.
Now, B is E. That is internal energy,
thermal internal energy E plus kinetic
energy U squared plus V squared by two.
So this
So B equal to MB.
So if you apply this, so what is that
you get is the rate of
rate of increase
of E
in CV.
That is this term.
Minus
rate at which
E
enters
the control enters through the
surface
of control
volume.
Plus
rate
at which
E leaves
through
the surface of
control volume. That means these two
terms
these two terms corresponding correspond
to this.
This should be equal to
This should be equal to I've written
already dE/dt
equal to dQ/dt
plus dW/dt.
So this which are acting on the control
volume. That is rate of
heat transfer rate of heat transfer into
control volume
into control volume
by conduction
by conduction.
Plus
rate of
surface. Who is doing the work here?
This This is
this term. The rate of surface rate of
surface and body forces surface and body
forces
doing the work.
So each of this block now we need to
expand that in the differential form.
That's the next effort what we are going
to doing the work on control volume.
That is what we are going to represent
now.
So, let us take up first the left-hand
side. So, if I take the left-hand side,
that is the rate of increase in the CV,
rate at which it enters, and the rate at
which the E leaves the control volume.
So, I take the control volume.
A simple parallelopiped.
Okay? And you put that here.
So, you have delta X, delta Y, delta Z.
So, what is entering is rho
E. It is entering with a velocity U
delta Y delta Z.
E U, the energy which is entering
is rho E U delta Y delta Z. This is this
term. The rate at which this term is
what I am writing. Okay? Rate at which
the E enters through the surface. That
is that. And what is it? It is leaving
rho E U delta Y delta Z plus do rho E U
by do X delta X delta Y delta Z.
And what about this? This is rho E V
delta X delta Z.
Then rho E V delta X delta Z plus
do of rho e v delta x delta y delta z.
rho e v
do y
delta y delta x delta z.
Okay? So, this
this will be this will become
This is negative because this is
entering. So, these two terms are
negative.
And these two terms are positive.
So, that means this term
this term gets canceled out with this
term. And this term gets canceled out
with this term. So, on the left-hand
side, I am left out with that I am left
out with
that is control volume rho e v.n
dA. Remem-
Remember, this is what we are trying to
write.
If you This This term This term is what
we are trying to write. So, that is why
if you have rho e v.n, so that's why you
have e into flow rate. E into flow rate,
that is rho into u into delta v delta z.
That's what it happened. So, you are
left with
do of rho e u by do x plus
do rho e v by do y
into delta x delta y delta z. And
another term which is
control sur- This is across Sorry, this
is control surface. So, do by do t
across control volume rho
e
dx dy DZ. That is do of row E by do T
delta X delta Y delta Z. So, whole of
the left-hand side, if you write the
left-hand side
divided by delta X delta Y delta Z is do
of row E by do T
plus do of row E U by do X plus do of
row E V by do Y.
Okay? So, now we need to expand this.
So, we will expand this. So, that is do
of
row E by do T plus do of row E U by do Y
plus do of row E V by Sorry, this is X
by do Y is equal to LHS divided by
delta X delta Y delta Z. Now, let us
expand this. I will keep first row
constant and then E constant. So, you
have row
do E by do T
plus
E do row by do T
plus
take out row U as row U into E you can
write. So, you can write this as row U
do E by do X
plus
E do row U by do X
plus
row V. Here you take row V and e. So,
rho v do e by do y plus e into do rho v
by do y. This is equal to LHS by delta x
delta y delta z.
So, now what we can do is you can just
take rho as constant here. Rho into
do e by do t plus
u into do e by do x plus v into do e by
do y
plus e into do rho by do t plus do rho u
by do x plus
do rho u v by do y
is equal to LHS divided by delta x delta
y delta z. Now, you see what happens.
So, this is nothing but the total
derivative
de by dt. So, you have rho de by dt.
And this is continuity equation.
That means the continuity equation this
becomes zero by conservation of mass.
Conservation
of mass. So, you get rho de by dt equal
to LHS divided by delta x delta y delta
z. This is means is also equal to d by
dt of capital e is nothing but e plus u
squared plus v squared by two.
Okay. Now, let us
keep this like this for a minute and get
to the other terms, that is the rate of
heat transfer, rate of heat transfer
into control volume by conduction.
So, let us take up that term. So, the
rate of the heat transfer by conduction.
Rate of
heat transfer
by
conduction.
Okay? So, you again take a
parallelepiped like what you did in the
heat diffusion equation if you recall.
You had done this in heat diffusion
equation while we derived for heat
diffusion equation,
which Professor Arun did.
So, you have delta X, delta Y, delta Z.
So, you have the conduction which is
taking place. That is QX dot into delta
Y delta dy you can even write dy dz.
dy dz and QX dot dy dz plus do QX dot
divided by do X into dx dy dz.
Similarly, the heat transfer in the Y
direction QY dot into dx dz.
into a another term QY dot dx dz plus
do QY dot divided by do Y into dy dx dz.
So, the negative sign
is heat transfer is
arises because the heat transfer is
counted in this is heat added. Heat
added is considered as positive. This is
positive term.
Because the heat added, this is heat
lost from the control volume.
So, this will be positive and this will
become negative.
This will become negative. So, in that
case, in that case, this term and this
term get cancelled out and this term and
this term get cancelled out. So, you
will be left with
you will be left with
do QX dot. So, do QX dot by do X
in the negative sign
plus do QY dot by do Y.
Both are negative. Okay? So, this is
into DX DY DZ.
Now, QX dot, the by Fourier's law of
conduction, the QX dot can be written as
- K do T by do X
plus do by do Y of - K do T by do Y.
Into DX DY DZ.
So, minus into minus, if you just push
this and pull out K, K is constant if
you assume that K is uniform here in
this case. It is not constant. Constant
term is used with reference to time.
With respect With reference to space,
you use the term uniform. In that case,
this equation reduces to K into del
squared T divided by del X squared plus
del squared t by del y squared into
dx dy dz. dx dy dz. So, that is about
the rate of the heat transfer by
conduction. Now, let us try to do the
rate of heat transfer, rate of work
done. Rate of work done. Rate of work
done by pressure force.
Rate of work done by pressure force. So,
this to do this, so let us draw
parallelepiped
which is having delta x delta y delta z.
So, you have
Work means
force into velocity. Force acting here,
pressure always acts inwards. So, p into
delta y delta z, this is the pressure
into area gives me the force. This into
velocity u x direction velocity gives me
the work done. So, here it is
Pressure is acting inwards, so u into p
into dy dz plus do of up by do x into dx
dy dz.
This is the at the other side. Now, here
again p into dx dz is the force into
velocity, velocity is v in the y
direction, and here
This is vp
dx dz plus do of vp by do Y into dy dx
dz. So, how to take them as positive
negative all that we shall try to
understand in the next class and
continue this derivation so that we will
achieve the energy equation and try to
see the temperature in our energy
equation. Thank you.
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