… power generation.
The basic idea is simple. If you have some
kind of a channel and if you allow a very
hot gas to flow through, hot gas in the sense
plasma that means the bulk of the gas is ionized,
the electrons have been stripped off, so you
have essentially electrons and ions flowing
and if you have a magnetic field in the direction
perpendicular to the sheet of paper that means
we normally denote it like this that means
it is like arrow going into the paper, then
the electrons will be deflected by this magnetic
field. The holes, not holes, the ions will
also be deflected by the magnetic field in
opposite direction, so that if you put electrodes
in the two sides, then some voltage will be
induced in these two electrodes and if you
connect it through some external circuit,
through some external circuit, a current will
flow. So, this is the essential idea of the
magnetohydrodynamic power generation. That
means here because of the magnetic field there
is a charge separation and that charge separation
is collected by electrodes and that is what
allows the current to flow through an external
circuit.
Now, you might ask how you actually do it,
what creates this plasma? So, the essential
idea of the magnetohydrodynamic power generation
is where you have a fuel which is either a
liquid fuel or a gaseous fuel that will be
burnt creating a very high temperature and
that high temperature gas will be right in
the beginning allowed to flow through this
kind of a channel. It is like a nozzle. That
means it passes at a high speed thus extracting
power that is contained in that, in that gas
and then after the power is generated, a part
of the energy is extracted.
The rest of the energy still contains a lot
of energy that goes through and that is then
further extracted by a conventional thermal
power plant. That means this MHD cycle becomes,
MHD cycle in conjunction with the conventional
steam based thermal power plant becomes what
is known as a combined cycle power plant,
where the MHD is the toping cycle and the
Rankine cycle is the bottoming. Is that clear?
Toping in the sense that where it is high
temperature that is when the toping part of
the energy is extracted and when it goes out
of this at a relatively lower temperature
that is when the bottoming part is effective.
But, there is some, something more to it.
Firstly, how much can we raise the temperature
simply by burning a gas? That means these
kinds of systems are considered where there
is a availability of natural gas. As you know,
in India there are places where natural gas
based power plants are coming up and those
places, the MHD cycle will be very effective.
It is also considered to use coal in the sense
that you have already learnt that coal can
also be converted into gas, gasification of
coal and then the coal can be burnt. So, if
you are considering that kind of a cycle,
then coal can also be used as a fuel. But,
essentially you have to have some kind of
a clean gaseous fuel as a fuel. But still,
the temperature that is attained here is not
sufficient to ionize most of the gas. So,
there has to be something additional put in
and in order to increase the level of ionization,
either caesium or potassium is injected. That
means a bit of caesium and potassium is injected
into the gas.
That is called seeding and that caesium or
potassium and potassium whatever it is, that
immediately gets ionized and that adds to
the overall level of ionization of the material.
That means the free electrons that are free
to move in this plasma is principally contributed
by the seed. So, even if the temperature is
not all that high, if it is sufficient to
ionize the seed that is sufficient. So, either
caesium or potassium is added as a seed. What
would actually contribute to the bulk of the
current? As you can see, there would be the
electrons and there would be the ions.
Ions would be, for example the caesium ion
or a potassium ion that will contain its mass.
Each of this ion’s mass will be far exceeding
the mass of an electron, right, many times
about 25 to 1000 times. As a result, the bulk
of the transport, they are in equal numbers,
the number of electrons assuming that one
is stripped off from one, would be in equal
numbers or other words in some cases, electrons
will be larger in number provided larger number
is stripped off from each ion. So, we are
considering the actual bending of the path
and then finally reaching that electrode,
right and the bending of the path will obviously
depend on the mass of that, the ion or the
electron taking part in the activity.
If it is heavy, then it will have a larger
probability of just going through. If it is
light, it will bend and take part in this
activity. So, then this lead us to conclude
that bulk of that current that is generated
is due to electrons, though theoretically
both would take part. Ultimately, the activity
that happens here is due to electrons. So,
let us consider the motion of electrons in
a bit more detail. Have you understood the
essential structure?
The structure of the MHD generator would be
something like this that you will have a chamber
in which you will have the injection of air
and injection of fuel. So, this is the combustion
chamber. From the chamber it will be sort
of going out like a nozzle and the electrodes
will be in the two sides of the nozzle and
the magnetic field will be perpendicular to
it and when it goes out of the nozzle, then
it is that only in this part the energy is
generated and after that it goes to a normal,
what is this? This is the normal boiler, where
in the normal way you will have, you already
learnt that there will be a boiler drum, there
would be the water walls like this, there
will be water walls like this and finally
there will be, this will go out, there will
be the super heater section, the economizer
section and all that. So, it is just before
that takes place, a part of the energy is
extracted at a higher efficiency through this
MHD process and at this point, there will
be the seed injection. So, if you have understood
this part, then let us go into 
what actually happens inside this, what happens
to each of the electrons?
Now consider one electron, say here and this
fellow is moving in this direction and you
have a magnetic field that is going into the
board like this. What will be the motion of
the electron like? That is obtained from the
left hand rule. So, apply the left hand rule
and tell me what will be the force felt by
this electron?
Downwards.
Downwards; so, it will feel a force downwards
and as a result, its motion will be deflected
away from this linear path to a path like
this. So, it will move like this and say it
has come here. At this point when it has come
here, then the motion is no longer like that.
The motion is now like this and what will
be the direction of the force? Again towards
the center, so it will be like this. It will
further move like this. So, do you see ultimately
what will its behavior be like? It will be
circular. So, it will be a circular motion
like this if nothing happens to it. That means
it is not colliding with anything or something
like that, then it will be a circular motion
and logically speaking it will go on moving
in the circle. But no, that cannot happen,
because we had assumed that there is a, there
is a, well, this fellow has a velocity in
the direction perpendicular to the magnetic
field, but there is no reason to assume that
there could be a component in the direction
of the magnetic field, which will be completely
unaffected by the magnetic field.
So, what will happen to that then? You have,
then you will have there is a magnetic field
and here is the electron. If there is no component
in that direction it will move like this,
but if there is a component which will be
unaffected, so it will move like a helical
path. It will move in a helical path. So far
so good, it will move like, just consider
it will move in helical path. Let me draw.
So, here is the magnetic field. So, if component
of the velocity in the direction of B is present,
then it will be like this.
Well, there need not be only the magnetic
field, there could also be electric field.
Because as you can see, there would be a pair
of electrodes and the pair of electrodes being
in two different voltage means each of these
electrons will face electric field also.
So, what will be the motion like if there
is an electric field also? It will drift in
the direction of the electric field. So, while
supposing I am not considering this that means
I am not presently considering the component
in the direction of B but just this and the
electric field. In that case it will move
like, move like what? It would have been a
circle if there is no electric field. If it
is a circle there will be a drift.
So, it will be, it is not really helical motion,
remember, but it will move like this. So,
there is a drift. So, this is, this was the
direction of B earlier and in the presence
of both, it will be a combination of all these
motions. So, first we considered what? If
there is only B and the component of the velocity
is only in the direction orthogonal to B.
Second, we considered if there is a component
of the velocity in the direction of B what
will be the motion and third we considered
what will happen if there is an additional
electric field. So, it will be like this.
But not really, because there will be a large
number of electrons and ions in that gas.
So, you cannot really say that an electron
will go just like that, because it will collide.
It will automatically collide. So, ultimately
the electron will traverse a path that will
go through various collisions.
But before we go into that, let us just figure
out what will be the radius of this? You can
easily figure that out from Newton’s law.
So, let us write that.
The acceleration will be, will be v perpendicular
square by R. I am denoting that perpendicular
because it is, it is related with the perpendicular
direction of that velocity. Then, Newton’s
law will say that the force is mass into acceleration,
so mv perpendicular square by R, mass into
the acceleration is the force and what will
be the force? Force will be electronic charge
times v perpendicular times B, provided I
am considering only one electron. But, if
you write it in a general sense that means
something that is applicable both to electrons
as well as the ions, then you will have to
multiply with the number Z representing how
many of the electronic charge is there in
that ion?
So, Z is the how many of the electrons charge
is there in that ion. So, in general, it will
be this expression. So, v cancels off and
you get, so this gives you v by R which is
nothing but the omega as Z e B by m, so that
is the omega, fine. So, that is the angular
velocity of the motion.
But now, as I told you, that it will actually
cover paths like this and then another collision,
another collision, another collision and so
on and so forth, right, it will go through
many collisions either this type or you might
say that if the angular frequency is much
higher, then it will be a collision like that.
There are two possibilities. What is the difference
between these possibilities? Here, the mean
free path was smaller and here the mean free
path was larger. In comparison to what? In
comparison to what?
Suppose your, the mean free path 
is lambda and the mean time is tau, then you
can say that lambda by tau is … Now, we
know that v is omega times R, the radius.
So, we have, if you substitute, we have omega
tau is equal to …, fine. Now, you would
notice that this, this term omega tau is what
actually makes the distinction between these,
this type and that type. If omega tau is large
that means this is large, this is the mean
free path is larger than the radius, far larger
than the radius, then it will be this type.
If it is small, then it will be this type.
So, the omega tau is the quantity that we
normally use in order to distinguish between
the types of the average. These are all average;
you will not be able to say that each one
will traverse paths like that, but on an average
if omega tau is small, then the mean free
path is small, then it will be like this,
sorry, in that case it will be like that.
So, this will be where omega tau is less than
less than 1 and this will be where omega tau
is greater than greater than1. So, these are
the two possibilities.
Now, what will actually be the result of all
these? Consider this.
You have, consider again one electron and
the magnetic field and this fellow as you
understood would be moving like this and suppose
it has come here.
Remember, we had initially considered that
this will be the direction of flow, so the
electrons will be flowing like this and you
have produced a B, due to which there will
be a voltage generated here. That means the
electrons will be actually moving like this.
That is producing what is the, what you know
as the Faraday effect. So, let us quantify
this Faraday effect.
It has come here. So, it was actually moving
like this. In this time it would have gone
here, but instead it has come here, right.
So, it has moved by an extent this in the
y direction. So, there would have been, otherwise
if this is not there, no motion in this direction;
now, it has the, it has motion y. That is
the Faraday effect, right. But, notice that
there is something in addition to that. That
is it would have traveled up to this point
in the absence of the magnetic field, but
now it has traveled up to this in the x direction.
So, here is something that is representing
what? The bulk of the gas is moving with the
velocity, v say, what does this represent?
It will represent the fact that the electrons
are falling back, electrons are failing to
travel with the same velocity in the x direction,
right. If electrons are falling back from
that direction what its effective result is
that the system will see a voltage in the
x direction. If the electrons are travelling
along with the rest of the mass, there will
be no voltage. But if they are falling back,
yes, there will be a voltage, there will be
current, which you normally did not anticipate.
From the idea of the Faraday motion, you would
not anticipate that there will also be a voltage
in the x direction, but there will be. If
you allow current to flow, there will be a
current in the x direction also. So, that
is called the Hall effect and that voltage
is the Hall voltage, the current is Hall current.
So, that is the Hall effect.
How much will that be? So, you had this, this
path and it has, I will, I will draw it correctly.
This center would be somewhere here, here.
So, if it travels up to this point before
the next collision then it has traveled by
this extent and what is this related to? That
is related to the omega tau; so, that is related
to omega tau, fine. So, the Hall effect, the
amount of Hall effect will be large if the
omega tau is large. For example, if the next
collision happens here, obviously it has not
fallen back much from the rest of the gas,
but it has, it has been allowed to travel
up to this. It has fallen back much from the
rest of the gas. So, the extent of Hall effect
will depend on the mean free path, the mean
time between the collisions, effectively on
this omega tau, fine.
So, it depends on omega tau which is expressed
in the unit of radiance. So, you have omega
tau quantifying the Hall effect. So, let us
us write down omega tau. So, now for example
if the omega tau is something like 0.1, you
would anticipate it to move on an average
only up to this extent. If it is of the order
of say 1, omega tau is equal to 1, then you
would on an average anticipate it to go up
to this extent. If it is say 5, you would
allow it to traverse almost the full cycle
before the next collision, like that and as
a result there will be the Hall effect.
So, now on this basis, let us try to understand
the actual working.
So, you see, initially the kind of picture
that we had of the functioning of the MHD
generator, now it is somewhat changing, right.
There is some more complication due to very
fundamental physical effects. So, we need
to understand and we need to do something
about that. How much will be the voltage induced?
The voltage induced will be the or that will
produce the current, so there will be two
currents, Faraday current. Faraday current
will be, is the conductivity times … So,
that will be the current. So, the motion of
these electrons or the ions will depend on,
suppose you have got, you have applied a field
E; there are two electrodes and you have applied
a field E and there is electron or an ion
here. How much will it move? That will depend
on its mobility. Obviously, the mobility of
the electrons is far larger than that of the
ions, but nevertheless it will depend on mobility.
So, that mobility is represented by mu.
So, you say that this v is equal to the mobility
times E, the electric field induced. But,
for the electrons it will have a negative
sign, because the electrons move in the direction
opposite to the electric field. Now, if this
is the velocity, then the current will be
J, as a vector, will be if each of the electrons
have this velocity, then what will be the
J? What will be the current induced? The current
will be how many electrons are flowing, n,
how many of these charged particles are flowing
and each charged particle has a charge of
z e, for the electron z is 1, but for the
ions it will be something else, times the
mobility times oh sorry, times the V or times
the mobility times E. So, n z e v and v is
this, so it will be this.
Now, let me, let me write it clearly; is equal
to n z e v in the negative direction, because
for the electrons flow it is, the current
is opposite, is equal to n z e mu. Now, this
has the appearance of the ohm's law. Here
is the voltage, here is the current and here
is then the effective resistance. So, effective
resistance will express it in terms of the
conductivity. Then, sigma zero will be our
notation for the conductivity on that condition,
that will be n z e mu. So, that will be the
conductivity. Effectively, that will be the
conductivity of the flow of charge and we
will write J is equal to …, fine. So, that
defines how much would be the motion.
Now, you see, let us consider both of them.
Here there is a magnetic field, here there
is an electron. This fellow is moving, but
now we are considering not only the motion,
now we are considering the, we are trying
to consider the voltage induced and the current
which are palpably evident quantities. So,
the v cross B will give you the electric field
that is produced by, because of this, fine
and in addition to that, there will be an
existing electric field. So, the effective
electric field will be the combination of
these two, right. So, we will say that E f
is equal to the electric field existing plus
v. So, the electrons drift velocity will be
v is equal to minus mu, I am not talking about
electrons, so let us put this subscript e
times this. So, you see, v appears in both
the sides. Is that clear?
So, effectively the electric field experienced
by this fellow is this much and due to that
there will be the, because this is the mobility,
this will be the velocity. So, the velocity
appears in both the sides. We need to separate
them out to obtain the actual effect. So,
we will consider this later. But, let us now
figure out these are the vectors and the vectors
will be somewhat difficult for us to handle.
So, what we will do is we will break them
out into three components - x component, y
component and z component and try to find
out this in all the different components and
in order to do that, we will draw 
x y z and B will be in the direction of z,
right, into the paper. x is the direction
in which the gas is flowing and y is the direction
perpendicular to it. So, there will be a component
of the motion in the x direction and that
when coupled with B, v cross B, there is a
component in the x direction v x. Cross B
will be in which direction?
Up, so the v x cross B will be 
in this direction. Similarly, there will be
a component of the velocity in the v y direction
and that when coupled with B will be giving
you is the minus x direction, clear. So, this
is v y cross B. So, now we can, on the basis
of this we can write down the equations in
each of the coordinates. Let us see. Let us
clearly draw the vector diagrams.
So, here you have x y z and you have B in
that direction. E x is in this direction,
v x is in this direction, E y the voltage
in the y direction here and v y is in this
direction and here whatever it is, I do not
really care, because that does not interact
with the B. These are the two things that
are of our interest and we had just computed
that v x cross B will be in this direction
and v y cross B will be in this direction,
fine. Based on that we can write the v x is
equal to, the way we have just written, mu
e times E x minus v y, we have times B because
now we are not writing vectors, we are writing
just the product of the two quantities. v
y is minus mu e, v y will be E y plus v x
B and v z is simply, as I told you, we need
not consider this, these are the two things
of our interest and meaning.
Now notice, very importantly that v x appears
here and here, v y appears here and here,
so you could solve these two equations to
obtain them individually, clear. So, do that.
Can you do that? Do it. To cut a long story
short, let me give you the results.
You will have v x is equal to minus mu by
1 plus beta square E x plus beta E y and v
y is equal to minus mu by 1 plus beta square
E y minus beta E x. This beta is B mu. So,
let me write on and then I will write that.
v z is minus mu E z, where beta is … This
is equal to omega tau. So, this is an important
quantity, as I told you, quantifying the Hall
effect. That is why we have put it in terms
of the beta. Now, if we know the velocities
we also know the currents.
So, we can write the currents as J x is equal
to, J is the current density in the x direction,
it will be sigma by 1 plus beta square, notice
the expression, so we are writing in terms
of this, beta square, E x plus beta E y and
here it is, J y is sigma naught by 1 plus
beta square E y minus beta E x and J z is
equal to minus mu E z, simple. So, we have
noticed that this is the thing like the ohm’s
law. This is how the current vector is related
to the voltage vector. But, you see that these
are somewhat involved and complicated, because
the current in the x direction depends on
the voltage in the, both the direction. Current
in the y direction depends on voltage on both
the directions and z direction is independent.
So, what is the, what is the meaning on the
effect of all this? Let us try to figure that
out.
Now, it is, it is not difficult to see that
supposing the electrodes are placed such that
there is no voltage in the x direction, E
x is zero. As we have seen already, here is
the channel and here is the electrode. That
means there is no voltage in the x direction,
because the electrodes will short it. If E
x is zero, then what will be the J x? It will
be this times beta E y and how much will be
J y? J y will remain, it will be, this is
zero, this, which means that even if there
is no voltage in the x direction, there will
be the current in x direction. So, you can
now draw sort of a vector diagram.
E will be in this direction here. There will
be a, there will be a current in this direction,
there will be also a current in this direction.
So, the current resultant will be like this,
here with J here with E. They will be moved
from each other; this way or that way. Probably
it will be the other side, because of the,
we will see that, see to that. But, the point
is that the voltage and the current will not
be in the same direction. Is that point understood?
Effectively, all these rotation of this things
and things like that we are trying to get
that extracted in terms of the voltages and
the currents and we have come to the conclusion
that the voltage and the current will not
be vectorially in the same direction, they
will be in different directions.
Now, let us consider what is happening inside
this fellow?
You have the plasma flowing with some velocity
and you have got some kind of an electrode
structure. There can be different types of
electrodes that is why I am not drawing it
right now. But, what happens is that as it
flows and as you extract the power, effectively
what you do? You slow it down; extract some
power, power out of it means that you make
it loose kinetic energy, so it looses. You
have some amount here, some amount, some other
amount there and it is something like, in
case of any motor, any generator, suppose
there is a generator, the kind of generator
that you have heard of like the DC generator,
like the industrial generator, those things
that you heard of.
Imagine that as you run it in no load, that
is you have not put in load, there will be
some voltage induced and if you load it that
voltage will drop, because there will be,
when in no load the generator voltage is equal
to the back EMF that maintains the balance.
When you load it, the back EMF is there, but
there is also a load, so due to which there
will be a new balance. So, always there is
a generated voltage and there is a back EMF.
Now, here also there will be something like
that. Here also there will be say a similar
thing. So, you will consider, we will consider
the E x E y and E z as the back EMF components.
The generated voltage is U times B. U is the
motion times B.
So, the generated voltage is U B. These are
the back EMF. So, the resultants will be E,
we will denote it as sx, E sy, E sz is the
resultant. So, we can write the equations
as E x in the x direction is equal to minus
E sx, because there is no induced voltage
in this direction, in the x direction. E y
is equal to UB, U is the velocity of the plasma
times the magnetic field, this is the voltage
induced, minus E sy and E z is equal to minus
E sz. So, these are the back EMF’s which
are expressed in terms of the generated voltage
minus the resultant holder that we actually
see that is E sy. So, all this voltage will
be related like this and the loading factor
K is the open circuit voltage oh, the closed
circuit voltage by open circuit voltage.
So, if you have, there are electrodes here,
you have not loaded it, whatever voltage induced
is the denominator. If we load it, then whatever
voltage induced is the numerator. That is
the definition of the loading factor K, because
you could put in different types of loads
and you will soon find that for here also
like in the photovoltaic, say like in the
wind generator, you will always find that
for a specific load it gives the maximum power.
So, we need to find that. That is why this
is important, loading factor K. So, if you,
if you have seen all these, then you would
realize these are the equations that we would
be using. So, just keep them very clearly
in your mind.
So, out of all this we have concluded that
if you have the channel like this and if you
have the electrodes like that and if the plasma
is flowing with a velocity U and you have
this, you might connect a resistance to the
external circuit which will lead to the voltage
being changed to the s quantities, resultant
quantities. If it is like this, this is called
the continuous electrode Faraday generator,
because these electrodes are continuous. Through
the channel, there is one electrode up and
one electrode down.
Now, the electric field is in this direction,
but the current flows in this direction. As
we have seen already, the current is not in
the direction of electric field, as a result
of which the effective power goes down. By
how much? The current, the transverse current,
the power is due to this voltage and this
current, but the actual current is that direction.
So, its component in this direction will be
less by a factor 1 by 1 plus beta square;
we have already seen that, see 1 plus beta
square here. So, this will make the effective
power somewhat less.
It would have been good if the current also
were in the same direction, but the current
is actually in this direction. That is why
there will be a reduction in the power. In
this case the boundary conditions are, boundary
conditions are E x is zero, there is no voltage
allowed in the x direction because we have
continuous electrode, E x is equal to zero,
E z is also is equal to zero and in the z
direction we have not, no electrode and therefore.
there is no current that can flow, J z is
zero. But, there is current that can be flow
in the x direction. Can you see? Through the
electrode there is current that can flow in
the x direction, so that is not zero. Current
will flow in this direction that is not zero.
So, these are the boundary conditions.
If you put that in these equations, what do
you get? First, let us see in this case the
K, the loading factor is E sy, the resultant
voltage in that y direction divided by the
generated voltage which is UB. So, this is
the generated voltage and this is the, this
is the generated voltage and this is the resultant
voltage after the loading and that is that
gives you the factor, loading factor K. It
is not difficult to see that if you multiply
this with J y and this with J y what does
the numerator say? The actual power that is
delivered to the load. What does the numerator
say? Actual power that is delivered to the
load. What does the denominator say? That
power U B is the is the voltage generated,
J y is the current flowing through that that
is the sort of that will be related to the
breaking that is produced in the gas. So,
this is also effectively the efficiency that
you have in the system.
So, if you have that then you can write E
sy is equal to K UB and E y, we already know
as, here K UB you substitute, you get UB 1
minus K. So, now in that terms we can write
J x is equal to sigma naught divided by 1
plus beta square UB K minus 1 beta and J y
is sigma naught 1 plus beta square UB 1 minus
K. Here there was a negative sign that is
why it has become K minus 1. So, the power
generated, power generated per unit volume
will be E sy is the voltage times J y. So,
if you substitute that you will get this will
become sigma naught divided by 1 plus beta
square UB 1 minus K times K UB. So, that is
what you got.
We can write that as P is equal to sigma naught
1 plus beta square U square B square K 1 minus
K. Interesting thing is that you would like
to maximize the power. So, in order to obtain
the maximization of power we will write, dP
dK equal to zero. Just do that. This fellow
is constant, only K 1 minus K, so you get
…
No, only this much will have to be differentiated
with respect to K.
K is equal to half.
Yes, K is equal to half. So, it will be K
minus K square. So, 1 minus twice K is equal
to zero or K is equal to …, which means
that the power maximizes for K is equal to
0.5. So, you have the arrangement like this
and there will be an external load. That external
load has to be variable and it has to be so
set that K becomes 0.5. K depends on the external
load and this can be done by a power electronic
controller in between. Today let us stop.
We will continue with that in the next class.
