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Hello everybody, in the last class we had gone through the discussion on DC motors and
we almost completed the discussion on the DC motors towards the end of the session.
We just very briefly discussed on how we go about performing the dynamic model of the
DC motor. In this session we shall carry that through that is make a dynamic model of the
DC motor which will complete our discussion on the DC motor and then we shall take up
a new topic and that is on 3 phase systems.
So first, the DC motor. So we have been studying quite at length about the DC motor. Now you
are familiar with symbology; now we have the armature and to the armature is attached the
brushes through the commutator and we have two non-idealities in the dynamic case: one
is R a and the other is L a; R a is the winding equal and winding resistance of the armature,
L a is the equivalent inductive reactants of the armature.
Now this is the DC motor which has a mechanical shaft and we said that in the case of the
motor the energy is moving from the electrical domain to the mechanical domain and the mechanical
loading gets reflected on the electrical domain as back emf across the brushes.
Now we apply a voltage E a and there is a armature current I a which is flowing through
it, this is R a and L a and then in the mechanical domain we have a torque T g which is getting
generated and because of the torque T g there is a speed angular speed of rotation omega
in radians per second or in rpm which gets generated. Now this is on the mechanical domain.
Now this mechanical domain also contains its own load or the reflected inertia J which
is like the mass of a vehicle or the mass of the load or the mass of a very big wheel
which has an inertia to it and then when it is rotating it gets stored in the kinetic
form and that is called the inertial energy which is stored by virtue of an object being
in motion which we know that it is stored in the kinetic energy form and therefore J
or the inertial parameter is similar to L by inductive parameter and therefore the state
variable in the case of the mechanical domain which is involved with the parameter J would
be omega this speed. There could also be at the bearing some friction associated and we
will call it as friction B.
Now there is this is the electrical domain and this on the right side we have the mechanical
domain, the link between the electrical and magnetic domain is the motor. There is a back
emf we said and this back emf can be expressed in terms of the mechanical domain parameter
or mechanical domain state variable and that is k phi omega we saw this in our earlier
discussion and that is equal to the back emf. The torque component here the torque that
is generated can be related to the state variable on the electrical side by the same k what
we have written phi i a.
Now as we are doing a dynamic model we do not want to take the rms values, let us take
the instantaneous values. So therefore I will replace these by the instantaneous values.
So we have e a we have i a and E b so this is the relationship. Now we can apply the
Kirchoff's voltage law to the electrical domain here. You see that the applied voltage is
e a, there is a voltage drop r a there will be a drop across the ℓ a inductor and then
of course there is a drop that gets developed across the brushes because of the rotation
of the armature that is the generated emf or the back emf and that is called e b. Therefore
we have we can apply the Kirchoff's voltage law in this area in this electrical area.
Likewise, on the mechanical area the mechanical side also; one can apply equivalent to the
Kirchoff's voltage law the potentials or the potential variable being torque. The speed
is common; like the current being common in the electrical circuit the speed is common
to all the things connected to the shaft of the armature. the sum of the potentials which
means in this case the torque should be equal to zero like the Kirchoff's voltage law.
So, generated the generated torque is T g should supply J d omega by dt and also d omega
and if there is any other load torque that component also. So we are applying the Kirchoff's
voltage law on both the electrical domain and the mechanical domain with the state variables
which are the state variables; on the electrical domain we have one dynamic element L a with
associated state variable i a and then we have J as another dynamic variable in the
mechanical domain which is the reflected inertia and then the associated state variable omega.
So we have two state variables at this point.
Now there is one more issue and that is phi. how are we generating phi; because the excitation,
the field is being generated differently with respect to different topologies of the motor
as we saw; in the case of the shunt motor, the series motor, the compound motor the phi
is being generated separately so which means as far as this motor is concerned there is
one more input which is phi that we need to take care.
So first let us consider the case of the shunt motor where phi is a constant or a separately
excited motor where phi is a constant, we are not touching phi. So under that condition
we can now write the equations. Now the electrical domain side first so we have L a dia by dt
so the voltage across that is e a minus R a e a minus i a R a minus e b. but e b can
be expressed in terms of the state variable which is given by k phi omega so which is
e a minus k minus i a R a minus k phi omega. So, for the moment we are considered considering
phi as a constant that is this.
Let us take this, copy, we go the next page, paste.
So now on the mechanical side there is a generated torque T g. Now this is going to supply the
inertial component d omega by dt which is similar to L di by dt plus i r a equivalent
to the resistive drop B omega in the mechanical domain.
Now plus any other load torque, any other load torque which may be applied to the mechanical
shaft, all these should be generated by T g. So J d omega by dt will be equal to T g.
Now T g can be expressed as k phi i a in terms of the state variables k phi i a minus B omega
minus T load. So, as a mechanical input to the system there is T load and as the electrical
input to the system we have e a.
So combining these two we have dia by dt equals............. look at this e a by L a R a by L a into i
a and k phi by L a into omega. So I will rewrite it as minus R a by L a into i a minus k phi
by L a into omega plus 1 by L a into e a so this is one equation. Then we have d omega
by dt which is the mechanical equation so we have for the state variable i a k phi by
J k phi by J into i a minus B by J into omega minus 1 by J into T load. This is the second
equation.
This equation has been rewritten this one. These two equations completely define the
dynamics of the motor. Now if it is a separately excited motor or a shunt motor where we can
assume phi to be a constant where we can assume phi to be a constant then we can write it
in the standard form of i a dot d phi by dt omega dot d omega by dt which is equal to
minus R a by L a minus k by L a k by J k phi by J this is and minus B by J; i a and omega
being the state variables in the state vector plus 1 by L a minus 1 by J sorry 0 and minus
1 by J and then we have two inputs e a and T load. So this is of the form x dot is equal
to A x plus B u the standard form.
Now if phi is not a shunt but let us say a series motor in that case phi itself is a
function of i a so which means here phi becomes proportional to i a which results in.........
let us say this becomes now k 1 i a k 1 i a so we do not..... because there is now straight
multiplication i a into omega i a into i a this is i a square this is a nonlinear system
therefore we do not put it in the standard matrix form we just leave it as differential
equation system so that would be become a nonlinear system but still that would give
you the entire dynamic model of the DC motor system.
So the phi should automa should appropriately be modified whether it is a constant or whether
it is a now a function of i a or so or any other variable that should automatically be
put in the........ and that will give you the complete dynamic model of the DC motor
in the form or differential equation form.
So we thought we conclude our discussion on DC motors, the DC generators and DC motors
are a class by themselves, they both are similar machines, similar in structure but the only
difference being that in the case of the DC generator the energy is being input on the
mechanical domain, you are taking out the energy in the electrical domain and therefore
from the electrical domain point of view it is a generator. And in the case of the DC
motor you are applying the energy input in the electrical domain and taking out the energy
in the mechanical domain.
But constructionally both the machines are very similar and a DC generator can be operated
as a DC motor and DC motor as a DC generator and vice versa.
Internally inside the machine inside the armature the currents that are flowing in the coil
are always AC but only by means of the device called the commutator which we have discussed
and the brush combination external to the motor in the electrical circuit or external
to the generator in the electrical circuit the signals are DC otherwise within the motor
it is always an AC signal.
Now the mot the idea the concept of energy being passed through many domains is the underlying principle in which most
of the machines in most of the applications applications are being used by many of the
applications for example the induction motor induction motor or in the case of the alternators
for generators or a combination of all these domains. And most of the in the electrical
technology the central what we call the central concept or the central theme is some prime
mover, propulsion, movement on the mechanical domain and the electrical domain is just the
currents and the potentials. The core of the transformation of the energy from one domain
to the other is always most of the cases in these set of equipments being done in the
magnetic domain. So the central..... what we call them is generally a magnetic domain
in most of our electro electrical technology applications magnetic domain. So you could
get power or energy input from the electrical domain, take out power in electrical domain,
you could take out the power in mechanical domain or you can put energy into the magnetic
domain from the mechanical domain and take out power into the electrical domain all these
are possible.
So the central mechanism which makes possible the conversion between many of these domains
is the magnetic domain in the case of most of the electrical equipments and all the all
these area including the interfaces between the electrical magnetic and within the magnetic
they are always AC and in many applications it is not single phase AC it is 3 phase AC.
So before we try to understand and learn about the equipments electrical equipments which
fall into these categories like the induction motor, the synchronous motor and generators,
3 phase transformers they all form into the fall into the category of this multi-domain
principle. In the case of 3 phase transformers of course this domain is not there, there
is only that is the mechanical domain is not there, there is only electrical transformation
between electrical and the magnetic domains.
In the case of the induction motor with shorted rotors we do not have you have only one electrical
domain which is inputting and this other electrical domain is not there and you have the mechanical
domain. In case of the wound rotor where you have the rotor windings not shorted but taken
out to a resist to resistor to a set of resistors then you have the input electrical domain
or output electrical domain, you also have the mechanical domain so on and so forth.
So like that you can have many combinations but all falling into this class of this type
of concept.
So, before we go further at all into the understanding of these equipments like induction motors,
3 phase transformers, synchronous motors, synchronous generators we should have a good
understanding of the 3 phase system and what is this 3 phase systems.
So the major part of this session is going to now deal with 3 phase systems. This is
also written as 3 phi for 3 phase system. So in the literature you will see that phi
being used for the term phase. So now what is this 3 phase system, how does it look like
and what is its character and what is its what are its features.
Now let me take a source a sinusoidal voltage source let me arrange it in this fashion.
So let me say that this source is connected to resistive load, let me connect it to a
resistive load as shown like this.
So this resistive load R is connected across this source which has two terminals here and
let me name that terminal a and let me name this terminal as n 1 and the voltage across
these two e a n 1. So this is the voltage and there is going to be a current through