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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

  • this one and we will call that one as i a.

  • Now I will put one more source one more source and let me connect it in this fashion; electrically

  • they do not have anything in common except that I am putting one within the other and

  • that is also having the same resistor R and let me call those terminals as b and n 2 and

  • this is e bn2 which also has a potential like that and it has a current i b which takes

  • this path along this resistor to end to back again to the other terminal of the source;

  • that is also an AC source.

  • Now I put one more the third so I have one more source which is going to also be supplying

  • a load resistance R and that is also not going to be electrically connected to the other

  • two sources and it has two terminals which I am going to name it as c and n 3 and the

  • voltage source itself is named between the terminals given the terminals names e c n

  • 3 and there is going to be a voltage and a current i c.

  • So you see there are three sources they are not at all linked electrically, they are absolutely

  • independent of each other each of them supplying their own loads each of them supplying their

  • own loads.

  • Now suppose let us apply some constraints. So what are the constrains? Let us make the

  • amplitudes of the voltages same for all the three. So what is the first constraint; we

  • make we make not the instantaneous mind you, we are making the peak amplitudes peak amplitudes

  • or peak values or the sign waves of each of these AC sources sinusoidal sources same for

  • all three that is e an1 this is n 1 e an1 peak will be equal to e bn2 peak will be equal to e cn3 peak

  • the peak values of all the three sinusoids will be the same that is the first constraint.

  • And then, of course we have made R the impedances of all the three resistive and all the three

  • are same, the loads that all the three see are the same.

  • Now we apply the second constraint. Each of the sources have a phase shift with respect

  • to other of 120 degrees phase shift with respect to each other. Each of these sinusoidal sources has a phase shift

  • of 120 degrees with respect to each other that is the second and the most important

  • constraint that we are going to apply.

  • So, if such a thing is applied what happens to the phaser diagram?

  • You see; now let me have E an1 rms value of the E an1 source is being taken as the reference. Now let another voltage

  • and let me call it let us say E bn2 lag E an1; note that the rms amplitude is the same

  • because the peaks are same the rms is the same is lagging by 120 degrees. The E b source

  • voltage is lagging the e a source voltage by 120 degrees. Now let me have the E c source voltage E cn3 lagging E a source

  • voltage by 240 degrees then the second constraint gets established. what does the second constraint

  • say; it says there should be a 120 degrees phase shift phase shift with respect to each

  • other that is all these three sources.

  • So we have 120 degrees phase shift between E an and E bn, 240 degrees phase shift between

  • E an and E cn on the negative side but this means that E cn is leading E an by 120 degrees

  • and between and between E b and E c because it is 240 with respect to E an and 120 with

  • respect to E a n for e b this becomes also 120. So you see that E c is 120 degrees displaced

  • from E a and 120 degrees displaced from E b, E b is 120 degrees displaced by E c and

  • 120 degrees displaced with respect to E an this satisfies the second condition.

  • The first condition of course being the amplitudes are all same the phaser amplitudes and the

  • second condition is they are equally displaced with respect to each other in a circle and

  • that is 120 degrees 360 by 3 which is 120 degrees with respect to each other so this

  • is the second constraint.

  • Now this is the nature of the sinusoids that will be applied; so which means that though

  • electrically they are not connected we have applied a constraint on the three source voltages

  • which is equal in amplitude equivalent in effect effective amplitudes or the peak amplitudes

  • and the second is the phase displacement between each other should be equivalent which is 120

  • degrees, this also implies that we have the effective values of all three equal the peak

  • and the effective values.

  • So now we can make some conclusions here.

  • Now all the return paths of all these things can be clubbed together because there is nothing

  • here in this no other components which comes but just plain wires, conductors, so let us

  • say we remove all these conductors and club them together like this and join them. So what have we done? The six

  • wire system has now become 1 2 3 and a 4 a four wire system. From a six wire system it