 Hello and welcome to this lecture on advanced electric drives. In the last lecture, we are discussing about the traction drives and more specifically the speed time curve of a traction drive. Now, we have seen that when a traction drive or a locomotive starts from rest, it goes to the various modes of operation. It starts and then goes to the various modes and then come to a halt or a stop. And the various modes are as follows. We are seeing this diagram in the last lecture. We will just see it once again. So, we have the constant power region, constant torque region that is from this point to this point, which can also be known as constant acceleration, because during the constant torque region, the acceleration remains constant. And then we have the constant power region. This is the constant power region from this point to this point. And in this region, the power remains constant. And then what we have is a free running region. So, this is a free running region, where the motor speed is constant and the vehicle speed is constant. That is the free running region. The vehicle is running at constant speed or the locomotive is running at constant speed. That is the free running period. And then before we stop, the power is switched off to the motor and that is called coasting region. The motor speed reduces. So, this is basically the coasting region. The motor speed reduces. And then before we come to stop, we apply break. And this could be electrical braking or mechanical braking, especially in electric drives. We prefer to apply electrical braking, so that there will be minimal wear and tear of the mechanical part. So, we break the motor electrically. And the kinetic energy of the whole locomotive is converted to electrical energy, which could be fed back to the supply in case of a regenerative braking mechanism. So, this is basically the speed variation. We have constant torque or constant acceleration. Then we have constant power, free running, coasting and braking. What about the torque variation? The torque variation as follows. During the constant torque region, the torque remains constant. And then when we go for constant power region, the torque decreases. Because here, what we have in this case, during this region, the torque into omega is constant. That is the power. So, we have constant power. T into omega is constant. And since the speed is increasing or this omega m is increasing, torque has to reduce to keep this T omega m constant. Omega m is a mechanical speed. So, T into omega m is kept constant. And when the speed increases, omega m increases, torque has to reduce. Then we come to the free running region. This is the free running region. And here, the torque requirement is minimal because the locomotive is running at constant speed. There is no need of acceleration. So, it has to only cater for the friction losses, the internal friction and also the external friction. So, the torque requirement is minimal there. So, the torque is minimum, but constant. So, that basically the constant torque. And during coasting, what we do? We switch off the power. So, the torque becomes equal to 0. And the motor runs because of the stored energy in the kinetic energy in the mechanical system. So, the vehicle has got a large inertia, it stores energy. And during the coasting region, this kinetic energy stored in the inertia of the vehicle caters to the motion. So, we have 0 internal torque. The motor torque 0 is 0. The vehicle speed decreases. And then we go for braking because we have to stop in a scheduled period. So, what we do here? We apply mechanical break in this case. And it means the torque becomes negative. So, we apply a negative torque here. And hence, there is deceleration, there is braking. We can say it is constant deceleration. And hence, the motor come to a stop at t 5. Now, if we see the power, the power diagram here. Here, we have constant torque and speed is increasing. So, we have power increasing here. And at this point, it is the base speed. And maximum possible power is raised. We cannot go beyond the rating of the motor. So, the power remains constant. So, the power here remains constant. And then, at the maximum possible value, that is the rated power. And then, when we go for the free running, the power requirement becomes really low because the torque is also low. So, we have reduced power here. And then, when the coasting starts, the power becomes equal to 0 because the motor is not giving any torque. It is basically coasting. The electric supply to the motor is switched off. So, the machine is running or the motor is running, the vehicle is running because of the inertia. And hence, the power requirement is 0 here. And then, when we start the braking process, we apply electrical break. And here, initially, the speed is quite significant. And we have large initial power. This is electrical power. And then, gradually, the power decreases. And when the motor comes to a stop, the power becomes equal to 0. So, these are basically the variation of the speed, torque and the power of a motor when the vehicle is moving from one place to the other place. Now, we would like to see how does the load torque get shared between the various motors. Ultimately, when we have a locomotive, it is basically powered by electric motors. And we cannot have a single motor because the power requirement is quite large, something like 6000 hp or close to 4.5 megawatt. So, what we do? We have multiple motors to supply the various axles of the locomotive. We could have 5 motors, 6 motors, 10 motors. And all these motors are attached to the axle. And the axle is connected to the wheel and these motors drive the vehicle. We can see a diagram which will be clear. Say, for example, if we have these wheels, these are the wheels here. This is one pair of wheel. And we have the axle which connects this wheel. And the wheels are running on the track. So, we have the track here. And where does the motor seat? The motor actually seats on the axle. The driving motor seats on the axle. So, we have the motor here, which is basically it seats on the axle. A mounting is on the axle. And then we have the shaft output of the motor. And then we have the axle. And these are the two gears. We have the motor gears and we have the axle gears. And these two gears help transmit the torque on the motor to the axle. So, this is this is the electrical motor that we have. And this is the axle. And these are the wheels, another wheel. Now, we can have multiple axles. We can have one more axle. So, if suppose we have a choice of motors. So, for example, let us talk about DC motors. We have the choice of using either a separately excited motor or a series motor. Now, if we use a separately excited motor, how is the load torque shared? Now, let us see the torque speed characteristic of a separately excited motor. So, we have the speed in the y axis and torque in the x axis. Now, this is for separately excited DC motor. So, the torque speed characteristic is drooping like this. So, it is a straight line, but it is drooping in this following fashion. Now, on the other hand we have a choice of using a series motor. Now, if we use a series DC motor, the characteristic is something like a hyperbola. At very low torque we have very high speed and at full torque we have low speed as follows. So, if we plot the speed versus torque for a separately or for a series DC motor, this is series DC motor. The origin here and we have the speed in the y axis and torque in the x axis and this is the nature of the torque peak artistic. So, at low speed we have a high torque. So, it means the starting torque is very high for a series motor the starting torque is very high and when we have full speed the torque is very low as we see here. When we increase the speed here the torque developed by the motor decreases. Now, if we see which one shall be choose, we have this wheels and we have multiple wheels. We have another wheel here. Similarly, we will have an axle here connecting these two wheels and we have the corresponding motor. Now, the wheels are supposed to run at constant speed. They run at constant speed provided the wheel diameter is the same, but unfortunately the diameter of the wheels are not same because of wear and tear. One wheel may run at a different speed, very minor different speed than the other wheel. So, in that situation the two motors will not be running at the same speed. Suppose, let us take the first situation separately excited DC motor and we are supplying this from the same converter. So, say for example, say the motor number 1 is running at N 1 speed this is N 1 and motor 2 is running at little over speed which is N 2. So, this speed variation is very minor, but no doubt there is a variation of the speed. So, N 1 and N 2. So, N 1 is higher than N 2 and N 1 minus N 2 is equal to delta N is a change in speed. So, if they are on at two different speeds how is the load torque said? Now, we see that the first motor N 1 having a higher speed is having a torque corresponding to T 1 and then the load speed N 2 will have a torque and that torque is T 2. So, N 1 torque is T 1 and N 2 torque is T 2 and there is a large difference between T 1 and T 2. So, it means the load is say at unequally it means the motor which is having higher speed is sharing a lower torque and the motor which is having a lower speed is having a higher torque that is not desirable. The two motors are having the same rating let us say at the same load torque. Now, on the other hand if you go for a series DC motor suppose we have the same situation there we have N 1 here and then we have N 2 N 2 is less than N 1 and the corresponding torque will be T 1 and the corresponding torque for N 2 will be T 2 and here we see that T 1 is approximately equal to T 2 very close to each other. So, it means although there is a minor speed variation between the two motors for a series DC motor the torque sharing are almost equal and that is why the series motors are sometimes used for traction applications. It means if you have multiple series motors the sharing of the torque will be almost equal, but if we have already understood the coefficient of adhesion for the coefficient of adhesion it is better to have a motor with a steep torque speed characteristic it means the speed regulation has to be low. So, the separately excited DC motor are good for coefficient of adhesion, but not good for load sharing on the other hand the series DC motor are good for load sharing, but not good for coefficient of adhesion. So, we can conclude here in this case good for coefficient of adhesion because adhesion will be good in this case the possibility of wheel slip will be less, but not good for load sharing. On the other hand the series DC motor we can say that these are not good for coefficient of adhesion, but good for load sharing. So, we have a choice. So, each one is having it cross and cons. So, sometimes we select depending upon the application we select separately excited DC motor sometimes we go for series DC motor. Now, let us see the type of connection if you see the type of connection if you have multiple motors the motors can be connected in series or can be connected in parallel which one is better. Now, say for example here we have two motors connected in series motor connection. Say we take let us say series motor. So, we have two motors which are connected in series this is one motor which is series filled then we have other motor with series filled and they are connected in series. And we have we have applied voltage here that is V this is series connection and then we can have another situation where the two motors are connected in parallel. So, these are the traction motors they could be connected in parallel because we have multiple motors. So, we apply some voltage here this is V 1 and this V 2 of course the voltage will be different here because the series voltage will be higher than the parallel voltage. And here we have the parallel connection. Now, if we connect two motors in series suppose due to the bad patch of the track one patch of the track is bad and one motor axle slips one wheel slips. So, if that wheel slips because of the bad patch of the track the speed increases. Wheel slipping means the wheel is running at higher speed not able to have the grip on the track. So, it means the speed increases suppose this motor speed will increase this slips. Now, if this motor speed increases torque will decrease because we know that the torque peak characteristic is like this for series motor this is speed and this is torque. So, if speed increases the torque is going to reduce this is the torque we have in success. So, if the torque increases I mean the speed increases torque will reduce. And if the torque reduces the current through this series combination will also reduce. So, when the slip the motor slips the speed increases the current will come down and the current will come down the torque will reduce. And the torque is going to reduce in both the motors because both the motors as it is connected. And hence when the wheel slips there will be net reduction in the torque. And what about the coefficient of adhesion? Now, the coefficient of adhesion if you if we have seen the definition the definition of C A is maximum tractive effort that can be applied without wheel slippage divided by weight on the driving wheel. C A as per the definition is basically ratio the maximum tractive effort that can be applied without the wheel slippage divided by the weight on the driving wheel. Now, if the wheel slips both the torque are going to decrease and hence we have a net reduction in the torque. And if we have a net reduction in the torque we have a lower value of coefficient of adhesion for this connection. So, we can say that for series connection C A is less. Now, on the other hand if we if you go for the parallel connection we have two motors in parallel. Suppose this wheel slips. So, if this slips only this current is going to decrease and this torque is going to decrease. So, this current and this torque of this of this motor which is which is slipping that torque is going to reduce. And other motor which is connected in parallel does not see any change. So, here effectively we have a higher torque. So, as per the definition the tractive effort which can be applied without wheel slippage divided by weight on the driving wheel here the coefficient of adhesion will be more. So, here the adhesion is better. So, we can say C A is more here. So, for this combination the adhesion is better we can say that here the adhesion is better compared to a series connection. So, we have we have the choice usually when you have the motors we can connect them in series or in parallel. So, we have a choice whether to go for series connection or parallel connection. Now, what is the drawback of parallel connection? Now, if we see the parallel connection this current is more the current which is drawn from the supply is more. Suppose, this is i source and this is i 1 and this is i 2 let us say. So, i s is going to be equal to i 1 plus i 2. So, the current requirement from the converter increases. On the other hand in this case the source current is same as i and i is the current flowing in the motor. So, we have the same current. So, if the motor is rated for 100 amperes the 100 amperes will be drawn from the source. Here we have a reduced current requirement from the source. So, we will have a choice between the two or we can go for a trade off. So, we can connect some motors in series and then the combination of the series in parallel to have optimized connection. So, with this background let us try to see how can we analyze the speed time curve in a better way and how to calculate the distance covered by the train and also the drive rating. So, we will again look at the speed time curve of a locomotive. This is traveled by a locomotive. So, we will again see the speed time curve. So, here we have the speed in the y axis and time t in the x axis. So, we go through this region constant acceleration or constant torque then constant power then we have the free running period then we have the coasting and then braking. So, this is a curve which is very well known, but to analyze this we will make some approximation. The approximation is as follows this can be approximated to be a trapezoid. So, if you approximate this to be a trapezoid the calculation will be simple. So, what we do here is the following we extend this and then we take a equivalent speed in this case and again we extend this. So, we will consider this to a trapezoid. So, the trapezoid is a b is this point now c is this point and d is here. And we have the various times here the acceleration is constant. So, we call that to be t 1. So, what we have here is a following now this time is a time for acceleration approximately. So, we call that to be t 1 that the duration of for which it accelerates. And then the free running time and we call that to be t 2 and then we have the braking time and we define the acceleration as alpha and deceleration as beta. So, this is our t 1. So, we call this to be constant acceleration and then we have braking in this case this is the braking region this is the free running period and this is coasting. So, here what we do this the area under this curve will give us the distance traveled. And we define the following variables and the variables are as follows this is the maximum velocity that we have here v m. And we have so drawn this curve. So, that here we have a little bit increase of area in this case. And then here we have some reduction in the area I mean increase of area here. So, we have increase of area here and increase of area here that is compensated by a reduced v m. v m is not actual v m v m is little less than actual v m because we are trying to approximate the curve to be a trapezoid. And hence we are equalizing the area of the actual curve and that of a trapezoid. So, we have the various variables in this particular graph as follows. So, d is the distance covered in kilometer distance covered between a and d in kilometer. And capital T is the time taken to move from a to d in second here we have a b c and d. So, we are starting from a and we are ending at d. So, this is basically the capital T is the time taken to move from a to d in second. And then we have v m is a free running speed in kilometer per hour and alpha and beta are acceleration or deceleration in kilometer per hour per second. So, acceleration is constant acceleration. So, we can say that acceleration is alpha here and breaking or deceleration and the rate of deceleration is beta. So, if we have seen this graph now the objective is to calculate the distance. So, the distance is found out as follows. So, the distance covered is the area under this curve. Now, we can find out this area as follows we have we will give this name again as e and f. So, the area under this curve will be the area of the triangles a b e plus the area b c f e plus the area f c d. So, we will we will calculate this as follows. So, this area the area a b e this area will be half the maximum speed is v m into t 1 and t 1 are in second. So, that is t 1 by if you convert this second into hour. So, we have 3600 what about area of this triangle we have the area is half v m is the speed and this is t 3 and we have to convert this into hour 3600 and then we have we have t 2 here. And since we know the acceleration we can find out what is the distance covered. So, we will we will calculate as follows. So, we can do this calculation d is equal to area of the trapezoidal curve that is equal to we will take this v m outside and this is half t 1 plus t 2 plus half t 3. So, if we if we see here this area the area of this rectangle is basically v 3 v m into t 3. So, we have v m is the maximum speed here into t 3. So, that is what we have we have shown here this is t 2. So, we have t 2 here. So, v m into t 2. So, we have shown that here half t 1 plus t 2 plus half t 3. Now, this we can simplify as follows if we simplify this we will have v m by 7200 we will take this 2 out. And then we will have this 2 t minus t 1 plus t 3 because we know that the total time is equal to t 1 plus t 2 plus t 3. So, what we will do from this we will basically eliminate t 2. Now, if we eliminate this t 2 t 2 is given as the capital t minus t 1 plus t 3. So, we will eliminate this t 2 time. So, we can substitute that here. So, v m the v m by 7200 into 2 t minus t 1 plus t 3 and that can be simplified as follows v m by 7200 2 t minus v m by alpha plus v m by beta. So, alpha and beta are the acceleration. So, if you know the acceleration you can find out what is t 1 and what is t 3 because we know to reach that particular speed we have to accelerate that and that is the constant acceleration. So, in fact we can say that v m is equal to alpha into t 1 t 1 is in second and also v m is equal to beta into t 3 that we can substitute here and we have the expression for the distance traveled from a to d. So, we have calculated the distance traveled by the locomotive and this d the unit is in kilometer. So, we can write that this is in kilometer because t 1 are in second. So, we will get the expression of the d in kilometer. Now, let us see how we can calculate the rating of the motor or the rating of the drive. So, if we see the locomotive there are so many forces coming on the locomotive we will take each force one by one and try to find out the tractive effort which will be required to overcome that particular force. Now, let us see calculation of drive rating and before we do that we have to find out the tractive effort calculation of various tractive efforts. Now, the train is having a mass and it is a huge mass. So, the train has to be accelerated linearly we have equivalent mass of the train that is usually in tons and it has to be accelerated linearly and we need to have some tractive effort for linear acceleration. So, the first one is to accelerate the train horizontally that is equal to f a 1 how is it calculated that is calculated as follows m is the mass in tons. So, we have 1000 m and we have the acceleration. Acceleration is alpha and if we convert that into meter per second square into 1000 by 3600 and that is equal to 277.8 m into alpha and the unit is Newton. So, here m is the mass of the train in ton. So, 1 ton is 1000 kg. So, we convert that into kg and then the acceleration into meter per second square and we found out the effective tractive effort in Newton. Now, this is basically to accelerate the train linearly and if we see the train has got so many rotating parts we have the wheels, we have the axles, we have the motors. So, all these rotating parts will also have to accelerate in angular fashion. So, we also have angular acceleration. So, we need to also find out the equivalent tractive effort or equivalent torque which is required to accelerate these rotating parts in angular way. So, the second part of the tractive effort is to accelerate rotating parts. Let us effectively find out the moment of inertia. What is the moment of inertia of the rotating components? We have two rotating components. One is the wheel or there is a motor. Now, suppose we have number of axles equal to n x. So, if n x is the number of axles and j w is the moment of inertia of a wheel. So, how many wheels we have? Each axle will have two wheels. So, if we have n x axles we have two into n x number of wheels. So, the effective inertia for all the wheels will be 2 n x into j w. So, we call that to be j 1. So, j 1 is the moment of inertia of the wheels that is equal to 2 into n x into j w and then let us try to find out the moment of inertia of the motors. Say for example, we have n number of motors and each motor is connected to the wheel by means of a gear and the torque is transmitted from the motor to the wheel. So, there is a transmission mechanism there and we need to find out the effective inertia or the equivalent inertia under this gear condition. So, we have the following parameters with us. n is the number of driving motors and then n 1 is the teeth on on the on the on the motor side gear, n 2 is the teeth on the axle side gear. So, we have we have a ratio that is equal to n 1 by n 2 is a gear ratio that is equal to the wheel speed by the axle speed by the motor speed. So, this is what we have. So, when we refer this total moment of inertia to the wheel side because the motor is not on the wheel motor is actually connected to the axle and it is having its own speed. So, this has to be converted this inertia has has to be referred to the wheel side. Now, since we have a gear ratio that is equal to a the wheel side equivalent moment of inertia of the motors is n into j m j m is the inertia of each motor divided by a square. So, we can we can calculate that. So, moment of inertia of the motors referred to the wheels is equal to j 2 we have this j 2 that is equal to n into j m is the inertia of each motor divided by a square and hence we have we have the total moment of inertia is equal to total moment of inertia that is equal to j 1 plus j 2 and this total moment of inertia has to be accelerated angularly. And we we know the linear acceleration that is alpha alpha the acceleration of the train. Now, this alpha which is which is actually the distance traveled per hour per second or kilometer per hour per second has to be converted to radian per second square or angular acceleration. And to do that we have to have the radius of the wheel because the wheel is rotating and the train is moving forward. So, we have linear acceleration and that has to be converted into angular acceleration. So, if we have the diameter of the wheel is r meters we can convert alpha into angular acceleration as follows. So, what is the linear acceleration? The linear acceleration linear acceleration is equal to alpha is in kilometer per hour per second. So, we have to convert into meter per second square into 1000 divided by 3600 that is meter per second square. And then we have the angular acceleration that is equal to alpha into 1000 divided by 3600 into r that is radian per second square. Now, what is r here? r is the radius of the wheel in meter. So, when we have the angular acceleration we can find out the torque that is necessary for accelerating the rotating parts. And the torque is the moment of inertia into the angular acceleration. So, we can find out that. So, T a 2 is equal to j 1 plus j 2 we have the moment of inertia of the wheel and also of the motor referred to the wheel side into alpha into 1000 divided by 3600 into r. And that is equal to we can calculate this to n x j w plus n j m by a square we are replacing this j 1 and j 2 into alpha by 3.6 r. So, this is basically effective torque and if you want to find out what is the effective tractive effort f a 2, f a 2 and this is basically in Newton meter f a 2 is the tractive effort which has to be applied due to I mean to accelerate the rotating parts that is basically the torque by the radius that is equal to T a 2 by r. So, we can calculate what is the total tractive effort f a is the total tractive effort to accelerate the train horizontally including the rotating parts is f a 1 which we have already evaluated plus f a 2 and f a 2 is obviously the tractive effort for angular acceleration of the rotating parts of the rotating parts. So, f a 2 is important because f a 2 is the component of the tractive effort which has to be spent to accelerate the rotating parts like wheels, the motors and so on. And then we have a linear acceleration also. So, the total tractive effort is the summation of the linear acceleration, the force because of the linear acceleration and the force required for the angular acceleration of the wheel and the motors. Now, if we add these two tractive effort that is f 1 plus f f a 2, we will obtain the final one. So, what about the total tractive effort? So, f a is equal to we have this expression f a 1 plus f a 2 and that is equal to we have the expression to 77.8 m into alpha that is in Newton plus we have the angular acceleration to n x j w is the momentum inertia of the wheel plus n j m by a square the gear ratio square into alpha by 3.6 r again we have 1 by r in this case. And that will be 2.77 into 8 into m e is effective mass into alpha. So, we can simplify that the actual mass of the train is not m, m is just the weight of the train or the mass of the train in k g or in turn. Now, the effective mass includes the rotating part also. So, this effective mass m e is 10 percent higher than the actual mass. So, this is basically for the acceleration which is the horizontal plus angular and then we have other tractive efforts also. So, let us see the other tractive efforts we have the gravity. Now, sometimes the track is actually not linear not horizontal that is having a inclination. So, if you see the track this is basically nature of the track here and here we have the g is the gradient and then this is 1000 meters. So, what we have here because of this we have a tractive effort which is required this is basically the mass of the train or the weight the weight is 1000 m into g. So, we have a component of the force which is opposing the motion it is trying to go forward here and that is basically tractive effort which has to supplied by the locomotive. So, that is evaluated as follows. So, we have f g is the tractive effort because of the gravity that is equal to 1000 m m is the mass in term then 1000 is basically to convert that into k g into g is the gravitational acceleration 8.9.8 1 meter per second square into the g is the gradient by 1000 g is defined as the change in the vertical inclination in 1000 meters. So, effectively what we have here this is basically the sin component if this angle is theta this is 1000 m g into sin theta that is approximated as 1000 m g into g by 1000 and that is equal to m small g capital g Newton or that is equal to m g in k g. So, this is the gravity or sometimes we call this to be gradient. So, in this lecture we have discussed the speed time curve of a locomotive and we have seen how we can calculate the overall distance traveled and we got an expression for the distance traveled by the locomotive from one place to the other place when it goes through constant torque mode constant power free running coasting and braking. And then we tried step by step to evaluate the various tractive effort primarily to include the train to accelerate the train horizontally to accelerate the angular or the rotating parts of the train. And also when the train is overcoming a gradient the track is not a horizontal track it is having a off gradient or a down gradient. So, we found out the tractive effort required to overcome a gradient. So, in the next lecture we will be discussing the other type of tractive effort which are required to overcome the air resistance and we will see how the total rating of the electric motor which is used for driving this train can be calculated. So, these things we will be discussing in the coming lecture.