 Welcome to class five on topics in power electronics and distributed generation. So, we have been discussing models of components of the distribution system. We have looked at the transformer model, the line, then DG models and then we are looking at models of what go in to protection equipment, fuse, reclosers, sectionalizers. We are starting to look at the circuit breaker, the inverse current characteristic. So, if you look at the circuit breaker, you have equations that actually model both it is tripping characteristic or it is a reset action. So, it trips whenever it is current flowing through the breaker or over current relay is greater than the pickup current. So, when the ratio m, when the ratio m becomes larger than 1 it initiates tripping action and the time required to trip depends on the value of this ratio m. So, the larger the ratio m is the shorter it requires to trip. The parameters A, B, P, T, R, E are constants based on the type of characteristics that it is following the breaker or the relay is following. So, for m greater than 1 it trips and for m less than 1 it resets. So, if you look at the parameters A, B, P, etcetera you can actually classify your over current protection devices as moderately inverse, very inverse or extremely inverse, inverse characteristics. Good reference for this is IEEE. So, if you look at the characteristics of the relay, if you for the extreme and the very inverse type of characteristics the value of P in the equation is close to 2. So, if you look at a large value of m much greater than 1 and the value of B which is close to 0 we have your trip time is given by A by m square minus 1 and for m being much larger than 1 this is roughly equal to A by m square and we know that m is I by I pick up. So, you have A into I pick up square is equal to I square into T, T, R. So, essentially what you get for P is equal to 2 is a characteristic of I square T is a constant which is similar to that of a fuse characteristic. So, essentially in a fuse your I square R T represents the energy that is dissipated in the fuse. So, you end up requiring time depending on the amount of energy that is deposited in the fuse. So, this sort of characteristic with P is equal to 2 for small values of B represents that of a is similar to that of a fuse. If you look at the B and the previous equation the B term over here this represents the definite time aspect of the IDMT characteristic. So, if you look at having a positive value of B. So, if you look at the B it essentially shifts the characteristic to a finite value which says that no matter how large your current is it needs a certain delay before the actual protective device can act. So, essentially if you look at this sort of characteristic it can allow a large amount of current to flow for a short time without tripping the breaker, but if that is large amount of current continues for much longer then the breaker would trip. So, this type of characteristic is useful when you have say components such as motors where your motors starting current might be large which means that you will be operating at M which is large, but then once the motors feeds up then the current would come down. So, you can actually operate your device without causing your breaker to trip when you are starting the machine. If you look at the reset time of the circuit breaker the reset time is also dependent on the ratio of current, but typically if you look at your nominal current that is flowing through your circuit it might be much lesser than your pickup current. So, you can take M in that particular case under normal condition to be operating to be much smaller than 1 which means that T R E can be taken as roughly equal to capital T R E for M close to 0. So, also for simplicity we can assume that the rates at which your reset or your trip action is occurring is at constant trip rates or reset rates. So, depending on the time required for tripping it is going from a point of normal condition to a trip condition and if the fault current goes away for some reason then it is going at a constant rate back to the normal condition. So, if you look at example of say a electromechanical type of relay or a circuit breaker essentially tripping action in such a breaker occurs when your electromagnetic torque generated by the coil for tripping exceeds the restraining torque of your actual spring or the spring in the relay. So, whenever your generated torque is exceeding your protection disc moves forward and it reaches a point where it would actually cause a circuit breaker to act. So, these sort of actions that I have just mentioned also emulates not just of a fuse, but also of a electromechanical type of protective device. So, if you look at an example of say a circuit breaker where we will assume that it is a reset time is of the order of say 5 seconds and it is having over current level IOC and at that particular value of IOC with your parameters of the breaker suppose you had a trip time of say 2 seconds. So, if you had a 2 seconds of trip time then we can look at a couple of situations say this could be a situation where you had a source you had a circuit breaker and you have current flowing in the line and that is what is shown in the plot below where at some particular point your current went up and you have a over current which caused your M to have a value much greater than 1 and at that particular value of over current your trip time is 2 seconds. But for some reason say if your over current lasted say for 1.99 seconds and the over current went away it means that the breaker nearly tripped, but it did not actually trip. So, it is still staying close at this point and over here it is actually having a reset action and suppose this duration is 1 second then during this particular 1 second the circuit breaker was going through a reset action. So, to fully reset it would have required 5 seconds. So, in 1 second it would have reset by 1 fifth of the reset value. So, it would be at this particular point it would have been at 4 fifths of the value to actually trip the actual tripping device. So, the question is now if your over current again came up it will not need to the same value of IOC it will still not need another 2 seconds to trip, but it will need a smaller amount of time to trip and the amount of time that it would need to trip is about it would have typically required 2 seconds, but you have only one fifth of the distance for the tripping action to occur. So, it would need about 0.4 seconds for it to trip. So, you can see that assuming constant say tripping rates and reset rates you can actually look at what happens when currents go up and low in some switched manner. So, if you had another situation where say you had in a similar level of over current and the over current lasted for say 1.99 seconds and then the current went away or came back to a level much lesser than 1 for say 5 seconds it means that at this point now your circuit breaker is fully reset. Now, if your over current level went up to a higher level it would need actually full 2 seconds to actually trip again. So, depending on your calculated trip time and your reset time you could actually evaluate for what duration your circuit breaker would need to actually operate with devices such as reclosers where your current level can actually potentially go high and low. Again this analysis of assuming say constant trip rates and reset rates is a simplified analysis. So, if you look at what we are doing we are trying to look at what are the implications of adding a distributed generator to a distribution system rather than actually doing a precise analysis on how to set protection levels. In fact, today there is a lot of software which is available which can be used to actually do precise coordination calculations, but this simplified analysis can give you a good feel of what are the implications of adding of coordinating protective devices and the implications of adding a distributed generation source to your actual system. So, now before we actually do a protection and coordination based on these models what we have to do next is actually evaluate what the fault current levels are. And to actually calculate your fault current level it depends on where the fault is occurring whether it is occurring at a zone that is closer to the substation or zone midway through the feeder or at the distribution transformer at the consumption point. It depends on the type of fault whether it is a three phase fault or a single phase fault, the fault impedance whether it is a dead shot or it is a impedance fault. So, the variety of such aspects that need to be considered when you are doing fault calculations. So, here you are looking at whether it is three phase single phase different types of combinations line to line etcetera then fault levels. So, to do these calculations what the tool that the analysis tool that is used is what you would have studied in a power systems analysis course one is per unit analysis. So, where you look at the different components on a normalized basis and then evaluate what the fault currents are and to look at situations of unbalanced faults you will do essentially a sequence models. So, this both aspects are important I mean if you look at a three phase type of fault it is easy from the calculation perspective, but as we discussed in previous class the most common type of fault is single line to ground fault. So, looking at different situations with sequence models become important. So, if you do a per unit type of calculation first you need to identify all the components in your system then even if the components are all provided to you on a per unit basis they might be on different base you might have items which are closer to substation at higher power levels at a higher voltage levels the systems closer to the consumption point would be at lower power levels lower voltage levels. And to do the analysis you need to bring them to a common base and you identify the location of interest to you it might be your consumption point where you are doing your calculation or some distribution transformer where you are doing your calculation depending on the region where you are trying to do your engineering analysis you bring it to the common base associated typically with that particular point. Then you transform all the values to this particular common base corresponding to this common point and essentially when you do that with transformers essentially your transformers become transparent on a per unit type basis. So, all your voltage levels and your transformers are transparent and you bring in all the impedances for the different locations at from different voltage levels to a consistent per unit per unit ice bases. I mean you will get a better feel for it when we look at the actual an actual example where we can go through and do what calculations we would need to do. If you look at so once you do the analysis you get your results on a per unit ice values. So, what you really need is the physical values that you need for sizing a circuit breaker or setting your relay. So, you make use of your base quantities to get back your actual values for your currents voltages etcetera. If you look at the use of per unit analysis compared to the past the use is reduced today because of again extensive availability of software to do these calculations. So, all the normalization etcetera can be handled at a numerical level rather than you having to handle per unitization etcetera. However, there are many applications where it gives you an intuitive feel for does it make sense if you do the calculations on a per unit ice bases. Also when you look at the power electronics design even though you do not need everything on a per unit ice bases for your controller design per unitization is a important aspect because often you are implementing your control on a fixed point processor. So, you need to actually normalize it to the fixed point floating point processors are more expensive than fixed point. And even if you are using powerful processors you need to do some level of normalization especially when you are interfacing A to D converters which have fixed voltage ranges and finite number of quantization levels etcetera D to A or PWM ports again which have finite bit resolution. So, having normalization and doing such an analysis is actually useful in multiple ways. So, if you then look at what the base quantities need to be the common base quantities are the power it can be the apparent power the S base on a if you are looking at distribution system you might be talking about MVA on a three phase bases. Your voltage can be you have to get a base voltage it can either be your kilo volts on a line to neutral bases or a line to line bases. In this course we will look at the voltage on a line to neutral bases, but often it can also be done on a line to line bases. In fact when people state a voltage of a three phase system when they talk about 415 or 690 or 11 k v they are actually referring to a line to line voltage. So, unless otherwise specified people give the physical units of a three phase system the voltage on a line to line bases that is the normal convection for the base quantity we will use the line to neutral voltage as v base. The frequency the is the fundamental frequency in India it is 50 hertz that is the nominal frequency your current base can be defined as your power base divided by three divided by your voltage base on a line to neutral bases. So, this gives your base current in kilo amps. Your impedance is z base is your v base by I base and you could have derived quantities these are useful especially when you are designing power converters when you are looking at filter design etcetera. Your omega base is 2 pi f base your l base can be obtained as your z base divided by omega base and your c base can be obtained as 1 by omega base z base. So, these are useful especially when you are looking at power converters filters etcetera. So, if you have 1 per unit capacitance connected in parallel it means that it is a filter drawing a lot of leading wars. So, you can get a feel for how much wars are being drawn by looking at these components on a per unit bases. So, if your l base is 0.3 it means that this could be a substantial drop across that inductor series connected inductor when rated current is flowing. So, looking at the impedances in per unit bases is useful. So, the next thing is to convert your from one base to another for actually your normalization calculations. One thing that I would like to just point out when you make a statement that 1 per unit equals say 230 volts you are committing multiple errors. One is per unit means it is does not have dimension 230 volts is in volts. So, you are talking about a quantity equation where the dimensions do not match and 1 is not 230 it will be to say 1 per unit corresponds to 230 volts or the base value is 230 volts. So, be careful when you say about quantities on a per unit base. So, if you look at change of base your actual physical quantity the idea of the change of base is that your physical quantity is not changing. If you change your base from one system to another system the per unitized value might change, but your physical quantity stays the same and using that concept you get your per unitized normalized impedance in a new basis to be the old value multiplied by the ratio of old to new base quantities square and the ratio of new power to old power to the power of 1. There are implications of having such an expression for change of basis. So, if you look at if you look at say change of basis 1 percent. So, example 1 percent. So, if you go to a new basis because your ratio was your new is equal to the old on a per unitized basis with the new value divided by old value. So, if you go up on a to a new basis which is at a higher power level your impedance quantity becomes larger and a implication of this is that say for example, if you have a load which is small. So, if you have a system where you have a source and you have some source impedance and you have say multiple loads. So, if you look at your load of 1 M a which on its particular basis it might be a 1 per unit load essentially now when you shift it to the 10 M a level this becomes your load resistance is now 10 per unit rather than 1 per unit once you bring it to a common basis. So, essentially what it means is the small individual loads becomes seen as smaller and smaller entities which is the reason why you might lump a lot of loads together and see it as a single load power p q load etcetera on at the consumption point rather than look at individual loads when you are looking at larger and larger systems to which it is connected. If you look at another example where you look at the implication of say change in voltage level. So, if you have say a load which is having a power level of 40 kilowatts and single phase and you have say two situations. So, case a you have a 1 kilometer overhead line at 400 volts and the impedance of the line is say 1 ohm. So, if you look at it we might define your v base as 400 i base as would be 100 amps because p is 40 kilowatts. So, your z base is 4 ohms. So, if you look at now the model of the source with your load your impedance of your source is now 1 ohm by 4 ohms which is your base. So, it has a value of 0.25 per unit. So, looking back you are looking at a weak source where this source impedance is fairly large. Now, if you take the same 1 kilometer line and if you increase the voltage level because of better insulation but you keep say the diameter of the line it is geometry etcetera similar. And say you consider case b where you have 1 kilometer line at 4 k v and say the magnitude of its impedance is again 1 ohm. So, your v base in this case is 4 k v 4000 volts i base is 10 amps z base is 400 ohms. So, if you look at the model of the system in this particular case load b now this particular physical line is now showing a source impedance of 0.0025 which means that if you look at it from the load perspective this is now a stiff source. And this is essentially the when you look at it on a per unit as basis you see that it goes as the square of the voltage. So, you can see that now if you just take the same 1 kilometer line it becomes important to look at it on a normalized basis to figure out whether you are actually encountering a situation where you are having a weak grid or a stiff system. So, for a given power level 1 percent impedance at 1 k v base is 100 percent impedance at 100 volts base. So, as one goes up to higher voltage levels similar impedance will appear actually much smaller. So, if you look at your transmission system your voltage levels tend to be very high. So, looking at it from the load side you can actually get closer to the assumption that your grid is getting more and more ideal when you are looking back into the system. So, if you look at the another example suppose you have now a transformer which is 1 k v 200 volt. So, you are actually having a transformer between going from 1 k v to 100 volt then essentially 1 percent impedance on the primary is 1 percent impedance on the secondary. So, this is because your base on either side of the transformer has to be consistent with your turns ratio of the transformer. So, essentially when you are handling transformers it becomes transparent in your per unitized analysis for this particular reason. So, with this we will actually look at an example of a system and look at how we could make use of what we have just covered to calculate what the ratings of some components such as fuses and circuit breakers could be on a system. So, we are looking at say a system coming in from a substation you have a substation grounded 10 m v a 11 k v feed are coming from the secondary side of the transformer at the substation you have impedance of the line. So, this is the distribution line feeder impedance and say you are adding a new facility on this particular feeder a fairly large facility with the interconnection transformer at 2 m v a. So, it is a fairly substantial facility and at such power levels may be you might consider say this particular point over here as the point of common coupling essentially the term point of common coupling is used as the point at which you connect with the public system. For example, if you look at a house the point of common coupling for the house into your street for your transportation coming in and going out is your gate. So, essentially in the distribution systems you can have different points of common coupling depending on where the responsibility of the utility ends and where your responsibility starts. So, if you take a home it might it would be at 230 volts or 450 volts after the meter because the meter is actually owned by the utility you do not go and make any changes to the meter your responsibility is whatever is downstream of the meter. So, depending on the size of the facility you might for large ones you might put your meter all the way up at 11 k v for smaller homes it would be at the consumption level. And here you say want to add a fairly large facility and it is rated at 2 m v a and you want to actually say size the protection components on this particular system. And for which you need to know the fault current levels one common way of explaining what the fault current level could be is in terms of your short circuit capacity. So, when someone says it is 120 m v a so that represents the product of the fault current and the nominal voltage of the system. And to evaluate what the short circuit capacity at that particular point is and the utilities can actually provided to you especially when you are connecting large systems this is be a important aspect of the information that you would need for sizing the interconnection. So, you have a fuse then you have a distribution transformer going from 11 k v to 415 volts it is impedance is 4 percent reactance and winding resistance of 1 percent is a delta y. And you have a circuit breaker at the secondary of the transformer and you have the internal winding within the facility. And you have a distribution burst before that and you have different circuit breakers feeding each load from your distribution burst within your facility. And say you have one particular large load which is the wiring to it is rated for 1 m v a with a given inductance and resistance impedance reactance and resistance. And then you connected to a load through another particular breaker to actually may be protect that load. So, the question is can we get a feel for what should be the values of the ratings for say f 1, c b 1, c b 2 and c b 3 in such a system. So, the first step is to evaluate the short circuit current at your PCC based on what is the equivalent impedance facing back. So, the impedance facing back from the PCC is this feeder distribution line impedance you might have the substation transformer impedance you might have the primary side of your substation transformer which is at a higher voltage which gets reflected back. So, you are looking at the overall impedance facing back. So, that can be calculated from this short circuit capacity at this particular point. So, if you look at the short circuit the source. So, the first thing is to look at the source side this source side impedance would consist of the feeder. And you are told its short circuit capacity is 120 m v a. So, your short circuit current is 120 m p a divided by 3 divided by 11 k v divided by root 3. So, this gives 6.3 kilo amps as your short circuit current level. So, if you look at your source you would it is v by i s c and it is 11 k v divided by root 3. And you are told that it is at this current is at 60 degree short circuit m v a at 60 lag. So, e to the power of minus j pi by 3. So, you can calculate the numbers this turns out to be 0.5 plus j 0.87 amps. So, you physical value of this impedance that this impedance now represents this total impedance as seen from the point of common coupling. And if you want to say look at it what does this mean on a per unit basis you know your source is 10 m v a source by source I mean the at the substation this is 10 m v a. So, your v b is 11 root 3 k v i base and your z base is your v b by i b which is 12.1 ohms. So, if you look at your value of your r s this would be 0.5 divided by 12.1 into 100 to give it in percent. So, this would be 4.2 percent and your x s is 0.87 which we calculated divided by 12.1 in percentage this would be 7.2 percent. So, now we could look at the fuse the fuse in this particular system should be rated assuming that you could have a fault right here and the primary of the transformer. So, the only element that is limiting the fault current level is the source impedance. So, that would determine what is the fault current level that would flow through the fuse. So, and the load of the fuse is a 2 m v a transformer. So, you know based on your 2 m v a rating that your nominal current for the fuse is the nominal current see that the fuse would see is 100 amps and your short circuit current from your previous calculation is 6.3 kilo amps r m s or 9 kilo amps peak. So, that gives you two of the quantities that you would need for actually selecting the fuse you know that your voltage rating your nominal operation is 11 k v your isolation voltage has to be much larger than that if you take it as twice that you are saying greater than 22 k v. So, this would help you in selecting the fuse. So, you look at the fuse in terms of the nominal current it would it needs to carry and also your peak current that it would need to interrupt if there is a fault on the distribution transformer on the primary side. So, the next element that we could look at is then the distribution transformer itself. So, once you look at the distribution transformer you could actually then calculate what is the fault current that could occur in your breakers C b 1 through the C b 5 its primary current is 1 kilo amp 0.1 kilo amp which we just calculated and the secondary current can be calculated with its turns ratio this is about 2.8 kilo amps. So, if you look at then your base quantities for this particular transformer it depends on whether you are looking at it from your 11 k v side or whether you are looking at it from your 415 volt side. So, if you calculate your z base or from your primary side this turns out to be 60.5 ohms and from your 415 volt side this turns out to be. So, when you say your x l is 4 percent and your r w is winding resistance is say 1 percent essentially if you are looking at it from the high voltage side this would correspond to 2.42 ohms on the low voltage side this would be 3.4 milli ohms and your r w would be 0.61 ohms and on the low voltage side you would have 0.86 milli ohms. So, to calculate now your fault current within the facility you need to combine both the per unit quantities of your distribution transformer and whatever you have from the source side which is the upstream feeder etcetera. And what we will do in the next class is combine it to the voltage level of the transformer because may be you are more interested in designing the particular facility and look at the components that go into the facility. So, we will make use of that in the next class to actually continue with this particular example of how to do the calculations and make use of the per unitized calculations for sizing your components protection components. Thank you.