 Welcome to class 19, topics in power electronics and distributed generation. In this class, we will look at a couple of examples. These are homework problems for the students in class. We will look at one problem on grounding and a second problem on distribution system, equipment, protection and coordination. So, in the first problem, we have a secondary distribution system, which is essentially at the consumption point, essentially 415 volts distribution of length 1 kilometer. We have a couple of major loads on this distribution, which is load 1 and load 2. The source is delta y transformer with the y point solidly grounded. Each grounding is solidly grounded, but the ground resistance to deep earth is about 4 ohms, which is essentially the resistance between the earth electrode and the soil to deep earth. You have parasitic capacitance to ground for the lines and that is 2 micro farads per kilometer. You have a DG now connected at load 2. Potentially, it could be for providing power or for power quality reasons. You have added a DG and that is grounded through a resistor R DG. So, if you look at this particular example, this is essentially if you ignore the DG for now. This is essentially a T T secondary distribution, where the source is grounded directly to tera earth. Each load has its own grounding electrode and load 1 and load 2 is connected. The earth, the ground wires are connected to earth, but if you look at the DG, the DG is connected at load 2 with a resistance. So, overall you can see that this is a system, which has both T T aspects to it and also high resistance grounding aspect to it. So, in a actual system, it is actually possible to have multiple grounding options present in the same system depending on what exactly your objectives are. So, the first problem is to look at what happens if you have a single line to ground fault in load 1 and what is the resulting fault current seen at circuit breaker 1 and circuit breaker 3, through which the DG is interconnected and you want to see whether there is any increase in fault current due to the addition of the DG. .. So, if you look at this problem, you have the system parameters. So, you have the source, which is 1 MVA 415 volt. So, its base impedance would be 172 milli ohms and we are told that the leakage inductance of the transformer is 4 percent. So, this corresponds to 0.04 into 1.722 or 6.89 milli ohms. So, 7 milli ohms is the impedance from the reactance of the transformer. If you look at the DG, the Z base for the DG is 200 kVA DG. So, its base resistance is base impedance is 0.86 ohms and you have the effective leakage of the DG being about 10 percent. So, that corresponds to 86 milli ohms. So, next we could look at the case for the first case we look at the situation when there is no DG connected. So, if you have a single line to ground fault, your equivalent sequence network would be and your source is grounded through and you have the earth impedance coming in between which is 4 ohms. So, 3 times 4 would be 12 ohms seen from the source side and when you have a fault occurring within load 1, essentially the path for the fault current is through this electrode through the system out into load 1 into the frame of load 1 connected through the grounding wires into the ground electrode that would also have a 4 ohm resistance equivalent grounding resistance. So, you could calculate your fault current we are neglecting the 7 milli ohms which is much lesser than the 12 ohms. So, you have a fault current of about 30 amps. So, then next case is when you have the DG also connected. So, when you have the DG connected you now have your sequence network which would have 2 sources one is your original source and you have the DG and if you look at the zero sequence network you have the grounding resistance of the DG which is 1500 ohms. So, in your sequence network you would have 3 times that which is 4500 ohms plus the grounding resistance of 12 ohms at load 1 and you have 4 ohms at the source which shows up as 12 ohms equivalent and you have I f by 3 flowing through the circuit and again you can see that the milli ohms can be neglected compared to the 12 ohms and 4500 ohms. So, I f by 3 which is the parallel combination of these 2 resistance circuits. So, you get I f equal to 30.04 amps which means that the increase in current seen because of the DG increase in fault current seen because of the DG is about 40 milli ohms. So, you can say that there is almost no change in the current level because now you have added the DG for a single line to ground fault. So, the question is to figure out what is the current seen by breaker 1 and 3 and so to calculate that we will have to look at your sequence quantities. So, if you look at your current flowing through the circuit you have this would be I a s 1 in circuit breaker 1 and this would be your I a in circuit breaker 3. So, similarly you can find the sequence currents in the branches of this particular circuit. So, you have the overall current. So, you have to find out how the current splits between these 2 branches. So, you could find out the share of current for the positive sequence, negative sequence and zero sequence. So, you could do that calculation. So, you get I a s 1 positive sequence through breaker 1 is. So, here you have 2 parallel reactance of 86 milli ohms and 7 milli ohms. So, the breaker would be 9.26 amps I a negative through C b 1 would be the same a similar calculation you will get 9.26 amps I a zero sequence through C b 1 would be. So, in case of the split over here again 12 ohms and 4500 and 512 ohms is much larger than the 7 and 86. So, you could take your current split as. So, you can then calculate your I a s 1 by going from the sequence to phase transformation. So, it is I a plus plus I a 0. So, this is 28.5 amps flowing through phase a of circuit breaker 1. So, to calculate the current flowing through phase a of circuit breaker 3 you have. So, if you then calculate what your I a through C b 3 is this is the sum of these 3 currents. So, you have about 1.5 amps flowing through the circuit. So, at this point you have calculated what would be the current flowing through C b 1 and C b 3 because of a fault in load L 1. So, the next problem is to see what would be the touch potential for a person say standing at load 1 or load 2. So, if some person is touching some electrical equipment whether that person would feel a shock when such a fault has happened. So, the first thing we look at is touch potential without the D g. So, at L 1 this would be the fault current going into the fault times your z of your grounding which is 30 amps into 4. So, 120 volts is what would be seen by a person at load 1. So, you have a fault over here and you have a person say touching the frame then because of the resistance between your ground wire and the earth electrode this frame gets elevated with respect to ground. So, the person touching at would see a shock of about 120 volts. If you have another person now at second location touching this particular frame with respect to the local earth over there there is no current flowing in which. So, the person would not see a shock at load 2. So, next we could look at what happens when the D g is there we saw that when the D g is there there is actually a small increase in current that flows through the fault. So, with D g the same about 120 volts. Now, if you look at at L 2 you have about if you look at the 0 sequence current it was about 30 milliamps into 12 ohms is what is flowing through the into the ground at load 2 when you had the D g. So, you get about 0.5 volts at the second location. So, you would hardly feel it this is slight increase in potential, but not much. So, the next question is to consider say the situation where because of the fault in load 1 if breaker C b 1 opened then what would be the current flowing through if you want to see whether C b 1 would open you want to look at the current flowing through switch S 1 considering the fact that you have. So, the question is what is the current flowing through the fault after C b 1 opens. So, what is the change in fault current and what is the effect of considering the parasitic capacitance of the of the line when you are doing such a calculation. So, if you look at the parasitic capacitance it is 2 micro farads. So, your x c is 1 by 2 micro farads to 2 pi 50 this is 1592 ohms. So, if you are modeling it you could model it as a as a y connected capacitors of reactants 1592 capacitive reactants 1592 ohms between each line to ground. So, in your sequence diagram for the single line to ground fault after the circuit breaker opens. So, you have now your positive sequence which is your negative sequence. So, you can calculate your fault current level with the parasitic capacitance and that would be 240 volts divided by 12 plus if you look at the effective circuit that is going to be limiting the current. Here you have 86 milli ohms in parallel with 1592 ohms. So, it is practically just 86 milli ohms. Here again you have 86 parallel 1592 practically it is 86 and then if you look at the 86 in series with numbers like 4500 and 1592 you have to consider just 1592 and the parallel combination of those two network. So, you would have you get a magnitude of your fault current to be about 0.48 amps about half an amp would flow through the fault point. Whereas, if you look at if x c is ignored your magnitude of your fault current is about 160 milli ohms. So, you can see that there is a substantial difference in the fault current especially if you have a ground fault measurement device on your on your equipment. If you neglect the parasitic capacitance you would see you would expect a smaller current, but because of the parasitic capacitance the actual current would be typically be higher especially when you have high resistance grounding. So, that is when your unbalance really shows up as unbalance currents flowing through your parasitic capacitors. So, some of these levels can actually be used to set how sensitive your ground fault trip devices ground fault fault current interrupting devices to what level of sensitivity you need to set them. So, the next problem is to look at. So, after circuit breaker opened what would be the subsequent voltage seen on a conductor to ground on a conductor to ground basis. If you look at the sequence network that we previously had. So, you could then look at what would be the voltage across the each of your sequence network points. So, you have 240 volts minus the drop is J 86, 86 milli ohms into 0.48 amps which is the fault current level divided by 3. So, this would be your current flowing through that particular network. So, this is roughly equal to 240 volts. So, I A plus is roughly 240 volts. If you look at V A minus this is the voltage across these two points because of the 0.48 amps fault current that is flowing and you can see that 0.48 amps will cause a very small drop across the negative sequence network. So, this is roughly equal to 0 and V A 0 is roughly equal to minus 240 volts because the voltage drop along that loop has to add up to 0. So, if you look at your voltage the voltage on which your fault has occurred that drops to 0 and V B is approximately your line to line voltage and we see a sub also having a magnitude of your line to line voltage on a conductive to ground basis. So, the next question is what type of relay can be used to actually detect the fault currents and respond in to operate under such a condition. So, if you look at the fault current level we saw that the fault current level is of the order of 30 amps. So, the order of 30 amps when the fault occurred in load 1 and if you look at the rating of C B 1 is it is 1 M V A divided by root 3 into 4 15. So, you are talking of something of the order of 1400 amps. So, it would definitely not trip under over current. So, if you want to trip in such a situation you would need a device which actually measures residual currents and trips in response to that. So, if you look at the ideal situation it may not be that C B 1 has to open in response to a fault in load 1 a better situation might be that you want to open the particular load at which the fault occurred. So, C B 2 should open in response to a fault in load 2 again it depends on whether C B 2 has the ability to measure residual currents. If you have C B 2 load corresponding to a power level of the order of 10 K B A then you might have circuit breakers which are rated for 15 amps and 30 amp short circuit current might cause a 15 amp breaker to trip may be in a few cycles not on an instantaneous basis. But, if you have load L 1 which is now 30 kilowatts or higher then again the current rating of the load would be higher than the fault current level. So, you have a possibility that you might have a load and because your breaker may not be equipped to measure residual current you might stay with fault for a longer duration and cause potentially overheating of conductors etcetera. So, you have to look at what your specific rating is what your fault current is and ensure that you have protection against overheating or sustained long term operation under faults. So, in the next problem you are looking at distribution system protection and coordination. So, we have system the source at say the substation end and we look at three devices one is a circuit breaker closer to the substation and then you have a recloser further downstream and then you have a fuse you have four buses and you have loads at each individual bus and you are for the protective devices you are given parameters for the circuit breaker it is extremely inverse with p equal to 2 you are given at a and b parameters b is the definite time delay that you might expect the pick up current level and the reset time of the circuit breaker. The recloser over here has been considered to include two aspects one is the recloser might have its own intrinsic circuit breaker type of operation programmed into its whether it is relay action and also it has on and off durations as soon as a fault occurs it stays on for 100 milliseconds then it shuts off then it recloses for a duration of t c 1 then it shuts off again for 6 seconds then it recloses again for a duration of t c 2 and looks at whether the fault has cleared by that particular duration of time. So, this is a two recloser recloser it also has a underlying circuit breaker c b r with a b and p values as given with the pick up current and the reset time for the fuse you are given melting current level of 105 amps i square t with some tolerance range around the nominal and minimum melt time of 100 milliseconds. So, that could be considered as its definite time delay you are also given that it takes 90 seconds for the fuse to cool down after it gets heated up you are told that upstream of bus 3. So, the conductors in the region between bus 1 and 2 and 2 and 3 has a i square t level of 30 mega ampere square second and the bus downstream of between 3 and 4 has a i square t of 7.5 mega ampere square second. So, that could be used to determine the settings of the components to see whether you are adequately protected you are also given the load current level. So, the load at bus 1 is 50 amps at bus 2 is 60 amps and 3 and 4 it is 70 amps each and based on the parameters of the circuit someone has done say a fault study and found out what the maximum or the typical maximum level of fault current is the minimum might corresponded to some particular fault configuration could be a single line to ground fault with some grounding impedances etcetera. So, you are given a fault current range to that could be expected at each of those buses. So, the problem is first to plot the trip time of the protection device versus current of the circuit breaker this circuit breaker at the street closer and for the fuse in a range from 100 amps to 10 kilo amps. So, to do that we have to look at the parameters of the protective devices. So, for the C B if the current is greater than the pickup current level of the circuit breaker which is 300 amps you have your trip time else when it is lesser than 300 amps during reset T reset is 20 by and what will take is if your nominal current is much lesser than 300 amps this is we can take it as just 20 seconds the pickup current. If you look at the particular breaker if you look at the total load that comes in from load 2 load 3 load 4 at this particular breaker that adds up to about 200 amps. So, if you look at that particular point your reset time is of the order of 36 seconds, but we will assume that the reset time is constant value of about 20 seconds. So, next for the C B R and finally for the fuse. So, for the nominal fuse we have for current greater than the melt current your I square T is a I melt square. So, you can evaluate what your equivalent A is and this is for the nominal fuse 2 into 10 power of 6 divided by. So, you could think of it as a equivalent A B C A B P parameters. So, your T melt as a function of current is plus a definite time of 100 milliseconds your cool down time and you could plot this. So, what is shown over here is the characteristics of the circuit breaker at closer to the substation than the recloser and for the fuse and you can see that the nature of the characteristics. So, you have almost a flat line over here, but then closer to the definite time characteristics it the curve starts flattening out. If you look at the fuse you have the nominal curve for the fuse and then plus 20 percent I square T or minus 20 percent I square T. So, it shows up as a band around the nominal curve in the range of interest. If you look at the reset curve you can see that the reset time also increases as you get closer to the melt time, but for many practical calculations you can take it to be roughly a constant number. So, the next question is to identify the zones of protection of each device and check the devices are coordinated in the specified current range for the fault in the zones that are being considered. So, you can see that the zones that would be considered would be from one protective device up to the next device where the next device would be expected to protect in the zone immediately downstream of that. So, for example, the circuit breaker one would protect up to the recloser plus may be including a fault at the recloser itself. Then R would be able to protect up to the fuse including a fault at the fuse itself. Then you would have the fuse protecting up to bus 4 and further downstream of bus 4 would be the whatever load is connected the load itself would have its own protective devices. It is not mentioned in the problem what it is, but it would have its zone corresponding zone. So, you could identify zone 1, 2, 3, 4 for systems such as this. So, the next question is to check if the protection systems are coordinated. So, what you have to do is you look at your trip curves corresponding to the maximum or the fault current in the ranges. So, for example, for CB 1 it was you are looking at 3500 amps to 9.2 kilo amps. So, you might be talking about some current in some range such as this. So, you are looking at each range would the curve in the zone upstream zone be lying above the curve corresponding to the downstream zone. So, actually plotting these protective characteristics in a time versus current curve gives you a good feel for whether it is coordinated what the margins are etcetera. So, if you gives you a overall picture of what the situation is. So, in this particular case for a fault in zone 1 if you look at your fault current level and your trip time of CB. So, for a fault in zone 1 obviously, the only device to protect over there is CB for a fault in zone 2 you can see that in the specified range CBR is much faster than CB. So, you have the adequate coordination. Similarly, you could then go further down. So, if you look at in the range that was being considered like you can see that whatever is the range you have sufficient clearance between these curves that is essentially what you are looking for. Similarly, you could look for other fault ranges. So, if you are looking at fault in zone 3 you are looking at about 1.4 to 2.1. So, again in this range you are looking at now the clearance between your fuse and your CBR and again you can see that even with the tolerance you have adequate clearance between the fuse and CBR. So, you have adequate coordination between your CB, CBR and the fuse. So, overall they are reasonably well coordinated. So, the next question is would these devices give adequate backup protection. So, what you could consider is if for some reason R fails would CB provide backup protection in the sense that would zone 1 itself be protected when R fails and would zone 2 also get some degree of protection or would is there a possibility that falls over here can propagate over to zone 1. Similarly, if for some reason fuse fails to melt and then would R provide backup protection to zone 3. So, if you look at fault in zone 2 and failure of CBR then you the device that is providing backup protection is CB. So, you could look at the maximum fault current level in for a fault in zone 2 is 3.5 and the minimum is 2 and you could look at what would be the trip time of circuit breaker for that particular level of fault. Then look at what the i square t level would be and you can see that this is now less than the 30 M mega ampere square second for the contactors of zone 1 and zone 2. So, you can see that the fault in this region and the failure of one particular device to operate is being backed up by a upstream device and not just that the fault will not propagate to zone 1 zone 2 itself is entirely protected by the CB also in this particular case. So, similarly if you look at the situation where you have a fault in this particular region and say the fuse F fails to melt then you can see that the device that would now give you the backup protection is R. So, you could look at the maximum and minimum fault current for a fault in this particular zone. Look at the trip times of this particular device protective device and look at your i square t level corresponding i square t level and this is now less than 7 mega ampere square second. So, it does provide backup protection for a fault in zone 3 also you could also look at a situation where say suppose you have a fault in zone 4 even though you do not know what is the protective device being used in zone 4. If that device fails to protect would you have backup because of the fuse F 1. So, you can look at the maximum and the minimum fault current range and look at what is the i square t level and you could again see that in this particular case the resulting i square t is less than the 7 mega ampere square second required for this particular zone. So, again you are having complete backup protection in this particular case. The next question is what is the margin available for between the tripping times for between the actual operation of the device and the upstream device. So, if you look at for a fault in zone 2. So, a fault in this particular region if you look at the tripping times of the circuit breaker it is about 1.89, 5.79 seconds of the CBR is 0.58. So, in terms of time you are having a margin of greater than 1.3 seconds. Similarly, for a fault in zone 3 your trip time of CBR is of the order of 1.56 and 3.49 seconds whereas, the melt time of the fuse is about 0.65 to 1.33 seconds. So, you have a margin of greater than say 4 cycles. So, it is more than a typical instantaneous trip settings of a circuit breaker. So, you have lesser possibility of some race condition causing a upstream device from operating when a downstream device should have operated. So, in the next class we will look at the question of what are the objectives of the coordination between the recloser and the circuit breaker. Similarly, between the recloser and the fuse and look at the settings calculations and what could be done for coordination and what would happen when DG unit is added to such a system. Thank you.