 Welcome to class 11 of the topics in power electronics and distributed generation. In the last class, we were talking about grounding of distribution systems. We looked at the need and the possible methods for grounding of distribution systems. In terms of the grounding of sources, you can have an ungrounded source or a delta type of source. In which case, in terms of the actual implementation, an IT type of secondary distribution grounding grounded network is similar to what would be for this type of system. The source is ungrounded, but the loads frames is connected to ground. Then you can have a solid grounded system, in which case the source neutral is connected solidly to ground. The ground point which is the neutral is now distributed to loads. The T n network is a example of such a secondary distribution network, grounding situation. The third scenario that we were talking about is a impedance grounded source. So, we are talking about having a impedance now connected between the neutral of the transformer on the y side and the earth, the earthing terminal. It can be a resistance or a reactance. You can have the possibility of high impedance or a low impedance grounding. We saw in the last class an example of a small value of reactance, which is connected between the neutral and the ground. We saw that in such a situation, when you have a single line to ground fault, then your phase voltages is somewhere. Now, in between that of a solidly grounded system and an ungrounded system, in terms of the fault current level, it is the current magnitude is high. It may be sufficient to trip a instantaneous time over current characteristic, but it is not so high that it might cause very severe arcs at the point of fault. These issues are important from the single line to ground fault perspective, because the majority of faults are actually occurring on a line to ground basis. Now, we will look at a case of a high impedance to ground fault. We are looking at a case where a large value of resistance is now added between say the neutral and the ground. The objective in this particular case would be to keep your fault current level quite small. So, we are talking about a high resistance grounding low say something like say 0.1 to 0.2 per unit range of the order of 10 percent. Typical current value is less than 10 amps typically, but it cannot be extremely tiny either. We saw that if an ungrounded system, if you have no physical path to ground, you would excite the parasitic capacitances of the line. So, typically you would keep the current through the resistor to be greater than 3 times the high capacitive current capacitive charging current through for that particular distribution network. So, that you get adequate damping for possible oscillations between your physical inductances of your components and the parasitic capacitance. So, in this particular case because the current level is so low, your single line to ground fault can continue for a long time after the fault inception. So, instead of waiting for an unanticipated point at which you would need to come in and do the repairs, you could maybe come in at night when everything is shut down, then go in and do the repair and clear the point at which you are having the line to ground fault. And your resistance is sufficient to damp parasitic capacitance and what we saw was you can have arcing at the point of fault and the arc the current through the arc need not be symmetric in the positive and negative half cycle. So, there is always a possibility of level shifting of your distribution feeder in case you are having an arcing fault and now your resistance provides DC path to ground to discharge possible level shifting. So, however in this particular case what you would see is that the after you have the fault the unfaulted phase would see the conductors would see voltage line to line voltage on a conductor to ground basis. So, your insulation and your O voltage protection and its coordination need to consider the fact that you would have sustained line to line voltage possibly occurring on a line to ground basis. So, if you look at the IT type of distribution network that we discussed earlier it is quite close to what the high resistance grounding situation is going to be in terms of its characteristics. So, you can actually look at that as a case where that value of the resistance is made to be really high just to look at your parasitic capacitance to prevent oscillations of your parasitic capacitance with your in your distribution network. So, you can think about the IT distribution secondary distribution network as a example of this particular situation. So, we will look at example of high resistance grounding. So, we will look at an example where you have a source and say this is delta y and the y is connected through resistance of say 5 per unit and you have a circuit breaker and say the impedance x l of the transformer is say 5 percent and you have a single line to ground fault occurring somewhere down the distribution line. So, if you look at what would be the corresponding fault current in this case you can draw the sequence network to analyze the fault current. So, you have your source with 1 per unit voltage your negative sequence network is just the impedance j 0.05 and your 0 sequence network is just the is a combination of the reactance plus 3 times the grounding impedance resistance which is 15. So, if you calculate the fault current your I f is now equal to 1 by. So, your fault current is is about point I f by 3. So, your I f is about point 2 per unit. So, you can see that with a point 2 per unit current you will not trip a circuit breaker because your load current itself might be 1 per unit. So, your over current trip will not function. So, if s 1 is a circuit breaker can see that s 1 will not trip and if you look at what the voltages are the voltage V a V b V c would be similar to what we calculated in the ungrounded case. So, one way in which then you could sense that a fault has happened is by looking at your voltages. So, if you see that your voltages on one phase has come down to 0 and the other phase has gone to line to line voltage with respect to ground then it means that there is a single line to ground fault somewhere on the system then the question is where is that particular fault there is another question, but you know that there is a fault when you see the shift in voltage. You could calculate. So, you can calculate the power dissipation in the resistor because after you have a fault potentially the resistor will stay connected in the circuit for a longer duration till the planned scheduled maintenance comes in. So, you are talking in this particular case of about 0.2 divided by 3 about 7 percent power is getting dissipated till your fault gets cleared in the grounding resistor. .. So, other ways of doing people have also looked at possibility of having high inductance grounding though it is not very common because you need to damp your L C oscillations and for damping you need a resistance. People have also looked at something called resonant grounding where you have a inductor with your parasitic capacitance tuned. So, that it is resonant at 50 hertz. So, in that particular case you are you might have some advantage that in a resonant circuit your parallel resonant circuit your current drawn from the source is in phase with your voltage because and the current is drawn only into your loss in the branch that contain losses. So, you have in phase voltage and current and having in phase voltage and current means your voltage 0 crossing and current 0 crossing is occurring close to each other. So, you have a possibility of self extinguishing the arcs in the fault because when the current goes to 0 even the voltage is 0. So, your arc might go away in your insulator, but it means that you have very good control of your parasitic capacitance which is not necessarily true people might switch in loads switch out loads. So, in a control situation you have such possibilities, but not in general. So, if you look at then the other possibility of when you look at the grounding of the source with the load you have the t t secondary distribution network where the first T corresponds to having a ground solid ground at your source and the second T corresponds to having grounding solid grounded electrodes at the loads. So, what is shown over here is a example of a t t network the y point is connected to power earth at the source the neutral is distributed to all the loads. So, you have 3 phase 4 wire distribution potentially each load can have its own grounding electrode. So, in this example we have shown what we have we see is load 1 and 2 has a common grounding electrode P e 2 load 3 has another its own grounding electrode it might be big housing structure load 1 and load 2 load 3 must be might could be the next house. So, it could be you can have different possibilities of your t t grounding network the one of the main thing to consider in such a while analyzing such a system is that when you have a grounding electrode inserted into the earth then you have actually the resistance earth resistance between your earth in conductor and your physical ground. So, you have what is shown over here as a connection to earth physically is actually you might have a point you might have some contact resistance to the deep earth and then you have the physical deep earth which would be considered to be at 0 potential. And you look at your typical earthing resistance you it is a number which can vary common values range from 4 ohms to 10 ohms it depends on a variety of conditions like the how much wetness is there for the soil the salinity of the soil whether the conductor is highly corroded or it is a relatively well maintained conductor. So, there can be a variation in the value of resistance, but these are some of the typical numbers that people consider for the grounding resistance. .. And if you look at the ground fault current in this particular case. So, the first thing to consider is the earthing earth contact resistance and that need not be uniform and what you would see is that the level of ground fault current is might be sufficient to create to clear low rated circuit breakers or circuit breakers of the order of 10 amps, but not high current circuit breakers. And when we say low voltage network we are talking about voltage less than 1000 volts is what people traditionally call low voltage network. And we will see this in an example we will do an example to see that the fault current level may not be large enough in all situations. And you can actually also make your ground fault current you can have ground fault current detection based protection which can be made sensitive. Say for example, when you are having situations where people might commonly use electrical appliance and there is a concern of safety you can have say areas like kitchen, bathrooms etcetera where the ground might be wet there is a chance of increase shock. So, you can try to set your ground fault detection current levels to be quite tight. So, that is to ensure that people do not get severe problem because of electric shock. So, we will look at an example of a T T distribution. And we are looking at say we will look at an example where say you have a 2 MVA source, 2 MVA transformer and say your leakage inductance is say 4 percent and the secondary network is 415 volts. And we will assume your resistance to earth is 5 ohms at each electrode. So, if you are looking at 2 MVA 415 volt network you are talking about I rated which is 2 MVA divided by 415 divided by root 3. So, you have about 2.8 kilo ohms. So, these are thick multi conductor cables that might be connected in such a situation. And we will look at an example where you have a single line to ground fault in load 1. So, you have a fault in load 1 and you look at your transformer reactance the 4 percent reactance in this particular case corresponds to J 7 milli ohms 7 into 10 to the power of minus 3 ohms is the reactance of the transformer. And then we could look at what is the resulting fault current in a situation such as this. So, we can draw the sequence network to figure out the fault current. So, you have your positive sequence network 415 by root 3 is a voltage your reactance is J 7 into 10 to the power of minus 3 your 0 sequences. And then if you look at your source on the source side you have the resistance between your grounding electrode and deep earth which is 5 ohms. So, that acts now as a effective grounding resistance of 15 ohms in your 0 sequence network. And then your fault has occurred on in load 1 and you also have now a electrode at load 1 which is also having a resistance of 5 ohms between the electrode and deep earth. So, that can be treated as a resistance. So, you could think of this point as P e 2 and this point as P e 1 and this point as your deep earth or your physical may be 0 voltage point. Then you can look at then your fault current and I f by 3 is now 240 volts divided by 15 plus 15 because the 7 milli ohms is quite negligible compared to your 15 ohms. So, your fault current level is about 24 amps. So, you can clearly see that 24 amps is a tiny percentage of your 2800 amps rating of your main circuit. So, obviously you will not see a over current tripping at the of your main breaker which is carrying your fault current. You could actually calculate what your touch potential is if say a person is over here and the person say touches load 2. Because you now have a current flowing through power of 2 and you have a resistance between your earth and your there is a contact resistance the frame the frame of load 1 and load 2 will get elevated when you have a single line to ground fault. And you could calculate what that particular voltage is and if you look at the sequence network that is essentially corresponding to the P e 2 point. So, touch potential at load 1 so with 120 volts you will definitely feel a shock if you actually touch a cabinet at that particular voltage level similar same thing at load 2. If you look at load L 3 so if you have a person now standing over here and touching load 3 with respect to power earth 3 the there is no current flowing into power earth 3. So, with that particular location the person will not get a shock. So, the all the loads connected parallely which has its own independent earth you do not see elevated touch potential. But within the particular load it may be a house or something like that you would see elevated touch potentials. So, you have seen the situation where you have some old aged wiring in a house and suddenly you go and touch different equipment you might get a shock because with in that particular establishment you might have now elevated touch potentials for your equipment. So, it is important to ensure that your quality of your wiring etcetera is good otherwise you end up having problems of possible shocks when you are touching equipment within a common ground load which has its common ground through which the current is actually flowing in. And you can see that just over current fault might be sufficient to clear may be 24 amps may be sufficient to clear a 10 amps circuit breaker, but definitely it will not clear 30 amps circuit breaker it will not clear 100 amps circuit breaker. So, in a situation where you are having high rated loads you cannot just depend on over current to give you protection you would need protection from ground fault current detection. And if you have something like 24 amps flowing through a circuit which is designed to carry may be 20 amps or 30 amps it means that your conductors would get hot. So, there is actually a potential for eventually having electrical fires. So, you have to ensure that the fault gets cleared or continuous over current that is flowing through the network can actually lead to possible fires. And that is something that one would need to avoid also if you if there may be some bad practices where instead of putting an actual fuse someone might have just a physical conductor strand of copper being used as a fuse. And in such situations you go and touch the fuse the fuse block will be very hot because it is carrying a high current, but not sufficient to actually interrupt the circuit and things are running hot because there is a fault somewhere in the system which you cannot identify. But for small facilities I mean for small establishment homes etcetera where your loads are typically 10 amps or less this would be a TT network would be low cost because the number of conductors that are need to be passed around is quite small and it is adequate for protection of smaller establishments. . So, at this point with this background on what can happen what type of possibilities are there in distribution systems and what can happen due to say or the issues in that are there when you have a common type of faults we can actually then look at the situation as of what would happen when you add a distributed generator source into such a network which has its own grounding characteristic. So, we look that again the IT grounding that is a type of grounding mechanism where you might have high priority for continuity of service. So, in a situation where you place greater priority for continuity of service you might put a IT type of network whereas, if you are very you feel that it is to interrupt the loads, but your fire is fire protection is something which is extremely important then you need to ensure that you have high current levels to ensure clearing of fault. So, a TN type of network will ensure that there is high fault current and it would clear the breakers in a short time frame. So, you do not have possibility of long lingering over currents and TT network is a balance you do not have the situation where one fault will cause elevated touch potential at multiple loads especially when they have independent grounding options and it is also sufficient when your establishment is drawing lower levels of current. In case it is drawing higher current levels of current you could look at additional options of actually measuring the ground fault current level and actually tripping your breakers. So, we will look at now an example where you have a distribution network and we look at the same distribution network that we studied when we looked at the issue of coordination where we had some initial fault current levels that the system was carrying and then we added the DG and we wanted to see what would happen to the fault current level. So, we look at the same network where you had the traditional the old network which is say solidly grounded and then you want to add a DG to this set up and you want to see what are the possible situations. So, when you have a fault and the fault that we would consider as say a fault f 1 just close to circuit breaker 1. So, to analyze this the fault situation we will assume that the single line to ground solid fault is occurring at f 1. So, to analyze this particular situation we will now again have to draw the sequence networks. So, you have the situation where you now have the possibility of the contribution from your main source which is a grid and also now the DG. So, your positive sequence network would consist of this reactance corresponds to the source reactance and then you have the DG which is also 1 per unit voltage and you have the reactance of the DG plus the interconnection J 0.1 and the feeder reactance which we took as J 0.03. So, this is your positive sequence network if you look at your 0 sequence your negative sequence network and if you look at your 0 sequence network on the source side you have now solid grounded network on your DG side you now see the delta side of the transformer which means that it is open. So, if you look at the example so the 0 sequence corresponding to this transformer would be a open circuit because it is delta on the on the on the this high voltage side as here you have a possible path for 0 sequence current. So, you could then calculate what what the fault current would be. So, we will I will give you the numbers your I f old what I mean by I f old is the fault current when there was no DG. So, assuming that this particular branch was not there and this particular branch was not there you would be able to calculate I f old. So, that was 71.4 per unit for a single line to ground fault a solid single line to ground fault and I f new is the fault current that you have when the DG is connected. So, that corresponds to this particular figure which is drawn over here and that was that is 76.4 per unit. So, if you look at this particular situation you see a increased fault current level at the point of fault and you can have your breakers operating in response to the fault because you have a solid grounding over here you will see a higher current over here which might say trips circuit breaker 1. So, we look at a situation that can potentially arise if the current the way it branches out because the impedance upstream is quite small compared to the impedance back towards the DG. If the current is such that it trips say circuit breaker 1 before circuit breaker 3 then you will end up with now a new situation where then the new sequence network would be 0 sequence and this new scenario is essentially a open circuit. So, your I f your fault current level is actually 0. So, as soon as your circuit breaker 1 opens your fault will fault current would go away which means that now you will potentially energize the feeder from the DG side and there is no fault current. So, if the DG is capable of supplying your power real power to the feeder corresponding to the level of demand of the feeder then you can continue in that situation for a really long time. So, the situation over here is say your C B 1 opened and C B 3 is still closed it means that even though you have a single line to ground fault you will not actually have any fault current because your system has now become a delta ungrounded system originally it was a Y grounded system when circuit breaker 1 opened it has now become a delta type of source which means that if C B if your DG over here is capable of supporting the loads on this particular feeder it will continue to operate for a long duration of time. So, if this is actually can lead to many problems and you say for example, because you are originally solid grounding your source you might assume that you are not going to have phase to ground voltage of that is going to be large, but now because you are having a delta connected source it means that now your conductor to ground voltage can potentially go to root 3 times what is the nominal. So, it will you will see higher over voltage and systems are not designed to actually experience over voltage of 70 percent etcetera. So, you can have damage to components the DG can support the power requirement and maintain the voltage it will continue to energize the feeder and original system is solid grounded whereas, after tripping there is a possibility of over voltage and. So, what it means is that your DG circuit breaker has to trip as soon as your upstream breaker trips and the upstream breaker might trip for reason other than for fault for some reason if the breaker is opened say C B 1 is opened then now what you have is the possibility of the grounding of the system changing from one strategy to another strategy it is to have one particular strategy, but it is difficult to have systems which are changing the grounding strategy being changed when your breakers open. So, again if you look at this particular requirement what it means is that for any reason if any fault happens in the system or if circuit breaker one opens your DG has to be disconnected quite rapidly which is similar to the conclusion that we saw in the other example when we looked at the protection coordination which means that you have to make your C B 3 trip potentially not just for a fault condition, but also for any reason if C B 1 opens you need to actually open C B 3 or your grounding strategy would not be consistent. So, now if you want to make C B 3 that sensitive it will become prone to a lot of nuisance trips. So, being able to meet such a requirement is actually quite challenging and the situation where C B 1 opens and C B 3 is closed is what people refer to as a situation of an unintentional island what you have done is you have disconnected from the main grid and you are actually operating your balance of your feeder as a island which is disconnected from your main grid. So, this unintentional island has many consequences safety problems liability problems etcetera and it is important then that your DG system is not just able to detect a fault, but actually now you need to generalize and be able to detect a situation where your system your distribution system might potentially have an unintentional islanding situation. And what we will do is in the next class we will discuss this islanding issue more closely because it is a important consideration for distributed generation sources and you have not just unintentional islands you also have intentional islands will look at why people do have intentional islands primarily from power quality perspective you may want to disconnect from the grid when the grid power quality is poor. So, islanding is a important consideration especially when you are operating a distributed generation source and you can have different situations of islanding and we will discuss that in the next class. Thank you.