 Welcome to class 20 on topics in power electronics and distributed generation. In the last class we were discussing example problems on fault protection with circuit breakers and fuse. And we looked at distribution system with 3 components circuit breakers, a recloser with a embedded circuit breaker and a fuse. And we plotted the trip time versus current characteristics of such a device. We had the nominal fault current ranges, the load current ranges. So, we looked at the properties of the individual devices. And then we looked at the time over current coordination between upstream device and the downstream device. We also checked whether we have backup protection in case 1 protective device fails. We also looked at the time margins in the settings of the protection, whether there is adequate margins available in the settings. So, today we will specifically look at. So, we looked at what are the curves, the trip time versus the current curves. Today we will specifically look at the coordination of the circuit breaker with the recloser and the recloser with the fuse. So, the first thing we are looking at what are the objectives of coordination of the circuit breaker with a downstream recloser. So, the objectives of the coordination of a upstream circuit breaker and a downstream recloser is we will call zone 1 to be the zone of protection for the circuit breaker and zone 2 for corresponding to the recloser. So, if you have a fault in zone 1 it is a circuit breaker that has to protect. If you have a fault in zone 2 the recloser with its underlying circuit breaker characteristics should provide the protection. If you are having temporary faults in zone 2 then the recloser attempts should clear temporary faults. If there is a permanent fault in zone 2 then one has to ensure that the recloser locks out before your circuit breaker upstream breaker trips. So, that is the overall objective you would have for this particular coordination problem. So, we are looking at then based on the time current curves for your underlying circuit breaker of the recloser and for the upstream breaker find out the timings required for the recloser. Again when we look at timings what we are looking at is essentially if you have the recloser carrying some nominal current and at some point of time a fault occurs and the current goes up and what is plotted is your IRMS versus time. You have some duration we called it as T 0 and for which the high current would flow. Then you have the first open duration T of 1 that is what we have called in this problem. Then we have a duration T where the recloser recloses which is T c 1 and then you have a duration where it opens again T of 2 and then it recloses does a reclose attempt for a second time T c 2 and if the fault is still persisting then it locks open. This occurs when the current level is higher than some I threshold and it is provided to you that the duration of T 0 is above 100 milliseconds 0.1 seconds T of 1 is 6 seconds and T of 2 is 6 seconds and you are required to find out what settings you could use for T c 1 and T c 2. .. So, for T c 1 and T c 2 we have we are asked to find what is T c 1 max. So, we have T 0 is 0.1 and if you look at it as a percentage of time required for your upstream breaker. So, 0.89 if you was the trip time of C B at the fault current level maximum fault current level which was I is equal to 3.5 kilo amps for a fault in zone 2. So, this is essentially the 1.89 and T 0 is corresponds to 0.01. So, if you are looking at essentially as a percentage this is this corresponds to 5.3 percent of the tripping of the circuit breaker C B's tripping. And if you look at the this duration how much it would correspond for the tripping of C B R. So, this would be 0.1 divided by 0.58 and if you remember the 0.58 in the last class discussion was the time of C B R at I is equal to 3.5 kilo amps and so this corresponds to about 17.2 percent of the way into tripping. If you look at T of 1 duration that corresponds to 6 seconds and we know the reset times for breaker C B and C B R is 20 seconds. So, it would reset by 30 percent in the 6 second duration. So, if you look at both C B and C B R it is fully reset by the end of your off duration. So, if you look at then the next duration available T C 1 max. If you allow for another duration of 0.58 duration then the C B R itself would actually trip. So, there is no point pushing your reclose duration to be larger than 0.58 seconds. So, or C B R. So, you need something less than this add something less than this for margin. If you take something longer it would the underlying C B R characteristics would cause the device to trip. So, you need it to be something that is smaller than 0.58. So, the next problem is to look at the objectives of coordination of the operation of the recloser and the fuse. So, if you look at the objectives of coordination between the recloser and the and a downstream fuse. So, if you have a fault in zone 2 then that has to be cleared by the recloser or its underlying circuit breaker. If you have a fault in zone 3 if it is a permanent fault in zone 3 that has to be cleared by the fuse. Whereas, if you have a temporary fault in zone 3 then the recloser attempts to clear the temporary faults, but before the recloser locks open or locks out essentially the fuse has to blow. So, this is this would be the requirement for coordination between the recloser and the downstream fuse. So, if you look at then what the duration of T C the T C durations are for this particular condition. So, if you make the time duration to be too small then there is a chance that the fuse will not blow and a recloser would lock open. So, you want the durations to be larger in this particular case. So, as to ensure that the fuse blows for a permanent fault before the recloser locks out. So, you look at the case for a couple of conditions. So, you are looking at what should be the settings for T C 2 for fault in zone 3 and you need to select T C 2 after or the melting of the fuse to be earlier. So, that the lock out does not happen. So, if you look at this particular condition you need to look at you also know that your fuses have a tolerance of plus or minus 20 percent. So, you consider say the larger plus 20 percent because you are looking at the worst case duration. So, you can look at the situation at the maximum current and the minimum current for the fault in the zone your T 0 would correspond to 0.1 divided by 0.65 is this would be 15 correspond to 15 percent where 0.65 is the T melt. The second is the T melt for fuse current I is equal to 2.1 kilo amps which is the maximum current level for fault in zone 3 and for the fuse with a plus 20 percent I square T. Then if you look at T of 1 that corresponds to 6 by 90. So, this is 6.7 percent. So, at the end of T of 1 fuse is. So, we could consider to be the fuse to be 8 percent melted and then we have your T C 1 that corresponds to 0.58 which we determined from the previous problem. If you then look at 0.58 as a percentage of the fuse blowing time this would be 90 percent. So, at the end of T C 1 now if you look at what would be the situation here we have considered a fuse with I square T of plus 20 percent above the nominal. If you took the fuse which was plus minus 20 percent below the nominal I square T. So, if you look at the fuse which was 80 percent of the I square T rating would already be melted. So, if you look at a desirable setting of that 0.58 it might be desirable to actually reduce it to actually prevent the melting of the fuse in the first reclose cycle itself because of the settings of the recloser. But, we will continue with this particular problem if you look at T of at this particular maximum current of zone 3 if you look at T of 2 which is 6 seconds and 90 seconds is the cool down time of the fuse. So, that corresponds to 6.7 percent. So, at the end of T of 2 uses about 92 percent melted. So, if you look at the remaining 7 to 8 percent melting you need a T C 2 which is greater than 0.05 seconds I max I f max of zone 2. Now, we look at what happens at the I f min current level of zone 3. So, you have now T 0 corresponding to 0.1 by 1.33. So, 1.33 is the melt time corresponding to the fuse with plus 20 percent I square T. So, that would now corresponds to 7.51 percent T of 1 corresponds to 6 by 90 this is 6.7 percent. So, fuse is 0.84 percent melted. So, then you could calculate T C 2 if it is 0.58 and 1.33 is the melt time at this particular 1.4 kilo amps current level that would be 43.7 percent and T of 2 is then one can calculate the condition of the fuse. So, fuses. So, if you look at the time that is required to melt it for the remaining remaining percentage for this number to reach 100. So, you can calculate what your T C 2 has to be it has to be larger than that. So, what you would say is T C 2 has to be larger than 0.83 seconds with some margin. So, that the fault current in zone 3 does not cause the fuse to still conduct when a permanent fault is still in that particular zone. So, we saw in the previous case that at the point 5 8 second settings this particular fuse is already melted. So, we can also look at the situation where T C 1 is modified to ensure that the fuse even at the lower end of the tolerance range does not melt by reducing the T C 1 duration. So, you could calculate what could be a T C 1 duration. So, if you look at the fuse with the point 8 level and zone 3 with I f max equal to 2.1 kilo amp you have T 0 corresponding to 0.1 by 0.46 that corresponds to 21.56 percent T of 1 corresponding to 6 out of the 90 seconds of the fuse cool down time that is 6.7 percent. So, if you look at so if T C 1 is 0.58 seconds this would correspond to 0.58 divided by 0.46 or this is 125 percent it means that fuse is already melted it will not see a secondary closed cycle. So, if T C 1 is reduced to a smaller value say we will take it as 0.38 seconds then. So, at the end of the T C 1 it is 82.4 percent melted which means it is not the fuse is not at damage at that particular point. So, the fuse will see a second reclose cycle and then we will have to then recalculate what T C 2 has to be for this new setting of T C 1 1.2 times the nominal value which is 2 into 10 power of 6 ampere square second. So, you can calculate T C 2. So, you can calculate T 0 which would correspond to 7.5 percent T of 1 would correspond to 6.7 percent T C 1 which is 0.38 now divided by 1.33 this corresponds to 28.7 percent. So, you can say fuse. So, for the remaining 77.1 percent melting 1.03 seconds duration. So, the new value of your T C 2 should be greater than 1.03 seconds with a margin. So, you can see that if you want to make sure that you do not you prevent fuse blowing by trying to make the maximum use of the recloser you will have to make sure that you are doing the operations. So, that you do not have unnecessary wastage of fuses and you do these calculations to ensure that you save the fuse as far as possible. The next problem is from the value of the T C 1 and T C 2 that has been calculated what should be the current threshold level for operation of the recloser to prevent nuisance lockout of the recloser before melting of the fuse and we saw in this problem the as the current level reduces you need more time for the fuse to melt. So, if you allow the threshold level to be too low then the amount of duration you need for the recloser to up for T C 2 becomes longer and longer. So, as to ensure that the fuse melts first before the recloser locks out. So, as to attain the objectives of the coordination that you had where a permanent fault over here causes this to fail before the upstream device locks out. So, suitable threshold level can be corresponding to the time that you have used for the I F min calculation and the I F min calculation that we had in the example that we looked at was I F min for zone 3 was 1.4 kilo amps we could use that as a threshold. So, that as for current levels below that you do not have nuisance lockout of a upstream device for a fault in the downstream zone. So, then you are given the next problem where you have D G that is connected and because of the connection of the D G. So, we are looking at the case where the D G is connected close to the substation itself at the very upstream which causes the fault current levels across the feeder to actually go up with the fault. This is a simpler possible configuration you could have D G is connected in where along the feeder. So, because the fault current levels have gone up you are asked to actually recalculate your T C 1, T C 2 settings etcetera for achieving coordination. So, we will look specifically at what the concerns are rather than redoing all the calculations. So, if you look at this particular situation with the D G unit in zone 2 4.2 kilo amps. So, the C B R trapping is 0.42 seconds and not the 0.58 seconds that we had as calculated previously. So, if you had set the C B R settings. So, there is a possibility that now because of your C B R setting because your C B R trips earlier you can have the C B R trip earlier in for a fault in zone 2 rather than allowing a reclose action which will do multiple attempts to clear a fault for a fault in zone 2. So, essentially if you look at the situation over here if you are having a fault over here now because of the reduction in the circuit breaker timing because of the higher fault current you will now have the possibility that the underlying breaker would cause it to lock out before the reclose attempts complete. So, there is a possibility that temporary fault would see an outage of the downstream zone rather than a reclose attempt. So, another concern is if you look at I F min of zone 3 is now 1.7 kilo amps. So, one could be the possibility that now an over current from zone 4 can actually actuate the recloser because your current levels you expect the recloser to protect temporary faults in zone 2 and zone 3. Now, even zone 4 can have faults which might trigger the recloser. So, if you look at a situation at I F max this may not be too bad a thing because if it is a temporary fault in zone 4 the recloser is helping out, but if you have a large load that is starting in zone 4 like a large induction machine a reclose attempt will cause the delay in startup of the machine which can potentially cause overheating etcetera. If you look at I F max of zone 3 and look at the T 0 1 duration corresponding to 0.1 by now 0.42. So, that is 23.8 percent and T of 1 which is 6.7 percent corresponding to 6.7 percent. So, the fuse is about 17 percent melted at the end of T of 1 then if you look at T C 1. So, the fuse has already melted at the end of the first reclose cycle. So, you can see that the previous setting of 0.38 seconds for T C 1 with the addition of the D G is now already causing the fuse to melt. So, your fuse saving strategy is not working. So, you need to further adjust T C 1 can be further adjusted say to 0.28 seconds further reducing T C 1 for ensuring reclose attempts. So, you can see that what you would have originally considered as the settings might have to change once you add the D G into the particular system which is one of the concerns where because the D G might be added by one party whereas the recloser etcetera might be set by some design engineer at the commissioning of that particular feeder. So, that person may not be available at the point when the D G is being added. So, there is a complexity of who will do the calculation what should be the new settings. So, there are concerns of protection coordination when you add a D G to the existing system. So, we have seen that there are concerns we were discussing about also the voltage profile on the feeder when you have the D G and when you have the normal operation of the feeder. So, we will today look at other possible configurations of a feeder you could have the majority of the feeders are radial distribution systems which we have been discussing so far in the class. .. So, if you look at the radial distribution systems they have say two nines of availability or 0.99 percent availability. So, if you look at what is 0.99 percent availability of the year it means that out of 365 days you might have 2.6 days of outage the remaining time your system could work normally assuming you have the power available to actually provide the load. You could expect about 3 to 4 days of outage per year on a radial distribution system and people look at more complex distribution systems. So, the next more complex network compared to just a radial structure might be a ring structure where you can actually go around in a loop or in a ring. And if you look at the reliability here you might have 4 nines 0.99 availability which means which should correspond to about 1 hour of outage per year. So, this is a more complex network, but your reliability is improving. So, if you say what is the next in terms of improvement of reliability is of just having a ring you can have a more complex meshed structure where your reliability can go to 6 nines. So, 6 nines of reliability would mean that your outage per year would reduce to less than a minute. And if you have a high end UPS dedicated to specific loads the reliability of such an UPS can be even of the order of can actually provide backup power with 9 nines of reliability. So, if you have large computer systems you can have the possibility of outage of less than a second per year with high quality UPS backup systems. So, if you look at a structure of such a system you can see that as you want better power quality you are going in a direction such as this. And this is actually going in the direction where it is more expensive where you are willing to pay that higher cost for lesser outage or higher power quality you are going in paying more for getting better power quality. And if you look at these the radial the ring or a network type of distribution all the loads connected to the ring or all the loads connected to that particular radial feeder or in a networks type of distribution system will see the same level of power quality. So, irrespective of whether the loads require such a complex network or not you provide that given level of power quality depending on the complexity of the network. Whereas, something like an UPS can be for dedicated loads. So, depending on which particular load needs the backup may be a computer system needs very high power quality. Whereas, building thermal system might it might be to actually have a slightly longer duration of outage you would not know if the heaters or air conditioners are switched off or may be a minute or so in terms of lights you might be willing to tolerate a few seconds of outages in the lights. But, so depending on what load is being connected you might have different levels of requirements. And ideally it should be possible to tune your particular requirement to the level of power quality and the cost you are willing to pay for it. So, we will first start with a net ring or a loop distribution system and see what would we look at it with an example where you have a loop. So, what is shown over here is instead of just a radial structure now you have a feeder which is in the form of a loop. And it is being divided in this particular example into four zones. Say you have this being fed from a substation where you have two transformers with level of redundancy. So, even if one is out you could provide power using the other transformer. And in the radial system if any point on the feeder has a problem then the whole feeder is down. So, in this particular example you could have a possibility because there is a possibility of getting power from both from two directions you have the possibility of localizing the fault into smaller sections. And ensuring that the entire feeder does not see the outage smaller sections might see an outage. And you have overall higher power quality to the average user across the network. So, we will look at a couple of cases how you would operate such a system. Say under normal conditions you might operate it almost like the traditional radial distribution system your breakers 1 A B C 2 3 6 and 5 might be closed. So, what is shown in green indicates a closed circuit breaker 4 might be open. So, essentially you have feeder A and feeder B. And in this particular example you have this potentially the ring or the loop can be covering a longer distance. So, you might have voltage regulators that are on the ring. So, you have now at the position of the ring you have a voltage boost being provided as say if one is going from let us call this the primary side and the secondary side. So, as you are going from the primary to secondary you are providing a boost on the voltage regulator on the feeder. So, you can have 2 SVRs you could have 1 SVR it could be a range of configurations depending on different types of systems. So, overall under normal conditions you have a feeder where the voltage is regulated based on what your nominal value is and it is the system is trying to regulate the voltage to the nominal value. So, you could think about couple of situations say if you have a fault in zone 1 A then essentially breaker 2 and breaker 3 would open. And then under normal if it was just a radial system then even zone 2 A would see the outage, but now because you can close breaker 4. So, 4 closes and essentially the fault is localized to Z 1 A and the power now flows 2 Z 2 A and and provides power to 3 sections even when you have a fault in one particular zone. So, you have power up to this point and essentially 2 and 3 isolates the fault. You could also have may be a fault in say transformer A or you might be servicing transformer A in which case you might say open 1 A 1 C and 2 and in which case the power is now flowing through all the way to all the 4 zones and ensuring that servicing can be done at the substation without deenergizing your customers or who are the loads on the feeder. You can immediately see that the feeder has to be rated each part of the feeder has to cater to the entire load on the feeder. So, the cost of the system can be higher than the traditional feeder, but then you have the advantage that you do not have outages under a variety of conditions. So, if you then look at this particular scenario where you have the series voltage regulator. So, if you look at the voltage profile in this particular configuration. So, here you have as you go from your left to the right of your regulator you get a voltage boost. Here in this particular configuration where you go from the right to left you have the voltage boost. If you compare that particular situation to what was the nominal configuration of under normal conditions when everything was under nominal conditions when you went from the left to right the boost was being provided. Here if you look at the different zones as you go through this particular as you go through this particular across the series voltage regulator on the series voltage regulator as you are going from left to right you are actually getting the boost as would be the nominal configuration. But, because now your loop is providing power in the other way as you are going over here from the close breaker 4 towards close breaker 3 it is actually going from right to left and you still need a boost which means that it is the opposite of the voltage what would be applied at the S V R compared to the nominal configuration where you had a boost going from left to right. Here you are having a boost going from right to left. So, if you see how this is being accomplished it means that essentially if the power flow is in this particular direction you get a boost from left to right. But, if the power flow in the previous example is in the opposite direction you need to actually change the polarity of the boost to ensure that your overall feeder voltage profile is appropriate. If you went the other way your voltage would further sag below and that would not be acceptable. So, you cannot just have because this series voltage regulators do not know what the status of these breakers are it has to make its decision on whether to provide the boost or a buck it does not have cables or signaling from status signals from the other breakers it is making its decision based on its local measurement and the local measurement that it can make is which direction is the power flow. So, you could then think about a situation where you might have a D G and say if you have a D G downstream of series voltage regulator and you are operating under normal condition of this drain and if the D G power is sufficient to send power the other way along the feeder. So, instead of providing the boost what would be the normal condition it would assume that now you have a configuration where the switchover has occurred and the power flow is going in the opposite direction which means that instead of providing a boost you could now have the voltage profile flipping over and now this whole section of the feeder will now potentially see a conflict in the voltage the S V R would try to apply a voltage of the opposite polarity. So, which could cause increasing circulating currents within the loop etcetera which can potentially damage the system. So, you can see that the issue of introducing the D G can actually cause potential problems in more complex networks such as radial distribution systems. Also, if you look at a network distribution type of system, if you look at the traditional network distribution system the network is not at the regular distribution system it is actually on the secondary distribution system is what is being networked. So, if you have transmission system coming to a substation plus feeders these the feeders would be at the medium voltage 11 k V etcetera the network is actually at the low voltage which might be at 415 volts or whatever is the consumption voltage of the network. So, here what we have shown is an example of a network you might have loads on the network connected to different points and essentially if you have a network such as this you are essentially having redundancy in the power flow into that particular network. So, if you have a fault on the network essentially the network protectors on such a distribution system what is shown over here are network protectors NP 1 through NP 4 would act to actually clear a fault within the network system itself. One functionality of a network protector is that it will allow power flow only from your feeder into your network system. So, it will if it detects that the power flow is happening in the opposite direction the network protector will open on a instantaneous basis. So, within a cycle it would try to open in case it detects a power flow in the opposite direction. So, if you have a fault on say feeder A because of the fault there is a the flow of power will be towards the fault and detecting the flow of power into the fault essentially the network protectors 1 and 2 would open and essentially the spot network would get immediately isolated from the fault. So, for any anything that happens on this particular feeder essentially it can be deenergized for servicing repair etcetera your network protectors would ensure that the power flow it does not go out of the network and the availability on the spot network would stay high. If you now think about a situation where now you have a DG connected to such a network system you have the possibility that the DG might try to send power back out into the into your feeder and your network protector might cycle open in response to seeing the power flow from the DG and you would you might still think that even if there is a one network protector that cycles cycling of the network protector is not desirable. But, even if you have a network protector that is cycling and say you have a fault now on feeder B then essentially your system would try to send out power and open up these two protectors and you might end up overloading the only network protector that is available. You might have redundancy that if one set of protectors go away you might have redundancy, but you might not have redundancy even when a reduced set is not non functional. So, you might be seeing that the loads in this overall secondary network might see a shutdown in case you have DGs that are now connected to a network distribution system a networked secondary distribution system. So, the potential for problems increase once you have more complicated configurations of distribution systems. So, with this we can actually look at the concerns of DGs and distribution systems and we will look at the concerns and the methods to address some of the concerns are primarily related to how quickly we can disconnect the DG, how quickly we can control the power flow through the DG etcetera and how many times how rapidly you can disconnect. So, the trends towards solving some of this are through power electronic means and so we will look at how effective power electronics means can be in addressing these issues and power electronics tend to be more expensive than traditional systems. So, we will have to look at first how cost effective it is which implies that you need to measure cost and see whether your engineering is cost effective to actually figure out whether you are moving in the right direction to address these problems. Thank you.