 Welcome to the class 8 and in the last class we were formulating the problem of coordinating an upstream and a downstream circuit breaker. So, we will continue with that particular problem. So, we had circuit breaker 1 and 2 and the question was how what do you mean by coordinating these two breakers. So, we discussed what should be the requirement and the points that are discussed was requiring to the requirements of the individual breakers. Then the next thing that you need for doing the coordination is to actually do you need to perform a fault analysis at the different zones in the bus and identify what is the range of fault currents that you could expect to the protective device and what is the nominal current and then you will decide on what would be the parameters that you would imply for your breaker to enable the appropriate coordination. So, we started with the data regarding the circuit breakers. So, if you look at the circuit breakers we had mentioned that the circuit breakers definite time delay of 40 milliseconds or 2 cycles. It has extremely inverse characteristic and research time of 30 seconds. We mentioned that the i square t for conductor in zone 1 is 2 into 10 to the power of 6 ampere square second and for conductors in zone 2 it is 1 into 10 to the power of 6 ampere square second and then you could calculate what is the A of requirement of C B 1 and based on our previous discussion you can relate that to the i square t divided by I pick up square and this case you needed an A less than 39.5 seconds for C B 1 and A for C B 2. We identified it should be less than 44.4 seconds and then we decide to actually have breakers C B 1 and C B 2. So, it has P equal to 2 B equal to 0.04 seconds and A is 30 seconds for C B 1 which is less than 39.5 for C B 2 your P and B is the same your A you could select it to be say 35 seconds and 35 is less than 44. So, next we will have to look at what are the fault conditions. So, if you have a fault in zone 1 your circuit breaker 1 has to operate. So, for circuit breaker 1 your trip time at I f max in zone 1 is now 30 divided by 1400 divided by 225 which is the pick up current for that particular breaker minus 1 plus rho 0.04. So, this is 0.84 seconds. If you look at the trip time at your I f min you can do the calculations it will be 1.64 seconds the current level is 1000 amps in this case. So, it is 30 divided by. So, if there is a fault in zone 1 C B 1 has to operate for fault in zone 2 essentially circuit breaker 2 has to operate. So, for circuit breaker 2 your A is 35 divided by 1000 the pick up current for circuit breaker 2 is 150 amps and if you look at your trip time at I f min the I f min is 600 amps pick up current of 150 amps. So, at 1000 amps the circuit breaker 2 trips in 85 milliseconds in at 600 amps it trips in 2.4 seconds. So, the next question is if for some reason circuit breaker 2 does not operate and circuit breaker 1 backs up circuit breaker 2 then you can ask what is that required trip duration for circuit breaker 1 during the fault range in zone 2. So, C B 1 in zone 2 is now 30 if you look at now at the minimum current level. So, if you look at now at the 2 current levels at 1000 amps C B 2 trips in 0.85 seconds and C B 1 trips in 1.64 seconds. So, C B 2 should obviously trip earlier if it does not trip then C B 1 acts at 1.64 seconds. If you look at 600 amps C B 2 trips in 3.37 and C B 1 trips in 4.9 seconds. So, if you look at this case you have a margin of about 1.64 minus 0.84 of 0.79 seconds and here you have a margin of 4.9 minus 2.37 of 2.6 seconds. So, the margin is greater than 20 to 30 percent tolerances that might occur. So, you may not have the nu sense trip problems of say breaker 1 for some reason operating before breaker 2 for a fault in zone 2. And if you can then calculate what would be the I square t level corresponding to the fault in zone 2 and breaker 2 failing to operate. So, if C B 2 fails your I square t is at 1000 amps it is 1000 into 1.64 is 1.64 into 10 to the power of 6 is less than 2 into 10 to the power of 6. So, the fault does not cause damage in zone 1. So, the damage does not spread to zone 1. So, if you look at the other range of current you have 600 square into 4.9 seconds equal to 1.78 which is also less than 2 into 10 to the power of 6 ampere square second. So, you might due to the failure of circuit breaker 2 you might have some damage in zone 2, but the damage does not now propagate into zone 1. So, the next thing that you would have in this particular situation is we had identified 3 zones. So, zone 1, zone 2 and zone 3 what happens for a fault in zone 3 ideally we have not indicated what is the protective device that is used in zone 3 it could be circuit breaker or it could be a fuse. The question is will the breaker C B 2 act as a backup and if it is acting as a backup what is the fault I square t that is happening in zone 2. So, if you look at fault in zone 3 the primary protection the primary protection is the local protection and C B 2 is backup and if you look at now your trip time for I f max in zone 3 you have 35 divided by and if you look at your trip time at I f min in zone 3 your current level is now 450 amps and you could then calculate what your I square t level is. So, your I square t at 600 amps that turns out to be 8.54 into 10 to the power of 6 or 10 to the power of 5 which is less than 1 into 10 to the power of 6. If you look at the 450 amp level you get 8.9. So, again in case there is a failure in the protection in its local zone then the C B 2 settings will not cause the fault to propagate into zone 2. So, there is backup protection however in many situations it may not be possible to apply to obtain all the possible conditions. You might have situations where if you look at the example that we have been looking at the fault current levels are not overlapping if you have overlapping ranges of fault current levels you may not be able to achieve all the conditions of your coordination requirement. Also you might have situations where your upstream source might have impedances that are varying over a wide range you might have transformers that are switching in and out you might have lines that might be tripping. So, the impedances might be varying over a wide range and over that entire range it may not be possible to achieve the full coordination. So, you might not be able to meet all these requirements but your objective is to try and meet as many as possible. So, the other thing that we did in this particular example is look at individual values of fault current levels and look at the maximum minimum and look at what the timing trip timings are of the breakers. Another way of looking at coordination is to plot your trip time versus current and see where the curve lies and we will look at this say in the case of a fuse and by looking at the position of the curve you can get a broader feel for is the coordination working or not is it possible for one device to trip before the other device. So, many times software that do the coordination calculations can actually give plots of time versus current coordination curves which also give you a visual feel of whether the coordination is happening over a range of conditions. So, in the second situation we will look at case where you are coordinating say between two fuses the upstream and downstream fuse. So, in case of circuit breaker you had you have the pickup current and the actual current flowing through the protective device in case of the fuse you have looking at the ratio between your actual current flowing through the fuse and the current required to melt the fuse. So, similar to what we defined for the circuit breaker you could actually now define a ratio M melt ratio as your actual fuse current divided by. So, as long as the current is below this melt current the fuse is not going to get damaged. Once you go above that particular melt current level you have a possibility of the fuse melting and the time required to melt would now depend on your value of the current that is flowing through the fuse. So, in this particular case again we can define say zone one and zone two and we can define what it means by coordination between fuse one and fuse two. For example, if you have a fault in zone one fuse one is the device which needs to operate by coordination if you have a fault in zone two you want fuse two to melt without damaging fuse one. So, that is essentially the requirement for a coordination between a upstream and a downstream fuse. So, you can now define in case of a upstream and downstream fuse if you have a fault say in zone two then the fuse that is actually protecting is actually F 2. So, F 2 might be called the protecting fuse and F 1 is actually the protected fuse F 1 is actually protected by F 2 and you want your protecting fuse to be below in terms of your fault melt time versus current characteristics you want it to be the curve to lie below that of the protected fuse. So, the protected fuse would typically have larger I square t and what you want to ensure is if you take a typical component you would have tolerance you would have the minimum melt time and the maximum melt time. So, in this particular case your solid lines might correspond to the minimum melt time required for the fuse to melt and your dotted lines might correspond to the maximum melt time. So, what you want to ensure is that the your some k times your t min of your protected fuse in this case F 1 is greater than your t max of your protecting fuse F 2 and k is a number which is less than 1. So, for fault current ranges in zone 2 you want a condition like this to be satisfied say you could take k as say 70 percent or 75 percent. So, you want to ensure that with some margin you have sufficient distance between this particular point t max of F 2 and t min of F 1. So, I think that gives you a picture about what it takes to actually do coordination between upstream and downstream fuse and here you can see that you the curve gives you a picture. So, this shows at one particular point the range of current that might be experienced by F 1 might be a range somewhere like this the range of current experienced by F 2 might be a smaller range. So, you want to make sure that within its range this coordination is being achieved. So, next we will look at the case where you are trying to achieve coordination between say recloser and circuit breaker say you have a upstream recloser and then you have a circuit breaker downstream. So, your recloser might be sitting at the substation and your circuit breaker might be somewhere further down the downstream and by a recloser it need not just be the timing of trying to stay close for some time then opening staying reclosing again etcetera the logic underlying the recloser might also have a circuit breaker characteristic. So, in addition to the recloser characteristic there might be a underlying circuit breaker characteristic. So, the circuit breaker in this particular R can be coordinated with CB 2 like what we have discussed previously. So, now we will focus on what the characteristic of the timing in terms of the cycling between the open and close of the recloser should be. So, that you can coordinate between R and CB 2. So, the first question is what you mean by coordination is this particular case if you have a fault in zone 1. So, obviously the device that need to protect that is your recloser. So, R has to clear faults in zone 1. Now, if you have a fault in zone 2 the question is that fault temporary or permanent. If you have a temporary fault in zone 2 then it should it can be cleared by the recloser by its recloser action. But if you have a permanent fault in zone 2 you want CB 2 to open before your recloser locks open. So, if you have a fault in zone 2 CB 2 would open then during the reclose the next reclose cycle your zone 1 would go back to normal operation and it would continue with this particular breaker being open and the section in zone 1 having its normal power feed. So, if you look at then what the logic in the recloser is if the current in the recloser goes above some particular threshold value it will open. So, once it opens the current goes down to 0 it will wait for some duration and then it will close again. So, after it recloses it will see whether its current is still going above the threshold and then it will go through such cycles of say closing for say n cycles after n cycles if the current is still above the threshold then it stays locked open. So, in terms of logic it is a fairly simple and straight forward logic in terms of what the recloser would do. So, if you look at the recloser logic if i r m s through the recloser is greater than i some threshold. So, one it opens it waits for some time then it recloses and if i r m s. So, depending on what the n is you can have 2 cycles 3 cycles number of cycles of the recloser if. So, with this logic we will make use of the same data for say zone 1 and zone 2 and look at an example of what can be done when you have a upstream recloser and downstream circuit breaker. So, we will look at the example where say the recloser if say you have a fault happening at some particular point of time for a short duration the current level goes high the duration t naught we have taken as 0.1 seconds then it opens for a duration of 5 seconds then it recloses it recloses now for 0.7 seconds if the current level stays high again for in this duration it opens again for 5 seconds. So, after 5 seconds it try reclosing a second time and there it stays close for 3 seconds if the current is continuing to be above your threshold level even at the end of this particular 3 seconds then it will lock open. If the current say came down at some point of time before this 3 seconds ended then essentially the recloser would go back to normal operation because it would not lock open it would go back to the normal conditions. So, that is essentially the logic. So, we will take these as the numbers in our example for the recloser for circuit breaker 2 we will use as use the same parameters as we had in our C B C B circuit breaker to circuit breaker coordination example and we look at the situation when you have faults in zone 2. So, what is shown over here is an example of fault in zone 1 it tried a permanent fault in zone 1 it tried 2 reclose cycles and it just locked open at the end of it. So, the current came down to 0 and stayed 0 at the end of such a situation. So, if you look at zone 2 fault what your objective is it use of reclosing to clear temporary faults and recloser. So, we will look at what happens in this particular situation when you have fault current in bit in the maximum and minimum current in that particular range in this particular example. So, at 1000 amps which is the maximum fault current level in zone 2 you have your true time of your circuit breaker is 0.85 seconds and if you look at now your T naught duration of 0.1 seconds then this corresponds to the breaker advancing by 0.1 divided by 0.85 which corresponds to 11.8 percent of your trip time for the circuit breaker. So, it is about 11.8 percent to trip at that particular condition and then you have actually the first half duration T of 1 of 5 seconds and the 5 seconds in terms of the overall reset time for the circuit breaker which is 30 seconds it means that it is resetting by 16.67 percent of T reset. So, obviously it is not going below 0 it is now resets back to 0. So, at the end of your cycle over here at this particular point your circuit breaker 2 is actually now fully reset by the fault current level it went up to 11.8 percent it came down to 0 it is fully reset at this particular point of your operation of the recloser. Now, if you look at your next action. So, if you look at your next action if you look at your close duration 1 T c 1 of 0.7 seconds. So, that corresponds to 0.7 by 0.85 of your circuit breaker. So, it is actually about 82.8 percent of your trip time. So, in this particular case it is advanced to 82.8 percent then if you look at your second open duration T of 2 of 5 seconds which is now 16.67 percent. So, if you look at your circuit breaker. So, at the end of your second T of duration. So, at this particular point over here at this particular point over here you are now about 66 percent of your trip level in your circuit breaker. Now, if you look at your close duration 3 of 3 seconds 3 seconds divided by 0.85 is 355 percent. So, obviously it has tripped before the 3 seconds has ended. So, your actual trip time for your circuit breaker that corresponds to 100. So, it did not required 3 seconds it tripped in 0.3 seconds in the second recloser. So, at this particular condition essentially zone 1 goes back to normal operation because the current level return back to the condition of the normal loading of that particular zone. And what the recloser did was it made 2 attempts to clear a temporary fault in zone 2. And because it was permanent it tripped the breaker before the recloser locked open. And if you look at the margin you had 3 seconds minus 0.29 seconds you have 2.7 seconds of margin at 1000 amps current level. So, next you could actually look at what happens at the lower current level because you will look at across the whole range. So, if you look at the current level of 600 amps your trip time is of the breaker. So, you know your trip time at now your reduced current level at the minimum current level. So, if you look at now your T naught in this case of 0.1 second this corresponds to 4.2 percent of TTR. If you look at T of 1 of 5 seconds it is 16.7 percent of T reset. So, again the breaker is fully reset at this particular point. If you look at now your reclose duration T c 1 of 0.7 second that corresponds to about 29 percent of your trip time. And if you look at your T of 2 of 5 seconds which is 16 percent your circuit breaker 2 at this particular point. So, if you look at the point over here at the reduce fault current level it is about 12.8 percent into tripping. So, that is the situation of the circuit breaker at that particular point of time. So, if you look at then the next duration of reclosing which is for 3 seconds. So, if you look at T c 3 equal to 3 seconds. So, this corresponds to 3 divided by 2.37 or 126 percent of trip time. So, again the circuit breaker is going to trip before it when it reaches 100 percent. So, it is going to trip definitely before the 3 second duration that you have. So, your actual tripping. So, if you look at you had 12.8 percent from your first reclose cycle. So, what remains in your second reclose cycle is actually 87.2 which is 100 percent. So, if you look at the corresponding time you get of 2.1 seconds would be that time required for your breaker to trip. So, the margin that you have in this particular case is 3 seconds minus 2.1 or about 0.9 seconds of margin at the lower current level. So, you can see that the margin is reducing as a current level becomes lower and lower. So, at way low current levels your breaker needs more and more time to trip. So, at some level there is a possibility that you might actually cause a lockout before your breaker operates. So, one thing you could do is you could actually decide on what your threshold level for your reclose needs to be. So, if you look at your fault current in your zone 2 or further downstream if you have a fault in zone 3 etcetera your current level can come down below 600 amps. So, there is a possibility that at some reduced current level your reclose might actually lock open before the downstream device acts. So, one thing you could do is you could verify now through calculations at my F of 524 amps your CB2 trip will coincide with our lockout. So, if your current level is above say 524 amps then your CB2 would trip before the recloser operates locks out. So, if you set your threshold current level for your recloser to be higher than 524 amps then the recloser would not operate at the reduced current level and ensuring that you do not have nuisance strips upstream because of reduced fault current levels that are occurring downstream. Though in typical protection coordination practice you would typically design the upstream protection before you decide on what you do for the downstream. In some of these situations you will have to look at the entire situation to decide on what should be the settings. So, that you do not have nuisance strips. So, another situation that you can potentially have is then looking at a situation where say you have recloser which is sitting downstream and fuse or a circuit breaker which is upstream. So, we will look at an example where you have a upstream fuse and a downstream recloser. It could be say a backup protection fuse operating with a downstream recloser and then the question is how do you then determine the coordination settings. What does it mean for coordination to be done in this particular case. So, obviously in this case again if you have a fault in zone 1 it has to be cleared by fuse f 1. If you have a temporary fault in zone 2 your recloser should clear the temporary fault by its recloser action. If you have a permanent fault in zone 2 then your fuse f 1 should not get damaged which means that your recloser should lock open before fuse f 1 gets damaged. In the previous example your recloser should not get locked open before your circuit downstream circuit breaker operates. In this case you want your recloser to lock open before your upstream fuse gets melted. So, we will look at this particular example in greater detail in the next class and the main idea behind looking at these situations in a distribution system is that now if you have if you are doing coordination of your protective devices in some particular manner. Now if you add say a DG to one of these buzzers what would be the new situation? What would be the new fault current level? What would be the changes that you would need in the system when you introduce the newer distributed generation device and this is something that we will continue with in the discussion in the next class. We will look at the example of the upstream fuse and a downstream recloser. Thank you.