 Welcome to class 29 in topics in power electronics and distributed generation. Today we will be discussing few problems, these are homework problems for the students in the class and example problems for the students watching on the net. So the first couple of problems are related to distribution systems and relaying and then the other two problems are related to economic decisions that you could make in making engineering choices for DG systems. So the first problem we are looking at a on-load tap changer which at a distribution substation and 66 slash 11 kV and the substation transformer has 9 taps and each tap step is of 0.02 per unit. So tap 1 corresponds to 0.92 per unit, tap 5 corresponds to 1 per unit. So if you are looking at single line diagram you are having taking the on-load tap changer as depending on where you are placing the taps. So here you are at tap 1 your secondary voltage would be 0.92 per unit at tap 5 your secondary voltage would be equal to 1 per unit at tap 9 where the primary turns is say the lowest then your secondary voltage would be the highest. So VS would be 1.08 per unit and so this is when you are having nominal voltages at the input and output and you are having a 11 kV overhead line of 6 kilometers going out from the substation and you are given the feeder line has a impedance resistance of 0.6 ohms per kilometer x by r ratio is equal to 1. The loading of the feeder is uniform 25 amperes per kilometer so total 25 into 6 so 150 amps total for this single 6 kilometer feeder and at a power factor of 0.9. So you want to make plots of voltage profile current profile etc. So first we look at the circuit implementation of the tap changer and so how the on-load tap changing is being achieved. So if you want to change taps in a structure such as this if you want to go from one say tap 1 to tap 2 if you do it without interruptions say if you simultaneously contact 1 and 2 then you are shorting a coil so you can end up with large currents in the shorted coil. So how is that achieved and what is the sequence of operations to actually limit the maximum current during tap change to be say 1 per unit the rated current when you are doing the switching action. So one possible configuration for so if you look at the tap changer you can think of it as on a per phase basis it is a 1 pole 9th row switch. So if you are on a 3 phase basis it is a 3 pole 27th row switch so it is essentially a switching action between multiple throw points. So how could the switching action be achieved while limiting the shorting current so you could include resistors resistance of the tap changer to actually ensure that when you are actually having a shorted loop you limit the current. So sequence of actions for the tap changing could be say originally you might be connected to tap 1 so this switch might be closed and under normal conditions your switch SA would also be closed so this switch would be in close position so your pole is connected to throw number 1 through the switch SA and through switch 1 so this would be your starting configuration you want to eventually go to a configuration where switch 2 is actually connected to the throw corresponding to switch 2 is the one that is connected. So the first step would be to open switch SA so once you open SA the power flow would then be through the resistance RTC so you want to have a small value of resistance so that you do not have large drops in voltage and we will see that a small value of resistance is sufficient. So once you have the SA open then at this point you could then close your throw 2 and with throw 2 closed and with SA open now you have the resistance 2 RTC in the loop so if you want to limit this particular current to be less than 1 percent we know that the voltage over here is 0.02 per unit so you want 0.02 per unit divided by twice RTC to be roughly 1 your 1 per unit current which means that your RTC is about 1 percent resistance so it also meets the criteria that when you open the switch you will not have a large drop across that particular resistor. So this would be your second condition where you have close switch 2 then under this condition you could now open the throw 1 in which case the current flow will actually change from the top it will actually now flow through this particular path and then you could close switch SA back and because switch SA is closed now your current diverts through the switch which means that the losses in the resistor will be happening only during the transition time you might be operating a transformer maybe in many tens of minutes maybe in R it is not like something some that switches very frequently you are trying to ensure that over the course of the day when the voltage levels over a longer time frame changes you want to ensure that your secondary sees the nominal voltage. So in the next problem you are asked to plot the voltage and current profile on the distribution feeder assuming the substation and voltage to be equal to 1 per unit and this is based on the expressions that we derived in class and will also be discussing it in part C of the problem. So we know that with uniform loading your current profile is going from 0 to 1 per unit and in a uniform manner with a constant slope because the loading is uniform and the nature of the voltage profile starting from 1 per unit at the substation end it drops to about point slightly higher than 0.96 per unit at the end and the expressions for this will actually look at in part C of the problem. So the question is if we have a voltage profile that looks like this we do not want just the loads connected at the feeder end to see close to 1 per unit voltage we want to the average user of this particular feeder to see voltage close to 1 per unit. So it would be nice to actually ensure that the deviation at both ends of the feeder stay roughly equal so that you are within as the same tolerance range across the entire feeder. So for that to happen what should be the distance from the substation that the voltage that is regulated to the nominal value so that the whole feeder voltage is kept minimal. So for that we the current distribution along the line is given by so I at position x is I s to 1 minus x by d where x belongs to 0 comma d and the voltage at x is the voltage at the substation minus r l prime I s. So we had derived this in class and I s is the feeder current at the substation and here r l prime is the total feeder effective resistance is given by r l which is the actual resistance into p with power factor plus the x by r ratio of the line into square root of 1 minus p square. So if we take the voltage at the end of the feeder to be v d and v s to be the voltage at the substation end and we want this to be equal to 2 delta v then this is given by r l prime I s by d square into d square minus d square by 2. So r delta v is r l prime I s by 4. So if you ensure in this particular so if you ensure that this delta v range is equal to that particular value then your average user of the on the feeder will see voltage in the appropriate range. So the distance at which you get this particular value for delta v is r l prime I s that is what we need to solve. So we want to ensure that the voltage drop at x d of x x square by 2 is equal to this particular value which is r l prime I s by 4. So this is a quadratic equation in x which can be solved to get the value of x. So you have x square minus 2 d x minus plus d square by 2 equal to 0 and a solution in the range is 0 to d is d into 1 minus 1 by root 2. So this is equal to 0.29 roughly at one third the distance from the substation. If you are regulating it to the nominal voltage then over the entire feeder you will be within the same tolerance range. Again this is assuming that you are having uniform loading you are also assuming that your power factor is the same across the loads on the feeders. But this gives you a feel for what would be the nature of the requirement to maintain the voltage along the feeder. So in the next problem we would like to see what could be an expression that could be used at the transformer to predict the reference voltage that is required at the substation end based on this loading that we have just derived to ensure that the voltage regulation across the feeder stays in the required range. So we want v at x to be at x is equal to d of 1 minus by root 2 need to be to 1 per unit and this is a open loop because you are not explicitly taking measurements from that particular point and using it to control the tap selection so you have v at x equal to v s minus R L prime is by d square into d square 1 minus 1 by root 2. So this simplifies to v s minus 0.25 R L prime is. So you look at the total current at the substation end so this is essentially your actual feeder current and this term over here is essentially a voltage boost. So if you look at the actual voltage that has to be commanded at the substation v star it would be your nominal voltage you want to keep this particular value v at x to be equal to the nominal. So this is v norm plus 0.25 R L prime is. So this is essentially a boost term that you are providing and here we are assuming that in actual substation there are multiple feeders so we are also assuming that all the feeders have similar characteristics over the day so it is not just that one particular feeder has load the other has no load so it is a similar type of consumption on depending on the time of the day etc and you want to actually provide a boost action across all the feeders that go out from the substation. So in the next part of the problem we are looking at the voltage at the high voltage and if it is 70 kV what would be the tap position of the on load tap changer that would provide closest to 1 per unit voltage on the low voltage side assuming there is no compensation provided for line loading. So without the boost action what would be the voltage just assuming the primary side voltage has now gone high. So what is the actual secondary voltage in this particular case so with so your v norm by your v at your high voltage actual would be 66 divided by 70 so this corresponds to 0.943 per unit so if your v actual was 66 kV you would be at 1 per unit so because your actual voltage is now lower so your ratio turns out to be 0.94 if you look at what would be the closest tap that provides this particular voltage it would be the tap it would be the second tap so closest to 11 kV voltage it would be tap 2. So if you look at your actual secondary voltage so you have 70 into 11 by 66 into 0.94 which is the tap that has been selected so you have 10.97 kV and then if you look at the corresponding feeder voltage profile so you would have so the one start the red one starting from 1 per unit is the one corresponding to the base case where you have 1 per unit at the substation at vs so at d equal to 0 so you can see that in this particular case even though the secondary voltage has gone higher to 70 kV the actual voltage at the substation end is actually starting at 0.997 and falls down in a similar trend because the impedance profile is the same along the feeder so it is not necessary that you have a higher voltage on your primary you will see the same effect on the secondary depending on the tap selection you could have a voltage which could be on either side again this is because of the finite quantization effects you have only 9 taps you do not have a smooth variation of the taps starting from say your minimum value to your maximum value. So, in the next problem you are asked to actually look at what is the tap position that the transformer the OLTTC would provide closest to 1 per unit at a distance that is obtained in part C where we are looking at at what position along the feeder you would like to regulate so here we are looking at the light loading compensation at a loading of 25 amps per kilometer and when the high voltage side is sitting at 70 kV. So, in this particular case the so one thing when we consider say if we take 11 kV as your primary plus your parameters of your substation current to be say 150 amps your resistance of RL prime based on the calculations we have just made one thing to note as this is on a phase basis. So, whereas your 11 kV is on a line to line so you need to make sure that you consider things on a line to line basis including the appropriate change in voltage. So, this turns out to be 11.31 kV. So, if you look at your V s star by V norm you get 1.028 so again it is between 1.2 and 1.4 the closest being 1.2 it means that you need to boost of one tap because of the loading effect. So, your actual voltage so previously you are sitting at tap 2. So, if you boosted by one tap you now go to tap 3. So, if you look at the feeder voltage profile with the line loading compensation you will see that so the original case B was this one in the next case we were looking at the effect of just the voltage going to 70 kV and here you can see that you are getting a boost effect of due to the line loading compensation ideally you would like to boost it to so that one third at one third distance you would get close to one per unit, but because use your actual selection was 0.28 if it was boosted by one more step the voltage at the substation end would have gone to high. So, this would give you the closest to the desired profile. So, in the next problem instead of boosting it at say 0.29 0.3 D so 0.3 D would correspond to about 1.74 kilometers. So, instead of boosting the voltage at the 1.74 kilometers if you want say you want to boost it at 2 kilometers distance so that you want to have a slightly higher boost action then what would be the voltage here for this particular problem we are considering a slightly lower loading of 20 amps per kilometer and we are considering a voltage on the high voltage side at of 62 kV. So, we are looking at this particular nature of the voltage profile. So, if you get V of x equal to V s minus R L prime I s by D square into 2 D minus 2 square by 2. So, essentially you get V s star is equal to V norm plus R L prime I s by 3.6. So, essentially you get a slightly different boost term and if you look at the situation where in this particular case I s is it would have been say 0.15 kilo amps at 25 amps per kilometer loading at 20 amps per kilometer loading you have 0.15 into 20 by 25 equal to 1.012. So, this is the current. So, if your high voltage primary side voltage is 62 kV your desired tap point. So, if you look at what the is the closest to 1.065. So, after 1.06 you have 1.08. So, which is tap nine. So, this would correspond to tap eight would be the closest tap and your V s star in this particular case would be 11 kV plus 4.801 which is in which is your R L prime into 0.12 root 3 by 3.6 this would correspond to. So, essentially again this would correspond to boost by 1 tap. So, your actual tap location would be tap eight plus 1 tap boost for your line drop. So, your tap your actual tap location is 9 and your secondary side voltage would be 62 into 11 by 66 into 1.08 which would be 11.16 kV. So, if you look at where it would lie it would essentially lie starting at somewhere around here. So, this would be case G. So, starting at around 1.015. So, at distance two it is again around 0.98 per unit. So, again because of the finite number of tabs you have these quantization effects which means that you cannot actually exact number that you are targeting, but it would be the one that is closest. So, in the next problem we are looking at the options for increasing the capacity of the line and we are considering the case where two conductors are placed in parallel on the particular feeder. So, in this particular situation what is the new voltage at the end of the line and the case that is being considered is parallel conductors and not the case of parallel lines. So, we will look at what could be this the possible situations that you could have when you are say paralleling conductors. So, if you look at the original line. So, your original line might have say a pole with say three conductors on top. So, this might be your original configuration. So, we are looking at say configuration two where you have a pole and you have now parallel conductors on each insulator. So, this would be the parallel conductor case. You could also have other configurations say you could have a configuration such as this where you put two cross arms and then you have say conductors on now both cross arms. So, or you could have two poles. So, you could look at what is the effect of these different configurations. So, here you could think about this as a parallel feeder line, parallel circuit. So, this say this is case 3 case 4. If you look at effect of you will see that your R line is half because you now have twice the effective cross sectional area. So, it becomes R line by 2 and your X lines stays roughly the same. So, your X value does not change because your loop areas are not significantly different for the current conductors the distances between your conductors are roughly the same. Whereas, if you now look at the case where you are having say parallel conductors you now have effectively two parallel impedances essentially you will be changing your X line also. So, here your loading capacity increases your X by R ratio changes. So, your X by R whereas, in option 3 and 4 your X line would shift to approximately X line by 2 which means that your X by R ratio would stay roughly the same. So, with this you could then calculate what is your effective say resistance of the new line or your voltage regulation calculation your R L double prime is R L 2 by 2 because your resistance is halfed, but your power factor is P X 2 by R L 2. So, if you look at your R L double prime this is 3.19 ohms whereas, if your original line this was 4.81 ohms. So, because you now have just parallel the conductors it does not half, but it falls by a by some percentage, but does not exactly half. So, if you look at your voltage at the end of the feeder assuming 11 kV at the substation your V of D. So, this turns out to be 0.962 times your 11 kV. So, this is 10.59 kV. So, if you look at the voltage profile you can see that the original voltage went down to say 0.94 that was around 10.3437 kV. Now, with the parallel conductors you have 10.59 kV. So, your voltage drop reduction is about 210 volts if this corresponds to 11 kV. So, there is a reduction in voltage drop, but it is not as much as just taking the entire R L prime and taking half of that. So, you could then ask what is the difference in power dissipation in the line, what is the power dissipation that happens in the line and what was it in the original configuration and what is it when you have the parallel conductor as in the case above. So, you can write an expression for the power dissipation in the line. So, we know the expression for the current which is I s into 1 minus x by D and R L is the total resistance of the line. So, your resistance per unit length is given by R L by D. So, we can write an expression for what is the power dissipated along an incremental length delta x along that particular feeder is given by I of x square into R L by D into delta x. So, if you integrate this along the line your power loss in the line would be and you can write the expression for that I s. So, you get I s square the D's cancel out R L by 3. So, this is the power loss per phase of the feeder line. So, you multiply by 3 to get the total loss on a three phase basis. So, power loss in case B is 3 into 150. So, this turns out to be equal to 81 kilowatts. So, 81 kilowatts happens to be around 2.8 percent of feeder power. So, you can see that there is quite a bit of loss along the feeder and when you have the parallel conductors it is half of this particular number. So, you get 40.5 kilowatts or 1.4 percent of feeder power and often the reason for putting parallel conductors is not from the power loss reduction perspective it may be to increase the capacity of the line. So, the number that would be used when you have parallel conductors would be not the same level of loading there would be increased loading because you know your purpose for adding the parallel conductor is to actually serve more loads often. And the limitations of how much participation is often related to what is a temperature rise that the conductors could see which would also limit how much loading can happen on a per kilometer basis. So, in the next problem you are asked if you have now 2 DGs that are installed on this feeder where would you locate it and there are if it is possible to install 2 DGs on the line at what distance and power level would the DGs lead to flattice voltage profile on the feeder and look at the resulting voltage profile. So, if you look at the case of adding the fewer the 2 DGs you can see that most of the voltage drop occurs because of the higher IR drops that are occurring on the feeder. So, an objective of adding DG would be to actually reduce the drop the R is the property of the conductor what you can alter is your current profile and by adding a DG at a distance of at 2.4 kilometers which is 2 fifth of the distance and at 4.8 kilometers you can actually bring down the current level amplitudes of the current to be the smallest along the feeder. So, if you look at the power flow in this particular case essentially your substation power is actually flowing in to a short distance over here say section 1, section 2 and say section 3 the power is flowing from say the DG in section 4 and 5 is flowing from say DG number 2. So, if you look at then what would be the power rating for your 2 DGs your DG each of this DG we would need to actually supply the power corresponding to this particular section of the feeder and the power level would be there are 5 identical sections and the total length is 6. So, 6 by 5 and the location of the first DG is at the between your second and third section. So, you have 2.4 kilometers the second is 6 by 5 and the location is between your fourth and the fifth. So, that is why you get 4.8 kilometers and your DG current rating would be 150 amps which would be the total overall, but you are supplying it now to 2 sections into 2 by 5. So, it is 60 amperes at power factor of 0.9, 0.9 lead because the loads are at 0.9 lag. So, if you look at your MVA level of your DG this is 0.6 into 11 it is about 1.14 MVA and your power level is is about 1 megawatt. So, you need about 2 1 megawatt DGs located appropriately to get the flattest possible voltage profile. If you look at the voltage drop now you can consider a short feeder of the length of say the triangle in section 1 and you can calculate what is the maximum voltage drop the lowest point that it goes to is 0.9988 volts. So, there is hardly any voltage drop you can see that in this particular case it was going to almost 0.94. So, it is the voltage drop has become much flatter compared to the original situation. The last part of the problem is what is the participation in this compared to the previous case case B then we could also consider the parallel conductors and the case when you have DG. So, you could essentially the procedure adopted could be you could calculate what is a power loss in one small triangular section and we know that there are 5 such sections to actually calculate the power loss. So, if you look at it on a percentage basis this is about 0.1 percent. So, you can see that compared to the 2.8 percent in the original case the participation has reduced almost by a very large factor and the power loss is reduced significantly, but it is not always possible for us to site DGs at ideal locations and supply power at the desired generating levels. So, the DG can provide some level of service, but to increase the capacity of the line you need to put in parallel conductors. So, having both options of strengthening your system and adding distributed generation can actually give you overall benefits. So, the second problem is on switching that we will discuss is on the switching of DG and the grid. Essentially you have a switch which interconnects a DG and the grid and you want to see what would be the transient currents. Under two conditions one say when you are operating the DG in parallel when the switch is closed and say you have for some reason there is a fault and the voltage on the grid becomes a dead shot. So, you will have a transient current flowing out you want to actually look at what is that value. Then the second situation that you want to see is in response to the fault the switch has opened and now you have voltages of the grid and the DG which are close by, but not exactly identical. So, when you close the switch what would be the transient currents that are flowing between your grid and the DG. So, for the first problem you are looking at a model. So, your DG is 100 kilowatts your grid voltage is 415. So, on a per phase space it is 415 by root 3 50 hertz the switch was closed and you are opening it your excel is 0.2 per unit your power is 100 kilowatts power base V base is 415 by root 3 239.6 volts I base is 1.139 amps Z base is 1.72 ohms. So, your excel of 0.2 per unit is 0.34 ohms. So, your I fault is 239 volts by 0.34. So, it is 696 amps RMS. So, you are including taking the peak of the sinusoid you gets a factor of square root of 2 including transient effects due to large x by r ratios your peak fault current can be about twice square root of 2 times 696. So, you are talking about 2 kilo amps can actually flow out when you have a dead shot on the grid and your switch S takes a finite amount of time to open. So, before it opens you might end up with a fairly substantial peak current. So, the second problem is about is about what would be the voltage what would be the maximum voltage error between your grid and the DG that can be allowed the voltage error amplitude. So, that you your surge current is kept below 200 amps here will ignore the DC transient switch effect. So, if you but the 200 amps is the peak value that you want to keep your current below. So, you could calculate what your your I RMS is 200 by root 2 is 70.7 amps. And if you look at the circuit the error between your grid voltage and your DG voltage is the voltage that is being applied across the inductor. So, you want to ensure that that voltage that is being applied it has a magnitude such that the resulting current RMS current stays below this 70.7 amps. So, you have your delta V by 0.34 which is your impedance is 70.7. So, your delta V RMS is 24 volts. So, as long as your DG voltage lies anywhere in this particular circle your you would be ok because your amplitude error will not lead to a current greater than 70.7 amps. And the next problem you are asked what is the maximum phase angle difference between your grid and DG voltage such that this condition is satisfied. So, you can see that the maximum voltage error could occur when the voltage is such that it is just touching the circumference of the circle. So, this particular tangent is just touching the circumference of the circle at this point. This particular tangent is touching at this particular point. So, you could then make use of that to calculate what would be the maximum angle error. So, you have delta theta is sin inverse. So, sin inverse delta V G delta V by your EG. So, this turns out to be 5.8 degrees or 0.102 radians. So, it gives you an estimate of how far you can actually let your angle be so that your surge current is kept below your desired 200 amps. Then in the next part of the problem you are told that instead of using 5.8 degrees your actual relay is set to close within a 2 degree angle. So, this particular angle is held at 2 degrees and you are asked what is the maximum frequency difference between your grid voltage and your DG voltage such that you do not exceed 200 amps. So, essentially your grid voltage and your DG voltage there might be 2 vectors. So, they are moving apart at a frequency given by 2 pi F DG minus FG into your delay time of operating your switch because it takes a finite time for a switch to close. Here you are told that your switch takes 100 milliseconds from your logic command to actual closing and you want to relate that to what is your maximum frequency erosion. So, you can then write an expression you can relate this particular angle to be equal to 5.8 minus 2 into pi by 180. So, you get essentially this which is your delta F. So, you get delta F to be equal to 0.106 hertz. So, if your grid voltage is at 50 hertz you want to ensure that your DG voltage is 49.9 hertz to 50.01 hertz 50.1 hertz in that particular range to ensure that the surge current during closing will not exceed 200 amps under the condition that you are closing now at an angle of 2 degrees rather than 5 degrees. If you are closing at this particular point because of the delay you might have gone out to something much further out. So, you need to actually ensure that when you are closing a switch your amplitudes are matched, your phase is matched and your frequency is matched appropriately. In the next class we will discuss couple of problems related to economics of operation of the DG which can help with your engineering design. Thank you.