 Welcome to class 18 on topics in power electronics and distributed generation. In the last class we were talking about relays for connecting DG for the DG operation for the interconnection and for protection. We linked it to these requirements to standard ANSI relay protection and usage and what you could do is then work out the logic required to say operate the interconnection protection or the generator protection. So, each of those relay functionalities can be linked to under what conditions should some particular breaker for example, 50 to 3 when should it open and when should it close. So, that logic can be done as a fairly simple combinatory logic exercise from the outputs of the relays I will leave that exercise to you. So, another aspect that need to be considered when you do such an interconnection is that what should be the actual settings of the protection relay values for example, in this 59 under voltage relay the question might be what would be the neutral voltage setting that has to be included or say for example, in the 51 g what is the ground current level, what is the over current level at under what frequency under frequency would the decision be made to disconnect the device. So, these are actually the engineering evaluations that you as a system engineer would have to decide on whether what setting is required and those settings would typically be based on the analysis based on the ratings of these units the sizing of the components the fault calculations etcetera which you have got a feel for at this point of how to do those calculations. So, another thing to keep in mind is that in these relays that are implemented all of them are quite are standard relays I mean you do not have something which is not already there available as a standard the only thing which could be considered as being something that is not already there is the issue of how to detect an intentional islanding especially when you have a wide range of situations for example, you could use this 32 as reverse power relay to actually detect whether the power is going out of your PCC to detect whether it is an intentional islanding, but then that would restrict you to have a DG where the power is not allowed to go out if you have a say a large solar farm or a large wind farm your actual intention of operating the device is to send power out. So, you do not have a standard anti islanding protection relay other than that everything else is actually standard protective protection equipment which you could actually program. The other thing to keep in mind is that scaling this functionality to a at a power level of 10s of kilowatts or 100s of kilowatts might be because you need to add say the protective devices for interconnection the sensors etcetera and in terms of the cost at the 100s of kilowatt power level it may not be a significant cost addition to the DG cost in terms of the protection additional equipment required for interconnection and DG protection. However, if you are now looking at a small 1 kilowatt solar panel that you want to interconnect to the grid and if you want all this functionality then the cost becomes a significant factor compared to your photovoltaic system cost. So, scaling it to lower power level is still a challenge to some extent this issue of scaling to lower power level can be overcome because now the digital controllers that are available today can do a lot of complex functionality within one single controller. So, potentially you could then have a lot of this functionality now embedded within the controller of your low power DG that you are trying to interconnect, but if you want to have separate relay packages to implement this then scaling to lower power level is a challenge. The other thing to keep in mind is what we have shown over here is just an example your protection requirements would change depending on whether it is for example, whether the transformer is delta y y y y delta or whether the grounding is different whether it is low impedance high impedance grounding etcetera. So, you would have to actually modify your protection schemes to accordingly to actually address those circuit and component issues that you are having in your actual system. This is just an example to give you a feel for the issues that you would need to address when you are doing interconnection of a distributed generator to the grid. So, today we will look at another issue which is of importance is that of the voltage along the feeder and the voltage at the feeder especially at the point of common coupling is a important parameter. So, suppose if in this case if this was a high power facility and this was a point of common coupling you want to ensure that that particular voltage stays close to the nominal value. So, you may not be able to have the voltage exactly at the nominal value there might be some acceptable range like plus 5 percent minus 10 percent and you want to keep it within that range and you want to do that not just for one particular location you want to actually do it all along the feeder. So, irrespective of whether you are close to the substation or whether you are sitting at the end of the feeder you want to actually stay within the tight range and physically it might mean that irrespective of whether you are sitting at a distance 0 or a distance of multiple kilometers away you want to actually make sure that your voltage is in a tight tolerance range how it is done and then we will also look at the issue of how adding a DG can actually affect how the voltage regulation can be accomplished. So, we will start with a simplified lump of the system will take the substation voltage the source voltage as V s and will lump the impedance of the line to a Z line which could consist of some resistance and reactance to get the overall impedance and again will lump the load together at one point and call it the load voltage and you have some load current being drawn and will assume that the load power factor is some the load current is at some power factor P. So, we would be able to estimate what the resistance of the line is what the reactance of the line is we saw in the exercise where we looked at calculating the X by R ratios of your line how to actually estimate these parameters and once you know the resistance and reactance you could then get what would be the voltage that you would see at the load if you have the source at given voltage amplitude you can then calculate what the regulation is. So, you could then do it what you do is solve do a Faser analysis. So, you have your source voltage V s you have your current I load lagging at some angle theta with respect to your source voltage then you can have a drop term corresponding to your resistance and you can have a drop term corresponding to your reactance and then find out what is the amplitude of your load voltage Faser we will we can actually simplify this further because we will assume that the drop is quite small. So, we could take your the load current to be consisting of the in phase component the real term I real and a reactive term and then you can calculate we can approximate your your load voltage as. So, we know that your I real is I load times p which is your power factor and your I reactive you can get your reactive component of your current and then you could then write down the expression that we had which was V L and substituting for I real and I reactive you would get this to be equal to V s minus some R L prime I load where R L prime is equal to R L into p. So, from the real voltage drop term plus X L by R L square root of 1 minus p square and X L by R L is now your X by R ratio of your feeder. So, you can see that what we have done is we have taken a solution which would have been in Faser domain and then we have simplified it into as a algebraic equation. So, instead of every Faser is a complex number. So, it carries two real values in it. So, here now we have reduced it to just one real equation. So, with that we can then calculate what is your voltage regulation at the load your voltage drop that you would see is essentially your amplitude of your source voltage minus the amplitude of what the voltage see seen at the load and then you could see what is the effect of now connecting loads all along the feeder. So, we will make an assumption that you have a feeder where you have the source sitting at some point X equal to 0 and you have a line of length D and you are assuming that the loading is uniform all along the feeder. So, the load per unit distance would be I s what is the source seen at the substation end divided by D. So, you could think of I s as the substation current. So, the loading is I s by D and you can then calculate what your current would be all along the feeder and your current would be the maximum at the substation end and reduce to 0 at the very end of the feeder. So, you can you have D I line by D X equal to I s by D and because it is reducing as you go forward. So, you have a negative slope. So, if you plot your I line as a function of distance versus the point along the feeder you will get a current profile along the feeder which is having a triangular shape. So, you can write an expression as I line of X. So, now that you know what the current is along the feeder you could then calculate what your voltage is going to be at different points along the feeder. So, to do that you have if you consider a point along the feeder at some point X I say this is D and this is 0. We can calculate V at of X plus some delta X ahead would be equal to V of X minus this. So, V at X plus delta X is V at X minus essentially the resistance per unit length times the small distance delta X times the current that is being carried at that particular point along the line. So, you can now substitute for I of X from the expression that we had previously. So, you could write this as V of X minus. So, you could then simplify it and then get delta V X by delta X to be equal to minus R L prime. So, we could then integrate this to get your voltage along the feeder. So, V at 0 is V S. So, you can write V S minus V of X. So, essentially what you get is you can write an expression for V of X you will get that to be equal to V S minus some term as you proceed along the feeder. So, you get an expression for your voltage that is actually parabolic. So, you get V of X. So, essentially you get a parabolic expression and if you look at I S by D this is essentially the loading per unit distance of the line and R L prime by D is essentially the effective resistance per unit length including the terms considering the power factor term. And essentially the objective then is that if you have a profile along the length of the feeder which is parabolic you want to fit this particular profile to be equal around a nominal value. So, if your nominal value is some V norm you want this profile to be sitting right across straddling the nominal in a symmetric manner such that one end doesn't see over voltage and the other end doesn't see under voltage. If you keep it such that you are straddling the nominal all ends of the feeder would give you a fairly close value to the nominal. So, you might say that your maximum voltage might be some V norm with some delta percentage. So, say some delta by 100 and your minimum might be V norm into 1 minus some delta by 100. So, depending on what your delta is you want to keep the entire feeder voltage to be within this window. Then the other thing that can be noticed is that you are not trying to keep the voltage at the substation end to be exactly equal to the nominal. You want the voltage which is some distance away to be the nominal. So, that your overall feeder voltage requirement is satisfied. So, typically if you look at trying to fit a parabola along the nominal you will find that roughly about 30 percent of the distance. If you keep that particular voltage to be the nominal then you would fit this requirement of being able to keep your overall voltage within the value of plus or minus V norm plus or minus delta percent range. And then you could ask what could would be done in a typical distribution system to adjust the voltage along the feeder. So, one thing you could have is a tap changer at the substation you could have a online tap changer OLTC which could adjust the tabs to actually get such a profile such as this. So, tap changer would do it for all the feeders at the substation. You could also have a series voltage regulator you could have a series voltage regulator one for each line. These regulators can be located either at the substation or if you have a really long feeder it could be further down the feeder to get your required voltage to be within a acceptable range. And typically that is at the substation or you could have a series voltage regulator to get your profile in a in a desired range. And then you could ask what would be the objective of the say the tap changing transformer. So, one objective would be to ensure that if the primary voltage coming in at the substation is high or low you need to ensure that the distribution voltage is independent of changes that might happen on the primary side. Because your transmission level voltages might go up and down depending on what is happening on the overall larger system. But you want to keep your distribution voltage constant the other thing is to regulate your voltage along the feeder. So, the second objective might be to regulate the voltage at say 30 percent point. So, you could adjust the taps on the high voltage side to meet this requirement. And one needs to keep in mind that you are not actually commanding your low voltage side your 11 k v voltage side to be exactly equal to nominal you are actually commanding your voltage on the secondary side to be some nominal value plus some boost voltage corresponding to the impedance up to the 30 percent distance times your current that is being drawn by at the substation. Because you physically cannot take a sensing wire put it few kilometers out to sense the voltage at that particular point what you physically have at the substation is the voltage at the substation. So, you are making use of the line currents that you know that is coming through the feeder to actually predict what is what is the voltage that would be there at 30 percent distance down along the feeder. So, essentially you are adding a boost term to the voltage. So, that the boost term would correspond to essentially this particular value above the nominal which you are trying to boost. So, in a transformer such as this if you are say operating at tap 1 then your voltage step down ratio is higher. And if you are operating say for example, at a tap n over here you will have a higher voltage now coming on the secondary side. So, by adjusting your tap points you can get your desired voltage at the secondary point. Then you could say for example, in a situation such as this you have tap changes where you might de-energize the transformer while changing taps, but you do not want to de-energize entire feeder the output of the substation every time you switch from one tap to the other. So, you would have on load tap changes where say for example, you can have a mechanism such as what is shown over here to actually ensure that when you are changing from one tap to the next tap you do not have over current. Say for example, if you directly close switch 1 and 2 you would end up with a short circuit in that particular loop whereas say for example, if you want to go from switch 1. So, you might what you might do is when your position of your pole is at the top point you might say close switch 2 and then essentially these resistors would limit your current that is flowing in this particular loop through that transformer. Then as you shift your pole from the top to the middle to the to the low point of your contact then essentially you will always ensure that you have energy flowing through at least one particular circuit it might be through this particular circuit or through this particular circuit, but you will not have a short because you always have a resistance in the path and eventually you will come to a point such as here and then at this particular point you have 1 and 2 closed and then you would open 1 and then you would continue with position 2. So, you could have tap changes when you are still energized your loads are energized and that is why it is called a on load tap changer. So, when you are trying to do such an action you could also have say for example, a series voltage regulator also functions in a similar manner you could have series voltage regulators at the substation end or further down the line. So, if you have for example, a really long feeder which might be may of the order of 10 kilometers a long feeder then if you just allow one particular the substation to regulate the voltage you might end up with a large drop in voltage all along from one end to the other end. So, to prevent that you could then add say a series voltage regulator say in between the line to obtain the necessary boost and then the question is what happens when you connect a DG to a system such as this. So, once that you connect a DG to a system such as this we will see that you could have a DG that is connected at different points it could be at the feeder it could be in the middle it could be at the end it could be just above the series voltage regulator or just downstream. So, we will look at the different possible cases and we will see that now if we have DG's that are of reasonable size. So, the power rating of the similar magnitudes of the feeder power then you could have situations where you might end up with may be under voltage at the end of the feeder or you might have voltages at different locations on the feeder based on your DG connection. So, we will look at the different cases. So, the first case we will look at is when there is no DG and this is what we discussed you have a triangular current distribution along the feeder starting from the substation end up to the end of the feeder and you end up with a parabolic voltage profile along the feeder. We will then consider the case when we are adding a DG. So, the first case that we will consider is adding a DG at the substation end. So, you have the substation and say you connect a DG right at this point at x equal to 0 and we will assume that the DG power equals to the overall feeder load. So, essentially you have the same current profile along the feeder as what you had except that now because your DG power is equal to the substation power essentially it means that what is now coming end from the substation end is actually 0 because your DG is actually providing the entire power. So, you can see in this particular case the voltage profile would still be parabolic, but now you do not have say a boost term at the substation because the current that the operational sensors see at substation would be 0 because the entire power is being provided by the DG which means that at the end of the feeder you now have a situation where your voltage is. So, the feeder end would see a reduced voltage compared to the case when there was no DG. So, this is because the OLTC in this case is not adding a boost term. You could then look at another situation where you have DG connected at the end of the feeder. So, you are looking at a situation where the DG is connected at the end and if you look at the shape of the current profile it is the same triangular current profile, but I have shown it as being negative because instead of the current now flowing from the substation towards the end of the feeder the current is now flowing in the other direction which means that essentially you would have the maximum current at the end of the feeder and because the DG power is equal to the entire feeder power and we are also assuming that the DG power is being injected at the power factor that is required by the feeder in this particular situation. So, essentially at x equal to 0 you would have 0 would be your actual current that is being drawn in drawn from V s. So, in this case again you do not have a boost term, but now instead of having a voltage drop you have a voltage boost which is now coming towards the end of the feeder and if you look at the voltage you will end up with essentially a higher voltage at the end of the feeder than what you had at in the nominal situation and so here you see increased voltage at the end of the feeder compared to the case when you did not have DG unit connected. So, we will look at another situation where you have a DG connected at the end of the feeder, but now the DG power is equal to essentially half the feeder power. So, essentially you are injecting I s by 2 at the end. So, if you look at it then again assuming the uniform loading and the conditions for the analysis now the second half of the feeder would be the power would be provided from your DG unit the first half of the feeder the power would be now be sourced from your substation. So, if you now look at your overall voltage profile on the feeder you would have essentially now a parabola not just a part of what we saw you would have two parts one corresponding to the portion from the substation end and the other considering the portion from the DG end. So, you could see that now there is also a reduction in the amplitude of the voltage that you are seeing across the voltage drop that you are seeing across the feeder. So, we will then look at another possibility where instead of the DG being connected at the end of the feeder we will look at a situation where may be the DG is connected further in between on along the feeder. So, you are injecting I s by 2 half the feeder power, but located at three fourth the distance along the feeder. So, you can look at then your current profile it will again be triangular in shape you have a jump of I s by 2 at the three fourth distance. So, in this particular location there is power flowing in either directions from the three fourth point for the first half you have the power flowing from your sub source V s. So, if you look at the overall voltage profile you will see that there is a further improvement in the voltage regulation in the second half of the feeder. So, then you could ask if you had a single DG unit and where your power could be appropriately it could be sized to the appropriate power level where could where would be the ideal location for locating such a DG and what would be the power level at which you could operate. And then you would see that if you have a DG unit which is two thirds of the rating of the feeder power. So, at the two thirds power level located two thirds of the distance away from the feeder then essentially you will see that your you will end up with a voltage profile which is quite flat you what you would see is compared to the case of your case where you had no DG you will see that the voltage regulation along the feeder compared to your this was your case 1. And this corresponds to case 6 compared to case 1 you have a factor of 9 reduction in your voltage along the feeder. So, you can see that it is become almost flat compared to the nominal condition by adding a single DG of appropriate size along the feeder. But again keep in mind that this is a static analysis the actual load on the feeder would be changing depending on the time of the day and citing a DG at a particular point will not be feasible in all situations. And you also know that the loading we have made assumptions on the loading etcetera. But you can get a feel for what is the possibility in terms of improvements and what are the situations where you have concerns there is a possibility of over voltage or under voltage in different locations. And you also have by appropriately citing or appropriately having your DG units it is also possible to get a flatter profile all along the feeder. So, if you look at the ideal situation where if you had a situation rather than 1 DG you could have multiple DGs then the ideal situation would be the where every facility on your feeder has a DG where whatever power is being consumed by the DG by the load at that particular location is being sourced from the DG which means that essentially your voltage all along the feeder would be totally flat not drawing any power which means that the losses along the feeder would be eliminated. And we will also look at what the loss comparison would be for the case 1 and case 6 what would be the distribution losses. So, if you look at the power dissipation on the feeder we saw that our line current is I s and if r is your total resistance of the line. So, R L by D is the resistance per unit length you can calculate the power dissipation at some over a section of delta x at x over that incremental length delta x is I line square of x times R L by D times delta x which is essentially the resistance of that particular section. And we can substitute for we could then calculate the total loss by integrating over the entire length of the feeder 0 to D P delta x P D x this would be equal to 0 to D. Again this is we have the same assumptions of uniform loading etcetera. And so you could evaluate this but integral as so this is the loss per phase of the feeder. So, you could multiply that by 3 to calculate the 3 phase losses. So, if you now compare with case 6 where you had a d g of two thirds ratings located two thirds of the way down the feeder you could then consider each of this case as one small feeder of this particular size. And three such sections 1 over here section over here and the section over here if you look at the profile of the current in each section it is similar to what you have in all three sections. So, you could calculate the losses in one section and take three times that particular amount. So, you can see that again you have a factor of 9. So, the reduction in losses is by almost an order of magnitude if you have a well size d g which is well sighted even a single d g can actually bring down the losses by quite a bit. So, you can see that there can be substantial improvement in terms of reduction in losses and possible improvement in voltage profile by appropriately sighting d g which you would have also heard from people who talk that there can be possibility in reduction in transmission distribution losses by d g by introduction of d g into the system. We will also look at in the next class at the case where what could happen in the case where you have other types of distributions. For example, one can look at the case where what we have studied so far is the case of a radial distribution system you could look at what would be the impact when you look at say a ring type of distribution system. You also have challenges when you have d g in network type of distribution systems. So, we will look at some of these issues in the next class and look at some of the power quality related issues and the concerns that again come up when you have distributed generation connected to the grid. Thank you. Thank you.