 Welcome to class 22 on topics in power electronics and distributed generation. We were looking at economic metrics for evaluating distributed generation systems. We are talking about four of them initial cost, payback period, cost of energy. So, we have been discussing the payback period, examples of payback period and cost of energy in the last class. So, one way of looking at cost of energy is you are purchasing some equipment, installing it and you are getting energy out of it and you are operating the equipment over long time frame may be multiple years. So, you want to look at the cost on an annualized basis for which you make use of things like interest rate, annual maintenance cost, annual fuel cost. If there are secondary benefits you look at it as a negative cost on an annualized basis and then you look at what is the energy produced on an annual basis and you can calculate the cost of energy. So, we looked at started with a simple example of cost of energy for a diesel generator producing electricity and we looked at the cost of the fuel to be diesel fuel to be 50 rupees a liter and the cost of electricity from your distribution supplier to be 7 rupees per kilowatt hour. When we work through the example we found that the cost of energy in this particular case from your DG set would work out to be rupees 15 per kilowatt hour. So, which is much greater than 7 so it does not make sense to produce electricity in this manner except when absolutely necessary. So, in a second situation we looked at when you are operating this gen set as a combined heat and power unit where the fuel in the diesel is being converted into electricity with a given efficiency and the waste heat is also being captured with some given efficiency. And then based on the fact that you could make use of this thermal output which might potentially be originally heated from electricity you can then calculate what your benefit is and in this particular case the cost of energy we worked out the example worked out to be rupees 6 rupees 10 paisa per kilowatt hour which is less than 7 so potentially this is can be a situation where producing electricity in this manner might be economically feasible. But we have neglected the capital equipment cost there is a cost in the internal combustion engine, the alternator, the heat exchangers etcetera the control system. So, we will look at an example we start with an example where we are looking at the capital and operation and maintenance cost. So, we will start in this example we will we are given the thermal load is 150 kilowatt thermal and the cost of your electrical part of your dg system is 4000 rupees per kilowatt hour kilowatt rating of the system and the cost of the heat exchanger part is again 4000 rupees per kilowatt thermal and we are assuming an interest rate of 15 percent and we will assume that the operation and maintenance cost is 10 percent of your capital cost of the equipment that you are purchasing. So, in this example we can calculate what would be the energy required to at the input because we know the fuel to thermal output efficiency is 45 percent. So, we can then calculate what your p input is because this is p thermal which is given. So, your p input is p thermal divided by 0.45. So, that works out to be 33 kilowatt and then you can calculate you know your efficiency from your input fuel to electricity which is 35 percent. So, you know your electrical output power electrical which is 0.35 times 333 this is 117 kilowatts. So, now you know your rating of your electrical system is 117 kilowatts you know your rating of your thermal system is 150 kilowatts you know the cost per kilowatt rating of your electrical and thermal system. So, you can calculate the capital cost. . So, we will work through the example. So, your capital cost is you have a system which is your thermal system is rated at 150 kilowatts and the cost of the heat exchanger equipment is 4000 rupees per kilowatt and your electrical system is 117 kilowatts and its cost is also 4000 rupees per kilowatt. So, you get 10.6 into 10 to the power of rupees and your operation and maintenance cost is 10 percent of your capital cost which is as which is assumption 1.06 and you could then also calculate your fuel use is 333 kilowatt is the energy in the fuel that is used on a kilowatt into on a per second basis. So, if you divided by 9.4 kilowatt hours per liter into 8.760 which is the number of hours per year and multiplied by 50 rupees which is the rupees per liter. So, this works out to be 155 into 10 to the power of 5 rupees per year and if you look at the number over here 155 into 10 to the power of 5 is a fuel cost per year is much larger than the capital cost which is just 10.6 into 10 to the power of 5. So, you will see that the fuel cost is much larger than the capital cost you could also calculate your thermal saving. So, 150 kilowatt is a thermal output and if it was being formally heated at 7 rupees per cost of power from the grid which was 7 rupees per kilowatt hour 760. So, this is 92 into 10 to the power of 5 rupees per year. So, then you could then calculate your annual electrical energy production electrical energy production which is 117 into 8.760 is 10.2 into 10 to the power of 5 kilowatt hours per year. So, then you can calculate your cost of energy capital cost into interest rate which is 15 percent plus 1.06 into 10 to the power of 5 which is your annual operation and maintenance cost 155 into 10 to the power of 5 which is your fuel cost and 92 into 10 to the power of 5 which is your thermal saving divided by 10.2 into 10 to the power of 5. So, this works out to be 6.39 rupees per kilowatt hour. So, you can see that even when adding terms for capital cost the cost of energy did not increase significantly over the fuel cost which. So, ignoring the capital cost it was 6 rupees 10 paisa with the capital cost operation and maintenance it moved up to 6 rupees 39 paisa. And in many fuel intensive technologies one may observe that the cost of fuel can be a major aspect of operating a system. Whereas, if you look at a comparative say other technology like renewable energy there the fuel cost is essentially 0, but then your capital cost would be the dominant term. So, also this implies that say in this example we assume that the diesel cost is 50 rupees per liter suppose the diesel cost went up by some amount then the that would reflect quite dramatically in the cost of energy. So, when you are generating power based on fuels you have to be having control over the risk of fuel cost going up and down whereas, if you look at a system such as renewable energy system the risk of operation over time is lesser the risk has to be evaluated up front whether the capital can be invested, but once done you have a fixed rate of return over a long longer time frame assuming that the reliability etcetera are as per design. Next what we will look at is an example of making use of this cost of energy concept in designing a d g system in this case is an example of a photovoltaic system. We will assume that the this is a 2 kilowatt photovoltaic system the module cost the PV modules cost 1.5 lakhs per kilowatt or you have a choice of power electronic inverters. So, you might have choice which might be a older design current design may be a more futuristic design it might have different efficiencies it might have different performance capabilities and also it might have different cost something which might be lower performance might be lower cost the 92 percent efficiency inverter cost 15 rupees per kilowatt and the 95 percent efficient inverter cost at 25 rupees per kilowatt and the question is which inverter would be an appropriate choice for your design and could you make use of the cost of energy as a possible basis for making a design decision. We will also assume that the installation and wiring cost for installing this photovoltaic system cost 50,000 rupees and there is no major annual maintenance you might have to clean the PV panels and we will assume that it cost 1000 rupees a year to for as annual maintenance cost. We will assume the interest rate of 15 percent we will assume that the power is being produced by this panel 12 hours a day obviously it is not going to produce 2 kilowatts early morning late evening. So, we will assume that there is a peak capacity factor of 0.45. So, on an overall day basis 0.45 of the peak power is being produced. So, the question is can we make use of the cost of energy calculation in making a design decision. So, we could actually go through the calculation of what would be the annual cost for these 3 technology options. So, you have the panel cost which is 2 into 1 and half lakh per kilowatt. So, 3 lakhs would be the panel cost for this 2 kilowatt rated unit and the panel cost is the same for the 3 inverter choice. The installation and commissioning cost is also the same it is largely wiring how you mount the panels the civil works or the mechanical structures etcetera associated with it. We will assume it is the same then if you look at the inverter cost for the 88 percent inverter it is 10,000 rupees per kilowatt. So, 2 kilowatts is 20,000 rupees for the 92 percent efficiency it is 30 kilowatts for the 95 percent it is 50,000 rupees. So, you take the capital cost and multiply it by your rate of interest rate which is 0.15 and then you add your annual maintenance cost which is 1000 rupees to calculate your annualized cost. So, the annualized cost for the 3 options works out to 6,000 rupees for option 1, 58,000 for option 2 and 61,000 for option 3 and then you could also calculate what is the annual energy production from these 3 units. So, we will assume that the panels from these 3 units are identical what is different is just the inverter. So, for the energy production it is 2 kilowatts is the rating of the unit it is working for 12 hours a day with a P capacity factor of 0.45 365 days a year and there may be some losses in the inverter. So, you multiply it with the efficiency. So, you can calculate what the annual energy production is it is 3469 units for option 1, 3627 units for option 2 and 3745 units for option 3. Then you do the cost of energy calculation you get 16 rupees, 29 paisa for option 1, 15 rupees and 72 paisa for option 2, 16 rupees and 2 paisa for option 3. So, if you are looking at what is the lowest cost of energy you will find that just going with the lowest cost unit or the highest performance unit may not be the most appropriate something appropriate might be what gives you most value for the technology that is being introduced. Again the numbers like the 1 and a half 1 and a half lakhs per kilowatt for the photovoltaic panel might be a little bit dated number if the cost might actually be lower today. So, these are numbers, but as these numbers some of these numbers change over time you could see a trend, but then you could see what would be an appropriate design decision based on what power electronic technology you are adopting. So, again you can see in the cost of energy from the cost of energy calculation there is no fuel cost the reason why you are having numbers like 15 rupees to 16 rupees per kilowatt hour is actually the capital cost. So, capital cost is critical for renewable energy or also true for things like nuclear technology where the cost of construction is quite massive your fuel cost is 0 in this case also a important factor is the interest rate if the interest rate is lower your cost can come down quite drastically. So, government policies towards these technologies is important factor because the interest rates can be affected by larger societal policies. A drawback with the cost of energy approach when you are doing such a design is that the differences between the performance matrix is really small you are trying to make use make a power electronic design, but then the power electronic numbers are buried by the larger numbers of the overall system. So, the differences appear to be quite small when you are looking at it from a cost of energy perspective and the cost of the power converter is actually a small fraction over of the overall system cost. But you can see another factor here if you look at the annual energy production for the 95 percent unit and compare it with this particular unit you have about plus 3 percent 3.25 percent energy production in this particular unit. So, you may not need a 2 kilowatt panel you can actually reduce the size of the panel by 3 percent and you can see that the panel cost is quite significant. So, there is benefit from other balance of system to because of the power electronic unit because it is actually sitting closest to the point of common coupling where the energy is being sent out. In this particular case even though the capital cost the cost of the inverter is low you are actually producing 4 percent lower power in this particular case in which case to get the same 2 kilowatt you need 4 percent more panel area. So, your panel cost would go up proportionately. So, you can see that there can be system implications for power electronic choices and it would also be good to have a method for evaluating a power converter choice which take into balance of plant benefits rather than just the power electronic cost. So, if you look at a few examples of say balance of plant benefits say for example, in a photovoltaic system it is a power conversion unit say for an example in a photovoltaic system you might have a power converter which has say MPPT that can be incorporated in it and because now because of your power converter with the MPPT you might be able to produce more energy which means again it reduces the size of the array required. So, you have balance of plant benefit also you might have a situation where maybe because of your topology choice maybe a string inverter or a module inverter rather than a centralized inverter can actually harvest more power which means again you can have a smaller array which is producing more power. So, you might in a similar way if you take a wind turbine some of the early wind turbines where fixed speed just a machine induction machine connected to the grid maybe with some power factor correction they were operating at fixed speed base depending on what the grid frequency is and maybe some limited speed change based on your pulled number changes in your machine. But, if you are able to operate it variable speed with a power converter you could actually operate your wind turbine at the maximum power point and harvest more energy. So, if you are able to capture more energy for a given size blade it might mean that for a fixed power output you can maybe reduce the size of the blade height of the tower etcetera which can actually give you a cost benefits. Also you say you have a power converter which can have a high say rate of change of power can be set at quite rapidly because of higher bandwidth etcetera then you might be able to damp out resonances in the blades tower etcetera which means that you are you can have lesser margin the amount of steel required in the tower etcetera can come down and you can actually reduce a cost of structural cost which is a major cost in things like wind applications. Or if you take an example of a diesel generator if you look at the traditional gensets they will be running at constant speed irrespective of the load. So, to produce a 50 hertz output at providing 230 volts or 415 volts you might operate at the same speed irrespective of whether the genset is fully loaded or half loaded or at no load which means that you are consuming quite a bit of fuel and you might be operating your genset at not the most optimum point. Whereas, if you have a power converter in between which allow you to slow down your genset and your inverter is providing your output waveforms then you can allow your genset to reduce its speed when the load goes down which means that now your fuel consumption can potentially come down and you saw that in systems such as gensets the fuel cost is a major cost. So, because you are now adding a power converter you can reduce a major cost such as fuel which can be attractive. So, such a systems are also used in hybrid vehicles etcetera where your genset your actual IC engine might operate at the maximum efficiency point and the additional torque required might be provided by a electric motor over and above your average torque level that needs to be produced. So, that way you can operate your combustion engine at its best efficiency point and giving you lower fuel consumption. So, similar principle applies to vehicles also. So, one would definitely like to incorporate some of these benefits to the larger systems when you are considering whether to take one particular power conversion technology or some other power conversion technology because one particular technology might be more efficient at capturing some of these benefits. So, we will just look at a quick number of say an example of a wind turbine blade or a wind turbine. So, if you look at the cost breakdown we say assume say 35 percent of the cost of a turbine may be your blade cost there may be some composite materials carbon fiber possibly your tower cost is a major cost balance of plant here you are looking at gear box the generator the yaw pitch systems transformer etcetera your power converter might be just might be just 10 percent of the cost. But then if you look at the flow of energy in such a system say you have the blades may be your gear box and say your objective is to generate one megawatt being injected to the grid and then you can calculate what would be the power input will assume that 40 percent of the energy in the wind is being captured by the blade say 88 percent efficiency in your gear box 98 percent efficiency in your generator and you have say a couple of converter options one might be 95 percent the other might be 96 percent and you can see what would be the energy in the wind that needs to be captured corresponding to the one megawatt output for these two options and so for the 95 percent efficient case your P N P wind is 1 megawatt divided by 0.95 0.98 into 0.88 into 0.4. So, you are talking about 3.05 megawatts and you are looking at the 96 percent case the number corresponds to 3.02 megawatt. So, if you look at the delta P in your wind divided by P output. So, you get the difference which is 3.05 minus 3.02 by 1 which is your output. So, this corresponds to about 3 percent. So, a 1 percent efficiency difference in your inverter implies a upstream efficiency difference of about 3 percent. So, it is magnified by 3. So, the amount of stresses in the blades etcetera can be reduced and you can because your power converter is now sitting at the end of a chain right next to the point of common coupling it forms a crucial part in the link in the overall system and that can be significant upstream benefits even though your cost of the power converter is only 10 percent it can have implications on the balance of plant which is 90 percent of the cost. And another aspect of issue with the cost of energy approach is when you do the cost of energy you may not know your overall system cost upfront. Say you are designing a power converter for a wind turbine you might know the power converter cost roughly, but you do not know your wind turbine cost which where its design may be happening in parallel. So, your overall cost to evaluate cost of energy may not be accurately known, but the power electronic designer might be aware that by adding some particular feature in the power converter operation you can get a benefits in the balance of plant and that might be easier to quantify. Also as we mentioned we do not want to get buried by the cost of the overall system which is much larger than the cost of the power conversion unit. So, before we look at the effective initial cost we look at what would be the ideal characteristics of say a power converter. So, if you look at the ideal power converter if you look at power loss ideal converter should have 0 power loss in reality you might be talking about a converter which might be 95 percent efficiency or something in that range. So, we are talking about power ranges of 100 kilowatt to megawatt level if you are talking about converters which are may be tens of watts 100 watts your efficiency numbers may be lower and if you are talking about power converters which are used in larger transmission systems in HVDC, STATCOM type of applications your efficiency might be as high as 97 99 percent, but here we are looking at an application which might be used in a distribution system. So, we are talking about something of the range of 95 percent. If you look at the ideal volume your ideal volume is 0 your ideal power converter should not occupy any space in say assuming that the volume of the converter depends largely on the type of cooling that you are going to use whether it is natural cooling liquid cooling force dare etcetera we will talk about say 3 to 9 meter cube per megawatt. So, you are talking about on a 3 to 9 into 10 power of 3 cc per kilowatt note that 1000 cc is 1 liter. So, it is about 3 to 9 liters per kilowatt. So, you might look around say maybe you might seen a UPS sitting next to your computer and you may have a 1 kilowatt UPS and its volume may not be much larger than 3 liters in the cabinet if you exclude the battery. So, you can get a range for what the volume would be of typical equipment around you and also it depends on the type of enclosure whether it is dust proof open frame etcetera. If you are looking at weight your ideal power converter has 0 weight whereas, your realistic converter you are talking about 5 tons per megawatt again these are making assumptions of say typical force dare cooling filters being present in the converter etcetera. So, you are talking again of the or if you are talking about a kilowatt you are talking about 5 kg per kilowatt again you may not be familiar with a megawatt power converter you may not you will definitely not lift it single handedly, but you could try lifting a UPS which is there next to your computer and you would say if you ignore the batteries it would cost may be weight may be 4 or 5 kgs if you look at the cost your ideal cost is 0. So, if you look at again the cost which might be realistic again it depends on the application you might have applications where the cost it might be of the form of some consumer goods which is being mass produced. So, the cost might be lower or it might be some custom made product for space application or some strategic military type of application where the cost might be higher, but we are talking about something typical for a megawatt range commercial type of system you are talking about again rupees 50 into 10 to the power of 5 per megawatt. So, again to scale it down you might say on a per kilowatt basis you are talking about 5000 rupees per kilowatt again if you look at a UPS that you might purchase with for your computer in your lab you are talking about 5000 rupees you might be able to purchase in UPS in roughly in that range with some variation around it. Again another important aspect is the failure rate and ideal power converter never fails. So, it is always working whereas, your practical power converter it is failure rate depends on the components the application in which it is being used. So, we will look at the you might have components such as DC link capacitors might have IGBTs which are being cycled thermally you might have inductors transformers etcetera which whose insulation might degrade at higher temperature you might have circuit breakers contactors who have a defined number of cycles before it needs servicing you might have connectors which might experience fatigue after so many mechanical cycles. So, you have equipment where the failure rate is not 0 it is finite. So, depending on the particular application one can actually try to back calculate what your failure rate is and what could be potential replacement durations of within your equipment. .. So, if you look at the ideal power converter it is a ideal middleman who provides all the service, but is invisible it does not take cost any money it is always available there is no failure etcetera, but you are trying to achieve that ideal, but it may not be possible to meet that ideal requirement in all situations, but as a engineer you are trying to make these numbers to get to be as close to 0 as possible. And one thing to also keep in mind is some of these numbers are actually present in a tradeoff form for example, if you try to reduce your power loss your cost might increase similarly if you try to reduce your failure rate again cost could increase if you try to make it more compact by reducing your weight and volume. So, again potentially your cost can increase. So, there is a tradeoff between cost and the performance numbers. So, just trying to make it ideal in terms of performance will magnify the cost and that may not be always going for the highest performance or conversely going for the lowest cost of the power converter may not be the ideal situation. You could also look at now given these previous numbers what these are fun exercises to carry out here you know the cost per kilowatt of a power converter and you also know the weight per kilowatt of the converter. So, you could ask what is the cost per kg of a converter you may not think of power converter as you purchase in kg's, but you could look at if I purchase a equipment what would be the cost per kg or cost per volume. And these are fun exercises to do you could look at equipment that are available around you electrical equipment that are there in your house say consumer items, microwave ovens, mixes, grinders look at their weight, their cost etcetera and look at a cost per kg number and see whether they correspond to some of some do they tend towards some number that you could actually make use of in making rough engineering decisions. So, next we will look at making use of some of these characteristics in deciding to look at it from your effective implications from a effective initial cost perspective when you are looking at a decision to make a take a particular approach and we will look at a simple example of may be purchasing a vehicle. If you look at purchasing a vehicle you might have initial cost the initial cost might be what is the cost of the vehicle in the showroom. So, the cost of the vehicle in the showroom might have the manufacturing cost the margins that the manufacturer wants for producing the equipment it might have some tax element that you have to add you might have some cost initial cost associated with the registration of the vehicle getting license etcetera. So, you might have ongoing cost there can be a number of ongoing cost one major one would be a fuel your petrol cost you might periodically need to go to a mechanic for maintenance you might need spare parts you might have annual insurance payments. So, you would have things that are done on an annual basis and you might also have a end of life cost in which case here it may not be a cost you might get some money back when you sell the vehicle after so many years of operating it. So, you would like to reflect all these cost to your initial point when you are making a decision rather than just making a decision based on what the initial cost is. Because in fact when you are making a decision you are actually not just committing to the initial cost, but actually all the other cost associated with it. So, people talk about it as a net present cost or a net present value effective initial cost etcetera they all imply the same thing and the design methodology by which these costs are evaluated people call it as cradle to cradle design where you are not just looking at the initial upfront cost or the manufacturing cost, but also eventually you might not just sell the used vehicle eventually after a number of sales it might have to be recycled or dismantled or disposed in some manner. So, if all the cost can be put together that particular way of looking at a design is what the effective initial cost is trying to approximate. So, if you look at what goes into the effective initial cost calculation you might have your capital cost which might have the material, labor, markup etcetera you also have upfront cost corresponding to installation commissioning of a DG unit. You might also have shipping cost say your wind turbine may be produced somewhere in Goa it might have to be installed in Tabernardo. So, depending on if you want to ship huge towers blades across multiple states it depends on how big it is how much it weighs if you are trying to install something in a urban area the cost of land in a urban area might be quite high. So, the size, weight, area etcetera can actually reflect on the initial cost. You might also have change in balance of plant cost as we discussed earlier you might have benefits in balance of plant which has to you should be reflected in your effective initial cost. Also we know that a practical power converter has losses. So, ideal power converter would have loss of 0. So, whatever loss is being encoded in the power converter you could look at it as one way as that is a loss that you are incurring or a cost that you are incurring by operating that power converter you are having a fixed annualized loss cost associated with the loss in the power converter. Also you would have the operation and maintenance which can be reflected which might be on an annual basis it can be reflected to the present time. Also depending on the power plant or the type of plant you might have a decommissioning cost. For example, if you are as I mentioned if you have a wind turbine you might be able to get some value above from the scrap steel or copper in the generators etcetera. But you may not be able to get a value from say old electronic e-waste which might be there in your system or suppose it is a nuclear power plant if you want decommission some part which is radioactive that might be a lot of expense in disposing say something which is radioactive. So, depending on what you are trying to dispose there might be different types of decommissioning cost. And the idea is that effective initial cost is does not reflect actual physical amount spent. For example, you are not spending an amount because of cost of losses it is unearned you could have earned more if it was better that is the idea behind it. So, it can be used for a design comparison or making a design and it is not actual number where you can say this is the actual so many lakhs or some amount is the actual physical amount that is there. So, these some of these values are values that are available up front this might be ongoing or future values which are reflected back to the present. And then you could say how could one reflect the cost to the present say if you have amount C today and you are assuming an interest rate R then at the end of one year you might have C into 1 plus R at the end of the second year you might have C into 1 plus R into 1 plus R. So, it would be C into 1 plus R the whole square at the end of your P it would be C into 1 plus R to the power of P. So, if you have a cost C of P in your P if you want to reflect it to the present you can you just have to divide it by 1 plus R to the power of P. And if you have a annualized cost every year you have an equal cost of C A N N and you are assuming that this is going to operate for N years you can reflect the first years cost as C A N N by 1 plus R second years by 1 plus R square. So, you have a summation over 1 to the N years over which you are operating the equipment and the summation can be simplified to a fairly simple expression such as what is shown over here. And then one could also look at how to incorporate other factors we will consider a simple interest rate number R which reflects on a annualized basis, but you could also include terms for depreciation inflation rate taxes etcetera which can be captured by using appropriate values for R depending on the particular application that you are dealing with. So, you with this framework in the next class we will look at an example of how to use the concept of effective initial cost in making use of it in deciding power converter whether it is appropriate what could be a design methodology to make a decision of what power converter use would be appropriate. Thank you.