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Now that we understand the need of giving EV batteries a second life, let’s check where is the problem, because there is always a problem, otherwise, what engineers are for? The first and main problem is that the cells that are manufactured for the EVs are different from the ones manufactured for the stationary storage systems. The EV market needs high energy density, batteries capable of fast charging and absorbing high power demands. Cycles are not that important, because the battery capacity is bigger than the daily needs of most drivers. Those cells are usually with nickel-rich cathodes. On the other hand, in ESS we need high number of cycles and we don’t care that much about energy density, since space is not that critical. What we want is a low cost per number of cycles, so the investment makes sense. Have in mind, that a typical installation is a PV+ESS for home applications, where the battery will have a full daily cycle. In addition, we don’t need that high C-rate. Ok, an extra stop here, what is the C-rate of a battery? It’s a unit to measure the speed at which a battery is charged or discharged. 1C means that the battery is charged/discharged in 1 hour. 2C is in 30 minutes. C/2 is 2 hours. C/20, 20 hours. Now that we understand what a C-rate is, we can conclude that the EV needs to be charged very fast, so we need high C-rates, like 5C (12 minutes) or 2C (30 minutes), but when we are charging the battery in our home from PV, we usually have 3 to 5 hours to charge it, which means C/3 or C/5, a much lower C-rate. In conclusion, if we use cells that are manufactured for high C-rates, we will have less cycles, and that is not what we want in stationary applications. To solve this issue, we will have to oversize the system. How? Imagine that you need a battery that gives you 10 kWh each day. If you buy a new LFP battery, you will probably use it between it’s 10% SoC and 90% SoC, so you will have to buy a battery of 12.5 kWh. 10% State of Charge means that the battery has 10% of its total capacity available: 1.25 kWh 90% State of Charge means that the battery has 90% of its total capacity available: 11.25 kWh Therefore, you are using 11.25 kWh – 1.25 kWh = 10 kWh This new battery, designed for stationary applications, will provide you around 5000 cycles. A cycle per day, you will have this battery for 13 years. But if you are going to use a second-life battery, you shouldn’t get to such low SoCs like 10%, because the deeper you discharge the battery (lower SoC), the less number of cycle you would get. If we have less number of cycles, let’s say 3000 cycles for a new EV battery, to reach a similar number of cycles you will need to stay in a minimum SoC of 50% or 60%. Which means that you have to oversize the battery. If you still need to use 10 kWh and your SoC will be moving between 90% and 50%, the second-life battery will have to be of 25 kWh. You can do the math as in the previous case with the new battery. That is the double. A second-life battery of 25 kWh and a new battery of 12.5 kWh will provide you with the same service. If the new battery costs 250 USD/kWh, the second-life battery should cost 125 USD/kWh to be competitive. Those numbers are very approximate and not precise. The objective is to provide you with a sense of scale. In the next articles, I will publish more specific cases with more detailed simulations. Inshallah. Other challenges The real life cycle The current estimation of the end of life of a new battery in the EV market is when the SoH reaches 70%-80% or the internal resistance is double [1]. But most probably, this threshold of SoH will be lower with time, since we will discover that the batteries are able to survive longer periods, they might even be able to outlive the car itself. You can check more info in this article [2]. EV batteries having longer life in an EV than reused in the stationary market is a downside for the business model of the second-life batteries, because the remaining SoH will be lower the moment that you will refurbish them. Future prices Let’s assume that an EV battery costs 150 USD/kWh to manufacture today and it is sold at 250 USD/kWh. In 10 years, the battery might cost 100 USD/kWh to manufacture and will be sold for 200 USD/kWh. Let’s assume that the same battery that is manufactured today will be sold again to a second-life batteries manufacturer in 10 years. It will be sold at least at 150 USD/kWh to cover the cost. Therefore, you will have to sell it at 180 USD/kWh, to be competitive. But what about the additional cost of refurbishment? The margin for a competitive business model is risky. Refurbishment cost When a second-life battery manufacturer receives the original pack from an OEM, the battery can be kept in the same case, analyzed and installed in a bigger system or it can be opened, and the different modules taken. The modules could also be opened and the cells taken. If you reach the cells or modules level, you will be able to group different cells/modules with different levels of degradation. Separating the cells will be better than the modules, because the module might have a degraded cell that could affect the remaining cells. But that is a lot of work, which will increase the final price. If we keep the battery pack as it is from the start, the cost will be much more competitive, but the chance of a faster degradation of the whole pack is higher. Warranty The second-life

In a series of articles, I am going to share my research on second life batteries. This is the first article. What are second life batteries? To reply to this question, I must start from the beginning. Obviously. Lithium batteries are being widely used for two main applications: Both markets are growing and competing for the resources. Prices, that we always expected to fall, are increasing. BloombergNEF data for this year shows that lithium-ion battery pack prices have gone up 7% in 2022 [1]. That is why we need to use these resources in the most efficient manner. Now that we understand the need, let’s connect the dots. Electrical vehicles use battery packs that ranges from 10 kWh (a Mitsubishi Minicab MiEV) to 120 kWh (a Mercedes-Benz EQS). All these batteries will be used until they reach 80% of their State of Health. Let’s take a break here and learn some technical words. State of Health refers to the available capacity of the battery in a certain moment. Let’s say you buy a new battery with 100 kWh capacity. The State of Health (SoH) in that moment is 100%. With time, the battery will lose capacity due to wearing and some chemical degradation. After a year and a daily use, let’s assume that the battery will be able to provide you 90 kWh. The SoH of that battery is 90%. You will probably be asking yourself, why are we changing the batteries of the cars if their SoH is still at 80%? Why don’t we wait until it is 10% or 0%? There are multiple reasons, one of them is related to your amygdala. Let me list them here: That is why, the US Advanced Battery Consortium decided that 70 to 80% is reasonable target for a battery to reach it’s end of life [2]. Some researchers [3] have shown that the driving needs of American drivers could be met with battery capacity as low as 30%. We still need more time and real tests to decide what is the appropriate threshold. Now that we have the battery of a car, with still 80% SoH, can’t we use it for another application, like stationary storage systems? Et voila, you managed by yourself to understand what second life batteries are. The use of batteries after they have reached the end of their useful life is termed as ‘second-life’. To summarize, let me share what is the challenge and what is the opportunity: The challenge Towards 2030, the yearly volumes of discarded EV batteries are estimated to be between 112 and 227 GWh [4]. At least one third of these are expected to be fully functional with more than 80% remaining capacity and that can live for up to 20 years in a second-life application. It will be the less sustainable act to throw, or even recycle these batteries without giving them a second opportunity: a second life application. The opportunity In order to build renewable grid and transportation we are seeking, energy storage is a necessity, not an alternative. That is why we need the most cost efficient and resource efficient solutions, like second life batteries. In the next article, I will focus on the possible limits for this business opportunity. — — — — — — Notes [1] https://www.energy-storage.news/lithium-battery-pack-prices-go-up-for-first-time-since-bloombergnef-began-annual-survey/ [2] U. ABC, ‘US ABC Electric Vehicle Battery Test Procedures Manual, Revision 2,’ principal author: Gary Hunt, Idaho National Engineering Laboratory, US Dept of Energy Idaho Field Office. [3] https://doi.org/10.1016/j.jpowsour.2015.01.072 [4] https://www.mckinsey.com/industries/automotive-and-assembly/our-insights/second-life-ev-batteries-the-newest-value-pool-in-energy-storage

First approach using Python to estimate the payback of a hybrid system. It is based on the feasibility study of a solar system to power an oil pump in a GCC country. Context Based on the available data, the Diesel Generator is consuming 2% of the tank every day. The tank capacity is of 5000 l. Therefore, the DG consumes 100 l/day. The price of Diesel is 0,329 EUR/l. $0,329*100=32,9$ Therefore, the fuel cost (only, without maintenance, additional cost of transport and others) is 12k EUR per year (13560 USD) The EUR/USD FX is 1.13. Load estimation The average load as per the user information is 12 kW. The hourly consumption of fuel is approximately: $100/24=4,16 l/h$ The datasheet of the DG informs that the consumption at 1/4 of the nominal capacity is 5,7 l/h. 1/4 of the power is 16 kW. Therefore, the power demand of 12 kW makes sense. Calculations We must assume that the cost of maintenance and renting the DG remains the same. The reason is that the DG will still be used to power and charge the batteries as well as a back-up in case of any failure. Let’s assume the power system in this case will cost approximately 35000 USD. Batteries The battery price per kWh is 250 USD. Let’s assume the battery no of cycles is 2500. If we manage to discharge and recharge the battery only once a day, the lifetime of the battery will never be longer than 7 years. Therefore, the payback period should be less than 7 years. PV modules The assumed PSH are 5. Let’s assume the PV will only run to power the load during the PSH. The price per KWn of PV is 3 EUR. If PV will be only used to power the load, the PV price will be: 12000 EUR * 3 = 36000 EUR equivalent to 40680 USD. With different battery capacities I can reduce the number of hours the DG is working. Therefore, I will reduce the fuel consumption and see in how many years I can cover the cost. Code Results The initial result shows, as expected, that the more we use the battery capacity the more effective is the investment, although, the bigger is the battery, the more shorter is the RoI. Unfortunately, and based on the used prices, the system will not be able to be cost effective. Conclusions and improvements The payback analysis as a method can be limited. First of all, it does not show all the technical advantages and the whole life cycle of the different components. However, LCOE is more effective since it can have in mind future investments to keep using the overall system. The analysis doesn’t take into account the curve of cycles vs. DoD. If included, it can show a better approach of how effective is. Variations in the fuel cost in the future are not reflected. If added, it can improve the payback of solar installations. The maintenance is reduced while using solar power since the overall hours that the DG reduce considerably. The clarity has to come from the renting price of the DG and if it includes maintenance and overhaul. Civil works cost and the increased need of more shelters for a bigger load are also not reflected. Financially, the same amount of cash to earn or to spend in the future vs now is different. That is why the discount rate must be introduced. The graph must include more details to understand what does it reflect. For more information, or if you are interested in the most advanced versions of this analysis, feel free to contact me.

Context I am presenting an alternative solution to a telecom company to power it’s off-grid towers. The sites are working 24 hours a day, with an average load demand of 2.75 kW. They are currently running with a DG of 16.2 kW, with an estimation consumption of 11 255 liters. (The consumption estimation is based on a calculation shown in the next chapter). The site is located in a GCC country. Assumptions DG fuel consumption Based on the provided datasheet of the DG, the fuel consumption is as follows: Fuel consumption l/h Prime power 5.3 75% of prime power 4.0 50% of prime power 2.9 The estimated lineal curve of the consumption would be: fuel_c = 4.8 * Gen_c + 0.47 Financial assumptions: The calculations on the LCOE can be tricky and varies a lot. For this reason, I suggest to read this article. Algorithm Code Batt_cap is the full capacity of the battery, and I think there is room for improvement here since the battery should be used only to the 90% of it’s capacity. In each condition of the algorithm I have added a unique value that will help me trace the conditions behind every result. This is the reason of creating a list called logic. After running all the values of the system in lists, I can now move them back to the dataframe. The reason to do it outside the dataframes is because dfs are designed to fastly apply a specific logic in a whole column. Unfortunately, the values that I create here are depending on previous values. If I use dataframes for this, the time is exponentially slower. That is why this is the fastest way. Also, the only reason that I continue with dataframes is because they are easier for searching values depending on time. Otherwise, I would only stay with lists. I am saying this because in a later stage, I will have to come back to lists to represent some graphs because I was not able to do it using dataframes. Plotting one day As mentioned before, I faced issues to integrate in a same figure different graphs, in this case a bar graph with some charts. The available solution was to convert the values into a list and use an object-oriented approach to show the graphs. I believe the main reason was that the two types of graphs created two different type of X-axis that were not compatible, even though it should be the same. X-axis in this case is time. For more colors, please check the list here: List of named colors – Matplotlib 3.5.1 documentation Calculating the no of cycles An important value to have in mind while finding the payback or the LCOE, is to know when should we replace the batteries. For now, we know the no of cycles that they can withstand, but we need to know how many cycles is estimated to be used per year. That is why I created a logic that checks every time the values of the SoC pass from increasing to decreasing.. The result is 734 cycles per year. Estimating the cashflow and the LCOE Calculating the LCOE is a formula that needs to be developed separately. Since time was pressing for this specific case, I decided to come back to EXCEL and use the tables I already have for a faster getaway. If you are interested in those tables, please feel free to contact me. Possible improvements For more information, or if you are interested in the most advanced versions of this analysis, feel free to contact me.

When I started working with storage, the need for a good understanding of the different types of batteries became imperative. With time I managed to know and compare the different chemistries, but I lacked a method to choose between them. That is why I decided to propose a simple ranking comparing the different chemistries & analyzing their suitability for different applications. Just to clarify, the batteries that I am talking about here are the ones used in storage combined with renewable energy, excluding mechanical storage (like pumped hydro storage, compressed air energy or flywheels). I will be comparing the chemistry functionalities for the following three main applications: 1. Grid services (e.g. frequency response, energy shifting)2. Behind-the-meter (e.g. solar self-consumption, peak shaving, community storage)3. Off-grid (e.g. nano-grid, village electrification, island grid) Types of batteries The first step in my analysis will be to list the most common types of batteries I have encountered during my years in the market: 1. Lead-based. Lead acid batteries were the first rechargeable battery for commercial use. Depending on some production specs, the battery can be designed as: 1.a. Flooded batteries (OPzS) whose main characteristic is that it requires a constant maintenance by regularly filling it with distilled water. 1.b. Sealed AGM batteries. AGM stands for Absorbent Glass Mat. The electrolyte is obviously held in the glass mass. Those batteries have the capacity of recombining hydrogen and oxygen into water and do not require maintenance. Not available in big capacities. 1.c. Sealed gel batteries, or commonly known as VRLA (valve-regulated lead acid) where the electrolyte is gelified. Same advantages as the AGM batteries, but with wider variety of capacities. This type will be our main focus among the lead-based batteries. 1.d. There are other technologies I have heard of such as lead-crystal or lead-carbon batteries but I do not have enough experience or knowledge about them to share more information. If you do, please do not hesitate to contact me and share with us the info that you might have. 2. Lithium Iron Phosphate. Lithium-ion batteries (to not confuse with lithium-metal) are based on the same concept as lead-based batteries. They contain a cathode, an anode and an electrolyte. The cathode is normally a Li-Metal-Oxide and the anode consists of porous carbon. Many metals have been used, but the most famous alloy is the Iron Phosphate (LFP or LiFePO4). 3. NMC. Another type of Lithium-based batteries, with Nickel Manganese Cobalt Oxide as the metallic alloy. 4. NiCd. Nickel-cadmium batteries were the next batteries invented after the lead acid ones. They offer several advantages, like good performance in high ambient temperatures, although they are expensive (compared to lead). 5. Flow. A redox flow battery is a battery based on two components dissolved in liquids separated by a membrane. The concept behind it is like fuel cells but the ionic solution or electrolyte is not stored in the cell itself and rather in other storage tanks. They are named redox batteries due to the electrochemical reaction of reduction-oxidation. The most known flow battery is the vanadium redox battery that uses vanadium ions for the redox reaction. Am I missing any other important storage chemistry? Please let me know. What should I compare? Now that I have defined the list of batteries that I would like to work on and analyze, the first question that arises is: what are the parameters that I should compare? And moreover, what is the purpose for choosing each parameter? 1. Number of cycles. A cycle is the process of discharging and charging the batteries. The no. of cycles directly affects the life span of a battery. For example, a good VRLA cell can reach up to 3 000 cycles. This value is highly dependable on the Depth of Discharge (DoD) of the battery and the temperature. The deeper we discharge a battery and the higher the temperature is, the fewer cycles the battery will have. 2. Specific power. Specific power is a power-to-weight ratio. Some batteries as we will see have a high specific energy but they are incapable of providing high currents in short times. Loads that require inrush currents like pumps or motor-based systems can damage batteries that are not suitable for high demand of power. 3. Energy density is the ratio between the energy that the battery can provide and its volume. It is also common to analyze the specific energy, which is the ratio between the energy and the mass of the battery. Many applications require our attention regarding the required space for the storage system, for example, in household applications or in cases when the logistical part of a remote rural area project. 4. Efficiency. Efficiency has an important role in two aspects: economical & environmental. Any Wh of energy that is not used to power a load is an amount of money that has not been earned. But it is also an amount of energy that is only transformed to heat. This in turn increases the energy demand which might result in increased CO2 emissions. 5. Cost. Price is and will always be a sensitive and crucial requirement. We need to provide solutions that can be affordable and competitive, with a reasonable return of investment (RoI) rate. Comparison Now that I have listed the parameters and the different chemistries, below you can find some graphs to visualize and compare the results. VRLA batteries Lead-acid batteries are the best solution for cost-sensitive projects. They are robust and with a long record of robustness and reliability. Unfortunately, their cycling is low compared with all new developed technologies like Li-ion or flow batteries. I would highly recommend them for rural areas electrification projects, remote areas and some small telecom systems. LFP Batteries As indicated in the graph below, LFP batteries have a wide advantage in the market and its pricing is constantly dropping thanks to the electric vehicles market. There are many reliable products with quality Battery Management Systems (BMS) in the market since most of the vendors are only integrators of cells. Their advantages in many parameters make them suitable for most applications. I might exclude CAPEX-sensitive projects like in rural electrification. NMC batteries I like the analogy to