 Welcome to the course on Cathodic Protection Engineering. We have seen the concepts of cathodic protection engineering in relation to electrochemistry and the electrical engineering. Today we look at some of the numericals that will enable us to understand this concept better. The talk will consist of the following. The first two problems in today we will talk about how the electrochemical parameters would decide the cathodic protection engineering viability. These are today's contents of the lecture. We will first look at the electrochemical concepts in relation to cathodic protection engineering. Then we will move on to the application of electrical engineering concepts in the selection of rectifier for cathodic protection engineering. Then we move on to the design of cathodic protection system using sacrificial anodes. Let me first start with the principles of cathodic protection engineering. We have seen this slide in one of the earlier lectures. We know that when you expose a metallic structure in a soil or any corrosive environment it attains a corrosion potential in the parlance of cathodic protection engineering and natural potentials that is given by this diagram here. Corresponding to that particular potentials the metal suffers a corrosion given by the corrosion current density in this particular Evans diagram. Let us look at the first concept which is cathodic protection engineering in relation to the electrochemistry. This is an Evans diagram you have seen before that is between the potential and the log current density for the corrosion reaction. The corrosion reaction involves the anodic dissolution of the metal and a cathodic reaction occurring on the metallic surfaces. So, when the metal is exposed to a corrosive environment or to a soil it exhibits a potential called as natural potentials or also called as the corrosion potentials given by here. Corresponding to this particular potentials the metal will exhibit a corrosion rate given by the corrosion current density given by this Evans diagram. Now, as you notice that the corrosion current density and the corrosion potential are governed by these parameters such as exchange current density for the cathodic reaction, the exchange current density for the anodic reaction and the Tafel relationship for the anodic reaction that is metal oxidation process and the Tafel equation for the cathodic reaction that is reduction of some species such as oxygen or H plus ions on the metallic surfaces. Now, depending upon the slope of either the cathodic reaction and the anodic reaction then the current that are required for cathodic protection changes and also this slope also decides whether the applied potential is adequate to protect a metal to the desired levels. So, that we will be seeing in one of the problems. Next is the rectifier selection. Now, the impressed current cathodic protection of the engineering structures involves the use of the rectifier. The capacity of this rectifier depends upon the current that is required to protect these structures and the resistance offered by the ground bed that is the anodes, the cables and the pipelines. Today, we will see a problem that is related more to the anode ground bed. We will look at this the other problem that is sacrificial anode cathodic protection system. In this we need to look at selection of appropriate number of anodes. Even the selection of appropriate anodes what do you mean by appropriate anodes? It could be magnesium or it could be zinc or it can be even aluminum for example. So, what decides the number of anodes and what decides the life of these sacrificial anode cathodic protection system? The problem that we are going to discuss today we listed three concepts we have just enumerated so far. Let us go to the problem number one. The problem number one we talk about the cathodic protection of a steel pipe using impressed current cathodic protection. Here we will look at application of 300 millivolt potentials on the metallic surface in order that the structure is adequately protected. Now, the 300 millivolt that we talk about is not of potentials it is essentially the IR drop discounted. So, this potential is the actual polarized potential applied onto the metal surface. So, we look at the condition in which if the tapered slopes for the anodic reaction and the cathodic reaction is 0.12 volt per decade. So, what will be the consequent reduction in the corrosion rate of the pipeline? So, how many times did the corrosion rate of the pipeline reduce because of the application of 300 millivolt? One question. If the tapered slope happens to be 0.6 volt per decade the what will be the corresponding decrease in the corrosion rate? So, this problem in turn gives us an understanding how the tapered slopes affect the intended corrosion rates for a given polarized potential. So, let us look at this problem now. Now, what is given here in the left hand side is the Eman's diagram corresponding to the steel that is corrosion of the steel anodic polarization curve that is seen here and the natural potential is given by this point that is corresponding to we call them as E car 1 and the corresponding corrosion current density as I car 1 here. Now, we are applying 300 millivolt in the native direction that means, the shifted voltage is from E car 1 to E car 2 and as a consequence the corrosion current density decreases from this value to this value. So, we shall now calculate what is the ratio of decreasing current density corrosion current density because of the application of 300 millivolt. So, this we start with the well known the Tafel relationship for the anodic reaction that is eta is given as beta this Tafel slope log I car upon X n condensity then eta is over voltage is given by corrosion potential minus the equilibrium potentials that is this is the equilibrium potential and this is the corrosion potential and that is the over voltage if a metal is immersed in the soil without the application of cathode protection. Now, what we are looking at is we are shifting so, the same equation is given here we are rearranging the equation over here it is E car minus equilibrium is given as beta Tafel slopes logarithm of X n condensity upon I car 1 that is the metal that suffers corrosion when buried in the soil. On the application of the 300 millivolt the potential sits from here to this and we can rewrite the Tafel equation as I car 2 upon I naught over here. So, we can solve this equation to find out the ratio between the I car in the unprotected condition and the I car in the cathodic protected conditions by application of 300 millivolt. So, we can solve these equations and you will you will get the resultant equation here E car 1 minus E car 2 is given as Tafel slope beta A logarithm of I car 1 upon I car 2 ok. So, that is a general equation right. Now, if you are considering a Tafel slope of 0.12 volt per decade and you can see and if the polarized potential is 300 millivolts which is converted into volt here and substituting these values in the equation and solving them you will find that the ratio between I car 1 and I car 2 is 0.0031. So, that corresponds to a increase in corrosion resistance increase in corrosion resistance by 323 times should the Tafel slope be 0.12 volt per decade. Now, let us look at the other alternative should that Tafel slope be 0.6 volt per decade. So, what happens? We can use the same methodology to solve the problem and we retain the applied voltage as same 0.3 volt which is the polarized value in the cathodic protection then as a consequence you will see that the ratio of the final corrosion rate to the initial corrosion rate turns out to be 0.3164 which amounts to just about 3.16 times ok. So, just compare these two values should the Tafel slope be 0.12 volt per decade for the same 300 millivolts polarized potentials one would expect a increase in corrosion resistance by 323 times as opposed to just 3.16 times if the Tafel slope is 0.6 volt per decade. So, when you talk about cathodic protection criteria let us say 100 millivolt criteria is very good ok, but we should also know that to what extent the pipeline can get protected from corrosion also depends upon the Tafel slopes. The Tafel slopes can change depending upon seasonal conditions when it is wet and dry seasons or it can also depend upon the soil chemistry it can happen, it can also depend upon the coating conditions of the pipelines. So, there remains some uncertainty in cathodic protection when we use 100 millivolt criteria. I am not saying that 100 millivolt criteria is not to be used, but it is necessary that we need to have a perspective how a given situation can differ in terms of how effective the cathodic protection is. So, let us go on to the next problem. This problem deals with a pipeline that suffers from diffusion limiting current density, diffusion control corrosion rate that is happening and especially when the oxygen reduction reaction is the predominant cathodic reaction and the happens when the pipelines have slightly neutral or relatively alkaline soil conditions and where the pH is is slightly higher at neutral for example, 7 or 8. The predominant cathodic reaction is oxygen reduction reaction. Of course, in addition to that there is also water reduction reaction, but the driving force for the oxygen reduction reaction is quite significant compared to the driving force for water reduction reaction. So, in this case we have a pipeline whose diameter is 30 centimeter. The length of the pipeline is about 1 kilometer long pipeline. The measure limiting current density that is cathodic current density is 1 into 10 power minus 4 ampere centimeter square that is about 100 micro amperes centimeter square. It is given that the thickness of the pipeline is such that the pipeline will last for 5 years should there be no cathodic protection. So, one would like to increase the life of this structure by 10 times. So, what should be the applied potentials and what should be the current required to protect the pipeline? Given the Tafel slope, the anodic Tafel slope for the metal dissolution is 100 millivolt per decade and the pipe to soil potential that is measured without cathodic protection or it is a natural potential I would call it is minus 0.5 volt with respect to copper saturated copper sulphate reference electrode. If these are the values given then what will be the applied potential that will provide us increase in life by 10 times and again what is the current required to protect the pipeline? This is the problem. So, let us look at how to solve this problem. Let us again go back to this basic concept of Evans diagram. It is potential against log current that you see here. The cathodic current is is diffusion control limiting current density here you can see that the this line corresponds to the anodic Tafel slope, the anodic this line corresponds to anodic kinetics right. And the intersection point between the cathodic reaction and the anodic reaction is over here and there is a corresponding potential right and there is a corresponding corrosion rate I car. And in this case I car the corrosion current density equals the limiting current density right. Now, if you look at this diagram the E car corresponds to 0.5 volt with respect to copper saturated copper sulphate electrode that is what is given here. And because it is limiting current density ok the corrosion is diffusion control ok. The limiting current density equals the corrosion current density. So, that is given over here. So, I L corresponds to I car which is 10 power minus 4 ampere per centimeter square and E car is given as minus 0.5 volt with respect to copper saturated copper sulphate electrode reference electrode. Now, you want to increase the life of the structure by 10 times. So, the corrosion current density which is I car currently is 10 power minus 4 ampere per centimeter square should be reduced to 10 power minus 5 ampere per centimeter square that will give you an increase in life by 10 times. So, what is the applied voltage required to reach this current of 10 power minus 5 ampere per centimeter square. So, we again use the Tafel relationship that is the governing relationship for determining the potential for a given current density. I have replaced here E applied and E car because we are moving from E car to E applied and E car here corresponds to the potential where the metal dissolves at a current density of I car here. So, we need to find out what will be the applied potentials if we need the final current density as 10 power minus 5 ampere per centimeter square. So, substituting these values here in this Tafel equation and one would obtain the applied potential is equal to minus 0.6 volt with respect to copper saturated copper sulphate electrode. So, we need about 100 milli volt polarization towards negative potentials. The next thing is what is the current requirement the calculations right. We have seen in the first or the second slide ok. The applied current required for cathodic production is the difference between the cathodic current density and the anodic current density ok. So, the applied current density is given as the difference between the cathodic current density and the anodic current density and we know the cathodic current density does not change because it is under limiting under diffusion control. So, the problem becomes somewhat simple because the cathodic current density is constant because of the fact that the corrosion is diffusion control ok. Now, the this difference of 10 power minus 4 10 power minus 5 is this size to 5 into 10 power minus 5 ampere per centimeter square is the current density required per centimeter square area of the sample. So, we need to find out the overall current required to protect the pipeline. To do that we need to know the surface area of the pipeline that is substituted here is pi d into length you can see this ok and then the current density which turns out to be about 847.8 amperes of current. You will also notice that the pipeline when it is not supplied with a coating the current required for cathodic protection is quite significant. So, we will see later that how the application of coatings would reduce the current requirement. Let us look at a third problem wherein the following parameters are given that is the soil resistivity the size of the anodes employed for cathodic protection using ISACP method and these anodes are also as you know they required to be buried in a backfill right. So, the backfill dimension is given here the number of anodes and the spacing between the number of anodes are given. The current required for cathodic protection is given as 4 amperes and the current required for cathodic protection is generally established by installing a temporary anode ground bed and measure the current required to reach a protection potential of minus 0.85 volt with respect to copper saturated copper sulphate electrode. And the backdrop voltage also can be measured by switching off the rectifier and we need to know the cable resistance and pipe to soil resistance. Given these parameters now one need to calculate what is the rectifier capacity and one capacity in terms of amperes is given here. But there is a small caveat that the current required for cathodic protection might vary with respect to time because if it is coated with a nice coating as a coating disintegrates the current requirement becomes more. So, what is given here is that with the aging the current requirement increases by about 25 percent. So, we need to calculate what is the capacity of the rectifier. The capacity of rectifier here refers to the voltage and of course, when we talk about 25 percent increase in current that you can always calculate what will be the overall current required the capacity of the rectifier is known. But what is to be calculated is the voltage of a rectifier used to be calculated. So, how do you look at it? We know that in the circuit there are various elements that offer the resistance if I go back to this the resistance offered by the cable resistance offered by the pipe to soil is known. So, we need to calculate the resistance offered by the anodes. We use this equation that is been given earlier the resistance offered by a single anode when they are placed vertically in relation to the pipeline is given by this equation that rho represents the resistivity of the soil and L represents the length of the anode in this case the length of the baffle we will talk about. And k is called a save function it can be obtained using the standard tables the k is related to the length upon D the ratio of the backfill ok. And based on these values you can calculate you can you can find out the corresponding k value from these tables and you can substitute this k in this equation. And one would get the resultant resistance which is 6.56 ohms offered by a single anode. Since we have a single rho multiple anodes it is necessary to calculate what is the resistance offered by the ground bed the ground bed consists of multiple anodes. And so, we use this equation which is the resistance offered by the multiple anodes is given by this equation here p represents parallelizing factor but that depends upon the number of anodes. And this can be obtained from the standard tables and you can substitute these values in this equation you will get that for the 3 anodes you will have about 2.59 ohms is the resistance offered by this anode actually. Now, the resistance offered by the overall cathodic protection system consists of resistance offered by the anode the soil and the cable. So, this is the overall resistance offered by the circuit. In addition to that there is also what is called as a backdrop voltage that also resist the flow of current. So, that should be taken into account right. So, that will come to you later right. So, the current required for cathodic protection including the deterioration that happens over a time period is about 5 amperes. So, we simply apply the ohms law and you obtain the voltage required to pass 5 amperes of current which turns out to be 12.95 volt. So, that is the capacity of the rectifier required to drive the current in the soil. But however, if there is a backdrop voltage now the rectifier has to overcome the backdrop voltage in addition to these resistances which are in series. So, ultimately you can substitute this 1.9 volt which is a backdrop voltage given here and the resistance I mean the rectifier should have a minimum capacity of 14.85 volts in order that the cathodic protection can be successfully established. We move on to the next problem that deals with designing the sacrificial anode cathodic protection system. Various parameters are given here in this problem ok. So, we need to know the dimension of the pipeline certainly. We also need to know what is the current required for cathodic protection and what the anode used in this case the proposed anode is zinc anode whose voltage is minus 1.1 volt given here and we also need to know the anode is positioning whether it is horizontal or vertical. So, in this case it is a vertical anode the resistance offered by the pipeline as you have seen in the earlier case you need to know here as well. If you are going to use multiple anodes what are the spacing between these anodes as opposed to ICCP systems ok. The sacrificial anode systems as you see that the anodes themselves provide the required current in the ICCP systems the current is provided by the rectifier ok. So, the dissolution of the anode provides a current. So, we need to know what is the amount of anode required for protection of the structure for a required time. Now, we look at the characteristics of the zinc anode. The zinc anode has a characteristics which is which can dissolve about 100 sorry 10 kilogram per ampere year. So, the anodic the anode consumption rate of the zinc anode is 10 kg per ampere year that means, if I had to pass 1 ampere current per 1 year I would need 10 kilogram of zinc. We also like to know whether the pipeline is coated if so what it is it is coated with effusion bonded epoxy and each of the anode each of the anode we are talking about calculation of the number of anodes required. So, each of the anode the dimension I mean the the weight of the anodes is 14 kilograms. The resistivity of the soil is 1000 ohm centimeter. The anode dimension is given here ok is a rectangular shape and it is kept in a back wheel of dimension given. It is kept in a back wheel of dimension given here. The leads used for establishing the sacrificial anode cathodic protection system is 0.04 ohms and you all know that whatever the anode that you are going to give not all of them are consumed we may have to discard the anode after a certain dissolution. In this case the utility factor that is 0.9 is utility factor that is when 90 percent of the anode is consumed we may have to replace a new anode that is the value here. The current deficiency of the zinc anode is 90 percent which is 0.9 fraction. So, given these parameters we need to establish a cathodic protection system using zinc as a sacrificial anode. Let us look at the problems. The first and foremost is we need to determine what is the total current required for protecting the given pipelines ok. So, we know the current density this 70 micro ampere per meter square is the current density. So, we need to know the total current which means we need to determine the total area of surface area of the pipeline that is surface area of the pipeline that comes in contact with the soil. From that we can calculate the total current which turns out to be 66 milliamperes of current required. Now, once we know the current then we need to look at the resistance offered by the anode bed. We follow the same example as the previous problem we determine the resistance offered by a single anode. From there we compute the resistance offered by the ground bed consist of multiple anodes. The equations is well known to us and so, when you substitute all these values the resistance offered by one anode turns out to be 2.36 ohms. And we need to now decide how many anodes are required so, that the sacrificial anodes can protect the pipeline satisfactorily. The deciding number of anodes is sometimes is iterative because you need to see that the anode ground bed resistance is sufficiently small so, that the anodes can drive the current onto the pipeline. So, the number of anodes that we need to substitute here we do by iteration method. So, that the resistance calculated based on this equation that is here is sufficient enough or sufficiently small enough for the current to be flown to the pipelines. So, in this case we did some iteration sometimes we took 3 for example, when you took 3 obviously, the resistance offered was more and that was not a satisfying the condition. So, we moved over to 4 of course, you can also go for 5, but you will see later that by increasing to 5 you will have excess amount of anodes which are not really required both in terms of driving the current and in terms of the life of the cathodic protection system. So, this is a bit of iteration that you do that. So, once we know the resistance offered by the anode ground bed, we look at the total resistance is offered by the cathodic protection system that consists of a pipeline anode on the leaks right. So, these values are already known the pipe to soil resistance the lead resistance are known the anode ground bed resistance has been calculated and you can find out the total resistance offered by the anode ground bed. So, the total resistance offered by the cathodic protection system is 2.97 volt that constitute resistance from the pipeline anode on the leads. Now, our next task is to determine whether the zinc anode is able to drive the current against this resistance ok. So, that will be seen in the next problem next. So, we look at the driving force that zinc can really offer. We need to understand there is a difference between the polarized potential of zinc and the actual potential of zinc when you buried in the soil. When you when you bury a zinc in the soil which is not cathodically connected which is not electrically connected to the pipeline as you seen before is minus 1.1 volt when zinc is connected to the pipeline zinc gets polarized and that is how the current is generated and that current passes through the pipelines. So, when you get polarized the ultimate potential is somewhere close to minus 1.05 volt. In fact, the advantage of zinc over other sacrifices anode can be seen here because there is hardly a polarization. You start from minus 1.1 volt and it is moved just to minus 1.05 volt. This is primarily because zinc is less polarizable anodically. In fact, that is one of the reason why zinc is also used as a reference electrode for cathodic protection of pipelines. So, assuming the polar is a potential to be minus 1.05 volt and the minimum requirement for cathodic protection is minus 0.85 volt with respect to pipe to soil. The driving force corresponds to 0.2 volt, right. So, now the question is is this 0.2 volt is sufficient to overcome the resistance which is 2.97. You can see here that the anode bed can give 67 milli amperes of current barely managed to give the current actually and barely managed to give the required current of 66 milli amperes. Should be very comfortable you want to have a better capacity. The one way to do is you go for 4 number of anodes. Instead of 4 number of anodes you want to go for a better capacity you can use 5 number of anodes, but you will see later what it can lead to if you are going to go for 5 number of anodes instead of 4 number of anodes. So, coming back to this problem that if you are going to have 4 number of anodes, these anodes can drive a current of 67 milli amperes given the resistance of 2.97 ohms, ok. Now, we talk about 4 number of anodes and and so what is the life of the anodes? The life of the anodes is given by this equation the anode weight, current efficiency, the utility factor, the consumption rate and current demand, ok. Now, all these values are known, we substitute them in the equation and you will get a value of 68.7 years of service life. So, if you are going to use 4 number of anodes, the structure in principle can be protected for 68.7 years. However, there are some issues that zinc anodes pass away over a time period, right. If you also look at the other factor we discussed that the 4 number of anodes barely gives a current, right a current of 67 milli ampere as against 66 milli amperes, right and we said that you want to increase the number of anodes. If you increase the number of anodes automatically you notice that you know instead of 4 you substitute let us say by 5 then the life of these anodes are going to be much higher. That means, we are wasting the zinc and so we should avoid having more number of anodes. In that case, what are the other possibilities? The other possibilities is that you can look for the magnesium anode. So, today's lecture we looked at the three important concepts and we use the calculations to clarify these concepts. The first was that the Tafel slope is very important even when we describe the cathodic protection criteria in terms of let us say 100 milli volt criteria, ok. You see that how the Tafel slopes can change whether a metal will be adequately protected or less adequately protected. In the second problem we looked at the what are the constituents of the cathodic protection system that decide the capacity of rectifier. In the last problem we discussed how to design a cathodic protection system when the sacrificial anodes are used. I do hope that these problems will give you better clarity on concepts.