Chapter Six transformer protection. transformers are a critical and expensive component of power systems. Due to the long lead time for repair and replacement of Transformers a major goal of transformer protection is limiting the damage to faulted transformers. The comprehensive transformer protection provided by multiple function protective relays is appropriate for critical Transformers of all applications. The type of protection for the Transformers varies depending on the application and the importance of the transformer. Any extended operation of the transformer under abnormal conditions such as faults or overloads compromises the life of a transformer, which means adequate protection should be provided for quicker isolation of the transformer under such conditions.

Looking at the affer mentioned in a little bit more detail, the reasons to provide transformer protection are, first of all to detect and isolate the fault of faulty equipment. Secondly, is to maintain system stability. Then thirdly, we want to limit the damage and that involves minimizing the risk of fire and of course minimizing the risk risk to personnel including workers and the general public factors that are affecting the transformer protection. First of all, the cost of repair of the equipment. As I've said the transformers are a extremely expensive inventory item and we didn't order to repair or replace them costs a lot of money. So we want to minimize the cost of repair.

Also, we don't want the damage to be too extensive. So we want to minimize the cost of downtime because we ultimately could be losing revenue as a utility because of equipment outage, not to mention the public relations if we have people in or out of power for any length of time. And then it also affects the rest of the system and how the system would be operating without that equipment. And of course, if there is a catastrophic failure, like in the case of a transformer, if it explodes, there's a potential for damaging adjacent equipment. And of course, it's I've also mentioned the length of time Have a repair or in order to replace the equipment. At times, some of the generating stations are in a very remote location and in order to get a heavy piece of transformer equipment up to that location would require a lot of time and a lot of logistical movement of inventory and personnel.

Protection requirements fall under four categories. selectivity selectivity means that the protective scheme should accurately identify the system element, which is at fault and initiate the tripping of only the minimal possible number of circuit breakers required to isolate the defective element or elements. Proper selectivity ensures that the minimum number of customers are interrupted and the minimum number of system elements are taken out of service. unnecessary interruptions cut into revenue and reduce customer satisfaction. Unnecessary removal system elements contributes to overloading of equipment remaining in service and could contribute to the instability of the entire system. There must be a correct coordination provided in various power system protection relays in such a way that a fault at one portion of the of the system should not disturb another healthy portion.

Fault current may flow through the part of the healthy portion since they are electrically connected, but relays associated with the healthy portion should not operate faster than the release of the faulty portion. Otherwise undesired interruption of healthy equipment will occur. Again, if really associated with The faulty portion is not operated in proper time due to any defect in it or other reasons, then only the next relay associated with the healthy portion of the system must be operated to isolate the fault. Hence, it should neither be too slow, which may result in damage to the equipment nor should it be too fast which may result in undesired operation. Next literal speed. Speed means the protective relays must operate at the required speed speed of a fault clearance is essential since the likelihood of a widespread system disturbance or complete system collapse increases with the duration of the fault.

Rapid clearance of fault minimizes the damage due to faulty equipment and also minimizes the hazard to personnel who may be in the vicinity. Role ability is the most important requisite of a Protective Relay. relays remain inoperative for long periods of time before a fault occurs. But if a fault occurs the relay must respond instantly and correctly. Reliability that is dependability and security is essential for any relay scheme. dependability is defined as the assurance that the relay will operate when required to security is the assurance that they will not operate when not required to traditionally, utility facilities have favored dependability rather than security.

That is false trips out number of failures to trip when our while false trips are obviously not desirable. The emphasis on dependability is preferred since the consequences of a failure to trip are generally much more severe. Then the consequences of a false trip. And lastly, there is sensitivity. sensitivity of the reeling equipment must be sufficiently sensitive so that it can operate reliably when the level of a fault condition just crosses a predefined limit. Having said all that, transformer faults could fall into one of three categories.

First of all, there's overloads these tend to not be extremely fast. In fact, they kind of slow slowly grow as the load is added to the system, but there is a predefined limit. That Transformers should not exceed both noise temperature and current flow and several other factors. But overloads could cause damage and when they cross a predefined limit, that's when reeling action must be taken in consideration and either trip the transformer off or adjust. feeders coming off the transformer so that the overload condition will not exist. Then there's internal false.

These are the major reasons why we have reeling protection on a transformer. These can be very catastrophic if they've left into the system too long and depending on the type of fault could actually destroy the whole transformer. So we definitely want to have good relaying in place for the detection and isolation of internal faults within a transformer. Then there are through faults and we talked about faults external to the transformer but through faults could cause the Transformers to become overloaded in themselves, and they certainly have to have protective equipment there. In case the outside elements fail to isolate the fault and through fault condition could still damage the transformer and you want to make sure that that does not happen. transformer overcurrent protection falls into two main categories, relays and fuses, the choice of which is largely dependent on the transformer size, voltage levels and downstream loads.

Small distribution transformers are commonly protected only by fuses. In many cases no circuit breaker is provided making fuse protection The only available means of automatic isolation. fuses are overcurrent devices and must have ratings well above the maximum transformer load current in order to carry without blowing the short duration overloads that may occur when starting large motors and Also to withstand the magnetizing inrush currents drawn when a power transformer is first energized. It should be noted here that HRC or high rupturing capacity fuses, although very fast in operation with large fault currents are extremely slow, with currents of less than three times the rated value. It follows that such fuses will do little to protect the transformer. When load limit thresholds are just slowly approached and then exceeded.

Serving only to protect the system by disconnecting a faulty transformer after the fault has reached advanced stages. Larger Transformers 100 kV a and over may be controlled by circuit breakers in which case protection can be provided by overcurrent relays. improvement in protection is obtained in two ways. One is The excessive delays of the HRC fuses for lower fault currents is avoided and an earth fault tripping element can be provided. In addition to the overcurrent features, time delay characteristics are chosen to discriminate with circuit protection downstream on the secondary side. Instantaneous trips may be added the current setting being chosen to avoid operation with short circuits on the secondary side.

Also HRC fuses may be used to backup the circuit breaker. Although full discrimination with secondary circuit protection is necessary. The fuse rating need not be as high as through fault current depending on the transformer reactance the source fault power and whether the secondary system is a single circuit or a sub divided into several branches with each having its own protection. The main function of such instantaneous protection is to give high speed clearance to close in an internal false. The settings therefore may be relatively high when determining the parameters or limits of the protection for the various types and sizes of transformers, the following I triple E standards are used as guidelines for the choice and settings of transformer protection. C three 7.91 the I triple E guide for protecting power transformers sci fi 7.1 2.00 the I triple E standard for general requirements for liquid immersed distribution power and regulating transformers and c 57 dot 109 which is the I triple E guide for learning Quit immerse Transformers through fault current duration.

For the choosing and setting of overcurrent protections the following must be known and if not supplied by the manufacturers transformer documentation can be estimated using the I triple E standards. rated current which is in VA or K VA, the inrush current short circuit currents and the transformer damage curves. These parameters are plotted either on graph paper which is the old method or on a computer the characteristics of the relay and, and or fuse protection is that then fitted inside the envelope created by these four variables. Let's walk through a simple example given a 3000 150 kV a transformer 13.8 kV 241 60 volts Delta Wye transformer we are to select the proper fusing on the primary side. The rated primary current can be calculated by the KV a rating of the transformer or the VA rating of the transformer, which is the VA divided by the line to neutral voltage and we don't have the line to neutral voltage so it must be calculated so it's VA all over root three times the line to line voltage which is 3750 kV a all over root three times 13.8 kV which is equal to 157 amps.

The rated secondary current and may be coming calculated in the same way, it's the VA over root three times a line a line voltage, which gives 520 amps. The inrush current is either supplied by the manufacturer or is given by the I triple E standard as 12 times the rated current. So that would be 157 times 12 or 1884 amps. The transformer damage curves can be obtained from the I triple E standard 57.109 which is the guide for liquid immersed Transformers through fault duration. These values are plotted on a time current curve or a TCC Using log log graph paper or on a computer today with coordination software a lot of the transformer information is inaccessible databases including manufacturers transformer information, and how it's related to the I triple E standards as well as most popular fuse and relay curves. From this information the transformer damage curves can be plotted along with the inrush current of 1884 amps.

The transformer primary full load amps of 157 can also be added to the graph. So, the fuse curve should lie between the transformer full load amps and the inrush current curves and its damage curves and it does this shape to miss the full load amps and of the transformer and not be affected by the inrush current Current, but will melt below or before the onset of the damage curves. As Transformers get larger, it gets more difficult to fit the fuse curves. Also, the need for speed increases in relays and breakers will give the speed and characteristics that are required. We now replace the fuses with instantaneous and timed overcurrent relays and breakers allowing us to fit the tripping characteristics of the relate to our transformer parameters perfectly. The instantaneous overcurrent is set to pick up just above the inrush current level with some margin of error of course, and the timed overcurrent characteristic is set to line up just below the damage curves.

Relay is usually chosen with a family of curves that will fit the situation and the one curve that perfectly matches the require parameters is then chosen. Transfer transformer differential protection schemes are ubiquitous to almost any power system. While the basic premise of transformer differential protection is straightforward, numerous features are employed to compensate for challenges presented by transformer applications. Although this type of protection provides the best selectivity and speed for the detection and clearing of system false, it's not without its challenges. Primarily current mismatches are often caused by different sikhi ratios or even Using different manufacturer CTS and mixing them in a system may cause a mismatch and they have to be carefully looked at can transformer configuration delta y y delta y zigzag type Transformers introduced current shifts that have to be accounted for when applying differential protection. And if the transformer has on load tap changers, as soon as the tap changer starts to move, you automatically introduce MIT's mismatches between current ratios on the primary and the secondary.

And you also run into current saturations and the current the history says remnants of current transformers which have to be allotted for and with the transfer power transformers. There's the inrush phenomena that also introduces a large amount of harmonic content that has to be accounted for. Current differential relaying is applied to protect many elements of the power system. The simplest example of current differential routing scheme is shown in this slide. The protected element might be a length of circuit conductor, a generator winding a bus section or more to the point in our case, a transformer. It can be seen that the current differential relaying is a basic balancing of current using the application of current jobs current law.

The relay operates on the sum of currents flowing in the CTX secondaries that is I one plus i two. Under normal conditions or faults outside the protected element I one equals I two. If the turns ratios of the CT are the same for both currents in the CT secondaries will be the same. Therefore, by virtue of the CT connections I one and I to add to zero through the relay that is, I subscript di FF or I diff is equal to i one plus i two is equal to zero. The secondary currents less appear to circulate in the CT secondaries only and not through the relay. When a fault occurs inside the protected zone, let's see what happens under fault conditions in the zone of protected equipment, I one does not equal i to the currents in the CP secondary May be quite different element and will flow out of the spot secondaries.

Therefore, by virtue of the CT connections I won and I to add to a value greater than zero through the relay that is I diff is equal to i one plus i two is not equal to zero. The secondary current does no longer appear to circulate in the CTE secondaries only, but will add to a substantial amount flowing through the differential relay coil. In power transformers the input power is equal to the output power. However, the voltage and the current in both the primary and secondary sides are different depending on whether the transformer is step up or step down. For instance, if the transformer is a step up transformer that means the input voltage is higher than the output voltage And conversely, while the input current is higher than the output current For this reason, and in order to balance the current for differential protection, the CTS in the primary and the secondary sides of the power transformer will not have the same turns ratio.

However, they are carefully selected in terms of turns ratio so, that they both have the same output current under normal conditions of operation. Even if of different turns ratios the same make and model of CTS should be used. If this is not possible, other ones with characteristics as close as possible should be chosen. However, this mismatch can and will lead to erroneous differential current flowing in the operating coil. The higher the current, the larger the mismatch will be generated for the differential current However, if the CTS are matched with the proper ratios and we consider per unit values for the current under normal conditions or for faults outside the protected equipment, I one per unit is equal to two per unit. If the turns ratio of the CTS are matched to the turns ratio of the transformer, both currents in the CT secondaries will be the same.

Therefore, by virtue of the CT connections I one and I to add to zero through the relay. The secondary currents thus appear to circulate only in the CP secondaries and not through the coil of the differential relay. This brings us to One of the challenges with transformer differential protections as mentioned previously, and that is CP saturation and reminisce. As was stated a few slides back in power transformers, the input power is equal to the output power in the voltage and the currents in both primary and secondary sides could be quite different. For a step down transformer, the input current is lower than the output current. However, under ideal conditions, if the CP ratios were selected correctly the output current from both CTS balance However, even though the CP ratios were selected to balance the secondary CT could go into saturation before the primary CP causing an imbalance in the secondary currents.

This of course, could result Often would result in the erroneous operation of the differential Relay for faults outside the protected zone. transient magnetizing inrush or exciting current occurs in the primary side of the transformer whenever the transformer is switched on or energized. This is the current that is required to set up the magnetic flux that links the primary to the secondary of the transformer. At this time the first peak of the flux wave is higher than the peak of the flux at the steady state condition. This current appears as an internal fault and it is sensed as a differential current by the differential relay. The value of the first peak of the magnetizing current may be as high as several times the peak at full load the magnitude and duration of the magnetizing inrush current is influenced by many factors.

Some of these factors are the instantaneous value of the voltage waveform at the moment the circuit breaker is closed the value of the residual or remnant magnetic flux, the polarity of that residual magnetizing flux, the type of iron laminations used in the transformer core, the saturation flux density of the transformer core the total impedance of the supply circuit the physical size of the transformer the maximum flux carrying capability of iron core laminations and the input supply voltage level the effect of the inrush current on the differential relay could result in false tripping of the transformer without any existing type of fault. From the principle of operation of a differential relay, the relay compares the current coming from both sides of the power transformer. However, the inrush current is flowing only in the primary side of the power transformer. So that the differential current will have a significant value due to the existence of current in only one side.

Therefore, the relay has to be able to distinguish or recognize that this current is a normal phenomenon and not to trip as a result of it. This current is usually taken as 12 times the rated kba current for 0.1 second and on Load tap changer is installed on a power transformer to control automatically the Transformers output voltage. This device is required whenever there are heavy fluctuations in the power system voltage. The transformation ratio of the CTS can be matched with only one point of the tap changer range. Therefore, if the onload type changer is changed, unbalanced current flows in the differential relay operating coil. This action causes a CTS to be mismatched.

This current will be considered as a fault current which could make the relay operate and send out a trip signal. Here again the relay has to be designed to recognize this type of current as a normal phenomenon and not to trip due to this current. So to summarize, there exists three main difficulties that tend to handicap the conventional transformer differential protection they induce the differential relay to release a false trip signal without the existence of any fault. They are the magnetizing inrush current during the initial energization of the transformer, CT mismatch and saturation and transformer ratio changes due to tap changer operation. A solution to these difficulties is the percentage of differential type relay. It's really has three coils one operating coil which operates the same as a simple differential relay coil, except that it is restrained from operating by two restraining coils.

These coils restraining strength depends on the amount of flow through current the differential current required to operate this relay is a variable quantity owing to the effect of the restraining coils. In the percentage restraint current differential relay the operating current is the vector sum of the CT currents i is equal to i one plus i to the operating current must be large enough to overcome the restraint provided by some percentage k of the flow through current which is derived from the sum of the magnitude of the individual CT currents. That current is equal to k times the magnitude of AI one plus the magnitude of AI to the operating characteristic of such a relay is shown here except for a slight effect The control spring at low currents over the minimum pickup value. The ratio of differential operating current to the average restraining current is a fixed percentage, which explains the name of the relay.

The advantage of this relay is that it is less likely to operate incorrectly for faults that occur external to the protected soul. The operating characteristic of the percentage restraint differential relay with a slope k one is shown here. The operating zone is indicated in green and the restraining zone is indicated in red restraint differential relays can have different restraint slopes, depending on the relay you choose and some relays actually have dial in features that you can control the percentage or the slope of The operating characteristic of the relay some percentage differential relays may have a dual slope characteristics as shown in this slide. The dual slope percentage restraint characteristic improves the security of the current differential really for faults external to the protected zone. This is a particular advantage because current transformers may not accurately reproduce the primary fault currents under trancing fault conditions. The dual slope restraint characteristic is a form of adaptive restraint in which the magnitudes of the restraint quantity is increased for high current conditions where CT accuracy is worse and CT saturation becomes more probable.

The minimum pickup region is used between zero and approximately point five per unit Current, it provides security against CT remnants and accuracy errors and is usually set between point three and point five per unit. The slope one region is used between the minimum pickup region and the next slope breakpoint slope one provide provide security against false tripping due to see key accuracy. Class ccps accuracy is about plus or minus 10%. Therefore, 20% should be the absolute minimum setting with a greater than 30% preferred. For on load tap changer applications another plus or minus 10% is added. slope to provide security against pulse tripping during through fault events where CT saturation is likely in this area, the signet a signal nificant DC current component will be present and therefore, saturation is likely slope to is normally set at 60 to 80%.

The differential protection method implies that for normal loads or through faults the vector summation of all currents entering and leaving the protected zone must be equal. Therefore, there will be no current flow in the operating coil of the protection Relay for loads or external fault faults, but there will be relay current proportional to fault current for internal faults. As I've already mentioned, a common problem that exists for differential bus or and or transformer protection is CT saturation during external fault conditions, as we have already seen one or more CTS in the differential scheme can become saturated in this case The saturated CT then acts as a low impedance short circuit. The saturated CT will not perform correctly in order to transform the effect of fault current. The other CTS that do not saturate will output secondary fault current accurately. As a result the differential relay receives some current that could trip incorrectly.

In the case of high voltage buses and transformers, the system cannot tolerate to protection time required to operate the low voltage bus protection. High Voltage internal buses and Transformers must be cleared of faults instantaneously with no time delay. If we could increase the impedance of the relay path substantially above that of the saturated CT two, we could then prevent the relay from operating with an extra Resistance added in series with the coil. The current output of CQ one will now be forced through the low impedance of CP two. Therefore no or very little current will pass through the relay coil and the relay will not operate. If we now consider an internal fault both CP one and CP two transform the fault current correctly as the protection will operate fast enough to prevent any CP saturation.

Both I won and it will add and flow through the resistor and relay combination to correctly produce an instantaneous trip. This modification to simple differential protection is known as high impedance differential protection. With high voltage buses and Transformers The Fault currents are normally very large With the addition of external resistance in the relay path the voltage developed across our relaying and the CT secondaries will therefore be very high. This high voltage could in turn damage the secondary winding of the CT circuit and the overcurrent relay. If we installed a non linear resistor in parallel with the external resistor relay combination, we can limit the voltage to a very safe level. This type of nonlinear resistor used is called a metro cell.

It will limit the voltage to approximately 300 volts for internal faults. At low voltage the metro cell will behave like an open circuit and for voltages greater than 300 volts it will begin to conduct and shunt some of the relay current. As a result the differential junction voltage is limited to a safe path. And still allows the relay to operate correctly. This is what an installed Metro cell looks like. It will limit the voltage to approximately 300 volts for internal faults.

Let's consider a three phase transformer protected by differential relay. This is a typical step down power transformer carrying normal load power flow is from the 115 kV system to the 12 kV system. This 30 MVA power transformer has a Wye connected primary winding with a Delta connected secondary winding. As a result, we will obtain a 30 degree phase shift across the transformer that is the secondary currents and voltage will lag the primary by 30 degrees. To account for the phase shift the protection CTS on 115 kV y side of the power transformer are connected in delta, while the CTS on the 12 kV Delta side are connected in y. Therefore, the current entering the relay restraint coils from 115 kV side will be equal and in phase with the current leaving the relay on the 12 kV side, disregarding the transformer ratios for the moment.

Let's have a look at that and see how that works. Let's follow the currents for convenience and let's assume that the CTS match the transformer ratios. The red face current enters the h1 bushing of the transformer and we'll call that the red phase lag of the transformer the white phase current enters the h2 bushing of the transformer white phase leg and the blue phase current enters the H three bushing of the transformer these three currents set up in phase currents in the secondaries of the transformer like so. So, the in phase currents in the secondary windings are in phase with the primary turns of the transformer the secondary are phase is made up of the sum of the red phase or the current flowing out of the red phase leg of the transformer minus the blue phase. We will call that I subscript r minus B and we will call the current flowing in the secondary of The CT I subscript r minus b, lowercase in as the secondary side of the CTS.

Similarly, the secondary white phase is made up of the sum of the white face current minus the red face current. And it sets up a similar current, it's a secondary i w minus R. The secondary blue phase is made up of the sum of the blue phase current minus await phase current on the secondary. And that sets up a secondary current flowing in that CT i b minus w moving Back to the high side we see that the AR and BCP secondaries are summed and that means the current flowing in to the read phase restring coils will be made up of i r minus V. The current flowing into the white phase differential relay restraint coils will be made up of the i w minus our phases. And similarly, the current flowing into the blue face differential relay will be made up of ay b minus W. These three currents flow through the restraint coils of the relay and match the light flow In the secondaries of the low voltage CPS as a result, we should have zero current current flowing through the relay operating coils.

Let's consider the phasor diagrams for this section secondary red to white voltage will be in phase with a primary red to neutral voltage as well as secondary white to blue voltage is in phase with a primary white to neutral voltage and the secondary blue to red voltage will be in phase with the primary blue to neutral voltage. For simplicity, let's assume a unity power factor load corresponding currents are in phase with the voltages. Therefore, the red phase current is drawn in phase with the primary red to neutral voltage and the white to blue currents are showing in phase with the way to neutral and the blue to neutral voltage respectively. So, what we are looking at here are the current phasers of the transformer. Moving the drawing out of the way for now but let's keep it visible at least. And let's move the Phasers to give us more room for plotting the results.

The secondary currents of the CTS on the high voltage side of the transformer are flowing through the read phase differential rally and they are made up of the sum of the red minus the blue phase currents from the CTS. This is matched up with the red minus the blue phase and the lo fi low voltage currents flowing through the CD secondaries. There The secondary currents flowing through the white phase differential relay is made up of the sum of the primary white minus a red face secondary CT currents. And this is matched up with the white minus red face of the low voltage currents. The secondary currents flowing through the blue phase differential relay is made up of the sum of the blue minus the white phase currents from the CTS and this is matched up with blue minus white in the low voltage currents that are flowing in the secondaries of the CTS there.

Now, for clarity, we have removed all of the phasers, accept the results by rearranging the phasers it can clearly be seen that the currents on both sides of the relay match for normal operating conditions or for faults outside the protected zone. So far we have disregarded transformer ratios. Therefore, let's look at ratio matching in order to calculate the CT ratios that are required. We know the transformer rating is 30,000 kV A, which is given by this formula where the KV a or the VA rating is equal to three times the line to neutral voltage times aligned current. We don't know the line to neutral voltage, but we do know the line to line voltage. So we will make the conversion in our formula because we know the relationship between the line to neutral voltage and the line to line current.

And that gives us the VA is equal to root three times the line the line times the line Current rearranging the formula in order to solve for what the line current would be at the rate of K VA gives us a KV a all over root three times the line the line voltage. Now we have to be careful here of the KV multiplier, but since we're using kV a in the numerator, we can use kV in the denominator and the thousand multiplier takes care of itself. So on the 115 kV side, the line current would be 30,000 kV a all over root three times 115 kV, which means the primary amps flowing in the Transformers 151 amps if we choose practice Marie CTS that is to place three 300 to five amp CTS one on each phase of the primary. And in fact, the these ratios have already been chosen for us so that we know that the primary CT ratios are 300 to five.

So then the current flowing in the secondaries of each one of those CTS is the primary amps 151 divided by 300 divided by five, which gives us 2.52 amps flowing in the secondaries of each one of the CTS. We now know the current that would be flowing in each of the primary CT secondaries but the currents flowing into the relay is the sum of two CP secondary currents. So Let's start with looking at the blue phase differential relay. The relay sees from the hundred kv 115 kV side CTS, the blue phase current minus the white phase current, which is the blue phase current added to minus the white phase current. And each of these currents we know is 2.52 amps. This would give us the or the resultant will label as I subscript b minus w which stands for the blue phase minus the white phase current.

Looking at the train Oh, we've seen this ratio before. So, we know the resultant I subscript b minus W is going to be equal to root three times the white face current therefore, the reason We'll see root three times 2.52 amps which is equal to 4.36 amps. We already know that the secondaries of the primary and secondary matching phase so now we have to make sure that the current magnitudes are equal in order to net zero flowing through the relays operating coil on the 12 kV side using the same formulas but this time the line to line voltage is 12 kV. Therefore, the secondary current would be given by 30,000 kV a over three times 12 kV, which is equal to 1443 amps that are flowing in the secondary of the power transformer lines under normal or rated load conditions. We have on the secondary side the power transformer bushing CTS, and there's one for each face.

So there's three of them with the multiple tap ratios of 2000 1600 1200 to five. So, if we wanted to calculate what the exact CT ratio should be in order to match up with the primary current flowing in the into the relay, it would be the the secondary amps that are flowing in the secondaries, the power transformer divided by those really amps which happens to be 4.36. And if we divided 1443 amps by 4.36 amps, that would tell us we'd like to have a CT ratio of 330.96 One, or 1655 to five. However, we don't have that luxury of exact ratios of the current transformer, we've been dealt with ratio that's the closest one is 1600 to five. So that's the one that we would pick. And that would give us our best fit the relay settings that would then be calculated and calibrated in order to give us enough leeway, that the 1600 to five tap ratio would work.

So this ends chapter six