Motor and to study the different testing methods to arrive at their performance. Stephen J. Ltd, II, Pearson, Nagrath, I. It is required that most of the flux is to be confined to a definite, high-permeability path linking the windings.
Now with air ore it is not possible but with core of iron or other ferromagnetic material coupling is effectively done. Because most of the flux in iron core is confined to a definite, high-permeability path linking the windings. As the iron core is also under the flux variation in transformer there is some voltage induced in the iron core. This voltage is called the eddy voltage and in result there is a current named eddy current flow in the iron core.
These results in heating up the core. With a solid iron core the eddy current is high. To avoid this solid iron core is not used. Thin iron core is laminated to make it non-conductive. Then this thin laminated core is stacked by several to get the complete iron core structure.
With this modification eddy current is reduced but magnetic property of the iron core is remained unchanged. Complete basics and theory of Electrical Transformer.
This site uses Akismet to reduce spam. Learn how your comment data is processed. Pin Share 5. Since typical values of magnetizing reactance are several orders of mag- nitude greater than typical values of leakage reactance, and since no-load losses are usually much smaller than load losses when the transformer is car- rying rated load, the elements G m and X m can be neglected in most calculations.
If voltages and currents are expressed in per unit instead of using actual system voltages and currents, the ideal transformer in the model will have a turns ratio and can thus be eliminated from the equivalent circuit. Since R p , R s , X p , and X s are now all directly in series with each other, the resistance elements can be combined into a single resistance R and the reactance elements can be combined into a single reactance X.
The equivalent circuits of transformers modeled in zero phase sequence networks are different from those in positive and negative phase sequence networks.
The zero-sequence equivalent circuit depends on the type of wind- ing connections used; these equivalent circuits will be developed a little later in this chapter. Before continuing with transformer equivalent circuits, it is necessary at this point to digress with a brief tutorial on symmetrical components. Therefore, a brief tutorial on symmetrical components is presented here that will provide an understanding of positive, negative, and zero phase sequence voltages and currents and how the concept of symmetrical components is applied to transformers.
When three-phase circuits are balanced, calculations of voltages and cur- rents are simple and straightforward. In these instances, the actual circuit and its positive phase sequence network are one and the same thing. The positive phase sequence network is nothing more than a network where the phase rota- tion is positive normally A-B-C and the phases are balanced; i.
Although the ideal set of balanced conditions may exist approximately under normal conditions, this is rarely the case for design-limiting conditions such as system faults. When voltages, currents, or impedances are unbalanced, the solutions become much more complex, especially when there is mutual induction between various parts of the circuit, as in the case of three-legged core form transformers with the three phases coupled magnetically through the core.
There have been several methods for attacking three-phase circuit prob- lems for unbalanced conditions, but the most successful method in use today is called symmetrical components. In this method, an unbalanced circuit is represented by a combination of three balanced circuits, called sequence net- works. The positive phase sequence network has already been described above. In the negative phase sequence network, the voltages and currents of all three phases are balanced but the phase rotation is opposite to the positive phase sequence; i.
It is usually assumed that there are no negative or zero phase sequence voltage sources. In the zero-phase sequence network, the voltages and currents of all three phases are identical and are all in phase with each other; i.
Since the phase currents are in phase, these add together at the neutral. Therefore, zero phase sequence networks always involve neutral or ground currents. Zero-sequence currents exist only if there are physical ground current paths in the circuit. A set of unbalanced phase currents, I a , I b , and I c , may be transformed into a set of symmetrical components I 0 , I 1 , and I 2 , where I 0 is the zero phase sequence current, I 1 is the positive phase sequence current, and I 2 is the nega- tive phase sequence current.
There is also a set of sequence currents that is referenced to B phase, and a set of sequence currents referenced to C phase as well. The symmetrical component method does not require explicitly solving all three sets of sequence currents, because these sets of sequence currents are identical to each other except that they are shifted in phase. There- fore, it is implicitly understood that the sequence currents I 0 , I 1 , and I 2 are always referenced to A phase.
So far, it may seem that all this has accomplished is additional work; however, the additional work usually makes calculations involving unbalanced circuits much easier than dealing with phase quantities alone. We will see a direct application of this method at the end of this chapter. The following steps summarize the technique of symmetrical compo- nents: 1. Transform the sequence voltages and currents at the point of the fault calculated in step 2 back into phase voltages and currents using Eqs.
The corresponding sequence network con- nections for these unbalanced conditions are shown in Figure 3. We shall soon see that in zero sequence networks, the transformer equivalent circuits can be very different depending on the connections and type of core construction.
This section will show how to develop the zero phase sequence equivalent circuits for the vari- ous types of transformers. The basic equivalent circuit consists of series impedance elements be- tween the primary and secondary windings, and a shunt impedance.
The shunt impedance is approximately equal to the open-circuit impedance Z oc measured at the primary windings with the secondary windings open-circuited, or mea- sured at the secondary windings with the primary windings open-circuited.
If per-unit values are used, Z oc is approximately the same whether it is measured at the primary or secondary winding. The basic zero phase sequence trans- former equivalent circuit is shown in Figure 3. As we will soon see, the value of the shunt impedance is radically differ- ent for different types of core construction and winding connections.
In most cases G m is practically zero and the magnitude of X m is much greater than the magnitudes of R p and X p. The arrows in Figure 3. A similar result applies to the open-circuit impedance of a three-phase, three-legged core form transformer with the winding Y connected. A three-phase, three-legged core form transformer is shown in Figure 3.
Zero phase sequence voltages excite both of these transformers. This same effect exists for three-phase shell form transformers when excited by zero-sequence voltages. This could easily drive the core into saturation. The exact value of Z oc is uncertain at best in these cases. The series transformer impedance for zero phase sequence networks also depends on how the transformer is connected.
The diagrams in Figure 3. As previously discussed, the mag- netizing impedance Z M varies depending on the core construction. This situation can occur when a Grd. Y- Grd. Y transformer is energized by a three-phase source with an open phase.
Example 3. It is desired that the types of core con- struction of these transformers be determined prior to putting them into ser- vice. Using a single-phase VAC 60 Hz source and an ammeter, devise a test to determine which types of core construction these transformers have. The neutral bushing H0 was connected to the grounded neutral leg of the VAC source.
The low-voltage bushings were open-circuited. Based on the measured currents, all the transformers have zero-sequence open-circuit currents less than mA with the exception of transformer 4, which had a zero-sequence open-circuit current of 1.
The measured open-circuit current is 1. It is easily seen in this example how the zero- sequence equivalent circuits are affected by the core design. This impedance has a very low power factor, consisting almost entirely of leakage reactance with only a small resistance. As discussed earlier, the transformer design engineer can control the leakage reactance by varying the spacing be- tween the windings. While leakage reactance can be considered a transformer loss because it consumes reactive power, some leakage reactance is necessary to limit fault currents.
On the other hand, excessive leakage reactance can cause problems with regulation. This requires the primary voltage to be greater than the rated primary voltage at full load.
Let E p equal the primary voltage and let E s equal the secondary voltage when the transformer is fully loaded. Using per-unit values instead of primary and secondary voltage values, the per-unit secondary voltage will equal E p with the load removed. For a near-unity power factor, the regulation is much smaller than the regulation for an inductive load with a small lagging power factor. The three-phase fault is a balanced fault, so the positive-sequence equiv- alent circuit applies.
The turns ratios of all of the transformers must be nearly equal. The phase angle displacements of all of the transformers must be identical. The primaries of the two transformers are connected to a 66 kV transmission line through a single air break switch. This switch is designed to interrupt magnetizing current only, which is less than 1 A. The transformers were being removed from service and the second- ary loads had been removed.
A switchman then started to open the air break switch, expecting to see a small arc as the magnetizing current was interrupted. Something was obviously wrong. Because the turns ratios were unequal, a circulating current was set up even without any secondary load. The open- circuit secondary voltage difference, assuming 66 kV at the transformer pri- maries, is calculated below.
But since the turns ratios are not equal, I c does not get transformed into equal and opposite currents at the primaries. The resulting current exceeded the interrupting rating of the switch, causing it to fail.
The conditions described in this example are diagramed in Figure 3. If single-phase transformers are connected in a Y-Y bank with an iso- lated neutral, then the magnetizing impedances should also be equal on an ohmic basis. Otherwise, the transformer having the largest magnetizing imped- ance will have a highest percentage of exciting voltage, increasing the core losses of that transformer and possibly driving its core into saturation.
The general case where all three transform- ers have different impedances is derived below. Note: In general, these ratios are complex numbers because the impedances are com- plex numbers. The currents i a , i b , and i c are multiplied by 1. The graph in Figure 3. The temperature rise of the winding is caused by all of the transformer losses that were previously discussed in this chapter.
There- fore, the winding temperature is a function of load losses and no-load losses. Thermal capability is the KVA load applied to the output of a transformer that causes the hottest area in the windings, called the winding hot spot, to reach some limiting temperature. The hot-spot tempera- ture determines the rate of loss of life of the transformer as a whole, which is a cumulative effect. Therefore, the hot-spot temperature limit is usually based on a loss-of-life criterion. The standard method for calculating the ther- mal capability of oil-insulated transformers will be covered fully at the end of this chapter.
When one purchase order is used to procure a number of identical transformers from the same manufacturer, some testing is performed on each transformer while certain other tests are performed only on one representative unit in that pur- chase order. There is a recommended format of a transformer test report [1]. Some transformer manufacturers follow the recommended format exactly while others prefer to use their own format, but the information contained in the test reports is standard throughout the industry.
A sample test report is shown in Figure 3. It was based on an actual factory test report for an order of four transformers used in a generating station. In the upper portion of the test report is the general design information, such as the cooling class, number of phases, frequency rating, rated temperature rise, voltage ratings of the windings, and the KVA rating.
The transformers shown in Figure 3. This means that heat is transferred from the core and coils to the insulating oil, which is forced through the transformer by oil pumps. The heat is then transferred from the insulating oil to the ambient air through a set of radiators which have air forced through them by fans. The voltage rating of the HV winding is Grd.
The full winding rating from phase to ground is , V. Additional HV winding taps are available at ,, ,, ,, and , V. The full winding rating is the same as the phase-to-phase system voltage, 23, V. The methods used in performing factory tests will be covered in more detail in Chapter 8. Since typical magnetizing current power factors are extremely small, core loss measurements require carefully calibrated test equipment.
At the bottom of the test report are the results of a heat run test. The heat run test applies a load equal to the nameplate KVA rating of the transformer at full voltage using an inductive load. Using an inductive load reduces the en- ergy use during this test.
It is an expensive and time-consuming test, so it is usually only performed on one of the transformers on a purchase order. In the case of this sample test report in Figure 3. Note from the test report, however, that the HV winding only attained a Therefore, the as-built transformer has a 5.
It is this increased loading, and not the nameplate rating, that is the true thermal capability of that transformer. The average winding temperatures are not di- rectly measured in this test. Instead, the winding temperatures are inferred from conductor resistance measurements taken immediately after the test is shut down. The last two entries on the test report are the hot spot temperature rise of the HV and LV windings.
These values are not actually measured by any test, but are calculated based on the average winding temperature and the oil temperature.
There is only about a 6. This is fairly typical of transformers in the FOA cooling class. Since cooling oil is forced through the windings by pumps, the oil velocity is quite high.
For transformers without oil pumps, such as the self-cooled OA class or forced-air cooled only FA class , the oil velocities are much slower and there is a much greater temperature gradient between the hot spot temperature and the average winding temperature.
This method can be applied to trans- formers of any MVA size. A generally accepted value for m is 0. Most engineers consider the winding time constant to be very short, and a value of zero is used by some.
Although the IEEE method seems rather complicated, it is straightforward enough with the use of a desktop computer. The IEEE standard includes a computer program in Basic code that performs the temperature calculations. Satisfactory results can be obtained with just a few step changes over the course of 24 h, but better accuracy will be achieved if the load cycle is broken into 1 h intervals.
The relation of insulation aging to time and temperature fol- lows the well-known Arrhenius chemical reaction rate model. The adaptation used in the IEEE standard [3] has the following form.
Normal life for most transformers is considered to be around 30 to 40 years. The constants A and B depend on the types of material used to insulate the windings. To calculate the equivalent aging of the transformer F EQA with a varying hot-spot temperature such as occurs for a cycling load or a seasonal ambient temperature, F AA is integrated over time and the integral is divided by the total time to obtain the average.
The per unit life and F AA are plotted vs. It should be stressed that most transformer failures are random events that occur for various reasons besides insulation loss of life. Therefore, the formula for per unit life cannot be used as a predictive model to determine when a given transformer will ultimately fail.
However, it is indeed certain that overloading a transformer will shorten its life, so it is a good practice from FIGURE 3. From the test report, it is clear that the HV winding has the higher tem- perature and is the limiting winding.
To calculate the thermal capability pre- cisely, a trial-and-error approach is used where the MVA loading is varied and the temperatures are calculated until the desired per unit life is achieved. The Y-connected generator is grounded through a resis- tance. A bolted phase-to-ground fault occurs at one of the HV terminals of the transformer. The three-phase diagram of the actual circuit is shown in Figure 3.
In this example, we will use the generator ratings for the common MVA base. Let us assume that the series winding impedance of the transformer bank is purely reactive all leakage reactance.
Convert the The positive phase sequence, negative phase sequence and zero phase sequence networks are connected in series for a phase-to-ground fault on A phase, as shown in FIGURE 3. Calculating the HV sequence currents, 1. The two-phase load current is 1 per unit in each secondary winding. Calculate the positive, negative, and zero phase se- quence currents in the three-phase circuit in per unit. The T connection uses a transformer between A phase and neutral to supply one-half of the two-phase load and a 1.
The ammeter reads 0. Calculate the open-circuit zero phase sequence impedance of this transformer, in per unit of the transformer rating. Calculate the per-unit load in each transformer on its own base. Using the method described in Section 3. Additionally, the current in transformer C must be divided by 1. What are the positive-, negative-, and zero-sequence voltages at the generator? The primary of the transformer is modeled as three ideal transformers with series impedances of j0.
The voltages across the windings are E a , E b , and E c. Y bank, so at the generator these voltages are: Zero sequence 0 Positive sequence 1. National Electrical Manufacturers Association. Transformers, Regulators and Re- actors.
TR , Part 7, pp. It is used quite extensively in bulk power transmission systems because of its ability to multiply the effective KVA ca- pacity of a transformer.
Autotransformers are also used on radial distribution feeder circuits as voltage regulators. The connection is shown in Figure 4.
In two-winding transformers, the primary voltage is associated with the primary winding, the secondary voltage is associated with the secondary winding, and the primary voltage is normally considered to be greater than the secondary voltage.
The output terminals operate at a higher voltage than the input terminals. The other possible connection for an autotransformer is shown in Figure 4. The autotransformer shown in Figure 4.
The key feature of an autotransformer is that the KVA throughput of the transformer, i. The output terminals operate at a lower voltage than the input terminals. The volts per turn in the common winding equal the volts per turn in the series winding.
The common winding voltage divided by the series winding voltage is equal to the number of turns in the com- mon winding divided by the number of turns in the series winding. The sum of the ampere-turns of the common winding plus the am- pere-turns of the series winding equal the magnetizing ampere- turns. The magnetizing ampere-turn are practically zero, so the mag- nitude of the ampere-turns in the common winding is approximately equal to magnitude of the ampere-turns in the series winding.
The series winding current divided by the common winding current is equal to the number of turns in the common winding divided by the number of turns in the series winding. This is illustrated in the following example. Example 4. The current low-voltage input current is A. Refer to Figure 4. This is a consider- able multiplication of KVA capacity.
The capacity multiplication factor F c is a function of the ratio of the output voltage to the input voltage r. F c for a boosting autotransformer is always greater than 1, and it is always greater than F c for a bucking autotransformer for a given value of r. The chart shown in Figure 4. The heavy curve is for a boosting autotransformer and the lighter curve is for a bucking autotransformer.
A boosting autotransformer begins to lose its advantage over a two-winding transformer as the voltage ratio increases. Consider a two-winding transformer having a turns ratio n connected as a boosting autotransformer. The transformer impedance, which is mainly leak- age reactance, is split between the common winding and the series winding as Z c and Z s , respectively.
The magnetizing impedance is neglected. To deter- mine the impedance of the autotransformer, the secondary low-voltage termi- nals are short-circuited and a voltage source E p is applied to the primary high- voltage terminals, as shown in Figure 4. This is obtained by dividing both sides of Eq.
For a large ratio n this impedance is much smaller than the series impedance seen at the primary side. In general, the series impedance of a small two-winding transformer is greater than the series impedance of a large two-winding transformer if the two impedances expressed as percent of the winding KVA bases are equal. However, when a small transformer is connected as an autotransformer the series impedance seen at the transformer terminals can be substantially smaller than the impedance of a much larger equivalent two-winding transformer.
Recall from Eq. Recall- ing the capacity multiplication factor of a boosting autotransformer in Eq. In Section 4. What is the impedance of the autotransformer ex- pressed as a percent of the KVA capacity of the autotransformer? In Example 4. This ratio is calculated below.
In other words, the This example clearly illustrates the tremendous savings in economy that can be realized by using the autotransformer connection. The transformer is connected as a single-phase boosting auto- transformer in a single-phase radial distribution circuit.
The input to the autotransformer is V single phase. Converting the impedance to the autotransformer KVA capacity base, r 1. For this rea- son, this autotransformer connection is seldom, if ever, used. The Grd. Y connection using autotransformers is indistinguishable from a Grd. Y connection using two-winding transformers. This is the most common connection using autotransformers. The phase shift is a function of the ratio of the primary to secondary voltages and it can be calculated from the vector diagram, such as the example shown in Figure 4.
The situation is very different when a short circuit is applied to an auto- transformer, as can be seen from the circuit diagram shown in Figure 4. This almost guarantees that the core will go into saturation, with two immediate results. First, the admittance of the magnetizing branch of the transformer can no longer be ignored, and the admittance of the magnetizing branch is very nonlinear with the core in saturation; therefore, short-circuit current calculations are some- what uncertain for autotransformers.
Second, with the core in saturation, the transformer essentially becomes an air-core transformer, further limiting the ability to calculate impedances. Fortunately, this also limits the magnitude of the induced voltage across the common winding.
So although the voltage across the series winding may be several times the normal operating voltage of this winding, the overvoltage across the common winding is not quite as severe. This is because the relatively large line-to-ground capacitance of the external second- ary circuit prevents the voltage to build up at the secondary terminals before the voltage surge has reached its peak at the primary terminals.
Because the turns ratio between the common and series winding is usually large, a very high voltage is also induced in the common winding.
The ASA Test Code, which was used in the past, recognized this particular disadvantage of auto- transformers and required full impulse test voltages to be applied across each winding in turn see Ref. One way that this problem can be addressed is to install surge protection in autotransformers. Thyrite blocks are connected between the windings and ground and across the series winding to limit applied and induced impulse voltages. Thyrite blocks have a nonlinear resistance characteristic with the current through the thyrite increasing sharply when the voltage exceeds a cer- tain breakdown voltage.
This tends to limit the voltage across the thyrite, thereby protecting the windings. In a boosting autotransformer, thyrite blocks are typically installed across the common winding, between the high-voltage terminals and ground, and across the series winding. As was seen in Section 4. The problem with using thyrite blocks or other types of surge protec- tion in autotransformers is that the surge protection may break down under short-circuit conditions unless their breakdown voltages are carefully coordi- nated.
This conceptualization is electrically correct but not physically accurate, how- ever. This web site Dos not show how to desing transformers any body need desing for new trans former Mail your volt Ammper in put and out put free send desing derlikonfon yahoo. Hi Lokanandan with the multimeter check the winding resistance. Using a step down transformer of say 3 to 6 volt feed it to the low resistance winding and check the voltage in the high resistance winding the voltage measured gives you the idea of the ratio of the transformer.
I have one transformer in that there is not label to know the primary and secondary ,so how to find the primary and secondary of transformer. In transformer primary and secondary are interchangeable. The winding which is connected to supply side is primary and load side is secondary. Accordingly in two cases you get as step-up or step-down transformer design.
Means less no of turns on primary and more on secondary gives us step-up voltage transformer and vice-versa Voltage per turn is same in both the winding.
Sir, How to calculate the current rating of output from a transformer? I need a Ferrite core transformer, 6V mA. Hi HB with 12volt transformer with volt tap will give around volts due to drop across final transistors. To get volts, volt or V tapping to be used. For charging with volt tapping the secondary will reduce to 11 volt, will not be sufficient for charging.
Your problem once the transformer fabricated with two winding you can use either way step up or step down remember to fix change over for vice versa work. If you want to input 12 volt ac and get out put volt AC that supply to small winding If you want to volt input and 12volt output supply to big winding However your problem normal lead acid battery art fully charging required I see them as different as they have more than two output connections.
Once the transformer fabricated you can use either way as your application transformer named Step —up or Step down Bat carful apply the correct voltage to as you desired winding say if you want input 12 volt and get out put Apply 12 volt twelve volt winding Vice visa the transformer works. Author john. Do you know how RFID wallets work and how to make one yourself? February 14, Manjunath GR 3 years ago.
Relay good information and details study the Transformer. Maxwel Ogot 5 years ago. Pritam Kumar Singh 5 years ago. Very simple explanation. I like this. Bharat 6 years ago. K Chidananda 6 years ago. Gangadhar Ginmav 6 years ago. Raghu 5 years ago. If transformer is Setup then the voltage is high and if it is low then it is stepdown.
I am waiting for your reply please reply as soon as possible thank you. Shinde 5 years ago. Nicky 7 years ago. Rai Tanveer Khan 7 years ago. I like it. Seetharaman 7 years ago. Mani 7 years ago. Thanks sir ur information gud exaplation thanks,thanks. Siraj soomro 7 years ago.
Allan carter kyazze 8 years ago. Bhavithralakshmanasamy 8 years ago. Nice , also we need clear explaination of losses. Gbenga 8 years ago. Enquiry 8 years ago. Its so easy to understand. Tittu Thomas 9 years ago. Your Tutorials are so Easy to understand…..
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