Unit 2: Transformers
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Topics Covered:
Transformers
Introduction to polyphase induction machines
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Transformers
Electrical transformer is a static electrical machine which transforms electrical power from one circuit to another circuit, without changing the frequency. Transformer can increase or decrease the voltage with corresponding decrease or increase in current.
Working Principle
The basic principle behind working of a transformer is the phenomenon of mutual induction between two windings linked by common magnetic flux. The figure at right shows the simplest form of a transformer. Basically a transformer consists of two inductive coils; primary winding and secondary winding. The coils are electrically separated but magnetically linked to each other. When, primary winding is connected to a source of alternating voltage, alternating magnetic flux is produced around the winding. The core provides magnetic path for the flux, to get linked with the secondary winding. Most of the flux gets linked with the secondary winding which is called as 'useful flux' or main 'flux', and the flux which does not get linked with secondary winding is called as 'leakage flux'. As the flux produced is alternating (the direction of it is continuously changing), EMF gets induced in the secondary winding according to Faraday's law of electromagnetic induction. This emf is called 'mutually induced emf', and the frequency of mutually induced emf is same as that of supplied emf. If the secondary winding is closed circuit, then mutually induced current flows through it, and hence the electrical energy is transferred from one circuit (primary) to another circuit (secondary).
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Basically a transformer consists of two inductive windings and a laminated steel core. The coils are insulated from each other as well as from the steel core. A transformer may also consist of a container for winding and core assembly (called as tank), suitable bushings to take our the terminals, oil conservator to provide oil in the transformer tank for cooling purposes etc. The figure at left illustrates the basic construction of a transformer.
In all types of transformers, core is constructed by assembling (stacking) laminated sheets of steel, with minimum air-gap between them (to achieve continuous magnetic path). The steel used is having high silicon content and sometimes heat treated, to provide high permeability and low hysteresis loss. Laminated sheets of steel are used to reduce eddy current loss. The sheets are cut in the shape as E,I and L. To avoid high reluctance at joints, laminations are stacked by alternating the sides of joint. That is, if joints of first sheet assembly are at front face, the joints of following assemble are kept at back face.
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Types Of Transformers
Transformers can be classified on different basis, like types of construction, types of cooling etc.
(A) On the basis of construction, transformers can be classified into two types as; (i) Core type transformer and (ii) Shell type transformer, which are described below.
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(I) Core Type Transformer
In core type transformer, windings are cylindrical former wound, mounted on the core limbs as shown in the figure above. The cylindrical coils have different layers and each layer is insulated from each other. Materials like paper, cloth or mica can be used for insulation. Low voltage windings are placed nearer to the core, as they are easier to insulate.
(Ii) Shell Type Transformer
The coils are former wound and mounted in layers stacked with insulation between them. A shell type transformer may have simple rectangular form (as shown in above fig), or it may have a distributed form.
(B) On the basis of their purpose
(C) On the basis of type of supply
(D) On the basis of their use
(E) On the basis of cooling employed
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Ideal & Practical Transformers
Ideal Transformer
An ideal transformer is one that has
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Although an ideal transformer cannot be physically realized, yet its study provides a very powerful tool in the analysis of a practical transformer. In fact, practical transformers have properties that approach very close to an ideal transformer.
Consider an ideal transformer on no load i.e., the secondary is open-circuited as shown in the figure. Under such conditions, the primary is simply a coil of pure inductance.
When an alternating voltage V₁ is applied to the primary, it draws a small magnetizing current Iₘ which lags behind the applied voltage by 90°. This alternating current Iₘ produces an alternating flux ϕ which is proportional to and in phase with it.
The alternating flux ϕ links both the windings and induces e.m.f. E₁ in the primary and e.m.f. E₂ in the secondary. The primary e.m.f. E₁ is, at every instant, equal to and in opposition to V₁ (Lenz’s law). Both e.m.f.s E₁, and E₂ lag behind flux ϕ by 90°. However, their magnitudes depend upon the number of primary and secondary turns.
Phasor Diagram of Ideal Transformer
The phasor diagram of an ideal transformer on no load is also shown above. Since flux ϕ is common to both the windings, it has been taken as the reference phasor.
The primary e.m.f. E₁ and secondary e.m.f. E₂ lag behind the flux ϕ by 90°.
Note that E₁ and E₂ are in phase. But E₁ is equal to V₁ and 180° out of phase with it.
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Practical Transformer
A practical transformer differs from the ideal transformer in many respects. The practical transformer has,
1. Iron Losses
Since the iron core is subjected to alternating flux, there occurs eddy current and hysteresis loss in it. These two losses together are known as iron losses or core losses.
The iron losses depend upon the supply frequency, the maximum flux density in the core, volume of the core, etc.
It may be noted that the magnitude of iron losses is quite small in a practical transformer.
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2. Winding resistances
Since the windings consist of copper conductors, it immediately follows that both primary and secondary will have winding resistance. The primary resistance R₁ and secondary resistance R₂ act in series with the respective windings as shown in the figure.
When current flows through the windings, there will be power loss as well as a loss in voltage due to IR drop. This will affect the power factor and E₁ will be less than V₁ while V₂ will be less than E₂.
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3. Leakage reactances
Both primary and secondary currents produce flux. The flux ϕ which links both the windings is the useful flux and is called mutual flux.
However, the primary current would produce some flux ϕ which would not link the secondary winding. Similarly, the secondary current would produce some flux ϕ that would not link the primary winding.
The flux such as ϕ₁ or ϕ₂ which links only one winding is called leakage flux. The leakage flux paths are mainly through the air. The effect of these leakage fluxes would be the same as though inductive reactance were connected in series with each winding of the transformer that had no leakage flux as shown in the figure.
In other words, the effect of primary leakage flux ϕ₁ is to introduce an inductive reactance X₁ in series with the primary winding as shown. Similarly, the secondary leakage flux ϕ₂ introduces an inductive reactance X₂ in series with the secondary winding.
There will be no power loss due to leakage reactance. However, the presence of leakage reactance in the windings changes the power factor as well as there is voltage loss due to IX drop.
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Note.
Although leakage flux in a transformer is quite small (about 5% of ϕ) compared to the mutual flux ϕ, yet it cannot be ignored.
It is because leakage flux paths are through the air of high reluctance and hence require considerable e.m.f. It may be noted that energy is conveyed from the primary winding to the secondary winding by mutual flux f which links both the windings.
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Equivalent Circuit of Transformer
Equivalent impedance of transformer is essential to be calculated because the electrical power transformer is an electrical power system equipment for estimating different parameters of the electrical power system which may be required to calculate the total internal impedance of an electrical power transformer, viewing from primary side or secondary side as per requirement.
This calculation requires equivalent circuit of transformer referred to the primary or equivalent circuit of transformer referred to secondary sides respectively. Percentage impedance is also a very essential parameter of the transformer.
Special attention is to be given to this parameter during installing a transformer in an existing electrical power system. Percentage impedance of different power transformers should be properly matched during parallel operation of power transformers. The percentage impedance can be derived from the equivalent impedance of the transformer so, it can be said that the equivalent circuit of the transformer is also required during the calculation of the % impedance.
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Equivalent Circuit of Transformer Referred to Primary
For drawing equivalent circuit of transformer referred to primary, first we have to establish general equivalent circuit of transformer then, we will modify it for referring from primary side. For doing this, first we need to recall the complete vector diagram of a transformer which is shown in the figure below.
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From the vector diagram in the previous slide, it is found that the total primary current I1 has two components, one is no – load component Io and the other is load component I2′. As this primary current has two components or branches, so there must be a parallel path with primary winding of transformer.
This parallel path of current is known as excitation branch of equivalent circuit of transformer. The resistive and reactive branches of the excitation circuit can be represented as
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The equivalent circuit is shown below:
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Since Io is very small compared to I1, it is less than 5% of full load primary current, Io changes the voltage drop insignificantly. Hence, it is good approximation to ignore the excitation circuit in approximate equivalent circuit of transformer. The winding resistance and reactance being in series can now be combined into equivalent resistance and reactance of transformer, referred to any particular side. In this case it is side 1 or primary side.
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Voltage Regulation
What is Voltage Regulation?
Voltage regulation is a measure of change in the voltage magnitude between the sending and receiving end of a component. It is commonly used in power engineering to describe the percentage voltage difference between no load and full load voltages distribution lines, transmission lines, and transformers.
Explanation of Voltage Regulation of Transformer
Say an electrical power transformer is open circuited, meaning that the load is not connected to the secondary terminals. In this situation, the secondary terminal voltage of the transformer will be its secondary induced emf E2.
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Whenever a full load is connected to the secondary terminals of the transformer, rated current I2 flows through the secondary circuit and voltage drop comes into picture. At this situation, primary winding will also draw equivalent full load current from source. The voltage drop in the secondary is I2Z2 where Z2 is the secondary impedance of transformer.
Now if at this loading condition, any one measures the voltage between secondary terminals, he or she will get voltage V2 across load terminals which is obviously less than no load secondary voltage E2 and this is because of I2Z2 voltage drop in the transformer.
Expression of Voltage Regulation of Transformer
The equation for the voltage regulation of transformer, represented in percentage, is
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Voltage Regulation of Transformer for Lagging Power Factor
Now we will derive the expression of voltage regulation in detail. Say lagging power factor of the load is cosθ2, that means angle between secondary current and voltage is θ2.
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Here, from the diagram in the previous slide:
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Angle between OC and OD may be very small, so it can be neglected and OD is considered nearly equal to OC i.e.
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Voltage regulation of transformer at lagging power factor,
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Voltage Regulation of Transformer for Leading Power Factor
Let’s derive the expression of voltage regulation with leading current, say leading power factor of the load is cosθ2, that means angle between secondary current and voltage is θ2.
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Here, from the above diagram,
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Angle between OC and OD may be very small, so it can be neglected and OD is considered nearly equal to OC i.e.
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Zero Voltage Regulation of A Transformer
‘Zero voltage regulation’ indicates that there is no difference between its ‘no-load voltage’ and its ‘full-load voltage’. This means that in the voltage regulation equation above, voltage regulation is equal to zero. This is not practical – and is only theoretically possible in the case for an ideal transformer.
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Maximum Efficiency Criterion
The Efficiency of the transformer is defined as the ratio of useful output power to the input power. The input and output power are measured in the same unit. Its unit is either in Watts (W) or KW. Transformer efficiency is denoted by Ƞ.
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Where,
Consider, the x is the fraction of the full load. The efficiency of the transformer regarding x is expressed as
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The copper losses vary according to the fraction of the load.
Maximum Efficiency Condition of a Transformer
The efficiency of the transformer along with the load and the power factor is expressed by the given relation:
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The value of the terminal voltage V2 is approximately constant. Thus, for a given power factor the Transformer efficiency depends upon the load current I2. In equation (1), the numerator is constant and the transformer efficiency will be maximum if the denominator with respect to the variable I2 is equated to zero.
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i.e Copper losses = Iron losses
Thus, the transformer will give the maximum efficiency when their copper loss is equal to the iron loss.
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From equation (2) the value of output current I2 at which the transformer efficiency will be maximum is given as
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If x is the fraction of full load KVA at which the efficiency of the transformer is maximum then,
Copper losses = x2Pc (where Pc is the full load copper losses)
Iron losses = Pi
For maximum efficiency
x2 Pc = Pi
Therefore
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Putting the value of x from the above equation (3) in equation (4) we will get,
Thus, output KVA corresponding to maximum efficiency
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The above equation (5) is the maximum efficiency condition of the transformer.
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Open and Short Circuit Test of Transformer
Open and short circuit tests are performed on a transformer to determine the:
The power required for open circuit tests and short circuit tests on a transformer is equal to the power loss occurring in the transformer.
Open Circuit Test on Transformer
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The connection diagram for open circuit test on transformer is shown in the figure. A voltmeter, wattmeter, and an ammeter are connected in LV side of the transformer as shown. The voltage at rated frequency is applied to that LV side with the help of a variac of variable ratio auto transformer.
The HV side of the transformer is kept open. Now with the help of variac, applied voltage gets slowly increased until the voltmeter gives reading equal to the rated voltage of the LV side. After reaching rated LV side voltage, we record all the three instruments reading (Voltmeter, Ammeter and Wattmeter readings).
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The ammeter reading gives the no load current Ie. As no load current Ie is quite small compared to rated current of the transformer, the voltage drops due to this current that can be taken as negligible.
Since voltmeter reading V1 can be considered equal to the secondary induced voltage of the transformer, wattmeter reading indicates the input power during the test. As the transformer is open circuited, there is no output, hence the input power here consists of core losses in transformer and copper loss in transformer during no load condition. But as said earlier, the no-load current in the transformer is quite small compared to the full load current so, we can neglect the copper loss due to the no-load current. Hence, can take the wattmeter reading as equal to the core losses in the transformer.
Let us consider wattmeter reading is Po.
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Where, Rm is shunt branch resistance of transformer.
If, Zm is shunt branch impedance of transformer.
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These values are referred to the LV side of the transformer due to the tests being conducted on the LV side of transformer.
These values could easily be referred to HV side by multiplying these values with square of transformation ratio.
Therefore it is seen that the open circuit test on transformer is used to determine core losses in transformer and parameters of the shunt branch of the equivalent circuit of the transformer.
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Short Circuit Test on Transformer
The connection diagram for the short circuit test on the transformer is shown in the figure below. A voltmeter, wattmeter, and an ammeter are connected in HV side of the transformer as shown. A low voltage of around 5-10% is applied to that HV side with the help of a variac (i.e. a variable ratio auto transformer). We short-circuit the LV side of the transformer. Now with the help of variac applied voltage is slowly increased until the wattmeter, and an ammeter gives reading equal to the rated current of the HV side.
After reaching the rated current of the HV side, we record all the three instrument readings (Voltmeter, Ammeter and Watt-meter readings). The ammeter reading gives the primary equivalent of full load current IL. As the voltage applied for full load current in a short circuit test on the transformer is quite small compared to the rated primary voltage of the transformer, the core losses in the transformer can be taken as negligible here.
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Let’s say, voltmeter reading is Vsc. The watt-meter reading indicates the input power during the test. As we have short-circuited the transformer, there is no output; hence the input power here consists of copper losses in the transformer. Since the applied voltage Vsc is short circuit voltage in the transformer and hence it is quite small compared to the rated voltage, so, we can neglect the core loss due to the small applied voltage. Hence the wattmeter reading can be taken as equal to copper losses in the transformer. Let us consider wattmeter reading is Psc.
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Where, Re is equivalent resistance of transformer.
If, Ze is equivalent impedance of transformer.
Therefore, if equivalent reactance of transformer is Xe.
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These values are referred to the HV side of the transformer as the test is conducted on the HV side of the transformer. These values could easily be converted to the LV side by dividing these values with the square of transformation ratio.
Hence the short-circuit test of a transformer is used to determine copper losses in the transformer at full load. It is also used to obtain the parameters to approximate the equivalent circuit of a transformer.
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Phasor Diagrams on No Load, Full Load, Lagging and Leading Power Factor Loads
We will draw the phasor diagram of transformer under no load condition in few steps, then by combining all the steps together, we will draw a final figure. This method will be very easy to understand, let’s start.
STEP 1
Take flux ɸM as the reference phasor, as shown in the figure below.
STEP 2
Emf equation of no load transformer is given by,
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To read about emf equation of a single phase transformer, click here
Since, the emf E1 and E2 are induced by the same flux ɸM, so they both will be in phase to eachother.
But, E2 will differ in magnitude from E1 because,
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From the above equations, we can also conclude that, both emf E1 and E2 lag behind the flux ɸM by an angle of 90°, as shown in figure below.
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STEP 3
The voltage drops in the primary windings are very negligible in this case, so we can neglect them, hence, E1 will be equal to the applied voltage V1 , and according to the lenz law, it is also opposite to the applied voltage V1. We can draw it as shown below,
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STEP 4
We have already discussed that, Iµ will be in phase with the flux ɸM.
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And, IW is in phase with the applied voltage V1.
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STEP 5
Now, take the phasor sum of Iµ and IW, it will be Io as shown below.
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Combine all the above steps to get the actual phasor diagram.
An approximate phasor diagram for a transformer under no load condition is shown below.
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We can conclude some results from the phasor diagram,
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As the angle is ɸ0, So power factor will be cosɸ0.
Also, core loss = V1I0cosɸ0 = V1Iw W
Magnetizing (reactive) voltamperes = V1I0sinɸ0 = V1Iµ VAr
We have discussed everything about no-load transformer, we will discuss about losses provided by the working component of no load current, later in this series of single phase Transformer.
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Transformer “On-load”
When an electrical load is connected to the secondary winding of a transformer and the transformer loading is therefore greater than zero, a current flows in the secondary winding and out to the load. This secondary current is due to the induced secondary voltage, set up by the magnetic flux created in the core from the primary current.
The secondary current, IS which is determined by the characteristics of the load, creates a self-induced secondary magnetic field, ΦS in the transformer core which flows in the exact opposite direction to the main primary field, ΦP. These two magnetic fields oppose each other resulting in a combined magnetic field of less magnetic strength than the single field produced by the primary winding alone when the secondary circuit was open circuited.
This combined magnetic field reduces the back EMF of the primary winding causing the primary current, IP to increase slightly. The primary current continues to increase until the cores magnetic field is back at its original strength, and for a transformer to operate correctly, a balanced condition must always exist between the primary and secondary magnetic fields. This results in the power to be balanced and the same on both the primary and secondary sides. Consider the circuit below.
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We know that the turns ratio of a transformer states that the total induced voltage in each winding is proportional to the number of turns in that winding and also that the power output and power input of a transformer is equal to the volts times amperes, ( V x I ). Therefore:
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Transformer Ratio
Where:
NP/NS = VP/VS – represents the voltage ratio
NP/NS = IS/IP – represents the current ratio
Note that the current is inversely proportional to both the voltage and the number of turns. This means that with a transformer loading on the secondary winding, in order to maintain a balanced power level across the transformers windings, if the voltage is stepped up, the current must be stepped down and vice versa. In other words, “higher voltage — lower current” or “lower voltage — higher current”.
As a transformers ratio is the relationships between the number of turns in the primary and secondary, the voltage across each winding, and the current through the windings, we can rearrange the above transformer ratio equation to find the value of any unknown voltage, ( V ) current, ( I ) or number of turns, ( N ) as shown.
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The total current drawn from the supply by the primary winding is the vector sum of the no-load current, Io and the additional supply current, I1 as a result of the secondary transformer loading and which lags behind the supply voltage by an angle of Φ. We can show this relationship as a phasor diagram.
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If we are given currents, IS and Io, we can calculate the primary current, IP by the following methods.
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Industrial Power Transformer
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Leading Power Factor
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Three Phase Transformer
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Delta-Star Connection
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Star-Star Three Phase Transformer Connection
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Introduction to Polyphase Induction Machines
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How a magnetic field is produced in 3 phase Induction Machine
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Vector Diagram
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Importance Of Air Gap
The amount of air or another non-magnetic material like a fibre plate or fibre board increases the reluctance of the circuit, thereby increasing the amount of current that we could put in a coil before we reach saturation. Also, the air gaps help the magnetic flux to expand outside the magnetic circuit.
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Difference between Induction Motor and Transformer
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Types Of Rotor
Types of three phase induction motor rotor
There are two types of induction motor rotors:
Squirrel cage rotor:
Squirrel cage motor works on the principle of Electromagnetism. It consists of Rotor, Stator and other parts like bearings, cylindrical laminated core, shaft, etc.
The function of bearings in cage rotor motor is to reduce friction between the rotating and stationary parts of the machine. The rotor of the motor consists of a cylindrical laminated core with parallel slots for carrying the rotor conductors. The rotor conductors are not wires, but it consists of heavy bars of copper, aluminum, or an alloy. The shaft is used in the motor to transfer mechanical power from or to the machine. The stator is the outer stationary part of the motor.
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The advantages of skewing of cage rotor conductors are:
Wound rotor or slip ring rotor:
The wound rotor consists of a slotted armature. Insulated conductors are put in the slots and connected to form a three-phase double layer distributed winding similar to the stator winding. The windings of the rotor are connected in star.
Rotor windings are distributed uniformly and usually connected in the star with here leads brought out of the machine by via slip rings placed on the shaft. The slip rings are tapped using copper carbon brushes. Wound rotor construction is generally used for large size machine, where the starting torque requirements are stringent. External resistance can be added in the rotor circuit through slip ring for reducing the starting current and simultaneously the starting torque.
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Difference between the cage and wound rotors:
The advantages of the cage rotor are as follows:
The advantages of wound rotors are as follows:
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