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AC MACHINES-1 (66761) Theory

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Book: AC MACHINES-1 (66761) Theory
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Table of contents

1. Understand working principle and construction of transformer.

Understand working principle and construction of transformer.

1.1 Define transformer.

1.2 Explain the working principle of a transformer.

1.3 Describe the construction of a transformer.

1.4 Identify the materials used for a transformer construction.

1.5 List different types of transformers.

1.6 Describe Core type, Shell type and Spiral core type transformer.

1.7 Compare between the core type and shell type transformer.

1.1. Define transformer

A transformer is an electrical apparatus designed to convert alternating current from one voltage to another. It can be designed to "step up" or "step down" voltages and works on the magnetic induction principle. ... A voltage is then induced in the other coil, called the secondary or output coil.

Electrical Transformers – All You Need To Know - Miracle ...

একটি ট্রান্সফর্মার একটি বৈদ্যুতিক যন্ত্রপাতি যা একটি ভোল্টেজ থেকে অন্য ভোল্টেজকে পর্যায়ক্রমে রূপান্তর করতে ডিজাইন করা হয়। এটি "স্টেপ আপ" বা "স্টেপ ডাউন" ভোল্টেজগুলির জন্য ডিজাইন করা যেতে পারে এবং চৌম্বকীয় আনয়ন নীতিতে কাজ করে। ... তারপরে একটি ভোল্টেজকে অন্য কয়েলে উত্সাহ দেওয়া হয়, যাকে মাধ্যমিক বা আউটপুট কয়েল বলা হয়।



1.2. Describe the construction of a transformer.

Transformer - Wikipedia

The transformer consists of a laminated steel core and the two coils. The two coils are insulated from each other and also from the core. The core of the transformer is constructed from laminations of steel sheet or silicon steel assembled to provide a continuous magnetic path.

ট্রান্সফরমারটিতে একটি স্তরিত ইস্পাত কোর এবং দুটি কয়েল থাকে। দুটি কয়েল একে অপরের থেকে এবং কোর থেকেও অন্তরক হয়। অবিচ্ছিন্ন চৌম্বকীয় পথ সরবরাহের জন্য ইস্পাত শীট বা সিলিকন ইস্পাত একত্রিত স্তরের স্তম্ভগুলি থেকে ট্রান্সফর্মারের মূলটি নির্মিত হয়।

1.3. Explain the working principle of a transformer.

...

1.4. Identify the materials used for a transformer construction.

  • Solid iron. Solid iron cores serve as an excellent pathway to provide magnetic flux and retain high magnetic fields without saturating the iron. ...
  • Carbonyl iron. ...
  • Amorphous steel. ...
  • Silicon steel. ...
  • Amorphous metals. ...
  • Ferrite ceramics. ...
  • Laminated magnetic cores.

  • সলিড লোহা। সলিড লোহার কোরা চৌম্বকীয় প্রবাহ সরবরাহ করতে এবং লোহাটিকে স্যাচুরেট না করে উচ্চ চৌম্বকীয় ক্ষেত্র ধরে রাখতে একটি দুর্দান্ত পথ হিসাবে কাজ করে। ...
    • কার্বনিল লোহা ...
    • নিরাকার ইস্পাত। ...
    • সিলিকন ইস্পাত ...
    • নিরাকার ধাতু। ...
    • ফেরাইট সিরামিকস। ...
    • স্তরিত চৌম্বকীয় কর।



    1.5. List different types of transformers.

    Types of Transformer - different types of transformer - Circuit Globe

    Transformer types

    The different types of transformer are Step up and Step down Transformer, Power Transformer, Distribution Transformer, Instrument transformer comprising current and Potential Transformer, Single phase and Three phase transformer, Auto transformer, etc.

    বিভিন্ন ধরণের ট্রান্সফর্মার হ'ল স্টেপ আপ এবং স্টেপ ডাউন ট্রান্সফর্মার, 

    পাওয়ার ট্রান্সফরমার,

     ডিস্ট্রিবিউশন ট্রান্সফরমার,

     ইনস্ট্রুমেন্ট ট্রান্সফর্মার সমন্বিত বর্তমান এবং 

    সম্ভাব্য ট্রান্সফরমার, সিঙ্গল ফেজ এবং থ্রি ফেজ ট্রান্সফর্মার, অটো ট্রান্সফরমার ইত্যাদি etc.

    Transformer types

    1.6. Describe Core type, Shell type and Spiral core type transformer.

    In core type transformer both the primary and the secondary windings are placed on the side limbs whereas, in shell type transformer, the windings are placed on the central limbs of the transformer. The core type transformer has two magnetic circuits whereas the shell type transformer has one magnetic circuit.

    কোর টাইপ ট্রান্সফর্মারটিতে প্রাথমিক এবং গৌণ উইন্ডিং উভয় পাশের অঙ্গগুলিতে স্থাপন করা হয়, শেল প্রকারের ট্রান্সফরমারে, উইন্ডিংগুলি ট্রান্সফর্মারের কেন্দ্রীয় অঙ্গগুলিতে স্থাপন করা হয়। কোর টাইপ ট্রান্সফর্মারের দুটি চৌম্বকীয় সার্কিট রয়েছে যেখানে শেল টাইপের ট্রান্সফর্মারটিতে একটি চৌম্বকীয় সার্কিট রয়েছে।

    Core type construction is used for larger low voltage transformer while shell type is used for small high voltage transformer. Natural cooling is not available for core type while for shell type natural cooling is available.


    1.7. Compare between the core type and shell type transformer.

    In core type transformer both the primary and the secondary windings are placed on the side limbs whereas, in shell type transformer, the windings are placed on the central limbs of the transformer. The core type transformer has two magnetic circuits whereas the shell type transformer has one magnetic circuit.

    Core type Transformer

    In core type transformers, winding is positioned on two limbs of the core and there is ONLY one flux path and windings are circumventing the core.

    Shell Type Transformer

    In shell type transformers, winding is positioned on the middle limb of the core while other limbs are utilized as the mechanical support.




    2. Perceive the emf equation, transformation ratio and Losses of transformer.

    2

    2.1. Define emf equation, transformation ratio of transformer

    EMF Equation of a Transformer
    When a sinusoidal voltage is applied to the primary winding of a transformer, alternating flux ϕm sets up in the iron core of the transformer. This sinusoidal flux links with both primary and secondary winding. The function of flux is a sine function.

    emf = turns x rate of change
    This is known as the Transformer EMF Equation. For the primary winding emf, N will be the number of primary turns, ( NP ) and for the secondary winding emf, N will be the number of secondary turns, ( NS ).

    As the magnetic flux varies sinusoidally, Φ = Φmax sinωt, then the basic relationship between induced emf, ( E ) in a coil winding of N turns is given by:

    emf = turns x rate of change

    transformer emf equation

    • Where:
    •   ƒ  –  is the flux frequency in Hertz,  = ω/2π
    •   Ν  –  is the number of coil windings.
    •   Φ  –  is the amount of flux in webers

    This is known as the Transformer EMF Equation. For the primary winding emf, N will be the number of primary turns, ( NP ) and for the secondary winding emf, N will be the number of secondary turns, ( NS ).


    The transformer transformation ratio or transformer turns ratio (K) is the quotient value obtained by dividing the number of turns of the primary winding (N1) and the number of turns of the secondary winding (N2). Then K = N1/N2.

    The turns ratio is defined as the ratio of turns of wire in the primary winding to the number of turns of wire in the secondary winding.

    spg

    2.2. Derive the emf equation of transformer.


    EMF Equation of a Transformer

    When a sinusoidal voltage is applied to the primary winding of a transformer, alternating flux ϕm sets up in the iron core of the transformer. This sinusoidal flux links with both primary and secondary winding. The function of flux is a sine function.

    The rate of change of flux with respect to time is derived mathematically.

    The derivation of the EMF Equation of the transformer is shown below. Let

    • ϕm be the maximum value of flux in Weber
    • f be the supply frequency in Hz
    • N1 is the number of turns in the primary winding
    • N2 is the number of turns in the secondary winding

    Φ is the flux per turn in Weber
    emf-eq-of-transformer-figureAs shown in the above figure that the flux changes from + ϕm to – ϕm in half a cycle of 1/2f seconds.

    By Faraday’s Law

    Let E1 be the emf induced in the primary winding
    emf-eq-1

    Where Ψ = N1ϕ
    emf-eq-2

    Since ϕ is due to AC supply ϕ = ϕm Sinwt
    emf-eq-3

    So the induced emf lags flux by 90 degrees.

    Maximum valve of emf
    emf-eq-4

    But w = 2πf
    emf-eq-5

    Root mean square RMS value is
    emf-eq-6

    Putting the value of E1max in equation (6) we get
    emf-eq-7

    Putting the value of π = 3.14 in the equation (7) we will get the value of E1 as
    emf-eq-8

    Similarly
    emf-eq-9

    Now, equating the equation (8) and (9) we get
    emf-eq-10

    The above equation is called the turn ratio where K is known as the transformation ratio.

    The equation (8) and (9) can also be written as shown below using the relation

    (ϕm = Bm x Ai) where Ai is the iron area and Bm is the maximum value of flux density.
    emf-eq-11

    For a sinusoidal waveemf-eq-12

    Here 1.11 is the form factor.


    2.3. Explain voltage ratio, current ratio and transformation ratio.

    What is difference between turns ratio and transformation ratio?
    The transformation ratio is defined as the ratio of the secondary voltage to primary voltage. And Turns Ratio would be the number turns of the primary winding to the secondary winding (edit: those were swapped originally). Some transformers have multiple secondary windings as well to create multiple outputs.
    How do you find the ratio of a transformation?
    A transformer turns ratio is the division of the number of turns in the primary winding by the number of turns in the secondary winding by the equation TR = Np/Ns. This ratio should also equal the voltage of the primary winding divided by the voltage of the secondary winding, as given by Vp/Vs.
    What is meant by transformation ratio?
    Transformation ratio is defined as the ratio of number of turns in the secondary winding to the number of turns in the primary winding. It also defined as the ratio of voltage at the secondary terminal to the voltage at the primary terminals.


    2.4. List the losses of transformer.


    There are various types of losses in the transformer such as iron loss, copper loss, hysteresis loss, eddy current loss, stray loss, and dielectric loss. The hysteresis ...

    2.5. Interpret Hysteresis loss, Eddy current loss, Core loss and Copper loss.

    What is eddy current loss and hysteresis loss?
    Eddy current loss= The ohmic losses in a metal body, due to the eddy currents flowing through it, induced by an alternating magnetic field. Hysteresis loss= The loss in form of heat when magnetisation of the material is made to alternate with respect to time.
    What is hysteresis current loss?
    Hysteresis loss is caused by the magnetization and demagnetization of the core as current flows in the forward and reverse directions. As the magnetizing force (current) increases, the magnetic flux increases. ... Therefore, when the magnetizing force reaches zero, the flux density still has a positive value.
    How is hysteresis loss calculated?
    dB‟ is the area of elementary strip of B –H curve shown in the figure above, Therefore, Energy consumed per cycle = volume of the right x area of hysteresis loop. The hysteresis loss per second is given by the equation[20]: Hysteresis loss, Ph= (Bmax)1.6f V joules per second (or) watts.
    Conclusion: hysteresis Losses can be reduced by special core material which reached to zero or near zero flux density after removal of current. Eddy current Losses can be reduced by making core by thin sheets by reducing the area of each Eddy current branch.
    What do you mean by eddy current loss?
    Eddy current loss is conductive I2R loss produced by circulating currents induced in response to AC flux linkage, flowing against the internal resistance of the core.




    2.6. Solve problems on emf equation.

    prob

    3. Interpret the principle of operation of transformer on no-load condition and load condition.

    1

    3.1. Explain no-load operation of transformer.

    Transformer on No Load Condition

    When the transformer is operating at no load, the secondary winding is open-circuited, which means there is no load on the secondary side of the transformer and, therefore, current in the secondary will be zero. While primary winding carries a small current I0 called no-load current which is 2 to 10% of the rated current.

    যখন ট্রান্সফর্মারটি কোনও লোড ছাড়াই চলমান থাকে, তখন সেকেন্ডারি উইন্ডিংটি ওপেন-সার্কিট হয়, যার অর্থ ট্রান্সফর্মারের দ্বিতীয় দিকে কোনও লোড নেই এবং সুতরাং, গৌণটিতে বর্তমান শূন্য হবে। প্রাইমারি ওয়াইন্ডিংতে একটি ছোট বর্তমান আই0 হয় যা নো-লোড প্রবাহ বলে যা রেট করা বর্তমানের 2 থেকে 10% হয়।



    3.2. Define no-load voltage, current, mutual flux, no load power factor.

    The no load voltage is the terminal voltage when zero current is drawn from the supply, that is, the open circuit terminal voltage. Power supply performance is measured in terms of percent voltage regulation, which indicates its ability to maintain a constant voltage.

    The no-load current is small because the primary links with its own magnetic field and electromagnetic theory explains that this will induce a back-emf to oppose the voltage applied externally to the coil. The open circuit transformer therefore acts as a highly inductive choke with a power factor of some 0.15 lagging.



    This current is responsible for supplying the iron losses (hysteresis and eddy current losses) in the core and a very small amount of copper losses in the primary winding. The angle of lag depends upon the losses in the transformer. The power factor is very low and varies from 0.1 to 0.15.

    The no-load current consists of two components:

    • Reactive or magnetizing component Im
      (It is in quadrature with the applied voltage V1. It produces flux in the core and does not consume any power).
    • Active or power component Iw, also know as a working component
      (It is in phase with the applied voltage V1. It supplies the iron losses and a small amount of primary copper loss).

    The following steps are given below to draw the phasor diagram:

    1. The function of the magnetizing component is to produce the magnetizing flux, and thus, it will be in phase with the flux.
    2. Induced emf in the primary and the secondary winding lags the flux ϕ by 90 degrees.
    3. The primary copper loss is neglected, and secondary current losses are zero as
      I2 = 0.
      Therefore, the current I0 lags behind the voltage vector V1 by an angle ϕ0 called the no-load power factor angle and is shown in the phasor diagram above.
    4. The applied voltage V1 is drawn equal and opposite to the induced emf E1 because the difference between the two, at no load, is negligible.
    5. Active component Iw is drawn in phase with the applied voltage V1.
    6. The phasor sum of magnetizing current Im and the working current Iw gives the no-load current I0.phasor-diag-of-transformer-on-no-loadFrom the phasor diagram drawn above, the following conclusions are madeTRANSFORMER-ON-NO-LOAD-EQ

    This is all about transformer in no-load condition.


    3.3. Draw the vector diagram of a transformer on no load condition.

    No load transformer

    An approximate phasor diagram for a transformer under no load condition is shown below.

    Phasor diagram of no load transformer

    We can conclude some results from the phasor diagram,

    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.


    3.4. Solve problems related to no load test.

    3.4

    3.5. Explain operation of a transformer on load condition.

    Transformer “On-load”

    transformer loading

    Transformer changes the voltage and current levels from one side (primary) to the second side (secondary). ... This equipment is the “load" since in any circuit “ load” is something connected to supply which draws current. This load, connected to secondary, is supplied by the voltage available at secondary terminals.

    vector diagram of transformer


    3.6. Draw the vector diagram of transformer on lagging, leading and unity power factor.

    Practical transformer on load

    The efficiency of a transformer at full load condition on 0.8 pf ...

    Transformer With Resistance And Leakage Reactance | Electrical ...

    3.7. Solve problems related to transformer on load.

    Probl.

    4. Understand equivalent circuit of transformer, magnetic leakage and leakage reactance of transformer.

    https://jpilms.gnomio.com/pluginfile.php/944/mod_book/chapter/73/ACM-1%284.1-4.3%29%20%206th%20Class%20Equivalment%20ckt.pptx

    4.1. Draw the equivalent circuit and vector diagram of a transformer.

    What is the equivalent circuit of a transformer?

    Equivalent circuit diagram of a transformer is basically a diagram which can be resolved into an equivalent circuit in which the resistance and leakage reactance of the transformer are imagined to be external to the winding. Where, R1 = Primary Winding Resistance. R2= Secondary winding Resistance.

    Equivalent Circuit diagram of single phase Transformer

    Equivalent circuit diagram of a transformer is basically a diagram which can be resolved into an equivalent circuit in which the resistance and leakage reactance of the transformer are imagined to be external to the winding.

    The equivalent circuit diagram of transformer is given below:-

    Where,
    R1 = Primary Winding Resistance.
    R2= Secondary winding Resistance.
    I0= No-load current.
    Iµ = Magnetizing Component,
    Iw = Working Component,
    This Iµ & Iw are connected in parallel across the primary circuit. The value of E1 ( Primary e.m.f ) is obtained by subtracting vectorially I1 Z1 from V1 . The value of X0 = E1 / I0 and R0 = E1 /Iw. We know that the relation of E1 and E2 is E2 /E1 = N2 /N1 = K , ( transformation Ratio )

    From the equivalent circuit , we can easily calculate the total impedance of to transfer voltage, current, and impedance either to the primary or the secondary.

    PHASOR DIAGRAM ( INDUCTIVE LOAD) FOR A SINGLE PHASE TRANSFORMER ...


    4.2. Explain the equivalent circuit of a transformer.

    The equivalent circuit diagram of transformer is given below:-

    Where,
    R1 = Primary Winding Resistance.
    R2= Secondary winding Resistance.
    I0= No-load current.
    Iµ = Magnetizing Component,
    Iw = Working Component,
    This Iµ & Iw are connected in parallel across the primary circuit. The value of E1 ( Primary e.m.f ) is obtained by subtracting vectorially I1 Z1 from V1 . The value of X0 = E1 / I0 and R0 = E1 /Iw. We know that the relation of E1 and E2 is E2 /E1 = N2 /N1 = K , ( transformation Ratio )

    From the equivalent circuit , we can easily calculate the total impedance of to transfer voltage, current, and impedance either to the primary or the secondary.

    The total equivalent circuit of the transformer is obtained by adding in the primary impedance as shown in – Fig-3 .

    fig-3

    And It can be simplified the terminals shown in fig – 4 & further simplify the equivalent circuit is shown in fig.- 5 ,

    fig-4

    At last, the circuit is simplified by omitting I0 altogether as shown in fig- 5 .

    fig-5

    From the equivalent circuit which is shown in fig.-3 , the total impedance between the input terminal is ,

    This is so because there are two parallel circuits, one having an impedance of Zm and the other having Z’2 and Z’L in series with each other.



    4.3. Derive the equivalent resistance of a transformer referred to primary.

    Let us consider the transformation ratio be,

    The voltage appeared across winding is countered by primary induced emf E1. So voltage equation of this portion of the transformer can be written as,

    The equivalent circuit for that equation can be drawn as below,
    equivalent circuit 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


    equivalent circuit of primary side of transformer

    The complete equivalent circuit of transformer is shown below.
    Equivalent Circuit of Transformer referred to Primary

    Now if we see the voltage drop in secondary from primary side, then it would be ′K′ times greater and would be written as K.Z2.I2.
    Again I2′.N1 = I2.N2

    Therefore,

    From above equation, secondary impedance of transformer referred to primary is,

    So, the complete equivalent circuit of transformer referred to primary is shown in the figure below:
    Equivalent Circuit of Transformer referred to Primary

    Approximate Equivalent Circuit of Transformer

    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.

    approximate equivalent circuit of transformer referred to primary

    Equivalent Circuit of Transformer Referred to Secondary






    4.4. Calculate the equivalent resistance of a transformer referred to secondary.

    https://jpilms.gnomio.com/pluginfile.php/944/mod_book/chapter/77/ACM-1%284.3-4.8%29%207th%20Class%20Equivalent.pptx?time=1592112657175
    Equivalent.pptxEquivalent Circuit 
    when all the quantities are referred to Secondary side

    The equivalent circuit diagram of the transformer is shown below when all the quantities are referred to the secondary side.

    EQUIVALENT CIRCUIT REFERRED TO SECONDARY SIDE

    Circuit Diagram of Transformer When All the Primary Quantities are Referred to Secondary Side

    The following are the values of resistance and reactance given below

    Primary resistance referred to the secondary side is given as
    EQUIVALENT-CIRCUIT-OF-TRANSFORMER-EQ6

    The equivalent resistance referred to the secondary side is given as
    EQUIVALENT-CIRCUIT-OF-TRANSFORMER-EQ7

    Primary reactance referred to the secondary side is given as

    EQUIVALENT-CIRCUIT-OF-TRANSFORMER-EQ8

    The equivalent reactance referred to the secondary side is given as
    EQUIVALENT-CIRCUIT-OF-TRANSFORMER-EQ9

    No-load current I0 is hardly 3 to 5% of full load rated current, the parallel branch consisting of resistance R0 and reactance X0 can be omitted without introducing any appreciable error in the behavior of the transformer under the loaded condition.

    Further simplification of the equivalent circuit of the transformer can be done by neglecting the parallel branch consisting of R0 and X0.

    The simplified circuit diagram of the transformer is shown below:

    Simplified-Equivalent-Circuit

    Simplified Equivalent Circuit Diagram of a Transformer

    This is all about the equivalent circuit of the Transformer.


    4.5. Explain magnetic leakage of transformer.

    Magnetic leakage can be defined as the passage of magnetic flux outside the path along which it can do useful work. The passage of useful and leakage magnetic flux.

    Magnetic leakage flux is reduced by carefully selection of grain-oriented silicon steel core material, a large core structure that provides a good closed-loop path for the magnetic field, and using an operating point that's significantly below core saturation.
    How is magnetic leakage reduced? 
    Ans: Magnetic leakage is reduced to a minimum by sectionalizing and interleaving the primary and secondary...
     Magnetic leakage flux is reduced by carefully selection of grain-oriented silicon steel core material, a large core structure that provides a good closed-loop path

    4.6. List the disadvantages of magnetic leakage.

    As leakage flux increases, efficiency of the transformer decreases as a portion of the leakage flux may induce eddy currents within nearby conductive object such as the transformers support structure, and be converted to heat. When represented as inductance, it only represents reactive power loss

    4.7. Calculate leakage reactance of transformer in terms of primary and in terms of secondary.

    What is meant by leakage inductance?
    Leakage inductance in a transformer is an inductive component that results from the imperfect magnetic linking of one winding to another. In an ideal transformer, 100% of the energy is magnetically coupled from the primary to the secondary windings. ... This series inductance is the "leakage inductance."
    Where equivalent impedance of transformer referred to secondary, can be derived as

    approximate equivalent circuit of transformer referred to secondary

    4.8. Solve problems on equivalent circuit of transformer, leakage reactance and impedance of transformer.

    4.8 Prob

    4.9. Define percentage resistance, reactance and impedance.

    “The percentage impedance is nothing but a measure of the volt drops when the transformer in on full load due to the winding resistance and leakage reactance ...

    The percentage impedance of a transformer (Z%) is the voltage drop on full load due to the winding resistance and leakage reactance expressed as a percentage of the rated voltage.

    The percentage impedance of a transformer is marked on most nameplates – but what is it and what does the Z% figure mean?

    When referring to the impedance of a transformer, it is the equivalent impedance that is meant.

    It is marked in percentage value on the nameplate of power transformers in every electrical substation.

    Percentage Impedance in Nameplate of Transforemer

    Percentage Impedance in Nameplate of an 11kV/415V Transformer

    Calculation of Percentage Impedance

    In order to determine equivalent impedance, one winding of the transformer is short-circuited. And a just enough voltage is applied to the other winding to create full load current to flow in the short-circuited winding.

    This voltage is known as the impedance voltage.

    Percentage impedance of transformer testing
    Percentage impedance of transformer testing

    Either winding may be short-circuited for this test, but it is usually more convenient to short circuit the low-voltage winding.

    The nameplate of a transformer shows its impedance value in percent. This means that the voltage drop due to the impedance is expressed as a percent of rated voltage.

    Read More Here: How to Test Percentage Impedance of Transformer?

    Example Calculation

    Example Calculation

    For example, if a 2,400/240-volt transformer has a measured impedance voltage of 72 volts on the high voltage windings, its impedance (Z), expressed as a percent, is:

    Z%  =  (Impedance Voltage / Rated Voltage)   x  100

    percent Z = (72/2400)*100 = 3 percent

    This means there would be a 72-volt drop in the high-voltage winding at full load due to losses in the windings and core. Only 1 or 2% of the losses are due to the core; about 98% is due to the winding impedance.

    If the transformer were not operating at full load, the voltage drop would be less. If an actual impedance value in ohms is needed for the high-voltage side (Ohms Law):

    Z = V/I

    where V is the voltage drop or, in this case, 72 volts; and I is the full load current in the primary winding.

    If the full load current is 10 amps:

    Z = 72V/10A = 7.2 Ohms

    Of course, one must remember that impedance is a combination of both resistive and reactive components.






    Definition

    The percentage impedance of a transformer is the volt drop on full load due to the winding resistance and leakage reactance expressed as a percentage of the rated voltage.




    It is also the percentage of the normal terminal voltage required to circulate full-load current under short circuit conditions.

    In other words, the percentage impedance of a transformer is the percentage of the rated voltage applied at one side (primary winding) to circulate rated current on transformer keeping its other side (secondary winding) under short circuit conditions.


    Formula Cheatsheet

    1. Impedance Z = R or XLor XC(if only one is present)
    2. Impedance in series only Z = √(R2 + X2) (if both R and one type of X are present)
    3. Impedance in series only Z = √(R2 + (|XL - XC|)2) (if R, XL, and XC are all present)
    4. Impedance in any circuit = R + jX (j is the imaginary number √(-1))
    5. Resistance R = ΔV / I
    6. Inductive reactance XL = 2πƒL = ωL
    7. Capacative reactance XC = 1 / 2πƒC = 1 / ωC

    4.10. Express the deduction of the equation for percentage resistance, reactance and impedance.

    4.10

    5. Realize the open circuit test, short circuit test and voltage regulation of transformer

    https://jpilms.gnomio.com/pluginfile.php/944/mod_book/chapter/84/ACM-1%20%285.1-5.7%29%20O.S.T%20V.%209th%20Class.pptx5

    5.1. Describe open circuit test.

    5.1These two transformer tests are performed to find the parameters of equivalent circuit of transformer and losses of the transformer. Open circuit test and short circuit test on transformer are very economical and convenient because they are performed without actually loading of the transformer.

    Open circuit or No load test on Transformer

    Open circuit test or no load test on a transformer is performed to determine 'no load loss (core loss)' and 'no load current I0'. The circuit diagram for open circuit test is shown in the figure below.


    Usually high voltage (HV) winding is kept open and the low voltage (LV) winding is connected to its normal supply. A wattmeter (W), ammeter (A) and voltmeter (V) are connected to the LV winding as shown in the figure. Now, applied voltage is slowly increased from zero to normal rated value of the LV side with the help of a variac. When the applied voltage reaches to the rated value of the LV winding, readings from all the three instruments are taken.
    The ammeter reading gives the no load current I0. As I0 itself is very small, the voltage drops due to this current can be neglected.
    The input power is indicated by the wattmeter (W). And as the other side of transformer is open circuited, there is no output power. Hence, this input power only consists of core losses and copper losses. As described above, no-load current is so small that these copper losses can be neglected. Hence, now the input power is almost equal to the core losses. Thus, the wattmeter reading gives the core losses of the transformer.
    Sometimes, a high resistance voltmeter is connected across the HV winding. Though, a voltmeter is connected, HV winding can be treated as open circuit as the current through the voltmeter is negligibly small. This helps in to find voltage transformation ratio (K).
    The two components of no load current can be given as,

    Iμ = I0sinΦ0   and    Iw = I0cosΦ0.
    cosΦ0 (no load power factor) = W / (V1I0). ... (W = wattmeter reading)

    From this, shunt parameters of equivalent circuit parameters of equivalent circuit of transformer (X0 and R0) can be calculated as

    X0 = V1/Iμ  and  R0 = V1/Iw.

    (These values are referring to LV side of the transformer.)
    Hence, it is seen that open circuit test gives


    5.2. Describe short circuit test.

    Short circuit or Impedance test on Transformer

    The connection diagram for short circuit test or impedance test on transformer is as shown in the figure below. The LV side of transformer is short circuited and wattmeter (W), voltmere (V) and ammeter (A) are connected on the HV side of the transformer. Voltage is applied to the HV side and increased from the zero until the ammeter reading equals the rated current. All the readings are taken at this rated current.
    short circuit or impedance test on transformer
    The ammeter reading gives primary equivalent of full load current (Isc).
    The voltage applied for full load current is very small as compared to rated voltage. Hence, core loss due to small applied voltage can be neglected. Thus, the wattmeter reading can be taken as copper loss in the transformer.
    Therefore, W = Isc2Req....... (where Req is the equivalent resistance of transformer)
     Zeq = Vsc/Isc.
    Therefore, equivalent reactance of transformer can be calculated from the formula  Zeq2 = Req2 + Xeq2.
    These, values are referred to the HV side of the transformer.
    Hence, it is seen that the short circuit test gives copper losses of transformer and approximate equivalent resistance and reactance of the transformer.

    5.3. Draw the vector diagrams.

    5.3

    The phasor diagram of the transformer at no load or when an open circuit test is performed is shown below

    OPEN-CIRCUIT-PHASOR-DIAGRAM

    Phasor Diagram of Open Circuit Test

    The iron losses measured by the open circuit test is used for calculating the efficiency of the transformer.

    The phasor diagram of the short circuit test of the transformer is shown below

    short-circuit-phasor-diagram

    Phasor Diagram of Short Circuit Test

    From the phasor diagram
    OC-SC-EQ10

    Equivalent impedance referred to the secondary side is given by
    OC-SC-EQ11

    The equivalent reactance referred to the secondary side is given by
    OC-SC-EQ12

    The voltage regulation of the transformer can be determined at any load and power factor after knowing the values of Zes and Res.

    In the short circuit test the wattmeter record, the total losses, including core loss but the value of core loss are very small as compared to copper loss so the core loss can be neglected.



    5.4. Solve problems related to open and short circuit test.

    5.4Solving Problems on Open and Short Circuit Tests of a Transformer ...

    Transformers - Open Circuit and Short Circuit Tests (Full Video ...

    Solved: Open Circuit/short Circuit Tests Transformer Speci ...

    Example 6.6 The open-circuit and short-circuit tests on a 4-kVA, 200/400-V, 50-Hz, single-phase transformer gave the following results: oc test on the LV side: 200 V, 1 A, 100 W sc test with the LV side shorted: 15 V, 10 A, 85 W (a) Determine the parameters of the equivalent circuit. (b) Draw the equivalent circuit referred to the LV side.


    5.5. Define voltage regulation.

    Voltage Regulation of a Transformer

    Definition: The voltage regulation is defined as the change in the magnitude of receiving and sending voltage of the transformer. The voltage regulation determines the ability of the transformer to provide the constant voltage for variable loads.

    সংজ্ঞা: ভোল্টেজ নিয়ন্ত্রণকে ট্রান্সফর্মারের ভোল্টেজ গ্রহণ এবং প্রেরণের প্রস্থের পরিবর্তন হিসাবে সংজ্ঞায়িত করা হয়। ভোল্টেজ নিয়ন্ত্রণটি পরিবর্তনশীল লোডের জন্য ধ্রুবক ভোল্টেজ সরবরাহের জন্য ট্রান্সফর্মারের ক্ষমতা নির্ধারণ করে।

    Mathematically, the voltage regulation is represented as:

    VOLTAGE-REGULATION-EQ1where,
    E2 – secondary terminal voltage at no load
    V2 – secondary terminal voltage at full load

    The voltage regulation by considering the primary terminal voltage of the transformer is expressed as,
    VOLTAGE-REGULATION-EQ2



    5.6. Express the deduction of the equation for voltage regulation at unity, lagging and leading power factor.

    Let us understand the voltage regulation by taking an example explained below:

    If the secondary terminals of the transformer are open-circuited or no load is connected to the secondary terminals,  the no-load current flows through it.

    If the no current flows through the secondary terminals of the transformer, the voltage drops across their resistive and reactive load become zero. The voltage drop across the primary side of the transformer is negligible.

    If the transformer is fully loaded, i.e., the load is connected to their secondary terminal,  the voltage drops appear across it. The value of the voltage regulation should always be less for the better performance of the transformer.
    volltage-regulation-ckt--diagram

    From the circuit diagram shown above, the following conclusions are made

    • The primary voltage of the transformer is always greater than the induced emf on the primary side. V1>E1
    • The secondary terminal voltage at no load is always greater than the voltage at full load condition. E2>V2

    By considering the above circuit diagram, the following equations are drawn
    VOLTAGE-REGULATION-EQ3The approximate expression for the no-load secondary voltage for the different types of the load is

    1. For inductive loadVOLTAGE-REGULATION-EQ4Where,
      VOLTAGE-REGULATION-EQ5

      • For Capacitive loadVOLTAGE-REGULATION-EQ6

    In this way, we define the voltage regulation of the transformer.

    Voltage Regulation of Transformer at Unity, Lagging, and Leading Power Factor

    Voltage Regulation of Transformer

    The voltage regulation of a transformer can be described as the change in the secondary voltage as the current varies from full load to no load while keeping the primary voltage constant.

    It is clear from the transformer equivalent circuit in figure 1 that the secondary current Is produces voltage drop IsRs and Is Xs across the resistive and reactive components respectively. Also, the primary current Ip causes primary circuit voltage drops IpRp and IpXp. Consequently, the effective primary voltage Ep is less than the input voltage Vi, and the output voltage Vo is less than the calculated value of Es.

    Complete Equivalent Circuit

    Fig.1: Complete Equivalent Circuit of Transformer

    Voltage Regulation of Transformer Formula

    The percentage change in output voltage from no-load to full-load is termed the voltage regulation of the transformer. Ideally, there should be no change in Vo from no-load to full-load (i.e., regulation = 100%). For the best possible performance, the transformer should have the lowest possible regulation. Mathematically, voltage regulation can be expressed as

    Voltage Regulation=Vo(NL)−Vo(FL)Vo(FL)(1)

    Where Vo(NL) is the transformer no-load output voltage, and Vo(FL) is the full load output voltage. Voltage regulation for a transformer is illustrated in figure 2.

    Transformer Voltage Regulation a

    Transformer Voltage Regulation b

    Fig.2: Transformer Voltage Regulation


    The equation for the voltage regulation of transformer, represented in percentage, is

    Voltage Regulation Equation

    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.

    voltage regulation lagging

    Here, from the above diagram,

    Angle between OC and OD may be very small, so it can be neglected and OD is considered nearly equal to OC i.e.

    Voltage regulation of transformer at lagging power factor,

    Voltage regulation equation with a lagging power factor

     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.

    voltage regulation leading

    Here, from the above diagram,

    Angle between OC and OD may be very small, so it can be neglected and OD is considered nearly equal to OC i.e.

    Voltage regulation of transformer at leading power factor,

    Voltage regulation equation with a leading power factor













    5.7. Solve problems related to voltage regulation.

    6.1. Derive the formula for calculation of efficiency of transformer.

    To calculate the transformer efficiency, divide the output power by the input power. Example: with an output power of 1254 watts for an input power of 1320 watts, divide 1254 by 1320, which equals . 95, or an efficiency of 95 percent.

    The ordinary transformer has a very high efficiency (in the range of 96—99%). Hence the transformer efficiency cannot be determined with high precision by direct ...

    Just like any other electrical machine, efficiency of a transformer can be defined as the output power divided by the input power. That is efficiency = output / input . Transformers are the most highly efficient electrical devices. Most of the transformers have full load efficiency between 95% to 98.5% .

    Efficiency of transformer is simply given as:

    • The output power is the product of the fraction of the rated loading (volt-ampere), and power factor of the load
    • The losses are the sum of copper losses in the windings + the iron loss + dielectric loss + stray load loss.

    6.2. Explain the factors affecting core loss and copper loss of the transformer.

    Types of Losses in a Transformer

    There are various types of losses in the transformer such as iron loss, copper loss, hysteresis loss, eddy current loss, stray loss, and dielectric loss. The hysteresis losses occur because of the variation of the magnetization in the core of the transformer and the copper loss occurs because of the transformer winding resistance.

    The various types of losses are explained below in detail.

    Contents:

    types-of-losses-in-transformer

    Iron Losses

    Iron losses are caused by the alternating flux in the core of the transformer as this loss occurs in the core it is also known as Core loss. Iron loss is further divided into hysteresis and eddy current loss.

    Hysteresis Loss

    The core of the transformer is subjected to an alternating magnetizing force, and for each cycle of emf, a hysteresis loop is traced out. Power is dissipated in the form of heat known as hysteresis loss and given by the equation shown below:
    transformer-losses-eq1

    Where

    • KȠ is a proportionality constant which depends upon the volume and quality of the material of the core used in the transformer,
    • f is the supply frequency,
    • Bmax is the maximum or peak value of the flux density.

    The iron or core losses can be minimized by using silicon steel material for the construction of the core of the transformer.

    Eddy Current Loss

    When the flux links with a closed circuit, an emf is induced in the circuit and the current flows, the value of the current depends upon the amount of emf around the circuit and the resistance of the circuit.

    Since the core is made of conducting material, these EMFs circulate currents within the body of the material. These circulating currents are called Eddy Currents. They will occur when the conductor experiences a changing magnetic field. As these currents are not responsible for doing any useful work, and it produces a loss (I2R loss) in the magnetic material known as an Eddy Current Loss.
    The eddy current loss is minimized by making the core with thin laminations.

    The equation of the eddy current loss is given as:
    eddy-current-eq

    Where,

    • Ke – coefficient of eddy current. Its value depends upon the nature of magnetic material like volume and resistivity of core material, the thickness of laminations
    • Bm – maximum value of flux density in wb/m2
    • T – thickness of lamination in meters
    • F – frequency of reversal of the magnetic field in Hz
    • V – the volume of magnetic material in m3

    Copper Loss Or Ohmic Loss

    These losses occur due to ohmic resistance of the transformer windings. If I1 and I2 are the primary and the secondary current. R1 and R2 are the resistance of primary and secondary winding then the copper losses occurring in the primary and secondary winding will be I12R1 and I22R2 respectively.

    Therefore, the total copper losses will be
    transformer-losses-eq3These losses varied according to the load and known hence it is also known as variable losses. Copper losses vary as the square of the load current.

    Stray Loss

    The occurrence of these stray losses is due to the presence of leakage field. The percentage of these losses are very small as compared to the iron and copper losses so they can be neglected.

    Dielectric Loss

    Dielectric loss occurs in the insulating material of the transformer that is in the oil of the transformer, or in the solid insulations. When the oil gets deteriorated or the solid insulation gets damaged, or its quality decreases, and because of this, the efficiency of the transformer gets affected.





    6.3. Deduce the equation for maximum efficiency.

    How do you calculate the maximum efficiency of a transformer?

    The value of transformer efficiency will be maximum when the copper losses will be equal to iron losses in the transformer. The value of maximum efficiency can be found by taking total losses equal to 2Pi. It also depends on load power factor and has the maximum value at a power factor of unity.

    Transformer Efficiency

    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.


    Where,

    • V2 – Secondary terminal voltage
    • I2 – Full load secondary current
    • Cosϕ2 – power factor of the load
    • Pi – Iron losses = hysteresis losses + eddy current losses
    • Pc – Full load copper losses = I22Res

    Consider, the x is the fraction of the full load. The efficiency of the transformer regarding x is expressed as


    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:

    TRANSFORMER-EFFICIENCY-EQ3The 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.

    TRANSFORMER-EFFICIENCY-EQ4i.e Copper losses = Iron losses

    Thus, the transformer will give the maximum efficiency when their copper loss is equal to the iron loss.

    TRANSFORMER-EFFICIENCY-EQ5

    From equation (2) the value of output current I2 at which the transformer efficiency will be maximum is given as


    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

    TRANSFORMER-EFFICIENCY-EQ7

    Thus, output KVA corresponding to maximum efficiency
    TRANSFORMER-EFFICIENCY-EQ8

    Putting the value of x from the above equation (3) in equation (4) we will get,

    TRANSFORMER-EFFICIENCY-EQ9The above equation (5) is the maximum efficiency condition of the transformer.







    6.4. Evaluate the variation of efficiency with power factor.

    It is important to appreciate that, since copper loss depends on current and the iron loss depends on voltage, the total loss in the transformer depends on the volt-ampere product, and not on the phase angle between voltage and current, i.e., is independent of the load power factor. The transformers are, therefore, rated in kilo-volt amperes (KVA) and not in kilowatts.

    এটি উপলব্ধি করা গুরুত্বপূর্ণ যে, যেহেতু তামার ক্ষতি ভোল্টেজের উপর নির্ভর করে এবং লোহার ক্ষতি ভোল্টেজের উপর নির্ভর করে, তাই ট্রান্সফর্মারে মোট ক্ষতি ভোল্ট-অ্যাম্পিয়ার পণ্যের উপর নির্ভর করে, এবং ভোল্টেজ এবং স্রোতের মধ্যে ধাপের কোণে নয়, অর্থাত্ স্বাধীন লোড পাওয়ার ফ্যাক্টরের। ট্রান্সফর্মারগুলি তাই কিলোভোল্টে নয় কিলো-ভোল্ট অ্যাম্পিয়ারে (কেভিএ) রেটেড হয়।

    Efficiency versus Power Factor:

    Transformer efficiency is given as:


    The variations of efficiency with power factor at different loadings for a typical transformer are illustrated in fig.

    How is power factor related to efficiency?

    If the load power factor is unity then output power will be maximised and the efficiency will be maximum. If the load power factor is less than unity, the true or real output power will correspondingly reduce as will the efficiency.
    What is the relationship between efficiency and the power factor ...



    6.5. Define all day efficiency and mention the formula of all day efficiency.

    All Day Efficiency of a Transformer

    Definition: All day efficiency means the power consumed by the transformer throughout the day. It is defined as the ratio of output power to the input power in kWh or wh of the transformer over 24 hours. Mathematically, it is represented as

    ALL-DAY-EFFICIENCY-EQ1

    All-day efficiency of the transformer depends on their load cycle. The load cycle of the transformer means the repetitions of load on it for a specific period.

    The ordinary or commercial efficiency of a transformer is defined as the ratio of the output power to the input power.


    TRANSFORMER EFFICIENCY EQ

    What is the need for All Day Efficiency ?

    Some transformer efficiency cannot be judged by simple commercial efficiency as the load on certain transformer fluctuate throughout the day.

    For example, the distribution transformers are energized for 24 hours, but they deliver very light loads for the major portion of the day, and they do not supply rated or full load, and most of the time the distribution transformer has 50 to 75% load on it.

    As we know, there are various losses in the transformer such as iron and copper loss. The iron loss takes place at the core of the transformer. Thus, the iron or core loss occurs for the whole day in the distribution transformer.

    The second type of loss known as a copper loss and it takes place in the windings of the transformer and is also known as the variable loss. It occurs only when the transformers are in the loaded condition.

    Hence, the performance of such transformers cannot be judged by the commercial or ordinary efficiency, but the efficiency is calculated or judged by All Day Efficiency also known as operational efficiency or energy efficiency which is computed by the energy consumed for 24 hours.

    All Day Efficiency = Output (in kWh)/Input (in kWh)
    To understand about the all day efficiency, must know about the load cycle i.e. how much load is connected, and for how much time (in 24 hours).

    https://www.electricaltechnology.org/wp-content/uploads/2013/12/Transformer-Efficiency-All-day-Efficiency-Condition-for-maximum-Efficiency.jpg




    6.6. Solve problems on efficiency, maximum efficiency and all day efficiency.

    Example 1. A 15 kVA, 2000/200 V transformer has an iron loss of 250 W and full-load copper loss 350 W. During the day it is loaded as follows :

    No. of hours

    Load

    Power factor

    9

    ¼ load

    0.6

    7

    Full load

    0.8

    6

    ¾ load

    1.0

    2

    No-load

    –

    Calculate the all-day efficiency.

    Solution. Rating of transformer = 15 kVA

    http://www.electrical-engineering-assignment.com/all-day-efficiency/#

    Solved: Problem 2: A 10 KVA 2200/220 V, 50 Hz Transformer ...3. Example: The transformer of example 18 operates with the following loads during a 24-hr period: 1 ½ times rated kva, power factor = 0.8, 1hr; 1 ¼ times rated kva, power factor = 0.8, 2hr; rated kva, power factor = 0.9, 3hr; ½ rated kva, power factor = 1.0, 6hr; ¼ rated kva, power factor = 0.8; no-load, 4hr. Calculate the all-day efficiency.

    . Solution: Energy output, kw-hr Energy losses, kw-hr W1 = 1.5 x 5 x 0.8 x = 6.0 (1 ½)2 x 0.112 x 1 = 0.252 W2 = 1.25 x 0.8 x 2 = 10.0 (1 ½)2 x 0.112 x 2 = 0.350 W3 = 1 x 5 x 0.9 x 3 = 13.5 1 x 0.112 x 3 = 0.336 W6 = 0.5 x 5 x 1.0 x 6 = 15.0 (1/2)2 x 0.112 x 6 = 0.168 W8 = 0.25 x 5 x 1.0 x 8 = 10.0 (1/4)2 x 0.112 x 8 = 0.056 ____ Total. . . . . . . . 54.5 Iron = 0.04 x 24 = 0.960 _____ Total. . . . . . . . .. . . . 2.122 All-day Efficiency = (1 – 2.122/54.5 + 2.122) x 100 = 96.25%


    6.7. Explain the necessity of cooling system of transformer.

    Cooling of Transformers

    Cooling of a transformer is the process of dissipation of heat developed in the transformer to the surroundings. The losses occurring in the transformer are converted into heat which increases the temperature of the windings and the core. In order to dissipate the heat generated cooling should be done.

    How to Cool the Transformer?

    There are two ways of cooling the transformer:

    • First, the coolant circulating inside the transformer transfers the heat from the windings and the core entirely to the tank walls and then it is dissipated to the surrounding medium
    • Second, along with the first technique, the heat can also be transferred by coolants inside the transformer.
    The choice of method used depends on the size, type of applications and the working conditions.

    Methods of Cooling of Transformer

    Based on the coolant used the cooling methods can be classified into:

    1. Air cooling
    2. Oil and Air cooling
    3. Oil and Water cooling

    1. Air cooling (Dry type transformers)

    • Air Natural(AN)
    • Air Blast (AB)

    2. Oil cooling (Oil immersed transformers)

    • Oil Natural Air Natural (ONAN)
    • Oil Natural Air Forced (ONAF)
    • Oil Forced Air Natural (OFAN)
    • Oil Forced Air Forced (OFAF)

    3. Oil and Water cooling (For capacity more than 30MVA)

    • Oil Natural Water Forced (ONWF)
    • Oil Forced Water Forced (OFWF)

    1. Air Cooling (Dry Type Transformers)


    Air Natural (AN)

    This method uses the ambient air as the cooling medium. The natural circulation of the air is used for dissipation of heat generated by natural convection. The core and the windings are protected from mechanical damage by providing a metal enclosure. This method is suitable for transformers of rating up to 1.5MVA. This method is adopted in the places where fire is a great hazard.

    Air Blast


    ir Blast (AB)

    In this method, the transformer is cooled by circulating continuous blast of cool air through the core and the windings. For this external fans are used. The air supply must be filtered to prevent accumulation of dust particles in the ventilating ducts.

    2. Oil cooling (Oil immersed transformers)

    In this method, heat is transferred to the oil surrounding the core and windings and it is conducted to the walls of the transformer tank. Finally, the heat is transferred to the surrounding air by radiation and convection.

    Oil coolant has two distinct advantages over the air coolants.

    • It provides better conduction than the air
    • High coefficient of conduction which results in the natural circulation of the oil.

    ONAN

    Oil Natural Air natural (ONAN)

    The transformer is immersed in oil and the heat generated in the cores and the windings is passed on to oil by conduction. Oil in contact with the surface of windings and core gets heated up and moves towards the top and is replaced by the cool oil from the bottom. The heated oil transfers its heat to the transformer tank through convection and which in turn transfers the heat to the surrounding air by convection and radiation.

    This method can be used for the transformers having the ratings up to 30MVA. The rate of heat dissipation can be increased by providing fins, tubes and radiator tanks. Here the oil takes the heat from inside the transformer and the surrounding air takes away the heat from the tank. Hence it can also be called as Oil Natural Air natural (ONAN) method.

    ONAF


    Oil Forced Air Forced (OFAF)

    In this method, the oil is cooled in the cooling plant using air blast produced by the fans. These fans need not be used all the time. During low loads, fans are turned off. Hence the system will be similar to that of Oil Natural Air natural (ONAN). At higher loads, the pumps and fans are switched on, and the system changes to Oil Forced Air Forced (OFAF). Automated switching methods are used for this conversion such that as soon as the temperature reaches a certain level, the fans are automatically switched on by the sensing elements. This method increases the system efficiency. This is a flexible method of cooling in which up to 50% of rating ONAN can be used, and OFAF can be used for higher loads. This method is used in transformers having ratings above 30MVA.

    3. Oil and Water Cooling

    In this method along with oil cooling, water is circulated through copper tubes which enhance the cooling of transformer. This method is normally adopted in transformers with capacities in the order of several MVA.

    OFWF


    Oil Forced Water Forced (OFWF)

    In this method, copper cooling coils are mounted above the transformer core. The copper coils will be fully immersed in the oil. Along with the oil natural cooling the heat from the core passes to the copper coils and the circulating water inside the copper coil takes away the heat. The disadvantage in this method is that since water enters inside the transformer any kind of leakage will contaminate the transformer oil. Since heat passes three times as rapidly from copper cooling tube to water as from oil to copper tubes, the tubes are provided with fans to increase the conduction of heat from oil to tubes. The water inlet and outlet pipes are lagged in order to prevent the moisture in the ambient air fro condensing on the pipes and getting into the oil.

    Oil Forced Water Forced (OFWF)

    In this method, hot oil is passed though a water heat exchanger. The pressure of the oil is kept higher than that of the water. Therefore, there will be leakage from oil to the water alone, and the vise verse is avoided. This method of cooling is employed in the cooling of transformers with very larger capacity in the order of hundreds of MVA. This method is suitable for banks of transformers. Maximum of three transformers can be connected in a single pump circuit. Advantages of this method over ONWF are that the transformer size is smaller and the water does not enter into the transformer. This method is widely used for the transformers designed for hydro electric plants.











    6.8. Describe the methods of cooling system the transformer.

    6.8

    6.9. Narrate the transformer oil and its properties.

    Properties of new insulating oil filled in new power transformer ...

    TRANSFORMER OIL MAINTENANCE - PDF Free Download

    Transformer Oil

    Properties of used transformer oil before and after activated Bentonite treatment.

    Which oil is used in transformers?
    Mineral oil and Synthetic oil are the majorly used transformer oil. These are the petroleum products, like Naphthenic based transformer oil and Paraffinic based transformer oil. Naphthenic based transformer oils are known for their heat distribution, which is one of the main problems with transformer.
    PROPERTIES: The transformers oil shall be of low viscosity and shall offer the minimum resistance and maximum convective assistance to the flow. ... The oil shall have high electrical strength, good impulse
     strength and good are quenching properties
    Transformer oil (also known as insulating oil) is a special type of oil which has excellent electrical 
    insulating properties and is stable at high temperatures. Transformer oil is used in oil-filled electrical power transformers to insulate, stop arcing and corona discharge, and to dissipate the heat of the transformer (i.e. act as a coolant).

    There are two main types of transformer oil used in transformers:

    1. Paraffin based transformer oil
    2. Naphtha based transformer oil

    Naphtha oil is more easily oxidized than paraffin oil. But the product of oxidation – i.e. sludge – in the naphtha oil is more soluble than the sludge from the paraffin oil. Thus sludge of naphtha-based oil is not precipitated in the bottom of the transformer. Hence it does not obstruct convection circulation of the oil, means it does not disturb the transformer cooling system.

    Although Paraffin oil has a lower oxidation rate than Naphtha oil, the oxidation product (sludge) is insoluble and precipitated at the bottom of the tank. This sludge acts as an obstruction to the transformer cooling system.

    The properties (or parameters) of transformer oil are:

    1. Electrical properties: Dielectric strength, specific resistance, dielectric dissipation factor.
    2. Chemical properties: Water content, acidity, sludge content.
    3. Physical properties: Interfacial tension, viscosity, flash point, pour point.

    Electrical Properties of Transformer Oil

    a.  Dielectric Strength of Transformer Oil

    The dielectric strength of transformer oil is also known as the breakdown voltage (BDV) of transformer oil.Minimum breakdown voltage of transformer oil or dielectric strength of transformer oil at which this oil can safely be used in transformer, is considered as 30 KV.

    b. Specific Resistance of Transformer Oil

    This is another important property of transformer oil. The specific resistance of oil is a measure of DC resistance between two opposite sides of one cm3 block of oil. Its unit is ohm-cm at a specific temperature. With increase in temperature the resistivity of oil decreases rapidly.

    the temperature will be very high and may go up to 90oC at an overload condition. So resistivity of the insulating oil must be high at room temperature and also it should have good value at high temperature as well.
    That is why specific resistance or resistivity of transformer oil should get measured at 27oC as well as 90oC.

    Minimum standard specific resistance of transformer oil at 90oC is 35 × 1012 ohm–cm and at 27oC it is 1500 × 1012 ohm–cm.

    Chemical Properties of Transformer Oil

    Water Content in Transformer Oil

    Moisture or water content in transformer oil is highly undesirable as it affects the dielectric properties of the oil adversely. The water content in oil also affects the paper insulation of the core and winding of a transformer. Paper is highly hygroscopic. Paper absorbs the maximum amount of water from oil which affects paper insulation property as well as reduced its life. But in a loaded transformer, oil becomes hotter, hence the solubility of water in oil increases.

    he water content in oil is allowed up to 50 ppm as recommended by IS–335(1993). The accurate measurement of water content at such low levels requires very sophisticated instrument like Coulometric Karl Fisher Titrator.

    Acidity of Transformer Oil

    Acidic transformer oil is a harmful property. If oil becomes acidic, the water content in the oil becomes more soluble in the oil. The acidity of oil deteriorates the insulation property of paper insulation of winding. Acidity accelerates the oxidation process in the oil. Acid also includes rusting of iron in the presence of moisture.

    Physical Properties of Transformer Oil

    Inter Facial Tension of Transformer Oil

    Interfacial tension between the water and oil interface is the way to measure the attractive molecular force between water and oil. in Dyne/cm or milli-Newton/meter. Interfacial tension is exactly useful for determining the presence of polar contaminants and oil decay products. Good new oil generally exhibits high interfacial tension. Oil oxidation contaminants lower the IFT.

    lash Point of Transformer Oil

    Flash point of transformer oil is the temperature at which oil gives enough vapors to produce a flammable mixture with air. This mixture gives momentary flash on the application of flame under standard condition. Flashpoint is important because it specifies the chances of fire hazard in the transformer. So it is desirable to have a very high flash point of transformer oil. In general it is more than 140o(>10o).

    Pour Point of Transformer Oil

    It is the minimum temperature at which oil starts to flow under standard test condition. Pour point of transformer oil is a valuable property mainly at the places where the climate is icy. If the oil temperature falls below the pour point, transformer oil stops convection flowing and obstruct cooling in a transformer. Paraffin-based oil has a higher value of pour point, compared to Naphtha based oil, but in India like country, it does not affect the use of Paraffin oil due to its warm climate condition. Pour Point of transformer oil mainly depends upon wax content in the oil. As Paraffin-based oil has more wax content, it has higher pour point.

    Viscosity of Transformer Oil

    In few words, the viscosity of transformer oil can be said that viscosity is the resistance of flow, in normal condition. Resistance to flow of transformer oil means obstruction of convection circulation of oil inside the transformer. Good oil should have a low viscosity so that it offers less resistance to the conventional flow of oil thereby not affecting the cooling of a transformer. Low viscosity of transformer oil is essential, but it is equally important that the viscosity of oil should increase as less as possible with a decrease in temperature. Every liquid becomes more viscous if the temperature decreases.

    The acidity test of transformer oil can be used to measure the acidic constituents of contaminants. We express the acidity of oil in mg of KOH required to neutralize the acid present in a gram of oil. This is also known as neutralization number.

    Transformer Oil Testing

    Transformer oil needs to be tested to ensure that it works for today’s standards. Testing standards and procedures are defined by various international standards, and most of them are set by the ASTM.

    Oil testing consists of measuring the breakdown voltage, and other chemical and physical properties of the oil, either through portable test equipment or in a laboratory. Through proper testing, the transformer’s lifespan is increased, reducing the need to pay for replacement.

    What Factors Are Tested

    Here are the most common things to look for when performing a transformer oil test:

    • Standard Specification for Mineral Insulating Oil Used in Electrical Apparatus (ASTM D3487)
    • Acid number (ASTM D664)
    • Dielectric breakdown voltage (ASTM D877)
    • Liquid power factor (ASTM D924-08)
    • Interfacial tension (ASTM D971)
    • Specific resistance (ASTM D1169)
    • Corrosive sulfur (ASTM D1275)
    • Visual examination (ASTM D1524)

    Note: ASTM stands for the American Society for Testing and Materials.

    Common Problems When Testing

    The table below shows the most common issues that can occur when testing transformer oil:

    FaultKey GasResults
    Corona dischargeHydrogenLow energy discharges create methane and hydrogen and smaller quantities of ethylene and ethane.
    ArcingAcetyleneLarge amounts of hydrogen or acetylene or minor quantities of ethylene and methane can be produced.
    Overheated CelluloseCarbon MonoxideIf cellulose is overheated, then it will produce carbon monoxide
    Overheated OilMethane and EthyleneOverheating oil will produce methane and ethylene (300 degrees F) or methane and hydrogen (1,112 degrees F). Traces of acetylene might be created if the unit has electrical contacts or if the problem is severe.


    7.1. Describe the construction of three phase transformer.


    A three phase transformer is a three-legged iron core. Each leg has a respective primary and secondary winding. Most power is dispersed in the form of three-phase AC. Basically the power company generators produce electricity by rotating 3 coils or windings through a magnetic field within the generator.
    Three Phase transformers are widely used as Power transformers, Distribution transformers and in Electrical Grids. When we generate the Power using an alternator(AC), the voltage at which it is generated is of mostly 11KV( Sometimes a bit more than that but not too high due to insulation constraints).

    7.2. List various methods of connection of 3-phase transformer and their applications.

    Three-Phase Transformer Connections

    The three phase transformer consists three transformers either separate or combined with one core. The primary and secondary of the transformer can be independently connected either in star or delta. There are four possible connections for a 3-phase transformer bank.

    1. Δ – Δ (Delta – Delta) Connection
    2. Υ – Υ (Star – Star) Connection
    3. Δ – Υ (Delta – Star) Connection
    4. Υ – Δ (Star – Delta ) Connection

    The choice of connection of three phase transformer depends on the various factors likes the availability of a neutral connection for grounding protection or load connections, insulation to ground and voltage stress, availability of a path for the flow of third harmonics, etc. The various types of connections are explained below in details.



    7.3. Describe the methods of star–star, delta–delta, star–delta and delta–star connection.

    1. DELTA-DELTA (Δ-Δ) CONNECTION

    The delta-delta connection of three identical single phase transformer is shown in the figure below. The secondary winding a1a2 is corresponding to the primary winding A1A2, and they have the same polarity. The polarity of the terminal a connecting a1 and c2 is same as that connecting A1 and C2. The figure below shows the phasor diagram for lagging power factor cosφ.



    The magnetising current and voltage drops in impedances have been neglected. Under the balanced condition, the line current is √3 times the phase winding current. In this configuration, the corresponding line and phase voltage are identical in magnitude on both primary and secondary sides.

    The secondary line-to-line voltage is in phase with the primary line-to-line voltage with a voltage ratio equal to the turns ratio.

    If the connection of the phase windings is reversed on either side, the phase difference of 180° is obtained between the primary and the secondary system. Such a connection is known as an 180º connection.

    The delta-delta connection with 180º phase shift is shown in the figure below. The phasor diagram of a three phase transformer shown that the secondary voltage is in phase opposition with the primary voltage.


    The delta-delta transformer has no phase shift associated with it and problems with unbalanced loads or harmonics.

    ADVANTAGES OF DELTA–DELTA CONNECTION OF TRANSFORMER

    The following are the advantages of the delta-delta configuration of transformers.

    1. The delta-delta transformer is satisfactory for a balanced and unbalanced load.
    2. If one transformer fails, the remaining two transformers will continue to supply the three-phase power. This is called an open delta connection.
    3. If third harmonics present, then it circulates in a closed path and therefore does not appear in the output voltage wave.

    The only disadvantage of the delta-delta connection is that there is no neutral. This connection is useful when neither primary nor secondary requires a neutral and the voltage are low and moderate.

    2. STAR-STAR (Υ-Υ) CONNECTION OF TRANSFORMER

    The star-star connection of three identical single phase transformer on each of the primary and secondary of the transformer is shown in the figure below.The phasor diagram is similar as in delta-delta connection.


    The phase current is equal to the line current, and they are in phase. The line voltage is three times the phase voltage. There is a phase separation of 30º between the line and phase voltage.The 180º phase shift between the primary and secondary of the transformer is shown in the figure above.

    PROBLEMS ASSOCIATED WITH STAR-STAR CONNECTION

    The star-star connection has two very serious problems. They are

    1. The Y-Y connection is not satisfactory for the unbalance load in the absence of a neutral connection. If the neutral is not provided, then the phase voltages become severely unbalance when the load is unbalanced.
    2. The Y-Y connection contains a third harmonics, and in balanced conditions, these harmonics are equal in magnitude and phase with the magnetising current. Their sum at the neutral of star connection is not zero, and hence it will distort the flux wave which will produce a voltage having a harmonics in each of the transformers

    The unbalanced and third harmonics problems of Y-Y connection can be solved by using the solid ground of neutral and by providing tertiary windings.

    3. DELTA-STAR (Δ-Υ) CONNECTION

    The ∆-Y connection of the three winding transformer is shown in the figure below. The primary line voltage is equal to the secondary phase voltage. The relation between the secondary voltages is

     VLS= √3 VPS.

    delta-star-connection-of-transformer

    The phasor diagram of the ∆-Y connection of the three phase transformer is shown in the figure below. It is seen from the phasor diagram that the secondary phase voltage Van leads the primary phase voltage  VAN by 30°. Similarly, Vbn leads VBN by 30º and Vcn leads VCN by 30º.This connection is also called +30º connection.


    By reversing the connection on either side, the secondary system voltage can be made to lag the primary system by 30°. Thus, the connection is called -30° connection.

    4. STAR-DELTA (Υ-Δ) CONNECTION

    The star-delta connection of three phase transformer is shown in the figure above. The primary line voltage is √3 times the primary phase voltage. The secondary line voltage is equal to the secondary phase voltage. The voltage ratio of each phase is



    Therefore line-to-line voltage ratio of Y-∆ connection is



    The phasor diagram of the configuration is shown in the figure above. There is a phase shift of 30 lead exists between respective phase voltage. Similarly, 30° leads exist between respective phase voltage. Thus the connection is called +30º connection.

    The phase shows the star-delta connection of transformer for a phase shift of 30° lag. This connection is called – 30° connection. This connection has no problem with the unbalanced load and thirds harmonics. The delta connection provided balanced phase on the Y side and provided a balanced path for the circulation of third harmonics without the use of the neutral wire.



    7.4. Outline open delta connection or V-V connection.

    OPEN DELTA OR V-V CONNECTION

    If one transformer of delta-delta connection is damaged or accidentally opened, then the defective transformer is removed, and the remaining transformer continues to work as a three phase bank. The rating of the transformer bank is reduced to 58% of that of the actual bank. This is known as the open delta or V-V delta. Thus, in open winding transformer, two transformers are used instead of three for the 3-phase operation.

    Let the Vab, Vbc and Vca be the voltage applied to the primary winding of the transformer. The voltage induced in the transformer secondary or on winding one is Vab. The voltage induced on the low voltage winding two is Vbc. There is no winding between points a and c. The voltage may be found by applying  KVL around a closed path made up of point a, b, and c. Thus,


    Let, 


    Where Vp is the magnitude of the line on the primary side.


    open-delta-connection-of-transformer

    On substituting the value of Vab and Vbc in equation, we get


    The Vca is equal in magnitude from the secondary terminal voltage and 120º apart in time from both of them. The balanced three phase line voltage produced balanced 3-phase voltage on the secondary side.

    If the three transformers are connected in delta-delta configuration and are supplying rated load and if the connection becomes V-V transformer, the current in each phase winding is increased by √3 times. The full line current flows in each of the two phase windings of the transformer. Thus the each transformer in the V-V system is overloaded by 73.2%.

    It should be noticed that the load should be reduced by √3 times in case of an open delta connected transformer. Otherwise, serious overheating and breakdown of the two transformers may take place.



    7.5. Describe Scott or T-T connection.

    SCOTT-T TRANSFORMER CONNECTION

    Definition: The Scott-T Connection is the method of connecting two single phase transformer to perform the 3-phase to 2-phase conversion and vice-versa. The two transformers are connected electrically but not magnetically.One of the transformers is called the main transformer, and the other is called the auxiliary or teaser transformer.

    The figure below shows the Scott-T transformer connection. The main transformer is centre tapped at D and is connected to the line B and C of the 3-phase side. It has primary BC and secondary a1a2. The teaser transformer is connected to the line terminal A and the centre tapping D. It has primary AD and the secondary b1b2


    The identical, interchangeable transformers are used for Scott-T connection in which each transformer has a primary winding of Tp turns and is provided with tapping at 0.289Tp , 0.5Tpand 0.866 Tp.

    PHASOR DIAGRAM OF SCOTT CONNECTION TRANSFORMER

    The line voltages of the 3-phase system VAB, VBC, and VCA which are balanced are shown in the figure below. The same voltage is shown as a closed equilateral triangle.The figure below shows the primary windings of the main and the teaser transformer.



    The D divides the primary BC of the main transformers into two halves and hence the number of turns in portion BD = the number of turns in portion DC = Tp/2.The voltage VBD and VDC are equal, and they are in phase with VBC.


    THE VOLTAGE BETWEEN A AND D IS




    The teaser transformer has the primary voltage rating that is √3/2 or 0.866 of the voltage ratings of the main transformer. Voltage VAD is applied to the primary of the teaser transformer and therefore the secondary of the voltage V2t of the teaser transformer will lead the secondary terminal voltage V2m of the main transformer by 90º as shown in the figure below.


    Then, 


    For keeping the voltage per turn same in the primary of the main transformer and the primary of the teaser transformer, the number of turns in the primary of the teaser transformer should be equal to √3/2Tp.

    Thus, the secondaries of both transformers should have equal voltage ratings.The V2t and V2m are equal in magnitude and 90º apart in time; they result in the balanced 2-phase system.

    POSITION OF NEUTRAL POINT N

    The primary of the two transformers may have a four wire connection to a 3-phase supply if the tapping N is provided on the primary of the teaser transformer such that

    The voltage across AN = VAN = phase voltage = Vl/√3.

    Since the voltage across the portion AD.

    scott-connection-of-transformer-equation-6

    the voltage across the portion ND

    scott-connection-transformer-equation-7

    The same voltage turn in portion AN, ND and AD are shown by the equations,

    scott-connection-of-transformer-equation-8

    The equation above shows that the neutral point N divides the primary of the teaser transformer in ratio.

    AN : ND = 2 : 1

    APPLICATIONS OF SCOTT CONNECTION

    The following are the applications of the Scott-T connection.

    1. The Scott-T connection is used in an electric furnace installation where it is desired to operate two single-phase together and draw the balanced load from the three-phase supply.
    2. It is used to supply the single phase loads such as electric train which are so scheduled as to keep the load on the three phase system as nearly as possible.
    3. The Scott-T connection is used to link a 3-phase system with a two–phase system with the flow of power in either direction.

    The Scott-T connection permits conversions of a 3-phase system to a two-phase system and vice versa. But since 2-phase generators are not available, the converters from two phases to three phases are not used in practice.



    7.6. Explain the application of V-V and T-T connection.

    This reduces the project cost. An open-delta connection may be used to provide three phase power up to 57.7 % of the actual capacity of the three phase bank. If the three phase load is not reduced while using open-delta, there is high chance that one transformer gets overloaded; which can be more dangerous
    The open-delta, also known as the V-V connection, is a 3-phase arrangement that makes use of only two, instead of three, single-phase transformers, as shown in Fig. ... In the case of three transformers, the line current is times the current in the transformers windings when connected delta and when the load is balanced.
    The main advantage of a V-V is a reduction of cost of equipment and ... In V-V connection the transformer primary and secondary winding are ..

    Open-Delta or V – V connection

    If one of the transformers of a D – D is removed and 3-phase supply is connected to the primaries as shown in Fig. 33.11, then three equal 3-phase voltages will be available at the secondary terminals on no- load. This method of transforming 3-phase power by means of only two transformers is called the open – D or V – V connection.

    It is employed :

    1. when the three-phase load is too small to warrant the installation of full three-phase transformer bank.

    2. when one of the transformers in a D – D bank is disabled, so that service is continued although at reduced capacity, till the faulty transformer is repaired or a new one is substituted.

    3. when it is anticipated that in future the load will increase necessitating the closing of open delta.

    One important point to note is that the total load that can be carried by a V – V bank is not two-third of the capacity of a D – D bank but it is only 57.7% of it. That is a reduction of 15% (strictly, 15.5%) from its normal rating.

    image

    Suppose there is D – D bank of three 10-kVA transformers. When one transformer is removed, then it runs in V – V. The total rating of the two transformers is 20 kVA. But the capacity of the V – V bank is not the sum of the transformer kVA ratings but only 0.866 of it i.e. 20 ´ 0.866 = 17.32 (or 30 ´ 0.57 = 17.3 kVA). The fact that the ratio of V-capacity to D-capacity is 1/ 3 = 57.7% (or nearly 58%) instead of 66 2 per cent can be proved as follows :

    image

    It is obvious from above that when one transformer is removed from a D – D bank.

    Applications of Scott Connection

    It is used to supply the single phase loads such as electric train which are so scheduled as to keep the load on the three phase system as nearly as possible. The Scott-T connection is used to link a 3-phase system with a two–phase system with the flow of power in either direction.

    A teaser transformer is the second transformer of a two transformer set in a Scott-T or T-T transformer bank. The teaser transformer is tapped at 86.6% of the main transformer on the HV set of windings. The two transformers convert a 3-phase input to a 2-phase,

    The two transformers convert a 3-phase input to a 2-phase, 5-wire, output.



    7.7. Draw the connection of 3-phase to 2-phase and vice-versa.

    Connection of 3-phase to 2-phase 

    Transformer,three phase:Three-phase to Two-phase Conversion and ...

    Transformer,three phase:Three-phase to Two-phase Conversion and ...

    Scott-T Connection of Transformer | Electrical Notes & Articles

    CONNECTION OF 2-PHASE TO 3-PHASE 

    Scott Connection - Electrical idea


    Three Phase Transformer





    8.1. Describe auto-transformer.

    Auto Transformer

    An Auto Transformer is a transformer with only one winding wound on a laminated core. An auto transformer is similar to a two winding transformer but differ in the way the primary and secondary winding are interrelated. A part of the winding is common to both primary and secondary sides.

    On load condition, a part of the load current is obtained directly from the supply and the remaining part is obtained by transformer action. An Auto transformer works as a voltage regulator.

    Explanation of Auto Transformer with Circuit Diagram

    In an ordinary transformer, the primary and the secondary windings are electrically insulated from each other but connected magnetically as shown in the figure below. While in auto transformer the primary and the secondary windings are connected magnetically as well as electrically. In fact, a part of the single continuous winding is common to both primary and secondary.

    AUTOTRANSFORMER-TWO-WINDING

    Figure A: Ordinary Two Winding Transformer

    There are two types of auto transformer based on the construction. In one type of transformer, there is continuous winding with the taps brought out at convenient points determined by the desired secondary voltage. However, in another type of auto transformer, there are two or more distinct coils which are electrically connected to form a continuous winding. The construction of Auto transformer is shown in the figure below.

    AUTO TRANSFORMER

    Figure B: Auto – Transformer

    The primary winding AB from which a tapping at C is taken, such that CB acts as a secondary winding. The supply voltage is applied across AB, and the load is connected across CB. The tapping may be fixed or variable. When an AC voltage V1 is applied across AB, an alternating flux is set up in the core, as a result, an emf E1 is induced in the winding AB. A part of this induced emf is taken in the secondary circuit.

    Let,

    • V1 – primary applied voltage
    • V2 – secondary voltage across the load
    • I1 – primary current
    • I2 – load current
    • N1 – number of turns between A and B
    • N2 – number of turns between C and B

    Neglecting no-load current, leakage reactance and losses,

    V1 = E1 and V2 = E2

    Therefore, the transformation ratio:
    auto-transformer-eq1

    As the secondary ampere-turns are opposite to primary ampere-turns, so the current I2 is in phase opposition to I1. The secondary voltage is less than the primary. Therefore current I2 is more than the current I1. Therefore, the resulting current flowing through section BC is (I2 – I1).

    The ampere-turns due to section BC = current x turns
    auto-transformer-eq2Equation (1) and (2) shows that the ampere-turns due to section BC and AC balance each other which is characteristic of the transformer action.

    Saving of Copper in Auto Transformer as Compared to Ordinary Two Winding Transformer

    The weight of the copper is proportional to the length and area of a cross-section of the conductor.

    The length of the conductor is proportional to the number of turns, and the cross-section is proportional to the product of current and number of turns.

    Now, from the above figure (B) shown of the auto transformer, the weight of copper required in an auto transformer is

    Wa = weight of copper in section AC + weight of copper in section CB

    Therefore
    auto-transformer-eq3

    If the same duty is performed with an ordinary two winding transformer shown above in the figure (A), the total weight of the copper required in the ordinary transformer,

    W0 = weight of copper on its primary winding + weight of copper on its secondary winding

    Therefore,
    auto-transformer-eq4

    Now, the ratio of the weight of the copper in an auto transformer to the weight of copper in an ordinary transformer is given as
    auto-transformer-eq5

    Saving of copper affected by using an auto transformer = weight of copper required in an ordinary transformer – weight of copper required in an auto transformer

    auto-transformer-eq6Therefore,

    Saving of copper = K x weight of copper required for two windings of the transformer

    Hence, saving in copper increases as the transformation ratio approaches unity. Hence the auto transformer is used when the value of K is nearly equal to unity.



    8.2. Explain the terms transformed power and conducted power.

    An autotransformer is a kind of electrical transformer where primary and secondary shares same common single winding. So basically it’s a one winding transformer.

    auto transformer


    8.3. List the advantages and disadvantages of auto-transformer.

    ADVANTAGES OF AUTO TRANSFORMER

    • Less costly
    • Better regulation
    • Low losses as compared to ordinary two winding transformer of the same rating.

    DISADVANTAGES OF AUTO TRANSFORMER

    There are various advantages of the auto transformer, but then also one major disadvantage, why auto transformer is not widely used, is that

    • The secondary winding is not insulated from the primary winding.
      If an auto transformer is used to supply low voltage from a high voltage and there is a break in the secondary winding, the full primary voltage comes across the secondary terminal which is dangerous to the operator and the equipment. So the auto transformer should not be used for interconnecting high voltage and low voltage systems.
    • Used only in the limited places where a slight variation of the output voltage from input voltage is required.

    APPLICATIONS OF AUTO TRANSFORMER

    • It is used as a starter to give up to 50 to 60% of full voltage to the stator of a squirrel cage induction motor during starting.
    • It is used to give a small boost to a distribution cable, to correct the voltage drop.
    • It is also used as a voltage regulator
    • Used in power transmission and distribution system and also in the audio system and railways.


    8.4. Convert a Two-winding transformer to auto-transformer.

    Conversion of 2-Winding Transformer into Auto-transformer

    Any two-winding transformer can be converted into an auto-transformer either step-down or step-up. Figure below as shows such a transformer with its polarity markings. Suppose it is a 20-kVA,

    You can convert a two winding transformer into a Auto transformer by connecting its winding in series .But while connecting the windings keep the polarity into account because it will change its capacity.suppose you have a 240/120V two winding transformer by connecting these two windings with additive polarity the voltage at output terminal will be 240+120=360V,then transformer will be step up transformer .Similiarly connecting these two windings in subtractive polarity output votage will be 240-120.


    Two winding transformer to auto-transformer




    8.5. Mention the uses of auto-transformer.

    Applications of Auto transformer

    • It is used as a starter to give up to 50 to 60% of full voltage to the stator of a squirrel cage induction motor during starting.
    • It is used to give a small boost to a distribution cable, to correct the voltage drop.
    • It is also used as a voltage regulator
    • Used in power transmission and distribution system and also in the audio system and railways.

    8.6. Solve problems related to auto-transformer.

    8.6 Prob.

    1. A 5kVA, 200 V/ 100 V, 50 Hz, single phase ideal two winding transformer is to used to step up a voltage of 200 V to 300 V by connecting it like an auto transformer. Show the connection diagram to achieve this. Calculate the maximum kVA that can be handled by the autotransformer (without over loading any of the HV and LV coil). How much of this kVA is transferred magnetically and how much is transferred by electrical conduction.
    Solution
    Two connect a two winding transformer as an auto transformer, it is essential to know the dot markings on the two coils. The coils are to be now series connected appropriately so as to identify clearly between which two terminals to give supply and between which two to connect the load. Since the input voltage here is 200 V, supply must be connected across the HV terminals. The induced voltage in the LV side in turn gets fixed to 100 V. But we require 300 V as output, so LV coil is to be connected in additive series with the HV coil. This is what has been shown in figure 28.10.

    Solution Two connect a two winding transformer as an auto transformer, it is essential to know the dot markings on the two coils. The coils are to be now series connected appropriately so as to identify clearly between which two terminals to give supply and between which two to connect the load. Since the input voltage here is 200 V, supply must be connected across the HV terminals. The induced voltage in the LV side in turn gets fixed to 100 V. But we require 300 V as output, so LV coil is to be connected in additive series with the HV coil. This is what has been shown in figure 28.10. 50 A Figure 28.10: Two winding transformer as an autotransformer. 300 V Load 100 V 25 75 A 200 V supply P A Here the idea is not to exceed the voltage and current rating of HV and LV coils of the two winding transformer. Now for the transformer having rating 5 kVA, 200 V/ 100 V, 50 Hz we have: Rated voltage of HV coil is = 200 V Rated voltage of LV coil is = 100 V Phase turns ratio is a = 200/100 = 2 Rated current of each HV coil is = 5000/200 = 25 A Rated current of each LV coil is = 5000/100 = 50 A Since the load is in series with LV coil, so load current is same as the current flowing through the LV coil. Thus a maximum of 50 A can be drawn by the load otherwise overloading of the coils take place. Output kVA = 300 × 50 VA = 15 kVA input kVA = Output kVA = 15 kVA ∵ transformer is ideal Version 2
    Autotransformer - an overview | ScienceDirect Topics
    Auto Transformer at Rs 5000/piece | Auto Transformers | ID ...
    Solved: Example 1 A1KVA 240/120 V Autotransformer Operates ...


    9.1. Describe the purpose of polarity test.

    What is the use of polarity test?
    Definition: It is a test conducted on both the primary and secondary windings of a transformer and also to know the direction of the current. It helps in inter-connecting the primary winding and secondary windings in the case of the parallel operation of the Transformer
    Two Types of  Polarity Testing of Transformer 
    • Additive, and
    • Subtractive
    Additive Polarity

    The voltmeters V1 and V2 are connected across the primary and secondary windings of the transformer. The voltmeter V1 measures the voltage across the primary winding whereas voltmeter V2 measures the voltage across the secondary winding respectively. Apart from these two voltmeters, a third voltmeter V3 is connected across both the windings I,e primary winding and secondary winding. If V3 reads voltage as the summation of both V1 and V2 then, it is an additive type.

    The polarity of the primary winding and the secondary winding is different. The terminals A1 and B2 have the same polarity whereas A2 and B1 have the same polarity. The figure which explains the Additive type polarity is shown in the figure below.

    Additive_polarity

    Additive Polarity

    Subtractive Polarity

    The voltmeters V1 and V2 are connected across the primary and secondary windings of the transformer. The voltmeter V1 measures the voltage across the primary winding whereas voltmeter V2 measures the voltage across the secondary winding respectively. Apart from these two voltmeters, a third voltmeter V3 is connected across both the windings I,e primary winding and secondary winding. If V3 reads voltage as the difference of both V1 and V2 then, it is as subtractive type.

    subtractive_polarity

    Subtractive Polarity

    Polarity Test of Transformer

    Polarity means the direction of the induced voltages in the primary and the secondary winding of the transformer. If the two transformers are connected in parallel, then the polarity should be known for the proper connection of the transformer. There are two types of polarity one is Additive, and another is Subtractive.


    Additive Polarity: In additive polarity, the same terminals of the primary and the secondary windings of the transformer are connected

    Subtractive Polarity: In subtractive polarity, different terminals of the primary and secondary side of the transformer is connected.
    Explanation With Connection Diagram

    Each of the terminals of the primary, as well as the secondary winding of a transformer, is alternatively positive and negative with respect to each other as shown in the figure below. Let  A1 and A2 be the positive and negative terminal, respectively of the primary side of the transformer and a1, a2 are the positive and negative terminal of the secondary side of the transformer.

    If A1 is connected to a1 and A2 is connected to a2 that means similar terminals of the transformer are connected, then the polarity is said to be additive. If A1 is connected to  a2 and A2 to a1, that means the opposite terminals are connected to each other, and thus the voltmeter will read the subtractive polarity.


    It is essential to know the relative polarities at any instant of the primary and the secondary terminals for making the correct connections if the transformers are to be connected in parallel or they are used in a three-phase circuit.

    In the primary side, the terminals are marked as A1 and A2 and from the secondary side, the terminals are named as a1 and a2. The terminal A1 is connected to one end of the secondary winding, and a voltmeter is connected between A2 and the other end of the secondary winding.

    When the voltmeter reads the difference that is (V1 – V2), the transformer is said to be connected with opposite polarity know as subtractive polarity and when the voltmeter reads (V1 + V2), the transformer is said to have additive polarity.

    Steps to Perform Polarity Test
    • Connect the circuit as shown in the above circuit diagram figure and set the autotransformer to zero position.
    • Switch on the single-phase supply
    • Record the values of the voltages as shown by the voltmeter V1, V2 and V3.
    • If the reading of the V3 shows the addition of the value of V1 and V2 that is V2 = V1+V2 the transformer is said to be connected in additive polarity.
    • If the reading of the V3 is the subtraction of the readings of V1 and V2, then the transformer is said to be connected in subtractive or negative polarity.

    This is all about polarity test of transforme

    9.2. Describe the subtractive and additive polarity.

    • If the X1 terminal is directly across from the H1 terminal, the transformer has subtractive polarity.
    • If the X1 terminal sits diagonally across from the H1 terminal, the transformer has additive polarity.

    We can categorise the polarity of the transformer to two types,

    1. Additive Polarity
    2. Subtractive Polarity
    Transformer Polarity

    Additive Polarity

    In additive polarity, the voltage (Vc) between the primary side (Va) and the secondary side (Vb) will be the sum of both high voltage and the low voltage, i.e. we will get Vc = Va + Vb

    Subtractive Polarity

    In subtractive polarity, the voltage (Vc) between the primary side (Va) and the secondary side (Vb) will be the difference of both high voltage and the low voltage, i.e. we will get Vc = Va – Vb

    In subtractive polarity, if Vc = Va – Vb, it is a step-down transformer and if Vc = Vb – Va, it is a step-up transformer.

    We use additive polarity for small-scale distribution transformers and subtractive polarity for large-scale transformers.


    9.3. Illustrate the test to determine the polarity of a transformer

    PROCEDURE OF POLARITY TEST OF TRANSFORMER

    Transformer Polarity Test

    1. Connect the circuit as shown above with a voltmeter (Va) across primary winding and another voltmeter (Vb) across the secondary winding.
    2. If available, take down the ratings of the transformer and the turn ratio.
    3. We connect a voltmeter (Vc) between primary and secondary windings.
    4. We apply some voltage to the primary side.
    5. By checking the value in the voltmeter (Vc), we can find whether it is additive or subtractive polarity.

    If additive polarity – Vc should be showing the sum of Va and Vb.
    If subtractive polarity – Vc should be showing the difference between Va and Vb.
    Caution: Be careful that the max. measuring the voltage of voltmeter Vc should be greater than the sum of Va (Primary winding) and Vb (Secondary winding) otherwise during the additive polarity, the sum of Va and Vb comes across it.



    Transformer Polarity test, Additive, Subtractive ,Procedure,diagram

    9.4. Explain the purpose of parallel operation.

    Parallel Operation of a Transformer

    The Transformer is said to be in Parallel Operation when their primary windings are connected to a common voltage supply, and the secondary windings are connected to a common load. The connection diagram of the parallel operation of a transformer is shown in the figure below.


    Reasons For Parallel Operation

    Parallel operation of a transformer is necessary because of the following reasons are given below

    • It is impractical and uneconomical to have a single large transformer for heavy and large loads. Hence, it will be a wise decision to connect a number of transformers in parallel.
    • In substations, the total load required may be supplied by an appropriate number of the transformer of standard size. As a result, this reduces the spare capacity of the substation.
    • If the transformers are connected in parallel, so there will be scope in future, for expansion of a substation to supply a load beyond the capacity of the transformer already installed.
    • If there will be any breakdown of a transformer in a system of transformers connected in parallel, there will be no interruption of power supply, for essential services.
    • If any of the transformer from the system is taken out of service for its maintenance and inspection, the continuity of the supply will not get disturbed.




    9.5. List the conditions for parallel operation.

    • Necessary Conditions For Parallel Operation

      For the satisfactory parallel operation of the transformer, the two main conditions are necessary. One is that the Polarities of the transformers must be same. Another condition is that the Turn Ratio of the transformer should be equal.

      The other two desirable conditions are as follows:-
    • The voltage at full load across the transformer internal impedance should be equal.
    • The ratio of their winding resistances to reactances should be equal for both the transformers. This condition ensures that both transformers operate at the same power factor, thus sharing their active power and reactive volt-amperes according to their ratings.
    • For the satisfactory parallel operation of the transformer, the two main conditions are necessary. One is that the Polarities of the transformers must be same. Another condition is that the Turn Ratio of the transformer should be equal. The voltage at full load across the transformer internal impedance should be equal.
    • Transformer Parallel Operation | Electrical4u

    9.6. Describe the parallel operation of transformers with equal voltage ratio.

    Why do transformers connected in parallel have the same voltage ratio?
    Explanation: The transformers must have the same voltage-ratio to avoid no-load circulating current when transformers are in parallel on both primary and secondary sides. Since the leakage impedance is low, even a small voltage difference can give rise to considerable no-load circulating current and extra I2R loss.
    Condition for Parallel Operation of Transformer

    For parallel connection of transformers, primary windings of the Transformers are connected to source bus-bars and secondary windings are connected to the load bus-bars.

    Various conditions that must be fulfilled for the successful parallel operation of transformers:

    1. Same voltage and Turns Ratio (both primary and secondary voltage rating is same)
    2. Same Percentage Impedance and X/R ratio
    3. Identical Position of Tap changer
    4. Same KVA ratings
    5. Same Phase angle shift (vector group are same)
    6. Same Frequency rating
    7. Same Polarity
    8. Same Phase sequence

    Some of these conditions are convenient and some are mandatory.

    The convenient conditions are: Same voltage Ratio and Turns Ratio, Same Percentage Impedance, Same KVA Rating, Same Position of Tap changer.

    The mandatory conditions conditions are: Same Phase Angle Shift, Same Polarity, Same Phase Sequence and Same Frequency. When the convenient conditions are not met paralleled operation is possible but not optimal.

    Conditions for Parallel Operation of Transformers

    When two or more transformers run in parallel, they must satisfy the following conditions for satisfactory performance. These are the conditions for parallel operation of transformers.

    1. Same voltage ratio of transformer.
    2. Same percentage impedance.
    3. Same polarity.
    4. Same phase sequence.

    Example Connecting two 2000 kVA, 5.75% impedance transformers in parallel, each with the same turn ratios to a 4000 kVA load.

    • Loading on the transformers-1 =KVA1=[( KVA1 / %Z) / ((KVA1 / %Z1)+ (KVA2 / %Z2))]X KVAl
    • kVA1 = 348 / (348 + 348) x 4000 kVA = 2000 kVA.
    • Loading on the transformers-2 =KVA1=[( KVA2 / %Z) / ((KVA1 / %Z1)+ (KVA2 / %Z2))]X KVAl
    • kVA2 = 348 / (348 + 348) x 4000 kVA = 2000 kVA
    • Hence KVA1=KVA2=2000KVA

    • Example Connecting 3000 kVA and 1000 kVA transformers in parallel, each with 5.75% impedance, each with the same turn ratios, connected to a common 4000 kVA load.

      • Loading on Transformer-1=kVA1 = 522 / (522 + 174) x 4000 = 3000 kVA
      • Loading on Transformer-1=kVA2 = 174 / (522 + 174) x 4000 = 1000 kVA

      From above calculation it is seen that different kVA ratings on transformers connected to one common load, that current division causes each transformer to only be loaded to its kVA rating. The key here is that the percent impedance are the same.


    • Example Two 2000 kVA transformers in parallel, one with 5.75% impedance and the other with 4% impedance, each with the same turn ratios, connected to a common 3500 kVA load.

      • Loading on Transformer-1=kVA1 = 348 / (348 + 500) x 3500 = 1436 kVA
      • Loading on Transformer-2=kVA2 = 500 / (348 + 500) x 3500 = 2064 kVA

      It can be seen that because transformer percent impedances do not match, they cannot be loaded to their combined kVA rating. Load division between the transformers is not equal. At below combined rated kVA loading, the 4% impedance transformer is overloaded by 3.2%, while the 5.75% impedance transformer is loaded by 72%.

      xample Two transformers in parallel with one 3000 kVA (kVA1) with 5.75% impedance, and the other a 1000 kVA (kVA2) with 4% impedance, each with the same turn ratios, connected to a common 3500 kVA load.

      • Loading on Transformer-1=kVA1 = 522 / (522 + 250) x 3500 = 2366 kVA
      • Loading on Transformer-2=kVA2 = 250 / (522 + 250) x 3500 = 1134 kVA

      Because the percent impedance is less in the 1000 kVA transformer, it is overloaded with a less than combined rated load.


    • Example Two 2000 kVA transformers connected in parallel, each with 5.75% impedance, same X/R ratio (8), transformer 1 with tap adjusted 2.5% from nominal and transformer 2 tapped at nominal. What is the percent circulating current (%IC)

      • %Z1 = 5.75, So %R’ = %Z1 / √[(X/R)2 + 1)] = 5.75 / √((8)2 + 1)=0.713
      • %R1 = %R2 = 0.713
      • %X1 = %R x (X/R)=%X1= %X2= 0.713 x 8 = 5.7
      • Let %e = difference in voltage ratio expressed in percentage of normal and k = kVA1/ kVA2
      • Circulating current %IC = %eX100 / √ (%R1+k%R2)2 + (%Z1+k%Z2)2.
      • %IC = 2.5X100 / √ (0.713 + (2000/2000)X0.713)2 + (5.7 + (2000/2000)X5.7)2
      • %IC = 250 / 11.7 = 21.7

      The circulating current is 21.7% of the full load current.


    • Example Two transformers connected in parallel, 2000 kVA1 with 5.75% impedance, X/R ratio of 8, 1000 kVA2 with 4% impedance, X/R ratio of 5, 2000 kVA1 with tap adjusted 2.5% from nominal and 1000 kVA2 tapped at nominal.

      • %Z1 = 5.75, So %R’ = %Z1 / √[(X/R)2 + 1)] = 5.75 / √((8)2 + 1)=0.713
      • %X1= %R x (X/R)=0.713 x 8 = 5.7
      • %Z2= 4, So %R2 = %Z2 /√ [(X/R)2 + 1)]= 4 / √((5)2 + 1) =0.784
      • %X2 = %R x (X/R)=0.784 x 5 = 3.92
      • Let %e = difference in voltage ratio expressed in percentage of normal and k = kVA1/ kVA2
      • Circulating current %IC = %eX100 / √ (%R1+k%R2)2 + (%Z1+k%Z2)2.
      • %IC = 2.5X100 / √ (0.713 + (2000/2000)X0.713)2 + (5.7 + (2000/2000)X5.7)2
      • %IC = 250 / 13.73 = 18.21.

      The circulating current is 18.21% of the full load current.


    9.7. Explain the specification on the name plate of a transformer.

    TRANSFORMER NAME PLATE - YouTube

    Transformer Name Plate Details Fully Explained | Electrical4u

    How Is The Transformer Nemeplate Viewed | Transformer Nameplate ...


    9.8. Solve problems related to parallel operation.

    Transformer,three phase:Parallel Operation of 3-phase Transformers ...

    Transformer,three phase:Parallel Operation of 3-phase Transformers ...

    10. Realize the principle and construction of 3-phase induction motor.

    Realize the principle and construction of 3-phase induction motor.


    10.1. Explain the general principle of induction motor.

       BASIC PRINCIPLES
    Motors work through the principles of electromagnetism. If you run electricity through a wire, it creates a magnetic field. If you coil the wire around a rod and run electricity through the wire, it creates a magnetic field around the rod. One end of the rod will have a north magnetic pole and the other will have a south pole. Opposite poles attract one another, like poles repel. When you surround that rod with other magnets, the rod will rotate from the attractive and repulsive forces.
    Working Principle of an Induction Motor

    The motor which works on the principle of electromagnetic induction is known as the induction motor. The electromagnetic induction is the phenomenon in which the electromotive force induces across the electrical conductor when it is placed in a rotating magnetic field.

    An induction motor (also known as an asynchronous motor) is a commonly used AC electric motor. In an induction motor, the electric current in the rotor needed to produce torque is obtained via electromagnetic induction from the rotating magnetic field of the stator winding. The rotor of an induction motor can be a squirrel cage rotor or wound type rotor.

    Induction Motors are the most commonly used motors in many applications. These are also called as Asynchronous Motors, because an induction motor always runs at a speed lower than synchronous speed. Synchronous speed means the speed of the rotating magnetic field in the stator.

    There basically 2 types of induction motor depending upon the type of input supply - (i) Single phase induction motor and (ii) Three phase induction motor.

    Basic working principle of an Induction Motor
    In a DC motor, supply is needed to be given for the stator winding as well  as the rotor winding. But in an induction motor only the stator winding is fed with an AC supply.
    Induction motor works on the principle of induction where electro-magnetic field is induced into the rotor when rotating magnetic field of stator cuts the stationary rotor. Induction machines are by far the most common type of motor used in industrial, commercial or residential settings. It is a three phase AC motor.
    An AC motor works by applying alternating current to stator windings, which produce a rotating magnetic field. Because the magnetic field rotates in this way, an AC motor does not need power or mechanical aid to be applied to the rotor.

    10.2. Distinguish between induction motor and conduction motor.

    Difference Between Conduction and Induction

    The major difference between conduction and induction is that conduction allows the charging of a neutral body with a charged body by forming a direct contact with it. While induction is the process of charging a neutral body by a charged body without making any direct contact.

    Both conduction and induction refers to the two different ways that causes charging of neutral bodies. Here we will see what major differentiating factors exist between the two.

    Definition of Conduction

    Conduction is a phenomenon of transferring of energy from a charged body to an uncharged body by direct contact.

    conduction

    So, due to direct contact, the current flowing through one conductor (i.e., rod), begins to flow through the neutral body placed in contact as well.

    Definition of Induction

    A phenomenon that causes an uncharged body to get electrically charged by placing it near a charged conductor is known as induction.

    For induction to take place, the two bodies must be placed nearer to each other.

    And according to Faraday when an uncharged body is placed in a region where the magnetic field is present then the electric field gets generated inside it.

    induction

    More simply we can say charges of opposite polarity present in the charged body get induced to the uncharged body.

    Thus the generated electric field in the second body causes flow of current through it in the opposite direction.

    Key Differences Between Conduction and Induction
    1. Conduction is the process of transferring charges from a charged body to a neutral body. While Induction is the process of inducing the charges on a neutral body by the use of a charged body.
    2. Conduction needs direct physical contact between the two bodies. However, no physical contact is needed in case of induction, but the two bodies must be placed close to each other.
    3. In the case of conduction, the current flowing through both bodies have the same direction. Whereas in case of induction, equal current flows through both the bodies but in different directions.
    4. A potential difference must exist between two bodies for conduction to take place thus requires a gradient path. As no direct contact is maintained at the time of induction thus the gradient path is not needed.
    5. As the charge gets transferred in case of conduction thus there is a reduction in the amount of overall charge present in the charged body. While in case of induction the charge on the initially charged body remains the same even after inducing, charges of opposite polarity in the other body.


    Conclusion

    So, this discussion simply concludes that conduction allows the charge flow due to electric field produced inside it. But induction permits flow of charge carriers due to change in magnetic field.





    10.3. List various types of induction motor with their applications.

    types-induction-motor

    There are mainly two types of induction motor on the basis of power supply, Single Phase Induction Motor and Three Phase Induction Motor. As their name suggests, 1 phase AC power supply is extended to single phase induction motor while three phase induction motor is connected to 3 phase AC power supply.

    Types of Single Phase Induction Motor:

    Again, single phase induction motor may be classified on the basis of their construction and starting methods. On this basis, they can be further categorized into following types:

    • Split Phase Induction Motor
    • Capacitor Start Induction Motor
    • Capacitor Start Capacitor Run Induction Motor
    • Shaded Pole Induction Motor
    Types of 3 Phase Induction Motor:

    A three phase induction motor (IM) has two major components, Stator and Rotor. Stator is the stationary part whereas Rotor is rotating part. Load is coupled to the rotor shaft of the motor. Three phase armature winding is wound on the stator. When balanced three phase current flows through this winding, a constant amplitude rotating magnetic field is created in the air gap. This armature winding is connected to the 3 phase power supply and carried the load current.

    Based on the rotor construction, it can be of two types: Squirrel Cage Rotor and Wound Rotor. On this ground, IM is also classified as

    • Squirrel Cage Induction Motor
    • Wound Rotor or Slip Ring Induction Motor
    Types of single phase induction motors

    The single-phase induction motor is started by using some methods. Mechanical methods are not very practical methods that is why the motor is started temporarily by converting it into a two-phase motor.

    Single-phase induction motors are classified according to the auxiliary means used to start the motor. They are classified as follows:

    1. Split-phase motor
    2. Capacitor-start motor
    3. Capacitor-start capacitor-run motor
    4. Permanent-split capacitor (PSC) motor
    5. Shaded-pole motor
    Types of three phase induction motor rotor

    There are two types of induction motor rotors:

    1. Squirrel-cage rotor or simply cage rotor.
    2. Phase wound or wound rotors. The motors that use this type of rotor are known as Slip-ring rotors.



    10.4. Mention different parts of a 3-phase induction motor.



    Basic Parts of a Three Phase (3-Փ) Squirrel Cage Induction Motor ...

    Parts of an Induction Motor

    The stator of the three-phase induction motor consists of three main parts :
    • Stator frame,
    • Stator core,
    • Stator winding or field winding.

    • The A.C. Induction Motor has three main parts, rotor, stator and enclosure. ...
    • The stator is the stationary part of the motor's electromagnetic circuit and is made up of thin metal sheets, called laminations. ...
    • Coils of insulated wire are inserted into slots of the stator core.
    3 phase induction motor construction


    10.5. Describe the construction of stator of an induction motor.

    Construction of Induction Motor - Circuit Globe

    Explain the construction of three phase induction motor with neat ...

    Electric motor - Construction of induction motors | Britannica

    Construction of 3 Phase Induction Motor

    The main body of the Induction Motor comprises of two major parts:

    Stator

    The stator is made up of a number of stampings in which different slots are cut to receive 3 phase winding circuit which is connected to 3 phase AC supply.3 Phase Induction Motor

    The three-phase windings are arranged in such a manner in the slots, that they produce a rotating magnetic field after AC supply is given to them.

    The windings are wound for a definite number of poles depending upon the speed requirement, as speed is inversely proportional to the number of poles, given by the formula:

    Ns= 120f/p

    Where Ns= synchronous speed

    f  = Frequency

    p = no. of poles

    Induction motors work on the principle of electromagnetic induction. Electrical energy from the stator winding is transferred to the rotor winding by electromagnetic induction. Therefore these are called as induction motors.

    Constructional Features of a Three Phase Induction Motor

    A 3-phase induction motor   consists of two main parts namely stator and rotor 

    http://upload.wikimedia.org/wikipedia/commons/thumb/5/58/Stator_and_rotor_by_Zureks.JPG/220px-Stator_and_rotor_by_Zureks.JPG

     Stator

    It is the outer body of the motor and consists of outer frame, stator core and windings.

    Outer frame

    The outer frame acts as housing for the motor and supports the stator core. It also protects the inner parts of the motor. Fins are provided on the outer surface of the frame for heat dissipation and cooling of the motor. Frame is provided with legs/base plate to bolt it on the foundation. Motor housing is the outer cover or frame of the motor which contains stator, rotor and other parts. Fins are provided on the outer frame to increase heat dissipation. Housing can be square ( Fig.)

     or round  (Fig.)

    square housing.jpg http://t1.gstatic.com/images?q=tbn:ANd9GcSA1K46za9f2wsnx0OI4sTG4v4w1VfZeDO0wkYQn6-47S3Ps560

    Depending on the application it can made of any one of the following material:

    a.       Aluminum/ Aluminum alloy 

    b.      Mild Steel 

    c.       Stainless Steel 

    Stator core

    It is made of high grade silicon steel stampings of thickness 0.3 to 0.6 mm which are insulated from each other by a varnish layer. To minimize the hysteresis and eddy current losses core is constructed of steel stampings of high magnetic permeability. The stampings are assembled one over the other under hydraulic pressure and are fixed into the frame. The function of stator core is to carry the alternating magnetic field. Slots are cut on the inner side of the stamping, as shown in fig. 17.9, to accommodate stator winding.


                       fig.  Outer frame

    stator.png

     fig.   Stator Stamping

     
    Stator winding

    Coils of insulated wires are inserted into the slots of the stator. Each grouping of coils, together with the core it surrounds, forms an electromagnet (a pair of poles) on the application of AC supply. The number of poles of an AC induction motor depends on the internal connection of the stator windings. The three phase stator windings are connected directly to the three phase power source. Internally they are connected in such a way, that on applying AC supply, a rotating magnetic field is created. There are six terminals of the stator winding; two for each phase are connected in the terminal box of the motor. Number of poles depends on the speed requirement. For lower speed more number of poles are required as,

     


    10.6. Narrate the construction of squirrel cage rotor, double squirrel cage rotor and phase wound rotor of induction motor.

    The Rotor: Squirrel Cage Rotor:
    A squirrel cage rotor has very low resistance bars cast into the rotor steel which acts as the secondary windings. A slip ring (wound rotor) motor has insulated wire rotor windings which feed out through slip rings to a bank of large variable resistors where the resistance of the rotor circuit can be varied.
    squirrel cage rotor

    Article Imagestator2


    Construction of double squirrel cage rotor
    double squirrel cage motor rotor
    cross section of double squirrel cage rotor
    Rotor of a double squirrel cage motor has two independent cages on the same rotor. The figure at left shows the cross sectional diagram of a double squirrel cage rotor.
    Working of double squirrel cage motor
    At starting of the motor, frequency of induced emf is high because of large slip (slip = frequency of rotor emf / supply frequency). Hence the reactance of inner cage (2πfL    where, f = frequency of rotor emf) will be very high, increasing its total impedance. Hence at starting most of the current flows through outer cage despite its large resistnace (as total impedance is lower than the inner cage). This will not affect the outer cage because of its low reactance. And because of the large resistance of outer cage starting torque will be large.
    As speed of the motor increases, slip decreases, and hence the rotor frequency decreases. In this case, the reactance of inner cage will be low, and most of the current will flow through the inner cage which is having low resistance. Hence giving a good efficiency.
    When the double cage motor is running at normal speed, frequency of the rotor emf is so low that the reactance of both cages is negligible. The two cages being connected in parallel, the combined resistance is lower.
    The torque speed characteristics of double squirrel cage motor for both the cages are shown in the figure below.


    CONSTRUCTION DETAILS

    A three-phase, wound-rotor induction motor consists of a stator core with a three-phase winding, a wound rotor with slip rings, brushes and brush holders, and two end shields to house the bearings that support the rotor shaft.

    ills 1, 2, 3, and 4 show the basic parts of a three-phase, wound-rotor induction motor.


    ill. 1 Parts of a wound-rotor motor


    ill. 2 Wound stator for a polyphase induction motor


    ill. 3 Wound rotor for a polyphase induction motor

    A wound rotor induction motor has a stator like a squirrel cage induction motor, but a rotor with insulated windings brought out via slip rings and brushes.

    However, no power is applied to the slip rings. Their sole purpose is to allow resistance to be placed in series with the rotor windings while starting (figure below). This resistance is shorted out once the motor is started to make the rotor look electrically like the squirrel cage counterpart.

     

    Wound rotor induction motor

    Wound rotor induction motor

    • In Wound Rotor Induction Motor, the rotor has a 3 phase winding similar to stator winding.
    • Rotor is also cylindrical in shape and has slots to carry winding
    • The winding is placed evenly on slots of the rotor.
    •  are connected to 3 slip rings.
    • These slip rings are mounted on the shaft.
    • Each phase is connected to one of the three slip rings. These slip rings are associated with brushes.
    • The three slip rings rotate with rotor, while brushes remain stationary.Slip Ring Rotor Or Wound Rotor In Three Phase Induction Motor ...

    10.7. Explain the purpose of skewing the rotor bars in a squirrel cage rotor.

    Why squirrel cage rotor is skewed?
    In Squirrel cage rotor, slots in lamination or rotor core is not made parallel to the rotor shaft. A slight angle is maintained due to some advantages. This is called the rotor Skew. Skew helps to make the motor run quietly by reducing the magnetic hum.It reduce rotor locking tendency.

    In Squirrel cage rotor, slots in lamination or rotor core is not made parallel to the rotor shaft. A slight angle is maintained due to some advantages. This is called the rotor Skew.


    Skew helps to make the motor run quietly by reducing the magnetic hum.It reduce rotor locking tendency. Rotor locking tendency occurs when rotor teeth remain directly under stator teeth thus they might be magnetically attracted.squirrel-cage-rotor-skew


    10.8. Define slip and slip speed.

    Definition: The slip in an induction motor is the difference between the main flux speed and their rotor speed. The symbol S represents the slip. It is expressed by the percentage of synchronous speed. Mathematically, it is written as


    The value of slip at full load varies from 6% in case of small motor and 2% in the large motor.

    The induction motor never runs at synchronous speed. The speed of the rotor is always less than that of the synchronous speed. If the speed of the rotor is equal to the synchronous speed, no relative motion occurs between the stationary rotor conductors and the main field.

    The no EMF induces in the rotor and zero current generates on the rotor conductors. The electromagnetic torque is also not induced. Thus, the speed of the rotor is always kept slightly less than the synchronous speed. The speed at which the induction motor work is known as the slip speed.

    The difference between the synchronous speed and the actual speed of the rotor is known as the slip speed. In other words, the slip speed shows the relative speed of the rotor concerning the speed of the field.

    The speed of the rotor is slightly less than the synchronous speed. Thus, the slip speed expresses the speed of the rotor relative to the field.

    • If Ns is the synchronous speed in revolution per minute
    • Nr is the actual rotor speed in revolution per minute.

    The slip speed of the induction motor is given as

    SLIP-IN-INDUCTION-MOTOR-EQ-1

    The fraction part of the synchronous speed is called the Per Unit Slip or Fractional Slip. The per unit slip is called the Slip. It is denoted by s.

    SLIP-IN-INDUCTION-MOTOR-EQ-2

    Therefore, the rotor speed is given by the equation shown below.

    SLIP-IN-INDUCTION-MOTOR-EQ-3

    Alternatively, if

    • ns is the synchronous speed in revolution per second
    • nr is the actual rotor speed in revolution per second.

    Then,

    SLIP-IN-INDUCTION-MOTOR-EQ-4

    The percentage slip in revolution per second is given as shown below.

    SLIP-IN-INDUCTION-MOTOR-EQ-5

    The slip of the induction motor varies from 5 percent for small motors to 2 percent for large motors.

    Importance of Slip

    Slip plays an essential role in Induction motor. As we know, the slip speed is the difference between the synchronous and rotor speed of the induction motor. The emf induces in the rotor because of the relative motion, or we can say the slip speed of the motor. So,

    IN-INDUCTION-MOTOR-EQ-6

    The rotor current is directly proportional to the induced emf.

    SLIP-IN-INDUCTION-MOTOR-EQ-7

    The torque is directly proportional to the rotor current.

    SLIP-IN-INDUCTION-MOTOR-EQ-8

    Therefore,

    SLIP-IN-INDUCTION-MOTOR-EQ-9Hence, torque is directly proportional to slip.

    SLIP-IN-INDUCTION-MOTOR-EQ-10

    The above equation shows that the torque induces on the rotor is directly proportional to the slip of the induction motor. The high value of slip induces the emf in the rotor. This EMF develops the heavy torque on the rotor conductors.

    The value of the slip is adjusted by considering the load on the motor. For full-load, the high value of torque is required. This can be achieved by increasing the amount of the slip and reducing the speed of the rotor. The slip of the motor is kept low when the induction motor is running at no-load. The small slip produces the small torque on the motor.

    The value of the induction motor slip is adjusted according to the requirement of the driving torque at the normal working condition.



    10.9. Express the derivation of the equation r f = s f and (1 s) P 120f

    What is "Slip Speed in an Induction Motor"? - its Importance 
    Slip Speed of the Induction Motor is defined as the difference between the synchronous speed and the actual rotor speed.



    SPEED The speed of a rotating fields depends upon the frequency of the source and the number of poles on the stator.
    Synchronous speed is given by: ns = 120f / p where: ns = synchronous speed (r/min) f = frequency of the source (Hz) p = number of poles This equation shows that the synchronous speed a) increase with frequency and b) decreases with the number of poles
     Example: Calculate the synchronous speed of a 3 phase induction motor having 10 poles when it is connected to a 60 Hz source:
    Solution: ns = 120f / p =120 X 60 / 10 = 720 r/min Slip and Slip Speed The slip s of induction motor is the difference between the synchronous speed and the rotor speed. Expressed in per unit: s = 𝑛𝑠− 𝑛 𝑛𝑠 where: s = slip ns = synchronous speed (r/min) n= rotor speed (r/min) The slip is practically zero at no load And equal to 1 when the rotor is locked.
    Example: A 0.5 hp , 6-pole induction motor is excited by 3 phase, 60 Hz source. If the full load speed is 1140 r/min, calculate the slip.
     Solution The synchronous speed is
     Ns = 120f / p = 120 X 60/ 6 = 1200 r/min
     (Slip) s = 𝑛𝑠− 𝑛 /𝑛𝑠 = 1200 – 1140/ 1200 = 0.05 or 5%


    10.10. Outline rotor voltage, rotor current and rotor power.

    In the case of an asynchronous machine where the winding of the rotor coupled in star or delta configuration, the rotor voltage is linked to the slip which is the difference between the speed of the flow and the speed of the rotor.

    Rotor as the name suggests it is a rotating part of an electrical machine, in which current is induced by transformer action from rotating magnetic field. Induction motor rotor is of two types: Squirrel Cage Rotor. Wound Type Rotor or Slip Ring Type Rotor.

    In a three-phase induction machine, alternating current supplied to the stator windings energizes it to create a rotating magnetic flux. ... The rotor circuit is shorted and current flows in the rotor conductors. The action of the rotating flux and the current produces a force that generates a torque to start the motor.


    The stator is the stationary part of the machine, whereas the rotor is the movable part of the machine. The stator core, stator winding and the outer frame are the three parts of the stator whereas the rotor core and field winding are the parts of the rotor. The three-phase supply is given to the winding of the stator.


    10.11. Solve problems related to slip. .

    The multiplication of 100 and speed difference divided by the synchronous speed is known as slip of an induction motor. It's always represented in percentage, generally it ranges from 0.4% to 6% based on its design factors.

    Formula for AC Induction Motor Slip

    The below mathematical formula is used in this calculator to determine the slip of an induction motor
    formula to calculate slip difference & slip of an induction motor

    Solved Example

    The below step by step solved example problem may helpful for users to understand how the input values are being used in slip percentage calculations.
    Example Problem
    Find the slip of an induction motor, whose actual speed n = 1450 rpm & synchronous speed ns = 1500 rpm.
    Solution
    The given data
    n = 1450 rpm
    ns = 1500 rpm

    Step by step calculation
    Formula to find speed difference = ns - n
    Formula to find slip = (ns - n) x 100/ns
    substitute the values in the above formulas
    speed difference = 1500 - 1450
    = 50

    slip = 50/1500 x 100
    slip = 3.33%



    11. Recognize the concept of development of rotating magnetic field and torque in rotor.

     Recognize the concept of development of rotating magnetic field and torque in rotor.

    11.1 Explain the development of rotating magnetic field for three phase induction motor.

    11.2 Express the deduction of the formula

    R

     5m 1. .

    11.3 Demonstrate the principle of rotation of a 3-phase motor.

    11.4 Clarify starting torque, running torque and maximum torque.

    11.5 Explicit the deduction of the equation of starting torque, running torque and maximum

    torque.

    11.6 Describe the condition for maximum torque at running and starting condition.

    11.7 Mention the relation between torque and rotor power factor.

    11.8 Explain the relation between torque and speed.

    11.9 Draw the torque speed curve.

    11.10 Explain the effect of changing the voltage on torque and speed.

    12 Perceive the concept of Power stages of induction motor.

    12.1 List the losess in 3-phase induction motor.

    12.2 Indicate different stages of power developed in an induction motor.

    12.3 Solve Energy stages related problems.

    13 Understand the equivalent circuit and maximum Power output of an induction motor.

    13.1 Explain the equivalent circuit of an induction motor.

    13.2 Clarify maximum power output of an induction motor.

    13.3 Express the deduction of the maximum power output

    RL

     Zo

    14 Realize the principle of starting of a 3-phase induction motor.

    14.1 Explain the purpose of starter.

    14.2 List the starters used for starting 3-phase induction motor.

    14.3 Describe the direct Online starter method, Start-delta starter method (manual and

    automatic), Auto transformer starter method of starting squirrel cage induction motor.

    14.4 Illustrate the rheostat method of starting slip ring induction motor. .

    15 Understand the principle of speed control of induction motor.

    15.1 List the methods of speed control of 3-phase induction motor.

    15.2 Describe speed control by changing applied voltage, changing applied frequency and

    changing stator poles.

    15.3 Describe rheostat control method, concatenation method & injecting emf in rotor

    circuit method.



    11.1. Explain the development of rotating magnetic field for three phase induction motor.

    Production of Rotating Magnetic Field

    Phasor diagram of three phase

    The production of Rotating magnetic field in 3 phase supply is very interesting.

    When a 3-phase winding is energized from a 3-phase supply, a rotating magnetic field is produced. This field is such that its poles do no remain in a fixed position on the stator but go on shifting their positions around the stator. For this reason, it is called a rotating field.

    It can be shown that the magnitude of this rotating field is constant and is equal to 1.5 fm where fm is the maximum flux due to any phase.

    A three-phase induction motor consists of three phases winding as its stationary part called stator. The three-phase stator winding is connected in star or delta.

    The three-phase windings are displaced from each other by 120°. The windings are supplied by a balanced three phase ac supply. 

    WP 20140123 17 23 00 Pro 1390480376 106.76.50.168

    The three-phase currents flow simultaneously through the windings and are displaced from each other by 120° electrical. Each alternating phase current produces its own flux which is sinusoidal.

    So all three fluxes are sinusoidal and are separated from each other by 120°.

    If the phase sequence of the windings is R-Y-B, then mathematical equations for the instantaneous values of the three fluxes ΦR , ΦY ,ΦB can be written as,

       ΦR =  Φmsin(ωt)                
    ΦY =  Φmsin(ωt – 120) 
    ΦB  =  Φmsin(ωt – 240)
    As windings are identical and supply is balanced, the magnitude of each flux is Φm .  

    Phasor diagram of three phase

    Waveforms of three fluxes and their positive directions

    Case 1 : Wt = 0

    ΦR =  Φmsin(0) =  0

    ΦY =  Φmsin(0 – 120) =  -0.866 Φm

    ΦB  =  Φmsin(0 – 240) = +0.866 Φm

    Case 2 : ωt = 60

    ΦR =  Φmsin(60) =  +0.866 Φm

    ΦY =  Φmsin(- 60) =  -0.866 Φm

    ΦB  =  Φmsin(- 180) = 0

    Case 3 : ωt = 120

    ΦR =  Φmsin(120) =  +0.866 Φm

    ΦY =  Φmsin(0) =  0

    ΦB  =  Φmsin(- 120) = -0.866 Φm

    Case 4 : ωt = 180

    ΦR =  Φmsin(180) =  0

    ΦY =  Φmsin(60) =  +.866 Φm

    ΦB  =  Φmsin(- 60) = -0.866 Φm

    Rotating magnetic field magnitude

    By comparing the electrical and phasor diagrams we can find the flux rotates one complete 360 degrees on the 180-degree displacement of flux. 

    WP 20140123 17 27 33 Pro

    In general, for P poles, the rotating field makes one revolution in P/2 cycles of current.

    \therefore \quad Cycles\quad of\quad Current\quad =\quad \frac { P }{ 2 } \quad \times \quad revolutions\quad of\quad field\quad

    or\quad Cycles\quad of\quad current\quad per\quad second\quad =\quad \frac { P }{ 2 } \quad \times \quad revolutions\quad of\quad field\quad per\quad second

    Since revolutions per second is equal to the revolutions per minute (N s ) divided by 60 and the number of cycles per second is the frequency f,

    f\quad =\quad \frac { P }{ 2 } \times \frac { { N }_{ S } }{ 60 } \quad =\quad \frac { { N }_{ S }P }{ 120 }

    \therefore \quad { N }_{ S }\quad =\quad \frac { 120f }{ P }

    The speed of the rotating magnetic field is the same as the speed of the alternator that is supplying power to the motor if the two have the same number of poles. Hence the magnetic flux is said to rotate at synchronous speed.









    11.2. Express the deduction of the formula

    Rotating Magnetic Field

    When we apply a three-phase supply to a three-phase distributed winding of a rotating machine, a rotating magnetic field is produced which rotates in synchronous speed.

    In this article, we will try to understand the theory behind the production of rotating magnetic field. For that, we will first imagine one stator of an electric motor where three-phase winding is physically distributed in the stator core in such a manner that winding of each phase is separated from other by 120o in space.

    stator of rotating machine
    Although the vector sum of three currents in a balanced three-phase system is zero at any instant, but the resultant of the magnetic fields produced by the currents is not zero rather it will have a constant non-zero value rotating in space in respect to time.

    The magnetic flux produced by the current in each phase can be represented by the equations given below. This is a similar representation of current is a three-phase system as the flux is cophasial with the current.
    Where, φR, φY and φB are the instantaneous flux of corresponding Red, Yellow and Blue phase winding, φm amplitude of the flux wave. The flux wave in the space can be represented as shown below.

    three phase flux wave
    Now, on the above graphical representation of flux waves, we will first consider the point 0.
    Here, the value of φR is
    The value of φY is
    The value of φB is
    The resultant of these fluxes at that instant (φr) is 1.5φm which is shown in the figure below.



    Now, on the above graphical representation of flux waves, we will consider the point 1, where ωt = π / 6 or 30o.

    Here, the value of φR is
    The value of φY is
    The value of φB is
    The resultant of these fluxes at that instant (φr) is 1.5φm which is shown in the figure below. here it is clear thet the resultant flux vector is rotated 30o further clockwise without changing its value.

    Now, on the graphical representation of flux waves, we will consider the point 2, where ωt = π / 3 or 60o.
    Here, the value of φR is
    The value of φY is
    The value of φB is
    The resultant of these fluxes at that instant (φr) is 1.5φm which is shown in the figure below. here it is clear thet the resultant flux vector is rotated 30° further clockwise without changing its value.

    Now, on the graphical representation of flux waves, we will consider the point 3, where ωt = π / 2 or 90o.
    Here, the value of φR is
    The value of φY is
    The value of φB is
    The resultant of these fluxes at that instant (φr) is 1.5φm which is shown in the figure below. here it is clear thet the resultant flux vector is rotated 30o further clockwise without changing its value.

    In this way we can prove that the due to balanced supply applied to the three pfase stator winding a rotating or revolving magnetic fiels is established in thew space.

    11.3. Demonstrate the principle of rotation of a 3-phase motor.

    What is the principle of operation of all three phase motors?
    Principle of Operation of 3-Phase Induction Motor. When the motor is excited with a three-phase supply,three-phase stator winding produces a rotating magnetic field with 120 displacements at a constant magnitude which rotates at synchronous speed.
    Working Principle of an Induction Motor

    The motor which works on the principle of electromagnetic induction is known as the induction motor. The electromagnetic induction is the phenomenon in which the electromotive force induces across the electrical conductor when it is placed in a rotating magnetic field.

    The stator and rotor are two essential parts of the motor. The stator is the stationary part, and it carries the overlapping windings while the rotor carries the main or field winding. The windings of the stator are equally displaced from each other by an angle of 120°.


    The induction motor is the single excited motor, i.e., the supply is applied only to the one part, i.e., stator. The term excitation means the process of inducing the magnetic field on the parts of the motor.

    When the three phase supply is given to the stator, the rotating magnetic field produced on it. The figure below shows the rotating magnetic field set up in the stator.

    working-principle-of-induction-motor

    Consider that the rotating magnetic field induces in the anticlockwise direction. The rotating magnetic field has the moving polarities. The polarities of the magnetic field vary by concerning the positive and negative half cycle of the supply. The change in polarities makes the magnetic field rotates.

    The conductors of the rotor are stationary. This stationary conductor cut the rotating magnetic field of the stator, and because of the electromagnetic induction, the EMF induces in the rotor. This EMF is known as the rotor induced EMF, and it is because of the electromagnetic induction phenomenon.

    The conductors of the rotor are short-circuited either by the end rings or by the help of the external resistance. The relative motion between the rotating magnetic field and the rotor conductor induces the current in the rotor conductors. As the current flows through the conductor, the flux induces on it. The direction of rotor flux is same as that of the rotor current.

    Now we have two fluxes one because of the rotor and another because of the stator. These fluxes interact each other. On one end of the conductor the fluxes cancel each other, and on the other end, the density of the flux is very high. Thus, the high-density flux tries to push the conductor of rotor towards the low-density flux region. This phenomenon induces the torque on the conductor, and this torque is known as the electromagnetic torque.

    The direction of electromagnetic torque and rotating magnetic field is same. Thus, the rotor starts rotating in the same direction as that of the rotating magnetic field.

    The speed of the rotor is always less than the rotating magnetic field or synchronous speed. The rotor tries to the run at the speed of the rotor, but it always slips away. Thus, the motor never runs at the speed of the rotating magnetic field, and this is the reason because of which the induction motor is also known as the asynchronous motor.

    Why Rotor never runs at Synchronous Speed?

    If the speed of the rotor is equal to the synchronous speed, no relative motion occurs between the rotating magnetic field of the stator and the conductors of the rotor. Thus the EMF is not induced on the conductor, and zero current develops on it. Without current, the torque is also not produced.

    Because of the above mention reasons the rotor never rotates at the synchronous speed. The speed of the rotor is always less than the speed of the rotating magnetic field.

    Alternatively, the method of the working principle of Induction Motor can also be explained as follows.

    Let’s understand this by considering the single conductor on the stationary rotor. This conductor cuts the rotating magnetic field of the stator. Consider that the rotating magnetic field rotates in the clockwise direction. According to Faraday’s Law of electromagnetic induction, the EMF induces in the conductor.

    production-of-torqueAs the rotor circuit is completed by the external resistance or by end ring, the rotor induces an EMF which causes the current in the circuit. The direction of the rotor induces current is opposite to that of the rotating magnetic field. The rotor current induces the flux in the rotor. The direction of the rotor flux is same as that of the current.

    Working-principle-of-induction-motor-fig-3The interaction of rotor and stator fluxes develops a force which acts on the conductors of the rotor. The force acts tangentially on the rotor and hence induces a torque. The torque pushes the conductors of the rotor, and thus the rotor starts moving in the direction of the rotating magnetic field. The rotor starts moving without any additional excitation system and because of this reason the motor is called the self-starting motor.

    production-of-torque-2The operation of the motor depends on the voltage induced on the rotor, and hence it is called the induction motor.






    11.4. Clarify starting torque, running torque and maximum torque.

    Torque Equation of Three Phase Induction Motor
    The torque produced by three phase induction motor depends upon the following three factors:
    Firstly the magnitude of rotor current, secondly the flux which interact with the rotor of three phase induction motor and is responsible for producing emf in the rotor part of induction motor, lastly the power factor of rotor of the three phase induction motor.
    Combining all these factors, we get the equation of torque as-

    Where, T is the torque produced by the induction motor,

    φ is flux responsible for producing induced emf,
    I2 is rotor current,
    cosθ2 is the power factor of rotor circuit.

    The flux φ produced by the stator is proportional to stator emf E1.
    i.e φ ∝ E1
    We know that transformation ratio K is defined as the ratio of secondary voltage (rotor voltage) to that of primary voltage (stator voltage).

    Rotor current I2 is defined as the ratio of rotor induced emf under running condition , sE2 to total impedance, Z2 of rotor side,

    and total impedance Z2 on rotor side is given by ,

    Putting this value in above equation we get,

    s = slip of induction motor

    We know that power factor is defined as ratio of resistance to that of impedance. The power factor of the rotor circuit is

    Putting the value of flux φ, rotor current I2, power factor cosθ2 in the equation of torque we get,

    Combining similar term we get,

    Removing proportionality constant we get,

    Where, ns is synchronous speed in r. p. s, ns = Ns / 60. So, finally the equation of torque becomes,


    Derivation of K in torque equation.
    In case of three phase induction motor, there occur copper losses in rotor. These rotor copper losses are expressed as
    Pc = 3I22R2
    We know that rotor current,

    Substitute this value of I2 in the equation of rotor copper losses, Pc. So, we get

    The ratio of P2 : Pc : Pm = 1 : s : (1 – s)
    Where, P2 is the rotor input,

    Pc is the rotor copper losses,
    Pm is the mechanical power developed.

    Substitute the value of Pc in above equation we get,

    On simplifying we get,

    The mechanical power developed Pm = Tω,

    Substituting the value of Pm

    We know that the rotor speed N = Ns(1 – s)

    Substituting this value of rotor speed in above equation we get,

    Ns is speed in revolution per minute (rpm) and ns is speed in revolution per sec (rps) and the relation between the two is

    Substitute this value of Ns in above equation and simplifying it we get

    Comparing both the equations, we get, constant K = 3 / 2πns

    Working Principle of Three Phase Induction Motor – Video


    Starting torque is the torque produced by induction motor when it starts. We know that at the start the rotor speed, N is zero.

    So, the equation of starting torque is easily obtained by simply putting the value of s = 1 in the equation of torque of the three phase induction motor,

    The starting torque is also known as standstill torque.

    Maximum Torque Condition for Three-Phase Induction Motor

    In the equation of torque,

    The rotor resistance, rotor inductive reactance and synchronous speed of induction motor remain constant. The supply voltage to the three phase induction motor is usually rated and remains constant, so the stator emf also remains the constant. We define the transformation ratio as the ratio of rotor emf to that of stator emf. So if stator emf remains constant, then rotor emf also remains constant.
    If we want to find the maximum value of some quantity, then we have to differentiate that quantity concerning some variable parameter and then put it equal to zero. In this case, we have to find the condition for maximum torque, so we have to differentiate torque concerning some variable quantity which is the slip, s in this case as all other parameters in the equation of torque remains constant.
    So, for torque to be maximum


    Now differentiate the above equation by using division rule of differentiation. On differentiating and after putting the terms equal to zero we get,

    Neglecting the negative value of slip we get

    So, when slip s = R2 / X2, the torque will be maximum and this slip is called maximum slip Sm and it is defined as the ratio of rotor resistance to that of rotor reactance.
    NOTE: At starting S = 1, so the maximum starting torque occur when rotor resistance is equal to rotor reactance.

    Equation of Maximum Torque

    The equation of torque is

    The torque will be maximum when slip s = R2 / X2
    Substituting the value of this slip in above equation we get the maximum value of torque as,

    In order to increase the starting torque, extra resistance should be added to the rotor circuit at start and cut out gradually as motor speeds up.
    Conclusion
    From the above equation it is concluded that

    1. The maximum torque is directly proportional to square of rotor induced emf at the standstill.
    2. The maximum torque is inversely proportional to rotor reactance.
    3. The maximum torque is independent of rotor resistance.
    4. The slip at which maximum torque occur depends upon rotor resistance, R2. So, by varying the rotor resistance, maximum torque can be obtained at any required slip.

    Losses and Efficiency of Induction Motor

    Deep Bar Double Cage Induction Motor

    Speed Control of Three Phase Induction Motor

    What is the condition for maximum starting torque?
    Maximum starting torque is obtained when the slip is equal to the ratio between the rotor resistance (r2) and the rotor inductive reactance (x2). This slip is also known as slip at maximum torque, labelled as Smt.
    Maximum Torque Condition of an Induction Motor

    In the article named Torque Equation of an Induction motor, we have seen the developed torque and its equation. Here, the Maximum Torque Condition of an induction motor is discussed. The torque produced in the induction motor mainly depends on the following three factors. They are the strength of the rotor current; the flux interacts between the rotor of the motor and the power factor of the rotor.The value of torque when the motor is running is given by the equation shown below.


    Maximum-torque-condition-of-an-induction-motor-eq-2

    The developed torque will be maximum when the right-hand side of the equation (4) will be maximum. This condition is possible when the value of the denominator shown below is equal to zero.

    Let,

    Maximum-torque-condition-of-an-induction-motor-eq-3

    Hence. The developed torque is maximum when the rotor resistance per phase is equal to the rotor reactance per phase under running conditions. By putting the value of sX20 = R2 in the equation (1) we get the equation for maximum torque.


    The above equation shows that the maximum torque is independent of the rotor resistance.

    If sM is the value of slip corresponding to the torque which is maximum then from the equation (5)

    Maximum-torque-condition-of-an-induction-motor-eq-5

    Therefore, The speed of the rotor at maximum torque is given by the equation shown below.

    Maximum-torque-condition-of-an-induction-motor-eq-6

    The following conclusion about the maximum torque can be drawn from the equation (7) are given below.

    • It is independent of the rotor circuit resistance.
    •  Torque at the maximum condition varies inversely as the standstill reactance of the rotor. Hence, for maximum torque, X20 and therefore, the inductance of the rotor should be kept as small as possible.
    • By varying the resistance in the rotor circuit, maximum torque can be obtained at any desired slip or the speed. It depends upon the rotor resistance at the slip (sM = R2/X20).

    To develop maximum torque at the standstill condition, the rotor resistance must be high and should be equal to X20. But to develop a torque which is maximum at the running condition the rotor resistance must be low.

    Torque Equation of an Induction Motor

    The developed Torque or Induced Torque Equation in a machine is defined as the Torque generated by the electric to mechanical power conversion. The torque is also known as Electromagnetic Torque. This developed torque in the motor differs from the actual torque available at the terminals of the motor, which is almost equal to the friction and windage torques on the machine.

    The developed torque equation is given as


    The above equation expresses the developed torque directly in terms of the air gap power Pg and the synchronous speed ωs. Since ωs is constant and independent of the load conditions. If the value of the Pg is known then, the developed torque can be found directly. The air gap power Pg is also called as the Torque in Synchronous Watts.

    Synchronous Watt is the torque that develops the power of 1 Watt when the machine is running at synchronous speed.



    These electrical powers are dissipated as I2R losses or copper loss in the rotor circuit.

    Input power to the rotor is given as


    Where,


    Starting Torque Of Induction Motor

    At the start condition the value of s = 1. Therefore, the starting is obtained by putting the value of s = 1 in the equation (6), we get

    Torque-equation-of-an-induction-motor-eq-6

    The starting torque is also known as Standstill Torque.

    Torque Equation at Synchronous Speed

    At synchronous speed, s = 0 and hence developed torque Ʈd = 0. At synchronous speed, developed torque is zero.

    Torque-equation-of-an-induction-motor-eq-7

    Since E1 is nearly equal to V1 the equation (12) becomes

    Torque-equation-of-an-induction-motor-eq-8

    The Starting torque is obtained by putting s = 1 in equation (13)

    Torque-equation-of-an-induction-motor-eq-9

    Hence, it is clear from the above equation that the starting torque is proportional to the square of the stator applied voltage.

    Also See: Maximum Torque Condition of an Induction Motor

    Maximum Torque Condition of an Induction Motor

    In the article named Torque Equation of an Induction motor, we have seen the developed torque and its equation. Here, the Maximum Torque Condition of an induction motor is discussed. The torque produced in the induction motor mainly depends on the following three factors. They are the strength of the rotor current; the flux interacts between the rotor of the motor and the power factor of the rotor.The value of torque when the motor is running is given by the equation shown below.

    Maximum-torque-condition-of-an-induction-motor-eq-1

    The total impedance of the RC network always lies between 0º and 90º. The impedance is the opposition offered by the electronic circuit element to the flow of current. If the impedance of the stator winding is assumed to be negligible. Thus, for the given supply voltage V1, E20 remains constant.

    Maximum-torque-condition-of-an-induction-motor-eq-2

    The developed torque will be maximum when the right-hand side of the equation (4) will be maximum. This condition is possible when the value of the denominator shown below is equal to zero.

    Let,

    Maximum-torque-condition-of-an-induction-motor-eq-3

    Hence. The developed torque is maximum when the rotor resistance per phase is equal to the rotor reactance per phase under running conditions. By putting the value of sX20 = R2 in the equation (1) we get the equation for maximum torque.

    Maximum-torque-condition-of-an-induction-motor-eq-4

    The above equation shows that the maximum torque is independent of the rotor resistance.

    If sM is the value of slip corresponding to the torque which is maximum then from the equation (5)

    Maximum-torque-condition-of-an-induction-motor-eq-5

    Therefore, The speed of the rotor at maximum torque is given by the equation shown below.

    Maximum-torque-condition-of-an-induction-motor-eq-6

    The following conclusion about the maximum torque can be drawn from the equation (7) are given below.

    • It is independent of the rotor circuit resistance.
    •  Torque at the maximum condition varies inversely as the standstill reactance of the rotor. Hence, for maximum torque, X20 and therefore, the inductance of the rotor should be kept as small as possible.
    • By varying the resistance in the rotor circuit, maximum torque can be obtained at any desired slip or the speed. It depends upon the rotor resistance at the slip (sM = R2/X20).

    To develop maximum torque at the standstill condition, the rotor resistance must be high and should be equal to X20. But to develop a torque which is maximum at the running condition the rotor resistance must be low.






    11.5. Explicit the deduction of the equation of starting torque, running torque and maximum torque.



    Putting this value in above equation we get,

    s = slip of induction motor

    We know that power factor is defined as ratio of resistance to that of impedance. The power factor of the rotor circuit is

    Putting the value of flux φ, rotor current I2, power factor cosθ2 in the equation of torque we get,

    Combining similar term we get,

    Removing proportionality constant we get,

    Where, ns is synchronous speed in r. p. s, ns = Ns / 60. So, finally the equation of torque becomes,



    11.6. Describe the condition for maximum torque at running and starting condition.

    What is the condition for maximum torque?
    To develop maximum torque at the standstill condition, the rotor resistance must be high and should be equal to X20. But to develop a torque which is maximum at the running condition the rotor resistance must be low.

    Maximum Torque Condition of an Induction Motor

    In the article named Torque Equation of an Induction motor, we have seen the developed torque and its equation. Here, the Maximum Torque Condition of an induction motor is discussed. The torque produced in the induction motor mainly depends on the following three factors. They are the strength of the rotor current; the flux interacts between the rotor of the motor and the power factor of the rotor.The value of torque when the motor is running is given by the equation shown below.


    The total impedance of the RC network always lies between 0º and 90º. The impedance is the opposition offered by the electronic circuit element to the flow of current. If the impedance of the stator winding is assumed to be negligible. Thus, for the given supply voltage V1, E20 remains constant


    The developed torque will be maximum when the right-hand side of the equation (4) will be maximum. This condition is possible when the value of the denominator shown below is equal to zero.

    Let,


    Hence. The developed torque is maximum when the rotor resistance per phase is equal to the rotor reactance per phase under running conditions. By putting the value of sX20 = R2 in the equation (1) we get the equation for maximum torque.


    The above equation shows that the maximum torque is independent of the rotor resistance.

    If sM is the value of slip corresponding to the torque which is maximum then from the equation (5)


    Therefore, The speed of the rotor at maximum torque is given by the equation shown below.


    The following conclusion about the maximum torque can be drawn from the equation (7) are given below.

    • It is independent of the rotor circuit resistance.
    •  Torque at the maximum condition varies inversely as the standstill reactance of the rotor. Hence, for maximum torque, X20 and therefore, the inductance of the rotor should be kept as small as possible.
    • By varying the resistance in the rotor circuit, maximum torque can be obtained at any desired slip or the speed. It depends upon the rotor resistance at the slip (sM = R2/X20).

    To develop maximum torque at the standstill condition, the rotor resistance must be high and should be equal to X20. But to develop a torque which is maximum at the running condition the rotor resistance must be low.

    So, when slip s = R2 / X2, the torque will be maximum and this slip is called maximum slip Sm and it is defined as the ratio of rotor resistance to that of rotor reactance. NOTE: At starting S = 1, so the maximum starting torque occur when rotor resistance is equal to rotor reactance.


    What is the condition for maximum starting torque?
    Maximum starting torque is obtained when the slip is equal to the ratio between the rotor resistance (r2) and the rotor inductive reactance (x2). This slip is also known as slip at maximum torque, labelled as Smt.




    11.7. Mention the relation between torque and rotor power factor.

     Mention the relation between torque and rotor power factor.




    What is the difference between torque and speed?
    Speed of a motor is the rate at which it spins and is often given in “rpm” or rotations per minute. While torque is the amount of twisting force generated by the motor (often through a shaft). Higher speed does not necessarily imply high torque. Two motors having same speed may deliver different torque values.


    11.8. Explain the relation between torque and speed.

    What is the relation between torque and speed?
    The relation between the torque of vehicle and speed is Power = Torque x Speed. So as the speed increases, the torque decreases. The torque and power produced by the engine is varied in following manner: As you can see, the maximum torque is obtained at around 50% of the engine speed.

    Relation Between Torque And Speed

    Torque is the rotational equivalence of linear force. Speed measures the distance covered in unit time. The relation between Torque and speed are inversely proportional to each other. The torque of a rotating object can be mathematically written as the ratio of Power and Angular velocity.

    Torque and Speed Formula
    Torque=Powerspeed

    Or

    τ=Pω

    Where,

    • P is the power (work done per unit time)
    • τ is the torque (Rotational ability of a body)
    • ω is the angular speed/velocity(rate of change of angular displacement)

    The above equation can be rearranged to compute the angular velocity required to achieve given Torque and Power. The torque injects power and it purely depends on instantaneous velocity.

    Connection Between Torque And Speed

    For any rotatory motion, to derive the relation between torque and Power, compare the linear equivalent. The linear displacement is the distance covered at the circumference of the rotation and is given by the product of angle covered and radius. And linear distance is given by the product of linear velocity and time.

    => Linear distance = radius × angular velocity × time

    Torque makes object undergo rotational motion. It is expressed as-

    Torque = Force × Radius

    => Force=TorqueRadius

    Thus,

    Power=Force×LineardistanceTime Power=(TorqueRadius)×Radius×Angularvelocity×TimeTime Power=Torque×Angularvelocity Torque=PowerAngularvelocity
    Motor torque Calculation

    The rotational speed in motors is measured using rotations per minute. Thus on one rotation distance and linear speed is increases proportionally by 2??.

    Torque=Power2π×Rotationalspeed Torque=Power2π×Rotationalspeed×ft.lbfmin×hp33000(ft.lbfmin) ≈torque×RPM5252

    Where, 330002π≈5252.113122

    Hope you understood the relation and conversion between the Power and the Speed of a rotating object.


    11.9. Draw the torque speed curve.

     Torque Speed Characteristic of an Induction Motor

    Torque Speed Characteristic is the curve plotted between the torque and the speed of the induction motor. As we have already discussed the torque of the induction motor in the topic Torque Equation of an Induction motor.The equation of the torque is given as shown below.


    At the maximum torque, the speed of the rotor is expressed by the equation shown below.


    The curve below shows the Torque Speed Characteristic.



    Torque Speed Characteristic of an induction motor

    The maximum torque is independent of the rotor resistance. But the exact location of the maximum torque Ʈmax is dependent on it. Greater, the value of the R2, the greater is the value of the slip at which maximum torque occurs. As the rotor resistance increases, the pullout speed of the motor decreases. In this condition, the maximum torque remains constant.

    Related Terms:
    1. Torque Slip Characteristic of an Induction Motor
    2. Maximum Torque Condition of an Induction Motor
    3. Slip Speed in an Induction Motor
    4. Rotor Resistance Control of an Induction Motor
    5. Torque Equation of an Induction Motor

    Torque Slip Characteristic of an Induction Motor

    The Torque Slip Characteristic is represented by a rectangular hyperbola. For the immediate value of the slip, the graph changes from one form to the other. Thus, it passes through the point of maximum torque when R2 = sX20. The maximum torque developed in an induction motor is called the Pull Out Torque or the Breakdown Torque. This torque is a measure of the short time overloading capability of the motor.

    The torque slip characteristic curve is divided roughly into three regions. They are given below.

    • Low slip region
    • Medium slip region
    • High slip region

    The torque equation of the induction motor is given below.Torque-slip-characteristics-of-an-induction-motor-eq-1

    Low Slip Region

    At the synchronous speed, s = 0, therefore, the torque is zero. When the speed is very near to synchronous speed. The slip is very low and (sX20)2 is negligible in comparison with R2. Therefore,

    Torque-slip-characteristics-of-an-induction-motor-eq-2

    If R2 is constant, the torque becomes

    Torque-slip-characteristics-of-an-induction-motor-eq-3

    When k2 = k1/R2

    From the equation (1) shown above, it is clear that the torque is proportional to slip. Hence, in the normal working region of the motor, the value of the slip is small. The torque slip curve is a straight line.

    Medium Slip Region

    As the slip increases, the speed of the motor decreases with the increase in load. The term (sX20)2 becomes large. The term R22 may be neglected in comparison with the term (sX20)2 and the torque equation becomes as shown below.

    Torque-slip-characteristics-of-an-induction-motor-eq-4

    At the standstill condition, the torque is inversely proportional to the slip.

    High Slip Region

    Beyond the maximum torque point, the value of torque starts decreasing. As a result, the motor slows down and stops. At this stage, the overload protection must immediately disconnect the motor from the supply to prevent damage due to overheating of the motor.

    The motor operates for the values of the slip between s = 0 and s = sM. Where, sM is the value of the slip corresponding to the maximum torque. For a typical induction motor, the pull-out torque is 2 to 3 times the rated full load torque. The starting torque is about 1.5 times the rated full load torque.

    The curve shown below shows the Torque Slip Characteristic of the Induction Motor.

    Torque Slip Characteristic of an induction motor

    Related Terms:
    1. Torque Speed Characteristic of an Induction Motor
    2. Slip Speed in an Induction Motor
    3. Slip Energy Recovery of an Induction Motor
    4. Torque Equation of an Induction Motor
    5. Maximum Torque Condition of an Induction Motor

    11.10. Explain the effect of changing the voltage on torque and speed.

    Effect of Change in Supply Voltage on Torque and Speed
    Obviously, torque at any speed is proportional to the square of the applied voltage. If stator voltage decreases by 10%, the torque decreases by 20%. Changes in supply voltage not only affect the starting torque Tst but torque under running conditions
    Effect of Change in Supply Voltage on Torque and Speed

    image

    Obviously, torque at any speed is proportional to the square of the applied voltage. If stator voltage decreases by 10%, the torque decreases by 20%. Changes in supply voltage not only affect the starting torque Tst but torque under running conditions also. If V decreases, then T also decreases.

    Hence, for maintaining the same torque, slip increases i.e. speed falls.

    image




    12. Perceive the concept of Power stages of induction motor.

    Power Stages in an Induction Motor



    Different stages of power development in an induction motor • Stator iron loss (consisting of eddy and hysteresis losses) depends on the supply frequency and the flux density in the iron core. It is practically constant. • The iron loss of the rotor is, however, negligible because frequency of rotor currents under normal running conditions is always small. Total rotor Cu loss = 3 I2 2R2.


    An induction motor develops gross torque Tg due to gross rotor output Pm. • Its value can be expressed either in terms of rotor input P2 or rotor gross output Pm as given below.


    Power Flow Diagram and Losses of Induction Motor

    Power Flow Diagram of Induction Motor explains the input given to the motor, the losses occurring and the output of the motor. The input power given to an Induction motor is in the form of three-phase voltage and currents. The Power Flow Diagram of an Induction Motor is shown below.



    The power flow is given by the equation shown below.

    POWER-FLOW-DIAGRAM-OF-INDUCTION-MOTOR-EQ-1

    Where cosϕi is the input power factor

    The losses in the stator are

    I2R losses in the stator winding resistances. It is also known as Stator copper losses.

    POWER-FLOW-DIAGRAM-OF-INDUCTION-MOTOR-EQ-2

    Hysteresis and Eddy current losses in the stator core. These are known as Stator core losses.

    POWER-FLOW-DIAGRAM-OF-INDUCTION-MOTOR-EQ-3

    The output power of the stator is given as


    This output power of the stator is transferred to the rotor of the machine across the air gap between the stator and the rotor. It is called the air gap Pg of the machine.

    Thus,

    The Power output of the stator = air gap power = input power to the rotor

    POWER-FLOW-DIAGRAM-OF-INDUCTION-MOTOR-EQ-5

    The losses in the rotor are as follows.

    I2R losses in the rotor resistance. They are also called Rotor copper losses and represented as

    POWER-FLOW-DIAGRAM-OF-INDUCTION-MOTOR-EQ-6

    Hysteresis and eddy current losses in the rotor core. They are known as Rotor core losses.

    POWER-FLOW-DIAGRAM-OF-INDUCTION-MOTOR-EQ-7

    Friction and Windage losses Pfw

    Stray load losses Pmisc, consisting of all losses not covered above, such as losses due to harmonic fields.

    If the rotor copper losses are subtracted from rotor input power Pg, the remaining power is converted from electrical to mechanical form. This is called Developed Mechanical Power Pmd.

    Developed Mechanical power = Rotor input – Rotor copper loss

    POWER-FLOW-DIAGRAM-OF-INDUCTION-MOTOR-EQ-8

    The output of the motor is given by the equation shown below.

    POWER-FLOW-DIAGRAM-OF-INDUCTION-MOTOR-EQ-9

    Po is called the shaft power or the useful power.

    Rotational losses

    At starting and during acceleration, the rotor core losses are high. With the increase in the speed of the induction motor these losses decreases. The friction and windage losses are zero at the start. As the speed increases the losses, also start increasing. The sum of the friction, windage and core losses are almost constant with the change in speed. These all losses are added together and are known as Rotational Losses.

    It is given by the equation shown below.

    POWER-FLOW-DIAGRAM-OF-INDUCTION-MOTOR-EQ-10

    The Rotational losses are not represented by any element of the equivalent circuit as they are purely mechanical quantity.



    12.1. List the losess in 3-phase induction motor.

    There are two types of losses occur in three phase induction motor. These losses are,
    1. Constant or fixed losses,
    2. Variable losses.
    Constant or Fixed Losses

    Constant losses are those losses which are considered to remain constant over normal working range of induction motor. The fixed losses can be easily obtained by performing no-load test on the three phase induction motor. These losses are further classified as-

    1. Iron or core losses,
    2. Mechanical losses,
    3. Brush friction losses.
    Iron or Core Losses

    Iron or core losses are further divided into hysteresis and eddy current losses. Eddy current losses are minimized by using lamination on core. Since by laminating the core, area decreases and hence resistance increases, which results in decrease in eddy currents. Hysteresis losses are minimized by using high grade silicon steel. The core losses depend upon frequency of the supply voltage. The frequency of stator is always supply frequency, f and the frequency of rotor is slip times the supply frequency, (sf) which is always less than the stator frequency. For stator frequency of 50 Hz, rotor frequency is about 1.5 Hz because under normal running condition slip is of the order of 3 %. Hence the rotor core loss is very small as compared to stator core loss and is usually neglected in running conditions.

    Mechanical and Brush Friction Losses

    Mechanical losses occur at the bearing and brush friction loss occurs in wound rotor induction motor. These losses are zero at start and with increase in speed these losses increases. In three phase induction motor the speed usually remains constant. Hence these losses almost remains constant.

    Variable Losses

    power flow iagram of motor

    These losses are also called copper losses. These losses occur due to current flowing in stator and rotor windings. As the load changes, the current flowing in rotor and stator winding also changes and hence these losses also changes. Therefore these losses are called variable losses. The copper losses are obtained by performing blocked rotor test on three phase induction motor. The main function of induction motor is to convert an electrical power into mechanical power. During this conversion of electrical energy into mechanical energy the power flows through different stages.

    This power flowing through different stages is shown by power flow diagram. As we all know the input to the three phase induction motor is three phase supply. So, the three phase supply is given to the stator of three phase induction motor.
    Let, Pin = electrical power supplied to the stator of three phase induction motor,
    VL = line voltage supplied to the stator of three phase induction motor,
    IL = line current,
    Cosφ = power factor of the three phase induction motor.
    Electrical power input to the stator, Pin = √3VLILcosφ
    A part of this power input is used to supply stator losses which are stator iron loss and stator copper loss. The remaining power i.e (input electrical power – stator losses) are supplied to rotor as rotor input.
    So, rotor input P2 = Pin – stator losses (stator copper loss and stator iron loss).
    Now, the rotor has to convert this rotor input into mechanical energy but this complete input cannot be converted into mechanical output as it has to supply rotor losses. As explained earlier the rotor losses are of two types rotor iron loss and rotor copper loss. Since the iron loss depends upon the rotor frequency, which is very small when the rotor rotates, so it is usually neglected. So, the rotor has only rotor copper loss. Therefore the rotor input has to supply these rotor copper losses. After supplying the rotor copper losses, the remaining part of Rotor input, P2 is converted into mechanical power, Pm.

    Let Pc be the rotor copper loss,
    I2 be the rotor current under running condition,
    R2 is the rotor resistance,
    Pm is the gross mechanical power developed.
    Pc = 3I22R2
    Pm = P2 – Pc
    Now this mechanical power developed is given to the load by the shaft but there occur some mechanical losses like friction and windage losses. So, the gross mechanical power developed has to be supplied to these losses. Therefore the net output power developed at the shaft, which is finally given to the load is Pout.
    Pout = Pm – Mechanical losses (friction and windage losses).
    Pout is called the shaft power or useful power.





    12.2. Indicate different stages of power developed in an induction motor.

     Torque Equation of Three Phase Induction Motor | Electrical4U

    Power flow diagram of 3-phase induction motor

    Power Stages in an Induction Motor

    Power Stages in an Induction Motor



     Power flow diagram

    From the power flow diagram we can define,


      =  Pm / P2


    12.3. Solve Energy stages related problems.

     Solve Energy stages related problems.


    13. Understand the equivalent circuit and maximum Power output of an induction motor.

    Understand the equivalent circuit and maximum Power output of an induction motor.


    13.1. Explain the equivalent circuit of an induction motor.

    Equivalent Circuit for an Induction Motor

    An induction motor is a well-known device which works on the principle of transformer. So it is also called the rotating transformer. That is, when an EMF is supplied to its stator, then as a result of electromagnetic induction, a voltage is induced in its rotor. So an induction motor is said to be a transformer with rotating secondary. Here, primary of transformer resembles stator winding of an induction motor and secondary resembles rotor.

    The induction motor always runs below the synchronous or full load speed and the relative difference between the synchronous speed and speed of rotation is known as slip which is denoted by s.


    Where, Ns is synchronous speed of rotation which is given by-


    Where, f is the frequency of the supply voltage.
    P is the number of poles of the machine.

    Equivalent Circuit of an Induction Motor

    The equivalent circuit of any machine shows the various parameter of the machine such as its Ohmic losses and also other losses.

    The losses are modeled just by inductor and resistor. The copper losses are occurred in the windings so the winding resistance is taken into account. Also, the winding has inductance for which there is a voltage drop due to inductive reactance and also a term called power factor comes into the picture. There are two types of equivalent circuits in case of a three-phase induction motor-

    Exact Equivalent Circuit


    exact equivalent circuit

    Here, R1 is the winding resistance of the stator.
    X1 is the inductance of the stator winding.
    Rc is the core loss component.
    XM is the magnetizing reactance of the winding.
    R2/s is the power of the rotor, which includes output mechanical power and copper loss of rotor.

    If we draw the circuit with referred to the stator then the circuit will look like-

    exact equivalent circuit

    Here all the other parameters are same except-
    R2’ is the rotor winding resistance with referred to stator winding.
    X2’ is the rotor winding inductance with referred to stator winding.
    R2(1 – s) / s is the resistance which shows the power which is converted to mechanical power output or useful power. The power dissipated in that resistor is the useful power output or shaft power.

    Approximate Equivalent Circuit

    The approximate equivalent circuit is drawn just to simplify our calculation by deleting one node. The shunt branch is shifted towards the primary side. This has been done as the voltage drop between the stator resistance and inductance is less and there is not much difference between the supply voltage and the induced voltage. However, this is not

    appropriate due to following reasons-

    1. The magnetic circuit of induction motor has an air gap so exciting current is larger compared to transformer so exact equivalent circuit should be used.
    2. The rotor and stator inductance is larger in induction motor.
    3. In induction motor, we use distributed windings.

    This model can be used if approximate analysis has to be done for large motors. For smaller motors, we cannot use this.

    Power Relation of Equivalent Circuit
    1. Input power to stator- 3 V1I1Cos(Ɵ).
      Where, V1 is the stator voltage applied.
      I1 is the current drawn by the stator winding.
      Cos(Ɵ) is the stator power stator.
    2. Rotor input =
      Power input- Stator copper and iron losses.
    3. Rotor Copper loss = Slip × power input to the rotor.
    4. Developed Power = (1 – s) × Rotor input power.
      Equivalent Circuit of a Single Phase Induction Motor



    13.2. Clarify maximum power output of an induction motor.

     Clarify maximum power output of an induction motor.


    Chapter 34 Induction Motor. Chapter 34 Induction Motor. - ppt download

    Chapter 34 Induction Motor. Chapter 34 Induction Motor. - ppt download

    Chapter 34 Induction Motor. Chapter 34 Induction Motor. - ppt download


    Chapter 34 Induction Motor. Chapter 34 Induction Motor. - ppt download


    Chapter 34 Induction Motor. Chapter 34 Induction Motor. - ppt download

    Chapter 34 Induction Motor. Chapter 34 Induction Motor. - ppt download

    13.3. Express the deduction of the maximum power output RL =Zo

    Express the deduction of the maximum power output RL  =Zo

    Useful Circuit Analysis Techniques - ppt download



    Max+power+theorem


    Maximum Power Transfer Theorem for AC & DC Circuits


    Electrical Systems 100 Lecture 3 (Network Theorems) Dr Kelvin ...

    Maximum Power Transfer Theorem - Lessons - Tes Teach

    14. Realize the principle of starting of a 3-phase induction motor.

    Starting of an Induction Motor

    A three phase Induction Motor is Self Starting. When the supply is connected to the stator of a three-phase induction motor, a rotating magnetic field is produced, and the rotor starts rotating and the induction motor starts. At the time of starting, the motor slip is unity, and the starting current is very large.

    The purpose of a starter is not to just start the motor, but it performs the two main functions. They are as follows.

    • To reduce the heavy starting current
    • To provide overload and under voltage protection.

    The three phase induction motor may be started by connecting the motor directly to the full voltage of the supply. The motor can also be started by applying a reduced voltage to the motor when the motor is started.

    The torque of the induction motor is proportional to the square of the applied voltage.Thus, a greater torque is exerted by a motor when it is started on full voltage than when it is started on the reduced voltage.

    There are three main methods of Starting of Cage Induction Motor. They are as follows.


    14.1. Explain the purpose of starter.

    The main purpose of a starter is to limit the starting current taken by the motor within twice full-load current, either by reducing the applied voltage or by adding ...


    14.2. List the starters used for starting 3-phase induction motor.

    Necessity of Starter

    The three phase induction motors are self-starting due to rotating magnetic field. But the motors show tendency to draw very high current at the time of starting. Such a current can be 6 to 8 times of full load or rated current and it can damage the motor winding. Hence there should be a device which can limit such high starting current. Such a device which limits high starting current is called a starter.

    Let us study in detail why starter is required for an induction motor.

    • In a 3 phase induction motor, the magnitude of an induced emf in the rotor circuit depends on the slip of the induction motor. Thus induced emf effectively decides the magnitude of the rotor current. The rotor current in the running condition is given by :

    I2r = s E’2 / (R22 + sX22)

    1. At starting time, the speed of the motor is zero and slip is at its maximum i.e. unity. So magnitude of rotor induced emf is very large at start. As rotor conductors are short-circuited, the large induced emf circulates very high current through rotor at start.
    2. The condition is exact similar to a transformer with short-circuited secondary. Such a transformer when excited by a rated voltage, circulates a very high current through short-circuited secondary. As secondary current is large ⇒ Primary current also draws very large current from the supply line. 
    3. Similarly in a 3 phase induction motor, when rotor current is high, consequently the stator draws a very high current from the supply line. Due to such increment in line current ⇒ Voltage starts deeping. ⇒ Whole system can collapse. ⇒ Hence even 3 hp rating induction motor is not allowed without Starter.

    List the starters used for starting 3-phase induction motor.

    Starting methods of Induction motor include:

    1. Direct –On– line (DOL) starters for less than 10 Kw motors.
    2. Star–Delta starters for large motors. The stator winding is initially connected in a star configuration and later on changed over to a Delta connection, when the motor reaches rated speed.
    3. Auto transformer.                                                  There are three main methods of Starting of Cage Induction Motor. They are as follows.
    starting-of-induction-motor-figure-1

    Direct on line starter

    The direct on line starter method, of an induction motor is simple and economical. In this method, the starter is connected directly to supply voltage. By this method small motors up to 5 kW rating is started to avoid the supply voltage fluctuation.

    Star delta starter

    The star delta starter method of starting three phase induction motors is very common and widely used among all the methods. In this method, the motor runs at delta connected stator windings.

    Auto transformer starter

    The Auto transformer is used in both the type of the connections, i.e., either star connected or delta connected. The auto transformer is used to limit the starting current of the induction motor.

    The above three starters are used for cage rotor induction motor.

    Slip Ring Induction Motor Starter Method of Starting Induction Motor

    In the Slip Ring Induction Motor starter, the full supply voltage is connected across the starter. The connection diagram of the slip ring starter induction motor is shown below.


    Full starting resistance is connected and thus the supply current to the stator is reduced. The rotor begins to rotate, and the rotor resistances are gradually cut out as the speed of the motor increases. When the motor is running at its rated full load speed, the starting resistances are cut out completely, and the slip rings are short-circuited.


    14.3. Describe the direct Online starter method, Start-delta starter method (manual and automatic), Auto transformer starter method of starting squirrel cage induction motor.

    Describe the direct Online starter method, Start-delta starter method (manual and

    automatic), Auto transformer starter method of starting squirrel cage induction motor.

    Direct On Line Starter

    Direct On Line Starter Method is a common method of starting of Cage Induction Motor. The motor is connected through a starter across the full supply voltage. The Direct On Line Starter Method figure is shown below. It consists a coil operated contactor C controlled by start and stop push button as shown in the connection diagram below.

    Direct On Line Starter figure

    The buttons which may be installed in a convenient place away from the starter. The start button is held open by a spring. On pressing the START pushbutton S1, the contactor C is energised from two line conductors L1 and L2.


    Theory of Direct On Line Starting of Induction Motor

    Let,

    • Ist be the starting current drawn from the supply mains per phase.
    • Ifl is the full load current drawn from the supply mains per phase.
    • Ʈest is the starting torque.
    • Sfl is the slip at full load.      
      Star Delta Starter

      The Star Delta Starter is a very common type of starter and is used extensively as compared to the other type of starting methods of the induction motor. A star delta is used for a cage motor designed to run normally on the delta connected stator winding. The connection of a three-phase induction motor with a star delta starter is shown in the figure below.

    • star-delta-starter

    When the switch S is in the START position, the stator windings are connected in the star as shown below.

    When the motor picks up the speed, about 80 percent of its rated speed, the switch S is immediately put into the RUN position. As a result, a stator winding which was in star connection is changed into DELTA connection now. The delta connection of the stator winding in shown in the figure below.

    Star-Delta-Starter-Fig-3

    Firstly, the stator winding is connected in star and then in Delta so that the starting line current of the motor is reduced to one-third as compared to the starting current with the windings connected in delta. At the starting of an induction motor when the windings of the stator are star connected, each stator phase gets a voltage VL/√3. Here VL is the line voltage.

    Theory of Star Delta Starter Method of Starting of Induction Motor

    At the starting of the induction motor, stator windings are connected in star and, therefore, the voltage across each phase winding is equal to 1/√3 times the line voltage.

    Let,

    • VL is the line voltage
    • Istyp is the starting current per phase with the stator windings connected in star.
    • Istyl is the starting line current with the stator winding in the star

    For star connection, the line current is equal to the phase current

    Therefore,


    Since the developed torque is proportional to the square of the voltage applied to an induction motor. Star delta starter reduces the starting torque to one-third that is obtained by direct delta starting.

    Auto transformer Starter

    An Auto transformer Starter is suitable for both star and delta connected motors. In this method, the starting current is limited by using a three-phase auto transformer to reduce the initial stator applied voltage. The figure below shows the motor with the Auto transformer starter.

    Auto transformer starter figure 1

    It is provided with a number of tappings. The starter is connected to one particular tapping to obtain the most suitable starting voltage. A double throw switch S is used to connect the auto transformer in the circuit for starting. When the handle H of the switch S in the START position. The primary of the auto transformer is connected to the supply line, and the motor is connected to the secondary of the auto transformer.

    When the motor picks up the speed of about 80 percent of its rated value, the handle H is quickly moved to the RUN position. Thus, the auto transformer is disconnected from the circuit, and the motor is directly connected to the line and achieve its full rated voltage. The handle is held in the RUN position by the under voltage relay.

    If the supply voltage fails or falls below a certain value, the handle is released and returns to the OFF position. Thermal overload relays provide the overload protection.

    Theory of Auto transformer Starter

    The figure (a) shown below shows the condition when the motor is directly switched on to lines and the figure (b) shows when the motor is started with the help of auto transformer.


    Let,

    When the full voltage V1 per phase is applied to the direct switching, the starting current drawn from the supply is given by the equation shown below



     Entire stator connection for a star-delta starter.

    Wiring diagram (left) and line diagram of star-delta starter (right).



    Elementary diagram of sizes 1, 2, 3, 4, 5, and 6 star-delta starters with transition starting.



    14.4. Illustrate the rheostat method of starting slip ring induction motor. .

    SLIP RING INDUCTION MOTOR STARTER METHOD OF STARTING INDUCTION MOTOR

    In the Slip Ring Induction Motor starter, the full supply voltage is connected across the starter. The connection diagram of the slip ring starter induction motor is shown below.

    Starting of an Induction Motor figure 2Full starting resistance is connected and thus the supply current to the stator is reduced. The rotor begins to rotate, and the rotor resistances are gradually cut out as the speed of the motor increases. When the motor is running at its rated full load speed, the starting resistances are cut out completely, and the slip rings are short-circuited.

    Know how to control the speed of slip ring motors - Bright Hub ...

    What is a slip ring induction motor? - Quora




    15. Understand the principle of speed control of induction motor.

    A three phase induction motor is basically a constant speed motor so it’s somewhat difficult to control its speed. The speed control of induction motor is done at the cost of decrease in efficiency and low electrical power factor. Before discussing the methods to control the speed of three phase induction motor one should know the basic formulas of speed and torque of three phase induction motor as the methods of speed control depends upon these formulas.

    Synchronous Speed


    Where, f = frequency and P is the number of poles

    The speed of induction motor is given by,

    Where,
    N is the speed of the rotor of an induction motor,
    Ns is the synchronous speed,
    S is the slip.



    15.1. List the methods of speed control of 3-phase induction motor.

    An induction motor is practically a constant speed motor, that means, for the entire loading range, change in speed of the motor is quite small. Speed of a DC shunt motor can be varied very easily with good efficiency, but in case of Induction motors, speed reduction is accompanied by a corresponding loss of efficiency and poor power factor. As induction motors are widely being used, their speed control may be required in many applications. Different speed control methods of induction motor are explained below.

    Induction motor speed control from stator side
    1. By changing the applied voltage:

    2. By changing the applied frequency
    3. Constant V/F control of induction motor
    4. Changing the number of stator poles
    Speed control from rotor side:
    1. Rotor rheostat control
    2. Cascade operation
    3. By injecting EMF in rotor circuit

    15.2. Describe speed control by changing applied voltage, changing applied frequency and changing stator poles.

    Describe speed control by changing applied voltage, changing applied frequency and

    changing stator poles.

    1. By changing the applied voltage:
    From the torque equation of induction motor,
    torque equation of an induction motor
    Rotor resistance R2 is constant and if slip s is small then (sX2)2 is so small that it can be neglected. Therefore, T ∝ sE22 where E2 is rotor induced emf and E2 ∝ V
    Thus, T ∝ sV2,  which means, if supplied voltage is decreased, the developed torque decreases. Hence, for providing the same load torque, the slip increases with decrease in voltage, and consequently, the speed decreases. This method is the easiest and cheapest, still rarely used, because
    1. large change in supply voltage is required for relatively small change in speed.
    2. large change in supply voltage will result in a large change in flux density, hence, this will disturb the magnetic conditions of the motor.
    2. By changing the applied frequency
    Synchronous speed of the rotating magnetic field of an induction motor is given by,
    synchronous speed of an induction motor
    where, f = frequency of the supply and P = number of stator poles.
    Hence, the synchronous speed changes with change in supply frequency. Actual speed of an induction motor is given as N = Ns (1 - s). However, this method is not widely used. It may be used where, the induction motor is supplied by a dedicated generator (so that frequency can be easily varied by changing the speed of prime mover). Also, at lower frequency, the motor current may become too high due to decreased reactance. And if the frequency is increased beyond the rated value, the maximum torque developed falls while the speed rises.
    3. Constant V/F control of induction motor
    This is the most popular method for controlling the speed of an induction motor. As in above method, if the supply frequency is reduced keeping the rated supply voltage, the air gap flux will tend to saturate. This will cause excessive stator current and distortion of the stator flux wave. Therefore, the stator voltage should also be reduced in proportional to the frequency so as to maintain the air-gap flux constant. The magnitude of the stator flux is proportional to the ratio of the stator voltage and the frequency. Hence, if the ratio of voltage to frequency is kept constant, the flux remains constant. Also, by keeping V/F constant, the developed torque remains approximately constant. This method gives higher run-time efficiency. Therefore, majority of AC speed drives employ constant V/F method (or variable voltage, variable frequency method) for the speed control. Along with wide range of speed control, this method also offers 'soft start' capability.
    4. Changing the number of stator poles
    From the above equation of synchronous speed, it can be seen that synchronous speed (and hence, running speed) can be changed by changing the number of stator poles. This method is generally used for squirrel cage induction motors, as squirrel cage rotor adapts itself for any number of stator poles. Change in stator poles is achieved by two or more independent stator windings wound for different number of poles in same slots.
    For example, a stator is wound with two 3phase windings, one for 4 poles and other for 6 poles.
    for supply frequency of 50 Hz
    i) synchronous speed when 4 pole winding is connected, Ns = 120*50/4 = 1500 RPM
    ii) synchronous speed when 6 pole winding is connected, Ns = 120*50/6 = 1000 RPM



    15.3. Describe rheostat control method, concatenation method & injecting emf in rotor circuit method.

    Speed control from rotor side:
    1. Rotor rheostat control
    This method is similar to that of armature rheostat control of DC shunt motor. But this method is only applicable to slip ring motors, as addition of external resistance in the rotor of squirrel cage motors is not possible.
    2. Cascade operation
    In this method of speed control, two motors are used. Both are mounted on a same shaft so that both run at same speed. One motor is fed from a 3phase supply and the other motor is fed from the induced emf in first motor via slip-rings. The arrangement is as shown in following figure.
    cascade operation speed control of induction motor
    Motor A is called the main motor and motor B is called the auxiliary motor.
    Let, Ns1 = frequency of motor A
           Ns2 = frequency of motor B
           P1 = number of poles stator of motor A
           P2 = number of stator poles of motor B
           N = speed of the set and same for both motors
           f = frequency of the supply

    Now, slip of motor A, S1 = (Ns1 - N) / Ns1.
    frequency of the rotor induced emf in motor A,   f1 = S1f
    Now, auxiliary motor B is supplied with the rotor induce emf

    therefore,  Ns2 = (120f1) / P2  =  (120S1f) / P2.

    now putting the value of  S1 = (Ns1 - N) / Ns1
     At no load, speed of the auxiliary rotor is almost same as its synchronous speed.
    i.e. N = Ns2.
    from the above equations, it can be obtained that
    With this method, four different speeds can be obtained
    1. when only motor A works, corresponding speed = .Ns1 = 120f / P1
    2. when only motor B works, corresponding speed = Ns2 = 120f / P2
    3. if commulative cascading is done, speed of the set = N = 120f / (P1 + P2)
    4. if differential cascading is done, speed of the set = N = 120f (P1 - P2)
    3. By injecting EMF in rotor circuit
    In this method, speed of an induction motor is controlled by injecting a voltage in rotor circuit. It is necessary that voltage (emf) being injected must have same frequency as of the slip frequency. However, there is no restriction to the phase of injected emf. If we inject emf which is in opposite phase with the rotor induced emf, rotor resistance will be increased. If we inject emf which is in phase with the rotor induced emf, rotor resistance will decrease. Thus, by changing the phase of injected emf, speed can be controlled. The main advantage of this method is a wide rage of speed control (above normal as well as below normal) can be achieved. The emf can be injected by various methods such as Kramer system, Scherbius system etc.

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