Top 10 Difference Between Core And Shell Type Transformers - Types of Transformers

Top 10 Difference Between Core  And Shell Type Transformers - Types of Transformers

In our previous articles we have discusses about differences between lap,wave winding this tutorial we are sharing major differences between core type transformer and shell type transformer. In this tutorial comparisons between shell type and core type transformers are discussed.There are two major types of transformers based on construction.
They are,
1.Core Type Transformers
2.Shell Type Transformers

Types of Transformers

Difference Between Shell And Core Type Transformers

Core Type TransformersShell Type Transformers
1. In core type transformer winding is placed on
two core limbs.
1. In shell type transformer winding is placed on mid arm
of the core.It is installed on mid-limb of the core.
Other limbs will be used as mechanical supporting
2. Core type transformers have only one magnetic flux path.2. Shell type transformers have two magnetic flux path.
3. It has better cooling since more surface is exposed to
atmosphere.
3. Cooling is not effective in shell type when compared
to core type transformer.
4. It is very useful when we need large size low voltage.4. It is very useful when we need small size high voltage.
5. In core type transformer output is less. Because of losses.
So efficiency will be less than shell type transformer.
5. In core type transformer output is high. Because of
less losses.So efficiency will be more.
6. The winding is surrounded considerable part of core.6. Core is surrounded considerable part of winding of
transformer.
7. It has less mechanical protection to coil.7. It has better mechanical protection to coil.
8. Core has two limbs.8. Core has three limbs.
9. This transformer is easy to repair,Easy to maintain.9. This transformer is not easy to repair.We need a
skilled technician to maintain this type of transformer.
10. In this type transformer concentric cylindrical
winding are used
10. In this type transformer sandwiched winding are used.

Tags:
1.difference between core type and shell type transformer pdf
2.difference between core type and shell type transformer ppt
3.difference b/w core type and shell type transformer
4.Types of transformers 
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Back EMF | Back EMF Significance in DC Motor

Back EMF | Back EMF Significance in DC Motor

What is Back EMF in DC Motor?


We know whenever conductor cuts the magnetic field,e.m.f will induce in conductor.This also applies for conductors in armature too.When the armature of a d.c. motor rotates under the influence of the driving torque, the armature conductors move through the magnetic field and hence e.m.f. is induced in them as in a generator. The induced e.m.f. acts in opposite direction to the applied voltage  V (Lenz’s law) and in known as back e.m.f or counter e.m.f. denoted with  Eb. 

The back emf  Eb(= PΦZN/60 A) is always less than the applied voltage V, although this difference is small when the motor is running under normal conditions.

Back EMF in DC Motor Circuit Diagram



Significance of Back EMF In DC Motor:


It is seen in the generating action, that when a conductor cuts the lines of flux, emf. gets induced in the conductor. The question is obvious that in a dc. motor, after a motoring action, armature starts rotating and armature conductors cut the min flux.So is there a generating action exiting in a motor ? The answer to this question is 'Yes'

After a motoring action, there exists a generating action.There is an induced e.m.f in the rotating armature conductors according to Faraday's law of electromagnetic induction. This induced e.m.f. in the armature always acts in the opposite direction of the supply voltage. This is according to the Lenz’s law which states that the direction of the induced e.m.f. is always so as to oppose the cause producing it. In a dc. motor, electrical input i.e. the supply voltage is the cause and hence this induced e.m.f. opposes the supply voltage. This e.m.f tries to set up a current through the armature which is in the opposite direction to that, which supply voltage is forcing through the conductor.

So as this e.m.f. always opposes the supply voltage, it is called back e.m.f. and denoted as Eb Though it is denoted as Eb, Basically it gets generated by the generating action which we have seen earlier in case of generators.So its magnitude can be determined by the emf. equation which is derived earlier. So,

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Long transmission lines,Analysis With Rigorous Method

Long transmission lines,Analysis By Rigorous method

What are called as Long transmission lines?

Answer: When the length of an overhead transmission line is more than 150 km and line voltage is very high (> 100 kV), it is considered as a long transmission line. For the treatment of such a line, the line constants are considered uniformly distributed over the whole length of the line and rigorous methods are employed for solution.

Long transmission lines:

It is well known that line constants of the transmission line are uniformly distributed over the entire length of the line. However, reasonable accuracy can be obtained in line calculations for short and medium lines by considering these constants as lumped. If such an assumption of lumped constants is applied to long transmission lines (having length excess of about 150 km), it is found that serious errors are introduced in the performance calculations. Therefore, in order to obtain fair degree of accuracy in the performance calculations of long lines, the line constants are considered as uniformly distributed throughout the length of the line. Rigorous mathematical treatment is required for the solution of such lines.



Above shows the equivalent circuit of a 3-phase long transmission line on a phase-neutral basis. The whole line length is divided into n sections, each section having line constants 1/n th of those for the whole line. The following points may by noted :

(i) The line constants are uniformly distributed over the entire length of line as is actually the case.
(ii) The resistance and inductive reactance are the series elements.
(iii) The leakage susceptance (B) and leakage conductance (G) are shunt elements. The leakage susceptance is due to the fact that capacitance exists between line and neutral. The leakage conductance takes into account the energy losses occurring through leakage over the insulators or due to corona effect between conductors.

Admittance =√G ²+B²

(iv) The leakage current through shunt admittance is maximum at the sending end of the line and decreases continuously as the receiving end of the circuit is approached at which point its value is zero.

Analysis of  Long Transmission Line (Rigorous method)

Below shows one phase and neutral connection of a 3-phase line with impedance and shunt admittance of the line uniformly distributed.


Consider a small element in the line of length dx situated at a distance x from the receiving end.
Let z = series impedance of the line per unit length
y = shunt admittance of the line per unit length
V = voltage at the end of element towards receiving end
V + dV = voltage at the end of element towards sending end
I + dI = current entering the element dx
I = current leaving the element dx
Then for the small element dx,
z dx = series impedance
y dx = shunt admittance
Obviously, dV = I z dx
or dV/dx = I z  ...(i)
Now, the current entering the element is I + dI whereas the current leaving the element is I. The difference in the currents flows through shunt admittance of the element i.e., 
dI = Current through shunt admittance of element = V y dx
dI/dx= V y  ...(ii)
Differentiating eq. (i) w.r.t. x, we get,










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[ PDF ] Analog Electronics Book Bakshi and Godse Free Download Pdf

Analog Electronics Book  Bakshi and Godse Free Download Pdf

We at ElectricalEdition.Com Always try to help electrical & electronics engineering students by providing articles,PDF books for free download,PPTs,Email Answering etc.As a part of this we are sharing Pdf Book: Analog Electronics Book  Bakshi and Godse For Free Download.


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Barkhausen Criterion of Oscillators

Barkhausen Criterion of Oscillators

Consider a basic inverting amplifier with an Open loop gain A.The feedback network attenuation factor B is less than unity. As basic amplifier is inverting. it produces a phase shift of 180° between input and output as shown in the below figure.

Now the input Vi applied to the amplifier is to be derived from its output Vo using feedback network.

But the feedback must be positive Le. the voltage derived from output using feedback network must be in phase with V‘. Thus the feedback network must introduce a phase shift of 180° while feeding back the voltage from output to input. This more: positive feedback.


Consider a fictitious voltage Vi applied at the input of the amplifier. Hence we get,

Vo=AVi    -----------------(1)

The feedback factor β decides the feedback to be given to input,

Vf=βVo    -----------------(2)

Substituting equation (1) into equation (2) we get,

Vf=βAVi    -----------------(3)

For the oscillator, we want that feedback should drive the amplifier and hence V, must act as Vi. From equation (3) we can write that, V, is sufficient to act as V, when.

Aβ | = 1

And the phase of Vf is same as Vi i.e. feedback network should introduce 180°phase shift in addition to 180° phase shift introduced by inverting amplifier. This ensures positive feedback. So total phase shift around a loop is 360°.  In this condition, Vf drives the circuit and without external input circuit works as an oscillator.  The two conditions discussed above, required to work the circuit as an oscillator are called Barkhamen Criterion for Oscillation.


The Barkhausen Criterion states that
1.The total phase shift around a loop,as the signal proceeds from input through amplifier, feedback network back to input again. completing a loop, is precisely 0° or 360°.
2. The magnitude of the product of the open loop gain of the amplifier (A) and Magnitude of the feedback factor-β is unity Le. I Aβ I =1.



In reality, no input signal is needed to start the oscillation. In practice, Aβ is made greater than 1 to start the oscillations and then circuit adjusts it self to get Aβ=1 finally resulting into self sustained oscillations. Let us see the effect of the magnitude of the product Aβ on the nature of the oscillations.

Condition I Aβ I > 1

When the total phase shift around a loop is  or 360° and l Aβ|] > 1,then the  output oscillates but the oscillations are of growing type. The amplitude of oscillations goes on increasing as shown in the below figure.

Condition I Aβ I = 1

As stated  by Barkhausen Criterion, when total phase shift around a loop is 0°or 360° ensuring positive feedback and I Aβ I = 1 then the oscillations are with constant frequency and amplitude called sustained oscillations.
Condition I Aβ I < 1

When total phase shift around a output loop is 0°or 360 °but I Aβ I< 1 then the oscillations are of decaying type i.e. such oscillation amplitude decreases  exponentially and the oscillations finally cease. Thus circuit works as an amplifier without oscillations.So to start the oscillations without input I Aβ I is kept higher than unity and then circuit adjusts itself to get | Aβ | = l to result sustained oscillations.


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Condition for Maximum Efficiency of DC Machine

We have discussed about construction & working of DC machine. In this article  condition for maximum efficiency in DC machine will be derived. 

Condition for Maximum Efficiency in DC Machine

In this DC generator is taken as reference to find out maximum efficiency .The DC generator efficiency is perpetual but varies with load. Think through a shunt generator supplying a load current IL at a terminal voltage V.


Then Generator output  =  VIL
Generator input  =  Output + Losses

=  VI+ Variable losses + Constant losses
=  VI+ I2R+ Wc
=  VIL + (IL + Ish2)Ra + Wc  ( ∵  Ia = IL + Ish)

The shunt field current Ish is generally small as compared to Iand, therefore, can be neglected.
Generator input   =  VIL + I2a Ra + Wc

Now                Efficiency   η  =  output / Input

=  VI/ (VIL + I2a Ra + Wc
=  1 / {1+[(ILRa/V)+(Wc/VIL)]}

The efficiency is always maximum when the denominator of above equation  is minimum i.e.,
                                              d/dI{( ILRa/V) + (Wc+VI2L)} =0
                                                        Or
                                             (Ra/V) – (Wc/ VI2L) =0
                                                        Or
                                              Ra/V = Wc/VI2L
                                                       Or
                                               I2LRa = Wc

I.e. Variable loss = Constant loss (IL ≈ Ia)
The load current corresponding to maximum efficiency is given by;
                                             
I= √ Wc/Ra

Therefore, the efficiency of a DC generator will be maximum when the variable loss is equal to the constant loss.
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Harmonics In Synchronous Machines

What Are ? Harmonics In Synchronous Machines

Harmonics: When the uniformly sinusoidally distributed air gap flux is cut by either the stationary or rotating armature sinusoidal emf is induced in the alternator. Hence the nature of the waveform of induced emf and current is sinusoidal. But when the alternator is loaded waveform will not continue to be sinusoidal or becomes non-sinusoidal. Such non-sinusoidal wave form is called complex wave form.

By using Fourier series representation it is possible to represent complex non-sinusoidal waveform in terms of series of sinusoidal components called harmonics, whose frequencies are integral multiples of fundamental wave. The fundamental wave form is one which is having the frequency same as that of complex wave.The waveform, which is of the frequency twice that of the fundamental is called second harmonic. The one which is having the frequency three times that of the fundamental is called third harmonic and soon. These harmonic components can be represented as follows.

Fundamental: e1 = Em1 Sin ( t ± θ1)
2nd Hermonic e2 = Em2 Sin (2 t ± θ2)
3rd Harmonic e3 = Em3 Sin (3 t ± θ3)
5th Harmonic e5 = Em5 Sin (5 t ± θ5) etc.
In case of alternators as the field system and the stator coils are symmetrical the induced emf will also be symmetrical and hence the generated emf in an alternator will not contain any even harmonics.

Slot Harmonics: As the armature or stator of an alternator is slotted, some harmonics are induced into the emf which is called slot harmonics. The presence of slot in the stator makes the air gap reluctance at the surface of the stator non uniform. Since in case of alternators the poles are moving or there is a relative motion between the stator and rotor, the slots and the teeth alternately occupy any point in the air gap. Due to this the reluctance or the air gap will be continuously varying. Due to this variation of reluctance ripples will be formed in the air gap between the rotor and stator slots and teeth. This ripple formed in the air gap will induce ripple emf called slot harmonics.

Minimization Techniques of Harmonics: To minimize the harmonics in the induced waveforms following methods are employed:
1. Distribution of stator winding.
2. Short Chording
3. Fractional slot winding
4. Skewing
5. Larger air gap length.

Effect of Harmonics on induced emf:

The harmonics will affect both pitch factor and distribution factor and hence the induced emf. In a well designed alternator the air gap flux density distribution will be symmetrical and hence can be represented in Fourier series as follows.

The RMS value of the resultant voltage induced can be given as

Eph2 =  [(E1)2+ ....+ …………… (En)2]

**(A)2 Means A Square 

Effect of Harmonics of pitch and distribution Factor:

The pitch factor is given by Kp = cos /2, where is the chording angle.
For any harmonic say nth harmonic the pitch factor is given by Kpn = cos n α/2
The distribution factor is given by Kd = (sin mβ /2) / (m sin β/2)

For any harmonic say nth harmonic the distribution factor is given by 
Kdn = (sin mn β/2)/(m sin nβ/2)

This is the detailed info about Harmonics In Synchronous Machines,Minimization Methods of Harmonics.Effect of Harmonics on induced emf.

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EMF Equation Of Alternator / 3 Phase AC Generator EMF Equation

EMF Equation of an Alternator

We know Synchronous Machines generates E.M.F.the amount of EMF generated can be calculated using below simple derivation.

Consider following
Φ= flux per pole in wb
P = Number of poles
Ns = Synchronous speed in rpm
f = frequency of induced emf in Hz
Z = total number of stator conductors
Zph = conductors per phase connected in series
Tph = Number of turns per phase
Assuming concentrated winding, considering one conductor placed in a slot
According to Faraday's Law electromagnetic induction,
The average value of emf induced per conductor in one revolution 
eavg = dΦ /dt
eavg = Change of Flux in one revolution/ Time taken for one revolution

Change of Flux in one revolution = p x Φ
Time taken for one revolution = 60/Ns seconds.
Hence eavg = (p x  Φ  ) / ( 60/Ns) = p x   Φ x Ns / 60
We know f = PNs /120
hence PNs /60 = 2f
Hence eavg = 2   Φ f volts
Hence average emf per turn = 2 x 2 Φ f volts = 4Φf volts
If there are Tph, number of turns per phase connected in series, then average emf induced in Tph turns is
Eph,avg = Tph x eavg = 4 f   Φ Tph volts

Hence RMS value of emf induced E = 1.11 x Eph, avg
= 1.11 x 4  Φ f Tph volts
= 4.44 f  Φ Tph volts

Eph,avg= 4.44 f  Φ Tph volts

This is the general emf equation for the machine having concentrated and full pitched winding.In practice, alternators will have short pitched winding and hence coil span will not be  180o(degrees), but on or two slots short than the full pitch.

***If we assume effect of 
Kd= Distribution factor
Kc or KP = Cos α/2

Eph,avg= 4.44Kc  Kd f  Φ Tph volts
This is the actual available voltage equation of an alternator per phase.If alternator or AC Generator is Star Connected as usually the case, then the Line Voltage is √3 times the phase voltage.

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