3.DESIGN OF YOKE

Taking 20% greater area than that of core for yoke

Area of yoke (A_y)  = 1.2*A_i (take assumption into consideration)

                                 = 1.2*0.0182

                                 = 0.2184m²

Cross area of yoke =Net area/stacking  factor

                                   =0.02184/0.9   =0.0243m²

Taking shape of yoke is rectangular

Here,depth of yoke is equal to depth of core

        D_y= a=0.1567m

Height of yoke(H_y)= A_y/D_y

                                  = 0.0243/0.1567

                                   =0.15507m

  2. DESIGN OF WINDOW

 We know, Q=3.33*f* _m*A_w*k_w* *10^-3

For 125KVA transformer ,k_w=10/(30+KV)

                                                    =10/(30+11)  =0.2439s

 Assuming the value of current density

 current density , = 2.3 N/mm²[for distribution transformer]

A_w= Q/(3.33*f* _m*k_w* *10^-3)

      =125/(3.33*50*0.0226*2.3*10³)

       =0.0577m²

Assume , window height to width ratio =3

           i.e  H_w/W_w=3

    where H_w=height of window

                W_w=width of window

Assuming window height to width ratio=3  Then, H_W=3*W_w

        Now A_w= H_w*W_w

               A_w=3* W_w²

W_w= A_w/3)

W_w= (0.0577/3)

W_w=0.139

And H_w=3* W_w

               =3*0.139

               =0.417

Distance between adjacent limb(D)= W_w+d =0.139+0.174 =0.312m

transformer design

Q .Design a 125KVA , 11000/433, 50Hz, 3 phase delta/star core type oil immersed natural cooled distribution transformer .The transformer is provided with tapping ±2½%, ±5½% on the HV side. Maximum temperature rise not to exceed 45 ̊C.

Solultion

1.FOR DESIGN OF CORE

We have,

E_t=k  Q

For 3-phase distribution transformer k=0.45

Q=125KVA

E_t=0.45* 125 =5.03115 V/T

Now, _m=E_t/4.44f =5.03115/4.44*50 =0.0227 wb

Selection of material of core   Hot rolled silicon steel of grade 92

Flux density (B_m) =1.25 wb/m                [data from page 5.59]

Now, area of core (iron),A_i= _m/B_m

=0.0226/1.25

=0.0182m²

Assuming 3-stepped,

Area, A_i=0.6*d²

D

d= A_i/0.6

d=0.174

And , a=0.9*d = 0.9*0.174= 0.1566           b=0.7*d = 0.7*0.174=0.1218

           c=0.42*d= 0.42*0.174=0.07308

Explanation of Voltage Regulation of Transformer

Say an electrical power transformer is open circuited, means load is not connected with secondary terminals. In this situation, the secondary terminal voltage of the transformer will be its secondary induced emf E₂. Whenever full load is connected to the secondary terminals of the transformer, rated current I₂ flows through the secondary circuit and voltage drop comes into picture. At this situation, primary winding will also draw equivalent full load current from source. The voltage drop in the secondary is I₂Z₂ where Z₂ is the secondary impedance of transformer. Now if at this loading condition, any one measures the voltage between secondary terminals, he or she will get voltage V₂ across load terminals which is obviously less than no load secondary voltage E₂ and this is because of I₂Z₂ voltage drop in the transformer.

Expression of Voltage Regulation of Transformer

Expression of Voltage Regulation of Transformer, represented in percentage, is

What is Eddy Current Loss ?

In transformer, we supply alternating current in the primary, this alternating current produces alternating magnetizing flux in the core and as this flux links with secondary winding, there will be induced voltage in secondary, resulting current to flow through the load connected with it. Some of the alternating fluxes of transformer; may also link with other conducting parts like steel core or iron body of transformer etc. As alternating flux links with these parts of transformer, there would be a locally induced emf. Due to these emfs, there would be currents which will circulate locally at that parts of the transformer. These circulating current will not contribute in output of the transformer and dissipated as heat. This type of energy loss is called eddy current loss of transformer. This was a broad and simple explanation of eddy current loss.

Hysteresis Loss in Transformer

Hysteresis loss in transformer can be explained in different ways. We will discuss two of them, one is physical explanation and the other is mathematical explanation.

Physical Explanation of Hysteresis Loss

The magnetic core of transformer is made of ′Cold Rolled Grain Oriented Silicon Steel′. Steel is very good ferromagnetic material. This kind of materials are very sensitive to be magnetized. That means, whenever magnetic flux would pass through, it will behave like magnet. Ferromagnetic substances have numbers of domains in their structure. Domains are very small regions in the material structure, where all the dipoles are paralleled to same direction. In other words, the domains are like small permanent magnets situated randomly in the structure of substance. These domains are arranged inside the material structure in such a random manner, that net resultant magnetic field of the said material is zero. Whenever external magnetic field or mmf is applied to that substance, these randomly directed domains get arranged themselves in parallel to the axis of applied mmf. After removing this external mmf, maximum numbers of domains again come to random positions, but some of them still remain in their changed position. Because of these unchanged domains, the substance becomes slightly magnetized permanently. This magnetism is called " Spontaneous Magnetism". To neutralize this magnetism, some opposite mmf is required to be applied. The magneto motive force or mmf applied in the transformer core is alternating. For every cycle due to this domain reversal, there will be extra work done. For this reason, there will be a consumption of electrical energy which is known as Hysteresis loss of transformer.

Mathematical Explanation of Hysteresis Loss in Transformer

Determination of Hysteresis Loss

Consider a ring of ferromagnetic specimen of circumference L meter, cross - sectional area a m² and N turns of insulated wire as shown in the picture beside,

Let us consider, the current flowing through the coil is I amp,

Magnetizing force,

Let, the flux density at this instant is B,
Therefore, total flux through the ring, Φ = BXa   Wb

As the current flowing through the solenoid is alternating, the flux produced in the iron ring is also alternating in nature, so the emf (e′) induced will be expressed as,

According to Lenz,s law this induced emf will oppose the flow of current, therefore, in order to maintain the current I in the coil, the source must supply an equal and opposite emf. Hence applied emf ,

Energy consumed in short time dt, during which the flux density has changed,

Thus, total work done or energy consumed during one complete cycle of magnetism,

Now aL is the volume of the ring and H.dB is the area of elementary strip of B - H curve shown in the figure above,

Therefore, Energy consumed per cycle = volume of the ring X area of hysteresis loop.
In the case of transformer, this ring can be considered as magnetic core of transformer. Hence, the work done is nothing but the electrical energy loss in transformer core and this is known as hysteresis loss in transformer.

Emf Equation of Transformer

EMF Equation of transformer can be established in a very easy way. Actually in electrical power transformer, one alternating electrical source is applied to the primary winding and due to this, magnetizing current flowing through the primary winding which produces alternating flux in the core of transformer. This flux links with both primary and secondary windings. As this flux is alternating in nature, there must be a rate of change of flux. According to Faraday's law of electromagnetic induction if any coil or conductor links with any changing flux, there must be an induced emf in it.

As the current source to primary is sinusoidal, the flux induced by it will be also sinusoidal. Hence, the function of flux may be considered as a sine function. Mathematically, derivative of that function will give a function for rate of change of flux linkage with respect to time. This later function will be a cosine function since d(sinθ)/dt = cosθ. So, if we derive the expression for rms value of this cosine wave and multiply it with number of turns of the winding, we will easily get the expression for rms value of induced emf of that winding. In this way, we can easily derive the emf equation of transformer. Let's say, T is number of turns in a winding, Φ_m is the maximum flux in the core in Wb.

As per Faraday's law of electromagnetic induction,

Where φ is the instantaneous alternating flux and represented as, As the maximum value of cos2πft is 1, the maximum value of induced emf e is, To obtain the rms value of induced counter emf, divide this maximum value of e by √2. This is EMF equation of transformer. If E₁ & E₂ are primary and secondary emfs and T₁ & T₂ are primary and secondary turns then, voltage ratio or turns ratio of transformer is,