BASIC OF PYTHON PROGRMMING

BASIC OF PYTHON PROGRMMING

TRANSFORMER

                                   Transformers

 • A transformer is a static device that transforms electric energy from one ac voltage level to another. It is this device that has made the electric system almost universally AC. The electric power is generated at relatively low voltages (up to a maximum of 33 kV) which then is raised to very high voltages (e.g. 756 kV) by means of a transformer and then transmitted.

 • High voltages are associated with low currents and reduced transmission losses.

• The core which supports the transformers mechanically and conducts their mutual flux, is normally made of highly permeable iron or steel alloy (cold-rolled, grain oriented sheet steel). Such a transformer is generally called an iron-core transformer.

 • However in special cases, the magnetic circuit linking the windings may be made of non-magnetic material, in which case transformer is referred to as an air-core transformer.

• Air core transformer used in radio devices and in certain types of measuring and testing instruments.

• The magnetic core of the transformer is made up of stacks of thin lamination of 0.35 mm thickness of CRGO lightly insulated with varnish, this material allow the use of high flux density [ 1-1.5 T] and its low loss property together with laminated core reduces the core loss to fairly low values.

Core Type transformer: -

• In core type transformer windings are around two legs or three legs (depend on their phases) of the rectangular magnetic core.

• Rectangular magnetic core is made by using the E and L shape of the sheets.

• Though most of the is confined to a high permeability core, some flux always leaks through the core and lies in air called leakage flux.

 • Leakage is reduced by bringing the two coils closer. In core type we achieve this by using L.V. and H.V. on each limb of the core.

Shell type transformers: -

• In this type of the transformer windings are wound on the central leg of a three legged core.

• Leakage in shell type transformer is reduced by sub-dividing each winding into sub-section and interleaved in L.V. and H.V winding.

Conservator: -

Power transformer is provided with a conservative through which transformer breaths into air. It is a small size tank placed on the top of main tank. It prevents fast oxidization and consequent deterioration of insulating properties of oil.

Explosion vent tube: -

 Purpose of this is to prevent damage of transformer tank be releasing any excessive pressure generated inside the transformer.

Radiator: -

•  When an electrical transformer is loaded, the current starts flowing through its windings. Due to this flowing of electric current, heat is produced in the windings, this heat ultimately rises the temperature of transformer oil. Hence, if the temperature rise of the transformer insulating oil is controlled, the capacity or rating of transformer can be extended up to significant range. The radiator of transformer accelerates the cooling rate of transformer. Thus, it plays a vital role in increasing loading capacity of an electrical transformer. This is basic function of radiator of an electrical power transformer.
•   The working principle of radiator is very simple. It just increases the surface area for dissipating heat of the oil. It is used in large size transformers

Breather: -

• When the temperature change occurs in transformer oil, the oil expands or contracts. There is an exchange of air also occur when transformer is fully loaded.

• When the transformer gets cooled then oil level gets down and when it goes down it absorbs air this process is called breathing.

• Silica gel breather controls the level of moisture. Silica gel is used to absorb moisture content from air.

• When silica gel absorbs moisture, it becomes pink. Generally, its colour is blue. 

E.M.F. equation: -

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 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.

As 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 is the e.m.f induced in the primary winding

E1 = − dλ /dt

Where λ= N1ɸ

Therefore, E1 = − N1/ dɸ dt

Since ϕ is due to AC supply ϕ = ϕm Sinωt

E1 = − N1 d /dt (ϕm Sinωt)

 E1 = − N1 ϕm ω Cosωt

 E1 = N1 ϕm ω Sin (ωt− 90°)

So the induced e.m.f lags flux by 90 degrees.

Maximum valve of e.m.f,

 E1 = N1 ϕm ω

But ω = 2πf

 (E1) max = 2πf N1 ϕm

Root mean square RMS value is

 E1= (E1)max √2 …………….(1)

Putting the value of (E1)max in equation (1) we get

 

Now, equating the equation (2) and (3) we get

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

Equivalent circuit of the transformer: -

The equivalent circuit diagram of any device can be quite helpful in predetermination of the behaviour of the device under the various condition of operation.

 It is simply the circuit representation of the equation describing the performance of the device.

The simplified equivalent circuit of a transformer is drawn by representing all the parameters of the transformer either on the secondary side or on the primary side.

 The equivalent circuit diagram of the transformer is shown below

 

Let the equivalent circuit of a transformer having the transformation ratio K = E2/E1

 The induced e.m.f E1 is equal to the primary applied voltage V1 less primary voltage drop.

This voltage causes current I0 no load current in the primary winding of the transformer. The value of no-load current is very small, and thus, it is neglected. Hence, I1 = I1’. The no load current is further divided into two components called magnetizing current (Im) and working current (Iw).

 The secondary current I2 is

 

 The terminal voltage V2 across the load is equal to the induced e.m.f E2 in the secondary winding less voltage drop in the secondary winding.

 

Equivalent circuit when all the quantities are referred to primary side: -

 In this case to draw the equivalent circuit of the transformer all the quantities are to be referred to the primary as shown in the figure below

 

The following are the values of resistance and reactance given below Secondary resistance referred to primary side is given as

 

 The equivalent resistance referred to primary side is given as

 Req = R1 + R2 ’

Secondary reactance referred to primary side is given as

 

 

The equivalent reactance referred to primary side is given as

 Xeq = X1 + X2 ’

Further simplification of the equivalent circuit of the transformer can be done by neglecting the parallel branch consisting R0 and X0. The simplified circuit diagram of the transformer is shown below.

 

 O.C. test and S.C. test: -

The aim of carrying out O.C. test and S.C. test on a transformer is to predict its performance without actually loading it.

 O.C. Test: -

  O.C. test is carried out at rated frequency and rated voltage to determine the core loss. The iron loss is thus is treated as constant, in spite of minor voltage variation in voltage and frequency during actual operation.

  This test is carried out with the instruments placed on low voltage side while the high voltage side is left open circuited.

 This is done because it is easier to manage rated voltage supply at low voltage level rather than at high voltage level. Also the instruments used are economic in cost and it is easier to work on low voltage side.

  Therefore no load current is limited to 5% of full load current, a primary winding copper loss is ignored, also the primary impedance drop at such low current is neglected.

 Because no load power factor is very low, it is recommended that a low power factor wattmeter to be used.

Then the iron loss of the transformer Pi = W0 and

 

The no-load power factor is

 

 

 

S.C. Test: -

S. C. test is carried out at rated current to determine the full load copper loss.

  This test is carried out with the instrument placed on high voltage side while the low voltage side is short circuited by a thin a conductor (so that wire will not burn). This is because a rated current is lower on low voltage side as compared to high voltage side.

  Consequently, the instruments are economic in cost since the voltage required to circulate full load current at circuited would be about 10% of rated voltage.

 The core loss under this low voltage condition is ignored. Also, the exciting current at such low value of voltage will be completely neglected.

  Short circuit test need not to be carried out strictly at rated frequency because the copper loss that depends upon the winding resistance is independent of the frequency of the supply as the skin effect in transformer at power frequency is negligible.

 Wattmeter used in this test is of high-power factor.

 

 

ELECTRICAL DESIGN

in substation, control unit, SCADA unit we use the battery bank for operation of relay and circuit breaker.

 At the condition of supply failure our SCADA system will not get shut down due the battery bank.

when supply will restore, we required supply to operate the panel to connect the load, which get form the battery bank

so for maintain this battery bank continuously charge we required the battery charger.

we used the float cum boost battery charger to charge the battery bank .

the diagram for float cum boost battery charger is as shown in fig.

CONSTRUCTION: -

It contains the three-phase ac supply with to battery charger is attached.

This three-phase ac supply is connected to the input of both float and boost battery charger.

Both the battery charger contains the rectifier circuit at output. Between the both charger there is VC contactor, which separate both from each other.

It also contain the battery bank of 110 v .

 WORKING: -

From fig, normally we charge the battery bank form the float charger.

The float charger input is connected to the 3-phase ac input supply. Then it converted this ac supply to dc and connected to the battery and also load because it performs two function:

1.sypply load.

2. charge the battery.

 When battery is completely dead at that condition battery is charged through boost charger which produce large voltage for small time of period to boost the charging.

 when we used boost charger at that time float charger is off and also VC contractor is disconnected.

 Boost charger can also supply the load but through diode to limit the voltage. because boost charger gives 2.6 voltage for each 2v battery, so for 110v it will be 143v and this voltage is not good to directly connected to the 110v load.

HOW TO CALCULATE FUSE RATING FOR MOTOR

Fuse rating for 40 HP  3 phase motor: -

Fuse rating for 30 kw 3 phase motor: -