OPEN CIRCUIT TEST ON TRANSFORMER

In the previous article, we have seen that there are four tests of the transformer.

1) Polarity test

2) Open circuit test

3) Short circuit test

4) Sumpner’s test or back to back test

In this article, we will explain the open circuit test of the transformer.

Open circuit test should be conducted at rated flux in the transformer core. The rated flux can be achieved by keeping other winding terminals open and keeping the existing transformer at rated voltage and frequency.

The objectives of open circuit test are:

a) To find the constant losses of the transformer.

b) To find shunt branch parameters of equivalent circuit R₀ and X₀.

c) To separate iron losses and find hysteresis loss and eddy current loss.

This test is conducted on the LV side of the transformer because at LV side low range volt meter and watt meter are sufficient to conduct the test. Another reason is that the magnitude of the no-load current is more on the LV side. So, we can measure no-load current more accurately compared to the HV side. Therefore, in most of the cases, this test is conducted on the LV side of the transformer.

Generally, for this test LPF wattmeter is used because the power factor is at nearly 0.2 lag. The measurements are primary voltage (V₁), no-load current (I₀) and no-load power (W₀). This entire test can be divided into three parts.

1) Find out R₀ and X₀:

Above circuit represent approximate equivalent circuit of the transformer. From this circuit,

R₀ = V₁/I_w

X₀ = V₁/I_u

Where, I_w = core loss current = I₀ cos ф_o

and I_u = magnetizing current = I_u sin ф_o

W₀ = No load power = V₁I₀cosф_o

Cos ф_o = W₀/V₁I₀

From the measurement, we have no-load current (I₀), no-load power (W₀) and voltage (V₁). First, we have to find Iw and Iu. From the above equation, we can find out R₀ and X₀.

2) To find out constant losses:

W₀ = losses in transformer under no load condition

W₀ = Iron loss + dielectric loss + No load primary copper loss

The iron loss and dielectric loss is considered as a constant loss. The dielectric loss is negligible in this case.

Iron loss = W₀ – I₀²R₁

Now, by test measurement, we have a value of W₀ and I₀. But we don’t have the value of R₁. Therefore, the Kelvin double bridge method is used to measure the value of resistance R₁. In this condition, we neglect the primary copper loss and dielectric loss. so, watt meter reading during open circuit test is approximately considered as an iron loss.

3) Separation of iron losses: The iron loss can be divided into two parts; the hysteresis loss and eddy current loss. This test should be conducted with variable frequency and voltage, in such a way that the ratio of voltage and frequency (V₁/f) remains constant.

The first thing to do here is to apply rated voltage and rated frequency and note down the reading of watt meter and frequency meter. After that, apply to reduce voltage and set frequency in such a way that V₁/f ratio remains same and note down the reading of wattmeter and frequency meter.

By repeating this procedure with changes in voltage and frequency, we can draw a graph for (W_i/f) vs frequency. The find constant A and B, where A is the y-intercept and B is the slope of the curve.

Now from this you can calculate the hysteresis loss and eddy current loss at rated frequency and voltage by below equations.

Hysteresis loss at rated frequency = A x rated frequency

Eddy current loss at rated frequency = B x square of the rated frequency.

W_i = Af + Bf²

W_i/f = A + Bf

m = slop of straight line = tan θ = B

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180-DEGREE THREE PHASE INVERTER WITH SIMULATION

The inverters are used to convert DC power into AC power. There are several types of inverter available. Click on below link to know more about different types of inverter.

https://circuitdigest.com/tutorial/different-types-of-inverters

According to the mode of operation, three-phase inverters are classified into two types; 180-degree mode and 120-degree mode. Before going to classification, revise the circuit diagram of a three-phase inverter and it is as shown in below figure.

Circuit diagram of three-phase inverter:

As we can see in the circuit diagram, three phase star connected load is connected and output is measured at the load. One arm for each phase means we have a total of three arms. For switching purpose, we can use MOSFET or IGBT or any other power electronics switches according to the application.

When the upper switch of any phase is in conduction mode (ON), at that time positive half cycle of that phase appears in the output. When the lower switch of any phase is in conduction mode (ON), at that time negative half cycle of that phase appears in the output. In one arm two switches are available and these two switches never conduct at the same time. If this happens than DC source will short-circuit.

In the 180-degree mode of operation of the three-phase inverter, all switches conduct for 180-degree. In other words, MOSFET will ON for 180-degree and OFF for 180-degree of one cycle and the pulse width is 50% of the period.

Simulation in MATLAB/SIMULINK:

This is the simulation of three-phase inverter. The circuit diagram of 180-degree and 120-degree is same. The only difference is in the GATE pulse of switches. Here in this blog, we will discuss 180-degree inverter.

The gate pulse of 1 & 4, 3 & 6 and 5 & 2 are totally opposite in this mode of operation. When MOSFET-1 is ON at that same time MOSFET-4 is OFF.

Switching of MOSFET:

How to generate this gate pulse?

The screenshot of MATLAB for gate pulse generator is below image.

Gate pulse generator:

Vab = Va0 – Vb0

Vbc = Vb0 – Vc0

Vca = Vc0 – Va0

The peak value of phase voltage (Va0) is 2Vs/3 and the peak value of line voltage is Vs. where Vs is the DC supply voltage. The waveform of the 180 degrees is as shown in below figure. This is a book image. We will compare this image with simulation results at the end of this blog.

Waveform of 180-degree inverter:

How to set pulse width in the pulse generator. You have to create gate pulse only for MOSFET 1,3 and 5 and toggle it with the help of NOT gate and give it to MOSFET 6, 2 and 4. Now, the question is how to generate gate pulse for 1, 3 and 5. For that, we will use the PULSE GENERATOR. For 50Hz output signal, the time period is 20 msec and amplitude is 1 for all pulse generator. For 180-degree inverter mode, the pulse width is 50%.

Phase delay for one signal is 0 sec and this signal is for phase-A and MOSFET-1. It is clear in the above table that, there is 120-degree phase delay between these two signals. So, phase delay for phase B and MOSFET-3 is 120-degree or (2e-3)*2/3. Now for the third signal, the phase delay is 60-degree or (2e-3)*1/6 and this signal is for MOSFET-5.

By this way, we have three signals. For the other three signals we will use NOT gate and inverse this signals as shown in the above screenshot.                                      

Output waveform:

from above figure we can see that, for DC supply of Vs = 100V, the peak phase voltage Va0 is (2*100)/3=66.6V and the peak phase voltage Vab = 100V. This result is same as book waveform. If you need this simulation you can contact me or comment in this blog.

Thanks for reading.

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POWER SYSTEM STABILITY

The study of the electrical power system is incomplete without the study of the power system stability. The definition of power system stability is the ability of the power system to return back to a steady state without losing synchronism.

The power system network is a very large network and millions of equipment are operating together in parallel. This all equipment is working in synchronism. If one of the equipment goes out of synchronism by an abnormal condition. It will affect the other equipment and thereby in the worst condition it is affecting the entire network.

Let us consider a small system having five generators. If one generator is going out of synchronism due to any fault, it will affect the other generators. Due to this, the entire system will face connivances. So, in this condition, the faulty generator immediately switched out from the system. If the faulty generator is not switched out, then it will affect the voltage profile of the system. There will be large fluctuation in the voltage.

The power system stability is categorized into a steady state, transient and dynamic stability. The stability limit is defined as the maximum power that can be transferred in a network between source and load without loss of synchronism.

1) Steady state stability: It is the ability of the power system when operating under given load condition to retain synchronism when subjected to small and slow disturbances. The load fluctuation and turbine governors are examples of slow disturbances. The steady state limit is defined as the maximum power that can be transferred without making the system unstable. When the load is increased gradually under steady state condition.

2) Transient stability: It is the ability of the power system when operating under a given load condition to retain synchronism when subjected to large disturbances. The loss of generation, sudden changes in loads, a short circuit in the transmission line and short circuit in the transformer are the examples of the large disturbances in the power system. The time period of the transient study is very less. The transient stability limit is the maximum power that can be transferred without making the system unstable in the large disturbance condition. The transient stability limit is almost lower than the steady-state limit.

3) Dynamic stability: It is the ability of the power system to remain in synchronism after the initial swing until the system has settled down to the new steady state equilibrium condition. If the oscillation does not acquire more amplitude and die out quickly, then the system is said to be a dynamically stable system. The dynamic stability can be improved by the power system stabilizer. The time period of dynamic system study is 5-10 sec and some time it may be up to 30 sec. With an increase of load from the use of automatic voltage regulators and speed governors, if the increase in field current or adjustments in speed setting occur, the stability limit would be increased significantly. The limit under this condition is called the dynamic stability limit.

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POLARITY TEST IN TRANSFORMER

If the transformers are used in the parallel operation or used in the polyphase circuit, the polarities of primary and secondary terminals must be known at any instant. Due to the potential difference between two points, the current can from the high voltage point to low voltage point. The polarity is nothing but it describes the direction of the current flow. The dot convention is used to identify the polarity of two windings.

- If a current enters to the dotted terminal of one winding, the polarity of the voltage on the secondary winding will be positive at the dotted terminal.

- If current leaves to the dotted terminal of one winding, the polarity of the voltage on the secondary winding will be negative at the dotted terminal.

Assume the primary side of the transformer at a particular instant of time the polarity is A1 which is positive with respect to A2. In this case, if the voltmeter reading V = E1 + E2 means additive polarity, then B1 is negative and B2 is positive at any instant of time A1 is positive and A2 is negative.

Replace dot marks at either positive polarity terminals or negative polarity terminals on both sides.

During this polarity test, if the voltmeter reads sum of two EMFs, the secondary side terminals should assign as opposite polarity with respect to assumed primary polarity.

If voltmeter reading is V = E1 – E2 means negative polarity (difference of two EMFs), then the secondary side terminal should be assigned with the same polarity as assumed primary side.

This test can be easily implemented on a low voltage range of the transformer and cannot be used for a high voltage range of transformer.

DC kick test:

This method is used for low voltage range transformer as well as a high voltage transformer. In this method, the transformer is excited with DC supply. This excitation is not continuously. It is momentarily excitation.

If you give continuous excitation of DC supply than it is possible to damage the transformer because very high current will flow in the primary. So, never excite transformer with DC supply. If you want to excite transformer with DC, excite HV winding because it has a high resistance compared to LV winding.

When DC voltage is applied to the primary winding terminal A1 which is positive with respect to A2. If the switch is closed and opened and momentary supply is given to primary winding of the transformer. In this case, the galvanometer observed a kick.

If the kick is in the forward direction, secondary side terminals should be assigned as the same polarity of primary winding terminals. If dot notation is to be applied, a pair of similar terminals should be assigned with dot marks.

If the kick is in the backward direction, secondary side terminals should be assigned as the opposite polarity of primary winding terminals. If dot notation is to be applied, a pair of dissimilar terminals should be assigned with dot marks.

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DIFFERENCE BETWEEN POWER TRANSFORMER AND DISTRIBUTION TRANSFORMER

In the previous article, we have discussed the total losses occurred in the transformer. For good experience, read this article in DESKTOP MODE in any browser.

Total losses in transformer 

In this blog, we will discuss the difference between the power transformer and distribution transformer with design and operational perspective. 

	Power Transformer		Distribution transformer
1	The power transformer is used in the transmission network	1	The distribution transformer is used in the distribution network
2	Voltage level > 33 kV	2	Voltage level < 33kV
3	Fewer Load fluctuations	3	More Load fluctuations
4	The consumer is not directly connected	4	Consumers are directly connected
5	While designing copper losses are kept the minimum.	5	While designing iron losses are kept the minimum.
6	CRGO steel is used in the core of this transformer.	6	Amorphous steel is used in the core of this transformer.
7	Copper losses and iron losses take place steadily throughout 24 hours.	7	The copper loss takes place based on the load cycle of the consumer and iron loss take place steadily throughout 24 hours.
8	Full load copper loss is almost the same as an iron loss.	8	Full load copper loss is twice of the iron loss
9	Efficiency maximum occurs at nearer to the full load	9	Maximum efficiency occurs at 70-75% of full load.
10	Specific weight is less	10	Specific weight is more
11	Average load on the power transformer is almost nearer to the full load.	11	Average load on the distribution transformer is 70-75% of the full load.

 Watch video of Transformer explosion.

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TRANSFORMER LOSSES

There are four types of losses occurs in the transformer and that is,     

      1) Copper loss

      2) Iron loss

      3) Stray loss

      4) Dielectric loss

Click here to know the difference between the power transformer and the distributed transformer.

Out of above mentioned losses, copper loss and iron loss can be considered as major losses and stray loss and dielectric loss can be considered as minor losses. Let’s explain this in details.

1) Copper loss: There are two winding in the transformer; primary winding and a secondary winding. The copper loss is nothing but, the losses occur due to the winding resistance. The copper loss is also known as I2R loss or cu loss. If primary winding resistance is R1 and secondary winding resistance is R2, then

Total copper loss in transformer = 

(I1^2) R1 + (I2^2) R2

Where I1 is full load primary current

I2 is full load secondary current

From the above equation, we can see that copper loss is directly proportional to the square of the load current. It means that, as the load current changes, the copper loss also changes. So, this loss is considered as a variable loss.

2) Iron loss: This loss takes place in the core of the transformer due to the time-varying nature of flux in the core. So, the iron loss is also known as core loss. The total iron loss is divided into two parts; hysteresis loss and eddy current loss.

          (A) Hysteresis loss:  Whenever it is subjected to alternating nature of magnetising force, the hysteresis loss occurs due to the reversal of magnetisation of the transformer core. In this case, after every half cycle, the domain present in the magnetic material will change their orientation. The power consumed by this change of orientation after every half cycle is known as hysteresis loss.

The magnetic reversal of the transformer core is plotted with the help of the BH curve and it represents as below figure.

The hysteresis loss occurs in one cycle is equal to the area enclosed with the one hysteresis loop.

The hysteresis loss can be determined with the Steinmetz's formula. Which is given as

Where n= Steinmetz coefficient = range 1.5 to 2.5

F = supply frequency

V = volume of core

X = Steinmetz exponent = 1.6 for silicon steel

          (B) Eddy current loss: The eddy current loss is nothing but, the I2R loss present in the core of the transformer due to the production of the eddy current in the core. The eddy current is produced because of the conductivity of the core.

The eddy current loss is directly proportional to the conductivity (σ) of the core.

Rse = the resistance offered by the core to flow of eddy current.

Rse is inversely proportional to the conductivity.

By reducing the conductivity, the eddy current can be reduced. It is possible to reduce conductivity without affecting magnetic properties by adding silica content and by using laminated core.

3) Stray loss: It can be divided into two parts; copper stray loss and iron stray loss.

          (A) Copper stray loss: This loss occurs only under full load condition due to the leakage flux. This loss is the additional I2R loss due to stray current within the conductor. This loss is considered as a variable loss. Instead of the solid conductor, the stranded conductors are used to reduce the copper stray loss and by this way, the skin effect of winding is also reduced.

          (B) Iron stray loss: The iron stray loss is the additional iron loss occurred due to the auxiliary iron parts like transformer tank, steel channels, and conservation tank. In this auxiliary iron parts, due to the leakage flux, this loss is produced. This loss is less in shell type transformer compared to the core type transformer.

The leakage flux is directly proportional to the load current. So, this loss is also considered as a variable loss. The iron stray loss is just 0.5% of full load output.

4) Dielectric loss: This loss is produced in the insulating material of the transformer. In insulating material, the free electrons are not available. When a voltage is applied, a small amount of current will flow through this due to the conversion of atoms. There is displacement of charges and the current produce due to this is known as displacement current.

The process of conversion of atoms into electric dipole is known as polarization. The dielectric loss is depending on applied voltage and it is independent of load current. So, this loss is considered a constant loss. This loss is 0.25% of full load output.

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