How differential relay not operate during transformer no load charging or inrush current

During the no load charging of transformer, inrush current phenomena takes place. High current up to 10 times of normal rated current flows into transformer. During no load charging, transformer is connected to power supply at one side keeping other side open for no load charging. One type of overfluxing phenomena takes place inside transformer. During no load charging, there is heavy flow of harmonics current inside transformer for few minutes. Modern relays are built such that it identifies this 2nd and 5th harmonic current. Relays are set such that it not generates any trip signal during presence of this harmonic current and due to this transformers don't trip during inrush current of no load charging.

Purpose of using Differential Protection in Transformer

Transformers are being protected from over current and earth fault protection, though we use differential protection. The purpose behind this is the over current and earth fault protection senses the fault that is out of the zone of transformer also, but differential protection only senses fault within its specified internal zone. This specified zone we can consider where protection type PS class current transformers are installed on high voltage and low voltage side of transformer. 

Suppose there is any internal fault in transformer tank where oil is filled, then there is buchholz relay protection is given for detection of fault and to isolate the transformer. But in case of fault at points of bushing up to placement of CT, buchholz relay can not detect that fault. Differential protection can detect fault internal faults as well as this types of faults occurring at bushing points. The time taken by differential protection compared to buchholz relay is also very less and it can quickly isolate the transformer in case of faults.

Generally this type of protection is being used for transformers above 5 MVA capacity.

What happen if we connect Capacitor in series with Bulb

Hello Dear Knowledge Seekers,

Many of us may have tried by connecting the capacitor in series with bulb. They will have experienced that when we connect capacitor in series with 230 volt bulb, it glows light. If we connect two capacitor in parallel with each other and connect them in series with 230 volt bulb, then now bulb now glows some more bright than it was at previous.

Question may now can arise that capacitor is not allowing bulb to glow then how capacitors are useful for series compensation of transmission line.

Here we will understand that why this type of phenomena we are experiencing while series connection of capacitor to a bulb.

To understand this topic, see picture as below:

Here I have connected 20 micro-ohm capacitor and bulb in series with 230 volt, 50 HZ AC supply. I have considered bulb resistance is 20 ohms.

So here, resistance R= 20 ohms, capacitance C=20 micro-ohms and voltage V=230 volts.

The brightness of bulb depends upon how much voltage it gets from supply mains. To calculate that, we will need circuit current and impedance. For calculating impedance, we will need capacitive impedance Xc. Its value is 159 ohms as shown above in picture.

Now from R and Xc, we can calculate circuit impedance Z, which is 160 ohms as per above in picture. Now we can calculate current I which value comes 1.44 amps as per above in picture. Now we can calculate voltage drop across bulb, which is 28.8 volts and voltage drop across capacitor which is 228.96 volts as per above mentioned calculation in picture.

So only approx 28 volt becomes available across bulb and it doesn't glow bright.

Now consider the second case where two capacitor in parallel with each other and in series with bulb is connected as shown in below picture.

Here resistance R and voltage V is same as previous example. Two capacitor of 20 micro-farad are connected in parallel with each other so its total capacitance value will be sum of two capacitor and will be 40 micro-farad.

Now we will calculate capacitive impedance Xc, impedance Z, current I, voltage drop across bulb and voltage drop across capacitor. Their values will come 79.58 ohms, 82.05 ohms, 2.80 amps, 56 volts and 222.82 volts respectively as per calculation shown above in picture.

So we come to know that by connecting two capacitor in parallel, voltage drop across resistance is increased and voltage drop across capacitor is decreased. So in this case as the voltage available to bulb is now increased, it will glow some what brighter than previous case.

Hope you have understood the basics of this phenomena.

Thanks for visiting my blog....