Wednesday, 31 August 2016

How are Overhead Line Insulators Cleaned?


Cleaning of overhead transmission line is not required because any dust and dirt that falls out of the sky onto them will be limited thickness and will not affect the electricity flow in any way nor will it decrease clearance (which is the insulation) in any meaningful way. Thus it is not necessary to clean overhead conductors but it is necessary to clean the insulators that support the conductors when they become contaminated with salt spray from the ocean or magnesium chloride used to melt ice on roadways in cold regions as well as from other contamination products.



One method for cleaning insulators is to use water through a pressure washer similar to what is used at a car wash. This cleaning can take place with the transmission line de-energized or energized depending on the utility’s operational practices. If the line is energized, the water used needs to be distilled and free of minerals and salts to reduce the risk of a flash-over back to the nozzle or phase-to-tower.


Thank you!

Why do Capacitor Block DC but Allows AC




All of us know that a Capacitor do not allow DC current to pass through it but allows AC current. In this post we will discuss this kind of behavior of Capacitor.

First we will consider DC supply connected to a parallel plate capacitor as shown in figure below.



Let the capacitance be C. So as we connect the one of parallel plate capacitor to the positive terminal of battery and another plate to the negative terminal of battery, a potential difference exists.

Here in our case, the potential difference between plate A of capacitor and positive terminal of battery = 5 V. Because of this potential difference, positive charge will start moving from the positive terminal of battery to the plate A of capacitor. Mind that charge is not supplied by the battery rather is the mobile electron of the connecting wire. Thus, the charge on the Plate A of Capacitor will be increasing from zero value to some finite value till the potential of plate A becomes equal to the potential of positive terminal of the battery. After this no further movement of charge will occur from positive terminal of battery to the plate A. Thus we can say at steady state, potential of plate A = 5 V and no further movement of charge i.e. no current.

Similarly, the initial charge on the plate B of Capacitor is zero but as soon as we connect the plate B to the negative terminal of the battery, a potential difference will exists due to which electrons will move from negative terminal of battery to the plate B (mind that potential of plate B is zero while that of negative terminal of battery is -5 V and electrons move from low potential to high potential.). This movement will take place till the potential of plate B becomes equal to the potential of negative plate of battery. Thus in steady state, the potential of plate B = -5 V and no further movement of charge i.e. no current.

But note that DC current is flowing through the capacitor till the transient state lasts i.e. till the time the potential difference between the plates of capacitor and battery becomes zero.
Thus we observe that in steady state, there is no potential difference between the plates of capacitor and the battery terminals to drive current. That is why a Capacitor is said to Block the DC current.

How Capacitor Allows AC?

Consider a parallel plate Capacitor connected with an alternating Voltage Source as shown in figure.



Let V = VmSinwt and its curve will be as shown in figure.



When AC voltage is applied across the plates of parallel plate capacitor, plate A will start to get charge till VPK and plate B of capacitor will get negative charge. But after peak voltage VPK, as the voltage of source is less than the voltage across the capacitor plates, the capacitor will start discharging till source voltage becomes zero. After that as the source voltage is going negative, the Plate A will now become negatively charged while plate B will be positively charged till negative peak of source voltage but once negative peak of applied voltage crosses, the capacitor will again start to discharge as the potential difference across the plates of capacitor is more than the source voltage. In this way Capacitor continuously charge and discharge for applied AC and hence we say Capacitor is allowing AC to flow.

Mind that when the applied voltage is at its peak, the capacitor is fully charged and therefore no movement of charge will take place at this instant and hence current through the capacitor is zero when the applied voltage is its peak. Similarly, when the applied voltage is zero, the capacitor is fully discharged and therefore as the voltage increases just from its zero voltage, charging current will start flowing from source to the plates of capacitor but as the charge gets accumulated on the plate, the potential of plate rises resulting in decrease in potential difference between the plates and the source. Because of this the magnitude of charging current decreases and becomes zero when the potential of plates of capacitor becomes equal to the source potential. This is why, we say Capacitor takes leading current.

Have Question? Please write in comment box. Thank you!

Tuesday, 30 August 2016

Understanding Induction Motor Stability


The concept of stability of Induction Motor is very important and vital for interview point of view or from selection of a particular Induction Motor for a given load. In all Motors, the speed drops if we increase the load torque as the power of Induction Motor is constant (Power = TorquexSpeed).

For getting a operating point for a given load torque, the point of intersection of load torque characteristics and Motor speed Torque characteristics determines the point of stable operation of Motor. But this does not guarantee that the point of intersection will be stable one.

Suppose the Motor characteristics is such that as the load torque increases the speed of Motor increases then obviously this Motor and load combination will not result into stable operation.

Thus it is very important for stable operation of the system consisting of Motor and Load that as the Load Torque increases Motor speed decreases and vice versa.

Normally an Induction Motor is designed to operate at low full load slip ranging from 0.02 to 0.05 under normal operating condition.



Figure above shows the slip-torque characteristics of an Induction Motor. Normal operating region is shown from s=0 to s=smT by solid line. The dotted region from s=1 to s=smT is not used for the operation.

Now we will consider some cases to have full understanding of stability of Induction Motor.

Case1:

Consider the slip-torque and load torque curve as shown in figure below.



As can be seen from the figure above, the load curve and the slip-torque characteristics intersect at point C, so this point C will be the operating point but will this operating point be stable operation? To answer this, we increase the load torque, as shown by the dotted line. As we see, as the load torque increases, the new operating point becomes D where the slip is more when compared with point C which in turn means that speed of Induction Motor has decreased. Hence we can say from the definition of Stability of Induction Motor that operating point C is stable.

Case2:

Consider the slip-torque and load torque curve as shown in figure below.



Carefully observe that, the point of intersection of load torque curve and the slip-torque characteristics is A. Now we decrease the load torque from TL1 to TL2. As we the motor torque is now more than the load torque TL2, the motor will accelerate till it reaches the operating point B. Mind that there is a far variation in the speed of Motor at point A and B. At point A the slip is near 1 while at point B, slip is near zero. Therefore we can say that operating point between s=1 to s=sMT are not as stable as operating point between s=0 to s= sMT.

Now, we increase the load torque, again from TL2 to TL1, the operating point shifts to point C. As the speed of motor at C is less than that of speed at B, hence operating point B is stable one. Also note that in this case, the variation in speed at B and C is less, that is why we say as operating point between s=0 to s= sMT is more stable.

Case3:

Consider the slip-torque and load torque curve as shown in figure below.



In this case the load torque requirement is more than the Induction Motor starting torque so the Motor will not start at all and hence no question for stability analysis of operating point.


Please write in comment box if you have any question. Thank you!

Sunday, 28 August 2016

Calculation of Reactive Power of a Capacitor


This post gives is a quick derivation of the formula for calculating the steady state reactive power absorbed by a capacitor when excited by a sinusoidal voltage source.

Given a capacitor with a capacitance value of C in Farads, excited by a voltage source V in volts, it will draw a current i amps into its positive terminal. If  V  is a steady state sinusoidal source with frequency of  ω=2πf , then we may use a Laplace Transform to solve the circuit in the frequency domain where  V  is the phasor rms voltage,  I  is the phasor rms current, and  Zc  is the capacitive reactance in ohms of the capacitor as follows:



Note that the negative sign means that the capacitor is absorbing negative reactive power VARs which is equivalent to stating that the capacitor is supplying reactive power to the external circuit or system. For a three-phase system, multiply Q by 3 to get the total reactive power supplied by the Capacitor.


Thank you!

What Happen if an Induction Motor Overloaded?


Carefully observe the curve below to have an insight into the situation.



The peak is the Pull-Out torque in the curve or slip torque characteristics. To the right of the peak is the Stable region of operation and to the left is the unstable region of operation i.e. the motor cannot operate in unstable region. The slant lines are the load torque lines taken for demonstration purpose only, they are different for different types of loads.

Under such a scenario, wherever the Motor torque curve and the load torque line meet, in the STABLE REGION, that is the point of operation of the motor i.e. the speed at which the motor operates for a particular value of load. Now, if we go on increasing the load, as we can see from the curve above, after the point of maximum torque, the load torque and the motor torque characteristics do not meet. This shows that there exists no point of operation for the motor. Thus, the motor decelerates and stops.


For this reason, it is called the Pull-Out Torque since the motor pulls out of operation.

Have Question? Write in comment box. Thank you!

Saturday, 27 August 2016

Auto-Transformer Starting of an Induction Motor


It can easily be seen from the slip torque characteristics of an induction motor that, there is some finite torque when the slip s=1 i.e. speed is zero. This simply means that Induction Motor is a self-starting motor and begins to rotate on its own when connected to a 3 phase supply.

At the instant of starting, a three phase Induction Motor behaves like a Transformer with its secondary winding shorted. Therefore, Induction Motor during starting takes a high current from the supply mains. To limit this high starting current of Induction Motor, different starting methods are used. In this post we will have a look at the Auto-Transformer Starting Method of Induction Motor.

The main philosophy of starting any Induction Motor is to start it at a reduced voltage and as soon as the motor reaches its rated speed, full supply voltage is applied to the terminals of Induction Motor. A schematic diagram for Auto-Transformer Starting of an Induction Motor is given below.



It shall be observed that, using Auto-Transformer we are only applying a reduced voltage xV1 to the Stator terminal of Induction Motor. Here x is less than 1. As soon as the Induction Motor reaches its rated speed, full supply voltage is applied to the terminals of stator.

Therefore, per phase starting current of Motor = xV1/Zsc = xIsc

Here Isc is the current through the stator during direct switching of motor.

Thus we observe that starting current of Motor has reduced and is x times that of current during DOL (Direct Online) starting.

Again,

Input VA of Auto-Transformer = Output VA of the Auto-Transformer

Ist.V1 = xV1(xIsc)

Therefore,

Per phase starting current from the Supply Mains Ist = x2Isc

Thus per phase starting current from Supply Mains has now became x2 times that of DOL current. Mind that it has reduced as x is less than 1 so x2 will be much less than 1. Thus the main advantage of using Auto-Transformer is that it reduces the starting current from the Supply Mains by x2 times.

Note that starting current is the motor winding is x times while the starting current from the supply mains has became x2 time of DOL starting current.

Now we will have a look at Torque of Induction Motor.

As the torque of an Induction Motor is directly proportional of square of applied voltage at the stator terminals, therefore

Starting Torque with Auto-Transformer Test = K(xV1)2 where K is constant of proportionality.

Starting Torque with DOL Test = KV12 where K is constant of proportionality.

Therefore,

Test with Auto-Transformer/ Test with DOL = x2

Thus starting torque with Auto Transformer is less than the starting torque with DOL starting by a factor of x2.


Have question? Write in comment box. Thank you!

Friday, 26 August 2016

How to Prepare for Mathematics for GATE-2018


Mathematics is an important subject for securing a good rank in GATE exam. If you see the statistics, you will note that every year around five questions are asked form Mathematics. Marks wise it will stand around 12. So once can secure his/her 12 marks by just preparing Mathematics.



There are three sections where math is asked. One in Aptitude Section (1 Question Carrying one marks and 1 Question of two marks), Section-1 (carrying one marks) and Section-2 (Carrying two marks). The math question coming in Aptitude question is quite simple and can be solved by having a good knowledge of mathematics up to class 12. It does not need any special preparation to solve mathematics question of Aptitude Section. Please have a look at some of the mathematics Questions from previous year in Aptitude Section.

Year 2015:

1 Marks:


2 Marks:


Year 2014:

1 Marks Question:


2 Marks Questions:


It is clear that the above two questions are simple can easily be solved using elementary knowledge of Mathematics.

Mind that every year one question from Maxima Minima comes for sure. So be ready to grab the marks for that Maxima Minima question without giving special attention.
Now, we will discuss about the main sections i.e. Section-1 and Section-2. For main sections you need to prepare for mathematics and that too in very selective manner. If you observe the previous year question paper, you will notice that question from a particular section comes like they are from Fourier Transform, Z-Transform, Integration (Line, Surface and Volume) etc.

Thus special attention needs to be given on these topics. I will suggest you to go through the whole mathematics as the syllabus is very short for mathematics while the marks for math question is handsome. You are going to get 12-15 marks just by going through few topics like Integration, Maxima Minima, Rolls and Lagranges Mean Value Theorem, Fourier Transform, Z-Transform etc.

Made Easy Study material for Mathematics is quite good and the best part of it is that it is very precise and can be complete fast. Please download the Study Material for Maths from the link below and start you preparation. Mind that don’t waste your much time on a single paper rather be smart and finish it within specified time say 10 days and practice as much as you can do.


All the Best!!!