The post PLC Applications: Primary Resistance Starter for AC Motor appeared first on .
]]>Induction motors are widely used in industry. They have lots of advantages however they inherit a problem of high starting current. As compared to full load current, the starting current is 6 to 10 times higher. This high inrush of current can damage the motor windings and other parts. High inrush current can also affect the source. Therefore it is required to limit the high starting current in induction motor. There are lots of methods for smooth starting of induction motors. Each one has its own advantages and disadvantages. Some of them are
In this article we will discuss the primary resistance starter with detail and also learn how to implement it in PLC. Motor starters are among of the basic PLC applications.
Primary resistance starting method is based on the reduced starting voltage concept. This method introduces the resistance in series with the stator windings of squirrel cage induction motor. Voltage drop occurs in the added resistors during starting which provides reduced voltage across the stator windings. With the reduced voltage, the starting current is lower than it would be if the motor is started on full line voltage. After a preset delay, the resistance unit is shunted out of the circuit and the motor continues to run on line voltage.
Above figure shows the primary resistor starter arrangement. It consists of a main contactor and a second contactor labelled as S2. Resistors are connected with each phase of stator winding in series to reduce voltage. The value of these starting resistors can be determined by the method explained in the article
By pass switches are connected in parallel to with the starting resistors which are controlled by S2 contactor.
The motor is started by closing the main contactor. This supplies power to the motor through series starting resistance. The contactor S2 is open thus the current has to pass through the resistors resulting in voltage drop. Due to voltage drop in these resistors, the motor gets reduced voltage.
A timer is installed with a preset time delay. When the delay is over, the contactor S2 is closed closing the parallel switches.
As it can be seen in the above figure that closing the S2 bypass all the resistors. As a result full line voltage is applied to the motor windings.
Following figure shows the PLC program for Primary resistance motor starter
Rung 1: It consist of start button normally open contacts I1, normally close contacts of stop button I2 and normally open contact of over load relay I3. Main contactor Q1 will only be energized if start is pressed, stop is not pressed and overload is not operated. An input normally open contact depending upon output coil Q1 is added in parallel with the start button contacts. It will ensure that power is supplied to rung 1 even if the start button is released. Adding this parallel path enables the input start button to work as a momentary push button.
Rung 2: It only consist of normally open input dependent upon output coil Q1 and a timer T1. When the main contactor coil Q1 is energized, the motor gets the voltage through the resistor and also the timer starts counting the preset delay. When the preset delay is over, the timer is activated.
Rung 3: It contains input depends upon main contactor Q1, an input relying on timer T1 and output coil Q3. The Output Q3 will only be energized when main contactor Q1 is energized and preset time delay is over.
It can be seen that very easily the primary resistance starter can be implemented with PLC and it is one of the basic PLC applications.
Working and program demonstration is further explained in this video
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]]>The post Theory, working and operation of Transformer Tap Changer appeared first on .
]]>Tap changer is a device which can increase or decrease the output secondary voltage by changing the turn ratio or Primary or secondary winding. Tap changer is generally installed on high voltage side of a two winding transformer due to low current in this side.
As a transformer ratio
We want to regulate the secondary voltage (V2)
Since V1 is the system voltage so we cannot change it. As tap changer is installed on the primary side of step down transformer hence N2 cannot be changed. So
N2 and V1 = constant (k)
$latex V2=\ \frac{K}{N1}\$
It means that
Frequent changes occur in the load of the power system, due to which the voltage of the system increases or decreases. So the Tap changer is used to keep the output voltage of a Transformer within prescribed limit.
(i) OFF LOAD TAP CHANGER
(ii) ON LOAD TAP CHANGER
If the required change in voltage is infrequent, then an offload tap changer is installed on a transformer and taps can be changed after completely isolating a transformer from the circuit. Such kind of a tap changer is usually installed on distribution transformer (with 5 steps).
With the expansion and interconnection of power system it often becomes necessary to change the transformer taps several times daily to obtain the required voltage on system as per load demand. The demand of continuity of supply does not permit to disconnect the transformer from system for offload tap changing. To meet this requirement, on load tap changers are installed on the majority of power transformers.
Range tapping ranges between +/ 10 % to +/16 % is used in normal conditions and +/22% is in special cases. The number of steps in the range is chosen so that the change in voltage between adjacent positions approximately 1 % to 1.5 % of rated voltage.
A physical, arrangement of on load tap changer in power transformer is shown below.
A typical on load tap changer is sketched in the attached diagram
Let suppose a current inter from point ‘A’ to main winding than enter in tap winding through reversing switch contact N—>L , Taps l & 2 both are connected with move able selectors but the current flows only in tap no. I (Because the diverter side1 is closed) so the current finally end on point B as shown in above fig.
Now if we want to change the tap position from tap1 to tap2, operates the drive unit motor through raise control switch .The motor starts operation (at this time only an action starts in diverter switch because tap 2 is already selected ).
Following operations take place for changing tap during on load
Diverter Switch 
Operation 
Description 
Operation position side 1  The load current flows via side 1 through the XContact  
Contact X. Break of side 1  Load current flows via side 1 through the Y contact.  
Contact Y. Breaks of side 1  The load current flows via side 1 through the ZContact
(Arching at Y contact due to break the easy path of current) 

Contact Z of side 2 makes

The load current is divided equally between side 1 tap and side 2 tap flowing via two equal transition resistors R and contact Z. In addition a circulating current flows via contacts Z of both sides  
Contact Z of side 1 breaks  The circulating current no longer flowing (circulating current flows 20ms to 40ms )
The load current flows via side 2 tap through contact Z and transition resistance R (Arching at contact Z of side 1) 

Contact Y of side 2 makes

Load current flows via side 2 tap and through the Y contact.  
Operation position side 2  The load current flows via side 2 through the XContact
Operation of tap changing is completed 
To see it live click here
For changing the tap1 to tap2 the voltage will be increased on secondary side of transformer because the numbers of primary turns have now reduced.
Similarly, if we want to go on tap no.3, the operational sequence will be same as above but in reverse (side 2 to side1) and now the operation will become at first in selector switch, as the odd moveable selector selects the tap no.3. For doing continuous changing the taps in step by step, we select the tap10. At this time only a main winding will be in circuited and current will flows from ‘A’ to ‘B’ via the main winding to diverter switch. The tap10 of this arrangement is called ‘principal’ tap. At this tap the voltage on both sides will be rated. Now if we want further changing in taps, operates the motor through the raise switch placed in drive unit.
Now the following operations will take place
The polarity of tap winding will he reversed which means the primary turns decreased , hence the secondary voltage would increased
The voltages increase, so tap1 is now called step no 11 and the reversing switch will convert the tapping range (2n1).
Where n = number of taps
If n=14
The number of steps becomes = 27
If you have any question relating this topic please ask in comments. Dont forget to share it with your friends.
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]]>The post Purpose and Procedure of Transformer Oil testing appeared first on .
]]>Insulating oil fill in the main tank of transformer. It has two purposes. One is to provide insulation between the HT and LT winding and between ground and windings. It also provides cooling to the transformer. Oil circulates between different parts of transformer. It carries heat from the windings to the radiator of transformer. Depending upon the type of cooling, heat is removed from transformer either by natural air or forced air cooling. Some Engineers equates the oil of transformer to the blood of human body. So if the condition of transformer oil is up to the mark, transformer can work properly and safely. Therefore some tests are conducted on regular basis to verify the condition of transformer oil.
Among different tests, testing the dielectric strength test of transformer oil is one of main test which was carries out to verify the insulating properties of oil under high voltage. Dielectric strength of Transformer must be above the threshold value. However this strength can be reduced by contaminants such as water, sediment and conducting particles. There can different reasons of these particles and contaminants which is beyond the scope of this article. These particles are mostly conducting which tend to reduce the dielectric strength of oil. Weakening of dielectric strength can result in leakage current between HT and LT winding. In extreme cases it can result in the spark and may be short circuit. One indication of healthiness of transformer oil is its dryness and cleanliness. However this does not necessarily indicates the absence of all contamination.
Usually students and yound graduate confuse these tests. Although both are insulation tests but meggering is done to find insulation between HT & LT winding and between both windings and body (Ground). On the other hand transformer oil testing is done to find dielectric strength of oil. To understand the procedure of meggering please refer to my article how to megger transformer.
The equipment is used to measure the dielectric strength of transformer must have the following provisions.
Typical transformer oil testing device is shown the figure.
It can be seen that it contains transformer oil cup, a voltmeter and high voltage provision. There are certain rules or requirements oil cup.
(Courtesy to Engr Akhtar hussain shah for his expert opinion and guidance on this topic.)
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]]>The post Design of three point starter of DC Motor appeared first on .
]]>In DC motors, the initial starting current is very large. According to rough estimate, starting current is 10 times the full load rated current. This high current can damage the motor windings therefore dc motor starter is necessary which can limit the current during starting of motor. Once the motor attains speed, the starter can be removed.
In order for DC motors to function properly, they must have some special control and protection equipment associated with them.
The purposes of DC motor starter are:
At starting condition, the armature (rotor) is not rotating. Therefore the internal generated voltage is zero. The armature current equation is
VT = Terminal voltage EA = Internal generated voltage RA= Armature resistance
In starting internal generated voltage EA = 0, therefore the starting current IA is very high.
For instance, for a 50 HP, 250 V DC motor with armature resistance RA of 0.06 Ω and a fullload current about 200 A. To find starting current put values in equation (i)
It can be seen that the value of starting current is almost 20 times the full load current. Such high current can damage the motor.
One solution to high starting current is to insert an external resistor in series with the armature. High resistance will prevent high inrush current. However there will be I^{2}R losses due to external resistor. Also the resistor will also decline the motor’s torque characteristics. Therefore external starting resistor should be removed from the motor when the motor attain speed and develop back emf.
If one big resistor is used and removed at once, a spike in current and voltage can be originated which can surpass the maximum limit and can damage the motor. Therefore the starting resistance can be removed in small steps to ensure smooth transition.
In practice, the starting resistor consists of a number of series resistors which are removed in steps when the motor attains speed. So designing of a starter consists of two parts
Case study To understand the design process, consider a 100 Hp, 250 V 350 A shunt DC motor with an armature resistance of 0.05Ω. Design a starter which can limit the starting current to twice its rated value and remove the resistors when motor attains its rated speed.
Three point DC motor starter is such starter which will remove the starting resistor in three steps.
Since the rated current of motor is 350A. According to given requirement, the starter should limit the starting current to 700A. As the motor speeds up, armature will generate internal voltage E_{A } which will reduce the current to rated value. As rated current is achieved, remove the first segment of resistor. As a result the current shoots up again however such values of each segment of resistors are selected so that current only shoots up to twice of the rated current. As the motor gains more speed, the back emf will increase, which will decrease the armature current to rated current. At this stage, second resistor should be removed which results in increase of starting current to twice of rated current again. Similar process is repeated for 3^{rd} segment of resistor.
Now to calculate the total starting resistance of DC motor including dc motor starter. Total resistance should be such that it should limit the starting current to twice of the total load current.
Consider
Rtotoal= Total resistance of starter and armature
Ea= Back emf developed by motor
VT= terminal voltage=250V
Imin= Irated= 350A
Imax= 2xIrated = 700A
Ra = armature resistance
Rototal1= Rtotal – R1
Rtotal2= Rtotal1R2 = Rtotal –(R1 + R2)
R1, R2, R3 = three resistors of motor starters
For the given case the rated current is 350A, so maximum current to which it should be limited is 700A.
When the current reaches rated current that is 350A, one segment R1 should be removed.
After One segment of resistor R1 is removed, the current increases to
Since back emf depends on speed of rotor. At particular instant, Ea will be constant and VT is constant throughout therefore
For n resistors
Where I min = Irated and I max = 2x Irated
The armature circuit will contain the armature resistance RA and three starting resistors. Initially when motor is static and there is no back emf, Ea = 0, maximum allowable current is 2x Irated that is Imax = 700 A, and the total resistance is (Vt/Imax) = 0.357 Ω. The total resistance will be in the circuit until the current drops to 350 A. This occurs when
When armature current drops to 350A, the resistor R1 should be removed from the circuit and current climbs again to 700A so we can find total new resistance required at that moment is
$latex Rtotal\ 1=R\ totalR1=R2+R3+Ra=\ \frac{VtEa\ 1}{Imax}\ =\ \ \frac{250\ 125\ }{700\ }=0.1786\$
This (new) total resistance Rtotal 1 will remain in the circuit until armature current drops again to 350 A. This occurs when
At this time, the starting resistor R2 will be taken out leaving
This total resistance will be in the circuit until the current drops again to 350 A. This occurs when
At this time, the starting resistor R3 will be taken out leaving only RA in the circuit. The motor’s current at that moment will increase to
which is less than the allowed value. Therefore, the resistances are
The resistors R1, R2, and R3 are cut out when EA reaches 125 V, 187.5 V and 218.75 V, respectively.
As it can be seen that with DC motor starter, the current is limited to safe limits.
To implement and verfity the design and calculation of three point starter of DC motor please refer to my article modelling of DC motor starter in MATLAB simulink.
The post Design of three point starter of DC Motor appeared first on .
]]>The post Modelling of DC motor starter in MATLAB appeared first on .
]]>The starting current of DC motor is very high because armature resistance is very low and there is no back emf to limit the starting current. Approximately the starting current of DC motor is 10 times higher than its rated current. Such high current can damage the motor internal parts. Therefore it is necessary to limit the starting current of DC motor using starter.
The starting current of DC motor can be reduced using just a resistor. However that resistor can be the cause of I2R losses as well as it will reduce the torque capability of the motor. Therefore starter is only required during starting when there is no back emf. Once the motor gains speed and back emf is developed, starter needs to be removed. It is recommended that starter should not be removed suddenly rather it should be detached in steps so that transition from maximum starting resistance to no external resistance is smooth. According to the rule of thumb, starting resistance should be removed in three steps. All the requirements to design and select appropriate values of starting resistors can be given in following article
Design of three point starter for DC motor
Consider 240V, 5HP, 16.2A 1220rpm DC motor. Since the rated current is 16.2A, the maximum allowable current is 32.4A. The starter should be designed according to these current ratings.
The DC motor should be connected as shown in above figure. 240V should be applied to both armature and field windings. Since the armature current needs to be controlled, the starter should be connected with the armature in series. Speed of motor (w) multiplied with 0.2287 is given as a feedback to the motor. Scopes are connected at different points to analyze different output values. The most important among them is the armature current scope which will show the performance of our designed starter.
To calculate the values of resistor for three point starter Consider
Rtotoal= Total resistance of starter and armature
Ea= Back emf developed by motor
VT= terminal voltage=240V
Imin= Irated= 16.2A
Imax= 2xIrated = 34.2 A
Ra = armature resistance
Rototal1= Rtotal – R1
Rtotal2= Rtotal1R2 = Rtotal –(R1 + R2)
R1, R2, R3 = three resistors of motor starters
Find the voltage drops for and the resistance of each segment of starter resistor and calculate the values of each step of starting resistor using my article how to design three point DC motor starter.
R1= 3.87 Ω
R2= 1.80 Ω
R3=0.875 Ω
The resistor R1 should be removed first and R3 should be removed last.
It can be seen that three resistor are connected in series. With each resistor, a switch (breaker) is connected in its parallel. In the case the breaker is open; the current will pass through the resistor. If a breaker is closed, the resistor will be bypassed and the circuit will not see the resistor. Each breaker is controlled by the output of the comparator block. Comparator compares each the value of armature current with the constant value written on it.
Ideally, each comparator should have the value of 16.25. However it is required that R1 should be removed first, therefore its threshold value is taken slightly above 16.25 that is 16.30. Next R2 should be removed so its threshold value is less than R1 and greater than R3. Resistor R3 should be removed last so its threshold value is taken least.
Initially all three breaker are open and all three resistors are included in the circuit. The motor will start with a high current 32.4A and will reduce slowly as the motor gains speed. Once the speed reaches the value of 16.3, the breaker across R1 will close as a result resistor R1 will be bypassed. Motor current will shoot again to maximum allowable current that is 32.4A.
The motor current starts declining since the back emf is increasing due to rise in its speed. When the current reaches 16.27A, breaker across R2 will close and resistor R2 will be bypassed. Similarly R3 is also removed from the circuit.
First, the starter is not connected to the system and the motor is started. The result is shown in the figure.
It can be seen that starting current is very high that is 400A. It is more than 10 time of the rated current.
Now when the starter is connected with the motor.
It can be seen that starting current is limited to nearly 32. Once the armature current reaches near the rated value, one resistor is removed and current shoots back. Three peaks shows that three resistors are removed and since the size of each peak is same, it means our calculations are correct.
Please post your questions in comments.
The post Modelling of DC motor starter in MATLAB appeared first on .
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