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How to choose the type of curve of relays?
In order to balance the right amount of overcurrent protection against optimal machine operation, we need different trip curves. Selecting a circuit breaker with a trip curve that trips too soon can result in nuisance tripping. Choosing a circuit breaker that trips too late can result in catastrophic damage to machine and cables. The X axis represents a multiple of the operating current of the circuit breaker and The Y axis represents the tripping time. The three major components of the Trip Curve are: 1. Thermal Trip Curve. This is the trip curve for the bi-metallic strip, which is designed for slower overcurrent to allow for in rush/startup, as described above. 2. Magnetic Trip Curve. This is the trip curve for the coil or solenoid. It is designed to react quickly to large overcurrent, such as a short circuit condition. 3. The Ideal Trip Curve. This curve shows what the desired trip curve for the bi-metallic strip is. Because of the organic nature of the bi-metallic strip, and changing ambient conditions, it is difficult to precisely predict the exact tripping point. Trip curves represent the predicted behavior of a circuit breaker in ambient room temperature. This means when the bimetallic strip is within the specified ambient operating temperature for the breaker. If the breaker has experienced a recent thermal trip, and has not cooled down to the ambient temperature, it may trip sooner. Trip Curves predict the behavior of circuit protection devices in both slower, smaller overcurrent conditions, and larger, faster over current conditions. Choosing the correct trip curve for your application provides reliable circuit protection, while limiting nuisance or false trips.
How to calculate the neutral current in a unbalanced system?
If the 3 phase currents are equal to one another, then the neutral current will be equal to zero. However, if one of the phase current is different than the others, the neutral current is then equal to the difference between them. The real and imaginary both parts of the current will be added. A detailed calculation of neutral and phase currents and voltages in an unbalanced 4 or 3 wire system can be done by applying Kirchhoff's voltage law on all 3 phase loops.
In a 220/132kV Three phase combined winding(Single Equipment) trafo, one of the LV side fails so what will be the effect to the lv current of remaining phases ? Also will the trafo can be operated with 2phases in LV side ?
per https://generalpac.com/transformers/open-phase-condition-in-transformers Amir Norouzi writes about the behavior of three phase transformers when one of the phases is lost. This condition is commonly called a "single phasing". Upon a single phasing condition, Amir describes how voltages and currents (on both the HV side and LV side) greatly depends on the transformer winding connection (delta, wye, wye-grounded etc) as well as the transformer's core construction. Single phasing condition is very difficult to detect by relays where the CTs are monitoring current on the LV side of the transformer. Amir then presents algorithms that identify and detect open phase conditions. If the HV side of the transformer (primary side) was connected Wye-grounded similar to Figure 6 and we had a 3-Leg Core Type transformer connection, the primary winding with the lost phase will have an induced voltage equal to the voltage before the open phase condition. This basically means that if we look at the voltage alone across the transformer windings before and after the phase was lost, it will be the same. We should expect no difference in voltage on the HV side for a loss of phase given the conditions above. This is counter-intuitive however, it has to do with the summation of fluxes for three-leg core type transformer that Amir eloquently described in Section III-A-1 (page 4). To summarize this point clearly: we should have expect no discernable difference in voltage across the HV winding for an open-phase condition (as opposed to balanced conditions) given the following scenario Utility grid is supplying strong three-phase balanced voltages prior to open-phase condition Transformer HV winding connection is Wye-grounded. The LV winding connection doesn't matter Transformer is a 3-Leg Core Construction Type Since the HV winding still develops voltage across the winding even though the phase was lost, the corresponding LV winding will also develop voltage across it, and the LV phase/line currents will be relatively balanced - as if the phase was never lost. From the load's prospective, it sees no difference between Open Phase vs balanced conditions. Remember, this is only for HV transformer connected Wye-grounded and the construction is a 3-Leg Core Type. For other construction-types such as a 4-leg, 5-leg core type or a shell construction type, (given a Wye-grounded primary) if we have an open phase condition, we will have very little voltage developed across the winding for which the phase was lost. For example, if phase A fuse was blown as shown in Figure 6, we will have very little voltage across winding A on the HV side which results into very little voltage across winding a on the LV side. To summarize this point clearly: we should expect very little (or even zero) voltage develop across the winding for which the phase was lost given the following scenario Utility grid is supplying strong three-phase balanced voltages prior to open-phase condition Transformer HV winding connection is Wye-grounded The LV winding connection doesn't matter Transformer is a 4-Leg Core Type, or 5-Leg Core Type, or Shell Type construction Since HV winding will not develop very much voltage across the winding for which the phase was lost, the corresponding LV winding will not develop very much voltage across it. So from the load's prospective, it receives abnormal voltages and currents from two phases but very little (or even zero) from the phase that was lost on the HV side.
What are the disadvantages of not having a power factor correction in a substations?
How to select and apply circuit breakers.
how to wire a ct in delta for a tx secondary star winding for a differential relay ?
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is VB is same as VS?
Vb is the voltage across the burden, on the other hand Vs is the voltage across the excitation branch. We can see in the CT model that Rs (Secondary winding resistance) and XL (leakage reactance) is placed in series, which will cause the drop in the voltage. Then by applying KVL we can say that Vb = Vs - Vdropped (across Rs and XL), which can further transform into the equation: VB= Vs – Is*(Rs + Xl).
What is the secondary connection of the connected CT to the Star / delta power transformer?
It is a normal practice to connect CT's in Star/Wye when the transformer is delta connected and CT's are Delta connected on the side where transformer is Y connected. Flipping the whole circuitry (including CT's connection) horizontally will make it Star-Delta configuration.
What is Zs in a system where fault level calculations are done. In pu value, Zs = V/Ifault
We can define Zs as the maximum impedance due to a single line to ground Earth Fault (also called Earth Fault Loop Impedance). For the maximum fault current I(fault) and nominal system voltage V, Zs can be calculated by the formula Zs = V/Ifault. Zs is calculated and used in Fault Analysis in power systems. Also, it can be calculated from the time-current curve of the protective device during protection coordination.
At (D/Y) transformer, If short circuit in one phase ia (earth fault) at secondary of transformer about 2000 amp, what the effect on primary (D) ?
Typically for any a earth fault that would occur on the LV (Wye) side of the transformer, the impedance reduces and the current on the secondary side will increase causing the current in the primary side (Delta) of the transformer to be increased due to the cascaded effect and the current feeding through the upstream devices will also increase. The overcurrent protection devices will come into play to mitigate the effect of the fault.
I would like to know what’s a Simocode.. And its Function?
SIMOCODE is the flexible, modular motor management system for low-voltage motors. It can easily and directly be connected to automation systems via PROFBUS or PROFINET. It combines in just one compact system all required protection, monitoring, safety and control functions. The motor management system thus increases plant availability and allows significant savings to be made for installation, commissioning, operation and maintenance of a system.
What will happen, when during operation of transformer delta connection open inside
What are some effect of open phase on primary side of transformer?
I would like to obtain the arc flash calculation spreadsheets. Is this possible?
The calculation spreadsheets for arc flash using IEEE 1584-2002 are easily available online. However, the calculation spreadsheets for IEEE 1584-2018 calculation method are a bit complicated to develop and not readily available for general users. You can try searching for the IEEE 1584-2002 calculation spreadsheets yourself otherwise we can provide it to you.1. Corrective: after failure occurs 2. Preventive: regular maintenance to prevent future failure For the next part of your question. In order to plan your budget for equipment maintenance following factors should be considered so that the equipment maintenance can be done in the timely and cost efficient manner: · Availability of skillful maintenance worker · Tools and equipment for maintainability · Spare stocks for backup operations · Maintenance policy · Environmental factor (weather, size, working conditions) · Systematic maintenance data collection, analysis and continued reliability study · GMS (generator maintenance schedule) and TMS (transportation maintenance schedule) using different algorithm When considering your overall maintenance plan and frequency of test intervals, the following ten factors should be included in that decision-making process: 1. Equipment criticality and device significance 2. Current condition 3. Lubrication life 4. Maintenance history 5. Operational history 6. Industry experience 7. Maintenance philosophy 8. Operating environment 9. Time allowed for maintenance 10. Manufacturer’s recommendations Having an understanding of these ten factors, and using them as critical decision points, can greatly enhance your ability to accurately judge the frequency at which you maintain your equipment.
What is the role of maintenance in Power system planning? What type of maintenance required and how to calculate cost and other factors related to it?
There are many objectives of equipment maintenance plans in power systems. It is done for cost optimization, increasing system reliability, increasing life span and durability of equipment & safety of personnel and equipment . There are two major types of maintenance: 1. Corrective: after failure occurs 2. Preventive: regular maintenance to prevent future failure For the next part of your question. In order to plan your budget for equipment maintenance following factors should be considered so that the equipment maintenance can be done in the timely and cost efficient manner: · Availability of skillful maintenance worker · Tools and equipment for maintainability · Spare stocks for backup operations · Maintenance policy · Environmental factor (weather, size, working conditions) · Systematic maintenance data collection, analysis and continued reliability study · GMS (generator maintenance schedule) and TMS (transportation maintenance schedule) using different algorithm When considering your overall maintenance plan and frequency of test intervals, the following ten factors should be included in that decision-making process: 1. Equipment criticality and device significance 2. Current condition 3. Lubrication life 4. Maintenance history 5. Operational history 6. Industry experience 7. Maintenance philosophy 8. Operating environment 9. Time allowed for maintenance 10. Manufacturer’s recommendations
Having an understanding of these ten factors, and using them as critical decision points, can greatly enhance your ability to accurately judge the frequency at which you maintain your equipment.
1-Does this mean that assuming a balanced power system if the impedance on one phase is given as a magnitude of 10 with angle 30 degrees. Does this mean that it is 10 with angle 30 degrees for each phase? Or is it relative to each phase so on the first phase it will have an angle of 30, second phase 120+30, third phase 240+30? Video URL: https://youtu.be/kxtXSTgeg7I
The magnitude of Impedance and it's angle remain the same for each phase. So, in your case, the impedance will remain 10 with angle 30 degrees for all the 3 phases.
2-Hi, how did you calculate the imaginary number j60 ? video URL: https://youtu.be/vbue6iV-7iw
The imaginary number j60 is the inductive reactance of the transmission line. This value was already given in the question so we have not calculated it.
3-the Vout of the region of 2 is not the same as Vin of region 2 which is 4800V since there is a voltage drop on Zline. Video URL: https://youtu.be/vbue6iV-7iw
The calculated value of (4800V) is the new base value that would be used in region 2 and has nothing to do with the actual voltage drop.
The base quantity remains the same throughout the entire network unless there is a transformer which would then split the network into different regions. When there is a transformer between 2 regions, the base quantities for Voltage, Current, Impedance changes (Base Quantity for Apparent Power remains unaffected by change in voltage level). The base quantities are then computed using the turns ratio to obtain the base quantities that would be used in the next region.
The base quantities are arbitrary values that are used to make computations of power systems easier. These base values are ultimately used in per unit calculations to make complex calculations easier. We have a video series on Per Unit Systems which explains in details how these quantities are calculated.
4-How to banking 1 transformer 167kva 3phase 280v in 4wires! Video URL: https://youtu.be/HmF20PHWyFo
a single 3phase transformer can be connected in Wye configuration to get a 4 wire system, 3 of the wires will be for each of the 3 phases, while the 4th wire will be used for a neutral connection.
5-I recently purchased a 50KW 3phase genset, after having gone through the motor side i am now faced with the daunting challenge of the ac generator end portion. This is amazing knowledge and your time and energy towards is greatly appreciated! I've learned so much in just a short period. Sincerely, Mark Video URL: https://youtu.be/1HfnxB8u4T0
This course will cover the requirements needed to design protective devices and the applications of these devices through a schematic diagram. Furthermore, this course will analyze the effects of all types of faults in the power systems along with the easy hand on calculations. To develop your strong concept on fault analysis we will discuss how faults can be identified by analyzing waveforms. Lastly, you will learn the most tedious and complex theory of symmetrical components that are found in different types of faults.
6- In the window type current transformer, how does the primary conductor passing through, make it 'one turn'. It doesnt make any turns right! Video Link: https://youtu.be/EiDGc0TvK8E
: In literal terms, the conductor passing through the window type CT does not have any 'turns'. However, the term 'turns' is extensively used in electrical engineering literature to refer to the number of times the current has to travel through a similarly shaped part of the circuit. See the image in the link below which shows a solenoid, where there are 6 'turns' --or say 'identical portions' of the circuit, so in a circuit where there are no turns, there will only be one 'identical portion' for the current to pass through.
The term 'turns' has been adopted for use in a number of equations, one of the most famous being the transformation ratio of a transformer. To maintain consistency with the existing literature, a single straight conductor is called as having 'one turn'.
7-please can you briefly explain why the non-polarity side of the input terminal C is grounded.. For Video: https://youtu.be/UXVRRLApuuw
The secondary CT circuit must be grounded at one singular point for many reasons which are explained here: https://youtu.be/1HfnxB8u4T0 -- it can be grounded at terminal c of the relay or another terminal point on the secondary circuit
8- This is actually incorrect....if you factor in the 120deg displacement you will not have simultaneous current flows in the same direction....more easily to understand is current flow basics in the delta line......if delta line current is 1.732 times the amount of current to delta phase current how can you sum out like you explained? It's actually 1.732 delta line current INTO the node, and .866 going OUT of the node in each direction.
Regarding your query, it seems like that you might have some confusion here. Actually, 120 degree phase shift will put no effect on the direction of current. Secondly, it is clear from the figure that line current A is larger than phase current A and it is actually 1.732 times larger (as it is obvious that line current is 1.732 times of phase current in delta connection). Also, it is understood from the figure that line current A is at 30 degree phase shift with respect to phase current A.
9-If in LV side any one phase grounded, so what will be the voltage & current in HV & LV side in delta star 11 transformer. For this video: https://www.youtube.com/watch?v=kJbjOLNZlQA
LV Side: When a line to ground fault occurs on one line on the LV side then all current will flow from the faulty line. Other two lines will have no current passing through them. Similarly the line which is faulted will have almost zero volts on it while having rated voltage on other two lines.
HV Side: Transformer will feed its maximum fault current which will eventually saturate the transformer with a slight decrease in the rated HV voltage.
10- explain the meaning of zero sequence in a system.
Zero sequence current provides a unique and various prospective for the power system. One prospective is that it tells how how much unbalance we have in the system. Another prospective is that it measure the fault current magnitude for a one-line-to-ground fault. We highly recommend watching https://generalpac.com/power-system-protection/fault-analysis-part-1 which will walk you through the various faults and how symmetrical components are used to interpret and analyze the faults.
11- found during installation that CT's are marked P1-P2 and S1-S2. Could you explain according to this symbol, as I am always confused which one should be grounded/connect to relay whether S1 or S2.
The CT's primary winding connection are denoted as "P1" and "P2" (in some cases "K" and "L") and secondary windings as "S1" and "S2" (or "k" and "l"). For primary, the connections are made in such a way that P2 is the load facing side and the energy flows from P1 to P2. For secondary remember to use the same rule for the directional flow of energy from S1 to S2. Also, you can either connect S1 and S2 to the ground terminal.
However, for protective relays, ground that terminal which is nearest to the protected equipment. Please see the diagram below for your reference:
12- How the fault current would reflect in the Y side if the source were in the Y side and the fault were in the Delta? For this video: https://www.youtube.com/watch?v=A79nq9JUMIU
By the rules of Delta, there is no ground which means there is no support of 0 sequence currents. But when there is an SLG or an LLG fault, ground is induced in the delta connected system and this would mean that the delta side will follow the same methodology as explained in the video.
13- How to calculate 1/j0.4 become to -j2.5? For this video: https://www.youtube.com/watch?v=8jNuSwnL7HE&lc=UgwUTu8kvULiFrPo4eF4AaABAg
The 'j' operator is used in engineering denotes the complex number operation if it is attached with any number. It can be simply considered as a basic imaginary number 'i' iota which is equal to the sqrt(-1) . In electrical circuits +j shows that a phase advance of 90 degrees and -j shows the phase retard of 90 degrees.
We can simply convert 1/j0.4 to -j2.5 by using the basic imaginary number laws and operation:
We know that:
j x j =j^2 = sqrt(-1) x sqrt(-1) = -1
Therefore, if we rationalize the denominator, then:
1/j0.4 = 1/j0.4 x j/j = j/ 0.4j^2
We know that j^2 = -1, hence:
j/ 0.4j^2 = j/-0.4 = -j2.5
14- I have problem with calculating negative current sequence of a delta svc, I'm reading a paper related to this topic and the formula of negative sequence do not make sense to me, anyone can help? https://www.youtube.com/watch?v=iLOfLIHLaqs&lc=UgzdjvUL_NKbwOAi5g14AaABAg
calculating negative sequence current is highly dependent on the application in question. The way it was presented in this video was for a very particular application (ABC Phase Sequence, CCW Rotation, etc). You'll need to look carefully with the application in question and adjust accordingly.
15- what if the Vbase and transistor rating is not given and we have to find the per unit reactance? For this video: https://www.youtube.com/watch?v=LGVovZBRs64&lc=UgxTlFCoIKtfkKonWqh4AaABAg
Vbase can be chosen as the voltage of any equipment in the network, it then has to be modified according to the transformer (not transistor) ratio, in this case 480 V of Generator will stepped-up by 10 to give 4800 V.
Zbase can be calculated using Z=V/I formula. Ibase can be acquired as shown in the video from 5m 20s to 5m 30s. It too has to be modified according to the transformer ratio. In this case, it will be 20.83 A / 10 to give 2.083. The resulting Zbase will be 2304.4 ohms. This calculation is shown in part c of this series (next video). The per-unit impedance (not reactance) of any part can now be calculated using this formula: per unit=actual/base.
The per-unit reactance is always calculated using X/R ratios, which are supplied by the OEM.
what is pre fault voltage in distance relay?
Pre-fault voltage is the voltage of the system (particularly at the location of the fault) when a normal load current is flowing through it. The system voltage will be at the pre-fault level until any fault current starts to flow through it. The term 'pre-fault voltage' has no particular meaning in the context of distance relays. The general concept applies everywhere in the power systems.
Is single derived from it's three phase system directly or per phase model of that three phase system? Which one is right?
The Y-Bus Admittance Matrix For Solving Power Flow Equations", if you consider the same example of a pi representation that includes shunts, how would you generate the equivalent bus impedance matrix directly? (not admittance matrix)
what is total current entering in the node?
I want to know power system fault analysis
Is it possible to toggle on / of the background sound, it's a bit distracting :)
Hey you guys, thanks for the feedback. For all our upcoming videos we won't add any music. Unfortunately, Youtube doesn't allow us to remove the music from previously uploaded videos. Thank you for your understanding.
Type 1,2, 3 and 4 can alternatively be called Type A,B,C and D surge protection devices:
Type 1 (Type A) SPD
The Type 1 SPD is recommended in the specific case of service-sector and industrial buildings, protected by a lightning protection system or a meshed cage. They are also intended for installation between the secondary of the service transformer and the line side of the service equipment overcurrent device, as well as the load side, including watt-hour meter socket enclosures, and are intended to be installed without an external overcurrent protective device.
It protects electrical installations against direct lightning strokes. It can discharge the back-current from lightning spreading from the earth conductor to the network conductors.
Type 2 (Type B) SPD
The Type 2 SPD is the main protection system for all low voltage electrical installations. Installed in each electrical switchboard, it prevents the spread of over voltages in the electrical installations and protects the loads. connected Type 2 SPDs are intended for installation on the load side of the service equipment overcurrent device, including SPDs located at the branch panel
Type 3 (Type C) SPD
They are installed as a supplement to Type 2 SPD and in the vicinity of sensitive loads. Type 3 Point-of-Use Surge Protection Point-of-use surge protectors such as surge receptacles are installed within 30 ft of conductor length from the service panel and are designed to offer premium surge protection for specific electronics while providing innovative features to enhance user convenience.
Type 4 (Type D) SPD: Type 4 SPDs may be intended as Type 1 SPDs, Type 2 SPDs, or Type 3 SPDs and must be considered based on their intended application.
For Type 1, Type 2, and Type 4 SPDs, the manufacturer shall specify (declare) the value of Nominal Discharge Current to which the sample will be tested. The Nominal Discharge Current value selected by the manufacturer shall be as follows:
⎯ 10 kA for Type 1 SPDs or Type 4 SPDs used for Type 1 SPD Applications ⎯ 3 kA, 5 kA, or 10 kA for Type 2 SPDs or Type 4 SPDs used for Type 2 SPD Applications and, optionally, for Type 3 SPDs
An additional Nominal Discharge Current value of 20 kA for Type 1, Type 2, and Type 4 SPDs is provided in other standards ANSI/UL 1449-2006 [B2]; however, due to the scope of this standard and harmonization with IEEE Std C62.41.2-2002, the values in this standard are limited to 10 kA maximum.
Type 4 SPDs generally have conditions of acceptability dictated by the national regulatory body that would require Type 4 SPDs to undergo this testing when installed in its final form in its intended application. However, it may be acceptable to submit Type 4 SPDs to these tests.
I want to know what is the current when both CTs are connected in series for a same primary as the sketch sending herwwith as a attachment.
Typically, It doesn't matter how many Current Transformers we connect in the circuit, whether in series or in parallel. It does not affect the overall system's voltage or current neither the CTs will affect each other. A simple example will clear your query.
Suppose that we have attached 2 CTs in series in a system one having the rating (scale down ratio) of 100:5 and other is 200:5, and a current of 50A passes through the primary of CT. Then following will be
the calculations involved:
1- 100:5 CT (5/100)*50=2.5A Secondary Amps of CT.
2- 200:5 CT (5/200)*50=1.25A Secondary Amps of CT.
We can clearly see that the secondary currents of both the CTs are different and it completely depends on the CT ratio itself.
Could you please explain me that would it be happened if two different ratio CTs are connected in series and what is the current value flowing in both CTs?
The Ans to your Question is that basically It doesn't matter how many CTs we attach in circuit whether in series or in parallel. It do not effect the overall system neither the CTs will effect each other.A simple example will clear your query. For example we attach 2 CTs in series in a system one having rating 100:5 and other 200:5 and a current of 50A passes through the primary of CT then following will be the calculations