Iacdrive|Automation Solution

Load Break Switch VS Circuit Breaker

There are two important different parameters as far as the disconnection/connection of Electrical utilities is concerned.
– Breaking capacity.
– Making Capacity.

Making capacity is more higher since it is considering the initial starting current, inrush and faults that might occur at switching on moment.

Accordingly, the Circuit breakers have both while the Isolator or LBS has only breaking capacity. Isolator therefore used to isolate/disconnect/break load. To be switched on only under no load.

Basically and LBS is only designed to make and break load currents. It can be closed onto a fault (has a making rating) and cannot break a fault current. A CB is designed to make and break fault currents and off course load currents. Need to be careful with switching long cables and long lines with LBS, due to its limitation with switching cable/line charging (highly capacitive) currents. Similar precaution with needed with switching reactive loads like large transformers. Normally a CB is recommended for the 2 latter cases. Check the manufacturer test specs.

In simple way the Load Break Switch used to cut off healthy circuits or to break / disconnect the load. As a precaution, normally the LBS / Isolator are to be switched on under no load, the connected load to be utilized after switching on the Isolator. This why it has no making capacity. Circuit Breakers are intended to operate under unnormal conditions in order to clear the fault & to isolate the defective circuits protecting its associated electrical equipment, therefore breaking and making capacities are considered as the most important criteria as far Circuit breaker are concerned.

The making current is not an RMS value it is the peak value ie. Impk = 2.5Irms. The peak value of 2.5 times RMS is the DC offset at point when the LBS closes on the fault and is taken as the worst case X/R ratio of the source (X/R of about 20). This peak decays to RMS value Ith (thermal withstand current) dependent on the X/R delay constant. The decay rate is exponential with time. There is a misconception that making is 2.5times breaking current, but making is normally quoted as a peak and breaking as RMS. The breaking current in a CB is an RMS value. Breaking fault current is far more difficult that making especially when the contacts open when current is not at zero crossing point on the sine wave. For HV systems 132KV and above, the restrike and TRV starts to become a major consideration in CB selection, especially for long cables and lines.

MCCB burn out in connection with 22kW motor

125A rated MCCB is connected with 22KW motor. The motor runs in normal condition, no overload. But the contacts of MCCB is burnt out. Why?

When the transition between wye and delta takes place, be very sure that that the mechanical interlocks on the contractors are properly adjusted. If one doesn’t completely open before the other closes, you have a line to line fault. That will eventually take out the upstream breaker. Be careful, this is a very dangerous starter. I have been done this road many times.

“Star/delta switching” is delicate maneuver.
* The engine has almost no force to push something more than itself at startup. When it’s time for mode switching start to delta, may not happen too quickly. The arcs that occurs when the star contactor switch off, needs a few milliseconds (minimum 20 ms). Typical of an MCCB to go off is 50 ms and on is 20 ms.
* If the load is too high during startup, the engine will get overloaded when switching from star to delta and arcs can become very powerful and devastate even the strongest Components. In case the engine does not start at idle, the start method is directly harmful to both motor and power grids. Instead of a soft and comfortable start, we get instead two powerful surges. (In that case it may be even better with only one MCCB).Should this be the case, there are two solutions: Make sure the engine starts without load or replace the Star/Delta switch to a soft starter.
3: Next step is a frequency converter but then it’s about something completely different.

Parallel connection transformers

Change autotransformer to VFD or soft starter

What the parameters needed if we change from autotransformer (6.6kv/3.3kv) to VFD or Soft starter?

Since you got Auto Transformer 6.6/3.3 kV mean you have inductive load with constant speed in that case you should go for Soft starter, best way is collect all the motor data and send it to Soft Starter manufacturer along with the motor starting curve if available also you may ask for capacitor bank for PF improvement.

If you’re planning to use a Soft Starter or VFD for motor starting, you need to specify the horsepower rating, voltage rating, rated frequency, the type of motor (induction wound rotor or induction squirrel cage), the type of motor load (compressor, pump), motor load starting torque requirement and motor loading cycle. This information is important in selecting the type or model of the Variable Frequency Drive (VFD) or Soft Starter unit.

Directional Numerical over current relay

If current will flow in positive direction then the relay will behave as a Normal over current relay and if current will flow in negative direction then the relay will behave as a Directional over current relay…..Why the angle between healthy line voltage and fault current is required for sensing the direction??

Suppose you have purely resistive circuit with a voltage source connected to it. Now take any arbitrary node X in the network Look at the current flow from the node. Now when the source voltage is positive the current flows from the say upper node of the source to the lower node of source. Now if you look at this current flow from the arbitrary node X mentioned above, the current will be moving towards the node X from one side and it is moving away on the other side of the arbitrary node X node in one half cycle.
Now in the next half cycle the same thing repeats but with one difference, that the direction of current flow changes.

Now you want to operate your relay when the current is moving away from the arbitrary node in the first half cycle. place a CT at the node with primary P1 towards the node X1 and p2 away and take Secondary S1 to the relay . Now when current flow is from P1 to P2 current flow in the secondary will be say S1 to S2 thru relay.

Now we can say that, when the arbitrary node voltage is positive current flows from P1 to P2 and we take this as our direction required for the relay to operate.

Again when you look back from the node you will see that when the arbitrary node is at higher potential current is flowing towards you. Now place another CT with P1 towards the arbitrary node P2 on the other side and connect another relay at s2 side (now you can visualise 2 CTs on either side of the node with P1 towards the node X in both CTs). The current will be flowing from P2 to P1 in this CT and hence S1 will be negative with respect to S2 and current flow in this relay will be in reverse direction as that of the first relay.

In the next half cycle the current direction reverses and first relay current will be s2 to s1 thru relay and second relay current is from S1 to S2 thru second relay.

Now we want the relay no 1 to operate and relay no 2 not to operate. how do you achieve it when you connect the current alone to the relay which is changing the direction in every half cycle .
To achieve this now you connect a PT at the same arbitrary node X and connect the voltage to both the relays. The same point of the PT secondary voltage is connected to both the relays. Now find the phase angle between the current in the first relay and the voltage. You will see that when node voltage is positive, the current flow in the first CT will be from P1- P2 in primary and S1 to S2 in the secondary. In this case let us say that the phase angle between voltage and current is Zero in the first relay.

What happens in the second relay and CT /PT. The voltage is same as first relay which is positive and the current flow in the CT is P2 to P1 and in relay it is S2 to S1
– meaning opposite direction to first relay . As the current is in reverse direction with respect to voltage we can say that they are out of phase (180 degree).
Now you set both relays to operate when the voltage and currents are in phase. observe the result in first half cycle . Relay 1 operates (phase angle between v and I zeo ) and relay 2 no operation ( phase angle 180 deg ).

In the second half cycle observe the phase angle of relay 1 . Voltage at node is negative. (voltage phasor reversed ) Current flows from P2 to P1 and S2 to s1 in the relay 1 (current phasor also reversed) still the phase angle is zero and hence relay 1 operates Similarly relay 2 restrains.

Hence we found that, the relay 1 operates in both half cycle and relay 2 restrains. This is the importance of Voltage for directional relay.

Phase rotation errors

Phase rotation errors are not as rare as they ought to be. I’ve seen more than one building with a systematic phase rotation error. This can be prevented by carefully following the color coding system (Yellow Orange Brown and Red Blue Black for 480 volt and 208 volt systems in the US for example) and tagging feeders at both ends to assure proper connections.

To check for proper phase rotation sequencing (ABC and not ACB) you can use a phase rotation meter. Without that you can bump a three phase motor that should be correctly connected to see if it turns in the right direction. If it’s wrong, reverse any two phase wires from the source to the distribution equipment. However, if you have a tie breaker and intend to operate the secondaries of two transformers in parallel by closing it that is not good enough. Both transformer distribution networks have to be connected correctly on all three phases. You have to check the voltage across each corresponding pair of terminals on the tie breaker and be certain they are all about zero volts. If you don’t and there is an error, closing the tie breaker if that is possible at all (some electronic breakers may lock you out) will result in a phase to phase bolted fault that can result in severe damage to your distribution equipment. Phase rotation errors are invariably the result of incompetent installation, inadequate specifications for feeder identification, and inadequate inspection.

There are times when the phase rotation error is made on the primary side of the transformer. If this happens it can be compensated for by reversing the phase rotation error from the secondary side. This is less desirable but it will work. If you have multiple phase rotation errors in the same distribution network you have quite a mess to clean up. It will be time consuming and expensive tracking all of them down to be certain you have eliminated them. False economies by cutting corners on the initial installation of substations and distribution equipment will result in necessitating very expensive and inconvenient repairs. If it is not corrected you risk severe damage to three phase load equipment.

Moving data around within memory of an individual PLC

The first question would have to be – why do want to do it? If the data already exists in one location that is accessible by all parts of the program, why are you going to use up more PLC memory with exactly the same data?

Well, there are a couple of candidate reasons. One might be recipe data. You have an area of memory with a set of stored recipes for different products, and at an appropriate moment you want to copy a specific recipe from the storage area to the working area. The first thing to be said about that is that if your recipes are at all complex and you have a requirement to have a significant number of different recipes, then PLC memory is probably not the right place to be storing them. The ultimate, these days, of course, is that recipes are created by techies on PCs away from the production area, in nice quite, comfortable labs or whatever, and are stored on a SQL server. Only the recipe for today’s actual production run gets transferred to the PLC. But there are some applications where there is a limited number of different recipes required and the recipes themselves are quite simple, when it can be reasonable to store the recipes in PLC memory.

A second reason for copying memory areas within the same PLC is for procedures, sub-routines or whatever. But again, these days, all PLC languages have some sort of in-built facility for procedures – what Rockwell uniquely call Add On Instructions, what everyone else calls UDFBs – user defined function blocks. In any case, the point is that these facilities usually make all that memory management stuff transparent to the programmer. You just configure the UDFB and call it as required. The compiler takes care of all the memory data moves for you.

Another reason for copying memory, actually related to the previous, is a technique much used by PLC programmers where they use an area of memory as a ‘scratch-pad’. So they will copy some unprocessed data to the scratchpad area, all of the operations performed on the data take place using the scratchpad, and at the end, they copy the processed data back again. Again, it is questionable how much this technique is actually required these days, I would suggest that it most cases, there probably is a better way using a UDFB. But I have seen some programmers who routinely include a scratchpad area within any UDFBs they define.

Sub-transmission network

Q:
What factors determine current distribution between two 33kV feeders feeding a 33/11kV Substation within a sub-transmission network.

A:
Try the voltage divider rule. Take the R + X of each feeder (resistance and reactance) and find the Z. Remember that square root of R² + X² = Z. Now that you have the Z of each feeder, now find the Z of the two in parallel. To do this we have Z = 1/(1/Z1 +1/Z2). So, if Z1 = 2.16 ohms and Z2 = 1.67 ohms, then our Z of the two in parallel is 0.94 ohms. Now we pass the current of the entire substation between these two feeders. Let’s say that the current is 240 amps. Now we have 240Ax0.94ohms = 226 volts (IxZ=V). And since voltage divided by impedance gives us current (V/Z = I) we now take the voltage drop across the two feeders in parallel and divide each of the feeder impedances to get the separate feeder current. So we get 226V/2.16ohms = 105 amps (feeder 1) and 226V/1.67ohms = 135 amps (feeder 2). I have not tried this with your exact situation. Having different voltages from two different substations will change things, but at least this way you have a good start on the problem.

Since one end is tied together and the two other ends are from different substations, then you will have the classic voltage sending and receiving formula. Since the load is the one substation, then their will only be one power factor of the one load, so I would think this formula would apply: Es = Square Root of ((ErCosƟ + IR)² +(ErSinƟ +IX)²), which is square root of ((Receiving voltage times the cosine of the current phase angle plus current times resistance of the line)² + (Receiving voltage times Sine of the current phase angle plus current times reactance of the line)²). The voltage drop across each line would be VD=I(RcosƟ +XsinƟ) where R is the line resistance and X is the line reactance and the Ɵ is the phase angle of the load.

System with difference neutral

Q:
I have one system with two source. One from genset and the other from PLN (national power supply company) that each system has neutral.
The question is
1. Is there any problem if I connect both neutral directly?
2. Is there any spark when I connect both neutral?
3. How is the best solution to connect both neutral?

A:
1. I understand that the Genset is dedicated for essential load as an Emergency power supply which will be operated by hand (only in Manual Mode).

2. The Control Philosophy for a Generator that intended to be connected to PLN as emergency power source depends on the local service provider regulations.

3. Usually in your case there should be Electrical as well as Mechanical interlocks between the mains incomer & genset main breaker. ie both Sources will never be in Synchronism ( will not feeding the same load simultaneously).This measures will ensure that there will be only neutral point to the system.

Simulation on EMI

As a mathematical tool eventually, simulation can help to quickly approach the results that we need. If everything is done in right way, simulation can give us reliable conductive EMI results at the low frequency range.

Differential mode conductive EMI can be simulated with good accuracy at the low frequency range. The accuracy of common mode conductive EMI depends on the accuracy of a few parasitic parameters that need to be measured.

Personally for research, I would like to use simulation as a validation tool for calculation, and test results of prototypes can be used as proof for simulation.

E.g. for EMI filter:

1. Do the calculation for the differential mode conductive EMI filter;
2. Do the calculation for common mode conductive EMI filter base upon the parasitic parameters in the hand or estimation;
3. Use the simulation to check and validate if the calculation is right or if something is wrong and needs to be corrected;
4. Use prototype test results to check and validate if the simulation results are right.

Some other issues that caused by EMI filter can be found during system level simulation before prototyping. E.g. audio susceptibility and EMI filter damping problems.