Iacdrive|Automation Solution

Motor short circuit protection

In motor protection panel when 3 fuses are provided for short circuit protection, Is it always a condition that during short circuit minimum 2 fuses should be blown? If yes or no then why?

Because fuse is a type of low resistance resistor that acts as a sacrificial device to provide overcurrent protection, of either the load or source circuit.

Its essential component is a metal wire or strip that melts when too much current flows, which interrupts the circuit in which it is connected.
Short circuit, overloading, mismatched loads or device failure are the prime reasons for excessive current.
2 fuses are must & should blowing in motor control panel due to motor each winding sets are connected to 2 phase in delta connection (each winding set is works on 440 Volts power supply).

We need to identify the types of short circuits which can be experienced in a motor and if you are talking of 3 fuses for short circuit protection, it means you are dealing with a 3phase motor. All things being equal, a three phase motor should be balanced in operation hence the current in each phase must be the same.

If a short cct occurs, it could be a phase to frame ( L-E),or phase to phase(L-L)or even 3phase (3L) short cct. In each case, the fuse that ruptures will depend on its condition, rating, type and class. If a fuse has been subjected to various loading and abnormal conditions, the tendency to rupture faster exists. If I should view your question from the perspective that all fuses are of the same type, class and subjected to the same condition and motor windings are same and in the same healthy condition, then a L-E fault should not necessarily cause other fuses in the motor cct to rupture immediately. It should be the defaulted line. And if it happens, the motor will not be balanced which in turn causes the other fuses to rupture in turn due to increase in current. It is always better to use a circuit breaker to isolate all the lines in the event of a fault.

How to connect 3 phase motor?

Making a connection of 3 phase motor the nameplate shows different voltages for delta it is 380-400 volt and 660-690 volt for star, what option should be selected? the supply Line to Line voltage is 380-400.

Each stator winding of the motor can withstand 380-400 V.
Thus, if you connect your motor (the stator of your motor) in delta, it should be connected to 380-400 V line-to-line.

On the other hand, if you connect the stator winding of your motor in Y, you’d be able to connect your motor to line-to-line voltage that is sqrt(3) x 380-400 V = 660-690 V.

The actual output power (for a standard squirrel cage 3-phase AC motor) is not determined by the motor itself, but by the load it is driving. The motor will attempt to run at a speed near its synchronous speed, and to deliver the power required by the driven machinery at that speed. This means that the current taken up by the motor at any given voltage, will be almost the same whether it is star, or delta connected. If you therefore connect the motor in star while supplying it by the voltage it is designed for when delta connected, the current through each winding will be sqrt(3) times the winding is designed for. This again means that the heat dissipation in the winding will be approximately 3 times what it is designed for, and therefore it will burn out if you load the motor with its nominal load.

We should be aware that the motor power as mentioned on its nameplate, in relation with the available power of the MCC panel to which it is connected, are the important factors in choosing the type of starting of the motor. Take into account the fact that starting the motor direct in Delta connection (which is the correct one based on your network voltage) the currents may be up to 8xInomianl of motor and if your MCC doesn’t have the capacity to withstand this current (by decreasing its supply voltage ) you may fail with DOL Delta starting type. Is that why, based on the power of motors, in order to avoid high currents during the starting time, it is recommended the Y/D connection. Limitations in starting currents by Y/D are considerable by decreasing the current first with sqrt3 because the feeding voltage is not 660V (you feed the motor with 380-400V) and the current initially in Y is sqrt3<I delta, so it is 3 times less than Delta DOL. Y/D is not the single one, there are a lot of solution to start AC motors.

Ground fault detection in a Delta system

We have system which is connected to 16kV/2.4 utility transformer (delta on secondary) and we are using 2.kV/480V transformer for loads after 2000ft. Utility wants to protect against ground fault in the system. I am planning to select a ground over voltage relay using a broken delta PTs on secondary. I am having problem with calculating the 3Vo value, How much voltage will you set to trip the relay for SLG or LLG fault.

Let’s say you have a system 16kV/2.4kV with more than one, say 5 transformers T1, T2, T3….T5 interconnected transformers throughout your network, with the broken delta arrangement to detect the residual voltage on each of the transformer’s delta side, If you have an earth fault say on the LV of T1, the voltage displacement gets picked up on the delta side of all 5 transformers and there is a very high probability that all 5 transformers get taken out. This is because this scheme does not look for the earth fault current or where it exists, as long as it is on the interconnected system where the source is able to support the earth fault. The moment it notices a voltage displacement, bang goes your CB to clear the fault irrespective of it’s location.

A better scheme is to use the zig-zag transformer which offers a low impedance for zero sequence currents, generally used with a neutral grounding resistor to limit the current to more often than not the rated current (or lesser than that) of the transformer LV. So in a similar situation for an earth fault on T1 LV, the earth fault completes it’s path through the earthed NGR resistor and back to the fault point. A sensitive earth fault relay connected to a CT connected between the grounded resistor and the neutral point of the zig-zag transformer, designed to take out both the HV & LV CBs of T1 will do the job, without fear of taking out the other transformers.

Of course if this is the only transformer you are talking of, then the voltage displacement method should work in principle, however I would still go ahead and install the zig-zag arrangement described above. Let me know your thoughts, and then we can start discussing about the magnitude of 3Vo or 3Io as the case may be.

When we talk about detection of Earth Faults, that means we want to know it without tripping, so, we should absolutely use “IT Earthing system” for LV side, then with this system we can use “IMD – Injection Monitoring Devices” that monitor when it happen, and if wanted send the tripping order to the installed Circuit Breaker.

By the way:
1- As the transformer’s connection of secondary side is “Delta”, we use one of 3 phase to be connected via a special impedance to the earth to have “IT” system. Noting that for some of “IMD” we don’t need to use the special impedance as it’s integrated inside the IMD.
2- In the most of “IMD” we can adjust the value of “Insulation level” where above of this value an Alarm signal by auxiliary contact will be sent.
3- Some of “IMD” have 2 levels can be adjusted “1st level for Alarm” and “2nd level for tripping”.

How to select a drive between motor and machine?

We should select a drive (direct/flexible, chain, flat/vee/ribbed belt, gearbox, soft start). The motor/starter/drive characteristics should match that of the load. Design and factors to be considered in selection.

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.

Insulating resistance measurement

Please remember that Insulating resistance (IR) measurement and associated polarization index tests is just one of the many tools used for insulation system integrity analysis. Its value and repeatability is dependent on the environmental condition at the time it is taken; as mentioned temperature, humidity contaminations all contribute/effect the reading.

The baseline figure should be obtained from either factory or during initial commissioning (as per factory condition). So performing commissioning in the rain, dirty surface, high humidity may result in low values for both dry type and oil filled equipment. Low reading in itself does not indicate bad insulation where the machine cannot be returned to service.

The bottom line is assessment lacking or other data would be:
1. The machine was running at it was running ok before the test.
2. The leakage value at operating voltage will be V/R; therefore the heat loss will be I^2R. Is that OK or warrant some corrective measure.
3. PI may approach 1, is that OK or not? Is this mtruly and indication of wet insulation or of resistive value but will still be OK when energized as per 3. above?

IR, PI measurement along with Cap bridge / dissipation tests, PF test and others are performed to ensure the insulation integrity for maintenance and commissioning.

If cable and equipment have gone through routine maintenance, it is good practice to perform these tests and making sure no ground are left before energizing.

Please read a “a stich in time” by Megger.

It by itself is just a test. The test is meaningful.

Excitation system in generator

The excitation system requires a very small fraction of the total power being generated. If we could simply increase the excitation (a very small amount of power) and increase the generator’s real power output, the world’s energy problems would be solved, because we would have a perpetual motion machine.

In the case of a generator connected to a large grid, the generator will inject any desired amount of power into the grid if its prime-mover is fed the desired power (plus a small additional amount of power to take care of losses). This is true, regardless of the total load on the grid, because the generator’s output is an extremely small fraction of the total grid power, and it alone cannot make drastic changes to the grid’s frequency.

Normally, the load varies by a very small fraction of the total grid power. If the load increases, the frequency of the entire grid (including the generator in question) lowers a very small amount, generally less than one-hundredth of one Hz. The frequency slew (that is, the rate-of-change of frequency) is very low, because there is a massive amount of energy that is stored as the kinetic energy of the rotors of all of the generators. At this point, nothing needs to be done; the system simply runs a little faster or slower.

Over time, as the load changes a greater amount, the frequency moves further from the nominal frequency (50 Hz or 60 Hz). When the difference between the actual frequency and the nominal frequency becomes greater than about 0.01 Hz, action is taken to make changes to the output of the grid’s generators.

The specific action may be determined by the regulating authority (for instance a power pool in the US) and it is usually based on economics, subject to other constraints. If the load has increased (and the frequency is less than the nominal frequency), the generators that have the lowest incremental cost of power will be asked to increase their output, or if all generators are near their limits, new generators (with the lowest incremental cost) are asked come on line. It’s important to note that a generator’s limit is usually 80% or 90% of its rating. The 10% or 20% of unused capacity is the system’s “spinning reserve”, which is used to maintain grid stability for sudden, large power variations.

The same thing happens with a generator connected only to its load or a weak grid with just a few other generators. However, because there is relatively little kinetic energy stored in the rotors of the one or few generators, the change in frequency associated with a load change is much greater, so frequency variations are much greater and corrective actions may not be implemented before the frequency varies by more than a few Hz.

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.