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AQ: 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.

AQ: What causes cables to fault during weather seasonal changes?

I currently work for a small utility with a small amount of underground installations but a lot of it is aging and failing during weather changes. I am curious as to why it happens during weather changes and if there are scientific facts that can support it? Is there a way to predict when a cable will fail based on weather patterns? I’ve heard of different opinions on the matter, but is there a proven reason why? It is my goal as a young engineer and Gonzaga T&D engineering graduate student to learn more about these phenomena’s and what better way than to hear it from industry professionals in a technical discussion?

Scenario #1: Lightning strikes during summer on power cable installations can cause voltage spikes on the line, which in turn doubles back when it finds the open point on an underground cable installation. The initial voltage spike can cause the insulation of the cable to deteriorate or fail, and the reflection of the surge can cause the voltage to spike which can then finish off the already deteriorating insulation if it hadn’t faulted from the initial surge. Side note: This is why it is good practice to have transformers with load on them at the end of a cable run, or lightning arrestor at the termination points of an underground run and not just an open switch. Faults, recloser operations and other switching events can also cause a voltage spike on underground installations which can break down the cable insulation, thus making it more susceptible to failing after future events.

Scenario #2: Cables could fail more during the weather change due to the stresses that are inserted in to the cables during heavy irrigation motor start-ups and operation. Today you will see more soft-starts on your pump motors. With older cable supplying energy to older pumps you may find an across the line motor starter at the end of it. Cross-line starters rapidly heat the cables that have laid dormant over the cold winter months. If there was any sort of treeing, insulation deterioration, rodents chewing on the dormant cable, dig-ins, or any other common cable damaging scenarios during the winter, the startup will speed the deterioration process up in these locations, which in turn lead to cable faults.

Scenario #3: In areas where older open concentric cable has been installed, you are most likely experiencing many faults if it hasn’t already been changed out for newer jacketed solid-dielectric cable. As the ground dries out in the spring/summer, you will see higher resistances on the return path of the old and deteriorating open concentric neutrals. Without the cable being in wet conditions as it was throughout the winter, the electric field around the cable is no longer uniform and in some cases is a complete loss of your neutral.

Regardless of the insulation you use on cables, you most likely have faults. Maybe you’ve been “lucky” and it’s only in your joints and terminations? Regardless of which type of cable insulation, temperature has a significant impact on the dielectric withstand of the cable (i.e. higher temperatures will result in lower dielectric strength properties). Drying conditions also equates to higher insulation temperature due to poor heat transfer characteristics of your cable.

There are some common points in each of the above three scenarios but there really isn’t any scientific proof, just observations. Does it depend on your system load factor, your power factor, your installation practices, or even your cable design? Is it all of the above or is it much more simple than that? Is it different between different manufacturers of cable? Are there different scenarios that you’ve seen or heard of?

AQ: 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”.

AQ: The effect of power failure on VFD

Q: I am planning to put some variable frequency drives on non-critical section of factory where there will be planned interruption of 30 seconds but 2 times a day.

A: Why are you going to use VFD, variable frequency drives are expensive. What is the application?
If the application require fix speed / rate Soft Start device is required.

However, if there might be frequent start stop (Power OFF/ON) AC Contractor Duty AC3 are recommended to be used to bypass the Soft Starter or Static device once the required motor speed is reached and then Start/stop have no impact on the installation.

The technique of using AC3 Contactors, is not applicable for VFD if the VFD is necessary for application. In this case the other advantage associated with VFD no longer will be valid (Protection, control and monitoring).

AQ: 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.

AQ: Question about start a 450kW pump

Can I start a 450KW pump from the grid using star-delta and then use a bypass contactor to switch to an already running generator of 500kVA in order to avoid the starting current?

In my opinion, this operation is very dangerous. 500kVA is usually Diesel generator and interaction between load and source is very high.

Although maybe reduced starting current by means of your proposed figure but following comment shall be take in to account:
• The distance between load and generator is important
• Difference phase angle between grid and 500kVA generator possible to generate torsional effect and it is harmful for rotor in transfer moment
• Reacceleration is very important situation and maybe stall the motor
• Voltage dip due to starting another motor can make disturbance and this network is very weak respect to transient phenomena
• De-rating of generator maybe cause to have 70% or less then nominal rating of name plate (based on site elevation, ambient temperature and humidity)
• Meanwhile power absorption by electrical motor (450 kW) is more than generator normal capacity (500kVA).
As wrap up it is not safe and operational case

Actually I think it won’t work:
1). At 450 kw of a load is already bigger that the capacity of the Generator which is 500kva. (considering the pf of 20% the genset capacity is 400 kw which is way below even the maximum continuous power consumption of the load -450kw).

If your client had say 550KW GENSET, then I would definitely give him a solution which is sustainable. He just doesn’t even have to start the pump with the grid power then cross to Genset. We can propose an equipment that can give a smooth start of the motor and ration supply of power to the motor depending on the load requirement (the energy required to do a certain activity)

Soft Stop – When starting, an AC Induction motor develops more torque than is required at full speed. This stress is transferred to the mechanical transmission system resulting in excessive wear and premature failure of chains, belts, gears, mechanical seals, etc.

Additionally, rapid acceleration also has a massive impact on electricity supply charges with high inrush currents drawing +600% of the normal run current. The use of Star Delta only provides a partial solution to the problem. Should the motor slow down during the transition period the high peaks are repeated and can even exceed direct on line current. THE EQUIPMENT WE CAN PROPOSE provides a reliable and economical solution to these problems by delivering a controlled release of power to the motor, thereby providing smooth, stepless acceleration and deceleration. Motor life will be extended as damage to windings and bearings is reduced.

-Less mechanical stress.
-Improved power factor.
-Lower maximum demand.
-Less mechanical maintenance.

Soft Start and Soft Stop is especially useful with pumping fluids where torque transients often cause water hammer effects, and in some instances, failure to gradually slow the fluid down before stopping, can cause the kinetic energy to rupture pipes and couplings.

AQ: 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.

AQ: Motor starting time to reach full speed

It is not easily answered since there are many variables at play which will affect the starting time. For a large medium voltage motor, it is recommended that a motor starting analysis be performed so that proper control and protection of the motor can be set. The motor manufacturer is a good place to start to find a motor data sheet and torque curve responses; that should give you some good starting point data. Such an analysis can provide inrush current, voltage dip, and starting time.

The time that any motor to run up will depend on the actual load on the shaft. In broad terms the larger the load (related to the rated output) the longer it will take to run up. I would have expected 2 – 2.5MW motors to be manufactured to run on 10-11Kv and DoL. The startup times of these motors would typically be between 45 seconds (No Load) and 3 or 4 Minutes (dependent on the type and magnitude of the load).
I also tend to agree if the feed value is shut the motor will not initially see a significant load and should run up quite quickly.

I would start with Te time constant of the motor as the starting time in the worst case. If you intend let your motor live for long, you should design its protection to avoid starting times longer than Te and nor even close to it. As for specific application, it’s always try and error, but the guiding line should be: start at minimum load and increase it gently (some motor protection relays guard load increase rate).

AQ: 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.

AQ: 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.