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VFD replace mechanical gearbox to drive the load

Can an AC drive to replace the mechanical gearbox that used to decrease motor speed in conveyor application i.e to use a motor that will drive the load directly throw a coupling, belt or chain, without gearbox, motor rated up to 18.5 kw.

Theoretical is true as far the speed variation is concerned. Practically is not recommended for your application if the conveyor is required to be used with constant speed, on the other hand the gearbox also used for Torque purposes.

For light conveyors used on packing lines on which rate of production varies in accordance to some industrial parameters (Automation & PID control), direct coupled motor controlled by variable frequency inverter drive may be feasible.

VFD is expensive (capital & running cost) its selectivity should be done carefully among the other available options.

By using a variable frequency drive we can change the speed of an AC Motor, and working for any time on any choosing speed, even in some case we can exceed the speed more than the normal one if the motor can withstand it. Noting that:
1- We should be careful when choosing the type of AC drive that should ne normally done according on the application “Conveyor, Fan, Pump, Compressor, ext ” to determine the torque’s level at running time.
2- In some special case when the motor runs at too low speed comparing by his normal one, maybe we need a forcing cooling for that motor.
3- Each VFD has a value of the Short Circuit’s level that can be withstanded, so, we should be careful of that point.

frequency inverter

How to find the KA rating of Circuit breakers?

Before breaker’s selecting for your electrical system, you need to calculate value of expected short circuit current at the place of breaker’s installation. Then you need to calculate value of heat pulse and 1s current (expected value of current during one second). After that you need to calculate power of breaker and finally, after all, you can select appropriate breaker. Values of characteristics of selected breaker need to be higher from calculated values of characteristics of your power system.

1. The fault level of the upstream NW (Source) to be known, normally 500MVA or 250MVA.
2. Upstream impedance (reactance and resistor, capacitor to be ignored for Short cct calculation) can be determined accordingly.
3. The LV System starting from the secondary of the distribution transformer.
4. Short circuit percentage voltage for Transformer is known (normally 4% for 1000kVA and 6% for 1500 kVA) and hence reactance and impedance can also determined.
5. Impedance of Cables also can be determined from manufacturer TD sheet.
6. Subtotal impedance to be determined by conventional way (Submit if are in series/ (Z1+Z2+….Zn)/(Z1XZ2X…Zn) if are in parallel.
7. divide voltage by the Impedance up to the required location, will give you the fault current at that location.
8. Determine Maximum and minimum fault current. By the former you can decide the breaking capacity of CB and by the later the setting can be achieved.
9. verify the thermal constraints of the conductors(cables). ie

I²t ≤ S²K² , I short cct current, t time( < 5 s valid), K cable material Factor and S cable section area.
I²t Known as let through energy. accordingly breaking capacity of CB should be > than Circuit Maximum fault.

The MCB, MCCB, & ACB are all Low Voltage Circuit Breakers, where SF6 is a Non-active gaze used in Medium Voltage Circuit Breakers.

Now, to determine the value of Breaking Capacity of any circuit breaker, we should, by calculation, the Maximum Short Circuit Current Value ” Isc3max ” at the installation point of that circuit breaker, where we can calculate it by assuming a ” Short Circuit between 3 phases at that point “, then after knowing ” Isc3max ” we can determine the Breaking Capacity value that should be ” equal or bigger than Isc3max “.

Further:
1- The value ” 250 … 500MVA ” is the short circuit power at Medium voltage side for up to 36kV.
2- About the Short Circuit Voltage percentage value:  we called ” Ucc or Usc ” and the value is ” 4& for up to 630kVA transformers “, and ” 6% for up to 2500kVA transformers “, but in all case, we can read it at the transformer’s name plate.
3- Sorry Mr. Omar, we can’t do, you mentioned, the sum of all Z, as these values aren’t on the same vector, so, we should first calculate ” R & X ” for each component, then do the sum of all R ” R total ” and all X ” X total “, then calculate the ” Z total “.
4- By knowing the Minimum Short Circuit Current value ” Iscmin “, we use it to determine the value of “Setting Value” of “Magnetic Protection or Short-time Protection”.

Motor fuses

Mainly Fuses are used for protection against short Circuits due its high rupurting capacity (breaking capacity) and fast response (less than 10ms).

As far the electrical drives are concerned, Fuses can be used to protect the feeders, while the Electrical Motors will be protected by Thermomagnetic CB to achieve Short Circuit as well as overload protection. At least thermal overload has to be provided for the Electrical Motors.

Accordingly, Plow of fuses depend on the type of Short Circuit, Single phase or 3 Phase fault (ie location of the fault) and the let through energy. In case one phase blown (say earth fault) -ve sequence and Zero sequence will be generated and subsequently the motor thermal overload will operate to protect the Motor.

It is worth to mention that, now a day proper protections for Electrical motors are commonly used, MCCB/MCB for overcurrent (short Circuit & overload), Single Phasing, Under voltage, Phase sequence relays… etc.
Thermal Protection can be achieved by many technique (ie Bi-metal, thermostat, resistance (NTC or PTC),,,etc.

The Fuse for Electrical Motor is efficient for O/C (Short Circuit, either L-G or L-L) or/and internal fault in the Motor windings. Taking into consideration the fuse rating considering the Starting current of the Motor. Therefore the fuse will not be effective for overload protection on the similar case.

However, I believe the motor was either subjected to an internal fault due to insulation failure (Humidity, water, bearing damages, high temperature rise, Stator/rotor gap,,etc) , subsequently the fuse blown or in prior the fuse blown due to an external factor and the thermal overload device associated with Motor Control panel not operated / out of order in the proper time -The heat rise impaired the winding accordingly or the Motor was subjected to stall current and prolong starting period.

Electrical equipment in hazardous areas

With regards to hazardous areas, Electrical equipment to be installed in those areas should comply with the zone classification. I believe the location where you are intending to install this motor would have been classified according to your local classification standards or IEC 60079 for Liquid/gas/vapour explosives OR IEC 61241 for dusts. Therefore your motor should is to be certified to be installed in those areas, to verify this information you can ask the manufacturer or supplier to provide the Certificate of conformity.

Other information to be looked at, when installing the hazardous motors with variable frequency drive etc, the IEC requirements state that
– both motor & VFD to be certified and type tested together
– IP ratings, protection technique, temp class, gas group to comply with zone classification

It is critical to remember that the starting torque is reduced by the square, as the voltage is reduced. So at 70% voltage, the torque is down to 50%. That is where I have experienced the most trouble with soft starts.

It’s probably important to model or have someone model your load versus the motor torque on the soft starter, to make sure the motor will start, and that it doesn’t take so long to accelerate the load that it causes excessive heating, or trips overloads.

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.

Power electronics design

If you are interested in power electronics design at the board or system level, I would recommend LTspice (note the correct spelling) by far above all the others. In addition to being superb for IC design (Linear Tech uses LTspice to design all their own ICs), it also has been specifically designed to run board level, switched mode simulations.

Because of its robust, excellent performance and because it is available at zero cost, LTspice has become the de facto standard SPICE with by far more engineers using it than any other flavor of SPICE. LTspice allows 100 percent transportability and work sharing, i.e., anyone, even those who have not been previous users, can open your files and run your simulations (the free download is well under 10Mb, installs very quickly and is very system friendly – not cookies, messy registry alterations, scattering of installation folders, etc. – removal, if you so choose, is easy and complete).

Like most versions of SPICE today, LTspice has a fine user interface, but that feature should be low on your list. Schematic entry is NOT where you will be spending most your time when doing serious design work. Beyond a point, desktop eye-candy does nothing to help you understand your design and see its flaws and weaknesses (in fact, too many layers of hand holding can just get in the way of that).

Personally, I never breadboard a design anymore until it has proven itself in LTspice (unlike with a breadboard, a simulated circuit’s internals are ALL easily viewable – a great boon for understanding tricky operation). For me, first hardware is always a complete layout (and matches the simulation every time). Of course, the old axiom “garbage-in, garbage-out” very much applies, which means I often spend a lot of upfront time verifying (and modifying and/or making) models to match their components’ data sheets. In fact, I would recommend doing that as a very worthwhile exercise and as something that should impress a potential employer.

When developing a design in SPICE, you will want to spend your time debugging your design, not your simulation or your simulator, therefor it is worthwhile to learn what a simulator needs to run smoothly (with LTspice, all that means is that the input has to be realistic). It was years of working with simulators and a lot of sweat and aggravation before the keys to problem-free simulations gradually crept into my understanding.

1. If possible, make all nonlinear circuit elements be functionally continuous with continuous derivatives (this is not possible for some component behaviors), and

2. *always* craft your simulations so that the nonlinear bits become linear at high frequencies (this is always possible). Non linear devices should never be strict voltage sources. They should be Nortonized and be shunted with small capacitances such that the capacitances (which are linear elements) dominate at small time steps.

3. Always verify that the building blocks of your simulation behave realistically (GIGO).

Follow these guidelines and you will never see the “time step too small” message (I have never met a simulation that couldn’t be made to run well). Note that many (if not most) vendor supplied models fail to meet these guidelines and will give you nothing but headaches if you try to use them “as is.”

Calculate current setting of overcurrent relay

You can calculate current setting of overcurrent relay by using next expression:

Isetting ≥ (ks*Imaxopam)/(a*pi)
Imaxopam=kam*Imaxoptr

where are:

Isetting-current setting of overcurrent relay
ks-safety coefficient
Imaxopam-maximum operational current under which overcurrent relay shouldn’t to act
a-coefficient of layoffs overcurrent relay (0,85-0,95)
pi-ratio of current transformer
kam-coefficient which describes influence of common starting of all asynchronous electrical motors in the appropriate power network after elimination of fault (1-6)
Imaxoptr-maximum operational current of power transformer

Besides I have already explained meanings of all appropriate sizes, I would like to underline differentiate between Imaxopam and Imaxoptr.
Imaxoptr is maximum operational current of power transformer in normal conditions while Imaxopam is maximum operational current of power transformer after interruption of fault and mentioned current includes influence of common starting of all asynchronous electrical motors in the appropriate power network after elimination of fault. It is very important to say that appearing of fault in power network leads to significant decreasing of voltage what has a consequence deceleration of all asynchronous electrical motors in appropriate power network. After interruption of fault, it comes to appearing of process during which all asynchronous electrical motors are starting in some parts of power network which are still turned on. This is situation which is significant different from situation where asynchronous electrical motors are starting one by one while in this case all asynchronous electrical motors are starting at the same time. Because of this fact, after elimination of fault, value of current isn’t same as value of current before appearing of fault. After elimination of fault, value of current, which I called Imaxopam, is higher than value of operational current of power transformer, which I called Imaxoptr, but under those conditions there is no fault, so current setting of overcurrent relay should to be set on that value of current and mentioned relay shouldn’t act under those conditions.

After calculation of current setting of overcurrent relay, you need to check coefficient of sensitivity of acting of overcurrent relay by using next expression:

ksens=Ifmin/(Isetting*pi)

where are:

ksens-coefficient of sensitivity of acting of overcurrent relay
Ifmin-minimum fault current (1 phase fault to the earth or 2 phases fault)
Isetting-current setting of overcurrent relay
pi-ratio of current transformer

Value of coefficient of sensitivity of acting of overcurrent relay should to be higher or equal with 1,5 in case when is fault at opposite busbars (busbars where isn’t overcurrent relay) or higher or equal with 1,2 in case when is fault at the end of the longest feeder which begins at those opposite busbars. Time setting should to be selected like that overcurrent relay needs to wait acting of another protection which is on the feeders (for example distance protection).

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.