Iacdrive|Automation Solution

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 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.

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.

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.

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.