Iacdrive|Automation Solution

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.

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.

Cable faults/fails

There are only 2 distinct types of Cable fault
1. Due to system Parameter or operating conditions – which you have to take care at the initial design stage for selection of the cable. There is a long list of checks – Design Engineers know.
For example, if 50 sq.mm cable is adequate for a particular Load, you may have to choose higher size in relation to the Fault Level. In urban Distribution system, you will find large size of cable is connected to a small size distribution Transformer.

2. Failure of Cable does not normally occur in the run of the cable unless it is damaged for external reasons. Damaged cable may not fail unless there is ingress of moisture / water through the damage. You will find cable failing during rainy season, high tide in the coastal areas.

Cable fails at the joints mainly because the (construction) characteristics of Cable are changed at the joints – joints become weak links in the run of the cable.

It is a well documented phenomenon that underground cables fail a week or so after lightning activity. Some of the can be attributed to lightning surges that enter the primary conductor and reflect off an open as you indicate. I believe the majority of the failures come from lightning strikes on adjacent structures or trees that reach the cable through ground and cause slight damages to the cable insulation. The maddening part related to customer service is the cables end up failing on a sunny day. Most customers understand outages during inclement weather but are not so understanding of outages on clear days.

Resistance Grounding System

Low Resistance Grounding:
1. Limits phase-to-ground currents to 200-400A.
2. Reduces arcing current and, to some extent, limits arc-flash hazards
associated with phase-to-ground arcing current conditions only.
3. May limit the mechanical damage and thermal damage to shorted
transformer and rotating machinery windings.
4. Does not prevent operation of overcurrent devices.
5. Does not require a ground fault detection system.
6. May be utilized on medium or high voltage systems. GE offers low
resistance grounding systems up to 72kV line-to-line.
7. Conductor insulation and surge arrestors must be rated based on the lineto-
line voltage. Phase-to-neutral loads must be served through an
isolation transformer.

High Resistance Grounding:
1. Limits phase-to-ground currents to 5-10A.
2. Reduces arcing current and essentially eliminates arc-flash hazards
associated with phase-to-ground arcing current conditions only.
3. Will eliminate the mechanical damage and may limit thermal damage to
shorted transformer and rotating machinery windings.
4. Prevents operation of overcurrent devices until the fault can be located
(when only one phase faults to ground).
5. Requires a ground fault detection system to notify the facility engineer that
a ground fault condition has occurred.
6. May be utilized on low voltage systems or medium voltage systems up to
5kV. IEEE Standard 141-1993 states that “high resistance grounding
should be restricted to 5kV class or lower systems with charging currents
of about 5.5A or less and should not be attempted on 15kV systems, unless
proper grounding relaying is employed”.
7. Conductor insulation and surge arrestors must be rated based on the lineto-
line voltage. Phase-to-neutral loads must be served through an
isolation transformer.

Conclusion:
Resistance Grounding Systems have many advantages over solidly grounded systems including arc-flash hazard reduction, limiting mechanical and thermal damage associated with faults, and controlling transient overvoltages. High resistance grounding systems may also be employed to maintain service continuity and assist with locating the source of a fault.
When designing a system with resistors, the design/consulting engineer must consider the specific requirements for conductor insulation ratings, surge arrestor ratings, breaker single-pole duty ratings, and method of serving phase-to-neutral loads.