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 inverter, 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.
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.
System with difference neutral
Q:
I have one system with two source. One from genset and the other from PLN (national power supply company) that each system has neutral.
The question is
1. Is there any problem if I connect both neutral directly?
2. Is there any spark when I connect both neutral?
3. How is the best solution to connect both neutral?
A:
1. I understand that the Genset is dedicated for essential load as an Emergency power supply which will be operated by hand (only in Manual Mode).
2. The Control Philosophy for a Generator that intended to be connected to PLN as emergency power source depends on the local service provider regulations.
3. Usually in your case there should be Electrical as well as Mechanical interlocks between the mains incomer & genset main breaker. ie both Sources will never be in Synchronism ( will not feeding the same load simultaneously).This measures will ensure that there will be only neutral point to the system.
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.
Soft start motor tripped in fuel oil suction and discharge
First of all check all the component i.e.CB, CT, Heat Element, and the O/L setting then megger the motor to be shore that there is no problem with the motor winding insulation.
After that let the mechanical check the vibration analyses during the start-up also measure the startup currant of the motor and diffidently you will find where is the problem.
It could be a relay setting; or problem in the insulation; or even a problem in the motor itself.
On the other hand, check the motor on No Load condition and tune it to the Soft starter before coupling it to the pump.
Auto Tuning feature is generally inbuilt to Advanced soft starters.
If the No load startup of the motor is perfect, 2 causes arise:
1) Improper design.
2) Viscosity _ this can be tackled if you can make some temporary arrangement for pre-heating to confirm if this is the culprit.
As using soft starter could result in reducing torque of the motor. Soft starter normally reduces starting current by reducing starting voltage. However, decreasing voltage will lead to starting toque reduction. Hence, the motor may take longer time, especially when driving high-inertia load, with somewhat high current until it reach its full speed. Using an inverter will help you get full starting torque or even boost up it to 150-200% while keeping starting current at 150-200% of full load. Installation of heat tracing might also help and economic.
Assuming it is an electrical problem. On a motor of this size it has separate overload protection from the ground fault and short circuit protection. There are tolerance levels for motor that you may not be within. However a megger will not answer all the possibilities with motors unless you are ready to perform polarization index test etc….A power analyzer will allow you to see the operation in real world application. Assuming you have confirmed this is an electrical problem your next step would be to use a power analyser. You should be able to confirm by the signature and different placements of the analyzer the problem. Analyzer should be around all three phases.