PWM is shorted for Pulse Width Modulation, it’s a variable frequency drive (VFD) regulate way to change the pulse width according to certain rules to adjust the output volume and waveform.
PAM is shorted for Pulse Amplitude Modulation, it’s to change the pulse amplitude according to certain rules pulse amplitude pulse train to adjust the variable frequency drive output volume and waveform.
Transformer nameplate vector group is YNd1. However, the nature of connection on both its primary and secondary side is such that:
Generator phase A = Transformer phase c
Generator phase B = Transformer phase b
Generator phase C = Transformer phase a
Also, on transformer HV (secondary connected to grid),
Transformer phase A = Grid phase C
Transformer phase B = Grid phase B
Transformer phase C = Grid phase A
The questions are:
1. How does this affect the vector group (YNd1) of the transformer? Will it be changed to YNd11?
2. Will it make any difference as far as the vector group is concerned if instead of phase A and C, phase B and C were swapped on both ends of the transformer?
3. The transformer protection relay is configured for YNd1 group, and it is reading negative phase sequence current (ACB instead of ABC). Changing the vector group configuration will solve the problem?
4. Relay is used for differential protection (percentage differential) of the transformer.
Will this negative phase sequence affect normal operation of the transformer in any way?
1. How does this affect the vector group (YNd1) of the transformer? Will it be changed to YNd11?
Yes, the name plate vector group of a transformer is only valid for a standard phase rotation ABC. for a phase rotation ACB the apparent vector group will be YNd11.
2. Will it make any difference as far as the vector group is concerned if instead of phase A and C, phase B and C were swapped on both ends of the transformer?
No, by swapping any two phases the rotation becomes no standard and the apparent vector group will become YNd1
3. The transformer protection relay is configured for YNd1 group, and it is reading negative phase sequence current (ACB instead of ABC). Changing the vector group configuration will solve the problem?
I think the way the relay is configured at the moment will give you problems, if I’m correct you should be able to see differential current when the transformer is loaded, and it is likely to trip on the first through fault (can you confirm this). To resolve this issue you have two options.
i) Set the vector group to YNd11 in the relay, this will remove the differential current but will mean the relays see’s 100% NPS current and 0% PPS current, this may give you problem if you have any NPS elements enabled in the relay ( inter turn fault detection, directional elements etc)
ii)Set the vector group to YNd1 and the phase rotation setting to non standard ACB this will get rid of the NPS currents and the differential current, so this is probably the best solution.
4. Relay is used for differential protection (percentage differential) of the transformer.
Will this negative phase sequence affect normal operation of the transformer in any way?
No, there will be no problem with the transformer itself just the relay protecting it.
As i said previously if I’m understanding the problem correctly, you should be able to see differential current at the moment when the transformer is loaded, is this correct?
In any electrical system, we limit the neutral grounding to 1 or 2 locations at the power source, eg, the star-points of generators or transformers. By keeping the grounded neutrals at the power source, earth leakage current will be flowing radially from the power source to the point of short-circuit at downstream. In this way the direction of earth fault current flow can be easily identified and the earth fault protection relays in the distribution system can easily be coordinated.
Grounding a motor star point will create an earth path for earth leakage current to flow through that motor’s star point. If there are 10 motors in a process plant and their star points are all grounded, there are 10 additional paths for earth leakage currents to flow through.
If all the motors’ star points are grounded in this way the earth fault current detections by the protection relays will be complicated and likely they will trip at the incorrect locations because earth fault currents are flowing in many directions toward multiple grounded neutral points.
Therefore the electrical consumers (ie the load, including the capacitor banks), even if they are star connected, are not to be grounded.
Grounding of neutral point is not being decided base on the presence of unbalance loads. It is decided for safety reason and for earth fault protection requirement. Unbalance 3-phase load will result in some current flowing through the neutral conductor but it doesn’t result in a (residual) current flowing through the neutral-ground connection.
Motor is a balanced 3-phase load, this I agree. However when the system supply voltage is unbalanced caused by unbalanced loads somewhere else or due to network conductors problem, the motor operating under unbalance voltage will result in unbalance current in the 3 windings. The same is true for the generator windings under that condition. The design engineer may then decide that individual machines should be fixed with negative phase sequence current protection.
Even if there is a neutral voltage shift in the induction motor, we should not ground the motor’s neutral point. If you ground it, it may create nuisance trip of earth fault protection relays (the motor’s EF relay, upstream EF relays, or the EF relay connected to transformer’s neutral-ground CT).
I am sure in reality, there is some neutral voltage shift in motor’s star point. However, there is no harm with that.
If you ground the star point, you still will not get rid of the unbalance current/voltage from the motor windings. There the negative sequence current is still present in the motor winding.
If you think an unbalance voltage supply is causing problem to the motors, you should solve the unbalance voltage problem elsewhere, not by grounding the motor’s star point.
When two transformers are in a parallel group, a transformer with a higher tap position will typically have a higher (LV side) no-load voltage than the other one with a lower tap position. These unequal no-load voltages (unequal tap positions) will cause a circulating current to flow through the parallel connected transformers. A transformer with higher no-load voltage (typically higher tap position) will produce circulating current, while a transformer with lower no-load voltage (typically lower tap position) will receive circulating current.
When load is connected on these two parallel transformers, the circulating current will remain the same, but now it will be superimposed on the load current in each transformer, i.e. for a transformer producing circulating current, this will be added to its load current, and for a transformer receiving circulating current, this will be subtracted from its load current.
Thus voltage control of parallel transformers with the circulating current method aims to minimize the circulating current while keeping the voltage at the target value.
In case of a parallel operation of transformers, the electric current carried by these transformers are inversely proportional to their internal impedance. Think of it as two parallel impedances in a simple circuit behind a voltage source, you will have equal currents through each impedance only if you have two identical impedances, in some cases as stated above, tapping could be a problem, the other one is the actual manufacturing tolerances which could diverge by almost 5-10%, if the transformers are manufactured by different suppliers or not within the same batch. So, the difference in current between the currents through these two impedances is basically the circulating current as it is not seen outside these parallel impedances.
The currents that are produces due to magnetic flux circulation in the core are called eddy currents and these eddy currents are responsible for core losses in transformer.
While the circulating currents are the zero sequence currents that may be produces due to following causes.
1- when there is three phase transformer the (3rd, 5th, 7th….) harmonic currents which are called zero sequence currents from all the three winding of three phase transformer add up and become considerable even in loaded conditions these currents have no path in Y/Y connection of transformer so a tertiary winding is provided co conduct these currents but in Y/d or D/y connection these currents circulate in delta winding.
2- Whenever there is unbalanced loading in transformer. In which with positive sequence, negative sequence and zero sequence currents are also produced which cause circulating currents.
3- When the transformer banks are used and the transformers have phase between them then circulating currents are produced between them, than transformers in the bank get loaded without being shearing the power to the load.
Q:
First i had carried out experiments with a single IGBT (IRGPS40B120UP) (TO-247 package) 40A rating without snubber and connected a resistive load, load current was 25A , 400V DC and kept it on continuously for 20min . Then i switched the same IGBT with 10KHz without snubber and the IGBT failed within 1min. Then i connected an RC snubber across the IGBT (same model )and switched at 10KHZ. The load current was gradually increased and kept at 10A. This time the IGBT didn’t fail . So snubber circuits are essential when we go for higher switching frequency.
What are the general guide lines for snubber circuit design in You are not looking close enough at the whole system. My first observation is that you are using the slowest speed silicon available from IR. Even though 10KHz is not fast, have you calculated/measured your switching losses. The second and bigger observation I have, is that you think your circuit is resistive. If your circuit was only resistive, any snubber would have no effect. The whole purpose of a snubber is to deal with the energy stored in the parasitic inductive elements of your circuit. Without understanding how much inductcance your circuit has, you can’t begin designing a cost and size effective snubber.
To echo one of the thoughts of Felipe, you need to know the exact purpose of the snubber. Is it to slow down the dV/dt on turn off or is it to limit the peak voltage? Depending on which you are trying to minimize and your final switching frequency, will dictate which snubber topology will work best for you. The reason that so many snubber configurations exist, is that different applications will require different solutions. I have used snubbers in various configurations up to 100KHz.
As per Torque/Slip characteristic for AC Motor, the value of the Max. Torque can be developed is constant while the Starting Torque occurs @ S=.1, (T proportional to r2 and S also proportional to r2 where r2 is the rotor resistance, the ratio r2/x2 when equal to 1 gives the max. Torque w.r.t Slip at Starting. Wound rotor motors are suitable and recommended for application for MV drive where it is required to be started on load such as ID. Fans, S.D Fans, Drill, etc.
As you aware the torque is directly proportional to the rotor resistance “r2” & varies with slip “S”, hence injection of resistance into the rotor via Slip Rings, High Starting Torque can be got while the Speed, efficiency and Starting current will be reduced. Therefore resistance is the most practical method of changing the torque (i.e. wound rotor Slip ring Motors). Moreover, the Max torque can be achieved at starting when rotor Resistance “r2” = The Stator impedance, at starting S=1.
On the other hand, the slip of the Induction Motor (speed) can be changed by “extracting” electrical power from rotor circuit, more extraction increases the slip. By using thyristorized Slip-Recovery Scheme “ i.e Kramer Scheme” feedback of Power from rotor circuit to supply circuit which also known as “the slip Power recovery scheme”. The scheme is simply consists of rectifier and an inverter connected between slip-rings and the A.C Supply circuit. The Slip Rings voltage is rectified by the rectifier and again inverted to AC by the inverter and feedback to supply via a suitable Transformer. Such arrangement gives good efficiency with high cost due to Rectifier and Invertor.
There are different types of system configuration for grounding like TT,IT,TN-C etc. How do we decide which configuration is suitable for the particular inverter (string or central). What are the factors that help us to decide the configurations?
One of the main concerns in a system is to avoid large low impedance ground loops.
These are created by the return signal path connected to the chassis (metal work) at multiple points. The large current loop allows noise currents to radiate H fields and hence couple into other electronics. The antenna effect will be proportional to loop area.
Single point grounding of the return path to chassis prevents this. However single point grounding conflicts with good RF practice where you want to ground to chassis at the sending and receiving ends of a signal path. There is therefore no universal best practice.
In my field, spacecraft, the standard practice has all primary power electronics galvanically isolated from the spacecraft chassis. Individual modules must maintain the isolation with transformer coupled DC/DC converters. The centre tap of each PSU secondary output is then single point grounded to the module metalwork. We talk about primary side and secondary side electronics where only secondary side is grounded to the metalwork.
Anything powered directly from the primary bus must be isolated with a maximum capacitance to chassis of 50nF to avoid excess HF loop currents forming.
In general it depends on country specific law and standards required by Power Supply Operators. From design point of view it all depends on which point of grid you are going to connect and what type of inverter is used.
Q: For now I am working on a mining project which involves starting two SAG mills, the method of starting these mills is by rotor resistance and likewise we are using an energy recovery system (SER), could someone tell me how this system works SER? Each mills have two motors of 8000 kW at 13.8 kV.
A: For large mills requiring variable speed, the wound rotor motor and SER drive are economical for a total rating of approximately 2MW to 16MW. Above 16MW, the gearless drive (cyclo-converter) is typically used because gearboxes and pinion gears reach their present limit in size. Around 2MW and below, the squirrel cage/VVVF drive is simple and cost effective.
Advantages of the wound rotor/SER drive are:
1. If the SER converter drive fails, the drive can be switched to fixed speed bypass – starting the usual way with the LRS.
2. The converter only needs to be sized for 15-20% of the total motor rating with associated reduction in floor space, air-conditioning etc. The converter is only sized for the feedback energy which is proportional to the speed difference from synchronous speed. The drives are typically set up to run between about 85% to 110% of synchronous speed for an optimized arrangement.
3. Relatively low capital cost when all things considered – including spare motor cost etc.
Brush/slip ring maintenance is one issue. However, when the brushes are specified correctly for the load, the wear is manageable. Once the maintenance program is set up for shutdowns, it is not a major issue.
I expect that this type of drive would be the most common large mill variable speed drive in the world’s minerals processing industry for the range mentioned above for the last 15 years (approximately).
The SER drive converter controls the voltage in the rotor. Motor speed is proportional to rotor voltage. Resistance in the rotor indirectly achieves the same thing (with a different torque curve shape), but energy is lost in the resistors which is very inefficient. The SER drive via a feedback transformer feeds energy back into the power supply. This returned energy is proportional to the speed difference from synchronous speed. So at say 85% speed, 15% of the motor rated power is returned from the rotor to the supply. At a hypersynchronous speed, the SER drive feeds power into the motor rotor allowing it to run faster than synchronous speed. So for a fixed torque and higher speed, the power obtained from the motor is higher than the motor nameplate rating.
Gearbox ratio is best set up to allow the speed range to be covered using the SER drive’s hyper-synchronous capability.
NGR stands for Neutral Grounding Resistor. When an earth fault current occurs on a plant, assuming that there is no external device presented to limit the earth fault current, the magnitude of the earth fault current is limited only by the earth impedance presented between the point of fault (to earth) and the return path (typically a star point of a transformer). If the earth impedance is low (type of soil being one of the reason amongst others), the fault current magnitude can be significantly high, and if left unchecked could damage the primary equipment. It is therefore mandatory that the earth fault current be limited to a suitable value, which is typically the rated value of the plant as a thumb rule. Why use the rated value? Because the plant has been designed to carry the rated current continuously.
Let’s take an example: say you have a transformer 60MVA, 132/33kV Star-Delta transformer. It is required to calculate the value of NGR to be connected to the zig-zag transformer on the 33kV Delta. the value of the resistor required to limit the earth fault current to the transformer’s LV rated value is (33 x 33) / 60 = 18.15 Ohms.
(Earth Fault current limited to rated value = (60 x 1000) / (1.732 x 33) = 1050A) When you go to a supplier you might find he supplies only 20 ohms resistor (as you might not get the exact value that you have calculated theoretically). No problem, use the 20 ohms and calculate what your new value of earth fault current would be (33 x 1000 / (1.732 x 20) = 952.6A, which is less than the transformer’s rated LV current. So you’re safe. This is how I would go about. In fact I would go a step further and introduce a safety factor of 20% i.e. I’ll bump up the value of the resistor from 20 ohms by an extra 20% and buy a resistor/ NGR of 1.2 x 20 = 24 ohms. So I am 100% sure that the earth fault current is way below the rated value and my transformer will be safe, even if the fault current goes undetected for any unforeseen reason say my earth fault protection has failed to pick up.
Make sure however that the earth fault setting that you choose is sensitive enough to pick up for the earth fault current calculated. I would generally put two relays a 64 or REF designed to pick up and operate instantly backed by a 51N with a sensitive setting but with a delay of a couple of seconds to pick up in case the 64 has failed to pick up.
So that’s it. I have described how I would go about calculating the earth fault current, selection of NGR value and how I would protect it.
Protection and related devices aiding protection don’t come cheap. Also I assume by your comment “this method is the most expensive option available since the cost of the transformer shall be astronomical”, you are referring to the Zig-Zag transformer and not the actual 132/33kV Star-Delta power transformer, under question.
I have taken a very generic example and tried to focus on how to arrive at a suitable value of an NGR, assuming an Star HV and Delta LV. My aim being to calculate how I could limit the fault current on the Delta LV. Being a Delta winding, I have to use a Zig-Zag transformer, for providing a low zero sequence path for the flow of earth fault current. It is really the Zig-Zag trafo. that bumps up the cost.
Note: If the above transformer is one of a kind, i.e. this is the only transformer in an isolated network, then I simply disregard the Zig-Zag transformer + NGR method and use the 3 PT broken delta method for 3Vo detection to drive a 59N. My cost here would be very low.
If the transformer is a Star-Star type with HV start solidly grounded, and LV star impedance (NGR) grounded, then I don’t need a Zig-Zag trafo. on the LV side. My cost is purely for the NGR alone.(Of course this transformer will have a Delta tertiary which may need it’s own protection depending on the whether one plans to load the tertiary or not. We could di
There is a ANSI/IEEE standard that defines the standard number identification for electrical devices. You will find that some of the more common ones are 50 over current, 51 short terms over current, 27 under voltage, 59 over voltage, and 50G ground over current detection relay.
Older installations may have the older electro-mechanical relays. Mst installations have converted to using sensing devices that transmit to PLC, DCS or protective relays. Today’s protective relays are essentially PCs that monitor a number of power system parameters for metering or protection purposes. They also have programmable outputs and settings.
As stated previously, search the web for some examples or guidelines on electrical schematics. There is also a reference standard develop by IEEE and ISA to define the symbols used to represent the hardware or software functions that input to PLCs or are the functions within the PLC (or DCS). The older form of schematics was drawn horizontally, but the same ladder logic used today is drawn vertically with each line numbered in sequential order. The line numbers are used in the device identification and as a reference in the PLC programming.
It can be confusing, but think of items in series as AND logic versus items in parallel being OR logic. Understanding the And / OR logic will enable you understand how the respective logic components function on an electronic card used in a PLC.
I recommend that you join IEEE and ISA. They usually have local chapters that meet to network and share information. It is also a good way to network and meet persons at other companies and of course meet vendors who are eager to meet persons who work for engineering firms that they market to.