Iacdrive|Automation Solution

Harmonic current

I hate to call them harmonic currents. The do submit to Fourier analysis, but you are probably dealing with AC to DC power supplies. If you look at the current pulses, you will see that each pulse is about 1-2 milliseconds in duration in alternating directions. If you sum these all in the neutral there is the appearance of what looks like 180 Hertz in the neutral. If you use different sized power supplies on each phase, you can see that it is just the addition of the three phases. So the neutral current when you have non power factor corrected power supplies is the sum of the three phases. Unless the current waveforms overlap, there is no cancellation of current in the neutral, hence the neutral current is the sum of the phase currents. The reasoning behind this is the rectifier diodes in the front of the power supply and the DC storage capacitors size relative to the DC load on the capacitor. The general rule of thumb is that the capacitor is about 800 to 1000 microfarads per amp of current in the capacitor.

Realize that the extra heating in the three phase delta-wye transformers is due to the extra circulating current in the primary delta causing excessive heating of the primary conductor. The world calls transformers designed to deal with this “K” factor transformers. Let the world of electrical engineers bury all this simple stuff behind the maze of Fourier analysis. Change the incoming voltage slightly and your Fourier analysis is garbage. The issue here is switches and storage caps— not some magical mathematical garbage.

By the way if someone wanted to use the wire sizing guidelines of the National Electrical Code in the US to size wire for 100% power supply load, the neutral wire would be 8 gauge sizes larger than the phase conductors. People need to start demanding PFC power supplies. Fix a switching problem with switches.

Transmission line low voltages and overload situations

Q: I want to know just what the surge impedance loading (SIL) is but its relevance towards the improvement of stability and reliability of a power network especially an already existing one with various degrees of low voltages and overload situations?

A: The surge impedance loading will provide you with an easy way of determining if your transmission line is operated as a net reactor (above SIL, so external sources of (2) line-voltage-drop limitation
(3) steady-state-stability limitation

In contrast with the line voltage drop limitation, the steady state stability limitation has been discussed quite extensively in the technical literature.

However, one important point is rarely made or given proper emphasis; that is, the stability limitation should take the complete system into account, not just the line alone. This has been a common oversight which, for the lower voltage lines generally considered in the past, has not led to significant misinterpretations concerning line loadability

At higher voltage classes such as 765 kV and above, the typical levels of equivalent system reactance at the sending and receiving end of a line become a significant factor which cannot be ignored in determining line loadability as limited by stability considerations, so surge impedance loading plays a fundamental role in reliability and stability.

Neutral current is less than phase current?

In a balanced 3-phase system with pure sine waves, the neutral current is zero, ideally.
If there is phase imbalance, it shows up in the neutral, so check for imbalance.

The other major cause of high neutral currents is full wave rectification, where the current of each phase is flowing only at its peak voltage. In this case, the neutral current can be as high as three times the phase currents, theoretically.

If you can see the frequency of the neutral current, line frequency currents indicate imbalance. Current due to full wave rectification is high in third harmonics, so it may show mostly 3 x line frequency, or be a ratty square wave at 3 x line frequency.

High neutral currents, and some resulting fires, are largely responsible for the adoption of power factor correction requirements. If your loads are balanced and pfc corrected, you should not have neutral currents.

The neutral current (In) is summation of the phase currents. And obviously, the three phases are decoupled now; and not loading Y makes Iy=0.
So In = Ir + Ib (vectorial sum). Now depending on the amount of loading, nature of loads and their respective power factors, a variety of possibilities (for neutral current magnitude and phase) arise; which may include the case of In being higher.
The statement “neutral current is usually less than phase currents” is naive and not universal.

Nonlinear loads (i.e. rectifiers as Ed mentioned above) draw significant harmonic current. In many cases the current Total Harmonic Distortion (THD) is >100%. In a 3-phase, 4-wire system, the triplen harmonic currents (3, 9, 15, 21…) sum in the neutral wire because they are all in-phase. This is why the neutral current can be much higher than the phase currents even on an otherwise balanced load application. If you can put a current probe on the neutral and look at the waveform – you can see how much fundamental vs. harmonic current there is.

What is the surge impedance load

The surge impedance loading (SIL) of a line is the power load at which the net reactive power is zero. So, if your transmission line wants to “absorb” reactive power, the SIL is the amount of reactive power you would have to produce to balance it out to zero. You can calculate it by dividing the square of the line-to-line voltage by the line’s characteristic impedance.

Transmission lines can be considered as, a small inductance in series and a small capacitance to earth, – a very large number of this combinations, in series. Whatever voltage drop occurs due to inductance gets compensated by capacitance. If this compensation is exact, you have surge impedance loading and no voltage drop occurs for an infinite length or, a finite length terminated by impedance of this value (SIL load). (Loss-less line assumed!). Impedance of this line can be proved to be sqrt (L/C). If capacitive compensation is more than required, which may happen on an unloaded EHV line, then you have voltage rise at the other end, the ferranti effect. Although given in many books, it continues to remain an interesting discussion always.

The capacitive reactive power associated with a transmission line increases directly as the square of the voltage and is proportional to line capacitance and length.

Capacitance has two effects:

1 Ferranti effect
2 rise in the voltage resulting from capacitive current of the line flowing through the source impedances at the terminations of the line.

SIL is Surge Impedance Loading and is calculated as (KV x KV) / Zs their units are megawatts.

Where Zs is the surge impedance….be aware…one thing is the surge impedance and other very different is the surge impedance loading.

Snubber circuit for IGBT Inverter in high frequency applications

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.

AC motor maximum torque

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.

System configuration of grounding

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.

Starter of SAG Mills with rotor resistance

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.

The basis of rating a NGR in electrical system

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

Read control wiring diagram of relays in substation

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.