Iacdrive|Automation Solution

How to select a breaker?

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.

You can calculate operational current of breaker using this expression:

Inp=SnT/((sqrt(3))*Un)

After that, you need to calculate expected value of surge current:

kud=1+e(-0,01/Tae)
Iud=(sqrt(2))*kud*I’

After that, you need to calculate expected value of heat impulse:

A=(sqr(I0″))*Tae*(1-e(-2*ti/Tae))+(sqr(I’))*(ti+Td”)

And finally, you need to calculate 1s current (expected value of current during 1s):

I1s=sqrt(A/1s)

So, current of interruption of your breaker and power of interruption of your breaker are:

Ii=I’
Si=(sqrt(3))*Un*Ii

Additional expressions that you can use during your calculation:

I0″=Un/((sqrt(3))*Ze”);
I”=1,1*Un/((sqrt(3))*Ze”);
I’=1,15*Un/((sqrt(3))*Ze’);

where are:

ti-time of interruption
Inp-operational current of breaker
SnT-rated power of transformer
Un-rated voltage
kud-surge coefficient
Tae-time constant of aperiodic component of short circuit current
Iud-surge current
A-heat impulse
I0″-short circuit current in subtransient period (generators are in no-load conditions)
I’-short circuit current in transient period
Td”-time constant of subtransient component of short circuit current
I1s-current during one second
Ii=expected value of current of interruption of your breaker
Si=expected value of power of interruption of your breaker
Ze”-equivalent impedance of power system in the place of fault (subtransient period)
I”-short circuit current in subtransient period (generators are in full-load conditions)
I’-short circuit current in transient period
Ze’-equivalent impedanse of power system in the place of fault (transient period)

For a branch circuit feeding a single pump, you would generally size the circuit at 125% of the pump’s full-load amperage. If you’re not using a variable frequency drive or soft starter (which have built-in overload protection), you would use a Motor-circuit protector (MCP) breaker that has both thermal and magnetic trip capability. Sizing would be according the breaker manufacturer’s recommendations for a motor of a given horsepower, but not larger than would be required to protect the circuit conductors.

“The total load of an area” is much too ambiguous to answer. If you have lighting and receptacles, you’re going to need a different type of breaker than if you have motors or mixed types of load. There is no general approach. Circuit breaker types are very specific to the application.

Safety should not be taken lightly. Installing the wrong type of breaker could result in equipment damage and/or physical harm.

There are instantaneous breakers as well as time delay breakers. For time delay breaker, for example, you go 250% maximum of the rated current based upon the HP of a motor (look in the NEC), not on the nameplate label. The nameplate current value is for overload protection. Also try to size the breaker so that the conductors are protected.

As we kn

Variable frequency drive Constant Torque/Variable Torque

A typical variable torque application would be a centrifugal pump. A typical constant torque application would be a conveyor, and there are positive displacement pumps that are also constant torque. Have a talk with a mechanical engineer, get them to show you curves and explain.

DBR stands for Dynamic braking resistor. Regeneration will happen when the motor rotates a speed higher than the speed which corresponds to the frequency setpoint ie.. the rotor speed is more than the speed of the rotating magnetic field.
Regeneration feeds back energy to the drive which results in DC bus overvoltage. To prevent the drive from tripping due to DC bus overvoltage the DBRs are used. The regenerative energy is discharged in the resistor as heat.

Regenerative Breaking – we used to have VFD on a vehicle rolling road. So when the car is travelling faster than the VFD, the VFD generate back into the power supply – causing a break effect. If you had a large mass- large inertia that you want to stop quickly, you need to break the load- you can do that with regenerative breaking. Otherwise, disconnecting the variable frequency drive, will mean your load just freely rotates, and that can mean it will take 30 minute to come to a stop for a large inertia.

Active Front end- I first came across this term with ABB. It is all to do with how to mitigate harmonics from VFDs. You can use phase shift transformers, but with modern electronics, you can use a opposite phase current to counter act the harmonics generated from the VFD. So the overall impact on the network is small.
In active front end technology the rectifier is basically an inverter with IGBTs.
The main advantage are:
1) Low current THD <5 %
2) It is basically a four quadrant rectifier .Referring my last post please note that you will not require a DBR with AFE. The increase in voltage of DC Bus due to regeneration can be fed back to the input AC supply in the form of energy. So you don’t require a DBR.
3) AFE drives have very good immunity to input voltage fluctuations.

Just an advice. Please go through variable frequency drive literatures (available in plenty) to have a good understanding of the different VFD technologies.
Selection of VFD requires proper understanding of the VFDs and the overall electrical system. There are lots of marketing gimmicks in the world of VFD. Always be careful before selecting a VFD specially higher KW drives.

For large drives, you need to speak with supplier to configure your machine correctly. There are many options, but yes active front ends are available. But there are other solutions; ASI Robicon use a current driven VFD, so harmonics are lessened in the first place, so an active front end is not the right terminology. It is a different solution. I used a 10MW version of that type of ac drive. I think Siemens have bought the company since.

VFD PWM and PAM definition

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.

Change transformer vector group

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?

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.