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AQ: Why the motor rotation speed is changeable?

r/min: motor rotation speed unit, the number of revolutions per minute, also can be expressed in rpm.
For example: 2-pole motor 50Hz 3000 r/min, 4-pole motor 50Hz 1500 r/min
Conclusion: The motor rotation speed is proportional to the frequency

Herein, the motor is induction AC motor which is used in most industries. AC induction motor rotation speed is approximate depend on the motor’s poles and frequency. As we know, the motor poles constant. Since motor poles are not continuous numbers (multiples of 2, for example, the number of poles is 2,4,6), so it’s not suitable to change this value to adjust the motor speed.

The frequency can be adjusted before supplying to the AC motor, then the motor rotation speed can be controlled freely. Therefore, motor speed controls.

n = 60f / p
n: synchronous speed
f: power frequency
p: number poles

Conclusion: change frequency and voltage is the best motor control method.

But, if just change the frequency without changing the voltage, it will occur overvoltage (over-excitation) when frequency decreases and may cause AC motor damaged. So, the voltage should be changed while the frequency inverter outputs different frequency. If the inverter output frequency exceeds rated frequency, the voltage can not continue to increase, the maximum voltage only can be equal to the motor rated voltage.
For example: In order to reduce the motor rotational speed by half, the inverter output frequency change from 50Hz to 25Hz, then the inverter output voltage should change from 400V to about 200V.

AQ: Change 230V to 460V for operating an Electric Motor

I have a generator of 3 hp, and it outputs 230 V, and I have a submersible Electric Pump, the motor of which is rated to operate at 460 V, Can I use a step up transformer to increase the voltage output from my generator and power the pump? What more parameters do I need to know of in this case?

Check to see if the generator has 3 phase power output. A typical home generator will provide 230 volt single phase output. You will not be able to step up to 460 volt and start a 3 phase motor with single phase. The only way at that point to generate 3 phase would be to use a VFD with single phase input capability and use the drive to generate 3 phase. You will still need to use a transformer. Variable frequency drives won’t normally behave well on generator power but may for an intermittent load like a submersible pump.

AQ: Measuring inductance in per coils of switched reluctance motors

Fundamentally, inductance is the proportionality constant relating flux linkage to current, i.e. lambda = L * i, where lamda is flux linkage, L is inductance, and i is current. Inductance is a property of the geometry. One can write L = N^2 *R, where N = number of turns in the coil and R = the magnetic reluctance “seen” by the coil. So here are a couple of comments. The reluctance varies in an SR motor as the rotor turns, so you will have to take measurements at several positions. If the current is small, L is a constant. If i is large then a small increase in i produces a small increase in lambda, which is smaller than when the current is small. So, you’ll need to decide on the level of current; or, you’ll need to make measurements at several levels of current. For example, at the rated value of current and half the rated value, etc. Now flux linkage is the product of turns and the effective flux through the coil, lamda = N * phi, where phi is the effective flux through the coil. In the case, you have two coils in series where the coils are inside a motor. You only have two leads, one each to the two coils in series. So, you will need to disassemble the motor and tap into a wire that connects the two coils in series. One way to find the flux linkage is the following. Apply a step of voltage to one of your coils. Take traces of the voltage and current. Then apply the formula v = R’ * i + d(lamdda)/dt, where v is the applied voltage to the coil, R’ is the coil resistance (you’ll have to measure this), and i is the current. Then, integrate to find lamda: d(lamda) = int(v – R’*i)dt. You’ll have to do this to both coils.

As you know the inductance of SRM depends on two parameters: 1.coil current 2.rotor position .it means that you have a lot of possible situation that each situation has particular value of inductance .if you want to measure inductance at particular position, I think you should excite one phase with ac supply and use circuit equations (kvl) to find inductance. if you use a dc supply you should measure the flux and it’s hard to do.

AQ: What is a soft starter?

Motor starter (also known as motor soft starter) is a electronic device integrates soft start, soft stop, light-load energy saving and various protection functions for motor controls. Its main components are the three phase reverse parallel thyristors between power supply and being controlled motor and related control circuits. Control the conduction angle of the three phase reverse parallel thyristors by different methods, to achieve different functions by the changeable of the input voltage on the controlled motors.

The difference between soft starter and frequency inverter

Soft starters and AC motor speed control, it can change output voltage and frequency at the same time; actually, soft starter is a regulator for motor starting, only changes output voltage but not the frequency. The frequency inverter has all the features of soft starter, but its price is much more expensive than the soft starter, and the structure is much more complex.

Frequency inverter allows the AC motor smooth start up, control startup current growing from zero to motor rated current, reduce impact to the power grid and avoid the motor being burned out, also provide protection in motor running process. Besides these functions, the main function of frequency inverter is adjusting the motor running speed according to actual operation conditions, to achieve energy saving effect.

AQ: Motor die-cast rotor non-grain-oriented VS grain-oriented

If the material is non-grain-oriented, the path of least resistance for the magnetic flux varies widely from point to point across the sheet: in one place it may go left-to-right across the sheet surface, in another top-to-bottom, and in still another through the sheet. Other points may be anywhere and everywhere in between.

If the material is grain-oriented, the material is aligned such that there is a significant reduction in the energy requirement for passing flux in one direction relative to any other.

Most machines work best with a uniform flux distribution at all points of the airgap surface: this is achieved by stacking both stator and rotor using non-grain-oriented laminations in any arrangement. However, for a grain-oriented material, each lamination has to be rotated by some angle with respect to the one above and below it in the stack (think of it like a spiral staircase).

Regardless of how the winding is made for the rotor (form wound, bar and ring, or die-cast), it is the STACKING process for the core steel that affects grain orientation.

As to skewing BOTH stator and rotor … why? It is a more costly and complex manufacturing process to produce a skewed core vs an unskewed one, regardless whether the skew is in the rotor or stator. Once the skew is begun, there is no real cost difference between a full slot skew and a fractional slot skew.

If you really want to skew both, though – opt for a half-slot skew in one direction in the rotor, and a half-slot skew in the opposite direction for the stator. Note that this means there is only ONE way to assemble rotor and stator together – with the skews opposing. (With the full slot skew on either rotor or stator and an unskewed opposite piece, the rotor can be inserted from either end of the stator with the same effect.)

AQ: Sensorless motor control with TI and Microchip

Question:
I need to learn about the sensorless control of permanent magnet AC (PMAC) motors. Can you recommend a tutorial and/or open source code for the sensorless motor control using the
a) TI TMS320 series processor, or
b) Microchip dsPIC33EP128 series processor?

Answer:
I have used Microchip and TMS320 to develop VFD. They provide you with a demo kit, PCB and a motor. It take me half a day to get the demo PCB running with my PMSM. Then I copy their design to my own.

The Microchip solution provides you with demo code. I used that before, but it require quite a bit of C programming, and motor tuning take even longer. The demo code and application note are no where near the performance of the Ti solution (I do not work for Ti -so I am not advertising). I take me a week to get my motor spinning with the demo kit from Microchip.

Then there are the International Rectifier solution that is available from many years. The IR sensorless motion control solution have implemented a FOC motor control in FPGA. So you don’t need to write code for motor control. In the chip, it also has a 8051 cpu. You write the program in C; 1 page of code will get a washing machine working. It takes me 1 day to get a PMSM motor running with this solution.

I will use the TI solution for high end motor control – such as a US$40,000 dollar, 100HP direct drive PCP used in the oil field.
I will use the IR solution for a water pump, washing machine – things that is a few kw.
I will use the microchip for solution for toys, because Microchip is so much fun to play with.

AQ: Die-cast rotor design

The method of creating a die-cast rotor is as follows:

1. An assembly of steel laminations (which may or may not be grain-oriented) containing the openings for both rotor bars and ventilation (as required) is made and clamped together to form a cylindrical iron core.
2. The assembly is inserted into a mold, which has space both above and below the core for the end (shorting) ring assembly.
3. The molten conductor material (aluminum or copper, usually) is injected into the mold and allowed to flow through the bar openings. It also fills the end ring spaces.
4. The entire assembly is allowed to cool so that the conductor solidifies.
5. The “cast” core is then shrunk onto a steel shaft.

Now we have a “cast” rotor assembly, ready for bearings and mounting into machine.

AQ: ACS800-104-0105-3 (ABB VFD Drives)

Question:
I have a problem with ABB ACS800-104-0105-3 drive model, the output current reading on the VFD is always double the reading of the clamp ampere(i.e. drive reading= 40 A, clamp ampere reading=20 A), what is the procedure that i can follow to detect the cause of this error?

Answer:
I don’t know about ABB drives, but hope this thing will help you.
1. The variable frequency drive may have problem with current sensor, just replace with another drive for comparison.
2. Make sure you use, true RMS type clamp meter.
3. If there is leakage current (through cable insulation and air) between each phase. This normally because of the cable insulation already degraded. Add output reactor and replace the cable with suitable insulation can fix this kind of problem.
4. If there is leakage current between this VFD drive and the other drives, that both motor cable is quiet long and run in parallel together.

To Collect more data and get more idea, you can do this:
1. Clamp all the 3 phase motor cable together using clamp. The reading will show you the leakage current. Normally about 10% of motor rated current at full load.
2. Check the current on each phase, and see if the current is balance for each phase.
3. Run the variable frequency drive without the motor cable, check the current reading and clamp meter.
4. Run the AC drive with the motor cable but without the motor, check again the reading and clamp meter.
5. Run the drive with motor, check if any oscillation in motor current.
6. Check current input to the AC drive inverter.
7. Turn of the other drive (if the motor cable run parallel together with other VFDs), and see if any change in current.

AQ: 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

AQ: Motor line starting and ramp starting with VFD

Variable frequency drives are important power electronic devices. When we start an electric motor, we are increasing from 0 speed to full operating speed. A VFD ensures that the motor accelerates (increases its speed) to its full speed in a smooth manner, without causing much irregularities. In other words, VFDs make the motor accelerate uniformly.
VFDs are also easy to install and use. VFD drives are not only for starting motors (like the normal starters), but for easy speed control as well.

The difference between line starting a motor and ramp starting the motor with a variable frequency drive is that the motor/load does not pull the 6-7 times rated current of the motor, because the motor winding are not saturated with the full EMF produced to get the motor to synchronous speed it is ramped to it. If you are not trying to control the motors speed from process control then a soft start will serve the same purpose. The VFD drive main purpose is to control the V/F of the motor.

You will have to adjust the ramp time on the VFD or soft starter to over the force required to turn whatever the motor is turning, this can be accomplished with both devices. Soft starter is less expensive than variable frequency drive, thus it has limitations.