Author: ABBdriveX

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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: 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: 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: Motor output torque in rotation speed (frequency) changes

Frequency power: power supplied by the power grid (commercial power).
Start-up current: frequency inverter output current in motor starts.

The motor starting torque and maximum torque by frequency inverter driving is less than direct frequency power driving.
Motor accelerates in constant frequency power supply has high impact, which can be reduced by using frequency inverter. Because there is a big starting current in motor acceleration if it’s powered by constant frequency power supply; when using frequency inverter, the inverter output voltage and frequency is increased gradually, so the motor starting current and impact is much lower.

Generally, the motor torque is decreased with frequency decreases (speed reduction). By using vector control frequency inverter, to improve output torque during motor running in low speed, and even output sufficient torque at motor low speed zones.

AQ: Output torque of variable speed drive running above 50Hz

Generally, electric motors are designed according to 50Hz power supply, its rated torque also in this frequency. Therefore, the speed adjustment under rated frequency called constant torque speed adjustment. (T = Te, P <= Pe).

If the variable speed drive outputs frequency exceeds 50Hz, the motor torque is inversely proportional to the frequency in linear relationship decrease.
When the motor running in above 50Hz frequency, we should consider the motor loads to avoid motor lacks of torque.

For example, the motor torque is about a half in 100Hz running against 50Hz. Therefore, the speed adjustment in above rated frequency called constant power speed adjustment. (P = Ue * Ie).

As we know, for a specified motor, the rated voltage and rated current is constant.

For example, the variable speed drive and motor rated values are: 15kW/380V/30A, motors can operate at 50Hz or above.
When the frequency is 50Hz, the variable speed drive output voltage is 380V, current is 30A. Then if we increase the output frequency to 60Hz, the variable speed drive maximum voltage and current also is 380V/30A, it is obviously that the output power is fixed, so it called constant power speed adjustment, what’s the torque status now?

Since P = wT (w: angular speed, T: torque), as P keeps same, w increases, so the torque will decrease accordingly.

From another point: motor stator voltage U = E + I * R (I is the current, R is the electrical resistance, E is the EMF)
Then we can see, U and I are constant, E is constant.
And E = k * f * X, (k: constant, f: frequency, X: flux), when f changes from 50 to 60Hz, X will decrease accordingly.

For the motor, T = K * I * X, (K: constant, I: current, X: flux), so the torque T will decrease along with the flux X.

And, if the frequency is less than 50Hz, as I * R is very small, so if the U/f = E/f is constant, the magnetic flux (X) is constant, the torque is proportional to the current, which is why use the variable speed drive overcurrent capability to describe its overload (torque) capability, and known as constant torque speed adjustment (rated current is constant -> Maximum torque is constant).

Conclusion: When the variable speed drive outputs frequency increases from 50Hz, the motor outputs torque will decrease.

AQ: Which factors will affect VFD output torque?

Heating and cooling capacity to determine the variable frequency drive output current capability, thus affect its output torque capability.

Carrier Frequency: generally the variable frequency drive rated current is the continuous output value under the highest carrier frequency, the maximum ambient temperature. Reduce carrier frequency won’t affect the motor current, but will reduce electronic devices heating.

Ambient temperature: like will not increase VFD drive protection current when detect relative low ambient temperature.

Altitude: altitude increases will affect both heating and insulating property of the variable frequency drive. Generally it’s fine in below 1000m, and derate 5% per 1000meters for above.

AQ: Synchronous generators inter-turn faults

For the MW range of Synchronous generators, there is no terminology of “interturn fault” on the stator winding. There could only be coil to coil fault on the stator for such size of machine design.

There are possibilities of having inter-turn faults on the rotor winding: when the insulation positioned between adjacent conductors break (electrically) over time under certain mechanisms. These mechanisms can include; turn to turn movement caused by thermal expansions (during starts/stops cycles), rotor coil shortening, end strap elongation, inadequate end-turn blocking or conductive bridging formed by contamination. The protection of avoiding the interturn insulation is a function of how well the machine is designed, maintained and operated. The OEM of the generator usually provides recommendations to avoid any inter-turn fault during the lifecycle of the machine. Saying this, there are ways to monitor the interturn fault indication; such as data acquisition (air gap flux probe, air gap search coil), as supportive monitoring (RSO, Shaft voltage, shaft vibration levels, excitation current etc.). Ideally, you have to be knowledgeable with the machine design to interpret the acquired data to make valuable predictions.

If you start by contemplating what kind of symptoms inter-turn faults could give rise to, you will be part of the way.
While machine is at standstill, you could do some reflected-wave analysis. All phases should show (near) identical responses.
During operation, you could have non-identical current and voltage waveforms on the three phases (you must compensate for unequal load).
You may experience strange sounds, in the supersonic range. Changing for different locations around the stator. You can continue the list, and settle on systems that may be able to detect any anomalies, so you can react accordingly.

AQ: How is Vector Control improving motor output torque capability?

1: Torque boost: this function is the variable speed drive increases output voltage (mainly in low frequency) to compensate the torque loss due to voltage drop in the stator resistance, thereby improving the motor output torque.

2: Improve the motor insufficient output torque in low speed
“Vector control” can make the motor output torque at low speeds, such as (without speed sensor) 1Hz (for 4-pole motor, the speed is about 30r/min), same as the torque output at 50Hz power supply (maximum is approx 150% of rated torque).

For the V/F control variable speed drive, the motor voltage increases relatively as the motor speed decreases, which will result in lack of excitation, and make the motor can not get sufficient rotational force. To compensate this deficiency, the variable speed drive needs to raise voltage to compensate for the voltage drop in motor speed decreases. This feature called “torque boost”.

Torque boost function is to improve the variable speed drive output voltage. However, even if the drive increases voltage, the motor torque and current does not increase corresponding. Because the motor includes the torque and other components (such as the excitation) which generated by the motor.

“Vector Control” allocates the motor current value to determine the motor torque current component and other current component (such as the excitation component) values.

AQ: 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.

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