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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: Soft starter protection features

1) Overload protection: the soft starter has current control loop to track and detect of the changes of the electric motor current. Achieve overload protection by increasing overload current settings and inverse time control mode, to cut down the thyristor and send alarm signals when motor is overload.
2) Phase loss protection: soft starter detects changes in the three-phase line current all the time, to make phase loss protection response once the current off.
3) Overheating protection: the soft starter detects the thyristors internal radiator’s temperature by its thermal relay, automatic cut down and send alarm signal once the radiator’s temperature exceeds the allowable value.
4) Other features: achieve lots of mixed protection functions by combination of the electronic circuits.

AQ: Frequency inverter failure analysis

Transistor frequency inverter has the following disadvantages: easy trip, difficult re-start, poor overload capacity. As the rapid development of IGBT and CPU, the inverter drive integrates perfect self-diagnosis and fault prevention features, improve the reliability greatly.

Vector control frequency inverter has “automatic torque compensation function” to overcome “starting torque inadequate” etc. This function is the inverter uses a high-speed microcomputer to calculate the torque required at current time, to modify and compensate the output voltage quickly to offset the frequency inverter output torque changed by external conditions.

In addition, because as the inverter software development more and more perfect, we can pre-set various failures parameters in the frequency inverter, to ensure continuous running after failure resolved. For example, re-start motor in free parking process; automatic reset internal failures and maintain continuous operation; adjust running curve if load torque is too high to detect the mechanical system abnormal.

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: What is the soft stop of an electric motor?

In electric motor stop, the traditional control ways are accomplished by momentary power cutting off. But in lots of applications, it’s not allowed the motor instant shutdown. For example: high-rise buildings, building’s water pump system, it will appear huge water hammer during instant shutdown, to damage the pipe, even the pumps. To reduce and avoid “water hammer” phenomenon, the pumps motor need be shut down gradually, that is soft stop. The soft starter can meet such requirements. In pumping station, soft stop technology can avoid the pump door damaged of the pumping station, to reduce maintenance costs and maintenance works. The soft stop function in soft starter is, when the thyristor gets stop instruction, decrease conduction angle gradually from full conduction, and achieve full closed after a certain time. Stopping time can be adjusted according to actual requirements within 0 – 120s.

AQ: Avoid variable frequency drive damaged in lightning

Sometimes

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: Soft starter energy saving principle

Induction motor is inductive load, the current lags the voltage, most electrical appliances are the same. In order to improve the power factor we need to use capacitive load for compensation, parallel capacitors or with synchronous motor for compensation. Reduce motor excitation current also can improve the power factor (HPS2 saving function, reduce excitation current by reducing voltage at light loads, to increase COS∮). Energy-saving operation mode: decrease voltage in light loads to reduce excitation current, the motor current divides into the active component and reactive component (excitation component), to increased COS∮.

Energy saving operation mode: when the motor load is light, the soft starter working at energy-saving conditions, PF switch to Y position, under the current feedback action, the soft starter reduces the motor voltage automatically, to reduce excitation component of the motor current. Thereby improving the power factor of the motor (COS∮). If the contactor in bypass state, this feature cannot works. TPF switch provides energy saving features with two reaction times: normal speed and slow speed. The soft starter operation in energy saving state automatic (In normal and slow speed), saving 40% energy in no-load and 5% with load.

AQ: Motor power cable – bigger or smaller?

When a choosing a power cable for a motor, we prefer using one larger diameter cable than two smaller diameter cables in parallel, although it would cost less to do so. Why?

1. Conductors/Cables/Feeders in parallel connection generally are not recommended unless there is no option, therefore it can be adopted under the following conditions:
i. Cables are of the same material and cross section area.
ii. Are of the same route and length.
iii. The sum of the current carring capacity of the parallel circuits after applying all necessary applicable correction factors should be greater than the nominal regulated current of the protective device.
iv. The current carrying capacity (before derating) shall be not less than 300A (according to the local authority/Service provider requirement/regulation).
v. Capability of addressing the Thermal & electrodynamics constraints in proper way.

2. Some designs call for parallel connection so as to:
i. Overcome the voltage drop.
ii. Avoid the difficulties of installing big size cables (bending, pulling) due corridor limitation,etc.
iii. Meet the Power demand.
iv. Mitigate the cost (Costwise).

3. For electrical Motors, two connections are normally required. One from MDB to Motor CP and other from CP to the Motor.
By virtue of the requirement of Delta/star starter, two cables are required (Mandatory) between CP & motor (one will be dead just after changing to delta connection).
While the connection from the MDB to CP will be one, sized according to the Motor rating.

However, Parallel connection of Feeders need an expert engineer(s) to meet the requirement since Short Circuit fault protection for parallel circuits require further evaluation from the Engineer that the impact of the short circuit current within the parallel section will have severe fault due to fault current path that can occur in addition subtransient contribution of the downstream system.

AQ: 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?