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2017-07-17
skimmed：3600

With
the rapid development of power technology, AC motor frequency control
technology has made a breakthrough, into the popular application stage. In China, the frequency converter is more and more widely used, with
the same place, how to correctly selected, with the user has become a
very prominent problem.

1. About capacity selection

In the manual of the frequency converter, in order to help the user
select the capacity, there are "with the motor capacity" column,
however, the meaning of this column is not exact enough, often lead to
misuse of the inverter.

In a variety of production machinery, the capacity of the motor is selected according to the principle of heat. That
is, in the premise of the motor with the move, as long as the
temperature rise within the allowable range, a short period of time is
allowed to overload. The overload capacity of the motor is generally set at 1.8-2.2 times the rated torque. The temperature rise of the motor, the so-called "short time" at least ten minutes or more. The frequency converter speed overload capacity: 150%, l minutes. This indicator, for the motor, only in the start process is
meaningful, in the course of running, is actually not allowed to carry.

Therefore,
the "matching motor capacity" column of the exact meaning is "equipped
with the actual maximum capacity of the motor." When the inverter is actually selected, the maximum current of the
motor can be selected according to the maximum current of the motor. For
the blower and pump load, it can be directly selected according to the
"motor capacity" for long term constant load.

2. Drive system for optimal design

AC induction motor by frequency control, the effective torque and effective power range. With the frequency converter, must be based on the mechanical
properties of production machinery and the speed range requirements and
other factors, the transmission system for excellent design, optimize
the design of the main content and the general method is as follows:

2.1 determine the maximum operating frequency of the motor

(1)
Blower and pump load, the resistance torque TL of this type is
proportional to the square of the speed n TL = KTn2, the output power PL
is proportional to the square of the speed PL = KPn3, (KT and KP are
constant) , It can be seen, if the speed exceeds the rated speed, the load
torque and power will be increased by the square law and the law,
therefore, in general, is not allowed to run above the rated frequency.

(2)
Under normal circumstances, a variety of mechanical strength, vibration
and wear resistance, are based on the motor speed does not exceed 3000r
/ min for the premise of the design. Therefore, in the absence of mechanical re-design of the case, the
maximum speed of the two motors do not exceed the rated frequency too
much.

(3)
When the asynchronous motor is running above the rated frequency, the
electromagnetic torque Tx is almost inversely proportional to the square
of the frequency adjustment ratio Kf, i.e., T≈TN / Kf2 (and TN), when
the supply voltage is constant Torque at rated frequency fN). Therefore, the maximum operating frequency should not exceed the rated frequency.

(4)
When the asynchronous motor is running at low frequencies, torque
compensation is often required in order to obtain sufficient torque. The torque compensation will make the motor magnetic circuit tends to
saturation, thereby increasing the additional loss, reducing efficiency,
so, as long as the situation permits, should be ruled to increase the
upper limit of operating frequency.

2.2 Determine the transmission ratio of the drive system and check the capacity of the motor

(1) blower and pump load, are generally direct drive, do not have to consider the transmission ratio of the problem.

(2) constant torque load, first, according to the effective torque
line and the required frequency adjustment range, determine the motor
running the highest frequency and minimum frequency.

Assuming
that the maximum operating frequency of the motor has been determined
as fmax, the minimum operating frequency is fmin and the corresponding
torque relative to tTL is the rated torque Tn = TL / qTL (TL load
torque) of the motor. If
the original motor does not leave the margin, then equipped with
frequency converter, the motor capacity should be expanded 1 / tTL
times. The transmission ratio of the drive system is equal to the ratio of
the maximum speed nLmax required by the motor at the maximum operating
frequency at nDmax.

(3) constant power load: and constant torque load similar to the first
according to the effective power line and frequency adjustment range,
calculate the motor operating frequency of the upper and lower limits.

Similarly, when the highest and lowest operating frequencies are
obtained, the corresponding power relative value tPL is obtained, and
the rated power of the motor PN? PL / tPL (PL is the load required
power).

In
the design of constant power load, should pay attention to two points:
(1) as much as possible to use the rated frequency above the part; (2)
when the adjustment range is large, try to use two gear ratio. Because when the drive ratio is divided into two columns, the frequency range αf and αn speed range between the relationship. It can be seen that in the case of the same rotational speed range,
the frequency range will be greatly reduced, so that the capacity of the
motor can be reduced.

The
mechanical characteristics of the load, because the constant power
load, so the curve of any point of the abscissa and the ordinate of the
product are equal, and the load power is proportional to, that PL =
KPTLnL = KPTLmaxLmin. All
the speed in the rated frequency below the effective torque line
adjustment, in this case, the required capacity of the motor PN =
KPTNnLmax "KPTLmaxLmax = αnPL. This shows that the required motor capacity than the load power of the On times even larger, is very uneconomical.

⑴
when the maximum operating frequency of 2 times the rated frequency,
the transmission ratio is only one file when the situation. In this case, the capacity of the required motor PN = KPTN1 / 2nLmax 1 / 2αnPL. Visible, the required capacity as long as the load power of the On / 2 times on it.

⑵
when the maximum operating frequency of the rated frequency of 2 times
the transmission ratio of two cases when the situation. At this time, the required capacity of the motor PN1 / 2 PL. It can be seen that for a constant power load, this solution is ideal when αn "4".

3. Self-equipped external braking resistor

A variety of frequency converter are allowed to external braking resistor, speed up the braking speed, external resistance. But the matching brake resistor is expensive, easy to buy, automatic
configuration, the resistance and power can be determined as follows:

The voltage value of the DC circuit is the maximum value of the rated
current IDN of the motor, that is, Is ≤ IDN, so the braking resistor Rs
≥UD / Is.

Because the current through the Rs only a few seconds, so the power PR
can work according to their work (1 / 10-1 / 8) choice, that PR =
(0.1-0.125) UD2 / Rs.

When Rs access circuit, should pay attention to the inverter within
the drive resistor cut off, if not removed, it should be appropriate to
increase the value of Rs, in order to avoid excessive braking current
situation.

In
the case of an external braking circuit, it is sometimes possible to
connect the entire brake electronics (ie, the braking resistor and the
discharge transistor) in order to avoid burning the high-power
transistor (GTR) in the inverter. In this case, the GTR should select
VCEX ≥ 700 Volts; ICN ≥ (1.2-1.5) IDN.