• Fuji Electric Model DCR2-2.2 DC Reactor 11A 3mH Brand New System 1
  • Fuji Electric Model DCR2-2.2 DC Reactor 11A 3mH Brand New System 2
  • Fuji Electric Model DCR2-2.2 DC Reactor 11A 3mH Brand New System 3
  • Fuji Electric Model DCR2-2.2 DC Reactor 11A 3mH Brand New System 4
  • Fuji Electric Model DCR2-2.2 DC Reactor 11A 3mH Brand New System 5
Fuji Electric Model DCR2-2.2 DC Reactor 11A 3mH Brand New

Fuji Electric Model DCR2-2.2 DC Reactor 11A 3mH Brand New

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1 kg
Supply Capability:
1620 kg/month

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DC reactor, also known as smoothing reactor, is mainly used on the DC side of
converter and is widely used in general frequency converter. There is a DC
current with AC component flowing through the reactor. The main purpose is to
limit the AC component superimposed on the DC current to a specified value,
keep the rectifier current continuous, reduce the current ripple value, improve
the input power factor, and suppress the harmonic generated by the converter.
effect
DC reactors are connected in series on each pole of the converter station. The
inductance is about 0.4 ~ 1.0h. Its main functions are as follows:


(1) Prevent inverter commutation failure.


(2) Reduce voltage and current harmonics in DC lines.


(3) Reduce ripple coefficient.


(4) Limit the current in the rectifier when the line is short circuited.


However, it should be noted that the value of inductance must ensure that the
DC circuit does not resonate at power frequency.


DC reactor can increase the power factor to more than 0.9. Because of its small
size, many frequency converters have installed DC reactors directly in the
frequency converter. In addition to improving the power factor, the DC reactor
can also weaken the impact at the moment when the power supply is just
connected. If AC reactor and DC reactor are equipped at the same time, the
power factor of variable frequency speed regulation system can be increased to
more than 0.95.
Structural design
The DC smoothing reactor is used for the filter circuit at the output side of
the rectifier circuit. Because there is a DC current component, the magnetic
flux of the iron core works in the first quadrant, resulting in magnetic bias.
In order to increase the inductance and prevent magnetic flux saturation, DC
smoothing reactors adopt cut iron core and add air gap to lengthen their
hysteresis loop laterally, which is a bit like the case of single ended flyback
transformer in principle. In this way, when there is a current of the same
size, B is small and not easy to be saturated.


Four factors should be considered in the structural design of open core
reactor:


1) The DC component I of the current flowing through the reactor, that is, the
average value of the inductive current IL, determines the cross-sectional area
of the winding conductor and the heating of the winding.


2) The peak to peak value of AC current component is expressed by △ IPP or
abbreviated as △ I, which is determined by the continuity requirements of
inductive current. Generally, without affecting the output voltage ripple,
twice the minimum load current ilmin output by the rectifier filter circuit is
the peak to peak of the required filter inductance current change. Thus: △ IPP
= 2ilmin


3) The inductance L required by the filter circuit is determined by the
frequency of AC current component and peak to peak △ IPP. It is an
intermediate calculated value in the process of reactor structure design,
rather than the primary factor to be considered in advance.


4) Power loss and temperature rise are factors that need to be considered in
advance in the process of reactor structure design.


There are more than one method to design DC smoothing reactor, and these
methods are unified with each other. A simple, clear and practical engineering
design method is introduced here. Its steps are as follows:


(1) Calculated energy storage


We know that if the inductance of the reactor is l and the current is I, the
stored energy is


W=1/2LI


Because the highest flux density BM is determined by the maximum current in the
reactor, I in the formula should be calculated according to the maximum
current. If the DC current is I, the peak to peak value of AC current is △
IPP.


(2) Select the material of iron core

According to the proportion of AC component △ I in I, the frequency of AC
component and the index requirements of inverter on its output noise, the
material of reactor core can be selected. After the material is selected, the
maximum flux density BM and the effective coefficient K of iron core section
can be determined. Note that when selecting BM, it is generally required to
leave a large margin, because the reactor is not saturated after half an
inverter cycle of output short circuit (i.e. a normal rectifier cycle), so as
to make the protection circuit act, so as to improve the working reliability of
the inverter circuit.


(3) Select the shape of the iron core and determine the window utilization
coefficient KC


According to the material, winding process and installation structure of the
selected iron core, the shape of the iron core is selected, and then the type
of conductor is selected according to the current level to determine the window
utilization coefficient KC.


(4) Select conductor current density coefficient kJ and current density J


According to the heat dissipation conditions, the required copper consumption,
the shape of the iron core and the allowable temperature rise of the wire
package, select the appropriate current density coefficient kJ (refer to table
4.2) and the current density J of the conductor.


(5) Calculate the cross-sectional area of the wire and select the wire


The cross-sectional area of the conductor is calculated by the following
formula: S = I / J; Then select the shape and size of the wire.


(6) Calculate the number of winding turns


When calculating the number of turns of reactor winding, we should consider
making full use of the window area of iron core, so as to maximize the number
of turns and leave a certain margin for BM. N=A·Kc/S

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mold can make people very sick or die. also moisture could iodize electrical equipment.
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sounds like you have a problem with your outlet.wires are fed from the source Breaker Boxto a junction box where it branches offyoull have an outlet that should have 4 or more wires connectedthis is because the wiring goes into the outlet on one side and out on the other to feed another recepticle, light, switch etcif half of your outlet is not working, the contact plate inside brokeit may arc and start a fire. because the plate is broken, there is no output to the other devices connected.replace that outlet and check any nearby junction boxesmake sure wire caps are tight. to replace an outlet will costs you around 80 bucks for an electrician, or about 23 cents to do it yourself if you learn about wiring.
Q: If this is maintained, but voltage changed from 480V to 240V, by what ratio will supply line losses change?
Power voltage x current Therefore current power/voltage. Thus, if power is constant at 2000kW current at 480 volts will be: I 2000,000/480 4166.67amps At 240 volts: I 2000,000/240 8333.33 amps However, line losses are current (squared) x resistance Thus the ratio will be 4166.67 (squared) / 8333.33 (squared), resistance remaining constant of course. I make that 0.25 (or 4, if you are thinking that way up!) That help?
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Master Electrician
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I would be very pissed, you should go complain to the people
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Are you in danger only if the water comes in contact with electrical wires yes have the pipe repaired call a plumber don't wait til Monday when it will be cheaper because it won't when the ceiling collapses some one may get hit with plaster board and the. You wind up with an even bigger hole . The longer you wait the more it will cost. Especially if a fire were to start remember water and electricity don't play well together some one can be electricuted
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google long distance movers you will find many suggest that you get an estimate from several as prices can vary greatly. you will need to estimate how much you need to move. --
Q: Context:Unrated electrical equipment in the dust handling equipment and area could act as an ignition source, which would result in a fire or an explosion for the dust handling equipment or rooms.
Yes

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