ABB UFC719AE01 3BHB003041R0001 3BHB000272R0001 PC board DCS control system

Description

UFC719AE01 sets the current flowing through the coil as 0.55A, and the windings in the coil are uniformly wound on the iron core with 200 turns. We can see that the section of the iron core is circular, its outer ring radius Rw=25mm, and the inner ring radius Rn=20mm, so the diameter of the iron core section is 25-20=5mm. Then we set the picture of the relative permeability of the iron core.
Step 1: Let’s find out the magnetic induction intensity B at the outer ring and inner ring of the iron core and the magnetic flux picture in the iron core when the air gap is not opened.

We know that UFC719AE01 can be seen from the magnetic field distribution of the close wound current carrying spiral ring and Ampere’s loop theorem that the magnetic induction intensity of the close wound current carrying spiral ring is:
But we can see from the picture that there is an iron core. Therefore, the permeability also changes, and the permeability in the expression should be multiplied by the relative permeability of the iron core. Now we can get the value of magnetic induction intensity:
The radius of the outer ring of the iron core is 25, so the magnetic induction intensity Bw at the outer ring is:
The radius of the inner ring of the iron core is 20, so the magnetic induction intensity Bn at the inner ring is:
What about the middle of the iron core? Of course, it is the average of the two, namely:
Now let’s find the magnetic flux. Since the magnetic field in the ring section is uniform, the average magnetic flux is calculated based on the magnetic induction B value in the middle of the ring. The calculation formula is: picture. S here is the area of the core section. So there are:
Step 2: Let’s find out the magnetic induction strength and magnetic flux after the iron core is grooved.
It can be seen from the figure that the width of the slot, that is, the width of the air gap, is 2.5mm.
Now, there is an important law to emerge, which is the second Kirchhoff law of magnetic circuit. The law is as follows:
Note that the left side of this equation is called magnetomotive force, also called excitation magnetomotive force. It is equal to the product of current and winding number N; The right side of the formula is the algebraic sum of the magnetic voltage drop of each section in the magnetic circuit.
At the same time, we also know that the magnetic voltage drop is equal to the product of magnetic flux and magnetoresistance, namely:
there Φ Is magnetic flux, and the picture is magnetoresistance. Then the expression of Kirchhoff’s second law of magnetic circuit becomes:
With the calculation basis, UFC719AE01, let’s consider the following problems:
Before the iron core ring is grooved, according to Kirchhoff’s second law, we have:
We call this equation Eq. 1.
Now the iron core is grooved. We know that the magnetic circuit is divided into two sections. One is the air gap magnetic circuit, and the other is the iron core magnetic circuit. The two magnetic circuits are connected in series. According to Kirchhoff’s second law, we have:
We call this equation Eq. 2.
In Equation 2, the first item on the right is the magnetic flux and magnetic resistance of the air gap magnetic circuit, and the second item is the magnetic flux and magnetic resistance of the iron core.
Since the two magnetic circuits are connected in series, the magnetic flux is of course equal to the picture. So equation 2 is changed to:
We call this equation Eq. 3.
Now, we divide the left and right sides of equation 3 by the left and right sides of equation 1 at the same time to obtain the ratio of magnetic flux before and after opening the air gap:
Looking at the coil diagram, we can find that the width of the air gap is almost negligible for the circumference of the iron core ring, so there are:
By substituting the above formula, we get:
We call this formula Eq. 4
So what is the magnetic resistance Rj of the iron core ring equal to? Its value is related to the sectional area S of the iron core ring, the perimeter L of the center line of the iron core ring, and the magnetic permeability of the iron core ring. It is noted that the radius of the outer ring of the iron core Rw=25 and the radius of the inner ring Rn=20, so the radius of the centerline is (Rw+Rn)/2. Let’s look at the expression:
We substitute these two parameters into Equation 4, simplify and bring in specific values to get:
Now, we finally get the result: when the air gap is opened in the iron core ring, the magnetic flux is only 2.75% compared with the previous magnetic flux.
If we want to keep the magnetic flux of the iron core ring after opening the air gap unchanged compared with the original, it can be seen from the picture that the current I and the air gap magnetic flux Φ Is proportional, so the current needs to be increased. The multiple of current increase is equal to the multiple of magnetic flux decrease.
Now, let’s answer and calculate the last question: if you want to keep the magnetic flux unchanged, how many times must the current I after opening the air gap be the current picture of the iron core ring before opening the air gap?
That is to say, for our example, after opening the air gap, the current must be increased to 36.4 times of the original, so as to keep the magnetic flux in the iron core ring basically unchanged.
Above, we take the annular iron core as an example to illustrate the problem.
For relays and contactors, their iron cores are generally U-shaped or E-shaped. The magnetic flux (i.e. air gap flux) before the pull in is only about 4% of the magnetic flux (i.e. pure ferromagnetic flux) after the pull in. After taking the reciprocal, the ratio of pull in current to maintenance current is about 25 times.
We can see from the ABB contactor parameters given above. If the voltage before and after the contactor is pulled in remains unchanged, it can be seen from the table that the coil power consumption is 50VA when the contactor is pulled in and 2.2VA when the contactor is held in, the ratio of the two is about 22.7 times. This is also the ratio of pickup current and holding current of this type of contactor coil

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