ИЭ / 6 семестр (англ) / Курсач / Курсач

Peter the Great St. Petersburg Polytechnic University

 Institute of Energy

Higher schools of High Voltage Energy

Report

on the course project

The magnetic field modeling

Student:

gr. 3231302/90201

Supervisor:

Alexander Kalimov

Saint-Petersburg

2022

Ferromagnetic cylinder in the field induced by a two-wire line.

Variant 76

Ferromagnetic cylinder with a constant magnetic permeability is immersed in the magnetic field induced by a transmission line. The line consists of two identical wires with the diameter of D.

Analysis of this system should be performed:

1. by the method of images;

2. by the method of variable separation;

3. by numerical calculations (Q–FIELD)

a = 125 cm

b = 35 cm

R = 50 cm

D = 1.6 cm – diameter of the wire.

I = 100 A

μ_r = 600

To find:

1. Inductance of the line per meter;

2. Inductance of the line per meter without a cylinder;

3. Force applied to each wire per meter;

4. Dependence H(x), y=0 (plot).

1. Analysis of the system by the method of images

We replace a cylinder by two wires in special position. This new problem will have the identical boundary conditions with our main problem. That’s why solution of this problem is identical with solution of our problem.

Firstly, we need to find positions of inverse points by next relations:

1) Point B:

2) Point B’:

Next, we shall determine the length of the segments:

For convenience, we will introduce a new designation.

1.1. Inductance of the line per meter

Initially we find the internal inductance of the 2 wires:

After that, define the magnetic flux per meter (dS=1·dx) produced by each point (b>>0.5D):

Similarly for wires inside the cylinder, but whose current is equal to

Divide the magnetic flux on the current and find the external inductance:

Sum up the internal and external inductance per meter:

1.2. Inductance of the line per meter without a cylinder

The external flux of a two-wire transmission line is the same as the sum of the previously calculated Φ_A and Φ_A':

Now we can find an inductance:

1.3. Force applied to each wire per meter

Calculating for the force relative to the point A’.

The force applied to wire A' equals to the sum of three ampere forces acting from a conductor with current and 2 images ( ):

After summation we get:

Calculating for the force relative to the point A.

After summation we get:

1.4. Dependence H(x), y=0

By H(x) we mean the projection of the magnetic field intensity on the y axis.

We will find magnetic field strength by summing up the magnetic field strengths in different spaces, inside the cylinder and outside.

Values of currents flowing in conductors and their images:

A':

A:

B:

B':

Inside the cylinder, we take the currents B and B’ equal to 0.

Magnetic field strength outside the wire:

Magnetic field strength inside the wire:

The magnetic field strength induced by the wire:

– induced by current A';

– induced by current A;

– induced by current B;

– induced by current B'.

Further expressions are made considering the x sign.

To the left of the cylinder (-∞ < x ≤ -R):

Inside the cylinder (-R < x ≤ R):

Between the cylinder and left wire (R < x ≤ (X_A-0.5D)):

Inside of left wire ((X_A-0.5D) < x ≤ (X_A+0.5D)):

Between wires ((X_A+0.5D) < x ≤ (X_A'-0.5D)):

Inside of right wire ((X_A'-0.5D) < x ≤ (X_A'+0.5D)):

To the right of wires ((X_A'+0.5D) < x <+∞)

.

The figure shows that the magnetic field intensity near the cylinder is continuous, but the shape of the curve changes at the boundary, what is not visible in Figure 4.

2. Analysis of the system by the method of the variable separation

2.1. Inductance of the line per meter

Total inductance of the two-wire transmission line per unit length:

where ∆𝐿 is the additional inductance created by the cylinder and 𝐿_{2𝑤𝑖𝑟𝑒} is the inductance of 2-wire line without a cylinder.

Initially we found the inductance of the wires without a cylinder, that is why we must find additional inductance for the two-wire current line:

L_2wire we have already found in paragraph 1.2.

2.3. Force applied to each wire per meter

We will calculate the force acting on a conductor with a current as the product of the current and the magnetic flux density in the center of the wire. To do this, we need to calculate the intensity of the magnetic field in a certain place. Let's use the formula for this, which will be explained in detail in paragraph 2.4.

2.4. Dependence H(x), y=0

We have the same areas. But for convenience, first we calculate the strength of the magnetic field induced by each wire and the image separately, considering the coordinate.

In method of the variable separation, we assume that the intensity of the magnetic field inside the cylinder induced by the wires has the following formula:

The intensity of the magnetic field outside the cylinder induced by the images has the following formula:

Further expressions are made considering the x sign.

And the total magnetic field intensity is:

As can be seen from Figures 8 and 9, the inductance values and the intensity graphs completely coincided when solving the problem by different methods.

3. Analysis of the system by numerical calculation (Q–FIELD)

Due to symmetry, we applied a half of our current in Q–FIELD. Because of it we must double forces and halve value of inductance.

With Q–Field we can visualize lines of magnetic field intensity.

3.1. Inductance of the line per meter

3.2. Inductance of the line per meter without a cylinder

3.3. Force applied to each wire per meter

Due to symmetry, we applied a half of our current in Q–FIELD. Because of it we must double our results.

3.4. Dependence H(x), y=0

The conclusion

Table 1. Results

	Inductance,		Force applied to each wire per meter,
	with cylinder	without cylinder	left wire	right wire
Methods of images	1.613	1.611	-5.790	5.758
Variable separations	1.613	-	-5.790	5.757
Q–FIELD	1.613	1.611	-5.789	5.750
Maximum difference	0%	0%	0.02%	0.14%

Results which were obtained during the work are correct. Having in account a fact that the first method gives us correct results, we analyzed divergence of other methods. Numerical values of inductance and force are similar with small error is about 0.14 %. Moreover, the induction value and the graphs obtained by the method of variable separation are absolutely identical to the results obtained by the method of images.

Main plot built by different methods in MATLAB also has the same appearance. But in the domain near the cylinder, we observe visible difference between curves. Methods of image and method of variable separation give us approximately the same result. But Q–FIELD have a substantial otherness. The reason of this is the numerical method which is used in Q–FIELD. That’s why we have big error near and on the bounder of the cylinder.

The Q–FIELD program has shown high accuracy of calculations and fast calculation time (about 22 seconds), so the program is well suited for solving tasks such as drawing a picture of the field, calculating various quantities using an integral calculator. Despite this, when constructing various curves (in our case, the curves of the dependence of the magnetic field intensity on the coordinate) differ from the theoretical ones (Figures 19 and 20), especially near the cylinder. This is due to the inaccuracy of the calculation and the choice of an insufficiently small step.