T(fp chi khoa hoc Tn6tng Bai hgc Quy NhOn - So 2, Tap IX nam 2015
CALCULATION OF ELECTROMAGNETIC FORCES ON ELECTRICAL EQUIPMENT BUSBARS BY FINITE ELEMENT METHOD
DOAN THANH BAO, HUYNH DUC HOAN*
1. INTRODUCTION
In the early 1900s, the French physicist Ampere showed that two parallel wires carrying electric currents attract each other if the currents are in the same direction and repel if the currents are in opposite directions. Thus, the busbar in the electrical conductivity of the electrical current running through also withstands the effects of this force.
When a current is placed in a magnetic field, the electric current is affected by magnetic force which depends on the shape of the power line and its position in the magnetic field. The force is determined by the Lorentz formula and the direction is determined by the left-hand [1]. In addition. Maxwell's equations which take into account the content of the law of Biot-Savart-Laplace are applied lo calculate the magnetic induction caused by a current element [2],
Besides using Maxwell's equations, we can use modem methods to study the design calculations [3]. In this paper, the finite element method (FEM) is used to compute the electromagnetic force between two busbars carrying currents in the circuit of electrical conductivity.
2. THE THEORY OF ELECTROMAGNETISM
Based on vector Umov method-Poynting to calculate the electromagnetic force [1], the current flow in the busbars is placed in a magnetic field with a mutual force which is proportional to the magnitude of the electric current and the magnetic field intensity. Where the elements are described in Fig 1, the electromagnetic force is calculated by the formula/ the result is:
dF = [dldy x B] (I) so the continuous analogue to the equation is
dl = Jdxdz = cEdxdz (2)
B = n. H (3) Where current density in the busbar is 7 = o £ , c is Electrical conductivity and E
is Electric field intensity.
DOAN THANH BAC, HUYNH DUC HOAW Due to (1), (2) and (3), we have the following equation
dF = [(aEdxdydz) (/tH)] (4)
M
J^
Fig.l. The busbar in a magnetic Fig.l. X-axis ofthe busbar It can be inferred from the elements vector that electromagnetic force which is perpendicular to the E and H has the same direction with x-axis.
Suppose the electromagnetic force can be expressed as a complex number, we have:
dF = Re [((jEdxdydz) (^iH)] =pGdV Re[E.H*] (5) Where dV is volume to consider
Then, we can also write this electromagnetic force as follows,
dF = pGdVS^. (6) Finally, we have the following formula for the electromagnetic force:
^ dF
dV = li<^S^ (7)
Dimensions of the same force energy flow, dimensions S coincide with the lower energy density and with tiie way of the electromagnetic force, a function of position.
If thai wire cross section coincides with the E and H plane, the force will be exerted on surface area dy.dz represented in Fig.2:
dF F.dV ^^
dp = = - ! — = F.dx dy.dz dydz
The total pressure exerted on the surface of conductors becomes P = \ Fdx = fitj \ S^dx
0 0
* Comments:
(8)
(9)
The electromagnetic force on the busbars carrying the current is calculated by means of the U-P vector metiiods, including Biot-Savart-Laplace Law of magnetic
CALCULATION OF ELECTROMAGNETIC FORCES ON ELECTRICAL EQUIPMENT BUSBARS 7
induction caused by an electric current element to compute the magnetic flux density.
To calculate the electromagiietic, we can use analytical methods which require complex numerical calculations but we achieve just some desired results. Instead of using these ones, we can use softwares with the FEM, one of which is Maxwell - a convenient and efficient software. Therefore, in this paper. Maxwell software is used to calculate the electromagnetic force acting on the busbars conductivity in the electrical devices.
3. THE FINITE ELEMENT METHOD OF MAXWELL 3.1. Parametric Adaptive Analysis [4] [5]
1) Parametric Model Generation - creating the geometry, boundaries and excitations.
2) Analysis Setup - defining solution semp and frequency sweeps.
3) Solve Loop - the solution process is fully automated.
4) Results - creating 2D reports and field plots.
To understand how these processes co-exist, examine the lllusti-ation shown below (Shown specifically for a Magnetostatic setup).
Fig.3. Overall setup process
3.2. Description
This problem will be solved in two parts using the 2D Eddy Current and 3D Eddy Current solvers. The model consists of two rectangular copper busbars size 5xlOmm placed by a center-center spacing of 16nim. The excitation frequency is 50Hz.
DOANTHANHBAO,HUYNHDUCHOAN»
Fig.4, Two busbars model L Maxwell 2D
Fig.5. Two busbars model in Maxwell 3D
3.3. Input data
- Setup the Design Model: 2D & 3D.
- Set Geometry Mode: Cartesian, XY and Eddy.
- The model consists of two rectangular copper busbars: Left busbar, right busbar with the same size of SxlOmm, the same space of 40mm and the length of 50mm.
- Assign the Boundaries.
- The Excitations: frequency is 50Hz.
+ Left busbar: - 5A at 0 degrees.
+ Right busbar: - 5A at 60 degrees.
- Assign the Parameters: Force.
- Solve the Problem.
- View the Results.
4. RESULTS FROM MAXWELL 2D MODEL
•ini
^ B
1
•:-•'««*,
'n^4s
V 1
-^ ~ '' "' -~ \-
Fig.6. Meshing of the two busbars
model in Maxwell 2D Fig 7 The magnitude and direction of the magnetic flux density B ofthe busbar left in 2D
CALCULATION OF ELECTROMAGNETIC FORCES ON ELECTRICAL EQUIPMENT BUSBARS 9 Maxwell 2D models offer three analysis of instantaneous force methods acting on the two busbars stimulation between the DC and AC components, as follows:
Case 1: Results from software
Using the export function from the results model (Results > Solution Data), w e get the magnitude of tiie force results in the following table:
Table I. The magnitude of the force DC and AC Parameters
(N) Force DC Force AC
F(x) Mag -9.283E-06 3.729E-05
F(x) Phase 0 -104.39
F(y) Mag -9.642E-09 9.661E-09
F(y) Phase 0 177.31
MagV 9.283E-06 3.729E-05 Case 2: The Lorentz force method
The Lorentz force formula [1]:
F„c={\\'>'B-Yv
F,c=\\'\J^^iy
(10)
(11) (12) Using the calculator function on software, w e set up the formula (10) (11) (12), in which there are the components of Average DC (F^^), AC (F^J and instantaneous components of the Lorentz (Fj^^^).
Force is calculated according to the formula and the results are displayed as in Fig.8.
Fig.8. The DC, AC and instantaneous components (INST) ofthe Lorentz
force method
Fig.9. F,^j^ and F^^ overlap components ofthe Maxwell method
DOAN THANH BAC, HUYNH DUC HOAW
Case 3; The maxwell stress Tensor force method The maxwell stress Tensor force formula [2]
FUST \ ^.n).H~- 0 .5 (B .Wy'^AV (13) Using the calculator function on the software, we set up the formula the Maxwell (13) and shows the result Maxwell: F„5.j and F,„5.j (case 2) to make comparison between the two with the result as in Fig.9.
* Results and discussion of 2D model:
Magnetic flux density B (T) vector on the left busbar: magnitude B = 2.436E-04 (T) and the corkscrew rules of vector B have the same clockwise because of the current I entering way.
From Fig.8, for a period the excitations, current 360° is instantaneous force components oscillating with the frequency twice the frequency ofthe power supply.
Through the results of the force DC, AC and DC + AC (Table 1; Fig.8), we see three ways of calculaUng, but we reach the same analysis results, the magnitude of the force DC: -9.2574E-06 (N); force AC: 3.7175E-05 (N) and phase 104°.
Fig.9: mstantaneous components ofthe Lorentz force method (FINST: 2,7917E-05) and The maxwell stress Tensor force method (FMST: 2,7788E-05) are almost equal.
5. RESULTS FROM MAXWELL 3D MODEL
FiglO. Meshing ofthe two busbars model pig.ll. The magnitude and direction of in Maxwell 3D .u • i, ,
the magnetic flux density B ofthe busbar left in 3D
From Maxwell 3D models, it offers three analyses of instantaneous force methods acting on the two busbars stimulation between the DC and AC components as follows:
CALCULATION OFELECTROMAGNFnCFORCESONELECTRICALEQUIPMENTBUSBARS 11 Table 2. The magnitude of the force DC and AC
Parameters (N) Force DC Force AC
F(x) Mag 1.193E-09 9.109E-10
F(x) Phase
0 62.66
F(y) Mag -4.648E-07 1.865E-06
F(y) Phase
0 -104.42
F(z) Mag 1.132E-09 1.136E-09
F(z) Phase -0.498
Mag F 4.649E-07 1.865E-06 - Usmg the calculator function on software, we set up tiie formula (10) (11) (12) and display the results of instantaneous components force as in Fig.12.
:&JO0' 2 4 0 ' M 3X3.33' Phase [iieg] 360" £»
Fig.l2. The DC, AC and instantaneous components (INST) ofthe Lorentz force method
* Results and discussion of 3D model:
Magnetic flux density B (T) vector on the left busbar Fig.l 1: magnitude B = 2.5072E-04 (T) and the corkscrew rules of vector B have the same clockwise because of the current I entering way.
From Fig.12, for a period the excitations current 360" is instantaneous force components oscillating with the frequency twice the frequency of the power supply.
Through tiie results of tiie force DC, AC and DC + AC (Table 2; Fig. 12), we see three ways of calculating, but we reach the same analysis results, the magnitude of the force DC: - 4.6819E-07 (N); force AC: 1.8595E-06 (N) and phase 104°.
6. CONCLUSIONS
If we use analytical methods to calculate the forces acting on the busbar carrying the current, we will have only one result with the magnitude of the force. However, if FEM with the Maxwell software are applied, besides the results of the magnitude