Download HMagn4: Coil with ferromagnetic core

Coil
This is an example of the coil simulation, performed with QuickField software.
Problem Type:
Axisymmetric problem of transient magnetics.
Geometry:
All dimensions are in millimeters.
Given:
Relative permeability of winding μ = 1;
Relative permeability of steel μ = 500;
Voltage U = 0.1 V.
Problem:
DC voltage is applied to the coil. Calculate the switching current.
Solution:
Download video.
View on-line.
Result:
Magnetic induction time distribution graph:
Download simulation files.
HMagn4: Coil with ferromagnetic core
Sinusoidal voltage is applied to the electric coil with ferromagnetic core.
Problem Type:
Axisymmetric problem of AC magnetics.
Geometry:
Due to the model symmetry only the right half of the coil is presented in the model (shown by upper half of its
cross-section). Therefore the circuit elements' values are defined twice less than in the real object.
Given:
Magnetic permeability of the steel core μ - nonlinear;
Conductivity of the steel core g = 10,000,000 S/m;
Magnetic permeability of the winding μ = 1;
Conductivity of the winding (copper) g = 56,000,000 S/m;
Number of turns w = 120;
Applied voltage value U = 13.33 V;
Frequency f = 50 Hz.
Problem:
Determine the electric current within the coil winding.
Solution:
Current vs time plot in the real coil has non-sinusoidal shape (see TECircuit1 example). With AC magnetics
module we can estimate RMS value much faster. The drawback of this approach is that we cannot estimate the
wave form.
Results:
Problem
Current (RMS value), A
Current (wave form), A
Transient
34.1
48.1·sin(wt + 108°) + 3.2·sin(3wt + 147°) + 1·sin(5wt + 177°)
AC magnetics
35.3
49.9·sin(wt + 104°)
Download simulation files:
Student's version Professional version
hmagn4_stud.zip
hmagn4.zip
TECircuit1: Coil with ferromagnetic core
This is an example of the coil with ferromagnetic core simulation, performed with QuickField software.
Problem Type:
Axisymmetric problem of transient magnetics.
Geometry:
Due to the model symmetry only the right half of the coil is presented in the model (shown by upper half of its
cross-section). Therefore the circuit elements' values are defined twice less than in the real object.
Given:
Magnetic permeability of the steel core μ - nonlinear;
Conductivity of the steel core g = 10,000,000 S/m;
Magnetic permeability of the winding μ = 1;
Conductivity of the winding (copper) g = 56,000,000 S/m;
Number of turns w = 120;
Applied voltage value U = 13.33 V;
Frequency f = 50 Hz.
Problem:
Determine the electric current within the coil winding.
Solution:
The current wave period is T = 1 / f = 0.02 sec. We choose the time step of 0.0005 second that guarantees a
smooth plot. The finishing time 0.2 sec includes 10 periods and is long enough to fade out initial transient
currents. To reduce the results file size only the last period is stored (starting from moment 0.18 sec).
Results:
The current in the winding is: I(t) = 48.1·sin(wt + 108°) + 3.2·sin(3wt + 147°) + 1·sin(5wt + 177°)
Download simulation files:
Student's version Professional version
tecircuit1_stud.zip
tecircuit1.zip
Current flow in conductor with various cross-sections
By Dariusz Czerwinski
Problem Type:
Plane problem of DC conduction.
Geometry:
Conductor dimensions: 121x293 mm;
Rectangular gap dimensions: 71x50 mm.
Given:
Electric resistivity of conductor ρ = 0.0097784 Ohm·m;
Upper edge potential: 10 V;
Bottom edge potential: 0 V.
Problem:
Determine the current distribution within the conductor.
Results:
Download simulation files.
Heat2: Cylinder with temperature dependent conductivity
A very long cylinder (infinite length) is maintained at temperature Ti along its internal surface and To along its
external surface. The thermal conductivity of the cylinder is known to vary with temperature according to the linear
function λ(T) = C1 + C2·T.
Problem Type:
Axisymmetric problem of heat transfer.
Geometry:
Given:
R1 = 5 mm, R2 = 10 mm;
T1 = 100 °C, To = 0 °C;
C1 = 50 W/K·m, C2 = 0.5 W/K·m.
Problem:
Determine the temperature distribution in the cylinder.
Solution:
The axial length of the model is arbitrarily chosen to be 5 mm.
Results
Temperature distribution in long cylinder:
Radius (cm)
Temperature ( °C )
QuickField Theory
0.6
79.2
79.2
0.7
59.5
59.6
0.8
40.2
40.2
0.9
20.7
20.8
Download simulation files:
Student's version Professional version
heat2_stud.zip
heat2.zip
ACElec2: Cylindrical capacitor
This is an example of the cylindrical capacitor simulation, performed with QuickField software.
Problem Type:
Axisymmetric problem of AC conduction.
Geometry:
Capacitor consists of ceramic tube with silver electrodes mounted at the surface.
Given:
Relative permittivity of air ε = 1.
Relative permittivity of ceramic ε = 6.
Conductivity of ceramic g = 10-8 S/m;
Voltage U = 10 V.
Frequency f = 1000 Hz.
Problem:
Determine capacitance and dissipation factor of the capacitor.
Solution:
The value of dissipation factor can be calculated as
The value of dissipation factor can be calculated as tg(δ) = PA/PR. Capacitance can be calculated as C = q/U,
where U is potential difference between electrodes and q - is a charge on electrodes.
Results:
Field distribution in cylindrical capacitor:
QuickField
q, C
2.77·10-11
C, F
2.77·10-12
PA, W
2.45·10-8
PR, W
8.17·10-7
tg(δ)
0.03
Download simulation files:
Student's version Professional version
acelec2_stud.zip
ACElec2.zip
Cylindrical magnetic device
The cylindrical magnetic device consists of a fixed magnetic part, cylindrical magnetic plunger and a coil.
Problem Type:
Axisymmetric problem of DC magnetics.
Geometry:
Given:
Number of turns N = 1000;
Current i = 5 A;
Plunger position g = 5 mm.
Problem:
Determine the magnetic force and the inductance.
Results:
The inductance of the multiturn winding:
L = N2 · Ψ / I,
where I - is an overall current: I = N·i = 1000·5 = 5000 A.
L = 10002 · 0.00025468 / 5000 = 0.509 Hn
QuickField Reference
B, T
1.094
1.097
L, Hn 0.509
0.524
F, N
865
881.5
Download simulation files.
Reference:
Nicola Bianchi, "Electrical Machine Analysis Using Finite Elements", ISBN 0849333997, page 89.
Magn4: Electric motor
A brushless DC motor with permanent magnets and three phase coil excitation.
Problem Type:
Plane problem of DC magnetics.
Geometry:
Axial length of the motor is 40 mm.
The four magnets are made of Samarium-Cobalt with relative permeability of 1.154 and coercive force of 550000
A/m. The current densities for the coil slots are as follows:
1,300,000 A/m2 on R+, -1,300,000 A/m2 on R-,
1,300,000 A/m2 on S+, -1,300,000 A/m2 on S-,
and zero on T+ and T-.
The inner and outer frames are made of Cobalt-Nickel-Copper-Iron alloy. The B-H curve for the Cobalt-NickelCopper-Iron alloy:
H (A/m)
B (T)
20
60
80
95
105
120
140
0.19 0.65 0.87 1.04 1.18 1.24 1.272
160
180
240
2500
1.3
1.32 1.34 1.36
200
1.45
The B-H curve for the steel:
H (A/m)
400
600
800
1000 1400 2000 3000 4000 6000
B (T)
0.73
0.92
1.05
1.15
1.28
1.42
1.52
1.58
1.60
Results:
Flux density in brushless DC motor with permanent magnets and three phase coil excitation:
Download simulation files:
Student's version Professional version
-
Magn4.zip
Exercise N 4. Electromagnetic shielding
Task
Find the level of magnetic field reduction inside the shield. Shields made of steel and copper of the same
geometry are analyzed.
Experiment
Uniformal external magnetic field is produced by the electric magnet. The shield with the measuring coil inside is
placed between its poles. EMF in the coil is measured: in case of DC current in the coil - by ballistic galvanometer
(in the moment of switching on), in case of AC current - by use of millivoltmeter.
Geometry
The shield consists of two halves. Possible positions of the slot in the shield is shown by dotted line. The slot
could be enlarged up to 2 mm by sheets of non-magnetic materials. Due to symmetry only right-upper quarter
aOb is analyzed, and at the axes of symmetry the boundary conditions are set.
Spherical shield
Cylindrical shield
Given data
Relative magnetic permeability of air and copper μ=1.
Relative magnetic permeability of steel μ=1000.
Magnetic field is uniformal, B=0.139 T (peak value for AC magnetic problem).
Boundary conditions
Due to symmetry at the line Ob Ht=0. At the line Oa Bn =0. Equation B=rot A in the cylindrical coordinate system
leads to A=const (0.0695) at the axis Oa. Field fades at the infinity, so due to continuity at the line Oa A=0 also.
The field is uniformal, and the right boundary has the same condition as the left one H t=0.
Results
Current density in spherical and cylindrical shielding:
Shielding coefficient - relation of magnetic flux densities outside and inside the shield.
 Time-harmonic magnetic field, f=50 Hz
Shield type
Peak value of B in point (0; 0),
mT
Shielding
coefficient
Steel sphere without slot
0.082
1695
lab4Fe.pbm
Steel sphere with slot
36.0
3.86
lab4Fe+.pbm
Steel cylinder without slot
0.013
10700
lab4cFe.pbm
Steel cylinder with slot
40.3
3.45
lab4c_Fe+.pbm
Copper sphere without
97.94
1.42
lab4Cu.pbm
Problem
slot
Copper sphere with slot
100
1.39
lab4Cu+.pbm
Copper cylinder without
slot
68.98
2.02
lab4cCu.pbm
Copper cylinder with slot
70.73
1.97
lab4cCu+.pbm
 DC magnetic field
Shield type
Value of B in point (0; 0), mT Shielding coefficient
Problem
Steel sphere without slot
1.52
91.45
lab4_f.pbm
Steel sphere with slot
39.0
3.56
lab4_f+.pbm
Steel cylinder without slot
2.08
66.83
lab4c_f.pbm
Steel cylinder with slot
45.5
3.05
lab4c_f+.pbm
Download problem files.
Electromagnetic simulation of Duffing lock
Example prepared by Dr. Wlodzimierz Kalat
Duffing's electromagnetic lock is a kind of elecromagnet used to permanent attraction of magnet keeper (5) and it
fast releasing when additional coil (3) will be energized.
A holding coil (1) is constantly supplied with properly chosen current that saturates relatively thin iron core in the
surroundings of releasing coil (3). If additional current flows there, a magnetic field will be significantly reduced in
parts (4)+(8) and only slightly increased in parts (2)+(7) - because of the saturation effected by holding coil. If the
force of keeper attraction is comparable with the stretching force of the spring, even small demagnetizing current
will release the magnet keeper.
Problem type:
Plane parallel nonlinear magnetostatic.
Geometry of the electromagnetic lock:
/
\
/ <-- spring
________\________
|
5
|<- magnet keeper
|_________________|
/ ___
| |
___ \
| 8| 3 |7 |6| 4| 3 |2 |
| | + | /
\ | - | |
| |__/ /
\ \__| |
|
/ air \
|<- core
|
/_________\
|
|
/|
+
|\
|
|
|_|____1____|_|
|
|
|
|_____________________|
|
|
|_________| <- holding coil
Depth of electromagnetic lock: 5 cm.
Given:
Core and keeper material: iron
magnetic permeability: nonlinear
Holding coil total (n*I)h = 100'000 A·turns;
Releasing coil total current (n*I)r = 50'000 A·turns;
Force of stretching spring F = 250 N/m
Results:
When releasing coil is not energized (I=0, edlock_1 case) then the force of keeper attraction is about 295 N/m.
Otherwise (I>0, edloc_2 case) the attraction force decreases to 225 N and becomes less then stretching force of
the spring, and magnet keeper will be unlocked. The force was calculated by means of "Integral calculator" tools.
Ih, A·turns Ir, A·turns F, N/m
100000
0
295
100000
50000
225
Student's version Professional version
-
duffing_lock.zip
Student's version mesh size limitations do not affect to its ability of analysing the results of any complicated problem defined
and solved with the Professional version. To view the results with Students' version:
1. Open problem file (*.pbm)
2. Use Problem>View Field Picture command or click the View Field Picture button.
Electronic device radiator
Example problem offered by Mr. Isaac Cohen, Lambda Americas Inc.
The semiconductor device attached to a sapphire substrate installed on a copper header. The thermal resistance
of interest is the resistance from the semiconductor device to the ambient.
Problem Type:
Plane problem of transient heat transfer.
Geometry:
The Rca block represents the known thermal resistance from the header to the ambient.
Given:
Thermal load of semiconductor device: 60 W/mm2;
Heat conductivity of sapphire: 28 J/K·m;
Heat conductivity of copper: 394 J/K·m;
Heat conductivity of Rca: 1 J/K·m;
Mass density of sapphire: 3985 kg/m3;
Mass density of copper: 8950 kg/m3;
Specific heat of sapphire: 3985 J/Kg·K;
Specific heat of copper: 8950 J/Kg·K;
Solution:
The transient thermal resistance is calculated as the instantaneous temperature increase divided by the heat flux.
A normalized curve can be obtained by dividing the instantaneous temperature by the final temperature.
The shape of the graph can be qualitatively explained by the presence of two very different thermal time
constants in the system. The first time constant corresponds to sapphire substrate (it works up to 0.005 sec.). The
second time constant corresponds to copper header. The second time constant will be better visible if the
simulation time is extended above 2 sec., so two simulations are necessary to visualize the system behavior.
Download simulation files:
Student's version Professional version
-
IsaacCohen.zip
Electrostatic simulation of 3 phase cable
By Dariusz Czerwinski
Problem Type:
Plane problem of electrostatics.
Geometry:
The diameter of the cable with insulation is 20mm (for all cables).
The insulation thickness is 2 mm.
Given:
Relative permittivity of individual insulation ε = 3.3;
Relative permittivity of common insulation ε = 3;
Voltage applied UA = 9.8 kV, UB = -4.9 kV, UC = -4.9 kV.
Problem:
Determine the field strength, potential in cable and in insulation.
Results:
Download simulation files
Magn3: Ferromagnetic C-magnet
A permanent C-magnet in the air. The example demonstrates how to model curved permanent magnet using the
equivalent surface currents.
Problem Type:
Plane problem of DC magnetics.
Geometry of the magnet:
Given:
Relative permeability of the air μ = 1;
Relative permeability of the magnet μ = 1000;
Coercive force of the magnet Hc = 10000 A/m.
The magnet polarization is along its curvature.
Solution:
To avoid the influence of the boundaries while modelling the unbounded problem, we'll enclose the magnet in a
rectangular region of air and specify zero Dirichlet boundary condition on its sides.
Magnetization of straight parts of the magnet is specified in terms of coercive force vector. Effective surface
currents simulate magnetization in the middle curved part of the magnet.
Results:
Flux density in ferromagnetic C-magnet:
Download simulation files:
Student's version Professional version
magn3_stud.zip
magn3.zip
Field distribution around two copper bars
This example is prepared by Didier Werke AG, InduCer Group, Abraham Lincoln Str. 1,65 189 Wiesbaden, Germany
At Didiers Quickfield has been used to calculate AC-electromagnetic and heat transfer problems in the course of
the development of the "InduCer"- technology. The sample given here is dedicated to pure electromagnetic
problem: field distribution around two copper bars.
Problem Type:
A plane problem of time-harmonic magnetic field.
Geometry:
Two copper bars with equal but opposite currents.
Given:
Magnetic permeability of air μ = 1;
Magnetic permeability of copper μ = 1;
Conductivity of copper σ = 56 000 000 S/m;
Current in the conductors I = 1880 A;
Frequency f = 10 000 Hz.
Results of two copper bars magnetic simulation:
Student's version
-
Professional version
copper_bars_magnetic_field.zip
Student's version mesh size limitations do not affect to its ability of analysing the results of any complicated problem defined
and solved with the Professional version. To view the results with Students' version:
1. Open problem file (*.pbm)
2. Use Problem>View Field Picture command or click the View Field Picture button.
Exercise N 2. Force of interaction of two cylindrical coaxial coils
Task
Find the force applied to the coils with current with and without the shield between them.
Experiment
Forces are measured by use of digital balance on which one of coils is installed (see the right picture below).
Tests are performed with currents varying, and different conductors and ferromagnetics are placed between coils.
Geometry
Due to symmetry of the formulation only upper half of the problem (above ab line) is defined, and at the axis of
symmetry (line ab) the boundary conditions are set.
Given data
Current density in the coil j=100000 A/m2.
Relative magnetic permeability of air, aluminum and copper coils μ= 1.
Relative magnetic permeability of the steel shield μ= 1000.
Electrical conductivity of steel σ= 10000000 Sm/m.
Electrical conductivity of aluminum σ=37000000 Sm/m.
Coils are wound by insulated wire, so cross-section conductivity in coils σ= 0 Sm/m.
Boundary conditions
Along the horizontal symmetry axis (line ab) Bn =0.
Equation B=rot A in cylindrical coordinate system leads to A=const at the axis ab. The field fades at the infinity, so
due to the condition of continuity of the potential A=0 at the line ab.
Results
Field distribution around coils:
 Time-harmonic magnetic field, f=50 Hz.
Shield type Interaction force, mN
Problem
No shield
0.533
lab2.pbm
Steel
0.098
lab2_Fe.pbm
Aluminum
0.282
lab2_Al.pbm
 DC magnetic field
Shield type Interaction force, mN
Problem
No shield
0.533
lab2c.pbm
Steel
0.138
lab2c_Fe.pbm
Download problem files.
Ground connector
This is an example of the ground connector simulation, performed with QuickField software.
Problem Type:
Plane problem of DC conduction.
Geometry:
All dimensions are in millimeters.
Given:
Ground resistivity ρ = 10 Ohm·m.
Overvoltage value U = 250 V.
Problem:
Determine the resistance of the ground connector.
Solution:
Download video.
View on-line.
Result:
Current density in ground connector:
Download simulation files.
THeat1: Heating and cooling of a slot of an electric machine
Temperature in the stator tooth zone of power synchronous electric motor during a loading-unloading cycle.
Problem Type:
Plane problem of heat transfer.
Geometry:
All dimensions are in millimeters. Stator outer diameter is 690 mm. Domain is a 10-degree segment of stator
transverse section. Two armature bars laying in the slot release ohmic loss. Cooling is provided by convection to
the axial cooling duct and both surfaces of the core.
Given:
1. Working cycleWe assume the uniformly distributed temperature before the motor was suddenly loaded.
The cooling conditions supposed to be constant during the heating process. We keep track of the
temperature distribution until it gets almost steady state. Then we start to solve the second problem getting cold of the suddenly stopped motor. The initial temperature field is imported from the previous
solution. The cooling condition supposed constant, but different from those while the motor was being
loaded.
2. Material Properties
The thermal conductivity values are the same as in the Heat1 example. For transient analysis the values
of specific heat C and volume density are also required:
Heat Conductivity Specific Heat Mass Density
(W/K·m)
(J/Kg·K)
(kg/m3)
Steel Core
25
465
7833
Copper Bar
380
380
8950
Bar Insulation
0.15
1800
1300
Wedge
0.25
1500
1400
3. Heat sources and cooling conditions
During the loading phase the slot is heated by the power losses in copper bars. The specific power loss
is 360000 W/m3. When unloaded, the power loss are zero. We suppose the temperature of contacting air
to be the same fro both phases of working cycle. In turn, the convection coefficients are different,
because the cooling fan is supposed to be stopped when the motor is unloaded.
Loading
Stopped
Convection
coefficient
(W/K·m2)
Temperature of
contacting air
(°C)
Convection
coefficient
(W/K·m2)
Temperature of
contacting air
(°C)
Inner stator surface
250
40
20
40
Outer stator surface
70
20
70
20
Cooling duct
150
40
20
40
Solution
Each phase of the loading cycle is modeled by a separate QuickField problem. For the loading phase the initial
temperature is set to 20°C, for the cooling phase the initial thermal distribution is imported from the final time
moment of the previous solution.
Moreover, we decide to break the cooling phase into two separate phases. For the first phase we choose time
step as small as 100 s, because the rate of temperature change is relatively high. This allows us to see that the
temperature at the slot bottom first increases by approximately 1 grad for 300 seconds, and then begins
decreasing. The second stage of cooling, after 1200 s, is characterized by relatively low rate of temperature
changing. So, we choose for this phase the time step to be 600 s.
For heating process the time step of 300 s is chosen.
Please see following problems in the Examples folder:




THeat1_i.pbm for initial state, and
THeat1Ld.pbm for loading phase, and
THeat1S1.pbm for the beginning of stopped phase, and
THeat1S2.pbm for the end of stopped phase
Results
Temperature vs. time dependence at the bottom of the slot (where a temperature sensor usually is placed).
Download simulation files:
Student's version Professional version
theat1_stud.zip
theat1.zip
Inductance of a pair of concentric cylinders
A coaxial cable has an inner core of radius 1.0 mm and an outer sheath of internal radius of 4.0 mm. Determine the inductance
length.
Problem Type:
Plane problem of AC magnetics.
Geometry:
Given:
Inner radius a = 1 mm;
Outer radius b = 4 mm;
Current I = 0.001 A;
Analytical solution:
The total inductance per meter at low frequency is given by L = μ/2π · (1/4 + ln(b/a)) H/m
L = 4π·10-7 · (1/4 + ln(4)) = 3.27259·10-7 H/m
QuickField simulation results:
Note: QuickField calculates the total current and the total flux. To get RMS we should divide the corresponding amplitude value
L = Flux / I = 3.272·10-10/1.4142 / 0.001/1.4142 = 3.272·10-7 H/m
L = 2·W / I2 = 2·8.181·10-14 / (0.001/1.4142)2 = 3.272·10-7 H/m
Mesh size
251 (Student version)
QuickField
L = Flux / I
L = 2·W / I2
3.23·10-7 H/m
3.22·10-7 H/m
-7
Discrepancy with theory
0.17% / 1.03%
-7
0.08% / 0.26%
-7
1301 (automatic refinement in Prof. version) 3.267·10 H/m 3.239·10 H/m
4003 (automatic refinement 2)
-7
3.27·10 H/m
-7
1% / 9.55%
-7
3.263·10 H/m
8814 (automatic refinement 3)
3.271·10 H/m 3.268·10 H/m
0.05% / 0.14%
26474 (automatic refinement 5)
3.272·10-7 H/m 3.271·10-7 H/m
0.02% / 0.05%
78585 (automatic refinement 10)
-7
-7
3.272·10 H/m 3.272·10 H/m
0.02% / 0.02%
Download simulation files:
Student's version
Professional version
concentric_cylinders_inductance_stud.zip concentric_cylinders_inductance.zip
References:
 John Bird, "Electrical circuit theory and technology", p.520. ISBN-13: 978 0 7506 8139 1.
© Copyright Tera Analysis Ltd.
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Inductively Heated Ceramic
This example is prepared by Didier Werke AG, InduCer Group, Abraham Lincoln Str. 1,65 189 Wiesbaden, Germany
Problem Type:
An axisymmetric problem of magneto-thermal coupling.
Geometry:
Problem:
Calculate impedances and spatial heat source distribution in the inductively heated ceramic.
Results:
More detailed description (in PDF format).
Student's version
-
Professional version
inductively_heated_ceramic.zip
Student's version mesh size limitations do not affect to its ability of analysing the results of any complicated problem defined and solved wit
version. To view the results with Students' version:
1. Open problem file (*.pbm)
2. Use Problem>View Field Picture command or click the View Field Picture button.
© Copyright Tera Analysis Ltd.
RSS channel
Follow us on Twitter
Privacy policy
Line-to-line short circuit
Line-to-line short circuit is one of the most widespread damages of the transmission lines in electrical networks
(15-20 % from total number of failures). This model shows how to simulate the line-to-line short circuit in
QuickField.
Problem Type:
Plane-parallel problem of transient magnetics.
Geometry:
Wire design
Line-to-line short circuit
Tangent tower design. All dimensions
are in meters.
Given:
Rated voltage (RMS) Urated = 110 kV;
Rated current (RMS) Irated = 440 A;
Line lengh l = 20 km;
Loading parameters Rload = 100 Ohm; Lload = 0.23 H.
Problem:
Determine the transient process currents after a short circuit occurred. Find the electromagnetic forces acting on
the lines during this process.
Solution:
To solve this problem it is necessary to simulate both field part and circuit part. The scheme of connection of
power supply phase windings, loading, elements of the circuit and short-circuit bridge is presented in the circuit
model, created by the QuickField Circuit Editor.
Short-circuit bridge is a QuickField block with the resistance which varies in time. This dependence is
implemented using the electrical conductivity as a function of the temperature, which is also function of time – as
σ(T(t)) function.
Results:
Short-circuit current (amplitude) Imax = 7000 A.
Electromagnetic force of interaction between phases A and B FA-B = 1780 N/km.
Illustration of the short-circuit currents:
Download simulation files line_to_line_short.zip.
Local Over-Heat Model in SF6 GIS
Evgeni K. Volpov
Sulphur hexafluoride was first synthesized in the laboratories of the Faculté de Pharmacie de Paris in 1900 by
Moissan and Lebeau.
The dielectric strength of SF6 is about 2.5 times higher than that of air under the same conditions. This unique
property of SF6 has led to its adoption for a number of industrial and scientific applications as an electrical
insulation. Switchgear manufacturers use the unique dielectric or/and breaking properties to design their
equipment.
Due to contact wear the contact resistance in SF6 GIS increase from the normal value of 5-7 μOhm up to
100 μOhm. The problem is to determine the local overheating.
The result is applicable to SF6 GIS up to 100 m of length. The current range is from 1 to 2 kA.
View the report in PDF format.
Download simulation files.
Long solenoid inductance
Problem type:
Axisymmetric problem of DC magnetics.
Geometry:
R1 = 30 mm, R2 = 35 mm.
Given:
Current in the conductor i = 10 A;
Number of turns N = 100,
(number of turns per unit length n = N/l = 200 m-1);
Problem:
Calculate inductance of the long solenoid.
Solution:
Inductance of the long solenoid can be obtained from the equation:
L = N 2 · μ0 A / l = N n · μ0 A,
where A is a cross-section area of the core (m2).
To calculate the inductance in QuickField you should divide magnetic flux by the source current.
L=N·Φ/i
Results:
Inductance L, μH/m
QuickField
Theory
Δ, %
83.55
83.40
0.18%
Download simulation files.
References:
 John Bird, "Electrical circuit theory and technology", p.77. ISBN-13: 978 0 7506 8139 1.
Exercise N 5. Magnetic field of the cylindrical coil
Task
Make a plot of magnetic flux density at the axis of the coil with and without steel core.
Experiment
Magnetic flux density is measured in laboratory by microwebermeter.
Problem type
Linear axisymmetrical problem of magnetostatics.
Geometry
Due to problem symmetry only upper-right quarter a0b is defined, and at the axes of symmetry boundary
conditions are set.
Given data
Current density in the coil j=100000 A/m2.
Relative magnetic permeability of air and copper μ=1.
Relative magnetic permeability of steel of the core μ=500.
Boundary conditions
At the vertical axis of symmetry (line Ob) Ht=0. At the horizontal axis of symmetry Oa Bn=0. From B=rotA in the
cylindrical coordinate system we have at the axis Oa A=const. Field fades at the infinity, so at the line Oa A=0
due to continuity of A.
Results
Dependence of the magnetic flux density upon the distance to the coil center:
Download problem files
Elec1: Microstrip transmission line
A shielded microstrip transmission line consists of a substrate, a microstrip, and a shield.
Problem Type:
Plane problem of electrostatics.
Geometry:
The transmission line is directed along z-axis, its cross section is shown on the sketch. The rectangle ABCD is a
section of the shield, the line EF represents a conductor strip.
Model depth l = 1 m.
Given:
Relative permittivity of air ε = 1;
Relative permittivity of substrate ε = 10.
Problem:
Determine the capacitance of a microstrip transmission line.
Solution:
There are several different approaches to calculate the capacitance of the line:
 To apply some distinct potentials to the shield and the strip and to calculate the charge that arises on the
strip;
 To apply zero potential to the shield and to describe the strip as having constant but unknown potential
and carrying the charge, and then to measure the potential that arises on the strip.
Both these approaches make use of the equation for capacitance:
C = q / U.
Other possible approaches are based on calculation of stored energy of electric field. When the voltage is known:
C = 2·W / U2,
and when the charge is known:
C = q 2 / 2·W
Experiment with this example shows that energy-based approaches give little bit less accuracy than approaches
based on charge and voltage only. The first approach needs to get the charge as a value of integral along some
contour, and the second one uses only a local value of potential, this approach is the simplest and in many cases
the most reliable.
Results:
Potential distribution in microstrip transmission line:
Theoretical result (model depth l = 1 m.)
C = 178.1 pF.
Approach 1
C = 177.83 pF (99.8%)
Approach 2
C = 178.47 pF (100.2%)
Approach 3
C = 177.33 pF (99.6%)
Approach 4
C = 179.61 pF (100.8%)
See the Elec1_1.pbm and Elec1_2.pbm problems for the 1,3 approaches and the 2,4 approaches respectively.
Download simulation files:
Student's version Professional version
elec1_stud.zip
elec1.zip
Microstrip line
This is an example of the microstrip line simulation, performed with QuickField software.
Problem Type:
Plane problem of electrostatics.
Geometry:
All dimensions are in millimeters.
Given:
Relative permittivity of air ε = 1;
Relative permittivity of substrate ε = 10.
Shield is grounded U = 0 V.
Problem:
Determine the capacitance of the microstrip transmission line.
Solution:
Download video.
View on-line.
Result:
Voltage distribution in microstrip line:
Download simulation files.
Modelling electric field in case of the storm
Example offered by Dariusz Czerwinski, Dept. of Fundamental electrical eng. Lublin technical university
Problem Type:
Plane problem of electrostatics.
Geometry:
Our model consists of clouds, buidings, trees and ground.
Hight of the building 80m,
hight of the tree 40m,
hight of the church tower 120m,
hight of the clouds 250m.
Given:
Potential of the first cloud U1 = 4000000 V;
Potential of the second cloud U2 = -5000000 V;
Potential of the buidings, trees, gorund U3 = 0 V.
Results:
Download simulation files:
Student's version Professional version
-
storm.zip
Heat3: Multi-layer coated pipe
A very long cylindrical coated pipe (infinite length) is maintained at temperature Tinner along its internal surface and
Touter along its external surface.
Problem Type:
Plane and axisymmetric problems of heat transfer.
Geometry:
Thin layer of 3 mm coating is placed between steel pipe and insulation.
Given:
Tinner= 85°C, Touter=4°C;
Thermal conductivity of steel λ=40 W/K·m,
Thermal conductivity of coating λ=0.4 W/K·m,
Thermal conductivity of insulation λ=0.15 W/K·m.
Problem:
Derive heat flux at the internal diameter and calculate the overall heat transfer coefficient (OHTC) of the system.
Solution:
The OHTC of 3-layered pipe can be calculated as
λtotal = 2πL / [ 1/λ1·ln(r2/r1) + 1/λ2·ln(r3/r2) + 1/λ3·ln(r4/r3)],
where L is the length of the tube.
For very long pipe the heat transfer in axial direction may be neglected, thus the temperature distribution in any
cross section will be the same. In this case the plane-parallel can be solved. In plane-parallel problems all integral
values are calculated per 1 meter of depth.
Because, the axial length of the axisymmetric model do not affect to results, it was set to some small value (0.05
m) to reduce the mesh size. To convert heat flux (and other integral values) calculated in axisymmetric problem to
flux calculated in plane-parallel problem the former should be multiplied by 20.
Results
Temperature distribution in multi-layer coated pipe:
QuickField
plane
Heat Flux, W
axisymmetric
178.2 8.89·20 = 177.8
Theory
-
Temperature difference, K
81
81
81
OHTC, W/K·m
2.200
2.195
2.211
Reference
Heat transfer for Engineers - H.Y. Wong Table 2.2 with the heat transfer coefficient set at infinity
Download simulation files:
Student's version Professional version
-
heat3.zip
Mutual capacitance of collinear cylinders
Find the mutual capacitance between two collinear infinitely long cylinders.
Problem Type:
Plane problem of electrostatics.
Geometry:
a = 5 mm, b = 10 mm.
Model depth l = 1 m.
Given:
Relative permittivity of vacuum ε = 1,
The charge q = 10-9 C.
Problem:
Find the mutual capacitance between two collinear infinitely long cylinders and compare it with analytical solution:
C = 2πεε0 * l / ln(a/b) [F]. *
Solution:
Each of the conductor's surfaces is marked as 'floating conductor', i.e. isolated conductors with unknown
potential. At some point on each of conductor's surface the charge is applied. The charge is then redistributed
along the conductor surface automatically.
Results:
Potential distribution between collinear cylinders.
The capacitance can be calculated as C = q / (U2 - U1). The measured potential difference is U2-U1 = 12.459 V.
The capacitance is C = 10-9 / 12.459 = 8.027·10-11 F.
QuickField Theoretical result
C, pF
(model depth l = 1 m)
80.27
80.26
Download simulation files.
*Wikipedia, Capacitance.
Exercise N 6. Mutual inductance of coils
Task
Find the dependence of mutual inductance of coaxial cylindrical coils upon the distance between them.
Experiment
Electro motive force (EMF) in the right coil is measured by ballistic galvanometer (at the switching on).
Problem type
Linear axisymmetrical problem of magnetostatics.
Geometry
Given data
Relative magnetic permeability of air and copper coils μ= 1.
Current density in the left coil j=0.1 A/mm2.
There is no current in the right coil, thus it has no affection to the field shape.
Solution
The field source is the lest coil. Due to the field symmetry only upper-right quarter aOb is defined. At the axes of
symmetry the boundary conditions are set.
At the vertical axis of symmetry (line Ob) Ht=0. At the horizontal axis of symmetry Oa Bn=0. From B=rot A in the
cylindrical coordinate system we have at the axis Oa A=const. Field fades at the infinity, so at the line Oa A=0
due to continuity of A.
Results
Flux density distribution around coils:
Mutual inductance M - relation of the flux connected with all turns of the right coil Ψ to the current in the left coil J
(which is the origin of the flux).
L=Ψ / J
Ψ=Φ · w
Here w is number of turns of the right coil, Φ - flux across the right coil.
x, mm Flux across the right coil, μWb Mutual inductance M, μH
70
2.656
0.0306*w
150
0.637
0.0073*w
210
0.285
0.0033*w
Download problem files
Non-concentric spheres capacitance
Problem Type:
Axisymmetric problem of electrostatics.
Geometry:
a = 100 mm, d = 500 mm.
Given:
Relative permittivity of vacuum ε = 1,
The charge q = 10-9 C
Task:
Find the mutual capacitance between two spheres and compare its value with analytical solution:
C = 2π·ε·ε0 · a
where D = d/ (2a).
[F] *,
Solution:
Sphere's surfaces is marked as 'floating conductor', i.e. isolated conductors with unknown potential. At some point
on each of sphere's surface the charge is applied. The charge is then redistributed along the conductor surface
automatically.
Results:
Potential distribution around spheres.
The capacitance can be calculated as C = q / (U2 - U1). The measured potential difference is U2 - U1 = 143.4 V.
The capacitance is C = 10-9 / 143.4 = 6.97·10-12 F.
QuickField Theoretical result
C, pF
6.97
6.99
Download simulation files.
*Wikipedia, Capacitance.
HMagn3: nonlinear ferromagnetic core in sinusoidal magnetic field
A sinusoidal current carrying conductor is surrounded by nonlinear ferromagnetic core.
Problem Type:
Plane problem of AC magnetics.
Geometry:
Given:
Total current I = 500 A.
Frequency f = 50 Hz.
Core BH-curve: H = 100·B2
Problem:
Calculate B and H distribution along the radius r.
Results:
To calculate the sinusoidal filed distribution in nonlinear core we should use the modified curve B'(H) (see more in
QuickField documentation).
In this particular case the modified BH-curve can be calculated as B`(H) = B(H) * (1 + 1/sqrt(180))
B,H distribution along the radius
Download simulation files:
Student's version Professional version
hmagn3_stud.zip
hmagn3.zip
Magn1: Nonlinear permanent magnet
A simulation of permanent magnet and a steel keeper in the air, performed with QuickField software.
Problem Type:
Plane problem of DC magnetics.
Geometry:
All dimensions are in centimeters.
Given:
The permanent magnets are made of ALNICO, coercive force is 147218 A/m. The polarizations of the magnets
are along vertical axis opposite to each other. The demagnetization curve for ALNICO:
H, A/m -14728 -119400 -99470 -79580 -53710 -19890
B, T
0
0.24
0.4
0.5
0.6
0.71
0
0.77
The B-H curve for the steel:
H, A/m
B, T
400
600
800
0.73 0.92 1.05
1000 1400 2000 3000 4000 6000
1.15
1.28
1.42
1.52
1.58
1.60
Problem:
Find maximum flux density in Y-direction
Solution:
To avoid the influence of the boundaries while modelling the unbounded problem, we'll enclose the magnet in a
rectangular region of air and specify zero Dirichlet boundary condition on its sides.
Comparison of results
Flux density in nonlinear permanent magnet:
Maximum flux density in Y-direction:
By, T
ANSYS
0.42
COSMOS/M 0.404
QuickField
0.417
Download simulation files:
Student's version Professional version
magn1_stud.zip
magn1.zip
PCB heating
Example problem offered by Mr. Nils Forsström, Permobil AB.
The voltage is applied to the sides of conducting sheet placed vertically and surrounded by the still air. The
flowing current heats the sheet due to resistive losses. The front and back surfaces of the sheet are cooled by the
air (natural convection).
Problem Type:
Plane problem of electro-thermal coupling.
Geometry:
Given:
Data for magnetic analysis:
Sheet thickness d = 0.035 mm;
Material resistance ρ = 2·10-8 Ohm/m;
Current I = 10 A;
Material heat conductivity λ = 380 W/K·m;
Convection coefficient α = 10 W/K·m2;
Problem:
Calculate the temperature and potential distribution in a conducting sheet.
Solution:
The resistive losses are calculated in the DC conduction problem. Then these losses are transferred to the linked
heat transfer problem.
The sheet is cooled by convection from the front surface F(T) = -λ·(T - T0), where λ is a convection coefficient,
and T0 is an ambient temperature. We put T0 = 0 K and calculate overheating. The convection from other surfaces
is ignored.
The convection is modelled by the volume heat sink: Q(T) = F(T) / d.
Results:
Current distribution in the pcb
Temperature distribution in the pcb (overheating)
The pcb_current.pbm is the problem of calculating the current distribution, and pcb_heat.pbm analyzes
temperature field.
Download simulation files:
Student's version Professional version
-
pcb.zip
Pair of parallel wires capacitance
Problem Type:
Plane problem of electrostatics.
Geometry:
a1 = 4.72 mm, a2 = 1.13 mm, d = 100 mm.
Given:
Relative permittivity of vacuum ε = 1,
The charge q = 10-9 C
Model depth l = 1 m.
Task:
Find the mutual capacitance between two parallel wires and compare its value with analytical solution:
C = 2π·ε·ε0 l / ln( d2/a1·a2 ) [F]. *
Solution:
Wire's surfaces is marked as 'floating conductor', i.e. isolated conductors with unknown potential. At some point
on each of wire's surface the charge is applied. The charge is then redistributed along the conductor surface
automatically.
Results:
Potential distribution.
The capacitance can be calculated as C = q / (U2-U1). The measured potential difference is U2-U1 = 135 V.
The capacitance is C = 10-9 / 135 = 7.41·10-12 F
QuickField Theoretical result
C, pF
(model depth l = 1 m)
7.41
7.37
Download simulation files.
*Power System Engineering, D.P.Kothari, I.J.Nargath.
Perio1: Periodic Boundary Condition
This very simple example demonstrates the effect of applying periodic boundary condition, which forces the field
potential to be the same on opposite sides of the model.
Problem Type:
Plane problem of DC magnetics.
Geometry:
Two regions, A and B, have the same shape, volume loading and are surrounded by Dirichlet boundary condition,
which does not allow the field to penetrate outside. Region B is also subdivided into B1 and B2, and the periodic
boundary condition is specified on two sides, which makes these regions the continuation of each other. As a
result, field distribution in both A and B must be equivalent.
Results:
This example also demonstrates that the mesh on the periodic boundary is not necessarily the same - please
notice that the mesh spacing settings in four corners of the model are all different!
Ther are 2 problems Perio1.pbm and Perio1odd.pbm in the archive. Perio1odd.pbm is almost the same, but for
one difference: odd periodic condition is applied, which forces the field potential to be opposite on two sides of the
region.
Download simulation files:
Student's version Professional version
perio1_stud.zip
perio1.zip
ACElec1: Plane capacitor
This is an example of the plane capacitor simulation, performed with QuickField software.
Problem Type:
Plane problem of AC conduction.
Geometry:
Due to symmetry only a small part of 1 mm heigh is used. The length of capacitor in z direction is L = 10 mm.
Given:
Relative permittivity of substrate ε = 10.
Conductivity of substrate g = 10-8 S/m;
Voltage U = 5 V,
Frequency f = 50 Hz.
Problem:
Find current and dissipation factor tg(δ) of the plane capacitor with nonideal dielectric inside.
Solution:
Capacitor with nonideal dielectric can be replaced by electric circuit with ideal capacitor C and resistivity R
connected in parallel. The capacitance of the plane capacitor is calculated by the equation
C = εε0S/d, where S is the plate area S = h·l.
Resistance of the substrate is calculated by the equation R = ρ·d/S.
Current I has two components: active IA and reactive IR. For parallel scheme IA = U / R, IR = U / XC.
tg(δ) = |PA / PR| = |U·IA / U·IR| = |XC|/R = 1/ωC·R,
tg(δ) = 1 / 2πf·C·R
Results:
Field distribution in Plane capacitor:
QuickField Theoretical result
IA, μA
0.05000
0.05000
IR, μA
0.13908j
0.13902j
tg(δ)
0.3595
0.3596
Download simulation files:
Student's version Professional version
acelec1_stud.zip
ACElec1.zip
Capacitance calculation with QuickField
Free webinar on April 3, 2011.
Capacitors, cables, transmission lines (both power and signal) design and analysis require capacitance
calculation. During this webinar you will learn how to effectively apply QuickField simulations for capacitance
calculation.
Read presentation in PDF format.
Download simulation examples:
spherical_capacitor
cylindrical capacitor
parallel wires capacitance
plane capacitor real
winding_capacitance
The webinar is over. You can
download the record from our site (159 Mb) or watch it on
YouTube:
Part 1. Single body capacitance, spherical capacitor.
Part 2. Plane capacitor, cylindrical capacitor. Ideal and real capacitor.
Part 3. Transmission line capacitance, winding capacitance.
Spherical capacitor
Problem Type:
Axisymmetric problem of electrostatics.
Geometry:
R = 100 mm, r=50 mm.
Given:
Relative permittivity of vacuum ε = 1,
The charge q = 10-9 C
Problem:
Find the capacitance of spherical capacitor and compare it with analytical solution:
C = 4π·ε·ε0 · r·R / (R - r), [F]. *
Solution:
Capacitor plate's surface is marked as 'floating conductor', i.e. isolated conductors with unknown potential. At
some point on spheres' surface the charge is applied. The charge is then redistributed along the conductor
surface automatically.
Results:
Potential distribution inside of the spherical capacitor.
The capacitance can be calculated as C = q / (U2-U1).
The measured potential difference is U2-U1 = 89.87 V.
The capacitance is C = 10-9/ 89.87 = 11.13·10-12 F
QuickField Theoretical result
C, pF
11.13
11.11
Download simulation files
Mutual capacitance of collinear cylinders
Find the mutual capacitance between two collinear infinitely long cylinders.
Problem Type:
Plane problem of electrostatics.
Geometry:
a = 5 mm, b = 10 mm.
Model depth l = 1 m.
Given:
Relative permittivity of vacuum ε = 1,
The charge q = 10-9 C.
Problem:
Find the mutual capacitance between two collinear infinitely long cylinders and compare it with analytical solution:
C = 2πεε0 * l / ln(a/b) [F]. *
Solution:
Each of the conductor's surfaces is marked as 'floating conductor', i.e. isolated conductors with unknown
potential. At some point on each of conductor's surface the charge is applied. The charge is then redistributed
along the conductor surface automatically.
Results:
Potential distribution between collinear cylinders.
The capacitance can be calculated as C = q / (U2 - U1). The measured potential difference is U2-U1 = 12.459 V.
The capacitance is C = 10-9 / 12.459 = 8.027·10-11 F.
QuickField Theoretical result
C, pF
(model depth l = 1 m)
80.27
80.26
Download simulation files.
Power busses in the air above the steel plate
By Dariusz Czerwinski
Problem Type:
Plane problem of DC magnetics.
Geometry:
Given:
Relative permeability of air and copper busses μ = 1;
Relative permeability of steel plate μ - non-linear;
Current density in the copper busses j = 3,000,000 A/m2;
Problem:
Find magnetic field distribution.
Results:
Magnetic field distribution.
Download simulation files:
Student's version Professional version
-
power_busses.zip
Power cable parameters calculation
QuickField, enforced by ActiveField technology may be effectively used for multi-physics analysis of various
engineering tasks. This analysis could be highly automated. It even can be implemented as a Microsoft Word
document equipped by the set of VBA macros for automatic creation of QuickField problem, solving,
postprocessing and report generation. Rather complicated example - analysis of tetra-core cable - is available as
ActiveField Cable Example. If you have working knowledge of Visual Basic, and understanding of QuickField
Object model - you are welcome to analyze source code of these macros.
This document displays the results of cable analysis based on specific modeling parameters. Pictures, tables and
graphs below have been automatically calculated by Professional version of QuickField, controlled by VBA code
implemented as MS Word macros. Corresponding QuickField problems can be analyzed by the Students version.
1.
2.
3.
4.
Model description
Input parameters
Calculated cable parameters
Field pictures
1. Model description.
Figure1. Cable sketch.
This high-voltage tetra-core cable has three triangle sectors with phase conductors and round neutral conductor
in the lesser area of the cross-section above. All the conductors are made of aluminum. Each conductor is
insulated and the cable as a whole has a three-layered insulation. The cable insulation consists of inner and outer
insulators and a protective braiding (steel tape). The sharp corners of the phase conductors are chamfered to
reduce the field crown. The corners of the conductors are rounded. Empty space between conductors is filled with
some insulator, possibly with an air.
It is often required to design a cable according to parameters of the conductor section areas. Conductor section
areas are defined in the Table 1. The tables 2 to 7 describe other input parameters.
2. Input parameters.
Table 1. Conductors' geometric parameters.
Phase conductor area
120 mm2
Neutral conductor area
35 mm2
Thread rounding radius (R) 2
mm
Table 2. Insulator geometric parameters.
Cable-core insulation thickness
2 mm
Inner cable insulation thickness
1 mm
Protective steel braiding thickness 1 mm
Outer cable isolation thickness
3 mm
Table 3. The precision.
Areas calculation reasonable error 0.001 mm2.
Table 4. Conductors' loading.
Current amplitude
200
Voltage amplitude (electrostatics)
6500 V
Frequency
50
Current phase (for static problems) 0
A
Hz
deg
Table 5. Conductors' physical properties.
Relative permeability
1
Conductivity
36000000 S/m
Thermal conductivity
140
W/K·m
Young's modulo
6.9e+10
N/m2
Poisson's ratio
0.33
Coefficient of thermal expansion 2.33·10-5
1/K
Specific density
kg/m3
2700
Table 6. Steel braiding physical properties.
Relative permeability
1000
Conductivity
6000000
S/m.
Thermal conductivity
85
W/K·m
11
Young's modulo
2·10
Poisson's ratio
0.3
N/m2
Coefficient of thermal expansion 0.000012 1/K
Specific density
7870
kg/m3
Table 7. Insulator physical properties.
Core
Inner
Outer
Relative permeability
1
1
1
Conductivity
0
0
0
Relative electric permitivity
2.5
2.5
2.5
Thermal conductivity
0.04
0.04
0.04
W/K·m
Young's modulo
10000000
10000000
10000000
N/m2
Poisson's ratio
0.3
0.3
0.3
Coefficient of thermal expansion 0.0001
0.0001
0.0001
1/K
Specific density
900
1050
kg/m3
900
S/m
3. Calculated cable parameters.
Cable physical parameters are presented in the next table.
Cable outer diameter is calculated using conductor and insulator geometrical parameters put into Table 1 and
Table 2. Cable linear weight per meter is calculated from geometrical parameters and specific densities of the
cable components. The whole cable specific density is a total density calculated by taking into account all cable
components.
Table 8. Cable physical parameters
Cable outer diameter
42.8
mm
Weight (per meter)
2.74
kg
Cable specific density 1.90e+03 kg/m2
"Conductors' capacitance" table holds self- and mutual-capacitances of the cable conductors. These values are
calculated in the QuickField electrostatics problem.
q1 = c11 * (U1 - 0) + c12 * (U1 - U2) + ... + c1n * (U1 - Un)
q2 = c21 * (U2 - U1) + c22 * (U2 - 0) + ... + c2n * (U2 - Un)
......
qn = cn1 * (Un - U1) + cn2 * (Un - U2) + ... + cnn * (Un - 0)
Table 9. Conductors' capacitance, pF/m (lumped capacitance)
Conductor1
Conductor2
Conductor3
Neutral cord
Conductor1 Conductor2 Conductor3 Null-cord
170
66.1
9.47
36.8
66.3
169
66.3
0.413
9.45
66.1
170
36.8
37.0
0.534
37.0
64.5
Conductors' inductances are represented in the Table 10. Values in the columns 2–5 are calculated in the
magnetostatic problem at the phase defined in the Table 4. Values in the columns 6–9 are calculated in AC
magnetic problem. All values are got using the flux linkage approach by the formula: Lij = j / Ii. The table diagonal
elements represent the self-inductance values.
Table 10. Conductors' inductance, uH/m
In magnetostatic problem
In AC magnetic problem
C-1
C-2
C-3
0-cord
C-1
C-2
C-3
0-cord
Conductor1
11.5
11.2
11.1
11.3
8.73
8.47
8.41
8.51
Conductor2
11.2
11.5
11.2
11.1
8.47
8.73
8.47
8.38
Conductor3
11.1
11.2
11.5
11.3
8.41
8.47
8.73
8.51
Neutral cord
11.3
11.1
11.3
117
8.51
8.38
8.51
8.87
Table 11 includes the impedance and impedance-like values. In the magnetostatics problem the conductor's
impedance (equal to the resistance) per meter is calculated by the formula: R = l / (ρ·S)
Joule heat per meter in magnetostatics problem is calculated by the formula: P = IA2 · R, where IA is the rootmean-square current and R is the conductor impedance.
The conductors' impedances in AC magnetics problem are calculated using the Ohm's law as a complex ratio of
the conductor's average potential divided by the conductor total current density. The real part of this ratio
represents the resistance, imaginary part — reactance and the modulus — impedance. The Joule heat in the AC
magnetic problem is calculated using the corresponding QuickField integral.
Table 11. Conductors' impedance.
In electrostatics problem
In AC magnetic problem
Conductors
Null cord
Conductor1
Conductor2
Conductor3
Impedance, ω/m
2.31e-04
7.94e-04
2.40e-04
2.55e-04
2.80e-04
Resistance, ω/m
2.31e-04
7.94e-04
2.15e-04
2.37e-04
2.59e-04
Reactance, ω/m
0.00
0.00
1.08e-04
9.41e-05
1.06e-04
Joule heat, W/m
4.63
0.00
4.71
4.74
4.71
The generated heat field is exported from the AC magnetics problem into the heat transfer problem. As a result of
QuickField simulation you can see the cable exterior surface average temperature, heat flow from the cable
surface and the average temperatures of all conductors. Average temperatures are relative numbers presented in
Celsius assumed that ambient space temperature is 20 °C.
Table 12. Cable heat parameters
Exterior surface average temperature
23.5
°C
Heat flow
14.2
W
Conductors average temperature, °C
Conductor1
Conductor2
45.9
46.8
Conductor3 Null-cord
45.9
39.3
Stress analysis problem is the utmost one, that imports the temperature field from the heat transfer problem and
the magnetic forces from the AC magnetic problem. Due to this magnetic and thermal loading the cable
components become deformed. The numerical values of these deformations are presented in the next table.
Table 13. Stress analysis problem results.
Maximal displacement
5.14e-02
Mm
Maximal Mohr criteria value 8.16e+07 N/m2
The strength value is important for the cable fault analysis.
Table 14. The strength.
Maximal peak strength value 8.78e+03 A/m
The "Strength" field is shown on a figure below as well as the "Total current density", "Energy density",
"Momentary flux density", "Temperature" and "Displacement" field pictures.
4. Field pictures.
Current density in the cable
Magnetic field energy distribution in the cable
Magnetic field strength distribution in the cable
Temperature distribution in the cable
Cable mechanical deformation
Download simulation files:
Student's version
Professional version
-
cable.zip (3131 nodes)
Exercise N 3. Proximity effect
Task
Find the current density distribution along the cross section of long parallel conductors. Two types of conductors
are analyzed: two copper rods and two steel tubes.
Experiment
Current density is defined by measuring the voltages on the conductor segments, the phase of current is
measured separately by the digital phasemeter.
Problem type
Linear plane-parallel problem of time-harmonic electromagnetic field.
Geometry
Due to problem symmetry only upper-right quarter aOb is defined, and at the axes of symmetry the boundary
conditions are set.
Pic.1. Copper rods
Pic.2. Steel tubes
Given data
Relative magnetic permeability of copper and air μ= 1.
Relative magnetic permeability of steel μ= 100.
Total current I=300 A. Frequency f=50 Hz.
Electric conductivity of copper σ=57000000 Sm/m.
Electric conductivity of steel σ=10000000 Sm/m.
Boundary conditions
At the horizontal axis of symmetry (line Oa) Ht=0. At the vertical axis of symmetry Ob Bn=0. Equation B=rot A in
the cylindrical coordinate system leads to A=const at the axis Ob. Field fades at the infinity so at the other
boundaries A=0.
Results
 Current density distribution along the line Oa for copper rods.
 Current density distribution along the steel tube perimeter (from the point e to point f clockwise).
Please, see problems lab3_Cu.zip and lab3_Fe.zip.
TECircuit2: Pulse transformer
This is an example of the pulse transformer simulation, performed with QuickField software.
Problem Type:
Plane problem of transient magnetics.
Geometry:
Given:
Magnetic permeability of the steel core μ = 500;
Conductivity of the steel core g = 0 S/m (stell core is laminated);
Magnetic permeability of the windings μ = 1;
Conductivity of the windings (copper) g = 56,000,000 S/m;
Number of turns of the primary winding w1 = 20;
Number of turns of the secondary winding w2 = 40;
Number of turns of the third winding w3 = 20;
Pulse voltage U1 = 0.5 V;
Impulse time t = 0.1 s.
Problem:
Square voltage impulse applied to the pulse transformer. Calculate the currents in the secondary winding.
Solution:
Due to model symmetry we leave the upper half of the transformer only. Therefore we should reduce the circuit
elements' values by two.
Results:
Download simulation files:
Student's version Professional version
tecircuit2_stud.zip
tecircuit2.zip
Relay dynamics simulation using Libre Office (Open Office) and Parametric Object
Interface
The relay consists of the solenoid with moving core, which disconnects the electric switch contacts when the
control current is running in the solenoid. The spring keeps the core in the pull out position with switch contacts
connected. When the current is turned on, the magnetic field acts on the ferromagnetic core, overcomes the
resistance of the spring, and pulls the core inside the solenoid to the pull in position where it is stopped by
damper ring, which absorbs the shock. Operating time of this relay and the plunger motion function should be
calculated.
To combine the electromagnetic field analysis with the moving core dynamics both LibreOffice and LabelMover
are used. Interaction between LabelMover and LibreOffice is performed using application programming interface
(API).
Problem Type:
Axisymmetric problem of DC magnetics.
Geometry:
Number of turns N = 2000;
Current I = 0.2 A;
Plunger pull out position xout = 10 mm.
Plunger pull in position xin = 6 mm.
Plunger weight m = 4.5 g;
Spring constant k = 4 N/m
Spring free position xspring.free = 15 mm.
Problem:
The electromagnetic and spring forces act on the plunger. Both forces depends on the plunger position. Calculate
plunger motion function.
Solution:
The multi-turn winding is replaced with the equivalent total current.
The motion function can be found from second-order differential equation
m · d2x/dt2 = f(x),
where m - is a plunger weight (kg),
x - is a plunger position (m)
f(x) - is the force acting on the plunger (N).
The force acting on the plunger is a sum of spring force fspring(x) = k·(xspring.free - x) and electromagnetic force.
The equations are solved in LibreOffice VBA. The dynamic link is used to invoke QuickField and calculate the
electromagnetic force at each step.
The calculations are stopped when x=xin (pull in position, plunger hits damper).
View LibreOffice Calc document in the separate window: relay_control_script.ods.
Results:
The plunger hits initial position between 11th and 12th steps (0.055 - 0.06 s).
Download simulation files.
Relay dynamics simulation using MATLAB and ActiveField (QuickField Object
Model)
The relay consists of the solenoid with moving core, which disconnects the electric switch contacts when the
control current is running in the solenoid. The spring keeps the core in the pull out position with switch contacts
connected. When the current is turned on, the magnetic field acts on the ferromagnetic core, overcomes the
resistance of the spring, and pulls the core inside the solenoid to the pull in position where it is stopped by
damper ring, which absorbs the shock. Operating time of this relay and the plunger motion function should be
calculated.
To combine the electromagnetic field analysis with the moving core dynamics both MATLAB and QuickField are
used. Interaction between QuickField and MATLAB is performed using ActiveField application programming
interface.
Problem Type:
Axisymmetric problem of DC magnetics.
Geometry:
Number of turns N = 2000;
Current I = 0.2 A;
Plunger pull out position xout = 10 mm.
Plunger pull in position xin = 6 mm.
Plunger weight m = 4.5 g;
Spring constant k = 4 N/m
Spring free position xspring.free = 15 mm.
Problem:
The electromagnetic and spring forces act on the plunger. Both forces depends on the plunger position. Calculate
plunger motion function.
Solution:
The multi-turn winding is replaced with the equivalent total current.
The motion function can be found from second-order differential equation
m · d2x/dt2 = f(x),
where m - is a plunger weight (kg),
x - is a plunger position (m)
f(x) - is the force acting on the plunger (N).
The force acting on the plunger is a sum of spring force fspring(x) = k·(xspring.free - x) and electromagnetic force.
The equations are solved in MATLAB. The dynamic link is used to invoke QuickField and calculate the
electromagnetic force at each step.
The calculations are stopped when x=xin (pull in position, plunger hits damper).
View MATLAB script in the separate window: relay_dynamics.m, getQfForceX.m.
Results:
The plunger hits initial position at 54 ms.
Download simulation files.
Relay dynamics simulation using Microsoft Excel and QuickField API
The relay consists of the solenoid with moving core, which disconnects the electric switch contacts when the
control current is running in the solenoid. The spring keeps the core in the pull out position with switch contacts
connected. When the current is turned on, the magnetic field acts on the ferromagnetic core, overcomes the
resistance of the spring, and pulls the core inside the solenoid to the pull in position where it is stopped by
damper ring, which absorbs the shock. Operating time of this relay and the plunger motion function should be
calculated.
To combine the electromagnetic field analysis with the moving core dynamics both Microsoft Office (Excel) and
QuickField are used. Interaction between QuickField and Microsoft Excel is performed using ActiveField
application programming interface.
Problem Type:
Axisymmetric problem of DC magnetics.
Geometry:
Number of turns N = 2000;
Current I = 0.2 A;
Plunger pull out position xout = 10 mm.
Plunger pull in position xin = 6 mm.
Plunger weight m = 4.5 g;
Spring constant k = 4 N/m
Spring free position xspring.free = 15 mm.
Problem:
The electromagnetic and spring forces act on the plunger. Both forces depends on the plunger position. Calculate
plunger motion function.
Solution:
The multi-turn winding is replaced with the equivalent total current.
The motion function can be found from second-order differential equation
m · d2x/dt2 = f(x),
where m - is a plunger weight (kg),
x - is a plunger position (m)
f(x) - is the force acting on the plunger (N).
The force acting on the plunger is a sum of spring force fspring(x) = k·(xspring.free - x) and electromagnetic force.
The equations are solved in Microsoft Excel VBA. The dynamic link is used to invoke QuickField and calculate the
electromagnetic force at each step.
The calculations are stopped when x=xin (pull in position, plunger hits damper).
View Microsoft Excel document in the separate window: relay_control_script.xls.
Results:
The plunger hits initial position (0.006 m) at 0.055 s.
Download simulation files.
Relay dynamics simulation using Microsoft VBScript and Parametric Object
Interface
The relay consists of the solenoid with moving core, which disconnects the electric switch contacts when the
control current is running in the solenoid. The spring keeps the core in the pull out position with switch contacts
connected. When the current is turned on, the magnetic field acts on the ferromagnetic core, overcomes the
resistance of the spring, and pulls the core inside the solenoid to the pull in position where it is stopped by
damper ring, which absorbs the shock. Operating time of this relay and the plunger motion function should be
calculated.
To combine the electromagnetic field analysis with the moving core dynamics both Microsoft VBScript and
LabelMover are used. Interaction between LabelMover and Microsoft VBScript is performed using application
programming interface (API).
Problem Type:
Axisymmetric problem of DC magnetics.
Geometry:
Number of turns N = 2000;
Current I = 0.2 A;
Plunger pull out position xout = 10 mm.
Plunger pull in position xin = 6 mm.
Plunger weight m = 4.5 g;
Spring constant k = 4 N/m
Spring free position xspring.free = 15 mm.
Problem:
The electromagnetic and spring forces act on the plunger. Both forces depends on the plunger position. Calculate
plunger motion function.
Solution:
The multi-turn winding is replaced with the equivalent total current.
The motion function can be found from second-order differential equation
m · d2x/dt2 = f(x),
where m - is a plunger weight (kg),
x - is a plunger position (m)
f(x) - is the force acting on the plunger (N).
The force acting on the plunger is a sum of spring force fspring(x) = k·(xspring.free - x) and electromagnetic force.
The equations are solved in Microsoft VBScript. The dynamic link is used to invoke LabelMover and calculate the
electromagnetic force at each step.
The calculations are stopped when x=xin (pull in position, plunger hits damper).
View VBScript file in the separate window: relay_control_script.vbs.
Results:
The plunger hits initial position between 11th and 12th steps (0.055 - 0.06 s).
Download simulation files.
Relay dynamics simulation using Tcl/Tk and Parametric command line
The relay consists of the solenoid with moving core, which disconnects the electric switch contacts when the
control current is running in the solenoid. The spring keeps the core in the pull out position with switch contacts
connected. When the current is turned on, the magnetic field acts on the ferromagnetic core, overcomes the
resistance of the spring, and pulls the core inside the solenoid to the pull in position where it is stopped by
damper ring, which absorbs the shock. Operating time of this relay and the plunger motion function should be
calculated.
To combine the electromagnetic field analysis with the moving core dynamics both Tcl/Tk and LabelMover are
used. Interaction between LabelMover and Tcl is performed using parametric command line interface.
Problem Type:
Axisymmetric problem of DC magnetics.
Geometry:
Number of turns N = 2000;
Current I = 0.2 A;
Plunger pull out position xout = 10 mm.
Plunger pull in position xin = 6 mm.
Plunger weight m = 4.5 g;
Spring constant k = 4 N/m
Spring free position xspring.free = 15 mm.
Problem:
The electromagnetic and spring forces act on the plunger. Both forces depends on the plunger position. Calculate
plunger motion function.
Solution:
The multi-turn winding is replaced with the equivalent total current.
The motion function can be found from second-order differential equation
m · d2x/dt2 = f(x),
where m - is a plunger weight (kg),
x - is a plunger position (m)
f(x) - is the force acting on the plunger (N).
The force acting on the plunger is a sum of spring force fspring(x) = k·(xspring.free - x) and electromagnetic force.
The equations are solved in Tcl. The dynamic link is used to invoke LabelMover and calculate the
electromagnetic force at each step.
The calculations are stopped when x=xin (pull in position, plunger hits damper).
View TCL file in the separate window: relay_control_script.tcl.
Results:
The plunger hits initial position (0.006 m) after 0.06 s. The plot was created using plotchart library.
Download simulation files.
Safety fuse
This is an example of the safety fuse simulation, performed with QuickField software.
Problem Type:
Plane problem of heat transfer.
Geometry:
All dimensions are in millimeters.
Given:
Conductor diameter: 2 mm;
Conductor melting point T = 1050 °C;
Nominal current I = 15 A.
Problem:
Determine the operating time at short-circuit current of 40 A.
Solution:
Download video.
View on-line.
Result:
Temperature distribution in safety fuse:
Download simulation files.
ACElec3: Slot Insulation
This is an example of the slot insulation simulation, performed with QuickField software.
Problem Type:
Plane problem of AC conduction.
Geometry:
Given:
Relative permittivity of coating ε = 4.
Relative permittivity of insulation ε = 3.
Relative permittivity of SiC lacquer ε = 4.
Conductivity of coating g = 10-4 S/m;
Conductivity of insulation g = 10-10 S/m;
Conductivity of SiC lacquer g = 10-8... 10-5 S/m;
Frequency f = 50 Hz;
Voltage Uf = 15000 V;
Breakdown voltage E = 25 kV/mm.
Problem:
In the overvoltage test the voltage U = 2Uf + 5 kV is applied. The field intensity along the surface of insulation
should be less then ES < 0.5 kV/mm. To reduce ES value the special semi-conducting SiC lacquer is applied. We
shold determine the distribution of the field intensity along the surface of isolation.
Results:
Download simulation files:
Student's version Professional version
acelec3_stud.zip
ACElec3.zip
Slot conductor
This is an example of the slot of an electric machine simulation, performed with QuickField software.
Problem Type:
Plane problem of heat transfer.
Geometry:
All dimensions are in millimeters.
Given:
Current dencity j=2 A/mm2.
Problem:
Determine the temperature of the slot.
Solution:
Download video.
View on-line.
Result:
Temperature distribution in slot conductor:
Download simulation files
HMagn1: Slot embedded conductor
This is an example of the slot embedded conductor simulation, performed with QuickField software.
Problem Type:
Plane problem of AC magnetics.
Geometry:
A solid copper conductor embedded in the slot of an electric machine carries a current I at a frequency f.
Given:
Magnetic permeability of air μ = 1;
Magnetic permeability of copper μ = 1;
Conductivity of copper σ = 58,005,000 S/m;
Current in the conductor I = 1 A;
Frequency f = 45 Hz.
Problem:
Determine current distribution within the conductor and complex impedance of the conductor.
Solution:
We assume that the steel slot is infinitely permeable and may be replaced with a Neumann boundary condition.
We also assume that the flux is contained within the slot, so we can put a Dirichlet boundary condition along the
top of the slot. See model file for the complete model.
The complex impedance per unit length of the conductor can be obtained from the equation
Z=V/I
where V is a voltage drop per unit length. This voltage drop on the conductor can be obtained in Local Values
mode of the postprocessing window, clicking an arbitrary point within the conductor.
Results
Current density in slot embedded conductor:
Re Z (Ohm/m) Im Z (Ohm/m)
Reference
0.00017555
0.00047113
QuickField
0.00017550
0.00047111
Reference
A. Konrad, "Integrodifferential Finite Element Formulation of Two-Dimensional Steady-State Skin Effect
Problems", IEEE Trans. Magnetics, Vol MAG-18, No. 1, January 1982.
Download simulation files:
Student's version Professional version
hmagn1_stud.zip
hmagn1.zip
Exercise N 1. Slot embedded conductor
Task
Draw a plot of current distribution within the conductor placed in the slot of electric motor.
Experiment
Current density is obtained by potential measuring along the part of conductor done by voltmeter. Phase is
determined by digital phasemeter.
Problem type
Plane-parallel time-harmonic linear magnetic problem.
Geometry
Problem area is to the right of symmetry axis (line ab), corresponding boundary conditions are set.
Given data
Relative permeability of steel μ=100.
Relative permeability of air, copper μ= 1.
Conductivity of copper σ=57000000 S/m .
The steel is laminated, so its conductivity along the bar is σ=0 S/m.
Frequency f=50 Hz.
The total current per the bar is 600 A, or 300 A if there are two bars in the slot.
Boundary conditions
At the vertical axis of symmetry (line ab) Ht=0.
The field fades within ferromagnetics, thus at outer boundaries the field is zero A=0.
Results
Vertical distribution of current density (phase=0) along the copper bar (L=0 at the bottom of the slot).
Laminated cores simulation
Laminated cores or laminations are used in the electrotechnical devices
of almost any type. The goal of lamination is to decrease the eddy
current and losses by splitting the ferromagnetic material into smaller
sections, insulated from each other electrically. Due to lamination the
amount of the flux conductive media (pure steel) is somewhat less then
total width of laminated core. It is taken into account by use of
lamination factor kst, which is equal to the ratio of the sum of pure steel
lengths to the design length of the laminated core.
kst = lst / ltotal
Electromagnetic simulations of the device or equipment, which has
laminated parts, may be performed using special approach, allowing
replacement of the layered materials in the laminations by homogenous
media with specially adjusted magnetic properties. Finite element mesh
density in the laminations may be decreased, which leads to decrease
of the computer resource requirements, and considerably increases the
speed of simulation of the device with laminations.
 In the linear problem involving laminated core you can simply
reduce the magnetic permeability of the core by kst factor:
μ1 = μ / kst.
More about laminated cores
simulation and design:
Saturable reactor.
Laminated cores simulation
and design in the recorded
webinar:
QuickField Analysis for Electric
machines design.
Additional links on this subject:
Magnetic core
from Wikipedia, the free
encyclopedia.
 In non-linear case it is necessary to define the magnetization curve. Instead of material original curve B(H)
the modified curve B1(H) should be used.
In the strong fields the difference between B and B1 becomes significant. For example, the induction B = 1.3 T
corresponds to magnetizing force H = 550 A/m (steel 2211). Taking into account lamination the actual induction is
B1 = 1.3/0.93 = 1.4 T.
This induction corresponds to the magnetizing force H1 = 1000 A/m, i.e. inductions difference 7% leads to
magnetizing force difference more than 80%. It is obvious that this difference will grow with saturation.
Download problem files.
Heat1: Slot of an electric machine
This is an example of the slot of an electric machine simulation, performed with QuickField software.
Problem Type:
Plane problem of heat transfer with convection.
Geometry:
All dimensions are in millimeters. Stator outer diameter is 690 mm. Domain is a 10-degree segment of stator
transverse section. Two armature bars laying in the slot release ohmic loss. Cooling is provided by convection to
the axial cooling duct and both surfaces of the core.
Given:
Specific copper loss: 360000 W/m3;
Thermal conductivity of steel: 25 W/K·m;
Thermal conductivity of copper: 380 W/K·m;
Thermal conductivity of insulation: 0.15 W/K·m;
Thermal conductivity of wedge: 0.25 W/K·m;
Inner stator surface:
Convection coefficient: 250 W/K·m2;
Temperature of contacting air: 40 °C.
Outer stator surface:
Convection coefficient: 70 W/K·m2;
Temperature of contacting air: 20 °C.
Cooling duct:
Convection coefficient: 150 W/K·m2;
Temperature of contacting air: 40 °C.
Problem:
Calculate temperature distribution in the stator tooth zone of power synchronous electric machine.
Results:
Temperature distribution in a slot of an electric machine:
Download simulation files:
Student's version Professional version
heat1_stud.zip
heat1.zip
Slot of electric machine
This is an example of the slot of electric machine simulation, performed with QuickField software.
Problem Type:
Plane problem of AC magnetics.
Geometry:
All dimensions are in millimeters.
Given:
Relative permeability of air μ = 1;
Relative permeability of steel μ = 1000;
Conductivity of copper g = 58·106 Sm/m;
Current I = 500 A;
Frequency f = 50 Hz.
Problem:
In this tutorial we will analyze the skin effect occurring at the industrial frequency 50 Hz. The resistance of the
conductor, carrying 500 A sinusoidal current, in the slot of the electric machine will be calculated.
Solution:
Download video.
View on-line.
Results:
Current density in slot of electric machine:
Download simulation files.
Magn2: Solenoid actuator
This is an example of the solenoid actuator simulation, performed with QuickField software.
A solenoid actuator is an example of electromechanical transducer, which transforms an input electric signal into
a motion. Solenoid actuator, considered here, consists of a coil enclosed into a ferromagnetic core with a moving
part - plunger inside. Actuator design requires calculation of the magnetic field and a force, applied to the plunger.
Problem type:
Axisymmetric problem of DC magnetics.
Geometry:
All dimensions are in centimeters.
Given:
Relative permeability of air and coil μ = 1;
Current density in the coil j = 1,000,000 A/m2;
The B-H curve for the core and the plunger:
H (A/m)
460
640
720
890
1280 1900 3400 6000
B (T)
0.80
0.95
1.00
1.10
1.25
1.40
1.55
1.65
Problem:
Obtain the magnetic field in the solenoid and a force applied to the plunger.
Solution:
This electromagnetic transducer magnetic system is almost closed, therefore outward boundary of the actuator
model can be put relatively close to the solenoid core. A thicker layer of the outside air is included into the
actuator model region at the plunger side, since the magnetic field in this area cannot be neglected.
Mesh density is chosen by default, but to improve the mesh distribution, three additional vertices are added to the
actuator model. We put one of these vertices at the coil inner surface next to the plunger corner, and two others
next to the corner of the core at the both sides of the plunger.
A contour for the force calculation encloses the plunger. It is put in the middle of the air gap between the plunger
and the core. While defining the contour of integration, use a strong zoom-in mode to avoid sticking the contour to
existing edges.
The calculated force applied to the plunger F = 374.1 N.
Comparison of results
Flux density in Z-direction in the plunger:
Maximum flux density in Z-direction in the plunger:
Bz (T)
Reference
0.933
QuickField 1.0183
Reference
D. F. Ostergaard, "Magnetics for static fields", ANSYS revision 4.3, Tutorials, 1987.
Download simulation files:
Student's version Professional version
magn2_stud.zip
magn2.zip
Solenoid actuator parametric analysis
This is an example of the solenoid actuator simulation, performed with QuickField software.
Problem Type:
Axisymmetric problem of DC magnetics.
Geometry:
Given:
Relative permeability of air and coil μ = 1;
Current density in the coil j = 4 A/mm2;
Core and plunger are made of non-linear ferromagnetic material:
Problem:
1. Calculate the force vs. position characteristic
2. Study the current and plunger position influence on the mechanical force
3. Maximize the force.
1. Calculate the force vs. position characteristic
The plunger position can vary from 0 to 10 cm. With LabelMover we perform automatic geometry modifications
with step of 1 cm. For each step the mechanical force is calculated.
View movie.
2. Study the current and plunger position influence on mechanical force
Nominal position of the plunger is 10+0.5 cm. Nominal current is 4 A/mm2+10%. With LabelMover Tolerance
Analysis we perform variations and calculate the force automatically.
View movie.
3. Maximize the force
We can maximize the force by increasing the core window size. Thus more coil turns can be added to provide
greater force. At the same time the core reluctance increases. Let's find the core window size that provides
maximal force.
This can be easily done with LabelMover Optimization. The optimized core window height should be 71.75 mm
(initially it was 40 mm).
View movie.
Download simulation files:
Student's version
Professional version
-
solenoid-actuator-parametric-analysis.zip
Steel keeper
This is an example of the slot of the steel keeper simulation, performed with QuickField software.
Problem Type:
Plane problem of DC magnetics.
Geometry:
All dimensions are in millimeters.
Given:
Relative permeability of air μ = 1;
Relative permeability of steel μ = 1000;
Relative permeability of magnet μ = 1;
Coercive force Hc = 500'000 A.
Problem:
Calculate the mechanical force acting on the steel yoke.
Solution:
Download video.
View on-line.
Results:
Flux density in steel keeper:
Download simulation files.
Stratified high voltage bushing
Example prepared by Dr. Wlodzimierz Kalat
Stratified high voltage bushing is composed of several insulation layers separated with very thin "floating"
conductors. Their constant but unknown potential results from the capacitive distribution of electrostatic field and
may be controlled by variation of their lengths, thickness and permittivity of the dielectric layers. From the
technological point of view only the lengths adjustment can be considered as reasonable way of getting the most
uniform distribution of radial component Er of electric strength. It assures the best utilization of insulating material
and moderates the field across the high voltage bushing.
In theory, the infinite number of free potential electrodes leads to the uniform and continuous distribution of
electric field across the bushing (E(r)=const) instead of logarithmic one. In reality, only the finite number of layers
can be considered, usually 10-12.
The main goal of the example is to model very simple (only 2 layers) high voltage bushing, find the E(r)
distribution and compare results with some mathematical formula. The condition of Ermax equality in every layer
(E1rmax = E2rmax) gives the possibility to find the length of "floating" conductor and their potential.
Problem type:
An axisymmetrical problem of free potential electrode in electrostatic field. Dirichlet boundary conditions with
given potential are placed at zero and high voltage electrodes.
Geometry of the high voltage bushing:
______________________________________zero potential tubular electrode
| |
air
| |<-- barrier (zero pot. electrode)
| |
|_|
/
\ <-- 2-nd ins. layer
/__2__\ <-- "floating" electrode
/
\ <-- 1-st ins. layer
_______________/____1____\______________ high voltage tubular electrode
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ axis of symmetry
All dimensions are in centimeters. The length (L) of "floating" electrode is to be calculated before the modelling of
the bushing arrangement.
- the radius of zero pot. electrode r3 = 30 cm
- the radius of the hole in the barrier r2 = 6 cm
- the radius of "floating" electrode r1 = 4 cm
- the radius of high voltage electrode r0 = 2 cm
Given:
The relative permittivity of dielectric material (epoxy resin) is assumed as 5.0 and high voltage potential - 110 kV.
In order to be closer with mathematical description of the model, the slope of dielectric edge was considered as
perpendicular to the axis of symmetry.
Results of field calculations:
1. Three capacities: C1, C2 - of the layers and C10 - the free potential electrode-to-ground are calculated by
means of integrals (stored energy or surface charge). The condition of charge balance Q1 = Q2 + Q10
leads to the equation for determining the length (L) of free potential electrode.
2.
3.
4.
____________________________________zero potential tubular electrode
|
| |
|
| |
5.
|_____| |
6.
| | | === C10
7.
C2-->| === | |
8.
__|__|__|__|
9.
|
|
|
10.
|
=== C1 |
11. __________|_____|_____|_____________high voltage tubular electrode
12.
13. _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ axis of symmetry
14. The max. value of E(r) along the middle-cross section of the bushing is found from the "XY-plot" option. It
is recommended to notice the E1rmax and E2rmax equality.
15. The potential U of "floating" electrode is found by means of "Average surface potential" calculation or by
"Local Value" tool.
C10[pF]
C1[pF]
C2[pF]
Er1[kV/cm]
Er2[kV/cm]
U[kV]
Theory
3.84
84.64
68.51
36.57
36.57
59.31
QF_1
3.97
81.37
70.10
30.97
31.16
57.51
QF_2
3.52
79.68
75.90
-
-
-
Remarks:
1. Two ways of determining the capacity value in QuickField were applied. QF_1: from the energy of electrical field (2·W / U2), QF_2
from the total charge at the electrode (Q/U).
2. The difference with theoretical results is caused mainly by limited number of nodes of the Student's QF version.
Student's version
-
Professional version
high_voltage_bushing_stud.zip
Stress control tube for cable termination
This is an example of the cable termination simulation, performed with QuickField software.
Problem Type:
Axisymmetric problem of transient electric.
Geometry:
Given:
Relative permittivity of air ε = 1.
Relative permittivity of insulation ε = 2.
Relative permittivity of shield ε = 1.2.
Relative permittivity of stress control tube - nonlinear.
Voltage U = 4 kV.
Problem:
Calculate the effect of electric stress reduction.
Results:
With stress control tube the maximal electric stress reduces from 1.6 to 0.5 kV/mm.
Download simulation files:
Student's version Professional version
telec3_stud.zip
telec3.zip
Circuit1: Symmetric double line of conductors
This is an example of the symmetric double line of conductors simulation, performed with QuickField software.
Problem Type:
Plane problem of AC magnetics.
Geometry:
Two copper conductors with square cross-sections carrying opposingly directed currents of equal magnitudes are
contained inside the rectangular ferromagnetic coating. All dimensions are in millimeters.
Given:
The same as in the HMagn2 example. The current source is specified by the circuit.
Problem:
The current distribution within the conductors and the coating, complex impedance of the line, and power losses
in the coating should be determined.
Results:
Parameter
Circuit1
Hmagn2
Impedance, Ohm/m 0.000484 + i 0.000736 0.000484 + i 0.000736
Power loss, W/m
0.0000427
0.0000427
Download simulation files:
Student's version Professional version
circuit1_stud.zip
Circuit1.zip
HMagn2: Symmetric double line of conductors
This is an example of the symmetric double line of conductors simulation, performed with QuickField software.
Problem Type:
Plane problem of AC magnetics.
Geometry:
Two copper square cross-section conductors with equal but opposite currents are contained inside rectangular
ferromagnetic coating. All dimensions are in millimeters.
Given:
Magnetic permeability of air μ = 1;
Magnetic permeability of copper μ = 1;
Conductivity of copper σ = 56,000,000 S/m;
Magnetic permeability of coating μ = 100;
Conductivity of coating σ = 10,000,000 S/m;
Current in the conductors I = 1 A;
Frequency f = 100 Hz.
Problem:
Determine current distribution within the conductors and the coating, complex impedance of the line, and power
losses in the coating.
Solution:
We assume that the flux is contained within the coating, so we can put a Dirichlet boundary condition on the outer
surface of the coating.
The complex impedance per unit length of the line can be obtained from the equation
Z = ( V 1 - V2 ) / I
where V1 and V2 are voltage drops per unit length in each conductor. These voltage drops are equal with opposite
signs due to the symmetry of the model. To obtain a voltage drop, switch to Local Values mode in postprocessing
window, and then pick an arbitrary point within a conductor.
The impedance of the line Z = 0.000484 + i 0.000736 Ohm/m.
To obtain power losses in the coating:
1. In the postprocessing mode, choose Pick Elements and pick the coating block to create the contour.
2. Choose Integral Values and select Joule heat from the list of integral quantities and choose Calculate.
The power losses in the coating of the double line P = 0.0000427 W/m.
Current density in symmetric double line of conductors:
Download simulation files:
Student's version Professional version
hmagn2_stud.zip
hmagn2.zip
Coupl3: Temperature distribution in an electric wire
Calculate the temperature distribution in a long current carrying wire.
Problem Type:
Axisymmetric problem of electro-thermal coupling.
Geometry:
Given:
Wire diameter d = 10 mm;
Resistance ρ = 3·10-4 Ohm/m;
Electric current I = 1000 A;
Thermal conductivity λ = 20 W/K·m;
Convection coefficient α = 800 W/K·m2;
Ambient temperature To = 20 °C.
Problem:
Calculate the temperature distribution in the wire.
Solution:
We arbitrary chose a 10 mm piece of wire to be represented by the model. For data input we need the wire
diameter d=10 mm, and the resistivity of material: R=ρ·πd2/4 Ohm·m, and voltage drop for our 10 mm piece of
the wire: ΔU=I·R·L = 3·10-3 V.
For the conduction problem we specify two different voltages at two sections of the wire, and a zero current
condition at its surface. For heat transfer problem we specify zero flux conditions at the sections of the wire and a
convection boundary condition at its surface.
Comparison of results
Temperature distribution in an electric wire:
Center line temperature:
T (°C)
Theory
33.13
QuickField 33.14
The Coupl3CF.pbm and Coupl3HT.pbm problems are corresponding DC conduction and heat transfer parts of
this problem.
Reference
W. Rohsenow and H. Y. Choi, "Heat, Mass, and Momentum Transfer", Prentice-Hall, N.J., 1963.
Download simulation files:
Student's version Professional version
coupl3_stud.zip
coupl3.zip
Coupl5: Temperature distribution in the conducting sheet
The voltage is applied to the sides of conducting sheet placed vertically and surrounded by the still air. The
flowing current heats the sheet due to resistive losses. The front and back surfaces of the sheet are cooled by the
air (natural convection).
Problem Type:
A plane-parallel problem of electro-thermal coupling.
Geometry:
Given:
Data for magnetic analysis:
Sheet thickness d = 1 mm;
Material resistance ρ = 10-7 Ohm/m;
Voltage applied U = 0.02 V;
Material heat conductivity λ = 380 W/K·m;
Convection coefficient α = 10 W/K·m2;
Ambient air temperature T0 = 20°C.
Problem:
Calculate the current and temperature distribution in a conducting sheet.
Solution:
The resistive losses are calculated in the DC conduction problem. Then these losses are transferred to the linked
heat transfer problem.
The side surfaces area is much smaller then the sum of face surface areas. We can ignore the convection from
side surfaces. For simplicity, let's assume that the convection coefficient is constant across the vertically
positioned flat sheet.
The convection from the faces is modelled by the heat sink Q(T) = -λ·T, where λ - convection coefficient. We
should take into account the convection from the back face of the sheet. Both front and back vertical sides are
washed by the air and subjected to the same cooling conditions. Thus we should multiply the convection
coefficient by two.
The losses in the sheet are calculated per 1 meter of depth, which is 1000 times greater than the losses in the
real sheet. Therefore we should increase the convection coefficient l by the 1000.
Results:
Current distribution in the conducting sheet
Temperature distribution in the conducting sheet
The Coupl5CF.pbm is the problem of calculating the current distribution in the sheet, and Coupl5HT.pbm
analyzes temperature field.
Student's version
-
Professional version
Coupl5.zip (2488 nodes)
Mesh size limitations of the Student's version do not affect its ability of analysing the results of any large problem defined and
solved with the Professional version. To view the results with Students' version:
1. Open problem file (*.pbm)
2. Use Problem>View Field Picture command or click the View Field Picture button.
Temperature field inside the crucible stove
By Dariusz Czerwinski
Problem Type:
Plane problem of heat transfer.
Geometry:
Stove height = 740 mm;
Stove width = 800 mm.
Given:
Temperature of heat source: 700°C;
Temperature of outer surface: 50°C;
Thermal conductivity of mineral wool: 0.0583 W/K·m;
Thermal conductivity of glass tape: 0.058 W/K·m;
Thermal conductivity of fire-clay: 0.837+0.582·t W/K·m;
Problem:
Calculate temperature distribution in the crucible stove.
Results:
Temperature distribution in the crucible stove.
Download simulation files:
Student's version Professional version
-
crucible_stove.zip
THeat2: Temperature response of a suddenly cooled wire
Determine the temperature response of a copper wire of diameter d, originally at temperature T0, when suddenly
immersed in air at temperature Ti. The convection coefficient between the wire and the air is α.
Problem Type:
Plane problem of heat transfer.
Geometry:
Given:
d = 0.015625 in;
Ti = 37.77°C, T0 = 148.88°C;
C = 380.16 J/kg·K, ρ = 8966.04 kg/m3;
α = 11.37 W/K·m2.
Problem:
Determine the temperature in the wire.
Solution:
The final time of 180 s is sufficient for the theoretical response comparison. A time step of 4.5 s is used.
Comparison of Results
Temperature in wire:
Temperature, °C
Time QuickField ANSYS Reference
45 s
91.37
91.38
89.6
117 s
54.46
54.47
53.33
180 s
43.79
43.79
43.17
Reference
Kreif F., "Principles of Heat Transfer", International Textbook Co., Scranton, Pennsylvania, 2nd Printing, 1959,
Page 120, Example 4-1.
Download simulation files:
Student's version Professional version
theat2_stud.zip
theat2.zip
Dirich1: Time- and Coordinate-Dependent Boundary Condition
Conductive cylinder in rotating magnetic field.
Problem type:
Plane problem of transient magnetics.
Geometry:
Given:
Relative magnetic permeability of air μ = 1;
Relative magnetic permeability of conductor μ = 1;
Conductivity of conductor σ = 6.3·107 S/m;
Magnitude of external field B0 = 1 T;
Number of poles 2p = 6;
Frequency f = 50 Hz.
Solution:
To specify rotating magnetic field on the outer boundary of the region, B n = B0 sin (ωt - pφ), we apply the Dirichlet
boundary condition, using the formula: A = cos (18000*t - 3*atan2 (y/x)) / 60
The coefficient A0 = 1/60 arises from consideration
Bn = (1/r)(∂A/∂φ) = A0p·sin(ωt - pφ) / r
and
A0 = B0·r/p
Due to periodicity of the problem, only half of the model is presented, and odd periodic boundary condition A1 = A2 is applied on the cut. In fact, it would be enough to simulate just 60° sector of the model. In time domain,
problem is simulated with automatic adaptive time step, up to 0.05 seconds (approx. 2.5 periods).
Results:
t = 0.0002 sec:
t = 0.048 sec:
t = 0.05 sec:
Download simulation files:
Student's version Professional version
dirich1_stud.zip
dirich1.zip
Coupl4: Tokamak solenoid
The central solenoid of the ohmic heating system for a TOKAMAK fusion device.
Problem Type:
Axisymmetric problem of magneto-structural coupling.
Geometry:
The solenoid consists of 80 superconducting coils fixed in common plastic structure. Due to mirror symmetry one
half of the structure is modeled.
Given:
Data for magnetic analysis:
Current density in coils j = 3·108 A/m2;
Magnetic permittivity of plastic, coils and liquid helium inside coils μ = 1.
Data for stress analysis:
Copper of coils:
Young's modulus E = 7.74·1010 N/m2;
Poisson's ratio ν = 0.335;
Maximum allowable stress: 2.2·108 N/m2.
Plastic structure:
Young's modulus E = 2·1011 N/m2;
Poisson's ratio ν = 0.35;
Maximum allowable stress: 109 N/m2.
The Coupl4MS.pbm is the problem of calculating the magnetic field generated by the solenoid, and
Coupl4SA.pbm analyzes stresses and deformations in coils and plastic structure due to Lorentz forces acting on
the coils.
Results:
Magnetic flux density destribution in tokamak solenoid:
Mechanical stress distribution in tokamak solenoid:
Download simulation files:
Student's version Professional version
-
Coupl4.zip
TEMagn1: Transient eddy currents in a semi-infinite solid
This is an example of the transient eddy currents simulation, performed with QuickField software.
Problem Type:
Plane problem of transient magnetics.
Geometry:
The surface of semi-infinite solid plate is suddenly subjected to a constant magnetic potential A0.
Given:
Magnetic permeability of material μ=1;
Conductivity of material g=2,500,000 S/m;
Loading A0=2 Wb/m.
Problem:
Determine current distribution within the conductor.
Solution:
Flux density destribution in a semi-infinite solid body:
Graph of flux density destribution in a semi-infinite solid body:
The model length 20 m is arbitrarily selected such that no significant potential change occurs at the end points for
the time period of interest. The final time of 0.25 s is sufficient for the theoretical response comparison. A time
step of 0.005 s is used.
Comparison of results
Time t = 0.15 s
Coordinate x,m QuickField ANSYS Theory
Vector Potential A, Wb/m
0.2517
0.822
0.831
0.831
0.4547
0.280
0.278
0.282
0.6914
0.053
0.044
0.05
Flux Density B, T
0.2517
3.692
3.687
3.707
0.4547
1.716
1.794
1.749
0.6914
0.418
0.454
0.422
Eddy Current Density j, A/mm2
0.2517
-8.06
-7.80
-7.77
0.4547
-6.56
-6.77
-6.63
0.6914
-2.34
-2.45
-2.43
Download simulation files:
Student's version Professional version
temagn2_stud.zip
temagn2.zip
TEMagn2: Transient eddy currents in a two-wire line
This is an example of the two-wire line simulation, performed with QuickField software.
Problem Type:
Plane problem of transient magnetics.
Geometry:
Transmission line consists of two copper conductors with equal but opposite currents. All dimensions are in
millimeters.
Given:
Magnetic permeability of air μ = 1;
Magnetic permeability of copper μ = 1;
Conductivity of copper g = 56,000,000 S/m;
Voltage applied U = 0.001 V;
Problem:
Calculate the transient currents within the conductors.
Solution:
The resistance of one conductor can be calculated as Rcond = l / (g · S),
where
S = πr2 - cross-section area of conductor,
r - radius of conductor,
l - length of the line.
Rcond = 1 / (56·106 · (π·0.00012)) = 0.5684 Ohm
The resistance of both conductors is R = 2·Rcond = 2·0.5684 = 1.1368 Ohm
The inductance of the transmission line can be calculated as L = μ0·l/π · (ln(D/r) + 0.25), where
D - distance between conductors.
L = 4π·10-7·1/π · (ln(0.02/0.0001) + 0.25) = 2.219·10-6 Hn
The transient current for equivalent electric circuit with lumped parameter is described by the
formula
I(t) = U/R · (1 - e—t/T), where T=L/R - characteristic time of the circuit.
Results:
Download simulation files:
Student's version Professional version
temagn2_stud.zip
temagn2.zip
THeat3: Transient temperature distribution in an orthotropic metal bar
A long metal bar of rectangular cross-section is initially at a temperature T0 and is then suddenly quenched in a
large volume of fluid at temperature Ti. The material conductivity is orthotropic, having different X and Y
directional properties. The surface convection coefficient between the wire and the air is α.
Problem Type:
Plane problem of heat transfer.
Geometry:
Given:
a = 2 in, b = 1 in
λx = 34.6147 W/K·m, λy = 6.2369 W/K·m;
Ti = 37.78°C, T0 = 260°C;
a = 1361.7 W/K·m2;
C = 37.688 J/kg·K, ρ = 6407.04 kg/m3.
Problem:
Determine the temperature distribution in the slab after 3 seconds at the center, corner edge and face centers of
the bar.
Solution:
To set the non-zero initial temperature we have to solve an auxiliary steady state problem, whose solution is
uniform distribution of the temperature T0
A time step of 0.1 sec is used.
Comparison of results
Temperature distribution in an orthotropic metal bar:
Temperature, °C
Point
QuickField ANSYS Reference
(0,0) in
238.7
239.4
237.2
(2,1) in
66.43
67.78
66.1
(2,0) in
141.2
140.6
137.2
(0,1) in
93.8
93.3
94.4
Reference
Schneider P.J., "Conduction Heat Transfer", Addison-Wesley Publishing Co., Inc, Reading, Mass., 2nd Printing,
1957, Page 261, Example 10-7.
Download simulation files:
Student's version Professional version
theat3_stud.zip
theat3.zip
Transmission line attenuation constant
This example prepared by Elena Bozhevolnaya, Development Department,
LK A/S Industriparken 32 2750 Ballerup, Denmark
Problem Type:
A plane-parallel problems of AC magnetics and electrostatics.
Geometry:
The shielded transmission line is considered. The line consists of 2 copper strip-like conductors (block labels Elec1 and Elec 2) that are rested on the polyethylene substrate (Diel). The whole structure, that includes partly an
air, is protected by a screen (Shield) of the complicated geometry.
Solution:
Attenuation constant of the shielded microstrip-like transmission line is obtained on the basis of the electrostatics
and time-frequency analysis. Problem is solved by use of two QuickField models (transmission line electrostatics
and transmission line ac magnetics).
Voltage distribution in transmission line:
See detailed description in Attenuation constant of the shielded microstrip-like transmission line (in PDF
format).
Student's version
-
Professional version
transmission line electrostatics
transmission line ac magnetics
Student's version mesh size limitations do not affect to its ability of analysing the results of any complicated problem defined
and solved with the Professional version. To view the results with Students' version:
1. Open problem file (*.pbm)
2. Use Problem>View Field Picture command or click the View Field Picture button.
Transmission line capacitance
3 phase transmission line above earth.
Problem Type:
Plane problem of electrostatics.
Geometry:
d = 100 mm, h = 5 mm.
Model depth l = 4 km.
Given:
Relative permittivity of air ε = 1,
Solution:
Transmission line of 3 conductors features self- and mutual- partial capacitances. It is convenient to use
capacitance matrix calculator addin to find all the partial capacitances.
Results:
Electric field distribution around the transmission line.
C, nF
Phase A
Phase A Phase B Phase C
13
3
Phase B
3
12.6
2.9
Phase C
2.9
2.9
12.9
Download simulation files.
2.9
Elec2: Two conductors transmission line
This is an example of the two conductors transmission line simulation, performed with QuickField software.
Problem Type:
Plane problem of electrostatics.
Geometry:
The problem's region is bounded by ground from the bottom side and extended to infinity on other three sides.
Model depth l = 1 m.
Given:
Relative permittivity of air ε= 1;
Relative permittivity of dielectric ε= 2.
Problem:
Determine self and mutual capacitance of conductors.
Solution:
To avoid the influence of outer boundaries, we'll define the region as a rectangle large enough to neglect side
effects. To calculate the capacitance matrix we set the voltage U = 1 V on one conductor and U = 0 on the
another one.
Self capacitance: C11 = C22 = Q1 / U1 ,
Mutual capacitance: C12 = C21 = Q2 / U1 ,
where charge Q1 and Q2 are evaluated on rectangular contours around conductor 1 and 2 away from their edges.
We chose the contours for the C11 and C12 calculation to be rectangles [-6<x<0], [0<y<4] and [0 <x<6], [0<y<4]
respectively.
Comparison of results
Potential distribution in two conductors transmission line:
C11, F
Reference
C12, F
9.23·10-11 -8.50·10-12
QuickField 9.43·10-11 -8.57·10-12
Reference
A. Khebir, A. B. Kouki, and R. Mittra, "An Absorbing Boundary Condition for Quasi-TEM Analysis of Microwave
Transmission Lines via the Finite Element Method", Journal of Electromagnetic Waves and Applications, 1990.
Download simulation files:
Student's version Professional version
elec2_stud.zip
elec2.zip
Two intersecting conducting planes electric field
Two conducting planes intersect at an angle β = 270°. The planes are assumed to be held at potential V. Find the
electric field distribution near the corner.
Problem Type:
Plane problem of electrostatics.
Geometry:
Given:
Relative permittivity of vacuum ε = 1;
Volgate V = 100 V.
Solution:
Open space problem is converted to the closed region problem using the "zero potential" boundary far away from
the studying object.
Comparison of results
Electric field strength distribution along the plane is shown in the table. The theoreticsl result is: E ~ r -1/3.
r, m QuickField
Theory
(trend 6.3*r -1/3)
0.1
13.46 V/m
13.57 V/m
0.2
10.69 V/m
10.77 V/m
0.5
7.89 V/m
7.94 V/m
1
6.3 V/m
6.3 V/m
1.5
5.52 V/m
5.50 V/m
2
5.03 V/m
5.00 V/m
3
4.43 V/m
4.37 V/m
Reference
J.D.Jackson, Classical Electrodynamics, 3rd edition, Sect. 2.11.
Download simulation files:
Student's version
Professional version
intersecting_planes_corner_stud.zip intersecting_planes_corner.zip
Circuit 2: Welding transformer
This is an example of the welding transformer simulation, performed with QuickField software.
Problem Type:
Plane problem of AC magnetics.
Geometry:
Given:
R1 = 1000 Ohm;
C1 = 1 nF;
Frequency f = 60 Hz;
Number of turns = 220;
Problem:
Calculate the output U(I) chart of the welding transformer.
Results:
R, Ohm
U, V
I, A
infinity
100
0
20
95.6
4.78
10
88
8.8
5
70.7
14.14
2
38.8
19.4
1
21
21
0
0
21.9
Download simulation files:
Student's version Professional version
circuit2_stud.zip
circuit2.zip
Wire ring capacitance
Problem Type:
Axisymmetric problem of electrostatics.
Geometry:
R = 100 mm, a = 8.74 mm.
Given:
Relative permittivity of vacuum ε = 1,
The charge q = 10-9 C
Task:
Find the wire ring capacitance and compare it with analytical solution:
C = 4πεε0 · πR / ln(8R/a), [F]. *
Solution:
The capacitance can be calculated as C = q /U, where U is a body potential and q is a body charge. Conductor's
surface is marked as 'floating conductor', i.e. isolated conductor with unknown potential. At some point on
conductor's surface the charge is applied. The charge is then redistributed along the conductor surface
automatically.
It is supposed that field fades to zero far away from the ring. The calculation area is limited by the outer boundary,
where zero potential is applied.
Results:
Potential distribution.
As the potential of the body is calculated with respect to the zero potential of the 'infinity', the result depends on
the distance to the outer boundary. The set of simulations where made using automation tool LabelMover.
Distance to the outer boundary Voltage, V Capacitance, pF
10*R
119.2
8.39
20*R
123.7
8.08
40*R
125.9
7.94
80*R
127.0
7.87
160*R
127.6
7.84
320*R
127.8
7.82
Theory
7.74
Download simulation files.
*"The Capacitance of an Anchor Ring" by Thomas, T. S. E, Australian Journal of Physics, vol. 7, p.347.
ZnO lighting arrester
This is an example of the ZnO arrester simulation, performed with QuickField software.
Problem Type:
Axisymmetric problem of transient electric.
Geometry:
The arrester consists of ZnO tablet placed inside ceramic tube. Electrodes are connected to the end tablets.
Given:
Relative permittivity of the air ε = 1.
Relative permittivity of the ceramic ε = 2.
Relative permittivity of ZnO element ε = 60.
Conductivity of ZnO element - nonlinear.
Nominal voltage U0 = 35 kV.
Maximal current Imax = 520 A.
Surge peak voltage Umax = 200 kV.
Problem:
Calculate the varistor current.
Solution:
The arrester can be represented by the equivalent schema with resistor R and capacitance C connected in
parallel.
At nominal voltage the arrester acts as a capacitance (IC>IR). At overvoltage (surge) IR prevails and varistor acts
like a resistor. QuickField can calculate both active (IR) and reactive (IC) currents.
Results:
Voltage Active current
35 kV
0.67 mA
197 kV
433 A
Surge impulse passing through the arrester
Download video.
View on-line.
Download simulation files:
Student's version Professional version
telec2_stud.zip
telec2.zip
Long solenoid inductance
Problem type:
Axisymmetric problem of DC magnetics.
Geometry:
R1 = 30 mm, R2 = 35 mm.
Given:
Current in the conductor i = 10 A;
Number of turns N = 100,
(number of turns per unit length n = N/l = 200 m-1);
Problem:
Calculate inductance of the long solenoid.
Solution:
Inductance of the long solenoid can be obtained from the equation:
L = N 2 · μ0 A / l = N n · μ0 A,
where A is a cross-section area of the core (m2).
To calculate the inductance in QuickField you should divide magnetic flux by the source current.
L=N·Φ/i
Results:
Inductance L, μH/m
QuickField
Theory
Δ, %
83.55
83.40
0.18%
Download simulation files.
References:
 John Bird, "Electrical circuit theory and technology", p.77. ISBN-13: 978 0 7506 8139 1.