A numerical problem wherein the total inductance of an electromechanical energy conversion device is calculated furthermore the effect of changing the airgap length on the static force is also observed

1. Electrical Machines Assignment
Hassan Tirmizi
Department of Electrical Engineering
Polytechnical Institute of Coimbra
Coimbra, Portugal
Email: hassantirmizi100@gmail.com
Arpan Koirala
Department of Electrical Engineering
Polytechnical Institute of Coimbra
Coimbra, Portugal
Email: Arpankoirala@gmail.com
Shruti Singh
Department of Electrical Engineering
Polytechnical Institute of Coimbra
Coimbra, Portugal
Email: Shruti8singh@gmail.com
Abstract—Electromechanical energy conversion can take place
by using magnetic field as the coupling medium. When a current
carrying conductor is placed in a magnetic field it experiences
a magnetic force that tends to move it. In this report we will
analyze an electromechanical system and find the DC excitation
current required to attract an armature at a distance of 10mm.
We will also find out the inductance of the electromechanical
system by finding out the total reluctance. Moreover we will also
find out the voltage induced in the case of AC excitation. Lastly
we will plot the magnetic force as a function of the air-gap length
and compare the analytical results with the experimental ones.
I. INTRODUCTION
Magnetic materials have the property of concentrating flux.
When a coil is wound on a magnetic material and is excited
by a current, the magnetomotive force sets up a flux in the
core which is opposed by the total reluctance of the system.
F = Ni = φRt (1)
The flux linkage is defined as the total flux linking the coil
and it is directly propotional to current.
Nφ = λ = Li (2)
where L is the total inductance of the system.
From (1) and (2) we can find out the total inductance of the
system as a function of the parameters of the electromechan-
ical system
L =
N2
Rt
(3)
[?]
Fig. 1. Various components of a magnetic circuit[?]
A. Magnetic force and field energy
When a source is connected to a coil wound on the core, the
incremental electrical energy input is stored as an incremental
field energy.
eidt = dWfld (4)
and recalling that the voltage is the rate of change of flux we
can find out the energy stored in the field.
Wfld =
λ
0
idλ (5)
We can also express the current in terms of an inductance that
depends on the airgap length.
Wfld =
λ
0
λ
L(x)
dλ
Wfld =
λ2
2L(x)
(6)
The mechanical force can be calculated by taking the deriva-
tive of the energy function
fm = −
∂
∂x
λ2
2L(x) λ=constant
(7)
fm =
1
2
i2 dL(x)
dx
(8)
[?]
B. Magnetic force in terms of the flux
The ampere turns are equal to the magnetic field intensity
times the total length of the magnetic circuit
Ni = Hl
dφ = AdB
from (5) we have
Wfld = N
φ
0
idφ =
B
0
HlAdB (9)
Al
B
0
HdB (10)

2. writing the magnetic field intensity in terms of the flux density
in the air gap we have
Wfld = volume of the airgap ∗
B2
2µo
(11)
Mechanical force in terms of the energy function is given by
fm = −
∂
∂g
B2
∗ A ∗ 2g
2µo
(12)
fm = −
B2
∗ A
µo
(13)
where A is the total cross area of one pole. In terms of the
flux we can write
fm =
φ2
A ∗ µo
(14)
[?]
II. UNDERSTANDING THE PROBLEM
The electromechanical system that we are analyzing has an
E-shaped core that attracts an armature at a distance of 10mm.
The other specifications that are given to us are as follows
B ≤ 1.2T
weight of the armature is
W = 1kg
N = 1798
d = 0.6mm
Coil plastic housing is given by
a1 = 41mm
a2 = 43mm
b1 = 13mm
length of the mean turn is given by
lm = 2(a1 + a2) + πb1
Fig. 2. Magnetic Circuit under test
III. THEORETICAL FORMULATION
In order to find the various parameters for this circuit such as
the inductance or exciting current required for varying airgaps,
it is essential to find out the the total reluctance of the circuit.
Since the reluctance can be found out from the physical
parameters of the circuit we thereby proceed to find out the
reluctances around the entire periphery of the circuit inorder
to formulate an equivalent electric circuit.
The lengths of the various parts of the circuit are as follows
1) length of the right core limb = lrc = 63mm
2) vertical length of the bottom right armature = lar = 9mm
3) vertical length of the armature = lar = lal = 9mm
4) vertical length of bottom left armature = lal = 9mm
5) vertical length of core center = lccv = 63mm
6) Airgap length = lgl = lgc = lgr = 10mm
while the area of the center limb and the outer limbs are as
follows
1) Area of the center
Ac = depth ∗ width of central limb
Ac = 0.038 ∗ 0.036m2
2) Area of the outer limbs
Ao = depth ∗ width of outer limbs
Ao = 0.018 ∗ 0.038m2
Using the following formulas the reluctances can be calculated
for each section of the circuit.
Rcl =
llc ∗ ν
Ao
= 6578At/Wb
Rcr =
lrc ∗ ν
Ao
= 6578At/Wb
Rccv =
lccv ∗ ν
Ac
= 4605.2At/Wb
Rlcv =
llcv ∗ ν
Ao
= 9210At/Wb
Rrcv =
lrcv ∗ ν
Ao
= 9210At/Wb
Ral =
lal ∗ ν
Ao
= 1315At/Wb
Rar =
lar ∗ ν
Ao
= 1315At/Wb
Rac =
lac ∗ ν
Ac
= 657.89At/Wb
Rgl =
llc
µo ∗ Ao
= 1164003400 ∗ xAt/Wb
Rgr =
llc
µo ∗ Ao
= 582001700 ∗ xAt/Wb
Rgc =
llc
µo ∗ Ao
= 582001700 ∗ xAt/Wb

3. Fig. 3. electrical equivalent of the circuit
where x is the airgap length in meters and ν is the reluctivity
in the linear region given to be 100mH−1
The total reluctance
can be viewed as being comprising of a constant part and a
variable one depending on the airgap length.
Rt = Rc + Rx
Rt = 13809 + 1164003400x
A. Finding the excitation current
For the first part of the problem the airgap length has been
fixed to 10mm. We can find out the total flux passing through
the circuit by using equ(14). The weight of the armature has
been given as 1kg so we can conclude that a force of 9.8N or
greater is required to lift the armature.
φ =
√
9.8 ∗ 0.036 ∗ 0.038 ∗ 4π ∗ 10−7
φ = 1.29 ∗ 10−4
Wb
Finding the total reluctance for an airgap of 10mm we get
Rt = 13089 + 11640034 = 11653123At/Wb
From equ(1) we can find the excitation current
i =
1.29 ∗ 10−4
∗ 11653123
1798
i = 0.836A
B. Inductance
Now that we have found the total reluctance we can find
out the inductance using equ(3)
L =
17982
11653123
L = 0.277H
C. AC Excitation
In this case the coil is excited by a 50Hz AC source and
we can assume the dc current found out in the first case to be
equal to the rms current.
irms = 0.836A
As we have the inductance already so now we can find out
the inductive reactive for a 50Hz signal
XL = 2πfL = 87Ω
so the voltage induced can be found out as
Eind = 2πfNφm
Eind = 72.29V
From the given data about the coil plastic housing we can find
out the mean turn length
lm = 2(41 + 43) + π ∗ 13 = 208.82mm
The total length can be found out by multiplying the mean
turn length with the total number of turns.
l = N ∗ lm = 375.49m
The wire diameter is given to be 0.6mm and knowing the
resistivity of copper we can find out the total resistance
R =
ρ ∗ l
π d2
4
R = 22.306Ω
The total voltage required can be found out as
V = Eind + R ∗ irms = 90.93V
D. Static force for different airgaps
In the last part of the exercise we are required to find out the
static force exerted on the armature for different airgap lengths.
For this purpose it is essential to express the inductance as a
function of the gap length. So we can write
L(x) =
N2
Rt
=
17982
13809 + 1164003400x
From equ(2) we can find out the flux as a function of the airgap
since in this case we have the excitation current as 0.78A
φ =
L(x) ∗ i
N
φ =
17982
13809+1164003400x ∗ 0.78
1798
Now we can calculate the force exerted on the armature by
using equ(14)
fm =
(
17982
13809+1164003400x ∗0.78
1798 )2
0.038 ∗ 0.036 ∗ 4π ∗ 10−7
IV. EXPERIMENTAL RESULTS
A. Objective
The purpose of the experiment is to check what amount of
ac and dc voltage and current is needed by E Coil to produce
magnetic flux required to attract the armature.
B. Hypothesis
It is assumed that when the E Coil is supplied with DC
voltage, it attracts magnet with lower value of current in
comparison with AC supply. When the E Coil is supplied with
AC supply, it is supposed to attract the armature at very high
voltage due to the presence of induced emf in the copper coil.

4. Fig. 4. Magnetic force for different gap lengths
C. Materials
E Coil, Remote Display Clamp Multimeter,Digital Multi-
meter,Connecting Probes, Auto-transformer,Supply.
D. Methodology
In the first step the E-Coil was provided with DC supply
with the help of variable DC source and with the help of
connecting probes. The remote Display clamp multimeter was
connected across the probe to measure the value of current.The
value of voltage was increased gradually and the value of
both current and voltage was observed when the armature got
attracted to the E Coil.
Fig. 5. Results with DC excitation
In the second step the E coil was provided with AC supply
with the help of auto transformer and connecting probes.The
voltage was gradually increased and the value of current was
measured by connecting Display Clamp Multimeter.
E. Results
During DC excitation the value of Voltage and Current
observed when the armature got attracted to the E Coil was
15.7 V and 0.73A respectively.
During AC excitation the value of Voltage and Current ob-
served when the armature got attracted to the E Coil was 248V
and 0.69A respectively. It was also observed that the moment
when armature got attached to the E coil there was voltage
and current drop because of the absence of air-gap
Fig. 6. Results with AC excitation
V. CONCLUSION
The results obtained by experiment and the analytical results
closely match. It is worth mentioning here that the values
obtained in the last part for the force may not exactly match the
experimental results because of saturation effects. As we go on
decreasing the air gap length the reluctance goes on decreasing
and with constant excitation the flux goes on increasing. As
the flux value goes on increasing the flux density may cross
1.2T and the material will no longer be operating in the
linear region. Anyways for understanding the behavior of the
material with varying gap lengths this exercise was important.
REFERENCES
[1] Principles of Electrical Machines by Vk Mehta
[2] Electrical Machines by A.E.Fitzgerald, Charles Kingsley.Jr, Stephen D
Umans