1.4 Electrodynamic Forces - University of New South Wales - Lec6... · ELEC9712: Electrodynamic Forces p. 1/40 ELEC9712 High Voltage Systems 1.4 Electrodynamic Forces Because of the

ELEC9712: Electrodynamic Forces p. 1/40

ELEC9712 High Voltage Systems

1.4 Electrodynamic Forces

Because of the magnitude of short circuit currents that can occur in electrical equipment, the forces that can occur on current carrying components due to the interaction of their current with either their own or an external magnetic field must be determined. Thus, we need to develop means of calculation for:

(i) Forces on isolated conductors

(ii) Forces on the windings of coils

(iii) Forces on conductors near magnetic materials

(iv) Forces on static and dynamic contacts

(v) Resonance frequencies of rigid busbar spans In all of these cases, the forces arise as a result of the interaction of the current in a conductor with a magnetic field, so that the force is derived from the general I B× vector interaction. There are quite different geometries that make the above require different approaches to the method of evaluation. In the case of (i), which is the simplest situation of a straight linear conductor of length L with current I in an external magnetic field of flux density B, the calculation requires only that the magnetic field B be determined so that the force is obtained from:


L

F I B L I B= × = ⋅ ×∫ (newtons)

for a linear conductor of length L carrying current I. I B× is the force per unit length.

B

I

F

B

I

F

For situation (ii) we either define and use a current sheet density and consider interaction of the sheet with an external magnetic field or do the calculation using energy considerations by using the inductive energy storage formula and taking force as the rate of change of that storage. Generally the latter is the easier to apply, although the current sheet method can be used in transformer windings with well-defined configurations.

A

F K B= ×∫∫ (newtons)

for a current sheet of current density K amps/m and surface area A.

or: 2x

dLF Idx

= newtons


For situation (iii), the calculation is usually difficult to do quantitatively because of the non-linearities and possible saturation of the magnetic material when there a magnetic field impinging on it. In general the simplest approach is to assume the material has infinite permeability and then to use the analogy with the electrostatic case and use a mirror conductor to calculate the force from the equations for a two conductor interaction.

In the case of (iv) the problem is made complex because it will depend, to a degree on the distribution of current in the contact area, particularly for dynamic (switching) contacts. The basic formula used is the I B× one, or more accurately the j B× interaction which is defined at a point. We need to know the distribution of j in the contact area if this equation is to be used.. In case (v) the calculation of force can be done using method (i), but what is of importance here is the resonant frequency of the busbar span when subjected to the ED force. If the resonant frequency is close to frequencies of the forces (eg 100 or 120 Hz) then resonant vibration can occur with disastrous results. If the frequencies are calculated to be similar then the busbar support spacing must be changed. This will change the resonant frequency of mechanical vibration and will thus allow separation of the mechanical resonance and electrical power force frequencies to prevent resonant effects. Note that the electrical force frequencies will be 100/120 Hz in most cases because of the I2 term that will arise in the force calculations.


1.4.1 Forces on isolated current carrying conductors

1B

1dl 2dl

1I2I

θ

d F

1B

1dl 2dl

1I2I

θ

d F

In the general configuration shown above, the force on the current element dl2, due to the magnetic field B1, generated by current I1 in l1 is:

122d F I dl B= × (N/m) or: 2 2 1 sind F I dl B α= × α is the angle between 1B and 2dl : it is π/2 if 1B and 2dl are in the same plane. Integrating over the whole length of conductor l2, we get the total force F :

2

122l

F I dl B= ×∫ (newtons)

It can be seen that the requirement to determine this force is that we need to calculate B1 over the whole length of conductor l2.


Determination of the magnetic field of a current element of finite length

P

Idlθ

r

Ir [dB]P

Idlθ

r

Ir [dB]

The starting point in determining the formula for such a field is the Biot-Savart Law. At the point P:

24o Idl rd B

rμπ

×=

or: 2

sin4

o Idld Br

μ θπ

=

Integrating of this gives:

24o

pl

I dl rBr

μπ

×= ∫ (Wb/m2 or tesla)

Note: another magnetic flux density unit is gauss where 1T = 104 gauss. For comparison, magnetic flux densities of:

Colour television: 6~ 10− T Earth: 4~ 0.5 10−× T A small bar magnet: ~ 0.02 T MRI scanner magnet: 2 T

This expression above gives the general equation that we will use for the calculation of the magnetic field for any straight


section of conductor. However for our purposes, we need to simplify it a little for easier application to straight line conductor segments. Consider the finite current element AB as shown below:

Bθ

A

B

Aθ

θ

l

dl

r

2π θ−

PI

R

Bθ

A

B

Aθ

θ

l

dl

r

2π θ−

PI

R

We want to calculate the magnetic field at the arbitrary point P, due to current flow in the element AB, with current magnitude I. We can write the elemental form of this as:

2

sin4

o IdldBr

μ θπ

=

But this form has too many parameters (3 in fact): we need to simplify the expression to have only one parameter so that we can integrate the result. We want to make θ that one parameter.


To do this we use the following geometrical relationships:

tan cot2

l R Rπ θ θ⎛ ⎞= − =⎜ ⎟⎝ ⎠

and:

cos sin2

R r rπ θ θ⎛ ⎞= − =⎜ ⎟⎝ ⎠

Thus:

cossin

Rr R ecθθ

= =

also: ( ) 2cot cosdl Rd R ec dθ θ θ= = − Substituting all of the above into the equation for dB, we get:

( )2

2 2

cos sin4 cos

oI R ec d

dBR ec

θ θ θμπ θ

−=

sin4

oI dR

μ θ θπ

−=

Integration then gives:

sin4

B

A

oAB

IB dR

θ

θ

μ θ θπ

= −∫

Thus:

[ ]cos cos4

oAB B A

IBR

μ θ θπ

= −[ ]cos cos4

oAB B A

IBR

μ θ θπ

= −


For a current element 2dl at point P we have:

22 ABd F I dl B= × and:

2

22 AB

l

F I dl B= ×∫

Application of the formula to particular common cases:

P

I

Aθ

Bθ

R

Semi-infinite

P

I

Aθ

Bθ

R

Fully-infinite

P

I

Aθ

Bθ

R

Semi-infinite

P

I

Aθ

Bθ

R

Fully-infinite

(a) A semi-infinite line current element with current I

The geometry is defined by 0y = to y = ∞ as shown above. In this case:

2Aπθ = and 0Bθ =

Thus:


cos0 cos4 2 4

o oP

I IBR R

μ μππ π

⎡ ⎤= − =⎢ ⎥⎣ ⎦

(b) A fully infinite line current element with current I

The geometry is defined by y = −∞ to y = +∞ as shown above. In this case:

Aθ π= and 0Bθ = Thus:

[ ]cos0 cos4 2

o oP

I IBR R

μ μππ π

= − =

It can be seen that this approach gives the correct results in the simple well-known cases. Using such a simple formula it is possible to break any system of straight conductor elements down into a number of simple linear current element sections, then apply the formula to each individual element and then superimpose the results algebraically (or vectorially if necessary) to obtain the total magnetic field at any point for the whole conductor system. For example in the case shown below, we divide the conductor system into the three separate sections A, B and C and apply the formula to determine the three components BA, BB and BC at some specified point P and then superimpose the results at the particular point P to get P A B CB B B B= + + . [This example has some relevance to old style bulk oil circuit breakers where the section B represents the moving arm of the circuit breaker during operation.]


A

C

B

I

I

PP A B CB B B B= + +

A

C

B

I

I

PP A B CB B B B= + +

Example Calculate the electrodynamic force acting on the section AB, in the diagram below, due to the interaction of the current flowing in AB with the magnetic field at AB due to the same current I flowing in the two sections A - ∞ and B - ∞. The conductors are assumed to be of circular and uniform cross-section, with radius r.

a

I

I

y

to ∞

to ∞

x

dx

A

Bx z

2r

a

I

I

y

to ∞

to ∞

x

dx

A

Bx z

2r

We need to find the magnetic field BP due to current I in the sections A - ∞ and B - ∞ at the arbitrary point P on the section AB: P is at a distance x from the axis of A - ∞ which we take to be the reference.


We thus need to find the magnetic flux density BP at P:

( ) ( )P P A P BB B B−∞ −∞= + Both A-∞ and B-∞ are semi-infinite line current elements and thus we have:

( ) ( )4

oP A

IB xx

μπ−∞ = for A-∞

and ( ) ( ) ( )4o

P BIB x

a xμ

π−∞ =−

for B-∞

Thus ( ) 1 14

oP

IB xx a x

μπ

⎡ ⎤= +⎢ ⎥−⎣ ⎦

The force acting on the element dx is then:

Pd F Id x B= ×

or Pd F Idx B y= ⋅ ⋅

or ( )2 1 1

4oIdF x

dx x a xμ

π⎡ ⎤= +⎢ ⎥−⎣ ⎦

newtons/metre

Then the total force on the section AB is obtained from integrating dF/dx over AB.

2 1 1

4

a r a ro

ABr r

IF dF dxx a x

μπ

− − ⎛ ⎞= = +⎜ ⎟−⎝ ⎠∫ ∫


( )2

ln ln4

a ror

I x a xμπ

−⎡ ⎤= − −⎣ ⎦

Thus: 2

ln2o

ABI a rF

rμ

π−⎛ ⎞= ⎜ ⎟

⎝ ⎠ newtons

In a similar way, we can calculate forces acting on “ T ” joints in conductors sections, such as those geometries shown below.

II

I1F I0

I2

FII

I1F I0

I2

F

It can be seen that whenever there any deviation of a current carrying conductor from a straight line configuration, there will be some electrodynamic force acting in the vicinity of the point where the change in direction occurs. It is for this reason, for example, that lightning conductor connections to earth, which may have to carry many tens of kiloamps for a short duration, have to be made as straight as possible to avoid any forces that may cause damage to the conductor or to its attachments.


1.4.2 Forces on Coil Windings For windings such as transformers, inductors, line traps, rotating machine stators etc, the conductors cannot always be broken up into linear elements and so the above method of analysis is not applicable. However, the force on such windings is very important and needs to be able to be calculated in some way. The distribution of the forces can be complex, as is shown below where the forces on a transformer winding are illustrated. Depending on where the force is to be calculated there are two possible components of the magnetic field that can interact with current in the windings. These are the axial field component Bz and the radial field component Br. When each of these interact with the (azimuthal) current in the winding conductor, there are two possible force components generated, an axial force and a radial force. This is particularly the case at the ends of the windings where the field has a significant radial component Br.

zF z rF I Bθ= ×

zF zF

r zF I Bθ= ×rF

r

zθ

zB zB

rB rB

zF z rF I Bθ= ×

zF zF

r zF I Bθ= ×rF

r

zθ

zB zB

rB rB


At the end of the winding where the magnetic field has strong curvature, the various components of current, magnetic field and force are as shown below:

zF

rB

B

F

rF

zB

Iθ

zF

rB

B

F

rF

zB

Iθ

Using F I B= × we get the force components acting on the winding:

Radial B: rI B× → Fz [an axial (compression) force on the winding]

Axial B: zI B× → Fr [a radial expansion (hoop stress) force on the winding]

At the centre of the coil winding structure the magnetic field is almost purely axial, so that the interaction of the azimuthal current and the axial B field gives a force in the radial direction (a hoop stress acting on the winding conductors).

r zF I Bθ= × (newtons/metre) At the end of the winding there is a very substantial radial component so that this gives a very substantial axial


compression force from the interaction of azimuthal current with the radial magnetic field component.

z rF I Bθ= × (newtons/metre) In a transformer, there are two coils wound on the core and in core-form transformers they are coaxial, with the primary and secondary coils wound overlaying each other. There will be a magnetic field between the two coils and this will act to generate forces on the two windings that are more complex that those acting on a single winding. The diagram below shows the configuration.

There are thus three important forces that act on transformer windings: a hoop stress on the outer winding that tries to


rupture the conductor and a radial force that compresses the inner winding. The third force is an axial compressive force that attempts to squash the coils down. If the two windings are not symmetrical on the axis, or if the windings are split and there is an unequal split along the axis then there will be an unbalanced force acting because of the asymmetry. This asymmetry in the two windings will cause an asymmetry in the magnetic field and the stronger field at one end will cause the unbalanced axial force. The axial force will attempt to increase the asymmetry of the two parts of the winding. A similar force acts on stator windings and the forces on the stator end winding represents a significant problem for rotating machines. At the end windings, the conductor geometries are complex as the windings bendaround to re-enter slots in the stator core. The electrodynamic forces resulting from the interaction, coupled with the effects of the rotor are a major source of potential problems in large motors and generators. The vibrations resulting are often the cause of insulation damage at the end winding sections. For the calculation of such forces on coils we can use a current sheet interaction with a magnetic field approach, or another approach is to calculate the overall force on the winding from the stored energy and its variation when forces act to try to distort the winding and its magnetic field distribution and thus change the stored magnetic energy. The first of these calculates the force distribution and total force is obtained by integration. The second approach is a global method with the total force only being calculated.


(a) Calculation of the force on windings using the current sheet approach In this approach we take the current to be a sheet with linear current density K where K I L= Amps/m, where I is total current and L is the (axial) length of the sheet. In this approach the force occurs from the interaction of the current sheet with the magnetic field:

F K B= × newtons per unit area For example, when applied to a solenoid structure, with total current I, length L, total number of turns N and diameter D (=2a):

NIKL

=

The axial flux density for a long solenoid is

o oNIB nIL

μ μ= =

where n N L= = no. of turns/metre. The force on the conductors of the solenoid will then be:

12 o

NI NIFL L

μ= × ×

( )2

2

12 o

NIL

μ= N/m2


(we assume B at the winding is half that at the axis) For the total force acting on the whole winding, we multiply the above by the total winding surface area ( )DLπ . Thus total force is:

( )2

2

1 22total o

NIF a L

Lμ π= × N

( )2

total o

NIF a

Lμ π= N

2 2o an I Lμ π= N

As well as allowing the calculation of the total (expansive or hoop stress) force on the winding of a single circuit solenoid, the above method will also allow the calculation of the total force acting on the separate windings of a transformer structure. The two windings on the core will be laid concentrically and the gap between them will have the leakage flux of the transformer in it and this will act on the two windings in the same manner as does the inner flux in a solenoid. The only difference given by the separation of the windings will be that, when the whole forces on the windings are considered, there will be an expansive (hoop stress) force on the outer winding and a compressive force on the inner winding, as shown in the previous diagrams.


(b) Force calculation from the change in stored energy of a winding (i) For a coil with self inductance L and carrying a current i,

the energy stored in the magnetic field of the coil is

212

W Li=

for an isolated coil.

(ii) If there are two coils of self inductance L1 and L2 with

currents i1 and i2, and they have an interacting magnetic field, with mutual inductance M, the total stored energy is given by:

2 21 1 2 2 1 2

1 12 2

W L i L i Mi i= + +

Any change in the stored magnetic energy (dW) in the coil(s) will require some change in the dimensions (dx) and hence the inductance of the coils and this will require the application of some force (F) to the coil(s):

dW Fdx= ⇒ dWFdx

=

And for the two cases above, this equation gives expressions that can be used to determine the total force F acting on the coils as a whole due to the currents flowing in them.


(a) 212

dLF idx

= newtons

(b) 2 21 21 2 1 2

1 12 2

dL dL dMF i i i idx dx dx

= + + newtons

This does not require the calculation of any magnetic flux densities and distributions, but does require that we know the functional (spatial) variations of the inductances L(x,y) and M(x,y). Then the application so the various spatial derivatives will give the total force acting in that particular spatial direction.

e.g. ( ) 212

dLF x idx

= and ( ) 212

dLF y idy

=

In many cases, the form of the inductance equation is quite well defined and calculable, but in many other cases, such as in transformer windings where the magnetic flux in question is the leakage flux, the inductance is not easily expressed in an analytic form and the method is thus not easy to apply.


Application to some common conductor configurations Example: Calculate the force between two parallel straight conductors. Distance x apart, conductor radius r.

x

dx

i1

i2

x

dx

i1

i2 (i) If the conductors are the same circuit (go and return),

such that: 1 2i i i= − =

then the inductance formula is well-known. The self-inductance is:

1 ln4

oi e

xL L Lr

μπ

⎡ ⎤= + = +⎢ ⎥⎣ ⎦ H/m

Thus:

212

dLF idx

= N/m

21 12

o rix r

μπ

= × × ×

2

2oi

xμπ

= N/m

This is a repulsive force because the result was positive for an assumed increase in the value of the separation x.


(ii) If the conductors are parts of separate circuits such

that: 1 2i i≠

In this case, we can assume that the self-inductances L1 and L2 are not changed in any significant way by the change dx because of the large size of the circuits.

2 21 21 2 1 2

1 12 2

dL dL dMF i i i idx dx dx

= + +

But:

1 0dLdx

= and 2 0dLdx

=

Hence:

1 2dMF i idx

=

If the conductor length is much greater than l x :

2ln 12

ol lMx

μπ

⎡ ⎤⎛ ⎞= −⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

Thus:

1 2 2

22 2

o x lF i i ll x

μπ

= − × × ×

1 22oF i i lx

μπ

= − × newtons

This is a force of attraction because of the negative sign. We assumed that the separation would increase and the negative sign means that it actually decreases.


Example: Calculation of the total radial force on a one-turn loop of coil radius R, and conductor radius r.

When R is much greater than r the inductance of such a coil is given by:

8ln 1.75oRL Rr

μ ⎡ ⎤= −⎢ ⎥⎣ ⎦ H

The radial force is thus given by:

212R

dLF idR

=

21 8 8ln 1.752 8o

R ri Rr R r

μ ⎡ ⎤⎛ ⎞= − + × ×⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

21 8ln 0.752 o

Rir

μ ⎡ ⎤= −⎢ ⎥⎣ ⎦ N

This is the total radial force in newtons over the whole loop. The magnitude of this force per unit length is the hoop stress.


Example: Radial force on a long solenoid We can calculate by this method the radial hoop stress acting on the turns of a solenoid and compare it with the equation determined in the previous section. The self inductance of a long solenoid of length Ls, radius a and total turns N is:

2 2o sL n a Lπμ= H ( )sn N L=

The radial force acting will then be given by:

212

dLF Ida

=

( )2 21 22 o sI n a Lπμ=

2 2o sn I Lπμ= newtons

This is the same equation as derived previously using the current sheet method of calculation.


1.4.3 Forces acting on conductors located near magnetic materials

In many practical situations, the conductors will be near magnetic materials and the effect of the magnetic materials will be to substantially modify the magnetic field distribution at the conductor and this will then cause substantial change in the force acting on the conductor. For an isolated current carrying conductor, the magnetic field distribution around it is uniform and symmetric and there is no net force acting on the conductor.

B

(a) no net force on isolated conductor

(b) force towards material wall

B

(a) no net force on isolated conductor

(b) force towards material wall

However, when there is a magnetic material nearby the effect will be to change the flux and make it asymmetric, because no flux will penetrate into the volume of the material (we assume that the magnetic permeability of the material is infinite, μ = ∞ and thus the skin depth is zero). The resulting asymmetry of the magnetic field means that there is an imbalance in the field which will generate a force towards the material surface as shown.


We need to consider two specific cases:

(a) when the conductor is outside the magnetic material

(b) when the conductor is within the magnetic material (e.g in a slot)

(a) Conductor outside the magnetic material The magnitude of the force can be calculated using the image field analysis, noting that the field distribution in the air is the same as that for a two wire system of conductors carrying the same current in the same direction. Using the image conductor the force can be calculated using the two-wire force formula. For example if the separation of the wire from the magnetic material is d, then the wire and its image are separated by 2d and the force is then

( )

2 2

2 2 4o oI IF

d dμ μπ π

= = N/m


If the material is not infinitely permeable (ie μ ≠ ∞), as is generally the case, the image position will not be exactly in the mirror position, in d from the surface. This must be taken into account and the effective d value changed to accommodate this. The effective current may also need to change. (b) Conductor is in a small slot within the material In this case (shown below) the magnetic flux will be completely retained within the material and none will escape into the air. The field and the resulting forces in such a situation as shown below can be analysed once again by the image method noting again that the flux distribution will be the same as for a two-wire conductor system with current flowing in opposite directions.

F

μ = ∞

F

μ = ∞


The direction of the resulting force will be such as to try to push the conductor further into the material volume. The force magnitude will be given by

( )

2 2

2 2 4o oI IF

d dμ μπ π

= = N/m (as before)


1.4.4 Forces acting on electrical contacts Any electrical contact, whether a static bolted contact or a dynamic switching contact will always have the (undesired) effect of distorting the current distribution in the contact interface region. This redistribution of the current will inevitably mean also a redistribution of the magnetic field within the material contact and this will then generate some force acting on the contact when the current and its magnetic field interact at the interface. Knowledge of such forces is very important for the determination of the required fastening strength needed for static contacts such as in busbars and for the required retaining spring strength in dynamic switching contacts. Welding of switching contacts during short circuits is a common occurrence when the electrodynamic forces open the contact against the latching springs and the resulting arcing then melts the contact which then closes again but onto molten surfaces with welding being the end result. Such redistribution of current in a contact interface also causes an increase in the effective contact resistance and this can lead to increased thermal generation and heating of the contact. This aspect will be covered in detail later. An idealised contact between two releasable contact surfaces will cause a constriction of current as shown below. The current is forced to constrict down to allow passage through the true electrical contact area as shown. In doing so it generates a radial component of current near the contact


interface region. Thus, at the contact there are two components of current present, a radial and the normal axial component. While this new radial component of current will not generate another component of flux density, which remains only azimuthal (Bθ) in nature, the magnitude of Bθ near the interface will vary with radius. The interaction of the radial current with the azimuthal field will generate an axial force on the contact faces tending to cause them to separate and thus open.

There is a radial force Fr at the contact (and along all of the conductor) which is the so-called “pinch force”, caused by interaction of the azimuthal field with the axial current. However this force is resisted by the metal structure of the conductor and has no mechanical effect on the system.


However the axial force Fz generated by the interaction of the magnetic field with the radial component of current acts on the contacts and will open it if the restraints are not strong enough. The magnitude of the contact opening force is given by:

2

ln4oI RF

rμ

π⎛ ⎞= ⎜ ⎟⎝ ⎠

N

where R is the overall total contact cross-section radius and r is the actual radius of the true electrical contact area in the contact interface. This is a very simplified view of an electrical contact as the contact will normally contain a number of such true contact areas over the whole cross-section and this will thus act to change and complicate the calculation. During operation the contact interface will heat and the metal will become a little more plastic and the thermal expansion will mean that the contact area will increase a little in size and reduce contact forces. While such forces can be a problem they can be avoided by proper design of the contact and in many cases the forces can be used to positive effect by keeping the contact closed and thus decreasing contact resistance by increasing contact force.


Some examples of using specific design modifications to use forces positively at electrical contacts are shown below:

(a) A cross arm on a circuit breaker The contact opening forces are adding to the main electro-dynamic force acting on the CB cross arm. The result may be premature opening against the springs.

cF

II

cF

cF

cF

EDF

cF

II

cF

cF

cF

EDF (b) Re-arrangement of CB crossarm The main electrodynamic force and the contact forces are now opposed to each other.

cF

II

cF

cF

cF

EDF

cF

II

cF

cF

cF

EDF


(c) Use of electrodynamic force to maintain contact pressure on a poker-type contact arm.


1.4.5 Calculation of natural (resonant) frequencies of rigid conductor systems

When rigid conductors such as low voltage busbars in bus ducts and HV busbars in outdoor substations are supported and clamped in a number of spans along their run, they will have some natural frequency of oscillation if they are deflected by some impulse such as an electrodynamic force from a short circuit for example. If this frequency is near the typical frequency of the applied force then resonance may occur with possible large amplitude oscillation and rupture of the busbars. To avoid such failure events, the natural frequencies of the conductor system must be known and steps must be taken to ensure that the frequency of the applied forces (e.g 50 or 100 Hz or 60 or 120 Hz) is not likely to activate resonance in the conductor motion. As the natural frequency is dependent to a large degree on the conductor support spacing, a change in the span length is a simple means of changing the natural frequency. The natural frequency of such a system can be obtained by analysis of the system as a beam clamped at each end. The natural frequency is given by:

2REJfm

γ=l

Hz

where:


γ = a geometrical factor [= 3.56 for fixed supports]

l = span length (support separation) [m]

E = Young’s modulus of the conductor material [Pa]

J = moment of inertia of the conductor system [m4]

m = conductor mass per unit length [kg/m] The Young’s modulus is a measure of the stiffness of a given material. Typical values of E are:

Aluminium 107 10× Pa Copper 1012 10× Pa Steel 1021 10× Pa Lead 101.6 10× Pa

The moment of inertia is a measure of the inertia of a rigid body with respect to rotational motion (c.f mass quantifies inertia of a body with respect to translational motion). Typical expressions for J are:

(i) For a rectangular bar such as a busbar of width b and thickness d, the moment of inertia of the busbar about an axis parallel to side d:

3

12b dJ ⋅

= [m4]

(ii) For a cylinder of radius r and length l , moment of inertia about axis perpendicular to the main axis:

2 2

2

12 4rJ rπ

⎛ ⎞= +⎜ ⎟

⎝ ⎠

l [m4]


1.4.6 AC effect on electrodynamic force magnitude From the general expression for force:

1 2F ki i= [ 2ki= if a single circuit] Then if the currents are sinusoidal, 2 sini I tω= and: 2 22 sinF kI tω=

( )2 1 cos 2kI tω⎡ ⎤= −⎣ ⎦

Sinusoidal current waveform and resulting double frequency force. The peak force is thus:

22peakF kI= The average force is :

2aveF kI=


and the frequency of oscillation of the force is twice power frequency (either 100 or 120 Hz). If there is a phase difference between the currents i1 and i2, this will then complicate the calculation further. For example:

( ) ( )1 22 sin 2 sin 120F k I t I tω ω⎡ ⎤ ⎡ ⎤= −⎣ ⎦ ⎣ ⎦ The peak force will be reduced somewhat because of the phase difference. Effect of DC offset in the current If the current has some DC offset, the force at the first half cycle will be enhanced substantially, as shown below.


If Ipk is the peak asymmetric current and Imax is the peak symmetric current, then typically:

max1.8pkI I≅

1.8 2 2.54rms rmsI I= = Thus:

2pk pkF kI=

2 22.54 rmsk I= ( )26.45 rmskI=

( )23.24 2 rmskI= Note that ( )2

rmskI is the average force of the symmetric

current waveform, and ( )22 rmskI is the peak force of the symmetric current waveform. Three-phase circuits Consider the case of a flat arrangement of three phase conductors:

Using the phase currents ia, ib and ic, and assuming they are all directed into the page, the various forces Fa, Fb and Fc can be calculated.


0

0

0

. . N/m2 2

. . N/m2 2

. . N/m2

The currents and their phasors are:i

a b a ca ab ac

c b c ac bc ca

b a b cb ba bc

i i i iF F Fx x

i i i iF F Fx x

i i i iF F Fx x

μπ

μπ

μπ

⎡ ⎤= + = +⎢ ⎥⎣ ⎦

⎡ ⎤= + = +⎢ ⎥⎣ ⎦

⎡ ⎤= − = −⎢ ⎥⎣ ⎦

a

b

c

0i ( 120) 120i ( 120) 120

a a

b a

c a

I Sin t II Sin t II Sin t I

ωωω

= = ∠= − = ∠ −= + = ∠

The peak forces that occur on each conductor under symmetrical balanced current conditions (Ia = Ib = Ic = I) can be shown to be:

20

( ) ( )

20

( )

0.808 ( ) N/m2 .

0.866 N/m2 .

a peak c peak

b peak

IF Fx

IFx

μπ

μπ

= =

=


Note that the forces on the two outer phase conductors are essentially unidirectional while that on the middle conductor is oscillatory in direction.

Forces on a three-phase conductor system with three linear parallel conductors.

Case (a) is with no offset [a resistive circuit]. Case (b) is with offset [an inductive circuit].

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