Electrical Power and MachinesLecture 2

<Dr Ahmed El-Shenawy><Dr Hadi El Helw>

MagnetismMagnetic fields are an essential element in the conversion of

mechanical

energy to electrical energy and vice versa

Sources of magnetic fields:

Permanent magnets

Current carrying conductors

In the region surrounding a permanent magnet there exists a magnetic field, which can be represented by magnetic flux lines. Magnetic lines exist in continuous loops, as shown in fig. The symbol for magnetic flux is the Greek letter Φ (phi).

(a) Permanent Magnets

If a nonmagnetic material, such as glass or copper, is placed in the flux paths surrounding a permanent magnet, there will be an almost unnoticeable change in the flux distribution. However if a magnetic material, such as soft iron, is placed in the flux path, the flux lines will pass through the soft iron rather than the surrounding air because flux line pass with greater ease through magnetic materials than through air. as shown in fig.

Magnetic lines of force that move from N to S

They form closed loops

Unlike magnet polesAttract

Like magnet polesRepulse

(b) Current Carrying Conductors

+

Direction can be determined using the right hand grasp rule

If the conductor is wound in a single-turn coil, the resulting flux will flow in a common direction through the center of the coil. A coil of more than one turn would produce a magnetic field that would exist in a continuous path through and around the coil which is quite similar to that of the permanent magnet.

The strength of the magnetic filed is determined by the density of the flux line. The filed strength of the coil can be effectively increased by placing certain materials, such as iron or steel, within the coil to increase the flux density within the coil or by increase the current in the conductor.

There are many application of the electro magnetic effect such as generator, transformer, Relay……..

Magnetic Flux and Magnetic Flux DensityMagnetic Flux and Magnetic Flux Density

(a) Magnetic Flux (φ)

Total number of magnetic lines of force present in a magnetic field

UnitUnit…….. Weber (.. Weber (wbwb))

(b) Magnetic Flux Density (B)

It is a measure of the concentration of the magnetic flux per unit areaUnit…… Tesla

B = φ/A (wb/m2) or Tesla

ExampleExample

The total magnetic flux at the pole face of a bar magnet is 3×10-4 wb, the bar

Magnet is rectangular and has a cross sectional area of 2 cm2. what is the

Flux density within the magnet?

SolutionSolution

B = φ/A = 3×10-4 / 2×10-4

= 1.5 Tesla (wb/m2)

(1)

(c) Magnetomotive Force (Fm)

In a magnetic circuit, a magnetic flux is created when a magnetomotive Force

(m.m.f ) acts on the circuit

Fm = N×I …… (AT)

N: number of turns I: Current flowing through coil (A)

(d) Reluctance (Rm)

In a magnetic circuit Rm, is the opposition offered by the magnetic circuit to the establishment of the magnetic flux by the m.m.f

Fm

φRmRm = Fm/φ …… (AT/wb)

(e) Permanence (Pm)

Permanence is defined as the ability of the magnetic circuit to permit the Establishment of a magnetic field

Pm = 1/Rm …… (wb/AT)

(it is more convenient to use Pm in making calculations in parallel magnetic circuits)

(f) Permeability (μ)

Permeability is a measure of the ease with which a magnetic field may be EstablishedIt could be given as the permanence per unit length and cross sectional area of a magnetic field

μ = Pm l /A = l /A Rm (wb / AT.m)(l: length of magnetic circuit, A: cross sectional area of magnetic circuit, Rm: Reluctance)

Note the followingNote the following

In electric circuits conductor materials are compared according to its Resistivity (δ), R = δ l/A Ω In magnetic circuits conductor materials are compared by means of its permeability (μ), μ = Pml/A wb/AT.m

Permeability of free space (μ0)= 4π×10-7 wb/AT.m For magnetic materials, permeability is given as follows:

μ = μ0 × μr

Where μr = 1 for air and non magnetic materials= (500-2500) for iron and steel

μr is dimensionless

(g) Magnetic field intensity (H)

It is the magnetomotive force per unit length

H = Fm/l (AT/m)

l: length of magnetic circuit

NoteNote

As Rm = Fm /φ, then Pm = φ/Fm and since μ=Pml/A

Then μ = (φ×l) / (Fm×A) = B/H

ExampleExample

What is the magnetic field intensity and the m.m.f. needed to produce a

flux of 2×10-4 wb in a steel ring whose mean circumferential length is 100

cm and Has a cross-sectional area of 5 cm2, assume μr = 500.

SolutionSolutionGiven: φ=2×10-4 wb, l =1 m, A = 5×10-4 m2 and μr =500

Required: H and Fm

Fm = NI = φRm = φl/μA = φl/μ0 μr A = 2×10-4 ×1 / 500 ×4π ×10-7 ×5 ×10-4

= 636.62 AT

H = Fm / l= 636.62/1 = 636.62 AT/m

(2)

Magnetization CurvesMagnetization Curves

Magnetic flux (φ) is produced by passing an electric current through a coil that creates a m.m.f

To have an effective comparison between materials , they are compared in terms Of unit quantities (Flux density “B” and Magnetizing force “H”)

1.Sheet steel

2.Silicon steel

3.Cast steel

4.Tungsten steel

5.Magnet steel

6.Cast iron

7.Nickel

8.Cobalt

9.Magnetite

B-H Curves for different materials

A magnetic material is said to be fully saturated when its permeability becomes almost the same as that of free space (H is so high)

If “H” is lowered by decreasing the current in the coil, B will not decrease as rapid as it increased (the B-H curve will not retrace itself), this irreversibility is called Hysteresis (B lags H)

when H is reduced to zero, residual flux density Br will appear meaning that the magnetic material has been magnetized

Analysis of magnetic circuits

c

cc A

l

g

gg A

l

gc

mmf

21 RR

VI

Magnetic Circuits versus Electric CircuitsMagnetic Circuits versus Electric Circuits

Magnetic Circuits versus Electric CircuitsMagnetic Circuits versus Electric Circuits

Electric circuit Magnetic circuit

Driving force (cause) emf mmf

Response (effect) current flux

opposition resistance reluctance

Equivalent circuit

Remf

I

Example 1: A ferromagnetic core is shown in the figure below. The depth ofthe core is 5 cm. The other dimensions of the core are as shown in the figure. Find the value of the current that will produce a flux of 0.005 Wb. With this current, what is the flux density at the top of the core? what is the flux density at the right side of the core? Assume that the relative permeability of the core is 1000.

There are three regions in this core. The top and bottom form one region, the left side forms a second region, and the right side forms the third region. If we assume that the mean path length of the flux is in the centre of each leg of the core, then the path lengths are l1=2(27.5 cm)=55cm,l2=30cm, and l3=30cm.

Solution

WbATA

lAl

WbATA

lAl

WbATA

lAl

or

or

or

/1049.95)05.005.0(1041000

3.0

/1075.47)1.005.0(1041000

3.0

/1036.58)15.005.0(1041000

55.0

373

372

371

WbATtotal /106.2011049.951075.471036.58 3333321

ATNImmf 1008106.201005.0 3

AN

mmfI 016.2500

1008

The total reluctance is thus:

and the magneto-motive force required to produce a flux of 0.005 Wb is

and the required current is

The flux density on the top of the core is

The flux density on the right side of the core is

TA

B 67.005.015.0

005.0

TA

B 205.005.0

005.0

Air Gaps and their effects

In many applications, magnetic flux must cross one or more air gaps

As the magnetic lines of force cross the air gap, they spread out because the Individual lines repel each other. This spreading out is called Fringing

Example 2: Determine the magnetic flux through the air gap in the geometryshown below. The structure is assumed to have a square cross section of area 10-6 m2, a core with μr = 1000, and dimensions l1=1 cm, l3=3cm, and l4=2cm.

Solution

WbATA

l

WbATA

ll

WbATA

ll

WbATA

ll

o

gg

o

o

g

o

/1058.79

/1066.6310002

/1044.151000

2001.0

/1083.311000

2

6

6433

642

6411

The magnetic flux through the source coil is the mmf divided by the total reluctance seen by the source:

Using “flux division” (analogous to current division), then:

WbNImmf

gtotal

6

3211 10286.0

//)(

Wbg

61

32

32 10115.0