5. energy conversion
Week 14
The Maxwell stress concept is another way of saying that the energy to the value
½ BH is stored in a unity cube of the space occupied by = magnetic, thus ½ BH is the energy density
[weber (metre
2
) x (amper/metre] = volt –second x ampere/cubic metre
= Joule /cubic metre.
Fig 2.6 magnetic energy .
Consider an airgap, initially unmagnetized. Apply a magnetic force to the gap, an increase of H from zero causes the flux density B = μoH to increase proportionately, Fig 2.6(o). The energy (m 3 is [HdB, and for and values Bi and H1 the final energy density (shaded area) is clearly ½ B
1
H
1
. The some summation applies to a filed set up by in a ferromagnetic matter, with similar result Fig. 2.6
(b0, if the permeability U is constant; but for the same and density B
1
a much smaller magnetizing force Hii is need and much less energy is stored.
If the ferromagnetic materials is subjected to saturation, the stored energy is as shown in Fig 2.6(c0 and is calculated by piece-wise approximation to composite area of DOAD plus area of trapezium
AB,CD.

5. energy conversion
Week 14
When the flux 4 associated with an electric increase in time at by amount d4, an emf, e = de/dt appears in the circuit. The minus sign implies that the direction of the emf is such that a current produced by it in the circuit opposes the change d4.
Flux-linkage 4 (weber- turn] is the product of a magnetic flux and the number of turns through which it passes in the same direction. Since the current is proportional the flux, then flux likage Ų = NQ
Since we are neglecting the coil resistance, then around the electric circuit loop formed by the voltage source and the N turns of coil on the toroid, the KVL gives u = +Dq/DT =-E. 2.17
The instantaneous electric power input to the coil
P =vi = (dw/dt)i
The total energy required to establish from zero a flux Q
1
and a linkage Q
1
, (corresponding to a current mmf F
1
=Ni) is
Wf = ( t
pdt = (
4
idQ =(
Q
fdQ
Since the core of the toroid has constant permeability.
Which is represented by the shaded area in Fig, 2. 6 (d). this magnetically stored energy can be assumed to be uniformly distributed through the active volume Al of the core. Then because
Q
1
= Q
IN
=NB,A and Nli =Fi =H il
The energy density = ½ 4ili/Al =1/2 Bi Hi
The total magnetic energy can be stated in several ways
[f =1/24I =1/2QF =1/2Q 2 S =1/2 F 2 S =1/2Q / 1/2F 2 =1/2Li 2 ]

5. energy conversion Week 14
Also, since Q BA and F =Hl and H =B/U
Then wf =1/2 QF =1/2 BAHL =1/2AL
U
But al = volume
Wf = Vol. B
2
2U
Any expression for the energy of the field wf in equation 2.32 and 2.23 may be employed depending on the parameters given.
Fig. 2.7 Energy change with position
Fig 2.7 (a) shows airgap region and existing coil of a magnetic circuit, the ferromagnetic core of which has a high. Permeability, the plane parallel polar faces. Have an area A and are spaced x apart. The n-turn coil carrying current I magnetic the system. The problem is to find the magnetic force of alteraction between the polar faces.
In the comparable system Fig 2.7 (b), with a rotatable part (rotor) port (stator), the problem is to find the tongue.
Insight into the inteplay of energy can be obtained from a study of finite mobvements, say from an initial position (1) to afinal direction position (2). At (a0. this movement -∆x (i.e

5. energy conversion Week 14 against the positive direction of x) of the right-hand member’ are ( b) it is a rotation -∆Q of the rotor. The static 4/I relations for the two positions are shown in Fig 2.7 © differ because the gap reluctance for (2) is less than for (1). Clearly the filed energy will differ too. How it changes depends on the conditions wholly in the gap, and the effect of coil resistance will initially be ignored.
(1)
Let the current be held constant at is throughout, as shown at Fig 2.7 (d). for position the linkage is 41 and the filed energy is ½ Fig 4. to reach position (2) a linkage 42 with constant current, an electrical energy input + ∆we =(Ų
2
-Ų
1
) i0 must be fig 2.7 (d). the current sources. This is represented by the hatched area at Fig
2.7(d0. Now, the increase in field energy is ∆WF =1/2 (Ų
2
-Ų
1
)io, which, comparing the expressions or the hatched areas at (d), is only one-half of∆we. What has happened to the other half?
Writing the energy balance and excluding the loss and mechanical storage terms:
∆we + ∆wf i.e (4
2
-4
1
)io + ∆wm =1/2 (Ų
2
-Ų
1
)io
Hence ∆wm =1/2 (Ų
2
-Ų
1
) i o
]
As 4
1
the mechanical input is negative: it is in fact an output work
(force x displacement). A precisely similar consideration gives for the rotary case (b) the output work (torgue x angular displacement). For constant current, therefore, the source provides ∆we, of which one-half is taken as energy into the filed and the other half is converted into mechanical energy output.

5. energy conversion
(2) CONSTANT FLUX
Week 14
Let the flux be kept constant so that linkage is always . the condition implies that the current fall from Li to i1 to compensate for the rise in permeance in Fig (e). there is no electrical energy transfer between the source and system for with constant linkage there is no induced e.m.f to be balance. Hence ∆we =o. but there is change of filed energy ∆wf =1/2Ų
0
(l2-l1) which is negative because l2 <l1. the energy balance is
O +∆wm =1/2 Ų
0
(l2 –l1)
Hence ∆wm =fm (-∆x) =1/2 Ų
0
(l2 –l1)
For constant flux, therefore, the mechanical work done comes from an equal reduction in filed energy.
(3)
In a practical device neither of condition (1) and (2) is likely to apply consistently.
The transition will follow some arbitrary contour, such as that in Fig 2.7 (f), with changes in both 4 and i. the energy balance is then some combination of cases (1) and
(2). The change in field energy is the shaded area ∆wf, and this will correspond, as before in magnitude to the mechanical energy ∆wm even, if owing to saturation effect, the Q/I relation is non- linear, they are ∆wf can still be found by graphical integration. Ni any case,
The mean force
The mean torgue
= ∆ wf
∆x
, =
∆wf, = f m m m
2.28
2.29

5. energy conversion Week 14
(4)
IF ∆ AND ∆Q are reduced to the infinitesimal differentials dx and dq, the force and torgue are obtained for a single position x or Q. Then
Force, fm torgue, m m
= dwf
dx
= dwf
dQ
2.30
2.31