Electrical Machines Notes 2 (Singly Excited Magnetic System) by Dr. P.S. Bimbhra

Electrical Machines by Dr. P.S. Bimbhra

1. Singly Excited Magnetic
System
by Dr. P.S. Bimbhra

2. Singly Excited Magnetic System
In this article we will derive the expression for:-
 Electrical energy input.
 Magnetic energy stored.
 Mechanical work done.

3. Electrical energy input
•Consider a simple magnetic system of toroid, excited by a
single source.
•The instantaneous voltage equation for the electric circuit can
written by applying kirchhoff’s law as: -
V
t
= I.r + e ----------Eqn(1)

4. •Where ‘e’ in the Eqn(1) is the reaction Emf taken as voltage drop in the
direction of the current ‘I’.
Therefore
-----Eqn(2)
Combining Eqn(1) & Eqn(2) we have:-


------Eqn(3)
Here ‘Ψ’ is the instantaneous flux linkage with the circuit.
Multiplying Eqn(3) by ‘Idt’ on both side we have:-
dt
d
e
Y
=
Microsoft
Equation 3.0
Y+= dIrdtIIdtV
t
...
2
dt
d
rIV
t
Y
+=.

5. As we know that—

----Eqn(4)
Assuming that ‘Φ’ links all the N-turns of the coil, the flux linkage ‘Ψ’ are
equal to ‘N.Φ’ Wb turns.
Therefore from Eqn(4)
------Eqn(5)
Y=Þ
Y=-Þ
Y=-
dIIdte
dIIdtIrV
dIrdtIIdtV
t
t
..
.)(
...
2
Y==
=
IdIdtedW
dWIdte
elec
elec
.
.
F=F=Y= dFNdIdIdW
elec
...

6. •‘Φ’ is instantaneous value of flux / coil.
•F=N.I, is the instantaneous coil mmf.
From Eqn(4) and Eqn(5) we can easily make out that for
a toroid to extract energy from the supply, the flux
linkage of the magnetic field must change. This change
in flux linkage cause the generation of reaction emf ‘e’.

7. Magnetic Field Energy Stored
•Consider a simple magnetic relay as shown:
•Initially the armature is in open position, when supply is given
current ‘I’ is established in N-turns coil.
•Magnetic field thus produced creates north and south poles.
•As a result, magnetic
Force is created which
Tends to shorten the
airgap.

8. •If the armature is not allowed to move, the mechanical
workdone, dWmech is zero.

Then by Eqn-
dWelec =dWmech +dWfld

Fig(a) dWelec =
0 + dWfld
Therefore
dWelec = dWfld ------Eqn(6)
From Eqn(5)-
dWfld =dWelec = I.dΨ = F.dΦ
Fig(b)

9. If the initial flux is zero, then the magnetic field energy stored
Wfld, in establishing a flux Φ1 or flux linkage Ψ1 is-

------Eqn(7)
When the armature is held in open position, then most of the mmf is
consumed in the air gap;
From fig(a)
=area OABO
From fig(b)
=area OABO
ò ò
F
Y
Y=F=
1
0
1
0
.. dIdFW
fld
òò
FF
F==
1
0
1
0
.dFdWW
fldfld
òò
YY
Y==
1
0
1
0
.dIdWW
fldfld

10. •In fig(a) & fig(b):-

OACO=
This area OACO is called “Co-energy”, W’fld.
---------Eqn(8)
“The term Co-energy has no physical significance, it is however
useful in calculating the magnetic forces.”
ò òò
Y=F=
1
0
1
0
..'
IF
fld dIdFdW
òò
Y=F=
1
0
1
0
..'
IF
fld dIdFW

11. •With no magnetic saturation:
Area OABO=Area OACO
Wfld = W’fld
And
Wfld + W’fld = Area OCABO = Φ1.F1 = Ψ1. I1
In general, for a linear magnetic circuit:
----Eqn(9) as [Ψ=LI]
----Eqn(9a)
It can also be expressed in terms of B & H as-
joules/m
3
F=Y== FIWW
fldfld
2
1
2
1
'
L
LIWW
fldfld
2
2
2
1
2
1
'
Y
===
HB
B
HWW
fldfld
2
1
2
1
2
1
'
2
2
====
m
m