Page 43 - Electric Machinery Fundamentals

INTRODUCTION TO MACHINERY PRINCIPLES 19

(b) Equation (1-28) states that
(1-28)
Since the flux ¢ = BA and tf = Ni, this equation becomes
Ni = BA'R
so
. BA'R
l=-;{
(0.5 T)(0.00126 m')(383,200 A· turns/Wb)
400 turns
= 0.602 A
Notice that, since the a.ir~gap flux was required, the effective air-gap area was used in the
above equation.

Example 1- 3. Figure 1-9a shows a simplified rotor and stator for a de motor. The
z
( mean path length of the stator is 50 em, and its cross-sectional area is 12 cm , The mean
path length of the rotor is 5 em, and its cross-sectional area also may be assumed to be
z
12 cm , Each air gap between the rotor and the stator is 0.05 em wide, and the crass-
2
sectional area of each air gap (including fringing) is 14 cm , The iron of the core has a rel-
ative permeability of 2000, and there are 200 turns of wire on the core. Jfthe current in the
wire is adjusted to be 1 A, what will the resulting flux density in the air gaps be?
Solution
To detemune the flux density in the air gap, it is necessary to first calculate the magneto-
motive force applied to the core and the total reluctance of the fl ux path. With this infor-
mation, the tota] flux in the core can be found. Finally, knowing the cross-sectional area of
the air gaps enables the flux density to be calculated.
The reluctance of the stator is
I,
'R =--
.1 /-L,. fLrjA,
0.5 m
(2000)(41T X 10 7)(0.0012 m')
= 166,000 A • turn51Wb
The reluctance of the rowr is
I,
CR,.=---
J-Lr fLrjA,
0.05 m
(2000)(41T X 10 7)(0.0012 m')
= 16,600A'IurnsIWb
The reluctance of the air gaps is
( I"
'R,,=--
J-Lr ,llQAa
0.0005 m
(1)(41T X IO 7)(0.0014 m')
= 284,000 A • turn5IWb