Page 40 - Electric Machinery Fundamentals

16 ELECTRIC MACHINERY FUNDAMENTALS

Solution
We will solve this problem twice, once by hand and once by a MATLAB program, and
show that both approaches yield the same answer.
Three sides of the core have the same cross-sectional areas, while the fourth side has
a different area. Thus, the core can be divided into two regions: (1) the single thinner side
and (2) the other three sides taken together. The magnetic circuit corresponding to this core
is shown in Figure 1- 7b.
The mean path length of region 1 is 45 em, and the cross-sectional area is 10 X 10
2
em = 100 cm . Therefore, the reluctance in the first region is
II 11
(1- 32)
'R, ~ f<A, ~ /" f.4JA I
0.45 m
2
(2500)(41T X 10 7)(0.01 m )
~ 14,300 A · turns/Wb
The mean path length ofregion 2 is 130 em, and the cross-sectional area is 15 X 10 (
2
em = 150 cm . Therefore, the reluctance in the second region is
(1-32)

1.3 m
7 2
(2500)(41T X 10- )(0.0 15 m )
~ 27,600 A · turns/Wb
Therefore, the total reluctance in the core is
CRe q = CR 1 + CR.}
~ 14,300 A · turns/Wb + 27,600 A · turns/Wb
~ 41,900 A · turns/Wb

The totalmagnetornotive force is
'if ~ Ni ~ (200 turns)(1.0 A) ~ 200 A · turns
The total fl ux in the core is given by
200 A · turns
41,900 A · turns/Wb
~ 0.0048 Wb

This calculation can be petformed by using a MATLAB script file, if desired. A sim-
ple script to calculate the flux in the core is shown below.

% M- f ile : exl 1 . m
% M- file t o ca l c u late t he f lux in Example 1 - 1.
11 0 . 4 5; % Length of region 1
12 1 . 3; % Le ngth o f region 2
a1 0 . 01 ; % Area of region 1
a2 0 . 01 5 ; % Area of regi on 2
ur 2500; % Relative permeabi lity