Page 41 - Electric Machinery Fundamentals

lNTROOUCTION TO MACHINERY PRINCIPLES 17

uO = 4 *pi *lE-7 ; % Permeability of free space
n = 200 ; % Numbe r of turns on core
i = 1; % Cur rent in amps
% Calculate the fi rst reluctance
rl = 11 I (ur * uO * al) ;
d i sp ([ ' rl = I n um2str(rl)]) ;

% Calculate the second re luctance
r2 = 1 2 I (ur * u D '" a 2 );
disp ([ 'r2 = I num2str(r2) 1) i

% Calculate the total reluctance
rtot = rl + r 2 ;
% Calculate t he mmf
mmE = n * i ;
% Finally , get the flux i n the core
flu x = mmf / reat ;

% Display r e sult
disp ([ 'Flux = I num2 str (flux ) 1 );

When this program is execmed. the results are:
» ex! 1
r l = 1432 3 . 94 49
r 2 = 27586 . 856 8
Fl ux = 0 . 0047 72
This program produces the same answer as our hand calculations to the number of signifi-
cant digits in the problem.

Example 1-2. Figure 1- 8a shows a ferromagnetic core whose mean path length is
40 cm. There is a small gap of 0.05 cm in the stmcture of the otherwise whole core. The
2
cross-sectional area of the core is 12 cm , the relative permeability of the core is 4000, and
the coil of wire on the core has 400 turns. Assume that fringing in the air gap increases the
effective cross-sectional area of the air gap by 5 percent. Given this information, find
(a) the total reluctance of the flux path (iron plus air gap) and (b) the cunent required to
produce a flux density of 0.5 T in the air gap.

Solution
The magnetic circuit corresponding to this core is shown in Figure 1-8b.
(a) The reluctance of the core is
ie ie
'l<,, = - = - - ( 1- 32)
.uAe ILr J.LoAe
0.4m
(4000)(417 X 10 7)(0.002 m')
= 66,300 A • turnslWb