Page 8 - Solutions Manual to accompany Electric Machinery Fundamentals


  P    50 N m 1500 r/min    1 min     2 rad     7854 W
 60 s   1 r 
P   7854 W   1 hp    10.5 hp
 746 W 

1-5. A ferromagnetic core is shown in Figure P1-2. The depth of the core is 5 cm. The other dimensions of
the core are as shown in the figure. Find the value of the current that will produce a flux of 0.005 Wb.
With this current, what is the flux density at the top of the core? What is the flux density at the right side
of the core? Assume that the relative permeability of the core is 800.

SOLUTION There are three regions in this core. The top and bottom form one region, the left side forms a
second region, and the right side forms a third region. If we assume that the mean path length of the flux
is in the center of each leg of the core, and if we ignore spreading at the corners of the core, then the path
l
l
lengths are = 2(27.5 cm) = 55 cm, l 2 = 30 cm, and = 30 cm. The reluctances of these regions are:
3
1
l l 0.55 m

 R    72.9 kA t/Wb

1  7 
 A   A  o 800 4  10  H/m 0.05 m 0.15 m 
 r
l l 0.30 m

 R    59.7 kA t/Wb


2  7 
 A   A  o 800 4  10  H/m 0.05 m 0.10 m 
 r
l l 0.30 m

 R    119.4 kA t/Wb

3  A   A  o 800 4  10  7   H/m 0.05 m 0.05 m 
 r
The total reluctance is thus




 R  R R R 72.9 59.7 119.4 252 kA t/Wb
TOT 1 2 3
and the magnetomotive force required to produce a flux of 0.005 Wb is

F  R    0.005 Wb 252 kA t/Wb   1260 A t 
and the required current is

i  F  1260 A t  2.5 A
N 500 t
The flux density on the top of the core is
2