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Problems
1.1 A motor’s shaft is spinning at a speed of 3000 rpm. What is the shaft speed in radians per second?
Solution: nm =3000 rpm
rad s
r
m /
min
=314.2 rad/s
1.2 A flywheel with a moment of inertia of 2 kg. m^2 is initially at rest. If a torque of 5 N. m (counterclockwise) is suddenly applied to the flywheel, what will be the speed of the flywheel after 5 s? Express that speed in both radians per second and revolutions per minute.
Solution: J 2 kg. m^2
5 N. m
t 5 s
s
kgm
Nm
t
J
t 5
2
12.5 rad/s
/min
/min 12. 5
12. 5 12. 5 r r
s
rad
n
^ 1194 rpm
1.3 A force of 5 N is applied to a cylinder, as shown in Fig. 1.3. What are the magnitude and
direction of the torque produced on the cylinder? What is the angular acceleration^ ^ of the
cylinder?
Fig. 1. Solution:
F 5 N
r 0.25 m
J 5 kg. m^2
rF CCW
ind
sin ,
m N CCW Nm CCW
ind
( 0. 25 )( 10 )sin 300 , 1. 25. ,
2 2
rad s
kgm
Nm
J
1.4 A motor is supplying 60 N. m of torque to its load. If the motor’s shaft is turning at 1800 r/min, what is the mechanical power supplied to the load in watts? In horsepower?
Solution: 60 N. m
m
n 1800 r/min
P Nm ) rad / s 11309. 73 W
hp
hp
P W 15. 16
1.5 Find the relative permeability of the typical ferromagnetic material whose magnetization curve is shown in Fig.1.5.
a. H 50 A turns / m
b. H 100 A turns / m
c. H 500 A turns / m
d. H 1000 A turns / m
Fig. 1. Solution:
a. H 50 A turns / m
r ?
Fig. 1. Solution: a. The direction of the voltage while the flux is increasing in the reference direction must be positive to negative as shown in the figure 1.6.
b.
t V
t V
dt
d
turns
dt
d
N
ind
e
1885 cos 377
( 1000 ) ( 0. 05 sin 377 )
t V
ind
e 1885 sin( 377 900 )
1.7 A ferromagnetic core is shown in Fig. 1.7. The depth of the core is 5 cm. The other dimensions of the core are as shown in the figure. Assume that the relative permeability of the core is 1000. a. Find the value of the current that will produce a flux of 0.005 Wb. b. What is the flux density at the top of the core? c. What is the flux density at the right side of the core?
Fig. 1.
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 lengths are: Mean path length of top and bottom region of core: l 1
Mean path length of left side region of core: l 2
Mean path length of right side region of core: l 3
0. 005 Wb
N=500 turns
r 1000
o 4 10 ^7 H / m
a. i ?
We know that:
f N i N
f i N
f i
N
R
i
( f R )
Now we have to find the total reluctance:
R R 1 R 2 R 3
kAt Wb H m m m
m
r oA
l A
l R 58. 36 ./ ( 1000 )( 4 10 7 / )( 0. 05 ))( 0. 15 )
1
1 1
1 (^1)
kAt Wb H m m m
m
r oA
l A
l R 47. 75 ./ ( 1000 )( 4 10 7 / )( 0. 05 ))( 0. 10 )
2
2 2
2 (^2)
kAt Wb H m m m
m
r oA
l A
l R 95. 49 ./ ( 1000 )( 4 10 7 / )( 0. 05 ))( 0. 05 )
3
3 3
3 (^3)
R R 1 R 2 R 3 ( 58. 36 47. 75 95. 49 ) kA. t / Wb 201. 6 kA. t / Wb
Now A t
Wb kAt wb N
R
i 2. 016 500
b. T m m
Wb A
B
Topof core
Top of core^0.^67 ( 0. 15 )( 0. 05 )
c. T m m
Wb A
B
Rightsideofcore
Right sideof core^2.^0 ( 0. 05 )( 0. 05 )
1.8 A ferromagnetic core with a relative permeability of 2000 is shown in Fig.1.8. The dimensions are as shown in the diagram, and the depth of the core is 7 cm. The air gaps on the left and right sides of the core are 0.070 and 0.050 cm, respectively. Because of fringing effects, the effective area of the air gaps is 5 percent larger than their physical size. If there are
c. (^) Center Total
Total Wb kAt Wb
t A R
f
Wb Wb R R R R
R R
Left Total 90. 1 108. 3 90. 1 77. 3 (^0.^0033 )^0.^00193
1 2 4 5
Wb Wb R R R R
R R
Right Total (^0.^0033 )^0.^00229
- 1 108. 3 90. 1 77. 3
1 2 4 5
d.
T m m
Wb A
B
eff
Left Left (^) ( 0. 07 )( 0. 07 )( 1. 05 )^0.^375
T
m m
Wb A
B
eff
Right Right^0.^445 ( 0. 07 )( 0. 07 )( 1. 05 )
1.9 A two-legged core is shown in Figure 1.9a. The winding on the left leg of the core ( N^1^ )
has 400 turns, and the winding on the right ( N^^2 ) has 300 turns. The coils are wound in the directions shown in the figure.
a. If the dimensions are as shown, then what flux would be produced by currents
i 1 0. 5 A and i 2 0. 75 A? Assume r 1000 and constant.
b. By using the magnetization curve shown in Fig, 1.9b, find the flux produced in the core by the currents specified. What is the relative permeability of this core under these conditions? c. Was the assumption in part ‘a’ that the relative permeability was equal to 1000, a good assumption for these conditions? Is it a good assumption in general?
Fig. 1.9a
Fig. 1.9b Solution: The two coils on this core are wound so that their magnetomotive forces are additive.
a. ? Total
Total R
f
f (^) Total N 1 i 1 N 2 i 2 ( 400 t )( 0. 5 A )( 300 t )( 0. 75 A ) 425 A. t
kAt Wb H m m m
m
r oA
l A
l RTotal 92. 0 ./ ( 1000 )( 4 10 7 / )( 0. 15 ))( 0. 15 )
Wb kAt Wb
At R
f
Total
Total (^) 0. 00462
- /
b. BA ?
Now in order to find the value of ‘B’ from the magnetization curve we will find ‘H’
At m m
At m
t A t A l
N I N I
l
Ni H C C
1 1 2 2 (^400 )(^0.^5 ) (^300 )(^0.^75 )
From the magnetization curve, B 0. 15 T
BA ( 0. 15 T )( 0. 15 m )( 0. 15 m ) 0. 0033 Wb
We know that (^)
f A
l A
l f R o
C r r o
C
At h m m m
m Wb f A
l
o
C
r
c. The assumption that r 1000 is not very good here. It is not very good in general.
Fig. 1. Solution:
a. This core can be divided up into four regions. R 1^ be the reluctance of the left-hand
portion of the core, R^ 2 be the reluctance of the center leg of the core, R 3^ be the
reluctance of the center leg of the core and R 4^ be the reluctance of the right-hand air
gap.
? Total
Total Total R
f
RTotal R 1 ( R 2 R 3 ) IIR 4
2 3 4
2 3 4 1
R R R
R R R
R Total R
kAt Wb H m m m
m A r o
l A
l R 127. 3 ./ ( 1500 )( 4 10 7 / )( 0. 09 ))( 0. 05 )
1
1 1
1 (^1)
kAt Wb H m m m
m A r o
l A
l R 24. 0 ./ ( 1500 )( 4 10 7 / )( 0. 15 ))( 0. 05 )
2
2 2
2 (^2)
kAt Wb H m m m
m A o
l A
l R 40. 8 ./ ( 4 10 7 / )( 0. 15 ))( 0. 05 )( 1 0. 04 )
3
3 3
3 3
kAt Wb H m m m
m A r o
l A
l R 127. 3 ./ ( 1500 )( 4 10 7 / )( 0. 09 ))( 0. 05 )
4
4 4
4 (^4)
kAt Wb kAt Wb
R R R
R R R
R Total R. / 170. 2. /
2 3 4
2 3 4
Left Total
Total Total (^) kAt Wb Wb
t A R
f
Wb Wb R R R
R
Center Total 24. 0 40. 8 127. 3 (^0.^00235 )^0.^00156
2 3 4
Wb Wb R R R
R R
Right Total 24. 0 40. 8 127. 3 (^0.^00235 )^0.^00079
2 3 4
b.
T m m
Wb A
B
eff
Left Left (^) ( 0. 09 )( 0. 05 )^0.^522
T
m m
Wb A
B
eff
Center Center^0.^208 ( 0. 15 )( 0. 05 )
T
m m
Wb A
B
eff
Right Right (^) ( 0. 09 )( 0. 05 )( 1. 05 )^0.^176
1.12 A core with three legs is shown in Figure 1.12a. Its depth is 8 cm, and there are 400 turns on the center leg. The remaining dimensions are shown in the figure. The core is composed of steel having the magnetization curve shown in Figure 1.12b. Answer the following questions about this core: a. What current is required to produce a flux density of 0.5 T in the central leg of the core? b. What current is required to produce a flux density of 1.0 T in the central leg of the core? Is it twice the current in part ‘a’? c. What are the reluctances of the central and right legs of the core under the conditions in part ‘a’? d. What are the reluctances of the central and right legs of the core under the conditions in part ‘b’? e. What conclusion can you make about reluctances in real magnetic cores?
Fig. 1.12a
Bouter 0. 5 T H (^) outer 70 A. t / m (From graph of Fig. 1,12b)
A
t
At m m At m m N
H l H l N
f i Total^ centercenter outer outer 0. 22 400
c. kAt Wb m m
At m m B A
f H l R center
centercenter center
center outer (^) ( 0. 5 )( 0. 08 )( 0. 08 )^5.^25. /
kAt Wb T m m
At m m B A
f H l R right
rightright right
right right (^) ( 0. 25 )( 0. 08 )( 0. 08 )^22.^5. /
d. kAt Wb m m
At m m B A
f H l R center
centercenter center
center outer (^) ( 1. 0 )( 0. 08 )( 0. 08 )^6.^0. /
kAt Wb T m m
At m m B A
f H l R right
right right right
right right (^) ( 0. 5 )( 0. 08 )( 0. 08 )^15.^75. /
e. The reluctances in real magnetic cores are not constant.
1.13 A two-legged magnetic core with an air gap is shown in Figure 1.13a. The depth of the core is 5 cm, the length of the air gap in the core is 0.06 cm, and the number of turns on the coil is 1000. The magnetization curve of the core material is shown in Figure 1.13b. Assume a 5 percent increase in effective air-gap area to account for fringing. a. If the flux density of air gap is 0.5 T then find the flux densities of the four sides of the core. b. How much current is required to produce an air-gap flux density of 0.5 T?
Fig. 1.13a Fig. 1.13b
Solution:
a. B^ air gap ^0.^5 T
We know that A
B
In order to find the flux densities of the four sides of core total flux is required.
(^) total airgap left top right bottom
total airgap BA ( 0. 5 T )( 0. 05 m )( 0. 05 m )( 1 0. 05 ) 0. 00131 Wb
T
m m
Wb A
Bleft left 0. 262 ( 0. 10 )( 0. 05 )
T
m m
Wb A
Btop top 0. 262 ( 0. 10 )( 0. 05 )
T
m m
Wb A
Bright right 0. 524 ( 0. 05 )( 0. 05 )
T
m m
Wb A
Bbottom bottom 0. 262 ( 0. 10 )( 0. 05 )
b. ? N
f i Total
N
H l H l H l H l H l N
f i Total^ leftleft toptop rightright airgap airgap bottombottom
Now
Bair (^) gap 0. 5 T kAt m H m
B T
H
o
air gap^398. / 4 10 /
B (^) left Btop Bbottom 0. 262 T (From graph H (^) left Htop Hbottom 240 A. t / m )
Bright 0. 524 T (From graph H (^) right 410 A. t / m )
Now again:
N
H l H l H l H l H l N
f i Total^ left left top top right right airgapairgap bottombottom
BA ( 1. 27 T )( 0. 25 0. 0254 m 0. 0254 m ) 0. 000205 Wb
1.15 The core shown in Figure 1.16a has the flux shown in Figure 1.16b. Sketch the voltage present at the terminals of the coil.
Fig. 1.16b Fig. 1.16a
Solution: Using the graph of Fig. 1.16b we can find the induced voltage as:
Time (t) dt
d N
eind
0 < t > 2 s s
Wb t 2
- 010 (^500 2). 50 V
2 < t > 5 s s
Wb t 3
- 020 500
3. 33 V
5 < t > 7 s s
Wb t 2
- 010 500 2. 50 V
7 < t > 8 s s
Wb t 1
- 010 (^500 5). 00 V
Table 1.
The resulting voltage (Table 1.16) is plotted in Fig. 1.16c.
Fig. 1.16c
1.16 Figure 1.17 shows a ferromagnetic core whose mean path length is 40cm. There is a small gap of 0.05cm in the structure of the otherwise whole core. The cross-sectional area of the core is 12 square 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 gap by 5%. Find the following:
a. The total reluctance of the flux path (iron plus air gap). b. The current required to produce a flux density of 0.5T in the air gap.
Fig. 1.
Fig. 1.18b
Fig. 1.18c Solution:
Note: This is a design problem, and the answer presented here is not unique. We can select the reasonable flux density using the graphs of Fig. 1.18b and 1.18c. From the graphs of Fig. 1.18b and 1.18c the reasonable flux density range is B 1. 0 T 1. 4 T. a. Suppose the reasonable maximum flux density would be about 1.2 T.
b. total BA (^1.^2 T )(^0.^04 m )(^0.^04 m )^0.^00192 Wb
c. i^ ^1 A B^ ^1.^2 T^ r ^3800 Fig.^1.^18 b & c
N ?
i
R
i
f N total total total
Rtotal Rstator Rairgap 1 Rrotor Rairgap 2
kAt Wb H m m m
m A
l R stator stator
stator stator^62.^8. / ( 3800 )( 4 10 / )( 0. 04 )( 0. 04 )
- 48 (^) 7
kAt Wb H m m m
m A
l R rotor rotor
rotor rotor (^) ( 3800 )( 4 10 / )( 0. 04 )( 0. 04 )^5.^2. /
- 04 (^) 7
kAt Wb H m m
m A
l R R airgap airgap
airgap rair gap airgap ( 4 10 / )( 0. 0018 )^221. /
- 0005 1 ^2 7 2
Rtotal Rstator Rrotor 2 Rair gap
R (^) total 62. 8 kA. t / Wb 5. 2 kA. t / Wb 2 221 kA. t / Wb 510 kA. t / Wb
t A
Wb kA t Wb
i
R N total^ total 979 1
( 0. 00192 )( 510. / )
N 1000 t
1.18 Figure 1.19 shows a simplified rotor and stator for a dc motor. The mean path length of the stator is 50cm, and its cross-sectional area is 12cm^2. The mean path length of the rotor is 5 cm, and its cross-sectional area also may be assumed to be 12cm^2. Each air gap between the rotor and the stator is 0.05cm wide, and the cross-sectional area of each air gap (including fringing) is 14cm^2. The iron of the core has a relative permeability of 2000, and there are 200 turns of wire on the core. If the current in the wire is adjusted to be 1A, what will the resulting flux density in the air gaps be?
