Solid works

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Chapter 1: Introduction to Machinery Principles
1-1. A motor’s shaft is spinning at a speed of 3000 r/min. What is the shaft speed in radians per second? SOLUTION The speed in radians per second is

ω = ( 3000 r/min ) ¨
1-2.

§ 1 min · § 2π rad · ¸¨ ¸ = 314.2 rad/s © 60 s ¹ © 1 r ¹

A flywheel with a moment of inertia of 2 kg ⋅ m2 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 The speed in radians per second is:
5 N ⋅m §τ · ω =α t = ¨ ¸ t = ( 5 s ) = 12.5 rad/s J¹ 2 kg ⋅ m 2 ©

The speed in revolutions per minute is:
§ 1 r · § 60 s · n = (12.5 rad/s ) ¨ ¸¨ ¸ = 119.4 r/min © 2π rad ¹ © 1 min ¹1-3.

A force of 5 N is applied to a cylinder, as shown in Figure P1-1. What are the magnitude and direction of the torque produced on the cylinder? What is the angular acceleration α of the cylinder?

SOLUTION The magnitude and the direction of the torque on this cylinder is:

τ ind = rF sin θ , CCW
τ ind = ( 0.25 m)(10 N ) sin 30° = 1.25 N ⋅ m, CCW
The resulting angular accelerationis:

α=
1-4.

τ
J

=

1.25 N ⋅ m = 0.25 rad/s2 5 kg ⋅ m 2

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 The mechanical power supplied to the load is
P = τω = ( 60 N ⋅ m )(1800 r/min ) 1 min 60 s 2π rad = 11,310 W 1r

1

P = (11,310 W )

1 hp= 15.2 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 ofthe core is 1000.

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 l1 = 2(27.5 cm) = 55 cm, l 2 = 30 cm, and l3 =30 cm. The reluctances of these regions are:
R1 =
R2 = R3 =

0.55 m l l = = = 58.36 kA ⋅ t/Wb −7 µ A µr µo A (1000) 4π × 10 H/m ( 0.05 m )(0.15 m )

(

)

l

µA
l

= =

l

µr µo A
l

= =

0.30 m = 47.75 kA ⋅ t/Wb (1000) 4π × 10 H/m (0.05 m)(0.10 m )

(

−7

)

µA

µ r µo A

0.30 m = 95.49 kA ⋅ t/Wb 1000) 4π × 10−7 H/m ( 0.05 m )( 0.05 m ) (

(

)

Thetotal reluctance is thus

RTOT = R1 + R2 + R3 = 58.36 + 47.75 + 95.49 = 201.6 kA ⋅ t/Wb and the magnetomotive force required to produce a flux of 0.003 Wb is

F = φ R = ( 0.005 Wb )( 201.6 kA ⋅ t/Wb ) = 1008 A ⋅ t
and the required current is
i=

F 1008 A ⋅ t = = 2.52 A N 400 t

The flux density on the top of the core is
B=

φ
A

=

0.005 Wb = 0.67 T 0.15 m )( 0.05 m ) (

2

Theflux density on the right side of the core is
B=

φ
A

=

0.005 Wb

(0.05 m )(0.05 m)

= 2.0 T

1-6.

A ferromagnetic core with a relative permeability of 1500 is shown in Figure P1-3. 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 0.020 cm, respectively. Because of fringing effects,the effective area of the air gaps is 5 percent larger than their physical size. If there are 4001 turns in the coil wrapped around the center leg of the core and if the current in the coil is 1.0 A, what is the flux in each of the left, center, and right legs of the core? What is the flux density in each air gap?

SOLUTION This core can be divided up into five regions. Let R1 be the...
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