Mechanical Engineering Principles

TORQUE 117

efficiency of the motor, (b) the speed of the

driven pulley wheel.

(a) From above, power output from motor

=(F 2 −F 1 )rxωx

Force F 2 = 400 N and F 1 = 50 N, hence

(F 2 −F 1 )=350 N,

radiusrx=

500

2

=250 mm= 0 .25 m

and angular velocity,

ωx=

1150 × 2 π

60

rad/s

Hence power output from motor

=(F 2 −F 1 )rxωx

=( 350 )( 0. 25 )

(

1150 × 2 π

60

)

= 10 .54 kW

Power input=15 kW

Hence,efficiency of the motor

=

power output

power input

=

10. 54

15

× 100 = 70 .27%

(b) From above,

rx

ry

=

ny

nx

from which,speed of driven pulley wheels,

ny=

nxrx

ry

=

1150 × 0. 25

0. 750

2

=767 rev/min

Problem 19. A crane lifts a load of mass 5

tonne to a height of 25 m. If the overall

efficiency of the crane is 65% and the input

power to the hauling motor is 100 kW,

determine how long the lifting operation

takes.

The increase in potential energy is the work done

and is given by mgh (see Chapter 14), where mass,

m= 5 t =5000 kg,g = 9 .81 m/s^2 and height

h=25 m.

Hence, work done=mgh=( 5000 )( 9. 81 )( 25 )

= 1 .226 MJ.

Input power=100 kW=100000 W

Efficiency=

output power

input power

× 100

hence 65 =

output power

100000

× 100

from which, output power

=

65

100

× 100000 =65000 W=

work done

time taken

Thus,time taken for lifting operation

=

work done

output power

=

1. 226 × 106 J

65000 W

= 18 .86 s

Problem 20. The tool of a shaping machine

has a mean cutting speed of 250 mm/s and

the average cutting force on the tool in a

certain shaping operation is 1.2 kN. If the

power input to the motor driving the

machine is 0.75 kW, determine the overall

efficiency of the machine.

Velocity,v = 250 mm/s = 0 .25 m/s and force

F= 1 .2kN=1200 N

From Chapter 14, power output required at the cut-

ting tool (i.e. power output),

P=force×velocity=1200 N× 0 .25 m/s.

=300 W

Power input= 0 .75 kW=750 W

Hence,efficiency of the machine

=

output power

input power

× 100

=

300

750

× 100 =40%

Problem 21. Calculate the input power of

the motor driving a train at a constant speed

of 72 km/h on a level track, if the efficiency

of the motor is 80% and the resistance due to

frictionis20kN.

Force resisting motion=20 kN=20000 N and

velocity=72 km/h=

72

3. 6

=20 m/s