Mechanical Engineering Principles

158 MECHANICAL ENGINEERING PRINCIPLES

(xii) Sound energy is converted to electrical

energy by a microphone.

(xiii) Electrical energy is converted to chemical
energy by electrolysis.

Efficiency is defined as the ratio of the useful
output energy to the input energy. The symbol for
efficiency isη(Greek letter eta). Hence

efficiency,η=

useful output energy

input energy

Efficiency has no units and is often stated as a per-
centage. A perfect machine would have an efficiency
of 100%. However, all machines have an efficiency
lower than this due to friction and other losses. Thus,
if the input energy to a motor is 1000 J and the
output energy is 800 J then the efficiency is

800

1000

×100%=80%

Problem 9. A machine exerts a force of

200 N in lifting a mass through a height of

6 m. If 2 kJ of energy are supplied to it,

what is the efficiency of the machine?

Work done in lifting mass

=force×distance moved

=weight body×distance moved

=200 N×6m=1200 J

=useful energy output

Energy input =2kJ=2000 J

Efficiency, η =

useful output energy

input energy

=

1200

2000

= 0. 6 or 60%

Problem 10. Calculate the useful output

energy of an electric motor which is 70%

efficient if it uses 600 J of electrical energy.

Efficiency, η=

useful output energy

input energy

thus

70

100

=

output energy

600 J

from which,output energy=

70

100

× 600 =420 J

Problem 11. 4 kJ of energy are supplied to

a machine used for lifting a mass. The force

required is 800 N. If the machine has an

efficiency of 50%, to what height will it lift

the mass?

Efficiency, η=

useful output energy

input energy

i.e.

50

100

=

output energy

4000 J

from which, output energy =

50

100

× 4000

=2000 J

Work done =force×distancemoved,

}

hence 2000 J=800 N×height,

from which, height=

2000 J

800 N

= 2 .5m.

Problem 12. A hoist exerts a force of

500 N in raising a load through a height of

20 m. The efficiency of the hoist gears is

75% and the efficiency of the motor is 80%.

Calculate the input energy to the hoist.

The hoist system is shown diagrammatically in

Figure 14.8.

Output energy=work done

=force×distance

=500 N×20 m=10000 J

Input

energy

Output

Motor energy

80% efficient

Gearing

75% efficient

Figure 14.8