116 MECHANICAL ENGINEERING PRINCIPLES

torque of 43.6 N m. Find the speed of the

shaft (a) after 15 s, and (b) after the first

four revolutions.

[(a) 380.3 rev/min (b) 110.3 rev/min]

9. The driving torque on a turbine rotor is
   203 N m, neglecting frictional and other
   resisting torques. (a) What is the gain in
   kinetic energy of the rotor while it turns
   through 100 revolutions? (b) If the moment
   of inertia of the rotor is 23.2 kg m^2 and the
   speed at the beginning of the 100 revolu-
   tions is 600 rev/min, what will be its speed
   at the end?
   [(a) 127.55 kJ (b) 1167 rev/min]

9.4 Power transmission and efficiency

A common and simple method of transmitting power
from one shaft to another is by means of abelt
passing over pulley wheels which are keyed to the
shafts, as shown in Figure 9.6. Typical applications
include an electric motor driving a lathe or a drill,
and an engine driving a pump or generator.

rx ry

ωx

F 1 F^1

F (^2) F 2
Driven
pulley wheel
Driver
pulley wheel Belt
Figure 9.6
For a belt to transmit power between two pulleys
there must be a difference in tensions in the belt on
either side of the driving and driven pulleys. For the
direction of rotation shown in Figure 9.6,F 2 >F 1
The torqueTavailable at the driving wheel to do
work is given by:
T=(F 2 −F 1 )rxNm
and the available powerPis given by:
P=Tω=(F 2 −F 1 )rxωxwatts
From Section 9.3, the linear velocity of a point on
the driver wheel,vx=rxωx
Similarly, the linear velocity of a point on the driven
wheel,vy=ryωy.
Assuming no slipping,vx=vyi.e.rxωx=ryωy
Hence rx( 2 πnx)=ry( 2 πny)
from which,
rx
ry

ny
nx
Percentage efficiency=
useful work output
energy output
× 100
or efficiency=
power output
power input
×100%
Problem 17. An electric motor has an
efficiency of 75% when running at 1450
rev/min. Determine the output torque when
the power input is 3.0 kW.
Efficiency =
power output
power input
×100% hence
75 =
power output
3000
× 100
from which, power output=
75
100
× 3000
=2250 W.
From Section 9.2, power output,P = 2 πnT,
from which torque,
T=
P
2 πn
where n=( 1450 / 60 )rev/s
Hence, output torque=
2250
2 π
(
1450
60
)
= 14 .82 N m
Problem 18. A 15 kW motor is driving a
shaft at 1150 rev/min by means of pulley
wheels and a belt. The tensions in the belt on
each side of the driver pulley wheel are
400 N and 50 N. The diameters of the driver
and driven pulley wheels are 500 mm and
750 mm respectively. Determine (a) the