Page 411 - Marks Calculation for Machine Design

P1: Naresh
January 4, 2005
Brown˙C09
Brown.cls
The work done can be related to the angular velocities and the inertia of the system by
modifying Eq. (9.55) as 15:28 MACHINE ENERGY 393
1 2 2
Work = I sys ω max − ω min (9.57)
1→2 2
The difference in the squares of the angular velocities in Eq. (9.57) can be expressed
algebraically as the product of two terms as shown in Eq. (9.58).
1 2 2
Work = I sys ω max − ω min
1→2 2
1
= I sys (ω max + ω min )(ω max − ω min )
2
(9.58)

ω max + ω min
= I sys (ω max − ω min )
2
= I sys ω o (ω max − ω min )
where (ω o ) is not the mean or average angular velocity (ω m ) as the torque curve is not
symmetrical about the horizontal axis.
If a coefficient of speed fluctuation (C f ) is defined as
ω max − ω min
C f = (9.59)
ω m
then the expression for the work done (Work) given in Eq. (9.58) becomes
1→2
Work = I sys ω o (ω max − ω min )
1→2
(9.60)
= I sys ω o (C f ω m )
Most designs call for a small coefficient of fluctuation (C f ), which means the angular
velocity (ω o ) will be approximately equal to the mean angular velocity (ω m ). Therefore,
Eq. (9.60) becomes
Work = I sys ω o (C f ω m )
1→2
(9.61)
2
= I sys C f ω m
Solving for the mass moment of inertia of the system (I sys ) in Eq. (9.61), and substituting
for the work done in terms of the mean torque (T m ) and the total angle of rotation (φ) from
Eq. (9.56), gives
Work
1→2 T m φ
I sys = = (9.62)
C f ω 2 C f ω 2
m m
Note that while it is desired to keep the coefficient of fluctuation (C f ) as small as possible,
it would take an infinite mass moment of inertia in the system to make it zero. Therefore,
the system will always have some variation in angualar velocity.
The mean torque (T m ) and mean angular velocity (ω m ) are related to the power (P)
delivered by the engine. The power (P), measured experimentally, is usually given at a