Page 415 - Marks Calculation for Machine Design

P1: Naresh
January 4, 2005
15:28
Brown˙C09
Brown.cls
397
MACHINE ENERGY
The torque (T 1 ) and angular velocity (ω 1 ) are usually taken to be the rated torque (T rated )
and rated angular velocity (ω rated ). Therefore, the torque (T 2 ) and angular velocity (ω 2 )
must be found using information associated with the punching process and the available
time for recovery.
Again, the electric motor does not do the actual punching; the inertial energy in the
system, mostly in the flywheel, does the punching. The system loses energy, and therefore
speed, during the punching process and must return to its design speed before the next cycle
begins. As might be expected, the inertia in the system must be just the right amount for
the punch press system to work properly.
During the punching process, the torque required (T punch ) draws energy from the system
and slows it down to an angular velocity (ω 2 ). Leaving out the details, the corresponding
torque (T 2 ) can be found from Eq (9.67) as
τ

T 2 T punch − T 1
= (9.67)
T 1 T punch − T 2
where the exponent (τ) is the ratio of the recovery time to the punching time, meaning
t 2 − t 1
τ = (9.68)
t 1
Unfortunately, the torque (T 2 ) must be determined from Eq. (9.67) by trial-and-error;
however, this is not a burden. Once the punching torque (T punch ) is known, the rated torque
(T 1 ) is found from Eq. (9.65) where the rated power (P rated ) and the rated angular velocity
(ω rated ) are obtained from the motor identification plate.
During recovery, the rated torque (T 1 ) adds energy to the system as it increases its angular
velocity from (ω 1 ) to (ω 2 ) in time for the next punching interval. The system resists this
increase in speed through its mass moment of inertia (I sys ). Leaving out the details, the
mass moment of inertia of the system (I sys ) can be found from Eq (9.69) as
a (t 2 − t 1 )
I sys = (9.69)
T 2
ln
T 1
where (a) is the slope of the motor torque curve in the linear region, given in Eq. (9.70) as

T 1
a = (9.70)
ω 1 − ω syn
The slope (a) will be negative; however, the denominator of Eq. (9.69) will also be
negative, so the mass moment of inertia of the system (I sys ) will come out positive.
For electric motors, a coefficient of fluctuation (C f ) can be defined as
ω max − ω min ω max − ω min
C f = = (9.71)
ω max + ω min ω m
2
where the maximum angular velocity is (ω 2 ) and the minimum is (ω 1 ). Also, as the coeffi-
cient of fluctuation is usually very small, the mean angular velocity (ω m ) can be assumed
to be (ω 1 ). Therefore, the coefficient of fluctuation (C f ) becomes

ω 2 − ω 1
C f = (9.72)
ω 1