Page 409 - Marks Calculation for Machine Design

P1: Naresh
15:28
January 4, 2005
Brown˙C09
Brown.cls
391
MACHINE ENERGY
If the flywheel and shaft assembly is accelerated from one angular velocity (ω 1 ) to another
angular velocity (ω 2 ), either speeding up or slowing down, the change in inertial energy
levels is the work done on or by the system, denoted (Work), and is given by the relationship
in Eq. (9.55) as 1→2
1 2 2
Work = I total ω − ω 1 (9.55)
2
1→2 2
If the system is speeding up, the work done(Work) is positive. Conversely, if the system
1→2
is slowing down, the work done (Work) is negative.
1→2
U.S. Customary SI/Metric
Example 2. For the flywheel and shaft Example 2. For the flywheel and shaft
assembly of Example 1, calculate the work done assembly of Example 1, calculate the work done
(Work) to increase its speed, where (Work) to increase its speed, where
1→2 1→2
ω 1 = 1,000 rpm ω 1 = 1,000 rpm
ω 2 = 1,500 rpm ω 2 = 1,500 rpm
2
2
I total = 29.82 slug · ft (from Example 1) I total = 39.89 kg · m (from Example 1)
solution solution
Step 1. Convert the given angular velocities Step 1. Convert the given angular velocities
from (rpm) to (rad/s). from (rpm) to (rad/s)
rev 2 π rad 1 min rev 2 π rad 1 min
ω 1 = 1,000 × × ω 1 = 1,000 × ×
min rev 60 s min rev 60 s
= 105 rad/s = 105 rad/s
rev 2 π rad 1 min rev 2 π rad 1 min
ω 2 = 1,500 × × ω 2 = 1,500 × ×
min rev 60 s min rev 60 s
= 157 rad/s = 157 rad/s
Step 2. Substitute the angular velocities from Step 2. Substitute the angular velocities from
step 1 and the total mass moment of inertia for step 1 and the total mass moment of inertia for
the system from Example 1 in Eq. (9.55) to give the system from Example 1 in Eq. (9.55) to give
the work done as the work done as
1 2 2 1 2 2
Work = I total ω − ω 1 Work = I total ω − ω 1
2
2
1→2 2 1→2 2
1 2 1 2
= (29.82 slug · ft ) = (39.89 kg · m )
2 2
2 2 2 2
rad rad rad rad

× 157 − 105 × 157 − 105
s s s s
1 1
2
2
= (29.82 slug · ft ) = (39.89 kg · m )
2 2
rad rad

× 13,624 × 13,624
s 2 s 2
slug · ft 2 kg · m 2
= 2.03 × 10 5 = 2.72 × 10 5
s 2 s 2
5
5
= 2.03 × 10 ft · lb = 2.72 × 10 N · m
= 203 ft · kip = 272 kN · m
As the work done is positive, work was done As the work done is positive, work was done
on the system. on the system.