Page 412 - Marks Calculation for Machine Design

P1: Naresh
15:28
January 4, 2005
Brown.cls
Brown˙C09
APPLICATION TO MACHINES
394
specific angular velocity in revolutions per minute (rpm). The relationship between power,
mean torque, and mean angular velocity is given in Eq. (9.63) as
P = T m ω m (9.63)
Solving for the mean torque (T m ) gives
P
T m = (9.64)
ω m
Once the mean torque (T m ) is found from Eq. (9.64), rather than graphically, over a total
angle of rotation (φ) for one cycle, and using the given mean angular velocity (ω m ) and the
desired coefficient of fluctuation (C f ), the required mass moment of the system (I sys ) can
be determined from Eq. (9.62).
Consider the following example where a four-stroke, single cyliner, internal combustion
engine is to deliver continuously a specified amount of power at a specified angular speed
to a centrifugal pump, and for a given coefficient of fluctuation.
(Note, the coefficient of fluctuation (C f ) will usually be given as a percentage, as it is
the ratio of the difference between the maximum and minimum angular velocities and the
mean angular velocity.)

U.S. Customary SI/Metric
Example 3. For the engine and pump arrange- Example 3. For the engine and pump arrange-
ment presented above, determine the required ment presented above, determine the required
mass moment of inertia for the system, where mass moment of inertia for the system, where
P = 10 HP P = 8.5 kW
ω m = 1,800 rpm ω m = 1,800 rpm
φ = 4π rad (four-stroke engine) φ = 4π rad (four-stroke engine)
C f = 5% = 0.05 C f = 5% = 0.05
solution solution
Step 1. Convert the given power (P) from Step 1. Convert the given power (P) from
horsepower (HP) to (ft · lb/s). kilowatts (kW) to (N · m/s).
ft · lb N · m
550 1,000
s ft · lb s N · m
P = 10 HP × = 5,500 P = 8.5kW × = 8,500
HP s kW s
Step 2. Convert the given mean angular veloc- Step 2. Convert the given mean angular veloc-
ity (ω m ) from (rpm) to (rad/s). ity (ω m ) from (rpm) to (rad/s).
rev 2 π rad 1 min rev 2 π rad 1 min
ω m = 1,800 × × ω m = 1,800 × ×
min rev 60 s min rev 60 s
= 188.5 rad/s = 188.5 rad/s
Step 3. Substitute the power (P) from step 1 Step 3. Substitute the power (P) from step 1
and the angular velocity (ω m ) from step 2 in and the angular velocity (ω m ) from step 2 in
Eq. (9.64) to give the mean torque (T m ) as Eq. (9.64) to give the mean torque (T m ) as
ft · lb N · m
P 5,500 s P 8,500 s
T m = = rad = 29.2ft · lb T m = = rad = 45.1N · m
ω m ω m
188.5 188.5
s s