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1
CHAPTER 5
Rotary Actuators
Fluid Power Circuits and Controls,
John S.Cundiff, 2001
INTRODUCTION
±
Concepts developed for pumps are
applicable to hydraulic motors.
±
Motors convert fluid energy back into
mechanical energy and thus are the mirror
image of pumps
±
Typical motor designs are gear, vane and
piston.
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INTRODUCTION
±
Motor performance is a function of
pressure.
±
As Pressure increases,
²
leakage increases
²
speed decreases
²
quantity of mechanical energy delivered
to the load decreases.
INTRODUCTION
±
Motor volumetric efficiency is
e
vm
=
Actual motor speed
Theoretical motor speed
±
Pump volumetric efficiency is
e
vp
=
Actual flow
Theoretical flow
±
Purpose of the pump is to produce flow. Load sets the
pressure.
±
Purpose of the motor is to receive this flow and
reproduce rotary motion.
3
INTRODUCTION
±
Simple example :
²
Suppose a motor has a displacement of 3.9
in
3
/rev. Measured flow is 10 GPM. The
theoretical output speed is
N
mth
= 231Q
V
mth
where N
mth
= theoretical motor speed(rpm)
Q
= flow (GPM)
V
mth
= motor displacement (in
3
/rev)
INTRODUCTION
²
Substituting,
N
mth
= 231(10)
=
592 rpm
3.9
Assuming the measured speed is 536
rpm , the motor volumetric efficiency is
e
vm
= 536
x 100 =
90.5%
592
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INTRODUCTION
±
The overall efficiency of a hydraulic
motor is
e
om
= Actual output power
Input power
±
Input power is hydraulic power
measured at motor inlet port and the
output power is mechanical power
delivered by the motor output shaft.
INTRODUCTION
±
The motor in previous example
(V
mth
=3.9in
3
/rev) operates at a 2000 psi
pressure drop across the ports.
Measured flow to the motor is 10 GPM,
then hydraulic power input is
P
in
=
∆
PQ
1714
Where P
in
= input hydraulic power (hp)
∆
P = pressure drop (psi)
Q
= flow (GPM)
5
INTRODUCTION
±
Substituting,
P
in
= 2000(10)
=
11.67 hp
1714
Measured torque is 1080 lb
f
in at 536 rpm.
±
Output mechanical power is
P
out
=
TN / 63025
Where P
out
= mechanical power (hp)
T
= measured torque (
lb
f
in)
N
= measured speed (rpm)
INTRODUCTION
²
Substituting,
P
out
= 1080(536)
63025
=
9.18 hp
M
otor overall efficiency is
e
om
= (
P
out
/
P
in
)
x 100
= (9.18 / 11.67) X 100
=
78.7 %
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INTRODUCTION
±
Torque efficiency describes hydraulic
motor performance.
e
tm
=
Actual output torque
Theoretical torque
Theoretical output torque is
T
mth
=
∆
P
V
mth
/ 2
π
where
∆
P= pressure drop across motor
V
mth
= displacement (in
3
/rev)
INTRODUCTION
±
Substituting,
T
mth
= 2000(3.9)
= 1241
lb
f
in
2
π
Α
ctual output torque is 1080
lb
f
in,
torque efficiency is
e
tm
= (1080/1241) x 100 = 87%
7
INTRODUCTION
±
Overall efficiency is product of
volumetric and torque efficiencies.
Volumetric efficiency is
e
vm
= N/
N
mth
Where
N
= actual output speed (rpm)
N
mth
= theoretical output speed (rpm)
INTRODUCTION
Substitution of
N
mth
= 231 (Q/V
mth
)
Into previous equation gives
e
vm
= N
V
mth
/ 231Q
±
Torque Efficiency defined by
e
tm
= T/
T
mth
where
T = measured output torque (lb
f
in)
T
mth
= theoretical output torque (lb
f
in)
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This note was uploaded on 11/22/2011 for the course ABE 5152 taught by Professor Burks during the Fall '08 term at University of Florida.
 Fall '08
 Burks

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