Download 3 Valve actuator system - simple control analysis

TEP4195 TURBOMACHINERY
5 VALVE ACTUATOR SYSTEMS
SIMPLIFIED ANALYSIS
1
Linear Actuators
Linear actuators operate by producing a force as a result of a pressure difference across the area of
the piston where the level of the pressures acting will be dependent on the method used to supply the
flow to the actuator.
Force F
P1, Q1
Velocity U
Displacement Y
P2, Q2
X
PS
Figure 1 Linear equal area actuator.
Referring to Figure 1, for an equal area, or double-ended, actuator, the force on the piston is:
( P1 - P 2 ) Aa = Force (F)
Therefore, so long as ( P1 - P2 ) has the same value, the force produced will be the same and will not
be affected by the level of the pressure.
The actuator converts pressure into force and flow into velocity where:
Flow (Q) = Velocity (U) x Area (A)
The power transmitted by the fluid due to pressure and flow is converted into mechanical power due to
force and velocity.
Power = P Q = F U
Equal area actuators are often used in closed loop control systems because of the force and velocity
symmetry when working in either direction. Unequal area, or single-ended, actuators are also used in
many applications but these have non-symmetry of force and velocity in relation to the direction of the
valve opening. This inevitably leads to an increase in the complexity of the dynamic equations for
such a valve operated system and the analytical methods used in this course relate to equal area
actuators only. The steady state performance of unequal area actuators will, however, be considered.
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2
Ratings
2.1
Pressure
The maximum pressure is limited by the:
• Bursting strength of the cylinders (to include the effects of fatigue - around 1/4 to 1/5 of the maximum
static burst pressure)
• Seals on the piston and the rod. These are also the main limitation to the maximum velocity. This is
usually in the region of 5 to 10 m/s.
2.2
Thrust
For a given piston size there will be a limit to the maximum permissible thrust which is dependent on
the actuator stroke. This is because, during extension against an opposing load, the rod is acting as a
strut.
In applications of unequal area actuators in which the load can change direction, the diameter of the
rod will limit the available reverse thrust because of the reduced annulus area compared to that of the
piston. The structural strength of the rod is considerably affected by the method employed to mount
the actuator and on how the driven load is located (eg possible generation of side loads etc...)
In applications where limited movement is required hydraulic actuators have distinct advantages over
other types which include:
• High thrust to weight ratio
• Ability to place the driving force where it is required
• Low wear of the moving components giving good life
• Reasonably high velocity
• Self braking capability with blocked ports
2.3. Valves for controlling actuators
2.3.1 Valve Characteristics
Valves are used to introduce a reduced area into the flow path that creates an increase in the fluid
velocity and a consequent reduction in its pressure. The fluid velocity is then reduced downstream as
the flow area is increased but the amount of pressure increase downstream of the restriction is
dependent on the way in which the area reduction has been effected.
In the simple orifice the equation for the flow in relation to the overall pressure is given by:
Q = CQ a (
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2 P

)1/2
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Here:
a is the area of the orifice
CQ is the flow coefficient (usually around 0.7)
 is the density of the fluid
P is the orifice pressure drop.
For the actuator extending against the opposing force the way in which flow varies with pressure drop
is shown in Figure 2.
P1
P2
Flow Q
x
PS/2
PS
Pressure
Figure 2 Valve pressure drop
Here the pressure drops across the valve metering lands are:
P1  PS  P1 and P2  P2  PR and assuming PR  0, P2  P2
2.3.2
Valve metering area
The metering area of the valve opening can be fully annular as shown in the Figure 3 or, by using a
slot, have reduced opening.
Port
d
x
Figure 3 Valve area
Valve metering area =  d x
or for a slot = b x where b is the total width of the slot.
The most common type of control valve is a four- way valve with a metering section to each service
outlet and for each service to tank. A standard test for the valve is with the A and B ports connected
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together. For the valve shown in Figure 4 with supply pressure PS the port pressures = PS / 2 because
the metering areas are the same as are the flows.
Supply
P port
Return
T port
A port
B port
Return
T port
Figure 4 – Valve test
This is shown from the pressure/flow characteristics for each of the two restrictions in Figure 5. The
pressure drop across each metering restriction is the same and = PS / 2 at the intersection of the two
flow characteristics for equal flow.
Q
P1
P2
PS/2
PS P
Figure 5 Pressure-flow characteristics
For a given supply pressure the relationship between flow and valve position will be as shown in
Figure 6.
Q
xmax
x
Figure 6 Flow variation with X (Valve gain)
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Here:
P = PS /2
and:
Q  CQ dx (
Q  Kx
Or:
2 PS
)
 2
PS
2
where:
K = CQ d
2

= a constant
CQ = valve metering flow coefficient (usually 0.65 to 0.7)
The flow gain is given by:
P
P
Q
 K S  KQ  CQ d S
x
2

If PS is constant then the flow gain (KQ) is constant.
2.3.3 Valve lap
T
P
U+X
U-X
QT
Q1
QP, PP
X
Figure 7 Valve lap
Valves can be arranged to have different lap conditions, which refer to the width along the valve
centreline of the valve land in relation to that of the port as shown in Figure 7.
• Underlap is created when the valve land width is less than that of the port. This causes leakage in
the central, or neutral position, and modifies the valve characteristics whilst the spool is operating in
the underlap region.
• Zero lap refers to valves having a land width which is the same as that of the port, the valve being
just closed in the neutral position.
• Overlap applies to valves in which the land width is greater than that of the port. This reduces
leakage in the neutral position but creates a deadband.
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These three conditions are shown in Figure 8.
Q
Underlap
Zero lap
Overlap
x
Figure 8 Valve lap
2.3.4 Pressure gain measurement
Valves are tested with both service ports blocked to assess the amount of leakage in the neutral
position and this is shown in Figure 9.
Supply
P port
PS
A port
B port
P2
P1
Ports blocked
Figure 9 Pressure gain test
This is an important characteristic of the valve as it has a strong influence on the accuracy of the
closed loop position control of a linear actuator. The force generated by the actuator is proportional to
the pressure difference P1 - P2. It can be seen, therefore, that if the pressure difference required by
the load force varies then the valve position must change to maintain the static pressure conditions.
Valves having zero lap will always have some leakage in the null position because of manufacturing
tolerances and wear of the metering edges during use. The pressure gain is measured with the A and
B ports blocked so that the pressure in each port will vary with the valve position as shown in the next
Figure.
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P1 – P2
PS
5 to 10% max
Valve Position
-PS
Figure 10 Valve pressure gain
3
Valve actuator system - simple control analysis
3.1
Open loop system
Force F
P1, Q1
Velocity U
Displacement Y
P2, Q2
X
PS, Q
Figure 11 Valve Actuator Circuit.
The valve actuator circuit in Figure 11 has an equal area, or double ended, actuator that is frequently
used for closed loop control systems because of its symmetry with regard to the hydraulic force and
velocity in both directions of movement. For a force, F, that is acting against the movement of the
actuator rod as shown in Figure 11 the pressures are given by:
Extend
Retract
P1 
PS
F

2 2A
P1 
PS
F

2
2A
P1 
PS
for F  0
2
P1 
PS
for F  0
2
P2 
PS
F

2 2A
P2 
PS
F

2 2A
P2 
PS
for F  0
2
P2 
PS
2
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for F  0
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F
2A
F
2A
P2
Flow Q
x
P1
PS/2
PS
Pressure
Figure 12 Valve Characteristics
The force F is positive for the direction shown in Figure 11. For the equal area actuator, the valve
flows are the same on each side of the valve and the pressure differences, (Ps - P1) and P2
respectively, will therefore be the same.
For zero force, P1 = P2 = PS/2 that, as can be seen from Figure 12, is the pressure at which the flow
characteristics intersect.
The flows through the valve is given by:
Q1,2  Q  CQ dX
2 PS
 2
For the simple system assume that the moving components have negligible inertia so that, as a
consequence, the actuator pressures will remain constant during transient changes caused by
displacement of the valve.
Thus:
Q  KQ X
K Q  CQ d
where:
(1)
PS

m2 / s
And, for an actuator area A the actuator velocity, U is:
U 
As U 
KQ
dY
, then Y 
A
dt
Q KQ X

A
A
(2)
 X dt
(3)
The use of the Laplace Transform allows the equation to be written as:
U
KQ
KQ
dY
 sY 
X or Y 
X
dt
A
As
Here the initial conditions for Y are zero. Thus s 
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d
1
and 
dt
s
(4)
 dt .
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The actuator displacement is the integral of the valve opening and, as a consequence, it is referred to
as an integrator. A step opening of the valve will, therefore, cause the actuator to move at a constant
velocity, stopping when the valve is closed. This describes the open-loop performance of the system.
The time response of the actuator, or integrator, to a step change in valve position is shown in Figure
13.
Y
Q
X
time
Figure 13 Open loop time response.
3.2
Closed loop system
Closed loop systems operate by comparing the output position with an input demand signal such
systems being referred to as feed back control. In electrohydraulically operated valve actuator
systems, the input signal is a voltage and the output position is fed back as a voltage signal from a
position transducer. The comparison of the two signals produces a voltage difference referred to as
the error signal that is amplified as a current signal and supplied to the electrohydraulic valve. In this
analysis the relationship between the input and output is obtained from the following equations.
The valve position is given by:
X  K A KV (Vi  KT Y )
KT = position transducer gain, V/m
KV = valve gain, m/A (or m/V)
KA = amplifier gain, A/V (or V/V)
Vi = input signal, V
Vi
(5)
+
KA
A
KV
x
_
V0
Equations 2, 3 and 5 give:
KT
Y
dY K Q

( K A KV (Vi  KT Y )
dt
A
Expressing
d
by the Laplace operator ‘s’ allowing this equation to be treated algebraically.
dt
Thus:
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( KT 
A
s )Y  Vi
K Q K A KV
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1
KT
Y

Vi
(1  Ts )
This gives:
This is a first order differential equation where the time constant T 
(6)
A
which has the units
K A K Q KV KT
of time (s).
Note that electrohydraulic manufacturer’s literature will usually specify the valve performance in terms
of flow for a given input current (or voltage) at a specific valve pressure drop, which will enable the
gain product KVKQ to be obtained.
3.3
System response
The time response from equation 6 can be obtained for a range of input signals by referring to a
dictionary of standard inverse Laplace transforms. The response to a step input gives the solution:
Y 
Vi
t
(1  e T )
KT
(6)
This exponential variation of Y with time reaches 63.2% of the final value in one time constant as
shown in Figure 14.
Y=Vi /KT = Ym
Y=0.63Ym
Y
t
T
Figure 14 Step response
We have seen that the flow through a valve depends upon the valve opening and the pressure drop
across the opening. Thus, under specified pressure conditions, the valve can be used as a flow
source when connected to an actuator.
For the situation where the supply pressure is constant and there is zero force acting on the
actuator, the valve pressure drop will be constant and consequently, the velocity of the actuator will
be proportional to the valve opening.
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