Servo, Hydraulic - Equations

Servo, Hydraulic - Equations For reference, a servo is shown in Figure 1. Parameters are also defined in Figure 1.

Figure 1

Basic Servo Showing Flow Geometry and Parameters

Using the turbulent orifice equation, the flow expression for flow through servo flow ports is (1) ( ) ( ) 2 ρ p1 − p2 where A (x ) is the area of the valve orifice (servo port). The flow area depends on port v Q = Q x , ∆p = α A x d v v

geometry, which varies with manufacturer, valve type, and spool position. Inspection of the equation (2) indicates that the flow rate varies proportionally with area if the ∆p is held constant, and that the flow rate varies with the square root of ∆p if the flow area is held constant. Figure 2 shows notional charts of the flow behavior for a servo which are similar to orifice flow graphs.

Figure 2

Flow Rate Behavior for a Servo

Many servos use rectangular flow ports so that flow area varies linearly with spool position. For servos with rectangular flow ports, (2) A( xv ) = wxv where w is the slot width and xv is the position of the valve. A similar equation can be developed for servos with circular flow ports but the flow area will depend on the flow port diameter and spool position. In this case, the area relationship will be nonlinear. Developing an equation governing a specific servo requires developing a relationship for flow area as a function of the input parameter and determining an appropriate discharge coefficient to use in equation (1). Flow Area Determination For a two position (open/close) solenoid servo, the flow area is either Amin (which is usually zero) and Amax. Changing between Amin and Amax takes less than 100 milliseconds in most applications (see Figure 3).

Figure 3

Flow Rate Behavior for a Two Position Servo

For a proportional valve (position is proportional to current), the flow equation can be stated in terms of an applied current, uv, to the valve,

2 Av, max Q = Q(u , ∆p) = α u ∆p ≡ c u ∆p v v vu v d ρ u v, max

(3)

In equation (3), the flow area is ratio’d by the applied current, so the inputs to the equation are the applied current and ∆p across the servo. For a mechanically actuated servo valve the flow area will be a function of the mechanical linkage, Flow Area = f(Input, Geometry). Figure 4 shows an example of a mechanically actuated servo.

Figure 4

Mechanically Positioned Servo

For the servo shown in Figure 4, the spool position can be computed through the a ratio of input link lengths, for small angle movements, i.e.,

x

=x = servo v L

L AB

BC +L

BC

x input

(4)

The flow area then becomes a function of xservo. Discharge Coefficient Determination Once the flow area geometry as a function of the input is defined, the next step is to compute the discharge coefficient for the servo. Like orifices, discharge coefficient estimation is different for laminar and turbulent flow. The equations to compute discharge coefficient for servos are empirical and have been developed during laboratory testing. The discharge coefficient for a servo will be in the 0.61 to 0.70 range. A value in this range can be used if no other data is available. Typically, for turbulent flow, nominal fluid properties and rectangular port valve geometry

α

2 = 104.5 d ρ

for MIL-H-5606

and the rectangular flow port servo equation [equation (1)] becomes

Q = 104.5x w ∆p v

(5)

where xv and w are in inches, ∆p is in psi and Q becomes in3/sec. Equation (5) is the equation typically used in simulation models for servovalves. For low Reynolds numbers, an empirical formula for laminar-turbulent servo flow is

  Q=c x  v v   where

A = π dx c = lt

 2 c  c   lt  + ∆p − lt   x  x  v v 

(6)

v

δυ ρ Re

crit 2 4 2δ

δ = slope of the curve α = δ Re for orifice flow d

Recrit values are same values used for turbulent-laminar orifice flow (see Orifice Flow, Hydraulic Equations). A typical value for clt is 0.6. Other means to estimate the discharge coefficient are empirical (based on test data). An empirical approximation for the discharge coefficient is given by α =α d

  x v  x = α 1 − k  d v d0  d,corr x v,max  

()

(7)

where αd0

basic discharge coefficient

kd,corr xv,max

(typically αd0 = .65) correction factor (typically kd,corr = .32) maximum stroke of the spool

Equation (7) could be used with equation (1) when no data is available for discharge coefficient and flow port geometry is not available. The flow coefficient may also be obtained experimentally or using catalog data from the manufacturer (QN, ∆pn, xv,max). For a 4 way valve,

c = v

Q ∆p

1

N N

where QN

x 2

(8)

v, max

nominal (rated) flow

∆PN xv,max

nominal (rated) pressure drop maximum stroke of the spool

The corresponding discharge coefficient is

α = d

Q

N

(v, max )

A x

∆p

N

ρ

(9)

which is based on rated flow at rated pressure and maximum servo stroke. Instead of using stroke, valve current can also be used. Using current, equation (9) becomes

αd =

imax

where QN

QN ∆pN ρ

(10)

nominal (rated) flow

∆PN imax

nominal (rated) pressure drop maximum (rated) current for the valve

When using equation (10), the flow equation for the servo becomes

(

)

i Q = Q x , ∆p = α 2 ρ p −p v 1 di 2 max

(11)

In equation (11), i is the applied current, which in effect ratios the flow area between zero and the maximum flow area.

Axial Spool Flow Forces The force of flow on a servo is caused by a change in the momentum of the flow, due to differences in jet angles between inlet and outlet flows (see Figure 1) (e.g., for flow in the direction shown p1 > p2). In Figure 1, the flow forces act in a direction to move the servo to the left. Spool flow forces have a steady state and dynamic component. The steady state axial flow force on the spool can be calculated using

F = 2 α A(x ) (cosθ )∆p ax, steady d v

(7)

The jet angle can be assumed constant at θ = 69°, which is the theoretical value if no radial clearance existed between spool and sleeve. As stated above, the direction of the flow force is opposite the flow direction. For unsteady flow, the fluid in the valve chamber is accelerated. This creates a dynamic force which is reacted on the spool valve lands with a magnitude given by

 Q&  Fax,dyn = ma = (ρlAs )  = ρlQ&  As 

(8)

Using the orifice flow equation, the dynamic force becomes

Fax,dyn = ρlcv x&v ∆p + ρlcv xv

∆&p 2 ∆p

(9)

Dynamic force is proportional to spool velocity and pressure changes. Pressure rate change is often small and can be neglected. Since the force, Fax,dyn, tends to close the valve this term provides damping (i.e., improves stability). Note that the 1st term of equation (12) contains l, which is the length between the ports, and therefore, by increasing the distance between the ports the effective damping in the servo is increased. Equations (12) and (14) are used in the dynamic model of spool. Dynamic Modeling of Servos The pressure derivative for a servo is similar to a pipe. If we let P2, outlet pressure, and Q1, inlet flow, be the inputs to the servo model, the outputs become P1, inlet pressure, and Q2, outlet flow, using

(

β P& (t) = Q −Q 1 1 2 V p

)

(10)

The outlet flow, Q2, is computed by the appropriate flow equation for the servo geometry. The area in the flow equation will, of course, be a function of spool position. Hence spool position will be an input to a servo component model. The flow equation for Q2 is based on equation (1) for the particular valve.

(

)

( )2 ρ

Q = Q x , ∆p = α A x d v v 2

p −p 1 2

(11)

where A(xv) is based on the valve characteristics discussed above. In some cases, the dynamics associated with servo position may be important. For analysis, the servo dynamics may be important for a proportional controlled servo or for a servovalve (see Servovalve, Hydraulic - Equations).

Figure 5

Basic Servo Showing Flow Geometry and Parameters

Referencing Figure 5, the dynamic equation for spool position is computed from a mass balance on the spool,

x + F ( x& ) + k x + F (x , P , P ) = F m && f v s v ax v 1 2 s v 0

(12)

where ks is the spring constant and F0 is the spring preload. Equation (17) is a second order equation, with a natural frequency given by

ω = n

k

s m s

(rad/sec)

(13)

Equation (16) then represents the bandwidth of the system, however, the servo may have low damping and therefore the servo should be operated at a maximum input frequency much less than the natural frequency. Flow Equations for a 4 Way Servo To expand on the flow curves shown in Servo, Hydraulic – Description and Servovalve, Hydraulic – Description, the equations for a 4 way valve are presented below. A 4 way valve is shown in Figure 6. Note that the servo is equivalent to 2 orifices connected in series with a load. For a symmetric valve, the orifice areas will be equal for any spool position.

Figure 6

4 Way Servo Schematic

For the Q1 and Q2 orifices,

Q = α A (x ) (2 ρ ) ( p − p = x C ∆p d 1 v 1 S A v 1 1

Q = α A (x ) (2 ρ ) ( p − p = x C ∆p d 2 v B R v 2 2 2 Rewriting

∆p = 1

Q2 1

(xvC1 )

2

and

∆p = 2

Q2 2

(xvC2 )

2

(14)

(15)

where ∆p1 and ∆p2 are the pressure drops through each valve orifice. The total pressure drop

through the valve is

∆P = ∆p + ∆p = 1 2

Q2 1

+

Q2 2

(16)

(xvC1) (xvC2 ) 2

2

For a symmetric valve with equal flow areas and equal flow through each port,

Q2 ∆P = ∆p + ∆p = 1 2 x2 v

 C2 + C 2  2 2=Q  1  C2 C2  x2  1 2  v

 2   2 C 

(17)

Rewriting in terms of flow

x C Q= v ∆P = C x ∆P v v 2

(18)

Equation (23) can be further reduced by assuming the return pressure, pR, is negligible so that

(

)

Q=C x p − p − p =C x p − p v v S A B v v S L

(19)

For terminology, the parameters in equation (24) are listed below Q Cv xv ps pL

flow through the load or load flow valve coefficient spool displacement supply pressure pressure drop across actuator piston or load pressure

A plot of flow versus load pressure using equation (24) at several valve positions is shown in Figure 7.

Figure 7

Flow versus Load Pressure for a 4 Way Servo

The square root in equation (24) leads to the nonlinear flow curve. However, flow will increase linearly with valve position assuming a rectangular flow port.