mathcad - hydraulic cylinder

7/29/2019 Mathcad - Hydraulic Cylinder

1/27

Hydraulic Cylinder Simulation

Simulating a hydraulic cylinder starts with the basic F=m*a and then is enhanced by

doing a series of substitutions. The net force is equal to the force applied on the piston

minus the opposing force required to push oil out the other side and the opposing force

due to friction.

Fnet Fa Fb Fl Ff=

Substitute for the for the force on both sides of the piston and the frictional and load

forces. Assume the friction is proportional to the velocity. Ignore static friction for now

and just model the kinetic friction

aPa Aa Pb Ab Fl Ff

mass= The acceleration that must be

integrated to provide velocity and

position.

Since a is the same astv

d

d

The acceleration or rate of

change in velocitytv

d

d

Pa Aa Pb Ab

mass=

tP

d

d

Q

V=

Expressed as a simple ODE

Since the moving piston changes the volume orV on both sides of the piston the velocityof the piston must also be taken into account

tP

d

d

VQ v A+( )=

Calculate how pressure changes as a function of net flowVa and Vb are the volumes of oil between the valve or pump and the cylinder piston. The

subroutines Qa(Pa) and Qb(Pb) are needed to calculate the net flow.

tPa

d

d

VaQ Pa( ) v Aa( )= Calculate the rate of

change in pressure as a

function of flow and

velocity.

tPb

d

d

VbQ Pb( ) v Ab+( )=

(c) 2010 Delta Computer Systems, Inc. 1/27 1/16/2013 10:10 AM


2/27


Simulate the Hydraulic Actuator Acceleration

Pt 14.7:= Tank pressure psia

Ps0 2000:= Initial system pressure, psiaCylDia 2:=

RodDia 1.375:=

Aa

4CylDia

2:= Aa 3.142=

Ab

4CylDia

2RodDia

2( ):= Ab 1.657=

CylLen 24:=

W 600:= Weight in lbf

Slinch. A slinch is 12 slugs. 1 lb force

accelerates a slinch at 1 in/s^2mass

W

386:= mass 1.554=

Fl W sin 0( ):= Fl 0=

Kf 5:=lbf sec

infrictional velocity constant

Vamin 5:= Cubic inches of oil between the valve and the pumpTry setting the volumes to 1000 and then 10000.

Vbmin 5 0.25 CylLen+:=

160000:= The bulk modulus of oil

Calculate the natural frequency of the cylinder and system

n

mass

Aa2

Vamin AaCylLen

2+

Ab2

Vbmin AbCylLen

2+

+

:=

n 181.497= Natural frequency in radians per second

Calculate the acceleration time using the natural

frequency method.

18

n0.099=



3/27


Pa00.5 Ps0 Ab Fl

Aa Ab+:= Pa0 345.269= The initial pressure on the A and B port

should be 1/2 the supply pressure if the

valve leakage from P->A and A->T is

identical. The same applies to the B port.Pb0

0.5 Ps0 Aa Fl+

Aa Ab+:= Pb0 654.731=

Pa0 Aa Pb0 Ab 2.274 1013

=

Calculate the valve flow coefficient

GPM 27:= Rated valve flow at rated pressure drop.

P 1015:= Rated pressure

Calculate the flow constant for

the powered land. Yields flow in

cubic inches per secondKv

GPM231

60

P

2

:= Kv 4.614=



4/27


Valve Spool ResponseThe valve spool response can be roughly simulated by a natural frequency and a damping

factor. For the purposes of this model the spool's natural frequency and damping factor are

assumed to be fixed but in reality the natural frequency drops as the amplitude increases. Thismay be due to the fact that is takes significantly more power to move at twice the frequency.

s 20 2 := s 125.664= A 20 Hz valve

s 0.7:= Spool damping factor. The closer to 1the better.

Assume the spool is a second order

system and possibly under damped.

The Bode plots show this to be the

case

Ts s( )s

2

s2

2 s s s+ s2

+

:=

n 0 320..:= Iterate the frequency from .01 HZto 1000 HZ

using 64 steps per decade for a

total of 256

iterations.hzn

10

n

642

:=

Mn

n

r Ts j 2 ( ) hzn

aarg r( )

deg

a a 360 n 0> a n 1 100>if20 log r( )

a

:=Calculate magnitude and phase for T(s)



5/27


0.01 0.1 1 10 100 1.10

380

60

40

20

0

20

0

50

100

150

200

Magnitude

Phase

Valve Spool Bode Plot

Frequency Hz

Magnitudedb

Phasedeg

The Bode plot looks something like the plots provided by the valve manufacturers. It

must be acknowledge that this approximation is better than nothing but there is much

more to modeling just the valve. For instance the valve has on board electronics and

closed loop control which alter the response. The controller may even have a PID

control but I best most have just a simple proportional controller. Still the magnitude

is down -3db at 20 Hz. This means that at 20 Hz one must provide a 20% signal to

get a 10% response from the valve. This works at low amplitudes but the controllercan at most provide a 100% signal and get 50% amplitude from the valve 20Hz.

There are two difference equations for the spool because it is a second order system.

There is one for the velocity of the spool and one for the position of the spool

txs

d

dvs=

Calculate the spool position

tvs

d

d2 s s vs s

2u t( ) xs( )+= Calculate the spool velocity



6/27


Verifying the valve spool response

y0

0

:= Initial spool position and velocity are 0

The control signal to the valve is asimple step signal of 90%

u t( ) 100:=

Time 0.1:= Simulation time

D t y,( )xs

vs

y

vs

s2

u t( ) xs( ) 2 s s vs

:=

S rkfixed y 0, Time, 100, D,( ):=

0 0.02 0.04 0.06 0.08 0.10

50

100

150

2000

0

2000

4000

6000

Spool Position

Spool Velocity

Spool Response to a Step Input

Time

SpoolP

osition,percent

SpoolVelocit

y,percentpersecond

One can see that the model is close to 100% open in 25 milliseconds. The valve is actually

open 101% at 25 milliseconds which is close to the valve documentation. The over shoot is

caused because the spool is under damped as indicated by the damping factor being less than

1. In reality the spool will probably hit an end of travel stop instead of going over 100%.



7/27


Target Generator for Position, Velocity and AccelerationGeneration motion profile using a cosine ramp target generator. This target

generator will calculate a position, velocity and acceleration at time t. The

motion profile assumes the ramp up, constant velocity and ramp down

sections each take 1/3 of the total move time. All that is required is themove time and the distance to move.

Cosine Ramps

MoveTime 2:= MoveDist 16:= x0 4:=

MaxVel3

2

MoveDist

MoveTime:= MaxVel 12= AvgAcc

9

2

MoveDist

MoveTime2

:= AvgAcc 18=

r t t, x,( )

3

4xt

t 1

3t

sin 3

tt

x0+

3

4

x

t1 cos 3

tt

9

4

x

t2

sin 3

tt

0 tt

3


8/27


Linear Ramps

r t t, x,( )

9

4

x

t2t2

x0+

9

2

x

t2

t

9

2

x

t2

0 tt

3


9/27


Define the valve control signal a function of time, u(t)The control signal can be ramped shut to simulate quick stops and check for pressure increases.

t 0.001:=

t0 0:= Start ramping up

t1MoveTime

3:= End of ramp up

t22 MoveTime

3:= Start of ramp down

End of ramp down. Make the difference between t3

and t2 bigger to get a ramp that doesn't oscillate.t3 MoveTime:=

N MoveTime 1+t

:= N 3000= Ending time and number of iterations

Open loop Control

u t( ) 0.5 1 cos t t0

t1 t0

t0 t t1A and

B->T.



10/27


Estimate Valve LeakageCalculate the leakage flow constant K l from the leakage specification The total leakage flow is the sum of the flow from P->A and P->B. Since the orifice is

assumed to be symmetrical we can assume the flow from P->A is equal to the flow fromP->B therefore the total leakage flow Ql is equal to 2 times the leakage from P->A, Q pa.

Ql 0.25231

60:= Ql 0.963= The specification for valve leakage for

my NG10 is about .25 gpm. Convert to

cubic inches per second.

The total valve leakage, Ql, is equal to the leakage from P->A plus the leakage from P->B.

Assuming the leakage is symmetrical then the total leakage is 2 times the leakage from P->A.

The leakage flow is 2 times

the flow from A->BQ

l2 Q

pa= 2 K

l

Ps

2=

Ql Kl 2 Ps= Combine some constants

Kl

Ql

2 Ps0:= Divide both sides by the

square root term

Kl 0.015= Leakage flow constant. This is verysmall relative to Kv.

Select valve over/under lap type.The valve overlap is determined by x l. The spool position, xs, is in percent of full spool travel.

xl,

should be a small number in per cent overlap or under lap. A value of 0 signifies a 0

lapped spool whereas negative value specify under lapped spools and positive values signify

overlapped spools.

x.lt allows one to specify a different leakage coefficient A->T and B->T that differs from P->A

and P->B.

Spool over/under lap variable in percent.

Set negative of under lap and positive for

over lap spools. For the pressure port

xl 0:=

xlt xl:= Spool over/under lap for the tank port.Normally this should be the same as the

pressure port over/under lap.



11/27


Calculate the flow coefficients as a function of spool position

The flow around each edge of the spool is modeled using a simple function. Basically there

are two states. Either the edge is in a normal flowing condition or it is blocked and in a

leakage state. This results in two linear segments that will determine the flow coefficient asa function of the position as it goes from -100% to +100%.

Kpa xs( ) Kv Kl( )xs xl

100 xl Kl+ xs xl( ) 0if

Kl

100 xs+

100 xl+

otherwise

:= Calculate the flow coefficient from theP to A ports as a function of the spool

position

Kat xs( ) Kv Kl( )xs xlt

100 xlt

Kl+ xs xlt( ) 0if

Kl

100 xs

100 xlt+

otherwise

:=

Calculate the flow coefficient from the

A to T ports as a function of the spool

position


P to B ports as a function of the spool

position

Kpb xs( ) Kv Kl( )xs xl

100 xl Kl+ xs xl+ 0if

Kl

100 xs

100 xl+

otherwise

:=


B to T ports as a function of the spool

position

Kbt xs( ) Kv Kl( )xs xlt

100 xlt

Kl+ xs xlt+ 0if

Kl

100 xs+

100 xlt+

otherwise

:=

Note, the spools are assume to be cut so that the flow changes linearly with spool position.

This is not always the case. There are 'curvilinear' and 'nick' spools. Modeling either one of

these spools would require modifications to the four flow coefficient functions above. For

instance, and 'nick' spool can be simulated by adding a third linear segment with a higher

gain when the spool reaches 20% or 40% depending on the spool.



12/27


Sanity check the flow constant formula by graphing the flow coefficients as afunction of the control signal. Note that the control signal goes from -1 to 1.

x 100 099.9, 100..:=

100 50 0 50 100

0

2

4

6

Flow Coefficient P to A

Control Signal %

FlowCoefficient

100 50 0 50 100

0

2

4

6

Flow Coefficient P to B

Control Signal %

FlowCoefficent

100 50 0 50 1000

2

4

6

Flow Coefficient A to T

Control Signal %

F

lowCoefficient

100 50 0 50 1000

2

4

6

Flow Coefficient B to T

Control Signal %

F

lowCoefficient



13/27


Simulating Pump and AccumulatorsA real hydraulic system does not have a constant supply pressure. The supply pressure drops

when a lot of flow is required. To offset this accumulators are used to supply energy. This allows

the designer to use a pump that supplies just a little more than the average flow. During slow or

dwell times the oil/energy is stored in accumulators and during times when the flow exceeds thepump capacity the accumulator can make up the difference. To start lets assume the pump is a

fixed displacement pump of 60 gpm and that any excess oil flows over a relief valve at 2000 psi.

Also, assume the accumulator capacity is 10 gal accumulator.

GPM 5:= GPM231

60 19.25= Pump capacity

PPB 200:= Pump proportional band

p 0.050:= Pump time constant

AccumSize 5:= Accumulator size in gallons

Va AccumSize 231:= Va 1155= Accumulator size in cubic inches

1.4:= Ratio of specific heat Cp/Cv

Calculate the Supply Pressure and Accumulator Volume As A Function of Time.

P0 0.9 Ps0:= P0 1800= Pre charge pressure

Calculate K, nrT, using pre

charge values calculated abovefor pressure calculations below.

K P0 Va

:= K 3.49 107

=

K Ps Vg

=solve Vg,

explicite

lnK

Ps

Vg0 e

lnK

Ps0

:= Vg0 1071.268= Initial gas volume in cubic inches

Ps Vg( )K

Vg

:= Ps Vg0( ) 2000= Supply pressure as a function of theaccumulator volume.



14/27


Total Flow Equations

Qa and Qb are the net flow to either side of the cylinder piston. The total flow equations are

important since they determine the rate of change in pressure,P=*V/V orP=*Q*t.The current flow equation has four parts. The top equation calculates the flow from the supplyto the A or B port and the second equation calculates the flow from the A or B port to the

tank.The leakage terms are very important. Without the leakage terms the system will tend

to oscillate for an unrealistically long time since there is little if any energy loss when the

valve is shut. When accelerating there is plenty of energy loss due to the pressure drop

across the valve. You can see there isn't any hint of oscillating because the damping factor

is very high when the spools is wide open.

Qpa Pa xs, Vg,( ) Kpa xs 100( ) sign Ps Vg( ) Pa( ) Ps Vg( ) Pa:=

Qat Pa xs,( ) Kat xs 100( ) sign Pa Pt( ) Pa Pt:=

Qpb Pb xs, Vg,( ) Kpb xs 100( ) sign Ps Vg( ) Pb( ) Ps Vg( ) Pb:=

Qbt Pb xs,( ) Kbt xs 100( ) sign Pb Pt( ) Pb Pt:=

Qa Pa xs, Vg,( ) Qpa Pa xs, Vg,( ) Qat Pa xs,( )+:=

Qb Pb xs, Vg,( ) Qpb Pb xs, Vg,( ) Qbt Pb xs,( )+:=

Force load as a function of time and distanceThe force can be changed as a function of time or distance.

Fl t x,( ) 2000t 0.40 MoveTime

0.03 MoveTime

0.40 MoveTime t 0.43 MoveTime


15/27


When extending against a load

v 1:= The spool is symetrical

Ape Aa:= Ape 3.142=

FL 0:= FL 0=

cAa

Ab:= c 1.896= Ratio of the powered end area

the exhausting end area

Compute maximum steady state velocity in each direction to estimate velocity feed forwards

Calculate the maximum velocity

when the control output is one

using the VCCM equation

vss_ext KvPs0 Ape FL

Ape3

1v

2

c3+

:= vss_ext 61.342=

Ape Ab:=

FL 0:= No load when retracting

cAb

Aa:= c 0.527= Ratio of the powered end area

the exhausting end area

vss_ret Kv Ps0 Ape FL

Ape3

1v

2

c3

+

:= vss_ret 44.545=



16/27


Closed Loop ControlDefine the valve control signal a function of error between the target and

actual position, velocity and acceleration.

t 0.0001:= Update period. The update period must be smalleras the system response get faster.

Controller gains.

Ki 0.5:= Integrator gain

Kp 0.1:= Proportional gain

Kd 0.001:= Derivative gain

Kv_ext1

vss_ext:= Kv_ext 0.016= Velocity feed forward

Kv_ret1

vss_ret

:= Kv_ret 0.022=

Ka 0.00018:= Acceleration feed forward

Control output goes from -1 to 1

PID u xt, vt, at, x, v, a,( ) Kv Kv_ext vt 0>if

Kv Kv_ret vt 0


17/27


The Simulation Using Runge-Kutta.Execute the equations simultaneously by putting the equations and results in a matrix.

There are now eight differential equations that are computed simultaneously.

ActAcc t Pa, Pb, x, v, xt, vt,( )Pa Aa Pb Ab( ) Kf v Fl t x,( )

mass:= Calculate the actual

acceleration

Initial spool position

initial spool velocity

Initial A port pressure

Initial B port pressure

Initial velocity

Initial position

Initial control outputInitial accumulator gas volume

y

0

0

Pa0

Pb0

0

x0

0

Qp 0

Vg0

:=



18/27


D t y,( )

xt

vt

at

r t MoveTime, MoveDist,( )

xs

vs

Pa

Pb

v

x

u

Qp

Vg

y

a ActAcc t Pa, Pb, x, v, xt, vt,( )u

u'

PID u xt, vt, at, x, v, a,( )

vs

s2

u xs( ) 2 s s vs

Vamin x Aa+Qa Pa xs, Vg,( ) v Aa( )

Vbmin CylLen x( ) Ab+

Qb Pb xs, Vg,( ) v Ab+( )

a

v

u'

1

pmax min

Ps0 Ps Vg( )

PPB1,

0,

GPM231

60 Qp

Qpa Pa xs, Vg,( ) Qpb Pb xs, Vg,( )+ Qp

:=

SimTime MoveTime 1+:= Simulation time

NSimTime

t:= N 30000= Number of iterations

S rkfixed y 0, SimTime, N, D,( ):=



19/27


t S0 := xs S

1 := Vg S 3

:= Pa S 3

:= Pb S 4

:=

v S5 := x S 6

:= u S 7

:= Qp S 8

:= Vg S 9

:=

n 0 N..:=

xtn

vtn

atn

r n t MoveTime, MoveDist,( ):=

an

ActAcc tn

Pan

, Pbn

, xn

, vn

, xtn

, vtn

,( ):=

COn PID un xtn, vtn, atn, xn, vn, an,( )0:=



20/27


t xs vs Pa Pb v x u Qp

S

0 1 2 3 4 5 6 7 8

0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25

26

27

28

0 0 0 345.269 654.731 0 4 0 0

0 -1111810 -61.8310 345.58 654.794 0 4 -8.75810 -82.9610

0 -1085510 -6.24310 345.892 654.857 0 4 -7.10610 -7.18310

0 -9.62510 0 346.203 654.92 0 4 -7.49610 -72.6610

0 -9.82110 0 346.513 654.983 0 4 -7.44910 -7.72610

0.001 -9.40310 0 346.822 655.046 0.001 4 -76.9710 -7.37910

0.001 -8.26910 0 347.131 655.109 0.001 4 -6.00610 -6.06210

0.001 -8.99810 0 347.438 655.173 0.001 4 -6.37210 -6.44410

0.001 -8.95810 0 347.744 655.236 0.002 4 -6.79610 -6.88510

0.001 -8.17610 0 348.049 655.3 0.002 4 -6.27810 -6.38410

0.001 -8.68110 0 348.352 655.364 0.003 4 -6.81810 -6.94110

0.001 -8.49810 0 348.653 655.429 0.003 4 -6.41610 -6.55710

0.001 -8.65210 0 348.952 655.494 0.004 4 -6.07210 -64.2310

0.001 -7.21710 0 349.25 655.559 0.005 4 -6.78610 -6.96110

0.001 -7.50710 0 349.545 655.625 0.005 4 -6.55910 -6.74910

0.002 -7.83710 0 349.837 655.691 0.006 4 -66.3910 -6.59510

0.002 -7.21110 0 350.128 655.757 0.007 4 -6.27910 -6.49910

0.002 -7.62910 0 350.415 655.824 0.008 4 -6.22710 -68.4610

0.002 -7.09410 0 350.7 655.892 0.009 4 -6.23210 -6.47810

0.002 -7.60710 0.001 350.982 655.96 0.01 4 -0 0

0.002 -74.1710 0.001 351.261 656.029 0.011 4 -0 0

0.002 -7.78510 0.001 351.536 656.098 0.012 4 -0 0

0.002 -7.45310 0.001 351.809 656.168 0.013 4 -0 0

0.002 -7.17610 0.001 352.078 656.239 0.014 4 -0 0

0.002 -7.95510 0.001 352.343 656.311 0.016 4 -0 0

0.003 -7.79210 0.001 352.604 656.383 0.017 4 -0 0

0.003 -7.68610 0.001 352.862 656.456 0.018 4 -0 0

0.003 -79.6410 0.001 353.116 656.53 0.02 4 -0 0

0.003 -6.06610 0.001 353.365 656.604 0.021 4 -0 0

=



21/27


The first Spool Response graph shows the control signal, valve spool position and as a function

of time.

0 0.5 1 1.5 2 2.5 30.05

0

0.05

0.1

0.15

0.2

0.25

5

0

5

10

15

Control Signal

Spool Position

Velocity

Spool Response to Control Signal

Time

ConrolSignalandSpoolPositon-1to+1

Velocityinin/s



22/27


0 0.5 1 1.5 2 2.5 30

5

10

15

20

25

0.05

0

0.05

0.1

0.15

0.2

0.25

Actuator Position

Spool Position

Hydraulic Cylinder Response to Spool Position

Time

PositioninInches

SpoolPosition-1to1



23/27


0 0.5 1 1.5 2 2.5 35

0

5

10

15

0

500

1000

1500

2000

2500

Velocity

Pressure Port A

Pressure Port B

Supply Pressure

Velocity and Pressure Response

Time

Velocityinin/sec

A,

BandSupplyPortPressure,ps

i

The load can induce a large pressure difference across the piston.

The supply pressure is plotted to compare with the A and B port pressure. When the valveshut quickly pressure spikes are induced and these pressure spikes can cause the oil toflow back into the accumulator. This is why the code for flow must use

sign Ps Vg( ) Pa( ) Ps Vg( ) Pawhen computing the flow. This avoids errors caused by taking the square root of a negativenumber.



24/27


0 0.5 1 1.5 2 2.5 3

1000

0

1000

2000

3000

Cylinder Force

Friction Force

Load Force

Net Force

Force Response

Time

Force,

lbf

0 0.5 1 1.5 2 2.5 34.6

4.65

4.7

4.75

4.8

1850

1900

1950

2000

2050

Gas Volume

Presssure

Accumulator Gas Volume and Accumulator Pressure

Time

AccumulatorGasVolume,ga

l

AccumulatorGasPressure,psi



25/27


0 0.5 1 1.5 2 2.5 330

20

10

0

10

20

30

40

50

0

500

1000

1500

2000

2500

A Port Flow

B Port Flow

Pump FlowA Port Pressure

B Port Pressure

Supply Pressure

A and B Port and Pump Flow and Pressures

Time

C

ubicInchesperSecond

Pressures,psi

The pump flow is not as great as the flow into the cylinders A port. This can happen because the

extra oil comes from the accumulator.



26/27


0 0.5 1 1.5 2 2.5 30

5

10

15

20

25

5

0

5

10

15

Target Position

Actual Position

Target Velocity

Actual Velocity

Control Output -10 to 10

Target and Actual Positions and Velocities

Time

Ta

rgetandActualPositionininch

es

Ta

rgetandActualVelocityinin/sec

Ki 0.5= Integrator gain

Kp 0.1= Proportional gain

Kd 0.001= Derivative gain

Kv_ext 0.016= Velocity feed forward extend

Kv_ret 0.022= Velocity

Ka 0.00018= Acceleration feed forward

Normalized sum of squared errors.

Adjust gains to minimize

n

r n t MoveTime, MoveDist,( )0

xn

( )

2

N 0.002674=



27/27



mathcad - hydraulic cylinder

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