Fluid Power Controls Laboratory (Copyright – Perry Li, 2004-2010)

M..E., University of Minnesota (updated 12.2010)

• Labs this week:• Lab 10: Bleed-off Circuit

• Lab 11/12: Asynchronous/Synchronous and Parallel/Tandem Operations

• Systems Review Homework (due 10/12)

• Research / internship opportunities:• CCEFP fluid power scholar program

• Participation is research lab

• Fluid Bulk modulus

• Fluid Inertance

• Component modeling• Pressure reducing valve

• Pressure compensated flow control valve

Lecture 5

119

Fluid Power Controls Laboratory (Copyright – Perry Li, 2004-2010)

M..E., University of Minnesota (updated 12.2010)

120

Fluid Compressibility: Bulk Modulus

• Hydraulic fluid is slightly compressible: • fluid volume decreases from V by dV on application of pressure dP

• is the bulk modulus -

• varies a lot with temperature and amount of aeration

• For β = 200kPsi, 5000psi pressure compresses fluid by 2.5%

• Increases with pressure

• Decreased significantly by entrained or dissolved air content

dPV

dV

β1=−

β

��

�=

��

�

Fluid Power Controls Laboratory (Copyright – Perry Li, 2004-2010)

M..E., University of Minnesota (updated 12.2010)

121

Bulk modulus: cont’’’’d

• For precise application, can be important, especially for long stroke, narrow, cylinders

• dL = pressure*L/β = force*L/(Area*β)

• D=0.5in bore, 6 in stroke • 0.15in/1000 lb-f

• D=1 in bore, L=1.5in stroke (same volume)• 0.01in/1000 lb-f

• 16 times smaller

Note that compressibility equation looks like equation for a spring

Fluid Power Controls Laboratory (Copyright – Perry Li, 2004-2010)

M..E., University of Minnesota (updated 12.2010)

Energy in Compressed Fluid (Li and Wang, 2011)

•

122

Meanbulkmodulus: WV�Pg, P0� = P�

�

2β

Fluid Power Controls Laboratory (Copyright – Perry Li, 2004-2010)

M..E., University of Minnesota (updated 12.2010)

123

Accounting for Bulk Modulus

• Compressibility increases spring like action

• Possibility of having resonance

Compressible cylinder

MSpring, K

M

Ideal cylinder (velocity generator)+ Spring

V = Q/A

QA Questions: How to pick K ?

Fluid Power Controls Laboratory (Copyright – Perry Li, 2004-2010)

M..E., University of Minnesota (updated 12.2010)

124

Modeling compressibility in cylinders

• Consider when the cylinder ports are blocked

• Suppose the load F is applied on the piston, how much does the piston move?

• Note: the chamber should include all the volumes between the piston and the valve (i.e. include hoses)

• Case 1: Double ended actuator

• Case 2: Single ended actuator

• Does the spring constant change with position of the piston?

F

Fluid Power Controls Laboratory (Copyright – Perry Li, 2004-2010)

M..E., University of Minnesota (updated 12.2010)

125

Equivalent Spring Constant

• Differentiating:

F

Data

x

Keq

Worst case

Finite if dead volumes(hoses) included

Fluid Power Controls Laboratory (Copyright – Perry Li, 2004-2010)

M..E., University of Minnesota (updated 12.2010)

126

Fluid Inertance (inertia)

• F = m * a for the accelerating fluid (transients)

• Normally, the pressure needed to accelerate the fluid is neglected.

• When is this important ?

• Momentum calculation for a hose: Length = L, Area = A

P1 P2

[ ] [ ]LQdt

dLAv

dt

dPA ρρ ==∆⋅

dt

dQ

A

LP

=∆ ρ

inertance

Important for long, narrow pipes – water hammer effect!

Fluid Power Controls Laboratory (Copyright – Perry Li, 2004-2010)

M..E., University of Minnesota (updated 12.2010)

How to make a big mass out of little mass?

• Total weight of device = M

• Kinetic energy = (100 M) v2/2

• What is in the pink box?

127

v v

Fluid Power Controls Laboratory (Copyright – Perry Li, 2004-2010)

M..E., University of Minnesota (updated 12.2010)

128

Power Computations

• A hydraulic device is generally an n-port system • each port interacts with its environment - hydraulic, mechanical, (electric)

• Hydraulic power:• energy flux through each port

• Power_in = F*vel_in = P*A*vel_in = P * Q

• For a hose filled with incompressible fluid, Q_2 = -Q_1

• Net hydraulicpower in = (P1 - P2)*Q

• How about a single ended or a double ended actuator, or a hydraulic motor?

P1, Q1 P2, Q2

hoses load

E-power

Fluid Power Controls Laboratory (Copyright – Perry Li, 2004-2010)

M..E., University of Minnesota (updated 12.2010)

129

Power Computation - cont’’’’d

• Hydraulic actuator / motor• has 2 hydraulic port and 1 mechanical (load) port

• Net hydraulic power input =

• Net mechanical power input =

• For passive components, net power input not greater than 0.

• Calculate hydraulic power for hydraulic motor

• Relationship between hydraulic power and mechanical power?

Mechanical power variables:• Force and velocity• Torque and angular velocity

Fluid Power Controls Laboratory (Copyright – Perry Li, 2004-2010)

M..E., University of Minnesota (updated 12.2010)

130

Computer circuit analysis

0.019

SliderGain2

0.403

SliderGain

0

Pump flow

Pressure Flow rate

Pump

0

Pressure

Flow

turns

Pressure

Needle2

Pressure

turns

Flow

Needle1

0

Flow1

0

Flow 2 3

Constant1

3

Constant

Fluid Power Controls Laboratory (Copyright – Perry Li, 2004-2010)

M..E., University of Minnesota (updated 12.2010)

131

Component Modeling -Pressure Reducing Valve

• How do we write equations for this valve?

• Spool• Force balance / Newton’s law

• Spring• Preload / Compression

• Orifice

Fluid Power Controls Laboratory (Copyright – Perry Li, 2004-2010)

M..E., University of Minnesota (updated 12.2010)

132

Modeling

• Function: Regulate pressure at B

• Operation: If P_B is too large (small), spool moves up (down) to reduce (increase) orifice size

A

B

D

x

Fspr i n g

Preload

x

A(x)

Possible Spring and area functions

PBAB − Fspring(x)− Fseat( ) − PDAD = M&&x