Post on 16-Apr-2018
transcript
Fluid Power Controls Laboratory (Copyright – Perry Li, 2004-2010)
M..E., University of Minnesota (updated 12.2010)
This week:
• Lab 13: Hydraulic Power Steering
• [ Lab 14: Integrated Lab (Hydraulic test bench) ]
• 4-way directional control valve; proportional valve; servo-valve
• Modeling / Analysis of a servo-valve
• Valve sizing
• Pumps and motors (part 1)
10/19/2012:
• Hydraulic Hybrids
• Guest lecture – Mike Olson (Eaton Corporation)
Lecture 6
133
Fluid Power Controls Laboratory (Copyright – Perry Li, 2004-2010)
M..E., University of Minnesota (updated 12.2010)
134
Directional Control Valves (DCVs)
• Controls direction of flow
• Fluidic symbol signifies function
• Manually or electrically • (solenoid or equivalent elements)
• Can also meter flow• proportional valve
• servo valve
Fluid Power Controls Laboratory (Copyright – Perry Li, 2004-2010)
M..E., University of Minnesota (updated 12.2010)
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Fluid Power Controls Laboratory (Copyright – Perry Li, 2004-2010)
M..E., University of Minnesota (updated 12.2010)
136
Way-Position-Port, centering etc.• # Position = number of possible
positions for the valve shifting mechanism
• 1 = no shifting, 2={left, right}, 3={left, right, center},
• infinite
• # Way = number of flow paths (including reverse flow)
• check valve = 1 way
• 1-way, 2-way, 3-way, 4-way most common.
• # Port = number of plumbing connections.
• Center = connection while neutral• closed, open, tandem ...
3way, 3position
3port
2wayinfinite-pos
2port
4way, 3position
4port
4way, infinite-pos
4port
Fluid Power Controls Laboratory (Copyright – Perry Li, 2004-2010)
M..E., University of Minnesota (updated 12.2010)
137
4 way valve center
• Influences rest of circuit
• Energy / pressure when this part of circuit not used
• Open center• unload pump, energy saving for
fixed displacement pump
• Can you use 2 cylinders?
• Closed center• high pressure / relief
• responsive
• Tandem center (lab)• allows 2 circuits in series
Fluid Power Controls Laboratory (Copyright – Perry Li, 2004-2010)
M..E., University of Minnesota (updated 12.2010)
138
Modeling of directional control valve • Spool is driven by solenoid (in a
single stage valve) or by pilot stage(s) valves
• xv > 0 (spool moves downward)
Assume
• matched (i.e. supply and return orifices are the same)
• Critically lapped
• symmetric, +/- spool displacement give same orifice area
• Flow conserving load• Qin = Qout
• e.g. double ended cylinder or hydraulic motor
Fluid Power Controls Laboratory (Copyright – Perry Li, 2004-2010)
M..E., University of Minnesota (updated 12.2010)
139
D.C. Valve modelingFlow paths: (neglect leakages)
• xv > 0: Orifices 1 and 3
• xv < 0: Orifices 2 and 4
• Area of each orifice • A1(xv), A2(xv), A3(xv), A4(xv)
• For xv > 0, [Cd = discharged coeff approx 0.6 (see notes on orifice modeling)
)(2
)( 111 PPxACQ svd −=ρ
)(2
)( 0233 PPxACQ vd −=ρ
LQQQ == 31
Fluid Power Controls Laboratory (Copyright – Perry Li, 2004-2010)
M..E., University of Minnesota (updated 12.2010)
140
D.C. Valve Modeling
• Now the pressure across the load is:
• Area gradient = w
A1(xv) = A3(xv) = w * xv
A2(xv) = A1(-xv) = -w*xv
• Solve for QL when A1(xv) = A3(xv) =w xv (matched) to get:
21 PPPL −=
)( vv
xdx
dAw =
)(1
),( LsvdLvL PPwxCPxQ −=ρ
Rw π2=
Area gradient, w, relates orifice areato spool displacement.
For circular orifices:
Top view
Annular orifice
land
Annular
circular
Area versus xv
Fluid Power Controls Laboratory (Copyright – Perry Li, 2004-2010)
M..E., University of Minnesota (updated 12.2010)
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DC Valve Modeling
• Similar analysis for xv < 0:
• Normally PL < 0 when xv < 0.
• Putting formulae for +/- xv together:
• Flow increases linearly with xv
• Flow decreases with pressure load
)(1
),( LsvdLvL PPwxCPxQ +−=ρ
))sgn((1
),( LvsvdLvL PxPwxCPxQ −=ρ
Load overrun
Load overrun
Positive power
Positive power
Fluid Power Controls Laboratory (Copyright – Perry Li, 2004-2010)
M..E., University of Minnesota (updated 12.2010)
142
Effect of lapping / centering
• For critically lapped (or sometimes called critically centered) valve• Orifice is shut when xv = 0.
• At least one orifice will open when xv !=0
• Critically lapped valves are desirable but expensive to make• flow is linear with displacement (constant gain)
• Overlapped (closed centered) valves • deadband => un-responsive
• Underlapped (open centered) valves• valve always open => power loss
• nonlinear gain => more difficult to control
Fluid Power Controls Laboratory (Copyright – Perry Li, 2004-2010)
M..E., University of Minnesota (updated 12.2010)
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Valve Sizing
• Valves are rated (i.e. their flow rate QR given) at 1000 Psi pressure drop across the valve.• i.e. Flow = QR is flow when the valve is fully open and with 1000 PSI
pressure drop
• There are two flow orifices (1 and 3) in flow path. It is assumed that pressure drop is 500psi across each orifice.
• Since one does not always use the valve when there is 1000 Psi pressure drop, one must translate the required flow to the equivalent flow at 1000 Psi.
• Basic equation:
)(2
),( 1max PPwxCPxQ sdLvL −=ρ
)(2
),( 02max PPwxCPxQ dLvL −=ρ
Fluid Power Controls Laboratory (Copyright – Perry Li, 2004-2010)
M..E., University of Minnesota (updated 12.2010)
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Valve Sizing• At rated condition, Ps - P1= 500Psi, P2 - P0 = 500psi
• Suppose actual operating (Ps-P1) is 700 instead, with required flow Qd, then we would like:
• So the rated flow should be at least:
• Similarly, calculate for the return orifice: pick the larger rated flow
5002
max ρwxCQ dR =
)(2
),( 1max PPwxCPxQ sdLvL −=ρ
)(2
),( 02max PPwxCPxQ dLvL −=ρ
7002
max ρwxCQ dd =
700
500dR QQ =
Pick this when picking valve
111
500
PPQQ
sdR −
=
Fluid Power Controls Laboratory (Copyright – Perry Li, 2004-2010)
M..E., University of Minnesota (updated 12.2010)
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Valve sizing example• Given Ps = 3000 psi; P0 = 0 psi
• Load is a 2:1 cylinder, capside Acap= 0.005 m^2.
• Required extension speed = 1 m/s
• Load F = 500 lb-f.
• Step 1: Find Qd1 = speed * Acap.; Qd2 = Qd1/2
• Step 2: find P1 and P2 during operating conditions
• Step 3:find rated flow based on each orifice
P2P1F
)(2)( 021 PPPPs −=−
FPPAcap =− )2/( 21
99
81
s
cap
P
A
FP +=
cap
s
A
FPP
9
2
9
22 −=
111
500
PPQQ
sdR −
=02
22
500
PPQQ dR −
=
21, RRR QQQ ≥
Why?
Fluid Power Controls Laboratory (Copyright – Perry Li, 2004-2010)
M..E., University of Minnesota (updated 12.2010)
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Other component sizing• Actuator:
• stroke: how far does it have to travel?
• Area: load/pressure or pump-flow/area
• load can be friction, gravity, brake force, acceleration load etc.
• Motor: Disp= Q/speed; Disp = Torque/Pressure
• Pump flow:• Supposed actuator area determined by load/pressure
• Qpump = actuator speed * area (beware of cap versus piston side)
• Multiple circuits:• Add required flows from all circuits
• add safety factor!
• Beware of units !!!
Fluid Power Controls Laboratory (Copyright – Perry Li, 2004-2010)
M..E., University of Minnesota (updated 12.2010)
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Single stage proportional valves
Solenoids
• Spool is stroked directly by solenoid actuator
• LVDT spool position feedback
• Spring (sometimes) for safety
Fluid Power Controls Laboratory (Copyright – Perry Li, 2004-2010)
M..E., University of Minnesota (updated 12.2010)
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Single Stage Proportional Valves• Advantages:
• Simple design• Reliable• Cost effective
• Disadvantages:• Poor dynamic performance (bandwidth)• At high flow rates and bandwidths, large stroking force is needed• Large (and expensive) solenoids / torque motors needed.• Low end market …..
Research: Using unstable flow force to improve spool agility• K. Krishnaswamy and P. Y. Li, On using unstable hydraulic valves for controlASME Journal of Dynamic Systems, Measurement and Control. Vol
124, No. 1, pp. 182-190, March, 2002. • Q.-H. Yuan and P. Y. Li,Using Steady Flow Force for Unstable Valve Design: Modeling & ExperimentsASME Journal of Dynamic Systems,
Measurement and Control. Vol 127, No. 3. pp. 451--462, 2005.• Q.-H. Yuan and P. Y. Li,Robust Optimal Design of Unstable ValvesIEEE Transactions on Control Systems Technology. Vol. 15, No. 6, pp. 1065-
1074, November, 2007
Fluid Power Controls Laboratory (Copyright – Perry Li, 2004-2010)
M..E., University of Minnesota (updated 12.2010)
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Multi-stage valves
• Use hydraulic force to drive the spool …..
Fluid Power Controls Laboratory (Copyright – Perry Li, 2004-2010)
M..E., University of Minnesota (updated 12.2010)
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Electrohydraulic servo-valve• Multi-stage valve• Typically uses a flapper-nozzle pilot stage• Built-in feedback via feedback wire• Very high dynamic performance• Bandwidth = 100-200+Hz
Pilot stage
Main stage
For a fun place to learn how a servo-valve works:R. Dolid “Electrohydraulic Valve Coloring Book”http://www.lulu.com/items/volume_67/7563000/7563085/3/print/Servovalve_Book__091207.pdf
http://tinyurl.com/8s2t3on
Fluid Power Controls Laboratory (Copyright – Perry Li, 2004-2010)
M..E., University of Minnesota (updated 12.2010)
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Servo-Valve
Fluid Power Controls Laboratory (Copyright – Perry Li, 2004-2010)
M..E., University of Minnesota (updated 12.2010)
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Servo-ValveNozzle-flapper pilot valve:1. Electromagnetic torque motor
moves flapper to left (or right)2. Nozzle and restriction at source
form two resistances in series3. Flapper differentially opens and
closes nozzle4. Pressure increases on side with
closed nozzle; decreases on side with open nozzle; creating pressure differential
Main stage:1. Four-way spool valve actuated by
differential pressure generated by pilot stage
Feedback spring:1. Regulates the position of the main-
stage by negative feedback on the flapper
Fluid Power Controls Laboratory (Copyright – Perry Li, 2004-2010)
M..E., University of Minnesota (updated 12.2010)
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Two spool servo valve
• Main spool consists of two spools – one for meter-in, one for meter-out
• Reduces tolerance requirement
• Pilot stage is a pressure servo valve
• Cost effective
• Bandwidth around 30 Hz
See below for a model of such a valve• R. T. Anderson and P. Y. Li, Mathematical Modeling of a
Two-spool Flow Control Servovalve Using a Pressure Control Pilot ASME Journal of Dynamic Systems, Measurement and Control Vol 124 No. 3, Sept 2002.
(Sauer-Danfoss)
Fluid Power Controls Laboratory (Copyright – Perry Li, 2004-2010)
M..E., University of Minnesota (updated 12.2010)
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Pumps
• Source of hydraulic power
• Converts mechanical energy to hydraulic energy• prime movers - engines, electrical motors, manual power
• Two main types:
• positive displacement pumps
• non-positive displacement pumps
Fluid Power Controls Laboratory (Copyright – Perry Li, 2004-2010)
M..E., University of Minnesota (updated 12.2010)
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Pump - Introduction
Fluid Power Controls Laboratory (Copyright – Perry Li, 2004-2010)
M..E., University of Minnesota (updated 12.2010)
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Positive displacement pumps
• Displacement is the volume of fluid displaced cycle of pump motion• unit = cc or in3
• Positive displacement pumps displace (nearly) a fixed amount of fluid per cycle of pump motion, (more of less) independent of pressure• leak can decrease the actual volume displaced as pressure increases
• Therefore, flow rate Q gpm = D (gallons) * frequency (rpm)
• E.g. pump displacement = 0.1 litre• Q = 10 lpm if pump speed is 100 rpm
• Q = 20 lpm if pump speed is 200 rpm
Fluid Power Controls Laboratory (Copyright – Perry Li, 2004-2010)
M..E., University of Minnesota (updated 12.2010)
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Non positive displacement pumps
Impeller PumpCentrifugal Pump
Fluid Power Controls Laboratory (Copyright – Perry Li, 2004-2010)
M..E., University of Minnesota (updated 12.2010)
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Non-positive displacement pump
• Flow does not depend on kinematics only - pressure important• Also called hydro-dynamic pump (pressure dependent)
• Smooth flow
• Examples: centrifugal (impeller) pump, axial (propeller) pump
• Does not have positive internal seal against leakage
• If outlet blocks, Q = 0 while shaft can still turn
• Volumetric efficiency = actual flow / flow estimated from shaft speed
= 0%
Fluid Power Controls Laboratory (Copyright – Perry Li, 2004-2010)
M..E., University of Minnesota (updated 12.2010)
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Positive vs. non-positive displacement pumps
• Positive displacement pumps• most hydraulic pumps are positive displacement
• high pressure (10,000psi+)
• high volumetric efficiency (leakage is small)
• large ranges of pressure and speed available
• can be stalled !
• Non-positive displacement pumps• many pneumatic pumps are non-positive displacement
• used for transporting fluid rather than transmitting power
• low pressure (<300psi), high volume flow
• blood pump (less mechanical damage to cells)
Fluid Power Controls Laboratory (Copyright – Perry Li, 2004-2010)
M..E., University of Minnesota (updated 12.2010)
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Types of positive displacement pumps
• Gear pump (fixed displacement)• internal gear (gerotor)
• external gear
• Vane pump • fixed or variable displacement
• pressure compensated
• Piston pump• axial design
• radial design
Fluid Power Controls Laboratory (Copyright – Perry Li, 2004-2010)
M..E., University of Minnesota (updated 12.2010)
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External gear pump
• Driving gear and driven gear
• Inlet fluid flow is trapped between the rotating gear teeth and the housing
• The fluid is carried around the outside of the gears to the outlet side of the pump
• As the fluid can not seep back along the path it came nor between the engaged gear teeth (they create a seal,) it must exit the outlet port.
Fluid Power Controls Laboratory (Copyright – Perry Li, 2004-2010)
M..E., University of Minnesota (updated 12.2010)
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Gerotor pump
• Inner gerotor is slightly offset from external gear• Gerotor has 1 fewer teeth than outer gear
• Gerotor rotates slightly faster than outer gear• Displacement = (roughly) volume of missing tooth
• Pockets increase and decrease in volume corresponding to filling and pumping
• Lower pressure application: < 2000psi• Displacements (determined by length): 0.1 in3 to 11.5 in3
Inlet portOutlet port
Fluid Power Controls Laboratory (Copyright – Perry Li, 2004-2010)
M..E., University of Minnesota (updated 12.2010)
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Vane Pump
• Vanes are in slots
• As rotor rotates, vanes are pushed out, touching cam ring
• Vane pushes fluid from one end to another
• Eccentricity of rotor from center of cam ring determines displacement
• Quiet
• Less than 4000psi
Fluid Power Controls Laboratory (Copyright – Perry Li, 2004-2010)
M..E., University of Minnesota (updated 12.2010)
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Pressure Compensated Vane Pump
Fluid Power Controls Laboratory (Copyright – Perry Li, 2004-2010)
M..E., University of Minnesota (updated 12.2010)
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PC Vane Pump (Cont’’’’d)
• Eccentricity (hence displacement) is varied by shifting the cam ring
• Cam ring is spring loaded against pump outlet pressure
• As pressure increases, eccentricity decreases, reducing flow rate
• Spring constants determines how the P-Q curve drops:
• small stiffness (sharp decrease in Q as P increases)
• large stiffness (gentle decreases in Q as P increases)
• Preload on spring determines
• pressure at which flow starts cutting off
Fluid Power Controls Laboratory (Copyright – Perry Li, 2004-2010)
M..E., University of Minnesota (updated 12.2010)
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Axial Piston Pump• Each piston has a pumping cycle
• Interlacing pumping cycles produce nearly uniform flow (with some ripples)
• Displacement is determined by the swash plate angle• Generally can be altered manually or via (electro-)
hydraulic actuator.
Displacement can be varied by varying swashplate angle
Fluid Power Controls Laboratory (Copyright – Perry Li, 2004-2010)
M..E., University of Minnesota (updated 12.2010)
Bent-Axis Piston Pump• Thrust-plate rotates with shaft
• Piston-rods connected to swash plate
• Piston barrel rotates and is connected to thrust plate via a U-joint
• More efficient than axial piston pump
(less friction)
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Fluid Power Controls Laboratory (Copyright – Perry Li, 2004-2010)
M..E., University of Minnesota (updated 12.2010)
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Radial Piston Pump
• Similar to axial piston pump, pistons move in and out as pump rotates.
• Displacement is determined by cam profile (i.e. eccentricity)
• Displacement variation can be achieved by moving the cam (possible, but not common though)
• High pressure capable, and efficient
• Pancake profile