Lecture 7. Coming week s lab: Integrative lab (your choice!)

Transcription

1 Lecture 7 Coming week s lab: Integrative lab (your choice!) Today: Systems review exercise due end of class Your feedback Review: sequencing and asynchronous circuit analysis Hydraulic hybrid vehicles Valve modeling and sizing 1

2 How Hybrid Vehicles Save Energy? With a secondary power source/storage, it is possible to: Manage engine operation Store/reuse braking energy Turn off engine Downsize engine for continuous power Example vehicle on EPA-UDDS cycle: Baseline (10% engine efficiency): 24 mpg Engine management (38% efficiency): 95 mpg Above with regeneration: 140 mpg Engine 38% Energy storage Drive-train 10-15% Required for maximum performance wheel Normal driving operating range 2

3 Why Hydraulic Hybrids? Why not stick with electric hybrids? Electric batteries / ultracaps (cost, reliability, recycling, power density) Electric motors & inverters (cost, power density) Affect overall cost, weight, and power Metrics Fuel economy Cost Performance Toyota Prius 3

4 Hybrid Hydraulic versus Hybrid Electric Vehicle Hydraulic pump/motor have significantly higher power density than electric motor/generator (16:1 by weight, 8:1 by volume) Hydraulic drives have much lower torque density than electric drives Accumulators are 10x more power dense than batteries limits acc/braking and hence regenerative Efficient power electronics are expensive braking Batteries have 2 order magnitude higher energy density than accumulators Current hydraulic systems tend to be noisy and leaky Overall tradeoff: Hydraulic hybrids can be significantly lighter and cheaper than electric hybrids if energy density limitation can be solved. Engine Accumulator Accumulator Hydraulic pump/motor Engine Battery & ultracapacitor Electric motor/ generator Parallel Hybrid Hydraulic Parallel Hybrid Electric 4

5 Hydraulic Hybrids Versus Electric Hybrids Fluid Power Controls Laboratory (Copyright Perry Li, ) Performance Electric Fluid Power acceleration regenerative braking Component efficiency + - Regenerative efficiency Weight Cost Realize opportunities for Both performance & efficiency Cost and reliability Overcome threats in Inefficient components Low density energy storage Noise, vibration, harshness Reliability Environmental impact Energy storage NVH

6 Parallel Architecture Example: HLA system for F150 & garbage trucks Regenerates braking energy Utilizes efficient mechanical transmission Does not allow full engine management Achievable engine op. points 6

7 Series Architecture Example: Eaton/UPS (truck), Ford/EPA (Escape), Artemis (BMW-5), Regenerates braking energy Allows for full engine management Independent wheel torque control possible All power must be transmitted through fluid power components 38% 38% Required for maximum performance 10-15% 10-15% Normal driving Normal operating driving range operating range 7

8 Power-Split: Hydromechanical Transmission (HMT) Power split between mechanical and hydraulic paths Hybridized HMT i.e. w/ regeneration Efficient Mechanical Transmission Regenerative Braking Full engine management 8

9 Hypothesized hydraulic & overall efficiencies 0.7 Urban 0.45 Highway Overall efficiency series parallel hmt Overall efficiency series parallel hmt Mean hydraulic efficiency Mean hydraulic efficiency Series / HMT at peak engine efficiency (38.5%) Parallel at lower engine efficiency (33%) 9

10 191 Sequencing Circuits Sequence valve cracking pressure settings to enable proper sequencing Too high never cracks Too low simultaneous motion instead of sequence P1 F Crack at P_R 10

11 Asynchronous Circuit Analysis

12 Component Modeling - Pressure Reducing Valve 193 How do we write equations for this valve? Spool Force balance / Newton s law Spring Preload / Compression Orifice 12

13 194 Function: Regulate pressure at B Modeling Operation: If P_B is too large (small), spool moves up (down) to reduce (increase) orifice size P B A B ( F spring (x) F seat ) P D A D = M x A D F spr i n g A(x) B Preload x Possible Spring and area functions x 13

14 195 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 14

15 196 15

16 197 Way-Position-Port, centering etc. # Position = number of possible positions for the valve shifting mechanism 2way infinite-pos 2port 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 4way, 3position 4port 4way, infinite-pos 4port 16

17 Four Way Directional Control Valve

18 199 Influences rest of circuit Energy / pressure when this part of circuit not used 4 way valve center 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 18

19 200 Single stage proportional valves Solenoids Spool is stroked directly by solenoid actuator LVDT spool position feedback Spring (sometimes) for safety 19

20 Single Stage Proportional Valves 201 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 control ASME Journal of Dynamic Systems, Measurement and Control. Vol 124, No. 1, pp , March, Q.-H. Yuan and P. Y. Li, Using Steady Flow Force for Unstable Valve Design: Modeling & Experiments ASME Journal of Dynamic Systems, Measurement and Control. Vol 127, No. 3. pp , Q.-H. Yuan and P. Y. Li, Robust Optimal Design of Unstable Valves IEEE Transactions on Control Systems Technology. Vol. 15, No. 6, pp , November,

21 Multi-stage valves 202 Use hydraulic force to drive the spool.. 21

22 Electrohydraulic servo-valve Multi-stage valve Typically uses a flapper-nozzle pilot stage Built-in feedback via feedback wire Very high dynamic performance Bandwidth = Hz 203 Pilot stage Main stage For a fun place to learn how a servo-valve works: R. Dolid Electrohydraulic Valve Coloring Book pdf 22

23 Servo-Valve

24 Feedback spring: 1. Regulates the position of the mainstage by negative feedback on the flapper Servo-Valve 205 Nozzle-flapper pilot valve: 1. Electromagnetic torque motor moves flapper to left (or right) 2. Nozzle and restriction at source form two resistances in series 3. Flapper differentially opens and closes nozzle 4. 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 24

25 206 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 (Sauer-Danfoss) 25

26 207 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 26

27 D.C. Valve modeling 208 Flow 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) Q Q 2 A1( xv ) ( Ps 1) 1 = Cd P 2 A3 ( xv) ( P2 0) 3 = Cd P Q 1 = Q 3 = Q L 27

28 D.C. Valve Modeling 209 Now the pressure across the load is: P L = P 1 P 2 Area gradient, w, relates orifice area to spool displacement. For circular orifices: w = 2 R Area gradient = w w = da dx A1(xv) = A3(xv) = w * xv A2(xv) = A1(-xv) = -w*xv v ( x v ) Annular orifice land Top view Solve for Q L when A1(xv) = A3(xv) =w xv (matched) to get: Q L 1 ( xv, PL ) = Cd wxv ( Ps PL ) circular Area versus xv Annular 28

29 DC Valve Modeling 210 Similar analysis for xv < 0: Q L 1 ( xv, PL ) = Cd wxv ( Ps PL ) Normally P L < 0 when xv < 0. Putting formulae for +/- xv together: Load overrun Positive power Q L 1 ( xv, PL ) = Cd wxv ( Ps sgn( xv ) PL ) Flow increases linearly with xv Flow decreases with pressure load Positive power Load overrun 29

30 211 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 30

31 Valve Sizing 212 Valves are rated (i.e. their flow rate Q R given) at 1000 Psi pressure drop across the valve. i.e. Flow = Q R 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 Q L( xv, PL ) = Cd wxmax ( Ps P1 ) 2 Q L( xv, PL ) = Cd wxmax ( P2 P0 ) 31

32 Valve Sizing 213 At rated condition, Ps - P1= 500Psi, P2 - P0 = 500psi 2 Q L( xv, PL ) = Cd wx max ( Ps P1 ) 2 Q L( xv, PL ) = Cd wx max ( P2 P0 ) Suppose actual operating (Ps-P1) is 700 instead, with required flow Qd, then we would like: So the rated flow should be at least: Q R1 = Q d1 Q d = 500 P P s C wx Similarly, calculate for the return orifice: pick the larger rated flow 1 d max Q R= Q d Q R = C d wx max Pick this when picking valve 32

33 Valve sizing example 214 Given Ps = 3000 psi; P 0 = 0 psi Load is a 2:1 cylinder, capside A cap = 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 8F Ps P1 = 2Ps 2F P2 = 9A 9 cap 9A Step 3: find rated flow based on each orifice 9 cap P1 P2 A cap ( 1 2 P P / 2) = F ( P s P1 ) = 2 ( P2 P0 ) F Why? Q R1 = Q d1 500 P P s 1 Q R2 = Q d P P 2 0 Q Q R Q R1, R2 33

34 215 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!!! 34