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Chap 2 Basics of hydraulics

Chap 2 Basics of hydraulics. Advantages of hydraulic control ： ― easy of control ― high power output ― good dynamic response ― good dissipation of heat. Disadvantages ： ― high energy losses ― possibility of leakages （ dirty ） § Work w = F×L where w ： work （ N˙m ）

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Chap 2 Basics of hydraulics

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  1. Chap 2 Basics of hydraulics Advantages of hydraulic control： ― easy of control ― high power output ― good dynamic response ― good dissipation of heat Disadvantages： ― high energy losses ― possibility of leakages（dirty） § Work w = F×L where w：work（N˙m） F：Force（N） L：Distance（m） § Pascal´s Law（Multiplication of force） p = = = constant w = F1L1 = F2L2 （assume no energy losses）

  2. Power P= = = F × =（p × A）（ ）= p × Q where P：power p：pressure Q：flow rate （ ） F1 F2 L2 L1 A1 A2

  3. (Application of law of Pascal ) A2=40 cm2 A3=50 cm2 Qp=50 l/min G=10 5N Pmax=45 bar h=25 cm Questions: (1) 將工件G舉起之最小壓力（P3min）＝？ (2) 此時相對應之面積 A1＝？ (3) 推力 F1＝？ (4) 速度 ＝？ (5) 速度 ＝？ (6) 工作油壓缸由底上升至頂端所需之時間 t＝？ HW #1 G h+d d A1 A3 P3 F1 P2 A2 P1

  4. P= （kW） 1W = Ex：How to derive？ HP（horsepower）= In the case of rotary actuator（motor） For pump： V1=A×πd Q1=n1×V1 Where Q: Flow rate V1: Displacement of pump T: Torque Q n1 V1 Pn n2 V2 T1 T2

  5. For motor： V2 = A ×πd Q2 = n2 × V2 If Q1= Q2 n1v1=n2v2 = Piston A

  6. Given : n1 =1450rpm n2 =400rpm T2=250N-m Pn=200bar Please calculate: (1) The optimal displacement of motor V2 = ? Note: as follows are different types to choose : V2 = 40 / 56 / 71 / 90 / 125 cm3/rev (2) Pressure Pn= ? , Flow rate Q= ? , Displacement of pump V1= ? and the Power P = ? (according to the chosen V2) HW #2 Q V2 V1 Pn n1 n2 T1 T2

  7. Torque T1= F × di =（A×ΔP） =（ ）×Δp× = = similarly： T2 = Torque transfer relation： = Assume no energy loss： P1= T1w1 = × 2π× n1 =V1× n1 ×Δp =Q1 ×Δp Output power of motor： P2=Q2×Δp Total efficiency=η= =1（ideal case）

  8. + =0 =ρ1Q1=ρ1A1V1 =ρ2Q2=ρ2A2V2 pipe if ρ1=ρ2（incompressible fluid） then V1A1=V2A2 Ex： A1ρV1－A2ρV2－Aρ =0 If ρ: constant V1A1 －V2A2 = A Conservation of Mass（Continuity Equation） ρ1A1V1 ρ2A2V2 A V2A2 V1A1

  9. Conservation of Energy（Bernoulli’s Equation） P2 V2 P1V1 h1 h2 1 2

  10. Total Energy at any point = Elevation＋Pressure＋Kinetic = mgh＋ ＋   Bernoulli’s equation：   ＋ ＋h1= ＋ ＋h2 Ex1：（continuity equation） Q=25 D=50 mm d=30 mm d1=d2=15mm A1 A2 d D d2 d1 Q

  11. Questions：   Piston velocity =？   Velocities of the fluid in the inlet and  outlet pipes = ? Answer： （1） A1= πD2=19.6㎝2 A2= π(D2－d2）=12.56㎝2 Cross section area of inlet and outlet pipes A3= πd12=1.77㎝2 Piston velocity V1= =0.21 （2）Velocity of the fluid in inlet pipe：   V3=   =2.36 From continuity equation： Where V4：velocity of the fluid in outlet pipe   Thus V4= =1.49

  12. ρ=0.8 =800 Assume：no work and no energy dissipation Question：P2=？ Answer： h1=h2； ρ1=ρ2 V1= = = 0.236 V2=（ ）2×V1= 8.496 From Bernoulli’s equation： + = + + = + P2=69.7×105 =69.7 bar (Pressure drop ： 0.3bar) Ex2（Application of Bernoulli’s equation） 70bar P1 P2 d2=0.5cm d1=3cm V1 V2 Q=10 l/min

  13. Flow through orifice ∵ A2«A1 ∴V1«V2 P1+ 1/2 ρV12 = P2 + 1/2 ρV22 V2= where Δp＊=P1－P2 Flow rate Q=A2×V2 =A2× 1 0 2 3

  14. Area A2=Cc×A0 Where Cc：contraction coefficient Flow coefficient Cf Q = Cf × A0 × where Δp=P1－P3 Cf = f（Reynolds number；geometry） =0.6~1.0 Valve geometry Q = Cf ×πd × Cf = 0.6~0.64 P0 d P1 Q

  15. = m （momentum） = = = + m Theory of Momentum (Principle of impulse) F A1 F1 Q F2 F 1 F1 CV F2 A2 (Vector) 2 Q

  16. Steady flow （ = 0） position 1： =ρ1Q1 = ρ1Q1 position 2： =ρ2Q2 = - ρ2 Q2 Flow force： = + = - = - - Ex: assume incompressible （ρ=const） ρ1Q1=ρ2Q2 = ρQ = -ρQ（ + ） unsteady part ： m CV

  17. Assume incompressible = const Thus Q1= Q2 = Q = - = （A：const） = ×m（∵ = - ∴steady part=0） =ρ A × pressure drop：P1-P2 = = × （Q = v‧A） F =A（P1-P2） Where ：hydraulic inductance

  18. Flow force on the directional control valve （a） Recall Fstr= F1- F2 = Fi - Fo (scalar) Fax= -Fstr = Fo- Fi Fax= cosε0 - cosεi - ρ Io εi ε2 εo ε1 Ii Io Ii Io Ii Fax Fstr 2 1

  19. = ρv2 × Q = ρv1 × Q εo= ε2= 90o εi= ε1 + 180o Fax=0 + ρv1Qcosε1 - ρ （b） Fax= cosε0 - cosεi + ρ =ρv1Q ε0=ε1 =ρv2Q εE=ε2+1800=2700 Fax=ρv1Qcosε1+ ρ εi Io ε2 Ii εo, ε1

  20. (a) Shown above is a directional control valve Given : Q = 40 l/min =70 0 Xk = 0.5mm cf =0.8 問題: (1) 反應力Fax中之unsteady part 是否隨流動方向不同 而改變其大小及方向？（此處流動方向指圖示 （1）及（2）） (2) 反應力Fax中之steady part 是否與流動方向有關？ 試寫其支持力（Fax ) 之方程式。 (3) 請一所給的數據，計算出支持力（Fax）之值。 HW #3-a Po , Q (1) (2) Fax X Xk

  21. （b ) Laminar flow through eccentric clearance d=2cm , =50 cst , =0.85 l= 1cm , , Questions : (1) Qe=0 = ? ( l / min) (2) Qe= = ? ( l / min) HW#3-b D e P1 P2 d

  22. Case 1：small orifice area （ ~const） Q= Cf × A × v= = Cf × Where A=orifice area =πd x x d

  23. by small orifice area Δp~const thus Q ~ X Fstr=ρv cosε （only steady part） =ρv（vA）cosε =2Cf2 ×πd x ×Δpcosε Cs×X Case 2：large orifice area （Q~const） Fstr= cosε Thus Fstr~ Fstr linear Pmax=const Q3 Q2 Q1 x Small A Large A Q3>Q2>Q1 Fstr P3max P2max P1max x P3max>P2max >P3max Fstr x Pmax Q

  24. Shear stress where : dynamic viscosity cf. : kinematic viscosity Unit : 1Poise=1 =0.1 Viscosity of fluid U F h

  25. 1cp=10-2 Poise =10-3 1 stoke =1 1 cst=1 η（or ν）= f（T，P） Flow in Pipes Eqs. derived from viscous flow condition （1）laminar flow NR＜2000 （2）turbulent flow NR＞4000 only empirical Eqs.   Reynolds number NR（dimensionless）   NR = where V：velocity of fluid   DH：hydraulic diameter   ν：kinematic viscosity   Hydraulic Diameter DH = where A：Area of pipe   U：circumference of pipe   For pipes：   DH= （diameter of pipe）

  26. For narrow clearance DH= (if h << b) Conclusion: Design in hydraulics  Laminar flow h Q b

  27. （laminar flow in a pipe） τ×2πy =（P1-P2）πy2 τ=η = × = v= （r2-y2） for y =0 then Q= = = （P1-P2） § Hagen-Poiseuille formula V y r P1 P2 Vmax

  28. Laminar flow through the clearance V y h Q b Vmax V= （h2-4y2） Vmax= h2 Q =2 =2 Q = （P1-P2）

  29. Laminar flow through eccentric clearance D P1 P2 e d Q Q= （P1-P2） Where η：dynamic viscosity（e.g. Poise； ） When e=Δr, which implies max. eccentricity Qe= 2.5 Qe=0

  30. (a) Resistance Flow through resistance and orifice 2r P1 P2 Q= Δp =k1 Δp Q~Δp ~ (b) Orifice P1 P2

  31. Q=CfA =k2 Q ~ Thus Q orifice Resistance Δp

  32. Darcy-Weisbach formula： Pf = λ（ ） or hf =λ（ ） where Pf：pressure loss hf ：head loss ：length of pipe d ：internal diameter of pipe λ ：friction factor（derived experimentally） Pressure losses through pipes

  33. Q= Δp Q=A×v=πr2v Thus Δp =P1-P2 = Pf = v λ λ= = （valid only if Re＜2000） For turbulent flow： Blasius formula： λ= （for pipes with smooth inside surface） （3000＜Re＜105） Example：Laminar pipe flow

  34. ΔPff = k × （for turbulent flow） or hff = k（ ） where ΔPff：pressure loss hff：head loss Sudden enlargement: k =（1- ）2 Sudden reduction: k = 0.5（1- ） k-values for pipe fitting, valves, and bends are determined empirically k-values for fluid through sudden expansion, contraction, pipe fitting, valves and bends Q D1 D2 D2 Q D1

  35. （P1+ +ρgζ1）-（P2+ +ρgζ2）=ΔPf = + Circuit Calculation k1 d1 k3 1 k2 2

  36. HW#4-a

  37. HW#4-b

  38. P1 ：泵之入口壓力 P2 ：泵內之最低壓力 Pa ：大氣壓力 Pg ：空氣分離壓力 (1) From Bernoulli’s Eq. (2) (3) Define (Net Positive Suction Head) (4) No cavitation EX: 泵油之空穴(Cavitation of oil pump) A B Pa P1 P2 he (head loss) (A)

  39. 問題： (1)式（A）中，有那些（Ha? , H1? , he , Hg ?）會影響泵 是否產生 cavitation? (2)若欲避免產生cavitation，則影響泵是否產生cavitation之Heads 應如何改變（例如：變大或減小等） (3)如何在設計油泵系統時，達到（2）中所述之改變?

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