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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 forc e p = = = constant w = F 1 L 1 = F 2 L 2 assume no energy losses 1 1 A F 2 2 A F

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Page 1: [PPT]Chap.2 Fundamentals - 國立雲林科技大學YunTech - …teacher.yuntech.edu.tw/~rennjc/download/Chap2 Basic of... · Web viewPascal´s Law(Multiplication of force) p =

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×Lwhere 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 )

1

1

AF

2

2

AF

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Power

P= = = F × = ( p × A )( ) = p × Q where P : power p : pressure Q : flow rate ( )

             

tLF

tW

AQ

minl

F1

A1L1

F2

A2

L2

dtdL

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HW #1

(Application of law of Pascal )

A2=40 cm2

A3=50 cm2

Qp=50 l/min

G=10 5NPmax=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 =?

y

x

G

A3

P3

P1

P2

A1

A2

dh+d

y,y

x,x

F1

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P= ( kW ) 1W =

Ex : How to derive ?HP ( horsepower ) =

In the case of rotary actuator ( motor )           

For pump : V1=A×πd Q1=n1×V1

600min lQbarP

secmN

746WattP

Where Q: Flow rate

V1: Displacement

of pump

T: Torque

Q

PnV1

T1

n1 V2n2

T2

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For motor : V2 = A ×πd

Q2 = n2 × V2

If Q1= Q2

n1v1=n2v2

=             

1

2

nn

2

1

VV

Piston

A2d

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HW #2

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)

QPnV1n1

V2n2

T2T1

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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 )

2d

dV

1

2d

dpdV

21

21 pV

22 pV

1

2

TT

1

2

VV

21 pV

1

2

PP

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Conservation of Mass( Continuity Equation)

+ =0

=ρ1Q1=ρ1A1V1

=ρ2Q2=ρ2A2V2 pipeif ρ1=ρ2 ( incompressible fluid )then V1A1=V2A2

Ex :

A1ρV1 - A2ρV2 - Aρ =0 If ρ: constant

V1A1 - V2A2 = A

A

V1A1V2A2

1

.m

2

.m

ρ1A1V1 ρ2A2V2

1

.m 2

.m

dtdl

dtdl

dAundtd

dv

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Conservation of Energy( Bernoulli’s Equation)

h2h1

P2 V2

P1V1

1 2

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Total Energy at any point = Elevation + Pressure + Kinetic = mgh + + Bernoulli’s equation : + + h1= + + h2

 Ex1 :( continuity equation )

minl Q=25 D=50 mm d=30 mm d1=d2=15mm

 

mp

2

2mv

gP

1

1

gV2

21

gP

2

2

gV2

22

D

A1 A2

d1d2

d

Q

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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

  

414

1

1AQ

41

AQ

3

sm

VAVA 4312

AVA

3

12s

m

sm

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Ex2 ( Application of Bernoulli’s equation )

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)

3cmg

3mkg

P1P2

V2V1d1=3cm

70bar

Q=10 l/min

d2=0.5cm

gP

1

gV2

21

gP

2

gV2

22

81.98001070 5

81.92236.0 2

81.98002

P

81.92496.8 2

1AQ

2

33

03.041

sec601010

m

21

dd

sm

sm

2mN

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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×

*p2

*p2

1 0 2 3

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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    

  

p2

102 PP

P0

P1

dQ

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Theory of Momentum (Principle of impulse)

= m ( momentum )  

= = = + m

F

A2

A1

Q

1

2 Q

F2CV

F

tI

tm

v

tvm

)(

tv

F1

F F2

F1

1

2 (Vector)

I

V

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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     

1F

tv

tm

1v

tm

F

2v

strF

1F

2F

2F

strF

2F

1F

F

2v

1v

tv

CV

VQ 222VQ 111

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 Assume incompressible = constThus 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

1v

2v

v

F

tv

2v

1v

tv

AF

A

dtdQ

A

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Flow force on the directional control valve

  ( a ) Recall Fstr= F1- F2 = Fi - Fo (scalar) Fax= -Fstr = Fo- Fi

Fax= cosε0 - cosεi - ρ

A

ovm

ivm

dtdQ

Fa

x

Io Ii Io Ii

Fstr2 1

ε2 εo

Io εi

ε1

Ii

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= 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

ovm

ivm

dtdQ

ovm

ivm

dtdQ

ovm

ivm

dtdQ

εi

ε2 εo, ε1

Io

Ii

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HW #3-a

(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。

385.0

cmg

oil

Fax

dk

(2) (1)

XkX

Po , Q

mmdk 8

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HW#3-b

( b ) Laminar flow through eccentric clearance

d=2cm , =50 cst , =0.85 l= 1cm , ,

Questions : (1) Qe=0 = ? ( l / min)

(2) Qe= = ? ( l / min)

3cmg

barp 200 cmr 002.0

r

dr2D

P1 P2

d

eD

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Case 1 : small orifice area ( ~const ) Q= Cf × A × v= = Cf ×

Where A=orifice area =πd x

P

P2

AQ

P2

d

x

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by small orifice area Δp~const

thus Q ~ X Fstr=ρv cosε ( only steady part ) =ρv ( vA ) cosε =2Cf

2 ×πd x ×Δpcosε Cs×X

Case 2 : large orifice area ( Q~const ) Fstr= cosε

Thus Fstr~   

 

   

Q~

dXQ

2

X1

x

Q

Fstr

Pmax=const

P3max

P2maxP1max

P3max>P2max > P3max

Fstr

x

Fstr

x

linear

Q1Q2

Q3

Pmax

Small A Large AQ3>Q2>Q1

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Shear stress where : dynamic viscosity cf. : kinematic viscosity

Unit :1Poise=1 =0.1

Viscosity of fluid

dydu

= )~(hU

scmg

2msN

h

FU

y

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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 )

2msN

sec2cm

sec2mm

HVD

UA4

dd

d

=2

42

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For narrow clearance

DH= (if h << b)

Conclusion:

Design in hydraulics Laminar flow

hhb

bh2≈

+24

Qh

b

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§ Hagen-Poiseuille formula

( laminar flow in a pipe ) τ×2πy = ( P1-P2 ) πy2 τ=η = × = v= ( r2-y2 ) for y =0 then Q= = = ( P1-P2 )

dydv

dydv

21 PP

2y

v

dv0 2

21 PP

ry

yy

ydy

421 PP

ry

y

vdA0

ry

y

ydyv0

2

8

4r

r y

P1 P2

V

Vmax

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Laminar flow through the clearance

V= ( h2-4y2 ) Vmax= h2 Q =2 =2 Q = ( P1-P2 )

821 PP

821 PP

dAhy

yx

2

0

bdyhy

yx

2

0

12

3bh

h

b

Q

Vy

Vmax

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Laminar flow through eccentric clearance

Q= ( P1-P2 ) Where η : dynamic viscosity ( e.g. Poise ; ) When e=Δr, which implies max. eccentricity Qe= 2.5 Qe=0 

12

3rd

3

5.11r

e

2mSN

P1 P2

Q

De

dr2D

r

d

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Flow through resistance and orifice

 (a) Resistance  

Q= Δp =k1 Δp Q~Δp ~ (b) Orifice

8

4r1

1

P1 P2

2r

P1 P2

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Q=CfA =k2 Q ~Thus

2 p

p

p

orifice

Resistance

Q

Δp

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Pressure losses through pipes

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 )

d

2

2v

d

gv2

2

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Example: Laminar pipe flow

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 )

8

4r

2

8r

d

2

2v

dV

64

eR64

25.0

3164.0

eR

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k-values for fluid through sudden expansion, contraction, pipe fitting, valves and bends

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 aredetermined empirically

2

2V

g2V2

22

21

DD

22

21

DD

Q D1 D2

D1D2 Q

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Circuit Calculation

  

  

( P1+ +ρgζ1 ) - ( P2+ +ρgζ2 ) =ΔPf

= +

 

2

21V

2

22V

2

2i

i

i

ii

Vd

2

2i

ii

Vk

d1

k1

k2

k321

3

1

2

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HW#4-a

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HW#4-b

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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

g2

vHr

Pg2

vrP 2

11

a2

11 )gr(

hr

Pheg2

vr

Pg2

vrP 2

222

211

hr

PheHr

P 21

a

hHheHH 21a

z1a HheHHhNPSH

g1a

g2g2

HheHHh

HHPP

Pa

P1 P2

A B

he (head loss)

(A)

rPH a

a

rPH z

z

rP

H gg

EX: 泵油之空穴 (Cavitation of oil pump)

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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 ?