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Chap 2 Basics of hydraulics

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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 – PowerPoint PPT presentation

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Title: Chap 2 Basics of hydraulics


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
FL where wwork(N?m) FForce(N)
LDistance(m) Pascals Law(Multiplication of
force) p constant w
F1L1 F2L2 (assume no energy losses)
2
Power
P F (p A)( ) p
Q where Ppower ppressure
Qflow rate ( )                  
       
F1
F2
L2
L1
A1
A2
3
HW 1
  • (Application of law of Pascal )
  • A240 cm2
  • A350 cm2
  • Qp50 l/min
  • G10 5N
  • Pmax45 bar
  • h25 cm
  • Questions
  • (1) ???G???????(P3min)?
  • (2) ???????? A1?
  • (3) ?? F1?
  • (4) ?? ?
  • (5) ?? ?
  • (6) ????????????????? t?

G
hd
d
A1
A3
P3
F1
P2
A2
P1
4
P (kW) 1W ExHow to
derive? HP(horsepower) In the case of rotary
actuator(motor)                       For
pump V1Apd Q1n1V1
Where Q Flow rate V1 Displacement
of pump T
Torque
Q
Pn
V1
n1
V2
n2
T1
T2
5
For motor V2 A pd Q2 n2 V2 If Q1
Q2 n1v1n2v2                          

Piston
A
6
HW 2
  • Given
  • n1 1450rpm
  • n2 400rpm
  • T2250N-m
  • Pn200bar
  • 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)

Q
V2
Pn
V1
n1
n2
T1
T2
7
Torque T1 F di (A?P)
( )?p similarly
T2 Torque transfer relation
Assume no energy loss P1 T1w1
2p n1 V1 n1
?p Q1 ?p Output power of motor
P2Q2?p Total efficiency?
1(ideal case)
8
Conservation of Mass(Continuity Equation)
  • 0
  • ?1Q1?1A1V1
  • ?2Q2?2A2V2
    pipe
  • if ?1?2 (incompressible fluid)
  • then V1A1V2A2
  • Ex

?1A1V1
?2A2V2
A
V2A2
V1A1
9
Conservation of Energy(Bernoullis Equation)
P2 V2
P1V1
h2
h1
1
2
10
  • Total Energy at any point
  • ElevationPressureKinetic
  • mgh  
  • Bernoullis equation  
  • h1 h2
  •  
  • Ex1(continuity equation)


Q25  

D50 mm  

d30 mm  

d1d215mm  
A1
A2
d
D
d2
d1
Q
11
Questions   Piston velocity ?  
Velocities of the fluid in the inlet and  outlet
pipes ?   Answer   (1) A1
pD219.6?2 A2
p(D2-d2)12.56?2 Cross section area
of inlet and outlet pipes A3
pd121.77?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
Ex2(Application of Bernoullis equation)

  • ?0.8  

  • 800
  • Assumeno work and no energy dissipation  
  • QuestionP2?

70bar
P1
P2
d20.5cm
V2
V1
d13cm
Q10 l/min
13
Flow through orifice
    ? A2A1 ?V1V2   P1 1/2 ?V12 P2
1/2 ?V22   V2 where ?pP1-P2
  Flow rate QA2V2  
A2  
1
0
2
3
14
Area A2CcA0 Where
Cccontraction coefficient Flow
coefficient Cf Q Cf A0
where ?pP1-P3 Cf f(Reynolds
numbergeometry) 0.61.0 Valve
geometry
Q Cf pd

Cf 0.60.64
         
P0
d
P1
Q
15
Theory of Momentum (Principle of impulse)
  • m (momentum)  
  • m

F
A1
F1
Q
F2
F
1
F1
F2
CV
A2
(Vector)
2
Q
16
Steady flow ( 0) position 1
?1Q1 ?1Q1 position 2
?2Q2 - ?2 Q2 Flow force
- -
- Ex assume incompressible (?const)
?1Q1?2Q2 ?Q -?Q(
) unsteady part m          
CV
17
  Assume incompressible const Thus Q1
Q2 Q - (Aconst)
m(? - ?steady part0)
? A pressure dropP1-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 cose0 -
cosei - ?
Io
ei
e2 eo
e1
Ii
Io
Ii
Io
Ii
Fax
Fstr
2
1
19
?v2 Q
?v1 Q eo e2 90o
ei e1 180o Fax0 ?v1Qcose1 -
? (b)       Fax cose0 -
cosei ? ?v1Q e0e1
?v2Q eEe218002700
Fax?v1Qcose1 ?
ei
Io
e2
Ii
eo, e1
20
HW 3-a
  • (a)
  • Shown above is a directional control valve
  • Given Q 40 l/min
  • 70 0
  • Xk 0.5mm
  • cf 0.8
  • ?? (1) ???Fax??unsteady part ?????????
  • ??????????(?????????
  • (1)?(2))

Po , Q
(1)
(2)
Fax
X
Xk
21
HW3-b
  • (b ) Laminar flow through eccentric clearance
  • d2cm , 50 cst , 0.85
  • l 1cm , ,
  • Questions
  • (1) Qe0 ? ( l / min)
  • (2) Qe ? ( l / min)

D
e
P1
P2
d
22
  • Case 1small orifice area ( const)
  • Q Cf A v Cf
  • Where Aorifice area
  • pd x

x
d
23
by small orifice area ?pconst thus Q X
Fstr?v cose (only steady part)
?v(vA)cose 2Cf2 pd x ?pcose CsX
Case 2large orifice area (Qconst) Fstr
cose Thus Fstr          
 
Fstr
linear
Pmaxconst
Q3
Q2
Q1
x
Small A
Large A
Q3gtQ2gtQ1
Fstr
P3max
P2max
P1max
x
P3maxgtP2max gt P3max
Fstr
x
Pmax
Q
24
Viscosity of fluid
  • Shear stress
  • where dynamic viscosity
  • cf. kinematic viscosity
  • Unit
  • 1Poise1 0.1

F
U
h
25
  • 1cp10-2 Poise 10-3
  • 1 stoke 1
  • 1 cst1
  • ?(or ?) f(T,P)
  • Flow in Pipes
  • Eqs. derived
    from viscous flow condition
  • (1)laminar flow NRlt2000
  • (2)turbulent flow NRgt4000
  • only empirical
    Eqs.  
  • Reynolds number NR(dimensionless)  
  • NR
  • where Vvelocity of fluid  
  • DHhydraulic diameter  
  • ?kinematic viscosity  
  • Hydraulic Diameter DH
  • where AArea of pipe  
  • Ucircumference of pipe  
  • For pipes  

26
  • For narrow clearance
  • DH   (if h ltlt
    b)
  • Conclusion
  • Design in hydraulics ? Laminar flow

Q
h
b
27
Hagen-Poiseuille formula
  •  (laminar flow in a pipe)  
  • t2py (P1-P2)py2  
  • t?  
  •  
  • v (r2-y2)  
  • for y 0 then  
  • Q (P1-P2)

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 ?dynamic viscosity (e.g. Poise
)   When e?r, which implies max. eccentricity
  Qe 2.5 Qe0  
30
Flow through resistance and orifice
  •  
  • (a) Resistance  

2r
P1
P2
Q ?p k1 ?p   Q?p   (b)
Orifice  
P1
P2
31
  • QCfA
  • k2
  • Q
  • Thus

Q
orifice
Resistance
?p
32
Pressure losses through pipes
  • Darcy-Weisbach formula
  • Pf ?( )
  • or hf ?( )
  • where
  • Pfpressure loss
  • hf head loss
  • length of pipe
  • d internal diameter of pipe
  • ? friction factor(derived
    experimentally)

33
ExampleLaminar pipe flow
  • Q ?p
  • QAvpr2v
  • Thus ?p P1-P2 Pf v ?
  • ? (valid only if
    Relt2000)
  • For turbulent flow
  • Blasius formula
  • ? (for pipes with smooth
    inside surface)
  • (3000ltRelt105)
  •  

34
k-values for fluid through sudden expansion,
contraction, pipe fitting, valves and bends
  • ?Pff k (for turbulent flow)
  • or hff k( )
  • where ?Pffpressure loss
  • hffhead loss
  • Sudden enlargement
  •  

  • k (1- )2
  •  
  • Sudden reduction
  •  

  • k 0.5(1- )
  • k-values for pipe fitting, valves, and bends are
  • determined empirically

Q
D1
D2
D2
Q
D1
35
Circuit Calculation
  •  
  •  
  •  
  •  
  • (P1 ?g?1)-(P2 ?g?2)?Pf
  •  

k1
d1
k3
1
k2
2
36
HW4-a
37
HW4-b
38
(No Transcript)
39
EX ?????(Cavitation of oil pump)

  • P1 ??????

  • P2 ???????
  • Pa ????
  • Pg ??????
  • (1) From Bernoullis Eq.
  • (2)

A B
Pa
P1 P2
he (head loss)
(A)
40
  • ??
  • (1)?(A)?,???(Ha? , H1? , he , Hg ?)????
  • ???? cavitation?
  • (2)??????cavitation,????????cavitation?Heads
    ?????(????????)
  • (3)??????????,??(2)???????
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