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
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
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
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
For motor : V2 = A ×πd
Q2 = n2 × V2
If Q1= Q2
n1v1=n2v2
=
1
2
nn
2
1
VV
Piston
A2d
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
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
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
Conservation of Energy( Bernoulli’s Equation)
h2h1
P2 V2
P1V1
1 2
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
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
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
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
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
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
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
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
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
= 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
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
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
Case 1 : small orifice area ( ~const ) Q= Cf × A × v= = Cf ×
Where A=orifice area =πd x
P
P2
AQ
P2
d
x
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
Shear stress where : dynamic viscosity cf. : kinematic viscosity
Unit :1Poise=1 =0.1
Viscosity of fluid
dydu
= )~(hU
scmg
2msN
h
FU
y
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
For narrow clearance
DH= (if h << b)
Conclusion:
Design in hydraulics Laminar flow
hhb
bh2≈
+24
Qh
b
§ 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
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
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
Flow through resistance and orifice
(a) Resistance
Q= Δp =k1 Δp Q~Δp ~ (b) Orifice
8
4r1
1
P1 P2
2r
P1 P2
Q=CfA =k2 Q ~Thus
2 p
p
p
orifice
Resistance
Q
Δp
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
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
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
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
HW#4-a
HW#4-b
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)
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 ?