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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Chap 2 Basics of hydraulics Advantages of hydraulic control : ― easy of control ― high power output ― good dynamic response ― good dissipation of heat.
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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×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
A
F
2
2
A
F
Power
P= = = F × = ( p × A )( ) = p × Q where P : power p : pressure
Q : flow rate ( )
t
LF t
W
AQ
minl
F1
A1
L1
F2
A2
L2
dt
dL
HW #1
(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 =?
y
x
G
A3
P3
P1
P2
A1
A2
d
h+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
sec
mN
746
WattP
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
n
n
2
1
V
V
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)
Q
PnV1n1
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
d
V
1
2d
d
pdV
21
21 pV
22 pV
1
2
T
T
1
2
V
V
21 pV
1
2
P
P
Conservation of Mass( Continuity Equation)
+ =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
A
V1A1V2A2
1
.
m
2
.
m
ρ1A1V1 ρ2A2V2
1
.
m 2
.
m
dt
dl
dt
dl
dAundt
d 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 ) min
l Q=25 D=50 mm d=30 mm d1=d2=15mm
mp
2
2mv
g
P
1
1
g
V
2
21
g
P
2
2
g
V
2
22
D
A1 A2
d1d2
d
Q
Questions : Piston velocity = ? Velocities of the fluid in the inlet and outlet pipes = ?