Embed Size (px)
Citation preview
1
Chapter 2
Fluid Power Basics
Courtesy : Fluid Power Circuits and Controls, John S.Cundiff, 2001
IntroductionOne of the underlying postulates of fluid mechanics is that, for a particular position within a fluid at rest, the pressure is the same in all directions.A second postulate states that fluids can support shear forces only when in motion. These two postulates define the characteristics of the fluid media used to transmit power and control motion.
2
Introduction (contd..)
Traditional concepts such as static pressure, viscosity, momentum, continuity, Bernoulli’s equation, and head loss are used to analyze problems encountered in fluid power systems.
The product of fluid mass density and the gravitational constant is the constant of proportionality between the depth of fluid in the container and the pressure acting at that depth in the fluid.P = ρ * g * h
Where P = pressureρ = fluid mass densityh = depth of fluidg = gravitational constant
Hydrostatic Pressure
3
Hydrostatic Pressure (contd..)
The density of a fluid is often referred to in terms of specific gravity.Specific gravity, by definition, is the ratio of the specific weight of the fluid in question to that of water at standard conditions.
Hydrostatic Pressure (contd..)Modify the static pressure equation and introduce specific gravity as:
P = Sg ρwghWhere Sg = specific gravityρw = density of water
Rearranging the terms we can solve for the specific gravity of fluid in the column
Sg = P / ρwgh
4
Hydrostatic Pressure (contd..)
Cavitation occurs when the pump does not completely fill with liquid.The incoming fluid is a mixture of gas and liquid.An important application of fluid power is the force multiplication that can be achieved with a hydraulic jack.
Example Problem 2.1
The small cylinder of the jack shown below has a bore of 0.25” and the large cylinder has a bore of 4”. How much can be lifted if the jack handle is used to apply 10 lbf to the small cylinder (Fs = 10 lbf)?
5
Example Problem 2.1 (contd..)
Pressure developed: Ps = Fs / As
Where As = area of small cylinder (in2)As = 0.252π / 4 = 0.049 in2
Ps = 10 / 0.049 = 204 psi
Example Problem 2.1 (contd..)Pascal’s law: Pl = Ps
Where Pl = pressure on large cylinder (psi)
Lift developed:Fl = PlAl
Where Al = 42π / 4 = 12.56 in2
Fl = (204)(12.56) = 2560 lbf
6
Example Problem 2.1 (contd..)
If the small cylinder bore is 0.125 in, how much can be lifted?
As = (0.125)2π / 4 = 0.0123in2
Ps = 10 / 0.0123 = 813 psiFl = (813)(12.56) = 10200 lbf
A 10 lbf produces a 10,000-lbf lift.
Fluid StaticsIt is important to distinguish between gauge and absolute pressures.A height of 760 mm of mercury represents standard atmospheric pressure at sea level.Gauge pressures are always measured relative to atmospheric pressure.The mean value taken at sea level is termed standard atmospheric pressure.
7
Fluid Statics (contd..)If we are merely concerned with pressure differences throughout a system, then ignoring atmospheric pressure has little effect on our analyses.However, if we are to apply the equation of state for an ideal gas, such as in the case of analyzing gas-filled accumulators, then it is essential to consider absolute pressure, which is the sum of gauge and local atmospheric pressure.
Pabsolute = Patmospheric + Pgauge
Figure 2.4
8
Conservation of MassAnalytical approaches applied to fluid power systems are based on the concept of conservation of mass. The following equation describes flow in and out of a control volume.
Where M = system massT = timeV = system volumeρ = fluid densityV = velocity normal to incremental area dAA = area perpendicular to flow streamlines
*∫∫ +∂∂
=CVCV
dAvdVtdt
dMsystem
ρρ
Conservation of Mass (contd..)
If fluid density remains constant and the boundaries of the control volume are fixed, meaning that the control volume does not expand or contract, the equation simplifies to:
dAvCS
*∫ ρ=0
9
Conservation of Mass (contd..)This equation means that the total flow of mass out of the control volume equals the total flow in.
Q1 = Q2 + Q3 (refer Fig.2.6)In a hydraulic circuit, where oil pressures can rise to 6000 psi, the fluid does compress (density is not constant) and hoses do swell (control volume boundaries are not constant).Flow out of a component (valve, actuator, etc) will always equal the flow minus any leakage to a drain line.
Functions of a Working Fluid
The function of the fluid is to:1. Transmit power2. Provide lubrication3. Provide cooling4. Seal clearances
10
Functions of a Working Fluid (contd..)
To provide lubrication and seal clearances between moving parts, the fluid must be able to establish and maintain a continuous film between the parts.High pressures and high relative velocities affect the establishment and maintenance of the film.Friction is unavoidable and the resultant heat raises the temperature of the fluid.
Functions of a Working Fluid (contd..)
The main heat source is not friction but the conversion of fluid energy to heat energy when there is a pressure drop across a restriction or along a conductor, and no mechanical work is done.Conversion of fluid energy to heat energy is a key issue in the design of fluid power systems.Often it is the “price” paid for the excellent control of load motion that can be achieved with a fluid power circuit.
11
Functions of a Working Fluid (contd..)
Elevated temperature changes the properties of the fluid and, because of thermal expansion, changes the clearances between moving parts.It is estimates that 80% of all failures are related to a fluid “failure.”
Fluid PropertiesViscosity –
Viscosity is the fluids resistance to shear.Fluid velocity profile
- Top plate moves with velocity v relative to stationary bottom plate.- The slope of velocity profile established between plates is
slope = ∆v/∆yy = distance between the plates
12
Fluid properties (contd..)If the moving plate has area A and a force F to keep moving it at velocity v, the Shear stress in the fluid between the plates is
J = F / ADynamic viscosity – Ratio between the shear stress and the slope.
μ = F * yv * A
Units of dynamic viscosity are g /(s . cm)The name given these units is the poise.
Fluid properties (Contd..)
Kinematic viscosity – Dynamic viscosity divided by the fluid density measured at same temperature as dynamic viscosity
v = μ / ρμ = dynamic viscosity [g/s-cm)], ρ = density (g/cm3)
Units for kinematic viscosity cm2/s or stroke.
13
Fluid properties (contd..)
Standards ASTM D 2422 and ISO 3448rates fluids based on viscosity at 400 c.
- ISO grades have the letter VG( viscosity grade)
- SAE grades are based on tests run at 1000 c.
14
Fluid properties (contd..)The following equation can be used for estimates of viscosity in centistokes.
v = 0.226t – 195 32<t<100t
v = 0.22t – 135 t>100t
where v= kinematic viscosity (cS) , andt= time in SUS, which stands for Saybolt
universal seconds. The time the liquid takes to drain through a Saybolt viscometer.
Fluid Properties (contd..)Problems caused by use of oil with too low viscosity :
Loss of pump efficiency.Component wear Breakdown of lubrication film, leading to “spot-weld” condition.
Problems caused by oil with too high a viscosity : Pump cavitation.High pressure drops.
15
Fluid properties (contd..)Bulk modulus – Degree of oil compressibility is expressed by the bulk modulus and is expressed as
β = -∆P / (∆V/V)β = bulk modulus (psi)∆P = change in pressure (psi)∆V = change in volume when ∆P is
appliedV = original volume.
Fluid properties (contd..)
Air and temperature effect on bulk modulus.
If air is entrained in the fluid, bulk modulus is reduced significantly.
Temperature causes fluid to increase in volume and affects bulk modulus .
16
Fluid properties (contd..)
Specific Gravity – The ratio of the density of the fluid to the density of water at 40 c and standard atmospheric pressure.
Sg = ρ / ρw
where ρ = density of fluid (slug/ft3)ρw= density of water (slug/ft3)
Fluid properties (contd..)Specific weight – Weight per unit volume
γ = mg / Vm = mass , V = volume, g = gravitational constant
Since ρ = m/V, γ = ρ g
Specific gravity can be written as Sg = ρ g / ρwg
17
Fluid properties (contd..)Oxidation : Reaction between the oil and oxygen. The end products are referred to as sludges and resins.Corrosion and Rust Resistance : Defined as chemical reaction between fluid and a metal surface.Fire Resistance : Key consideration in hydraulic systems.
Parameters : Flash point , Fire point, AIT.
Flow in Lines
Flow is laminar if layers of fluid particles remain parallel as the flow moves along the conductor.
Flow is turbulent if the fluid layers break down as the flow moves along the conductor.
18
Flow in Lines (contd..)
Flow is laminar if layers of fluid particles remain parallel as the flow moves along the conductor.
Flow in Lines (contd..)
Flow is turbulent if the fluid layers break down as the flow moves along the conductor.
19
Flow in Lines (contd..)Flow dominated by viscosity forces is laminar and inertia dominated flow is turbulent. Reynolds number : Laminar flow is a function of dimensionless parameter known as Reynolds number.
NR = v D ρμ
v = fluid velocityD = conductor inside diameterρ = fluid mass densityμ = dynamic viscosity
Flow in Lines (contd..)Reynolds number is a dimensionless ratio of inertia force to viscous force.
Rules obtained from Reynolds tests:1. If NR < 2000, flow is laminar.2. If NR > 4000, flow is turbulent.3. The region 2000 < NR < 4000 is defined
as transition region between laminar and turbulent flow.
20
Flow in Lines (contd..)When fluid flows, friction causes fluid energy to get converted to heat energy.
Fluid power at the inlet to a conductor is Phyd 1 = P1Q1
P1 = pressure at the inletQ1 = flow at the inlet
Fluid power at outlet is Phyd 2 = P2Q2
Flow in Lines (contd..)Friction results in a pressure drop in the line.Darcy’s Equation provides a means to calculate pressure drop :
hL = f (L/D) (v2/2g)
hL = head loss (ft)f = friction factor (dimensionless), for laminar f = 64/NR
D = conductor inside diameter (ft)L = conductor length (ft)V = average fluid velocity (ft/s)G = gravitational constant (ft/s2)
21
Flow in Lines (contd..)Hagen – Poiseuille Equation:
hL = 64/NR (L/D) (v2/2g) (laminar flow)
Another form of Hagen – Poiseuille Equation:
∆P = 128 μ L Qπ D4
∆P = pressure dropμ = absolute viscosity (lbs/s-ft)L = length (ft)
Q = volume flow rate D = diameter (ft)
Flow in Lines (contd..)
When flow is turbulent, the friction factor is a function of
Reynolds number Relative roughness of conductor
Relative roughness :Rel. roughness = / D
= conductor inside surface roughness.D = conductor inside diameter.
22
Flow in Lines (contd..)Blasius equation:
Friction factor in turbulent flow range can also be calculated using Blasius equation.
f = 0.1364NR
0.25
for smooth conductors and a Reynolds number less than 100,000.
Flow in Lines (contd..)The Moody diagram below is often used for turbulent flow
in smooth pipes.
23
Flow in Lines (contd..)In applications like mobile machines, pressure drops through other components (valves, fittings etc.) are more significant than pressure drop in the conductor itself.Losses in Fittings:- Tests show that head losses in fittings are proportional to the square of the velocity of the fluid.
hL = Kv2 / 2ghL = head loss (ft)v = fluid velocity (ft/s)g = gravitational constant (ft/s2)K = fitting factor
Flow in Lines (contd..)Typical fitting factors are:
Standard tee K= 1.8Standard elbow K= 0.945 degree elbow K= 0.42Return bend (u-turn) K= 2.2
24
Flow in Lines (contd..)Example problem 2.4The circuit from example problem 2.3 has an elbow (K = 0.9) at the motor. Fluid flows through a hose to an elbow and on into the motor. What is the pressure drop in this fitting for Q = 15 GPM?
Q = 15 GPMv= 10.9 ft/shL = Kv2 / 2g
= 0.9(10.9)2 / 2 (32.2)= 1.66 ft
∆P = 0.433 hL Sg = 0.6 psi
Leakage FlowTo understand leakage flow, consider the following cases.
Spool-type valves have a cylinder with grooves at intervals along the length. The cylinder slides back and forth in a bore to open and close passageways through the valve. There may be leakage through the annulus of spool and the bore.Leakage between the piston and the cylinder inner wall of a piston pump.
25
Leakage Flow (contd..) Consider an example of spool-type valve: (Fig 2.13)- The annulus has been “unwrapped” from around the spool.- Distance a is the clearance between spool and bore.- w is circumference of the bore.- L is distance between adjacent grooves machined in the spool.- Sections between grooves are called lands
Leakage Flow (contd..) Expression for leakage flow as a function of pressure drop across the land (along L):
Q = wa3 ∆P12 μ L
Q = leakage floww = width of rectangular openinga = height of rectangular openingμ = absolute viscosityL = Length of leakage pathway∆P = pressure diff. across the land (along the
length L)
26
Leakage Flow (contd..)
width w = π D and absolute viscosity μ =ρv
Substituting this is the previous equation, we get,
Q = π Da3∆P12 ρ v L
Orifice Equation The orifice equation states that flow through an orifice is proportional to square root of the pressure drop across the orifice (∆P = upstream pressure – downstream pressure)
Q = flow (in3/s)C= orifice coefficient (decimal)A= area (in2)g = gravitational constant (in/s2)∆P = pressure (psi)γ = specific weight of fluid (lbf/in3)
γP)/2(CA Q Δ= g
27
Orifice Equation (contd..)The previous equation simplifies to
where
Solving for ∆P, Q becomes an independent variable , a form that is useful for analysis of fluid power circuits.
∆P = Q2/k2 = keq Q2
where keq = 1 / k2 .If Q is doubled, then the pressure drop will increase fourfold.
Pk Q Δ=
γ /2g CA k =
Orifice Equation (contd..)
Technical data sheets (tech sheets) supplied by almost all valve manufacturers have a curve similar to the fig 2.14.
28
Orifice Equation (contd..)Classic problem to illustrates influence of an orifice in a circuit. (fig 2.15)
Fluid from accumulator flows into cylinder. How fast does cylinder extend?
- Principle – Inert gas is compressed as fluid enters into the accumulator.
- Boyles law : P1V1 = P2V2
Orifice Equation (contd..)
Accumulators are devices used to store fluid volume under pressure for later use.The accumulator is precharged to some pressure P1 and a control volume of inert gas oocupies volume V1. Fluid is pumped into the accumulator changing the gas’s pressure and volume to P2 and V2 ,respectively.
Boyles law : P1V1 = P2V2
29
Orifice Equation (contd..)Analysis to illustrate use of orifice equation(refer fig 2.15). The accumulator is charged to 2725 psi, and it takes 500 psi to lift load.
Directional control valve (DCV) has a keq = 32.5 x 10-4. So,
Using the orifice equation,
5.17 / 1 k == eqk
GPM214/in82550027255.17k Q
3 ==
−=Δ=
sP
Orifice Equation (contd..)Suppose the cylinder bore is 3 in. If flow from the accumulator is 825 in3/ s, the resultant cylinder velocity is
v = Q / A= 825 / (π 32 / 4) = 116.7 in/s
- This velocity is very high, it might be described as an “explosive” extension of the cylinder.
30
Orifice Equation (contd..)If we decreased the cylinder bore by halve, Flow was reduced , but cylinder velocity increased 2.3 times.Another option is installation of a flow control valve in the circuit, shown in the fig 2.17
Orifice Equation (contd..)
(Refer fig 2.17)
What keq value must be set for the flow control to limit cylinder velocity to 2.4 in/s?
Q = A c v = (π 32 / 4) 2.4 = 4.4 GPM- So, the required flow is about 2% of the
flow available when the DCV is opened rapidly.
31
Orifice Equation (contd..)Pressure drops across the DCV for a flow of 17 in3/s is obtained using the orifice equation.
∆P = keq Q2
= 32.5 x 10-4 (17)2
= 1 psi. Pressure drop is so low that DCV can be neglected in analysis.Assume that full ∆P to limit flow to 17 in3/s is developed in the flow control valve.
Orifice Equation (contd..)Then required kfc
Flow control valve converts some of the fluid energy to heat energy , thus reducing the energy that flows in the cylinder. When the energy stored in accumulator is required frequently, the energy loss due to the flow control valve is accepted.
36.0
5002725/17
/
=
−=
Δ=
fc
fc
fcfc
k
k
PQk
32
Orifice Equation (contd..)
We can also use the orifice equation to analyze a pressure reducing valve.
Orifice Equation (contd..)Operation of Pressure reducing valve. (Fig 2.19)
Spring provides downward force (Fs) on spool.Equal hydraulic force exerted in both directions by the inlet pressure. Hydraulic pressure is applied to the bottom of the spool by downstream pressure. (Fhb)Higher the downstream pressure, the more the valve displaces upward and more the spool closes the orifice.
33
Orifice Equation (contd..)Pressure reducing valve works well in this circuit, because it automatically changes it’s spool position as accumulator pressure drops.
Pressure reducing valve opens the orifice, thus reducing pressure drop, as inlet pressure falls.
To ensure enough pressure is available for breakout, pressure reducing valve should be chosen properly.
When cylinder velocity is low, a breakout force is required to overcome friction and move the gate valve off it’s seat.
Flow control valve is included to provide a means for “fine tuning” the flow to the cylinder.
END OF CHAPTER 2
THANK YOU
