Fluid Power Engineering Week 6/Lesson 2 Fluid flow in conduits Plan for today Today well talk about how fluid behaves when it flows through pipes/tubes/hoses Fluid flow friction We have covered most of the components that well encounter in hydraulic systems We need to see how fluid behaves when it flows through pipes/tubes/hoses, fittings, and valves Basically, whenever theres movement there is friction, both within the fluid itself and also between the fluid and the internal metal surfaces of the flow path
Two flow regimes Fluid Dynamics tells us that there are two flow regimes Laminar layers of fluid (laminae) The molecules in one layer stay in that layer Turbulent With greater fluid velocities the layering is disturbed The laminae interact with each other in a chaotic manner Laminar vs. turbulent flow The drawing at right illustrates the difference between the two types of flow We want laminar flow In general, we want to design the flow path for laminar flow Laminar flow has less fluid friction, generates less heat Less friction less energy lost
How do we ensure we stay in the laminar flow regime? We use the Reynolds number to design our fluid conduit , where v is velocity in m/sec D is the conduit diameter in m r is the fluid viscosity in kg/m3 m is the fluid viscosity in Nsec/m2 Viscosity (m and n), briefly We shall discuss viscosity soon in a little more detail It is a measure of how sticky or gooey a fluid is Molasses is more viscous than water Viscosity generally decreases with increasing temperature Cold oil flows more slowly than hot oil You always warm an engine up before changing its oil Absolute (m) vs. kinematic (n) viscosity There are two kinds of viscosity
Absolute (m) Kinematic (n) They are related by m has units of dynsec/cm2 where 1 N = 105 dyn 1 100 2 105 ( )
2 =0.1 2 We shall look at viscosity more closely later Laminar flow velocity profile If fluid flows over a flat surface, its velocity profile is Free-stream velocity
The fluid sticks to the surface; here its velocity is 0 In a round conduit, the profile is bullet-shaped We just use vavg Because the velocity varies over the conduit radius, we simply use vavg for all calculations =
= Mass continuity In a flow conduit with steady flow, the flow rate of mass past any point must be the same Mass continuity says But and If the fluid is incompressible (r = constant), So
But So for an incompressible fluid, mass flow rate continuity means volume flow rate continuity Volume-flow-rate continuity Note that if the cross-sectional area of the flow conduit changes as shown below Since = Since in this case AB > AA , then vB < vA As the fluid arrives at an area with a
bigger cross section, its velocity must decrease As we shall see, hydraulic fluid is not completely incompressible, but for almost all fluid-power calculations, treating it as incompressible leads to negligible error This, as we shall see, if the biggest difference between hydraulic and pneumatic calculations With compressible fluids, like air, r = r(p) Example Select conduit diameter for laminar flow Determine the minimum diameter of fluid conduit for laminar flow for water with n = 5.0 E5 m2/sec flowing at 20 lpm 2 =20 = = 4 =
Also 4 2 for laminar flow So From the table given in the last lesson, this is a small tube; thus achieving laminar flow with this small flow rate is relatively easy Example Flow velocity?
For a tube with this diameter, what would be vavg? = 4 2 4 20 3 =
= 24.6 2 1000 60 ( 0.0042 ) From the previous lesson, recommended by Esposito are vmax 1.2 m/sec (pump suction lines, to avoid cavitation) vmax 6.1 m/sec (pressurized lines) So his recommendations are very conservative, probably to reduce noise, heat build-up, and power consumption Pressure
losses Because of the friction encountered by the flowing fluid, pressure is lost as the fluid moves downstream; the amount of pressure loss depends on How thick the fluid is (viscosity) How rough the conduit surface is Well get to this part of an analysis soon; first we need a bit more development Incompressible flow, changing D Below is shown a reducer, a conduit section where the diameter gets smaller Because the diameter is decreasing and the volumetric flow rate is constant, the fluid has to speed up so that the same amount that passed point A in a period of time can pass point B during the same period Q continuity says =
> Since A AB, the fluid A speeds up Kinetic energy of fluid (Ev) But, like any physical object in motion, it has kinetic energy due to the motion If we look at a little packet of fluid, dm the kinetic energy of the packet is
Thus, each packet of fluid at B has more kinetic energy than it had at point A due to its increase in speed Where did this extra energy come from? Pressure energy of fluid (Ep) The answer is that the mass packet also has energy due to pressure The pressure energy can be expressed as This is not proven here; its proof is a standard part of the development of Fluid Mechanics So for our example, there has been an energy conversion from Ep to Ev Pressure energy has become kinetic energy If we ignore for the moment any frictional losses between A and B, the energy at the two points is the same; this can be expressed E v-A + E p-A = E v-B + E p-B Pressure energy of fluid (Ep) Continuing with this E v-A + E p-A = E v-B + E p-B
1 1 2 2 + = + 2 2 If we write this for a unit mass of material (dm = 1 kg) then Gravitational potential energy of
fluid (EZ) Besides these two types of energy, a packet of fluid can also have energy due to its position in a gravitational field The higher up it is, the more energy it has The expression for gravitational potential energy per unit mass of fluid is where Z is the position ofifthe above some reference position In our previous example, themass conduit is oriented
horizontally, then ZA = ZB and the terms cancel out Gravitational potential energy of fluid (EZ) But take the case of a constant diameter pipe oriented vertically Since DA = DB, the flow cross-sectional area is the same So vB = vA ; i.e. E v-A = E v-b But the pressure at B is greater than the pressure at A due to the weight of the fluid between A and B So here we have an exchange of gravitational energy for pressure energy E Z-A + E p-A = E Z-B + E p-B + =
+ In the general case Consider the case shown below pB < pA because B is higher than A vB > vA because AB is smaller than AA But since E v-B > E v-A , and this is not as a result of gravitational energy transfer to kinetic energy, pB is even less because E p-A provides also the increase in kinetic energy between points A and B Bernoullis equation E v-A + E p-A + E Z-A = E v-B + E p-B + E Z-B
1 1 2 2 + + = + + 2 2 If we divide by g 2
2 + + = + + 2 2 2 2
+ + = + + 2 2 This principle is known as Bernoullis equation after the Swiss scientist Daniel Bernoulli, who discovered this relationship in 1738 Bernoullis equation relates the energy states of two points in a fluid circuit, ignoring any frictional losses between the two states Bernoullis equation E v-A + E p-A + E Z-A = E v-B + E p-B + E Z-B
1 1 2 2 + + = + + 2 2 If we divide by g 2
2 + + = + + 2 2 2 2 +
+ = + + 2 2 This principle is known as Bernoullis equation after the Swiss scientist Daniel Bernoulli, who discovered this relationship in 1738 Bernoullis equation relates the energy states of two points in a fluid circuit, ignoring any frictional losses between the two states Bernoullis equation - Details 2 2 + + = +
+ 2 2 If we look at these terms in detail, we see something interesting The Z terms have as units meters That would mean the other terms need to be in terms of length too has units , also length has units , also length Thus the energy terms in the Bernoulli equation are expressed in terms of length This is simply an artifact of writing the equation on a per-unit-mass basis and also dividing the equation by g Bernoullis equation - Head We then speak of energy as head
Pressure head , for example, turns out to be the pressure that would occur if the packet of fluid considered were at the bottom of a column of fluid that high Velocity head is just kinetic energy on a per-unit-mass basis and then divided by g So we re-write the energy equation in terms of heads: H v-A + H p-A + H Z-A = H v-B + H p-B + H Z-B 2 = 2 = + =
Bernoullis fairyland world But the Bernoulli equation is applied to a system that has no losses To make it real, we need to also include Frictional losses in conduits, fittings, and valves Energy extracted by an actuator Energy added by a pump If between points A and B the frictional losses are small, there is no energy extracted by a cylinder or a motor, and there is no energy added by a pump, then the Bernoulli equation expresses the principle that no energy is created nor destroyed, but only changes form between these three typesvelocity, pressure, and height Outside learning To better understand this subject matter, view the following videos on the Big Bad Tech Channel Dont forget to turn the closed-captioning on to be able to understand better the details of the lectures
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