course number: me 321 fluid mechanics i 3 credit hour ...teacher.buet.ac.bd/mmrazzaque/fluid...
TRANSCRIPT
COURSE NUMBER: ME 321
Fluid Mechanics I
3 credit hour
Basic Equations in fluid Dynamics
Course teacher
Dr. M. Mahbubur Razzaque
Professor
Department of Mechanical Engineering
BUET 1
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Description of Fluid Motion
In the description of a flow field, it is convenient to think of
individual particles each of which is considered to be a small mass of
fluid, consisting of a large number of molecules, that occupies a
small volume that moves with the flow.
If the fluid is incompressible, the volume does not change in
magnitude but may deform. If the fluid is compressible, as the
volume deforms, it also changes its magnitude. In both cases the
particles are considered to move through a flow field as an entity.
There are two approaches of mathematical description of the physical
quantities of a fluid motion as a function of space and time
coordinates :
Lagrangian Descriptions of Motion
and
Eulerian Descriptions of Motion
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Lagrangian Descriptions of Motion
In the study of particle mechanics, where attention is focused on
individual particles, motion is observed as a function of time. The
position, velocity, and acceleration of each particle are listed as s(x0,
y0, z0, t), V(x0, y0, z0, t), and a(x0, y0, z0, t), and quantities of interest
can be calculated.
The point (x0, y0, z0) locates the starting point—the name—of each
particle. This is the Lagrangian description, named after Joseph L.
Lagrange (1736–1813), of motion that is used in a course on
dynamics.
In the Lagrangian description many particles can be followed and
their influence on one another noted. This becomes, however, a
difficult task as the number of particles becomes extremely large in
even the simplest fluid flow.
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Eulerian Descriptions of Motion
The region of flow that is being considered is called a flow field.
In Eulerian description, points in a flow field is identified in space
and then the velocity of fluid particles passing each point is observed;
we can observe the rate of change of velocity as the particles pass
each point, that is,
and we can observe if the velocity is changing with time at each
particular point, that is, .
In this Eulerian description, named after Leonhard Euler (1707-
1783), of motion, the flow properties, such as velocity, are functions
of both space and time.
In Cartesian coordinates the velocity is expressed as V = V(x, y, z, t).
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Lagrangian and Eulerian Descriptions
An example may clarify these two ways of describing motion. An
engineering firm is hired to make recommendations that would
improve the traffic flow in a large city.
The engineering firm has two alternatives: Hire students to travel in
automobiles throughout the city recording the appropriate
observations (the Lagrangian approach), or
hire students to stand at the intersections and record the required
information (the Eulerian approach).
A correct interpretation of each set of data would lead to the same set
of recommendations, that is, the same solution. In an introductory
course in fluids, however, the Eulerian description is used
exclusively since the physical laws using the Eulerian description are
easier to apply to actual situations.
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Acceleration
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Acceleration
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Acceleration
This derivative is called the substantial derivative, or material
derivative. It is given a special name and special symbol (D/Dt
instead of d/dt) because we followed a particular fluid particle, that
is, we followed the substance (or material).
It represents the relationship between a Lagrangian derivative in
which a quantity depends on time t and an Eulerian derivative in
which a quantity depends on position (x, y, z) and time t.
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Acceleration
The time-derivative term on the right side of Eqs. 3.2.8 for the
acceleration is called the local acceleration and the remaining
terms on the right side in each equation form the convective
acceleration.
Hence the acceleration of a fluid particle is the sum of the local
acceleration and convective acceleration.
In a pipe, local acceleration results if, for example, a valve is being
opened or closed; and convective acceleration occurs in the
vicinity of a change in the pipe geometry, such as a pipe
contraction or an elbow. In both cases fluid particles change speed,
but for very different reasons.
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Basic Laws in Fluid Dynamics
The basic laws in fluid dynamics are expressed either in terms of a
system or a control volume.
A system is a fixed collection of material particles. For example, if
we consider flow through a pipe, we could identify a fixed quantity
of fluid at time t as the system (Fig. 4.1); this system would then
move due to velocity to a downstream location at time t + Dt. Any of
the basic laws could be applied to this system.
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Our interest is most often focused on a device, or a region of space,
into which fluid enters and/or from which fluid leaves; we identify
this region as a control volume.
An example of a fixed control volume is shown in Fig. 4.2a. A
control volume need not be fixed; it could deform. However, only
fixed control volumes will considered from now on.
The difference between a control volume and a system is illustrated
in Fig. 4.2b.
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System to control volume transformation
System to control-volume transformation, or equivalently, the
Reynolds transport theorem is written as,
Here h represents the intensive property (the property of the system
per unit mass) associated with Nsys. The relation between h and Nsys
is expressed as
The first integral of Eq. (4.2.9) represents the rate of change of the
extensive property in the control volume.
The second integral represents the flux of the extensive property
across the control surface; it may be nonzero only where fluid crosses
the control surface.
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CONSERVATION OF MASS
A system is a given collection of fluid particles; hence its mass
remains fixed:
Using Reynolds transport theorem and recognizing that Nsys = msys
i.e. the mass of the system (an extensive property for which h =1),
we can write
This is called the continuity equation. If the flow is steady, there
results
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CONSERVATION OF MASS
For a uniform flow with one entrance and one exit, Eq. (4.3.4) takes
the form
Where we have used
If the density is constant in the control volume, the continuity
equation reduces to
This form of the continuity equation is used quite often, particularly
with liquids and low-speed gas flows.
The above derivation assumed uniform velocities at the inlet and exit
sections. What happens when the velocities at the inlet and the exit
sections are non uniform?
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Continuity equation
Suppose that the velocity profiles at the entrance and the exit are not
uniform, such as sketched in Fig. 4.7. Furthermore, suppose that the
density is uniform over each area. Then the continuity equation takes
the form
Where, are the average velocities at sections I and 2,
respectively. In examples and problems the overbar is often omitted.
It should be kept in mind, however, that actual velocity profiles are
usually not uniform.
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Continuity equation
Two fluxes are useful in specifying the quantity of flow. The mass
flux or the mass rate of flow is
and has units of kg/s; Vn is the normal component of velocity. The
flow rate Q, the volume rate of flow, is
and has units of m3/s. The mass flux is usually used in specifying the
quantity of flow for a compressible flow and the flow rate for an
incompressible flow.
In terms of average velocity, we have
Where uniform density and normal velocity is assumed.
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Solution
The control volume is selected to be the inside of the nozzle as shown. Flow
enters the control volume at section 1 and leaves at section 2. The
simplified continuity equation (4.4.6) is used since the density of water is
assumed constant and the velocity profiles are uniform:
The flow rate or discharge is found to be:
Example 4.1
Water flows at a uniform velocity of 3 m/s into a nozzle that reduces the
diameter from 10 cm to 2 cm. Calculate the water’s velocity leaving the
nozzle and the flow rate.
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or
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Continuity equation in differential form
Consider the mass flux through each face of the infinitesimal control
volume shown in Fig. 5.1. We set the net flux of mass entering the
element equal to the rate of change of the mass of the element; that is,
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This takes the form
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Then the continuity equation becomes,
Or, in vector form,
The continuity equation in cylindrical coordinate is
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These are the called Navier-Stokes equations, named after C. I. M.
H. Navier (l785-1836) and Sir George G. Stokes (1819-1903), who
are credited with their derivation. These are second-order nonlinear
partial differential equations.
Navier-Stokes equations
In an incompressible flow the density remains constant or the change
in density is very negligible. In an incompressible flow of a
Newtonian Fluid with constant viscosity, the momentum equations
(derived from Newton’s second law of motion) are:
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Which may be interpreted as:
Inertia force per unit vol. = gravity force per unit vol. + pressure
force per unit vol. + viscous force per unit volume
There are four unknowns: p, u, v, and w. So, for solution, it should be
combined with the incompressible continuity relation to form four
equations in these four unknowns.
The continuity equation for incompressible flow in steady state is:
Or,
Navier-Stokes equations ….
In vector notations, the Navier-Stokes equations may be written as:
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Incompressible Inviscid flow….
Computation Fluid Dynamics (CFD) is the branch of fluid
mechanics that uses numerical methods to solve the Navier-Stokes
equations and continuity equation with appropriate boundary
conditions and without any analytical simplification.
However, analytical solution for many viscous flow problems are
also possible through some simplifying assumptions.
If viscosity is assumed to be zero (inviscid flow approximation), the
second order terms in the Navier-Stokes equations are lost and we
get the Euler’s Equation, as:
Integration of this equation along a streamline results in Bernoulli’s equation.
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Energy Equation
Many problems involving fluid motion demand that the energy
equation, be used to relate quantities of interest.
If the heat transferred to a device (a boiler or compressor), or the work
done by a device (a pump or turbine), is desired, the energy equation
is needed.
It is also used to relate pressures and velocities when Bernoulli’s
equation is not applicable; this is the case whenever viscous effects
cannot be neglected, such as flow through a piping system.
Let us express the energy equation in control volume form. For a
system it is
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where the specific energy e is,
In terms of a control volume, Eq. 4.5.1 becomes
This can be put in simplified forms for certain restricted flows. The
rate-of-heat transfer term ˙Q represents the rate-of-energy transfer
across the control surface due to a temperature difference. (Do not
confuse this term with the flow rate Q.)
The work-rate term results from work being done by the system. Work
of interest in fluid mechanics is due to a force moving through a
distance while it acts on the control volume. The The work-rate term
is discussed in detail in the following section.
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Considering different situations in fluid problems, the work-rate term
becomes:
We should note that third and fourth components in the work-rate
term are seldom encountered in problems in an introductory course
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In many fluid flows, useful forms of energy (kinetic energy and potential
energy) and flow work are converted into unusable energy forms (internal
energy or heat transfer). However, in an introductory fluid mechanics
course the sum of these effects is lumped together. We define losses as the
sum of all the terms representing unusable forms of energy:
For a steady-flow situation in which there is one entrance and one exit
across which uniform profiles can be assumed, the energy equation
becomes.
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Here, Ws is shaft work and hL is head loss. If the losses are negligible,
if there is no shaft work or heat flow and if the flow is incompressible,
we can write the energy equation in its most useful form as,
Note that this form of energy equation is identical with Bernoulli's
equation. However, we must understand that the Bernoulli's equation
is a momentum equation applicable along a stream line and the
equation above is an energy equation applied between two sections of
a flow.
When the energy equation is applied to any steady uniform flow, the
control volume is usually selected such that the entrance and exit
sections have a uniform total head.
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For example, if the energy equation is applied to a flow passing a
gate, the appropriate control volume may be chosen as shown below
The total head at the entrance and exit can be evaluated at an point at
the entrance and exit, respectively. However, a convenient choice
would be the points at the water surface. Thus the energy equation
becomes
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where we have introduced the head loss hL, often written in terms of a
loss coefficient K as
A final note to this section regards nomenclature for pumps and
turbines in a flow system. It is often conventional to call the energy
term associated with a pump the pump head Hp, and the term
associated with a turbine the turbine head HT . Then the energy
equation takes the form
In this form we have equated the energy at the inlet plus added energy
to the energy at the exit plus extracted energy (energy per unit weight,
of course). If any of the quantities is zero (e.g., there is no pump), the
appropriate term is simply omitted.
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The power generated by a turbine with an efficiency of hT is
The power required by a pump with an efficiency of hp would be
Usually power in expressed in kilowatts or horsepower. One
horsepower is equivalent to 0.746 kW.
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Momentum Equation
Newton's second law, often called the momentum equation, states
that the resultant force acting on a system equals the rate of change of
momentum of the system. If a device has entrances and exits across
which the velocity is assumed to be uniform and if the flow is steady,
the momentum equation can then be written as
Where,
If we consider the nozzle
shown and determine the x-
component of the force of
the joint on the nozzle,
the momentum equation x-
direction becomes
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An example of a free-surface flow in a rectangular channel is shown
in Fig. 4.12. If we want to determine the force of the gate on the flow,
the following expression can be derived from the momentum
equation:
where F1 and F2 are pressure forces.
Momentum Equation Applied to Deflectors
The application of the momentum equation to deflectors forms an
integral part of the analysis of many turbomachines, such as turbines
and pumps.
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Momentum Equation Applied to Deflectors
The analysis is separated into two parts: jets deflected by stationary
deflectors and jets deflected by moving deflectors. For both problems
we will assume the following:
-The pressure external to the jets is everywhere constant so that the
pressure in the fluid entering the deflector is the same as that in the
fluid exiting the deflector.
-The frictional resistance due to the fluid-deflector interaction is
negligible so that the relative speed between the deflector surface and
the jet stream remains unchanged.
- Lateral spreading of a plane jet is neglected.
-Body forces are small and will be neglected.
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Stationary Deflector
Let us first consider the stationary deflector, illustrated in Fig. 4.13.
Bernoulli's equation allows us to conclude that V2 = V1 since the
pressure is assumed to be constant external to the fluid jet and
elevation changes are negligible.
Assuming steady, uniform flow the x- and y- components of the
momentum equation takes the form of Eq. 4.5.15.
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For given jet conditions the reactive force components can be
calculated.
Moving Deflectors
The situation involving a moving deflector depends on whether a
single deflector is moving (a water scoop used to slow a high-speed
train) or whether a series of deflectors is moving (the vanes on a
turbine).
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Moving Deflectors
Let us first consider that the single deflector shown in Fig. 4.14 is
moving in the positive x-direction with the speed VB. In a reference
frame attached to the stationary nozzle, from which the fluid jet
issues, the flow is unsteady; that is, at a particular point in space, the
flow situation varies with time.
A steady flow is observed, however, from a reference frame attached
the deflector.
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From this reference frame, moving with the constant velocity VB, we
observe the relative speed Vr entering the control volume to be V1 -
VB, as shown in Fig. 4.14. It is this speed that remains constant as the
fluid flows relative to the deflector. Hence the momentum equation
takes the forms
Where represents only that part of the mass flux exiting the fixed
jet that has its momentum changed.
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Since the deflector moves away from the jet some of the fluid that
exits the fixed jet never experiences a momentum change; this fluid is
represented by the distance VB Dt, shown in Fig. 4.14. Hence
Where the relative speed (V1 - VB) is used in the calculation; the mass
flux rAVB is subtracted from the exiting mass flux pAV1 to provide
the mass flux .
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Series of vanes
For a series of vanes (a cascade) the jets may be oriented to the side,
as shown in Fig. 4.15. The actual force on a particular vane would be
zero until the jet strikes the vane; then the force would increase to a
maximum and decrease to zero as the vane leaves the jet.
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Assume that, on the average, the jet is deflected by the vanes as
shown as viewed from a stationary reference frame; the fluid jet enters
the vanes with an angle b1 and exits with an angle b2. What is desired,
however, is that the relative velocity enter the vanes tangent to the
leading edge of the vanes, that is, Vr1 in Fig. 4.16b is at the angle a1.
The relative speed then remains constant as it travels over the vane
with the exiting relative velocity Vr2 leaving with the vane angle
a2.The relative and absolute velocities are related with the velocity
polygons of Fig. 4.16 b and c.
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Assuming that all of the mass exiting the fixed jet has its momentum
changed, we can write the momentum equation as
Interest is usually focused on the x-component of force since it is this
component that is related to the power output (or requirement). The
power would be found by multiplying the x-component force by the
blade speed for each jet; this takes the form
where N represents the number of jets.
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EXAMPLE 4.11
Water flows through a horizontal pipe bend and exits into the
atmosphere. The flow rate is 0.01 m3/s. Calculate the force in each of
the rods holding the pipe bend in position. Neglect body forces and
viscous effects.
Solution
We have selected a control volume that surrounds the bend as shown.
Since the rods have been cut, the forces that the rods exert on the
control volume are included. The pressure forces at the entrance and
exit of the control volume are also shown.
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The flexible section is capable of resisting the interior pressure but it
transmits no axial force or moment. The body force (weight of the
control volume does not act in the x- or y- direction but normal to it.
Therefore, no other forces are shown. The average velocities are
found to be
Before we can calculate the forces Rx and Ry we need to find the
pressures p1 and p2. The pressure p2 is zero because the flow exits into
the atmosphere. The pressure at section 1 can be determined using the
energy equation or the Bernoulli equation.
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Neglecting losses between sections 1 and 2, the energy equation gives
Now we can apply the momentum equation in the x-direction to find
Rx and in the y- direction to find Ry:
Note that we have assumed uniform profiles and steady flow. These
are the usual assumptions if information is not given otherwise.
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EXAMPLE 4.12
A deflector turns a sheet of water through an angle of 30o as shown.
What force per unit width is necessary to hold the deflector in place if
the mass flux is 32kg/s?
Solution
The control volume we have selected includes the deflector and the
water adjacent to it. The only force that is acting on the control
volume is due to the support. This force has been decomposed into Rx
and Ry.
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The velocity V1 is found to be
Bernoulli’s equation shows that if the pressure does not change, then
the magnitude of the velocity does not change, provided that there is
no significant change in elevation and that viscous effects are
negligible; thus we can conclude that V2 = V1 since p2 = p1.
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Next, the momentum equation is applied in the x-direction to find Rx
and then in the y-direction for Ry:
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EXAMPLE 4.13
The deflector shown moves to the right at 3 m/s while the nozzle
remains stationary. Determine (a) the force components needed to
move the deflector, (b) V2 as observed from a fixed observer, and (c)
the power generated by the vane. The jet velocity is 8 m/s.
Solution
(a) To solve the problem of a moving deflector, we will observe the
flow from a reference frame attached to the deflector. In this moving
reference frame the flow is steady and Bernoulli's equation can then
be used to show that Vr1 = Vr2 = 5 m/s, the velocity of the sheet of
water as observed from the deflector. Note that we cannot apply
Bernoulli's equation in a fixed reference frame since the flow would
not be steady.
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Applying the momentum equation to the moving control volume,
which is indicated again by the dashed line, we obtain the following:
When calculating the mass flux, we must use only that water which
has its momentum changed; hence the velocity used is 5 m/s.
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(b) Observed from a fixed observer the velocity V2 of the fluid after
the deflection is V2 = Vr2 + VB, where Vr2 is directed tangential to the
deflector at the exit and has a magnitude equal to Vr1 (see the velocity
diagram above). Thus
Finally,
(c) The power generated by the moving vane is equal to the velocity
of the vane times the force the vane exerts in the direction of the
motion. Therefore,
= 3 x 26.8 = 80.4 W
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EXAMPLE 4.17
Find an expression for the head loss in a sudden expansion in a
pipe in terms of V1 and the area ratio. Assume uniform velocity
profiles and assume that the pressure at the sudden enlargement
is p1.
Solution
A sketch is shown of a sudden expansion with the diameter
changing from d1 to d2. The pressure at the sudden enlargement
is p1, so that the force acting on the left end shown is p1A2.
Newton's second law applied to the control volume yields,
assuming uniform profiles,
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The energy equation provides
To express this in terms of only V1, we can use continuity and
relate
Then the expression above for the head loss becomes