Conservation of Mass and Momentum

1
Conservation of Mass and Momentum

• Using the Reynolds Transport Theorem, it is a
  simple matter to derive mass- and
  momentum-conservation laws for a control volume,
  i.e., the integral forms
• Global view
• Indirect computation of forces
• Focusing on a differential-sized control volume,
  we can deduce the differential form of the
  conservation laws, which hold at every point in
  the flow
• Detailed view
• Direct computation of forces and all flow
  properties
• From the differential forms we can derive
  Bernoullis Equation mechanical energy
  conservation

2
Conservation Principles

• The basis of our conservation laws is as follows,
  where B is an extensive variable and ß is the
  corresponding intensive variable
• We will concentrate on mass and momentum first,
  and do energy later (Chapter 7)

3
Mass Conservation-1

• We consider a general control volume whose
  bounding surface has velocity ucv
• Definition of a system
• Mass, M, is constant
• By definition,
• Hence, B M and ß 1
• dM/dt 0the Reynolds
• Transport Theorem yields
• In words,

the sum of the instantaneous rate
of change of mass in the control volume and

the net flux of mass out of the
control volume is zero
4
Mass Conservation-2

• Recall that for a differential-sized control
  volume
• Again, since dB/dt 0 and dß/dt 0, the
  Reynolds Transport Theorem tells us that
• This is called the continuity equationit holds
  at every point in the flow CV asymptotes to a
  point
• Using the chain rule

? Conservation form
? Primitive-variable form
?
5
Mass Conservation-3

• We can use our results to further simplify the
  Reynolds Transport Theoremas noted above, for a
  point we derived the following
• Therefore, the Reynolds Transport Theorem for a
  point simplifies to

? Zero according to mass conservation
6
Momentum Conservation-1

• We consider a perfect fluid so that only normal
  surface forces act
• Momentum of a system
• By definition,
• Hence, B P and ß u
• Newtons 2nd Law says
• We consider two types of forces
• Surface force, Fs transmitted across surface S
• Body force, Fb acts at a distance

7
Momentum Conservation-2

• The only surface force acting is pressure
• The magnitude of the force on a differential
  surface element is pdS
• Since n is an outer unit normal, the force
  exerted by the surroundings on the control volume
  is p n dS
• Thus,
• Typical body forcesgravity, electromagnetic
• Express in terms of the specific body-force
  vector, f
• f is body force per unit massfor gravity, f -g
  k
• Thus,

8
Momentum Conservation-3

• Using the Reynolds Transport Theorem
• The terms in this equation, from left to right,
  are

• Instantaneous rate of change of momentum in the
  control volume

• Net flux of momentum out of the control volume

• Net pressure force (e.g., buoyancy) acting on the
  control volume

• Net body force (e.g., weight) acting on the
  control volume

9
Momentum Conservation-4

• Focusing now on a differential-sized control
  volume, Newtons second law of motion tells us
  that
• Again, our corresponding extensive/intensive
  variable pair is B P (momentum) and ß u
  (velocity)
• From the Reynolds Transport Theorem at a point
• In Chapter 3, we showed that the net pressure
  force is

10
Momentum Conservation-5

• Finally, for the obvious reason
• Collecting all of this, we arrive at the
  differential form of the momentum-conservation
  principle
• This equation, valid for a frictionless fluid, is
  called Eulers Equationin words
• Eulers Equation is F ma per unit volume!

(Mass per unit volume) ? (Acceleration) S
(Forces per unit volume)
11
Mass Conservation for Incompressible Flows-1

• In Cartesian coordinates, the continuity equation
  for a general three-dimensional flow is
• When the flow is incompressible, so that ? is
  very nearly constant, this simplifies to
• Given the velocity vector for a flowfield, we can
  use this equation to determine whether or not the
  flow is incompressible

? Valid for both steady and unsteady flow
12
Mass Conservation for Incompressible Flows-2

• Example Show that two-dimensional flow
  approaching a stagnation point is incompressible,
  given that the velocity vector is u A(xi yj),
  where A is a constant
• Solution Taking the divergence, we find
• Thus, the flow is indeed incompressible
• Note By inspection, this result would still be
  true even if A were a function of time

13
Eulers Equation-1

• Vector notation conceals the complexity of this
  set of quasi-linear, partial differential
  equations
• Closed-form solutions exist only for simple
  geometries (cannot use superposition
  (nonlinear!) existence and uniqueness theorems
  do not exist)
• Excellent computer programs are available to
  solve for general flowfields
• Potential-flow theory (see Chapter 11) yields
  exact solutions for various interesting flows

? Terms like ?u?u/?x make the equations
quasi-linear It makes them much harder to solve
than linear equations
14
Eulers Equation-2

• As a simple example of an Euler-equation
  solution, consider flow in a tank that has been
  rotating long enough that transients have ceased
  (?/?t ? 0)
• Rigid-body rotation
• Gravitational field, f g -g k
• Incompressible flow
• Using cylindrical coordinates, the 3 components
  of the Euler Equation are

In general, u u(r,?,z) and p p(r,?,z) Much
simplification is possible for this flow
15
Eulers Equation-3

• Since the velocity is uq Or and ur w 0
• Integrating the first of the 3 equations, we find
• Using the second equation

? Centrifugal force balances ?p/?r
Gravity balances ? ?p/?z
? f(?,z) is a function of integration
?
16
Eulers Equation-4

• Finally, using the vertical component of the
  Euler Equation
• Therefore, the solution for the pressure at every
  point within the rotating tank is
• Rearranging terms, we have

Real live constant of ? integration
17
Galilean Invariance of Eulers Equation-1

• Consider the following descriptions of a moving
  body
• Flow (a) is unsteady ? Solution depends upon x,
  y, z, t
• Flow (b) is steady ? Solution depends only upon
  x, y, z
• QUESTION Does the linear Galilean transformation
  apply to the Euler Equation, which is not linear?
• ANSWER Yessee text for algebraic details
• THEREFORE We can measure forces on a stationary
  wind-tunnel model and apply to a moving prototype

? A Galilean transformation puts the body at rest
The body is ? moving at constant velocity u -U i
18
Galilean Invariance of Eulers Equation-2

• Example If a body moves at constant speed U 30
  ft/sec through water with ? 1.94 slug/ft3 and
  pstag pA 4.91 psi, determine UA (pstag is the
  pressure at the stagnation point)
• Solution After doing a Galilean transformation,
  we have
• Hence, substituting the given values

Original Problem
After Galilean transformation
ustag 0 ?
19
Derivation of Bernoullis Equation-1

• Using the following vector identity (Appendix D)
• we can rewrite Eulers Equation as follows
• If a flow is
• Inviscid (so that Eulers Equation holds)
• Steady (?u/?t 0)
• Incompressible (? constant)
• Subject only to conservative body forces (f
  -?V)
• Irrotational (? 0)
• then we have

? ? ? ? u vorticity
20
Derivation of Bernoullis Equation-2

• The only way the gradient of a scalar can be zero
  everywhere is if the scalar is a constant, so
  that
• This is Bernoullis Equation
• It represents conservation of mechanical energy
  (per unit volume), the 3 types of energy being
• Pressure potential, p
• Kinetic energy, ½? u u
• Potential energy, ?V
• Can relax the irrotational requirement and show
  that it holds on a streamline (see text)not very
  useful though since we dont know in advance
  where the streamlines are

21
Leaky Tank

• Consider a tank with a small hole
• Assuming d tank is negligible
• From Bernoullis Equation applied between the top
  of the tank and the jet of fluid

?
22
Flow Past a Vertical Tube-1

• It is often helpful to determine the constant
  first
• We know everything at the free surface far
  upstream, i.e., p pa, u U1i, z 0

?Applies at all points in the flowfield
23
Flow Past a Vertical Tube-2

• At the tube outlet, pressure is p pa, velocity
  is u U2 i and height is z z2, wherefore
• Solving for the velocity, we find
• When we computed constant in advance, we
  matched every point in the flow to a single
  reference point

24
Velocity-Measurement Techniques

• Recall that we used the hydrostatic relation to
  indirectly measure pressure (from height
  measurements)
• We can use Bernoullis Equation to indirectly
  measure velocity (from height or pressure
  measurements)
• The two most common devices for measuring
  velocity are the Pitot tube and the Pitot-static
  tube

25
Stagnation Points

• Must first introduce the concept of a stagnation
  point, defined as a point where u 0
• There are 2 stagnation points for flow past a
  cylinder, for example

26
Pitot Tube-1

• For a Pitot tube, fluid enters the tube and rises
  to a height h above the surface
• Select a reference point at the surface far
  upstream so that, in Bernoullis Equation

27
Pitot Tube-2

• After transients settle out,the flow is stagnant
  in the tube so that u 0
• The tube is open to the atmosphere so that p pa
• Applying Bernoullis Equation
• Solving for U1 gives

? Measuring height h gives an indirect
measurement of U1
28
Pitot-Static Tube-1

• A Pitot-static tube does not require a point in
  the flow where everything is known
• Two pressure measurementsone at the stagnation
  point, the other 10 or more tube diameters
  downstream of the stagnation point

29
Pitot-Static Tube-2

• Tube diameter is sufficiently small that
  potential-energy differences are negligible
• Stagnation point p pstag, u 0
• Downstream p pstatic, u ? U i
• Applying Bernoullis Equation tells us
• Thus, the velocity is

? Measuring two pressures gives an indirect
measurement of U
30
Conservation Equation Summary-1

• Integral forms For completely general
  geometries
• Mass
• Momentum
• State
• Differential forms At every point in a
  flowfield
• Continuity
• Eulers Equation

31
Conservation Equation Summary-2

• The equation of state and the mass- and
  momentum-conservation principles provide five
  equations
• Mass 1 scalar equation
• Momentum 1 vector equation with 3 components
• State 1 scalar equation
• Unknowns for a Liquid ?, u, v, w, p a total of
  5
• Unknowns for a Gas ?, u, v, w, p, T a total of
  6
• If temperature variations are important, liquids
  also have 6 unknowns in general, energy
  conservation provides a sixth equation
• For now, we will use Bernoullis Equation, which
  is okay provided thermal effects are unimportant