Download momentum principle

Chapter 6
Review: Last time, we derived the Reynolds
Transport Theorem:
MOMENTUM PRINCIPLE
• where B is any extensive property (proportional to
mass),
• and b is the corresponding intensive property (B / m ).
Fluid Mechanics, Spring Term 2010
a)
Recall:
• The system (e.g., Bsys ) is a fluid volume that moves
along with the fluid particles.
• The control volume (cv) may move relative to the fluid;
for example, it may be fixed in space.
• The control surface (cs) is a closed surface that contains
the control volume.
b)
c)
Meaning:
• Term a) describes the physical change that takes place
for a set of fluid particles.
• Term b) describes the change observed in a given
control volume.
• Term c) describes the difference: For example, if the
property B has not changed for the fluid particles, but B
has changed inside the control volume, then this must be
because new fluid particles have entered the cv which
carried a different amount of B with them.
Still review…
Today, we use the Momentum Principle:
We then applied the Reynolds Transport Theorem to the
property of mass (M ).
We will do 2 things:
Different ways of writing the results:
1) Use the law of conservation of momentum (Newtonʼs
2nd Law).
with
2) Set B = mv, (where B is the extensive property) and
substitute this into the Reynolds Transport Theorem.
Notice the similarities to last lecture:
1) Use conservation of mass, dMsys / dt = 0
2) Set B = m
The momentum conserved is that of the system:
Newtonʼs 2nd Law:
Particles move about in a box.
or
Definition of (linear) momentum:
Conservation of momentum:
No particles leave or enter,
hence the box is a system.
Particles can exchange momentum with each other, but
that does not affect the total momentum of the system.
The only way to change the total momentum is to apply a
force to the box.
Therefore:
Notice the similarities and differences between
conservation of mass and momentum:
(conservation of mass)
(conservation of momentum)
1) Both equations are applied to the system.
2) Mass is conserved absolutely (never changes in
classical physics); Momentum is conserved unless a
force is applied.
3) Mass conservation is a scalar equation; Momentum
conservation is a vector equation (3 equations).
Reynolds Transport Theorem with B = mv becomes
Notice the little “v ” and the big “V ” in the last term.
The big “V ” comes from the Reynolds Transport Theorem.
It is the fluid velocity relative to the control surface.
and using Newtonʼs 2nd Law
we get
The little “v ” comes from the momentum definition.
It is the fluid velocity relative to the reference frame.
“v ” and “V ” are the same only if the control volume is at
rest relative to the reference frame.
Vwall
Forces acting on a control volume (Fig. 6.1)
Vfluid
The integral over cv is the change in
momentum in the control volume.
Vwall
Vfluid
Vcv
The last term is again a convective
correction in going from the
Lagrangian to the Eulerian frame:
In the channel flow illustrated,
nothing changed between 1 and 2;
only the control volume moved.
cv consists
of fluid only.
Cv includes the
section of the
pipe.
But the momentum in cv increases because new particles with
greater momentum have flowed into cv.
Choice 1: control volume
is entirely within the fluid.
where ! is a viscous shear stress acting between the fluid
and the pipe wall (its direction depends on whether the
fluid is moving up or down).
Choice 2: The control volume
includes the entire section of
the pipe.
where Wp is the weight of the pipe.
Momentum equation (general):
Momentum Accumulation (Fig 6.2)
Steady flow through
a nozzle.
The momentum of
each fluid particle
passing through the
nozzle changes with
time.
If v is constant across inlet and outlet:
Recall (Chapter 5) the mass flow rate
For steady flow, the integrated momentum inside the cv does
not change with time.
There is no accumulation of momentum in the cv.
Momentum Diagram:
(Momentum equation for inlet and outlet ports)
Momentum Equation in Cartesian Coordinates
(inlet and outlet ports with constant velocities)
The change in momentum in the cv
may be visualized with a momentum
diagram.
(Notice that weʼre drawing changes
in momentum, i.e., accelerations!)
(inflow)
(outflow)
Net outflow of momentum:
(a fairly obvious, special form of the momentum equation…)
Systematic Approach to Solving Problems:
1) Problem Setup
• Select appropriate control volume.
• Select inertial reference frame.
2) Force analysis and diagram
• Sketch body forces on force diagram (gravity).
• Sketch surface forces: pressures, shear stresses,
supports and structures…
3) Momentum analysis and diagram
• Evaluate momentum accumulation term. If the flow
is steady, this term is zero. Otherwise evaluate
volume integral and add to momentum diagram.
• Sketch momentum flow vectors on momentum
diagram. For uniform velocity, each vector is
Example 6.1:
Rocket on test stand.
Exhaust jet has:
Diameter d = 1 cm
Speed v = 450 m/s
Density " = 0.5 kg/m3
Assume p in jet is atmospheric p.
Neglect momentum changes inside
rocket motor.
What is the force Fb acting on
support beam?
Example 6.1: Solution
Only involves vertical
momentum.
1) Forces:
2) Momentum change:
There is no momentum
accumulation because the
structure is stationary and
because we neglect momentum changes in the rocket motor.
Example 6.1 (continued)
Substitute forces and
momentum into the
momentum equation:
or
Example 6.2: Concrete flowing onto cart on a scale
Example 6.2: Solution
Stream of concrete:
Given:
Density = "
Area
=A
Speed = v
Cart + concrete:
Weight = W
Forces and momenta involve x and z directions.
Determine tension in cable and weight recorded by scale.
Example 6.2
(continued)
Forces in z:
Example 6.2
(continued)
Since flow is steady and cart does not move, there is no
accumulation of momentum inside the cv.
Momentum changes in x and z: (with
Forces in x:
The momentum equations in x and z give
)
thus providing T and N as the answer. (Notice in this
problem the cv is stationary; hence, the velocities v and V
are the same).
Example 6.4: Water jet deflected by a vane
Example 6.4: Solution
Given:
Speed of incoming jet = v1
Speed of outgoing jet = v2
Diameter of jet = D
1) Forces:
2) Momenta:
Note that mass flow rates have to be equal:
Net momentum change:
Example 6.4 (continued)
Example 6.7: Water flow through a 180o reducing bend
Again, substitute into momentum equation:
Given:
Discharge
=Q
Pressure at center of
inlet
= p1
Where
is given by
Volume of bend = V
Weight of bend = W
What force is required to hold the bend in place?
Example 6.7: Solution
Example 6.7 (continued)
Additional difficulties:
1) We donʼt know the velocities.
2) We donʼt know the outlet
pressure.
1) Get velocities from continuity equation (incompressible!):
Notice that there is no momentum change in the vertical.
2) Get p2 from Bernoulli equation:
Forces in x-direction:
Momentum in x-direction:
Momentum balance in x:
Using continuity and Bernoulli equations (previous slide),
all variables are known except Rx.
In the vertical we only have the force balance