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# Fenômentos de Transporte

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```of change per unit volume of kinetic energy as
one follows the fluid motion. The terms on the right-hand side represent,
respectively,
rate of work done per unit volume by pressure forces
198 Chapter 4: The Momentum Balances
rate of reversible conversion to internal energy per unit
volume
rate of work done per unit volume by viscous forces
rate of irreversible conversion to internal energy per unit
volume
rate of work done per unit volume by gravitational forces
To illustrate a common application of Equation (4.2-7), for a Newtonian
fluid, where x is the position vector
dv = = - 5 i
Substituting this in Equation (4.2-7) and using the definitions
yields a differential equation called the Navier-S tokes equation.
(4.2- 1 1)
(4.2- 12)
(4.2- 13)
I p g = - v p + p v 2 v + p g I (4.2-14)
The investigation of solutions of this equation for various system where
momentum is transferred by viscous motion is a large part of the topic of
momentum transport as a transport phenomenon.
For example, the axial flow of an incompressible Newtonian fluid in a
horizontal pipe is described in cylindrical coordinates by
(4.2- 15)
Chapter 6 is devoted entirely to the topic of momentum transfer in fluid
flow and utilizes the momentum balance in both its integral and differential
forms.
At this point it is useful to summarize both the macroscopic and
microscopic equations of mass, energy, and momentum.
Chapter 4: The Momentum Balances 199
4.3 Summary of Balance Equations and
Constitutive Relations hips
The balance equations in Table 4.3-1 are not applicable to systems where
there is interconversion of energy and mass.
Table 4.3-1 Tabulation of balance equations
MOMENTUM
ENERGY -
TOTAL,
ENERGY6 -
THERMAL
ENERGY -
MECHANICAL
MASS -
TOTAL
MASS -
SPECIES
- v p + p v 2 v + p g
+\$ l ;vpdV = ZF
I, (A + 6 + R) p (v . n) dA
- v . [P v] + P (v . g) = Q - W
Do = - ( v . q )
- p ( v . v ) - ( 5 : v v 1
Dt
- (v. p v) - p (- v . v)
+ P (v . 8)
= - 1 w - w - (v . [r. v]) - (- [5 : vv])
v . ( p v ) + \$ = 0 I
= k r i d V
For one-dimensional flow of an incompressible fluid.
200 Chapter 4: The Momentum Balances
We summarize the common constitutive relationships - those equations that
relate material properties to the fluxes - in Table 4.3-2.
Table 4.3-2 Tabulation of common constitutive relationships
4.4 The Momentum Equation in Non-Inertial
Reference Frames
The momentum equation as we have written it applies to an inertial
reference frame system. By an inertial frame we mean a frame which is not
accelerated.s (We exclude rotation entirely and permit only uniform translation.)
These frames are also referred to as Gallilean frames or Newtonian frames. The
acceleration of frames attached to the earth's surface, however, is slight enough
that our equations give good answers for most engineering problems - the error
is of the order of one-third of 1 percent. For problems involving relatively large
acceleration of the reference frame, however, Newtonian mechanics must be
reformulated. An example of such a problem is the use of a reference frame
attached to a rocket as it accelerates from zero velocity on the pad to some orbital
velocity. For such a problem, observations from an inertial frame and from the
accelerating frame can differ greatly.
For example, if an astronaut in zero gravity but in an accelerating rocket
drops a toothbrush, he or she sees it apparently accelerate toward the &quot;floor&quot;
(wherever the surface opposite to the direction of the acceleration vector of the
spacecraft might be at the time). This observation is with respect to an
accelerating frame (the spacecraft) within which the astronaut is fixed. To an
observer in an inertial frame, however, the toothbrush simply continues at the
One-dimensional.
Fanger, C. G. (1970). Engineering Mechanics: Statics and Dynamics. Columhus,
OH, Charles E. Merrill Books, Inc.
Chapter 4: The Momentum Balances 201
constant velocity it had when released, and the spacecraft accelerates past the
toothbrush.
In other words, observed acceleration depends on the motion of the reference
frame in which it is observed. If this is true, Newton's second law must change
in form when written with respect to an accelerating reference frame, because the
force experienced by a body should not depend on the motion of the observer.
Here we will consider only a simple example of this - that of a system of
constant mass.g
Return to the astronaut's toothbrush. There is no net external force on the
toothbrush (we neglect any drag from the atmosphere within the spacecraft).
Newton's law becomes
=
=
atoothbrush, observed from inutial frame
atoothbrush. observed from iautial frame (4.4- 1)
which implies that the acceleration of the toothbrush with respect to an inertial
coordinate fiame is zero.
With respect to an accelerating frame (one attached to the spacecraft),
however, there is an apparent acceleration of the toothbrush. This apparent
acceleration is equal to the negative of the difference in acceleration between that
of a frame attached to the spacecraft and that of the inertial frame. Since the
acceleration of the inertial frame is zero, the relative acceleration is
Newton's law will still be operationally valid in the accelerating coordinate frame
if we
interpret the acceleration on the right-hand side as the
acceleration of the object as observed from the accelerating
frame and also
add an appropriate force (the inertial body force) to the
left-hand side.
The inertial body force is the negative of the product of the mass and the
acceleration of the accelerating frame (the spacecraft)lo
We neglect the drag force from the atmosphere in this discussion.
l0 Hansen, A. G. (1967). Fluid Mechanics. New York, NY, Wiley.
202 Chapter 4: The Momentum Balances
Another example is a camera sitting on the ledge above the rear seat of your
car. If you hit a cow which happens to be crossing the road, and therefore
decelerate rapidly from a velocity of 60 mph, the camera flies forward with
respect to the car as well as to the ground. The sum of forces on the camera is
zero (again neglecting friction from the air). This time the coordinate frame is
decelerating, not accelerating.
To the hitchhiker standing at the side of the road, the acceleration of the
camera is zero - it continues at 60 mph until the intervention of a force (perhaps
(in this case, decelerating) reference frame, there is still no external force on the
object, but there is an inertial body force of (- m aautomobilc), where aautomobilc is
the relative acceleration between car and ground (inertial frame). This is balanced
on the right-hand side of the equation by the same mass times the relative
acceleration of the camera to the car, which is the same in magnitude as and
opposite in direction to the relative acceleration of the car and the ground.
(4,4-6)
In the case of rotating frames things become more complex, and the reader is
referred to texts such as that by Fanger,&quot; which cover this subject in more
detail.
I 1 Fanger, C. G. (1970). Engineering Mechanics: Statics and Dynamics. Columbus,
OH, Charles E. Merrill Books, Inc., p. 562.
Chapter 4: The Momentum Balances 203
Chapter 4 Problems
4.1 A fire hose has an inside diameter of 1 1/2 in. and a nozzle diameter of 0.5
in. The inlet pressure to the hose is 40 psig at a water flow of 8 ft3/min. If a
flat velocity profile is assumed, what is the value of and the direction of the force
exerted by tbe fmman holding the hose?
4.2 Your company is planning to manufacture```