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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 "floor" 
(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 
from the windshield - hopefully not the back of your head). In your accelerating 
(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," 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