Fenômentos de Transporte
1047 pág.

Fenômentos de Transporte


DisciplinaFenômenos de Transporte I12.466 materiais111.646 seguidores
Pré-visualização50 páginas
with the velocity gives the work passing into that 
surface, including the flow work. 
(3.2.1-5) 
IS2 Chapter 3: The Energy Balances 
Proceeding to use the above results in replacing first Q in the macroscopic 
energy balance 
(0 + p v + 6 + R) p (v n) dA+ #jv (0+ &+ R) p dV 
(3.2.1-6) = -[ A ( q . n ) d A - \ k 
and then W by moving the flow work contribution to the other side of the 
equation 
I (0 + db + R) p (v n) dA + $1 (0 + 6 + R) p dV 
V 
= -1 ( q . n ) d A - w - j p g p ( v . n ) d A (3.2.1-7) 
Defining C2 = (U + @ + K) and using the fact that 
bF 1 
gives 
I, sip (v . n) d A + * j f i p dV 
= -! ( q . n ) U - \ k - j A p b p ( v . n ) d A 
5, f ip (v . n) d A + $ l fhp dV 
= -IA (9 . n) dA-lA (T". v) dA 
Combining the area integrals 
(3.2.1 -8) 
(3.2.1-9) 
I A [ n p ( v . n ) + ( q . n ) + ( T " . v ) ] d A + $ I n p d V = 0 
(3.2.1-10) 
Chapter 3: The Energy Balances 153 
and substituting for the surface traction in terns of the stress tensor13 
(3.2.1-1 1) 
Applying the divergence theorem to the first term on the left-hand side to change 
the area integral to a volume integral, and applying the Leibnitz rule to the 
second term on the left-hand side to interchange differentiation and integration 
dV = V (a p v) + (V q) + (V . [T . v]) +at (3.2.1-12) 
Since the limits of integration are arbitrary, the integrand must be identically 
zero. 
Eulerian forms of the microscopic total energy balance 
This gives the microscopic total energy balance in Eulerian form (which 
takes the point of view of an observer fixed in space) 
,T a(n 4 + V 9 (n p v) + (V q) + (V . [ T . v]) = 0 
l 3 Writing [T. ' in rectangular Cartesian gives, since Ti* is symmetric: 
Til i 'i = I TU 'i = I TY 'i . But, in symbolic form, this is . r' T. 1 . 
154 Chapter 3: The Energy Balances 
a ( p + t5+ R] P) 
+ V - ([o+&+ R]p v) = -(V.q)-(V. [T. v]) 
(3.2.1-13) 
This is a scalar equation. The first term represents accumulation of energy per 
unit volume at the point in question; the second, energy input per unit volume 
by convection; the third, energy input per unit volume by conduction; and the 
fourth, surface work (by both pressure and viscous forces) per unit volume done 
on the system. 
This, however, is often not the most convenient form to use for many 
applications. To obtain a more convenient form, we first substitute for the 
kinetic energy term and then isolate the terms involving the potential energy 
a(& P) 
- + v . ( a p v ) + a t a t 
(3.2.1 - 14) 
= - (V q) - (V - [T . v]) 
Examining only the potential energy terms 
W + V . ( & p v ) at 
= [p&$+&i$]+[pv.v&+&v.(pv)] (3.2.1-15) 
= p [ $ + v. v6J + 6 [ g + v . (p v)] 
But the last term in brackets is zero as a result of the continuity equation. 
Substituting 
&+V.V& + ++ at v21 ) +v.([o+$]pv) 
p [ x ] 
(3.2.1 - 16) 
= - (V - q) - (V - [T . v]) 
Chapter 3: The Energy Balances 155 
Change in potential energy with time at a p i n t in a stationary fluid ensues 
from changing the potential fiela however, the only potential field of concern in 
our problems is gravitational, which is constant for problems of interest here. 
(We shall ignore all other field effects such as those engendered by 
electromagnetism, including any effects caused by fields that are unsteady. Our 
models therefore will not be valid for phenomena involving such field effects.) 
We therefore set the partial derivative of the potential energy with time equal 
to zero. Using this and the fact that 
(3.2.1 - 17) V@ = - g 
and moving the remaining term to the other side of the equation yields 
at + v * ([o + $1 p V) = - (V. q) - ( V . [T - v]) + p (v. g) 
(3.2.1 - 18) 
Examining the last two terms 
(3.2.1- 19) 
We see that the first of these terms is work resulting from the action of 
surface forces (in cases of interest here, mainly the thermodynamic pressure 
acting opposite to the normal direction, and viscous forces acting in the shear 
directions - remembering, however, that, in general, it is possible to generate 
normal forces through the action of viscosity in addition to the thermodynamic 
pressure) 
and the second 
P (v * 9) (3.2.1-21) 
is work resulting from the action of body forces (in cases of interest here, 
gravity ) . 
I56 Chapter 3: The Energy Balances 
We have chosen by this classification to regard the potential energy term as 
work rather than potential energy - the classification is, of course, arbitrary, but 
results from our regarding the potential energy term as a force acting through a 
distance rather than as a difference in two values of the potential function. 
This gives the alternative form of the Eulerian equation as 
The terms on the left-hand side represent, respectively 
the net rate of gain of internal and kinetic energy per unit 
volume at the point in question 
the net rate of input of internal and kinetic energy per unit 
volume to the point by convection. 
The terms on the right-hand side represent, respectively 
the net rate of input of energy per unit volume to the point 
as heat (conduction) 
the rate of work done on the fluid at the point per unit 
volume by viscous forces (including any normal forces from 
viscosity) 
the rate of work done on the fluid at the point per unit 
volume by the thermodynamic pressure 
the rate of work done on the fluid at the point per unit 
volume by gravitational forces. 
Lagrangian f o m of th2 microscopic total energy balance 
To obtain the Lagrangian forms of the microscopic total energy balance, 
that is, from the viewpoint of an observer riding on a \u201cparticle\u201d of fluid, we 
expand the derivatives on the left-hand side 
Chapter 3: The Energy Balances I57 
(3.2.1 -23) 
P X [ '* + v VQ] + h [$ + V . (p v)] = - V - q - V - [T - v] 
(3.2.1-25) 
But the factor in brackets in the fust term is the substantial derivative by 
definition, and the factor in brackets in the second term is zero as a consequence 
of the continuity equation, so we have the Lagrangian form of the microscopic 
total energy balance as 
(3.2.1 -26) 
Performing the same manipulations as with the Eulerian form we can 
obtain 
Dt = - (v. 9) - v * [v v3 - v * [p "1 + p (v . g) 
(3.2.1 -27) 
3.2.2 The microscopic mechanical energy 
balance 
By taking the dot product of the microscopic momentum balance with the 
velocity v, we obtain a microscopic balance that involves only mechanical 
energy - that is, energy directly convertible to work. This is the microscopic 
mechanical energy balance, below and in Equation (4.2- lO), whose development 
is shown in Section 4.2. 
258 Chapter 3: The Energy Balances 
= -p(-v.v)-v.[pv]-v.[.r.v]-(-.r:vv)+p(v.g) 
(3.2.2- 1) 
3.2.3 The microscopic thermal energy balance 
For many engineering applications, the energy equation is more useful 
written in terms of lhermal energy, using the temperature and heat capacity of 
the fluid in question. The differential total energy balance can be converted to an 
equation expressing only thermal energy by subtracting from it the microscopic 
mechanical energy balance, which has the form14 
P D t = p (v . v) - v . ['F . v] + ('F : v v ) - v . [p v] + p (v . g) 
(3.2.3- 1) 
Subtracting gives 
I 1 
DO = - ( v . q ) - p ( v . v ) - ( . r : v v ) 1 P D t (3.2.3-2) 
where the left-hand side of the equation represents the net rate of gain of internal 
energy per unit volume of a "point" of the fluid, and the terms on the right-hand 
side represent, respectively, 
the net rate of thermal energy input per unit volume by 
conduction 
the reversible work input per unit volume by compression 
the irreversible thermal energy input per unit volume from 
dissipation of kinetic energy into thermal energy via the action 
of