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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&quot;. v) dA
Combining the area integrals
(3.2.1 -8)
(3.2.1-9)
I A [ n p ( v . n ) + ( q . n ) + ( T &quot; . 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 &quot;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 &quot;point&quot; 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