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# Fenômentos de Transporte

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```of the pole or vaulter's
body.
Take the body of the vaulter as the system and take the initial state as the
point where the vaulter is running and the pole just begins to slide into the
socket, the final state as the vaulter at the height of the bar. Applying the
mechanical energy balance, we see that there are no convective terms, no heat,
no work, and no change in internal energy of the system (at this point the
chemical energy of the body has been converted in to kinetic energy), leaving us
with
\$SV(b+R)pdV = 0
\$ l ( g z + \$ ) p d V = 0 (3.1.3-9)
Neglecting the variation of z over the height of the vaulter's body and assuming
uniform density
+ [ ( g z + g ) p v ] = 0
146 Chapter 3: The Energy Balances
g(gz+\$)m] dt = 0 (3.1.3-10)
where m is the mass of the vaulter\u2019s body. Integrating with respect to time from
the time the pole begins to slide into the socket to the time when the body of
the vaulter is at maximum height
d ( g Z + T \u201d) = JI\u201d 0 dt
[(gz+?ii)]
m
(gz+g) = g z
\u2018gz-\$) i = 0
V2 z g z =
v2 = 2 g z
v = 02
Substituting known quantities
= 30 \$ = 20.5 mph
(3.1.3-11)
(3.1.3-12)
(3.1.3- 13)
( 3.1.3 - 14)
This is the equivalent of running 100 yards in 10 s - a pretty good pace,
particularly when carrying an unwieldy pole - and this is without considering
energy lost because of wind resistance, friction with the ground, etc., which also
must be supplied by the vaulter. Obviously, legs, arms, shoulders, etc. (which
contribute to the vertical component of velocity) are important.
Chapter 3: llze Energy Balances 147
Examble 3.1.3-2 Calculation o f lost work in p b e
Water is flowing at 230 gdmin. through 500 ft of fouled 3-inch Schedule
40 steel pipe. The pipe expands to a 4-inch pipe as shown. The gauge pressure at
point 1 (in the 3-inch pipe) is 120 psig and at point 2 (in the 4 in pipe) is 15
psig. If the pipe rises in elevation 100 ft what is the lost work?
Solution
2
100 ft I
120 L psig
length = 500 ft 1
Figure 3.1.3-1 Pipe system
The system, if chosen to be the water in the pipe between points 1 and 2, is
single inlet, single outlet
constant g, V
no variation of v, p, 6 , or R across inlet or across outlet
Applying the above assumptions and recognizing that there is no work term we
can use the simplified form of the macroscopic mechanical energy balance
P2 - P1 p + g (z2 - ZI) +
&+gAz+A(\$) P = -1%
(3.1.3-15)
Evaluating term by term, choosing zl = 0 and substituting known information
plus the values from the pipe table for cross-sectional flow aTea
.p+gAz+*(;) P = (3.1.3-16)
148 Chapter 3: The Energy Balances
Computing tbe velocities
gal
v , = - - Q - (230)iiiG ft3 min
(0.0513) ft2 (7.48) gal (60) s *2
= 9.994
gal
v2 = Q (230)iii6 ft3 min
= (0.0884) ft2 (7.48) gal (60) s
= 5.80s ft
Computing the individual terms of the equation
(15- 120) 144 i n 2
ft lbf ( = -242.3- p (62.4) Ibm ft2 lbm
- = AP
ft3
g Az = (32.2) 4 (100) ft lbfs2 = loo 7i.f S (32.2) lbm ft
ft lbf
lbm = - 1.03 -
lbf s2
(32.2) lbm ft
-1% = - 2 4 2 . 3 ~ + 1 ~ ~ - 1 . 0 3 - ft lbf ft lbf lbm lbm
ft lbf
lbm = -143-
(3.1.3-17)
(3.1.3- 18)
(3.1.3-19)
(3.1.3 -20)
(3.1.3-21)
(3.1.3-22)
Notice that the kinetic energy change is negligible compared with the other
terms. Also note that the lost work term is positive itself, which means that
according to our sign convention that net mechanical energy is leaving the
system, as we know it must for any real system.
Chapter 3: The Energy Balances 149
3.1.4 The macroscopic thermal energy balance
We now take the macroscopic energy balance
(R + & + I?) p (v . n) dA + %j (0 + db + R) p dV
V
= 0 - w (3.1.4- 1)
substitute the definition of enthalpy, and box the terms involving mechanical
energy, regarding the remaining terms as thermal energy
f* (0 +-I) p ( v . n) dA+ \$Iv (o+ ml) p dV
= Q-\u2022
(3.1.4-2)
The conversion of portions of these mechanical energy terms to thermal
energy represents a generation of thermal energy (not of total energy - total
energy is still conserved under processes of interest here). If we define
ye = rate of thermal energy generation
per unit volume (3.1.4-3)
Probably the most common generation of thermal energy comes from the
action of viscous forces to degrade mechanical energy into thermal energy
(viscous dissipation), the thermal energy so generated appearing as internal
energy of the fluid. However, it is also possible to generate thermal energy via
forms of energy that we have classified as work by default - microwave energy,
changing electrical fields, etc. - which are classified as work by our definition
because they cross the system boundary not associated with mass but not
flowing as a result of a temperature difference. Note that chemical reactions do
not generate thermal energy in this sense because the heat of reaction is
accounted for in the internal energy term.
By the above definition we adopt the Same sign convention for ye as that for
Q; that is, input is positive, output is negative. Substituting the definition of
thermal energy generation and writing the relation in the form of the entity
balance yields
150
output - input
rate of
thermal energy
Chapter 3: The Energy Balances
accumulation
thermal energy
+ rateof
I generation I
(3.1.4-6)
This is the macroscopic thermal energy balance. It is common in
practical applications for the heat transferred (Q) or the sensible and latent heat
terms (the left-hand side of the equation) to be much larger than the thermal
energy generation unless exceptionally large velocity gradients andor unusually
large viscosities are involved, so models using this equation will frequently
neglect Ye dV .
3.2 The Microscopic Energy Balances
3.2.1 The microscopic total energy balance
The macroscopic total energy balance was determined above to be
I, (R + 6 + R) p (v - n) dA + \$l (o + 6 + l?) p dV
= Q - W (3.2.1-1)
We wish to develop a corresponding equation at the microscopic level.
0, the rate of heat transfer to the control volume, can be expressed in terms
of the heat flux vector, q.
Chapter 3: The Energy Balances 151
Q = - j - ( q . n ) d A (3.2.1-2)
where the negative sign is to ensure conformity with our convention that heat
into the system is positive.
At the microscopic level W will not involve shafts or electrical leads
crossing the boundary of the system (a point); instead, it will involve only the
rate of work done per unit volume on the system at the surface via the total
stress tensor T, which is related to the viscous stress tensor T by
(3.2.1-3) T = ~ + p 6
Notice that the viscous stress tensor is simply the stress tensor without the
thermodynamic pressure, a component of the normal force to the surface. This is
desirable because the viscous forces go to zero in the absence of flow, while the
thermodynamic pressure persists.
For the fluids which will be our primary concern, the viscous forces act
only in shear and not normal to the surface, so the only normal force is the
pressure; however, particularly in non-Newtonian fluids, the viscous forces often
act normal to the surface as well as in shear.
The total stress tensor T incorporates the flow work from the enthalpy
term. The dot product of the total stress tensor with the outward normal to a
surface (a tensor operation which was defined in Table 1.6.9- l), yields the vector
Tn (the surface traction) representing the total force acting on that surface (this
resultant force need not - in fact, usually does not - act in the n direction).
Ten = T&quot; (3.2.1-4)
The dot product of this vector```