Fenômentos de Transporte
1047 pág.

Fenômentos de Transporte


DisciplinaFenômenos de Transporte I11.455 materiais110.403 seguidores
Pré-visualização50 páginas
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 
steady state 
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" (3.2.1-4) 
The dot product of this vector