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# Fenômentos de Transporte

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```if we know total flow rates and the
change in mass of the system with time; otherwise, we must evaluate the
integrals. In some cases, where density is constant or area-averaged velocities are
known, this is simple, as detailed below. Otherwise, we must take into account
the variation of velocity and density across inlets and outlets and of density
within the system
Simplijied forms of the macroscopic total mass balance
Many applications involve constant density systems. Further, our interest is
often only in total flow into or out of the system rather than the manner in
which flow velocity varies across inlet and outlet cross-sections.
Such problems are usually discussed in terms of area-average velocities.
The area-average velocity is simply a number that, if multiplied by the flow
cross-section, gives the Same result as integrating the velocity profile across the
flow cross-section, i.e., the total volumetric flow rate. (So it is really the area
average of the normal component of velocity.)
In such cases
which can be written in terms of mass flow rate as
or in volumetric flow rate
(2.1.1-10)
(2.1.1-1 1)
(V)Aout-(v)A,,,+% = 0
78 Chapter 2: The Mass Balances
(2.1.1 - 12)
m D l e 2.1.1-1 Mass balanc e on a surge ta&
Tanks3 are used for several purposes in processes. The most obvious is for
storage. They are also used for mixing and as chemical reactors. Another purpose
of tankage is to supply surge capacity to smooth out variations in process flows;
e.g., to hold material produced at a varying rate upstream until units downstream
are ready to receive it. Consider the surge tank shown in Figure 2.1.1-1.
10 f t
Figure 2.1.1-1 Surge tank
Suppose water is being pumped into a 10-ft diameter tank at the rate of 10
ft3/min. If the tank were initially empty, and if water were to leave the tank at a
rate dependent on the liquid level according to the relationship Qout = 2h (where
the units of the constant 2 are [ft2/min], of Q are [ft3/min], and of h are [ft]), find
the height of liquid in the tank as a function of time.
So 1 u t ion
This situation is adequately modeled by constant density for water, so
(2.1.1-13)
\u201cTank\u201d is here used in the generic sense as a volume used to hold liquid enclosed or
partially enclosed by a shell of some sort of solid material. In process language, a
distinction is often made among holding tanks, mixers, and reactors.
Chapter 2: The Mass Balances 79
2 h [ e] min - 10 [ g] + \$ [ &] (,q h) [ft3] = 0
h-5+39.3\$ = 0
- 39.3 I, ___ (-4 = dt
5-h
t = - 39.3 In [v]
h = 5[Lexp(-&)]ft
(2.1.1 - 14)
(2.1.1 - 15)
(2.1.1-16)
(2.1.1 - 17)
(2.1.1-18)
(2.1.1-19)
Note that as time approaches infinity, the tank level stabilizes at 5 ft. It is
usually desirable that surge tank levels stabilize before the tank overflows for
reasons of safety if nothing else. The inlet flow should not be capable of being
driven to a larger value than the outlet flow.
From a safety standpoint it would be important to examine the input flow
rate to make sure it is the maximum to be expected, and to make sure that the
exit flow rate is realistic - for example, could a valve be closed or the exit piping
be obstructed in some other way, e.g., fouling or plugging by debris (perhaps a
rag left in the tank after cleaning), such that the exit flow becomes proportional
to some constant with numerical value less than 2. Overflow could be serious
even with cool water because of damage to records or expensive equipment; with
hot, toxic, andor corrosive liquids the hazard could be severe.
nle 2.1.1-2 Volumetric flow rate o f flurd rn l&ar flow . . . . rn circular DW
The (axially symmetric) velocity profile for fully developed laminar flow in
a smooth tube of constant circular cross-section is
80 Chapter 2: The Mass Balances
v = v,[l- ( 3 2 1 (2.1.1 -20)
Calculate the volumetric flowrate.
Solution
Choose the flow direction to be the z-direction.
a) Since we are not referring the volumetric flow to any particular system or
control volume, we use the absolute value
Noting that
Q = 2rrvm,4 R [ l - (&)2]rdr
Q = 2 a v m a [ \$ - q
(2.1.1-21)
(2.1.1 -22)
(2.1.1-23)
(2.1.1-24)
(2.1.1-25)
R
0
(2.1.1 -26)
(2.1.1-27)
Chapter 2: The Mass Balances 81
(2.1.1-28)
Dle 2.1.1-3 Air stoLgpe t a d
A tank of volume V = 0.1 m3 contains air at 1000 kPa and (uniform)
density of 6 kg/m3. At time t = 0, a valve at the top of the tank is opened to
give a free area of 100 mm2 across which the air escape velocity can be assumed
to be constant at 350 m / s . What is the initial rate of change of the density of air
in the tank?
Solution
(2.1.1-29)
Examining the second term
But, since the tank volume is constant, dV/dt = 0 and
\$ l V p d V = V- dP
dt
The first term yields, since the only term is an output
J^*p(v.n)dA = p ( v . n ) l A d A = p v A
Substituting in the mass balance
p v A + V - dp = 0
dt
(2.1.1-31)
(2.1.1-32)
(2.1.1-33)
82 Chapter 2: The Mass Balances
kg = -2.1 -
m 2 s (2.1.1-34)
l jxamd e 2.1.1-4 Water man ifold
Water is in steady flow through the manifold4 shown below, with the
indicated flow directions known.
The following information is also known.
A manifold is a term loosely used in engineering to indicate a collection of pipes
connected to a common region: often, a single inlet which supplies a number of
outlets; e.g., the burner on a gas stove or the fuel/air intake manifold for an multi-
cylinder internal combustion engine. If the volume of the region is large compared to
the volume of the pipes, one does not ordinarily use the term manifold - for example,
a number of pipes feeding into or out of a single large tank could be called but would
not ordinarily be called a manifold.
Chapter 2: The Mass Balances 83
Assuming that velocities are constant across each individual cross-sectional
area, determine v.4.
Solution
We apply the macroscopic total mass balance to a system which is bounded
by the solid surfaces of the manifold and the dotted lines at the fluid surfaces.
j A p ( v . n ) d A + & l v p d V = 0
At steady state
g I v p d V = 0
so
S A p ( v - n ) d A = 0
and, recognizing that at the solid surfam
(&quot;..) = 0
(2.1.1-35)
(2.1.1-36)
(2.1.1-37)
(2.1.1-38)
the macroscopic total mass balance reduces to
84 Chapter 2: The Mass Balances
(2.1.1-39)
Examining the first term,
jA 1 p ( v - n ) d A = lAl ( 6 2 . 4 ) [ y ] ( 8 i [ # n ) d A
but the outward normal at A1 is in the negative x-direction, so the velocity
vector at A1 and the outward normal are at an angle of 1800, giving
( i .n ) = II(IlIcos(nradians) = - 1 (2.1.1-41)
and therefore
p ( v . n) dA = - (62.4) [T] (8) [%] 1 dA
= - (62.4) [T] (8) [g] (A1)
= - (62.4) [T] (8) [\$] (0.3) [ft2]
1 bmass
(2.1.1-42)
= - 1 5 0 . 7 -
Proceeding to the second term, and recognizing that this also is an input
term so will have a negative sign associated with it as in the first term
p ( v - n ) d A = p! (v .n )dA = - p Q 2
2 A 2
Chapter 2: The Mass Balances
lbmass
(2.1.1-43) = - 5 6 7
The third term, which is an output (therefore the dot product gives a positive
sign), then gives
The fourth term simplifies to
p (v a n) dA = p (v n) IA dA = p (v4 - n) A 4
4
p (v . n) dA = (62.4) [T] (v4 + n) [!] (0.4) [ft2]
4
= (25) [v] (v4 - n) 
= 25 (v4.n)-
(2.1.1-44)
(2.1.1-45)
Substituting these results into the simplified macroscopic total mass
balance
- (1 50) [ -1 - (56) [ 91 + (250) [v]
(2.1.1-46)
+ { (25) [y] (v4- n) [\$I} = o
Solving
ft v4 .n = -1 .768
(2.1.1-47)```