Fenômentos de Transporte
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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 
(2.1.1-1 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 
\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 
t = - 39.3 In [v] 
h = 5[Lexp(-&)]ft 
(2.1.1 - 14) 
(2.1.1 - 15) 
(2.1.1 - 17) 
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. 
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 -22) 
(2.1.1 -26) 
Chapter 2: The Mass Balances 81 
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? 
Examining the second term 
But, since the tank volume is constant, dV/dt = 0 and 
$ l V p d V = V- dP 
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 
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. 
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 
S A p ( v - n ) d A = 0 
and, recognizing that at the solid surfam 
("..) = 0 
the macroscopic total mass balance reduces to 
84 Chapter 2: The Mass Balances 
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) [$] (0.3) [ft2] 
1 bmass 
= - 1 5 0 . 7 - 
Proceeding to the second term, and recognizing that this also is an input 
term so will have a negative sign associated with it as in the first term 
p ( v - n ) d A = p! (v .n )dA = - p Q 2 
2 A 2 
Chapter 2: The Mass Balances 
(2.1.1-43) = - 5 6 7 
The third term, which is an output (therefore the dot product gives a positive 
sign), then gives 
The fourth term simplifies to 
p (v a n) dA = p (v n) IA dA = p (v4 - n) A 4 
p (v . n) dA = (62.4) [T] (v4 + n) [!] (0.4) [ft2] 
= (25) [v] (v4 - n) [9] 
= 25 (v4.n)- 
Substituting these results into the simplified macroscopic total mass 
- (1 50) [ -1 - (56) [ 91 + (250) [v] 
+ { (25) [y] (v4- n) [$I} = o 
ft v4 .n = -1 .768