Chapter 2 the Mass Balances

2

THE MASS BALANCES We are concerned in this course of study with mass, energy, and

momentum. We will proceed to apply the entity balance to each in tun. There also are sub-classifications of each of these quantities that are of interest. In terms of mass, we are often concerned with accounting for both total mass and mass of a species.

2.1 The Macroscopic Mass Balances

Consider a system of an arbitrary shape fixed in space as shown in Figure 2.1-1. V refers to the volume of the system, A to the surface area. AA is a small increment of surface area; AV is a small incremental volume inside the system; v is the velocity vector, here shown as an input; n is the outward normal; a is the angle between the velocity vector and the outward normal, here greater than ~t radians or 180 because the velocity vector and the outward normal fall on opposite sides of AA.

V \

Figure 2.1-1 System for mass balances

73

74 Chapter 2: The Mass Balances

Using the continuum assumption, the density can be modeled functionally by

(2.1-1)

2.1.1 The macroscopic total mass balance

We now apply the entity balance, written in terms of rates, term by term to total mass. Since total mass will be conserved in processes we consider, the generation term will be zero.

output input accumulation

The total mass in AV is approximately

P AV (2.1 0 1-2)

where p is evaluated at any point in AV and At.

The total mass in V can be obtained by summing over all of the elemental volumes in V and taking the limit as AV approaches zero

(2.1.1-3)

Accumulution of mass

First examine the rate of accumulation term. The accumulation (not rate of accumulation) over time At is the difference between the mass in the system at some initial time t and the mass in the system at the later time t + At.

J I duringAt

Chapter 2: The Mass Balances

r volumetric flow rate

through AA

7.5

The rate of accumulation is obtained by taking the limit

rate of accumulation of total mass

in V

(2.1.1 -4)

(2.1.1 -5)

Input and output of mass

The rate of input and rate of output terms may be evaluated by considering an arbitrary small area AA on the surface of the control volume. The velocity vector will not necessarily be normal to the surface.

The approximate volumetric flow rate' through the elemental area AA can be written as the product of the velocity normal to the area evaluated at some point within AA multiplied by the area.

(2.1.1 -6)

Notice that the volumetric flow rate written above will have a negative sign for inputs (where a > n) and a positive sign for outputs (where a c x ) and so the appropriate sign will automatically be associated with either input or output terms, making it no longer necessary to distinguish between input and output.

The approximate mass flow rate through AA can be written by multiplying the volumetric flow rate by the density at some point in AA

We frequently need to refer to the volumetric flow rate independent of a control volume. In doing so we use the symbol Q and restrict it to the ahsolute value

Q = k I ( v * n ) l d A


(2.1.1-7)

[F] - [$][$I The mass flow rate through the total external surface area A can then be written as the limit of the sum of the flows through all the elemental AAs as the elemental areas approach zero. Notice that in the limit it no longer matters where in the individual AA we evaluate velocity or density, since a unique point is approached for each elemental =a.*

mass

through A

(2.1.1 -8) = l A p (v n) dA

Substitution in the entity balance then gives the macroscopic total mass balance

Total mass output rate

Total mass -1 input rate

] + [ accumulation To:;as] = 0

We Erequently need to refer to the mass flow rate independent of a control volume. In doing so we use the symbol w and restrict it to the ahsolute value

w = I A p l ( v - n ) l d A

For constant density systems w = p Q

Chapter 2: The Mass Balances 77

l /Ap(v-n)dA+$lvpdV = 0

w o u t - w i n + 9 = 0

(2.1.1-9)

We can use the latter form of the balance 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


(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)

Tank 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.


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


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)


(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)


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.


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

("..) = 0

(2.1.1-35)

(2.1.1-36)

(2.1.1-37)

(2.1.1-38)

the macroscopic total mass balance reduces to


(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


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) [9] = 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)

The negative sign indicates that the velocity vector is in the opposite direction to the outward normal. At A4 the outward normal is in the positive y-direction; therefore, the velocity vector is in the negative y-direction, or


v4 = - 1.76g j (2.1.1 -48)

Water flows into the system at b.

Notice that if had emerged from the bottom of the manifold rather than the top, the outward normal would have been in the negative rather than the positive y-direction, giving a velocity vector in the positive y-direction, so the resultant flow would still have been into the system - we cannot change the mass balance by simply changing the orientation of the inlet or outlet to a system (we can, of course, change the momentum balance by such a change). The reader is encouraged to change the orientation of A4 to horizontally to the left or right and prove that the mass balance is unaffected.

2.1.2 The macroscopic species mass balance

As noted earlier when we discussed conserved quantities, total mass can be modeled as conserved under processes of interest to us here. However, individual species are not, in general, conserved, especially in processes where chemical reactions occur. Chemical reaction can give a positive or negative generation of an individual species.

For example, if we feed a propane-air mixture into a combustion furnace there will be a negative generation of propane (since propane disappears) and there will be a positive generation of C02.s In such a case, to apply the entity balance in a useful way we need a way to describe the rate of generation of a species by chemical reaction.

Let us consider for the moment applying, to a control volume fixed in space, the entity balance (again, as for total mass, written in terms of rates) now to mass of a chemical species, which is assumed to be involved in processes for which the generation is by chemical reaction.

We can, of course, at the same time generate elemental carbon (in soot, etc.), CO, and other byproducts of combustion. There will also be a negative generation of 0,.


Generation of mass of a species

Consider the generation rate term. Call the rate of species mass generation per unit volume of the ith chemical species r,, which will be a function of position in space and of time

[%I (2.1.2-1) The rate of generation of mass of species i in a volume AV is then

approximately equal to

1 rate of generation of mass of species i = ri AV in volume AV (2.1.2-2) To get the rate of generation throughout our control volume, we add the contributions of all the constituent elemental volumes in V. Taking the limit as

approaches 0, we obtain the rate of generation term in the entity balance

rate of eeneration 1 v ofmassofspeciesi = lim Z r i 1 AV+O 1 in volume V V (2.1.2-3) ,AV = Jv ri dV

Accumulation of mass of a species

In the same manner as for total mass, the amount of accumulation of mass of species i (not rate of accumulation) over time At is the difference between the mass of species i in the system at some initial time t and the mass of species i in the system at the later time t + At. The mass of species i in the system at a given time is determined by taking the sum of the masses of species i in all of the elemental volumes that make up the system and then letting the elemental volumes approach zero, in a similar way to that shown in Equation 2.1.1-5 for total mass.

88

r

of speciesi mass flow rate

through AA G pi ( v . n)AA I


I accumulation 1 of mass of species i = (m,) t+b - (m,)

in volume V I (2.1.2-4)

The rate of accumulation of mass of species i is obtained by taking the limit as At approaches 0

rate of accumulation of mass of species i

in volume V I = lim At 0 1 At (2.1.2-5)

Input and output of mass of a species

The input and output for mass of a species are readily obtained in the same manner as for total mass. The approximate mass flow rate of species i through an elemental area AA can be written by multiplying the volumetric flow rate6 by the mass concentration of species i, pi, (rather than the total density, as was done for total mass) at some point in AA

(2.1.2-6)

Strictly speaking, we must replace v by vi, since the individual species may not move at the velocity v in the presence of concentration gradients. The difference between v and vi is negligible, however, for most problems involving flowing streams. We discuss the difference between v and vi in our treatment of mass transfer.


r

mass flow rate of speciesi through A

89

[r] - [$][q The mass flow rate through the total external surface area A can then be

written as the limit of the sum of the flows through ail the elemental AAs as the elemental areas approach zem.

(2.1.2-7)

Rearranging the entity balance as

rate of output

of species i

rate of input

of species i 1 ofmass I-[ ofmass j

rate of accumulation (2.1.2-8) rate of generation _I ofmass 1 = [ ofmass ] of speciesi of speciesi

entering the various terms as just developed

(2.1.2-9)

which is the macroscopic mass balance for an individual species, i.

If there is no generation term, the macroscopic total mass balance and the macroscopic mass balance for an individual species look very much the same except for the density term.


BxarnDle 2.1.2-1 acroscopic sbecies mass balance with zero- order irreversible reaction

Water (whose density p may be assumed to be inde ndent of the

pipe of inside diameter 0.05 m into a perfectly mixed tank containing 15 m3 solution and out at the same rate through a pipe of diameter 0.02 m as shown in the following illustration.

concentration of A) with p = loo0 kg/m3 is flowing at 0.02 m F /min through a

The entering liquid contains 20 kg/m3 of A. The zero-order reaction

2 A + B

takes place with rate rg = 0.08 kg B/(min m3).

(2.1.2- 10)

If the initial concentration of A in the tank is 50 kg/m3, find the outlet concentration of A at t = 100 min.

Solution

Applying the macroscopic mass balance for an individual species

(2.1.2-11)

Evaluating term by term, starting with the first term on the left-hand side


But, since the volumetric flow rates in and out are tbe Same in absolute value

The second term on the left-hand side is

But a total mass balance combined with constant density and the fact that inlet and outlet volumetric flow rates are the same gives

-- dVtank - 0 dt (2.1.2- 15)

so

Examining the right-hand side, we note that the reaction rate of A is twice the negative of that of B by the stoichiometry

(2.1.2-1 7) rA = -2r,


= -2r,V,,--&- mol A

Substituting in the balance

but, since the tank is perfectly stirred

'tank

(2.1.2- 18)

(2.1.2-19)

(2.1.2-20)

(2.1.2-21)

Integrating and applying the initial condition

dp A = j i d t (2.1.2-22) (- Q [P A - P Ain] - 'B 'tank)


Evaluating the constant term

(2.1.2-23)

(2.1.2-24)

(2.1.2-25)

(2.1.2-26)

Introducing the required time

- (2) [ "1 min - ( O e o 2 ) [ $i] (P A) [ 51 - (2) [-%I min - (0.02) ["I min (50) [$]

m3

(0.02) (PA) = [ (2) + (0.02) (50)] exp [ - (F) ( ICNI)] - 2 (2.1.2-27)

(PA) = 31.3- kg m3


Bxamde 2.1.2-2 Macroscooic soecies mass balance with first- order irreversible reaction

Suppose that a reaction B 3 C takes place in a perfectly mixed tank. (By perfectly mixed we mean that Pi is independent of location in the tank, although it can depend on rime. Although no real vessel is perfectly mixed, this is often a good model.)

Further suppose that the initial concentration of B in the tank is 5 lbm/t3, that the inlet concentration of B is 15 lbm/ft3, that the volume of liquid in the tank is constant at 100 ft3, and that flow in and out of the tank is constant at 10 ft3/min.

If Q = (- k PB) o( = 0.1 m i d ) what is PB as a function of time?

Solution

Figure 2.1.2-1

9 - 10 ft3/min Perfectly mixed tank with reaction

Applying the macroscopic mass balance for an individual species

(2.1.2-28)

Evaluating term by term


= (pS),[ lbmysB] (10) ["I mm ft

Substituting in the balance

but, since the tank is perfectly stirred

(2.1.2-30)

(2.1.2-31)

(2.1.2-32)

(2.1.2-33)

which gives


(-5)[ln(15-2Ps)15 P B =

(15 - 5 e(-o.2d) = 2.5 (3 - e(-*s20) P B = 2 (2.1.2-34)

2.2 The Microscopic Mass Balances

The microscopic mass balance is expressed by a differential equation rather than by an equation in integrals. In contrast to the macroscopic balances, which are usually applied to determine inpudoutput relationships, microscopic balances permit calculation of profiles (velocity, temperature, concentration) at individual points within systems.

2.2.1 (continuity equation)

The microscopic total mass balance

The macroscopic total mass balance was determined (Equation 2.1.1-9) to be


j A p ( v * n ) d A + $ j v p d V = 0

which can be written as

(2.2.1-1)

(2.2.1-2)

In Chapter 1 from the Gauss theorem we showed

sv ( V . U) dV = js (n - U) dS (2.2.1-3) where U was an arbitrary vector (we have changed the symbol from v to U to make clear the distinction between this arbitrary vector and the velocity vector).

The macroscopic mass balance was derived for a stationary control volume, and so the limits in the integrals therein are not functions of time. Application of the Leibnitz rule therefore yields

(2.2.1-4)

Applying these two results to the macroscopic total mass balance gives (note that multiplying the velocity vector by the scalar p simply gives another vector)

I V ( V . p v ) d V + l v % d V = 0 (2.2.1-5) Combining the integrands, since the range of integration of each is identical

The Leibnitz rule is useful in interchanging differentiation and integration operations. It states that for f a continuous function with continuous derivative

Kaplan, W. (1952). Advanced Calculus. Reading, MA, Addison-Wesley, p. 220.


(2.2.1-6)

Since the limits of integration are arbitrary, the only way that the left-hand side can vanish (to maintain the equality to zero) is for the integrand to be identically zero.

i ( v . p v ) + - & = o I (2.2.1-7) This is the microscopic total mass balance or the equation of continuity.

Special cases of the continuity equationa

For incompressible fluids, the time derivative vanishes, and by factoring the constant density the continuity equation may be written as

p ( v * v ) = 0 (2.2.1-8)

but we can divide by the density since it is non-zero for cases of interest, giving

1 v . v = 01 (2.2.1-9)

For steady flow, the density does not vary in time, so the continuity equation becomes

(2 0 2.1 - 1 0) p p v = 01

These relationships often prove convenient in simplifying other equations.

These forms, as well as being used in and of themselves, are often used to simplify other equations by eliminating the corresponding terms.


Continuity equation in different coordinate systems

For computation of numerical results, it is necessary to write the continuity equation in a particular coordinate system. Table 2.2.1-1 gives the component farm in rectangular, cylindrical, and spherical coordinates.

Table 2.2.1-1 Continuity equation (microscopic total mass balance) in rectangular, cylindrical, and spherical coordinate frames

RECTANGULAR COORDINATES

CYLINDRICAL COORDINATES

D l e 2.2.1-1 Velocitv c o w n t s in two-dimasional dv incorn-le flow. r e c t k l a r coordinates

The x-component of velocity in a two-dimensional steady incompressible flow is given by

(2.2.1-1 1) v, = c x

where c is a constant.

What is the form of the y-component of velocity?

Icx) Chapter 2: TIre Mass Balances

Solution

The continuicj equation reduces to

v - v = 0

which is, for a two-dimensional flow

But from the given x-component

Integrating with respect to y

[ ( 2 ) d y = -J c cdy= - c y + f ( x ) = vy J

(2.2.1-12)

(22.1-13)

(2.2.1-143

(2.2.1 - 15)

(2,2.1-16)

Notice that since we are integrating a partis! derivative of a function of x and y with respect to y, we can determine the result of such an integration only to within a function of x rather than a constant, because any such function disappears if we take the partial derivative with respect to y. Consequently, the result shown would, as is required, satisy the inverse operation

(3) (a[-cY+f(.)i X ) x = c

- - = - ay (2.2.1 - 17)

We have added the subscript x to the partial derivative to emphasize that x is being held constant during the operation. We should write such a subscript in conjunction with all partial derivatives; however, the subscripts are usually obvious so we customarily omit them for conciseness.

Therefore, the most genera! form is


vy = - c y + f ( x )

This form includes, of course

f (x) = constant

ard

f (x) = 0

(2.2.1 - 18)

(2.2.1-19)

(2.2.1-20)

Examrrle 2.2.1 -2 Velocity comDonents in two-dimensional steady incomuressible flow. cvlindrical coordinates

Water is in steady one-dimensional axial flow in a uniform-diameter cylindrical pipe. The velocity is a function of the radial coordinate only. What is the form of the radial velocity profile?

Solution

The continuity equation in cylindrical coordinates is

which, for steady flow reduces to

(2.2.1-21)

(2.2.1-22)

which then becomes, for velocity, a function of r only

I d TZ(P'Vr) = 0 (2.2.1-23)

where we have replaced the partial derivative symbols with ordinary derivatives because we now have an unknown function of only one independent variable. This then becomes, for an incompressible fluid

I d P i z ( r v r ) = 0 (2.2.1-24)


We multiply both sides by the radius, and, since the density is not equal to zero, we can divide both sides by the density

a(rvr) d = o

Integrating

/ -$(r vr) dr = 1 0 dr rv , = constant

which gives as the form of the radial velocity profile

Vr = constant r

(2.2.1-25)

(2.2.1-26)

(2.2.1-27)

Examde 2.2.1-3 Compression o f air

Air is slowly compressed in a cylinder by a piston. The motion of the air is one-dimensional. Since the motion is slow, the density remains constant in the z-direction even though changing in time. As the piston moves toward the cylinder head at speed v, what is the time rate of change of the density of the air?

Figure 2.2.1-1 Air compression by piston

So 1 u t ion

We begin with the continuity equation

Chapter 2: The Mass Balances I03

( V . p v ) + % = 0

at r &(P TV,) ++&P ve) + Z(P vz) = 0

at +z(P vz) = *

JP % aP - 0 a t + P ~ + V Z ~ -

-+p- aP h z = o

In cylindrical coordinates

*+I& a

In one dimension (the z-direction)

a~ a

Expanding the derivative of the product

However, density is constant in the z-direction, so

(2.2.1-32) at aZ

(2.2.1-28)

(2.2.1-29)

(2.2.1-30)

(2.2.1-31)

2.2.2 The microscopic species mass balance

We can develop a microscopic mass balance for an individual species in the Same general way that we developed the microscopic total mass balance, but, as in the macroscopic case, individual species are not conserved but may be created or destroyed (e.g., by chemical reactions), so a generation term must be included in the model.

The macroscopic mass balance for an individual species took the form

Models based on the above equation (which are adequate for many real problems) are restricted to cases where the velocity of the bulk fluid, v, and the velocity of the individual species, Vi, are assumed to be identical.


As was just noted, individual species may not move at the same velocity as the bulk fluid, for example, because of concentration gradients in the fluid. This becomes of much more a concern in certain classes of problems we wish to model at the microscopic level than it was at the macroscopic level.

We begin with the more general equation, which includes the species velocity rather than the bulk fluid velocity

jAp, (v i .n)dA+&jvp,dV = 1 V ridV (2.2.2-2) By incorporating the species velocity v, instead of the velocity of the bulk

fluid v we can model cases where Vi and v are significantly different. This is the whole point of mass transfer operations.

The Leibnitz rule and the Gauss theorem can be applied as in the case of the microscopic total mass balance, giving

Again the limits of integration are arbitrary, so it follows that

(2.2.2-3)

(2.2.2-4)

which is the microscopic rnass balance for an individual species. There is one such equation for each species. In a mixture of n constituents, the sum of all n of the species equations must yield the total mass balance, and therefore only n of the n+l equations are independent.

The microscopic mass balance in the form above is not particularly useful for two reasons:

First, it does not relate the concentration to the properties of the fluid. This is done via a constitutive equation (such an equation relates the flow variables to the way the fluid is constituted, or made up).

Chapter 2: The Mass Balances 10.5

Second, it is written in terms of mass concentration, while most mass transfer process operate via molar concentrations (diffusive processes tend to be proportional to the relative number of molecules, not the relative amount of mass).

DifSusion

An example of a constitutive equation relating fluxes and concentrations is Fick's law of diffusion

(2.2.2-5)

where Nj denotes the molar flux of the ith species with respect to axes fixed in space, Xi is the mole fraction of species i, c is the total molar concentration, and Dm is the diffusion coefficient.

The molar flux can be written in terms of the velocity of the ith species, V i , which is defined as the number average velocity

(2.2.2-6)

It is not difficult to see that substituting this relationship in Fick's law to obtain velocity as a function of concentration and then putting the result in the microscopic species mass balances (adjusting the units appropriately) is not a trivial task. We will return to this later in Section 8.2.

Chapter 2 Problems

2.1 For the following situation: a. Calculate (v n) b. Calculate vx c. Calculate w

Use coordinates shown.


W

2.2 For the following

A=f d

a. Calculate (v n) b. Calculate w (p = 999 kg/m3) c. Calculate(*P(v.n)dA if Ivl = ( l -*)v-

where vmax = 4.5 m / ~ .


2.3 Calculate I, P (v * n) dA for the following situation

\ A , , circular; diameter * 1 I t

Use coordinates shown.

2.4 A pipe system sketched below is carrying water through section 1 at a velocity of 3 ft/s. The diameter at section 1 is 2 ft. The same flow passes through section 2 where the diameter is 3 ft. Find the discharge and bulk velocity at section 2.

2.5 Consider the steady flow of water at 7OOF through the piping system shown below where the velocity distribution at station 1 may be expressed by

v ( r ) = 9.0 ( 1 - - ;)


Find the average velocity at 2.

2.6 Water is flowing in the reducing elbow as shown. What is the velocity and mass flow rate at 2?

p = 62.4- ft3

to 2 ft/sec

2.7 The tee shown in the sketch is rectangular and has 3 exit-entrances. The width is w. The height h is shown for each exit-entrance on the sketch. You may assume steady, uniform, incompressible flow.

a. Write the total macroscopic mass balance and use the assumptions to simplify the balance for this situation.

b. Is flow into or out of the bottom arm (2) of the tee. c. What is the magnitude of the velocity into or out of the bottom arm (2) of the tee?

"2

2.8 The velocity profile for turbulent flow of a fluid in a circular smooth tube follows a power law such that


Find the value of /vmax for this situation. If n = 7 this is the result from the Blasius resistance formula.

(Hint: let y = R - r)

2.9 Oil of specific gravity 0.8 is pumped to a large open tank with a 1.0-ft hole in the bottom. Find the maximum steady flow rate Qin, ft3/s, which can be pumped to the tank without its overflowing.

2.10 Oil at a specific gravity of 0.75 is pumped into an open tank with diameter 3 m and height 3 m. Oil flows out of a pipe at the bottom of the tank that has a diameter of 0.2 m. What is the volumetric flow rate that can be pumped in the tank such that it will not overflow?

2.11 In the previous problem assume water is pumped in the tank at a rate of 3 m3/min. If the water leaves the tank at a rate dependent on the liquid level such that Qout = 0.3h (where 0.3 is in m2/min and Q is in m3/min, h is in m. What is the height of liquid in the tank as a function of time?

2.12 Shown is a perfectly mixed tank containing 100 ft3 of an aqueous solution of 0.1 salt by weight. At time t = 0, pumps 1 and 2 are turned on. Pump 1 is

I10 Chapter 2: The Mass Balances

pumping pure water into the tank at a rate of 10 ft3/hr and pump 2 pumps solution from the tank at 15 ft3/hr. Calculate the concentration in the outlet line as an explicit function of time.

2.13 Water is flowing into an open tank at the rate of 50 ft3/min. There is an opening at the bottom where water flows out at a rate proportional to

where h = height of the liquid above the bottom of the tank. Set up the equation for height versus time; separate the variables and integrate. The area of the bottom of the tank is 100 ft2.

2.14 Salt solution is flowing into and out of a stirred tank at the rate of 5 gal solution/min. The input solution contains 2 lbm saldgal solution. The volume of the tank is 100 gal. If the initial salt concentration in the tank was 1 lbm saltlgal, what is the outlet concentration as a function of time?

2.15 A batch of fuel-cell electrolyte is to be made by mixing two streams in large stirred tank. Stream 1 contains a water solution of H2SO4 and K2SO4 in mass fractions XA and x g l lbm/lbm of solution, respectively (A = H2SO4, B = K2SO4). Stream 2 is a solution of H2SO4 in water having a mass fraction XA2.

If the tank is initially empty, the drain is closed, and the streams are fed at steady rates w l and w2 lbm of solutionlhr, show that XA is independent of time and find the total mass m at any time.

(Hint: Do not expand the derivative of the product in the species balance, but integrate directly.)


Cqn;;ntration in tonk

2.16 You have been asked to design a mixing tank for salt solutions. The tank volume is 126 gal and a salt solution flows into and out of the tank at 7 gal of solutiodmin. The input salt concentration contains (3 lbm salt)/gal of solution. If the initial salt concentration in the tank is (1/2 lbm &)/gal, find:

a. The output concentration of salt as a function of time assuming that the output concentration is the same as that within the tank.

b. Plot the output salt concentration as a function of time and find the asymptotic value of the concentration m(t) as t + - .

c. Find the response time of the stirring tank, i.e., the time at which the output concentration m(t) will equal 62.3 percent of the change in the concentration from time t = 0 to time t + - .

Chapter 2 the Mass Balances

Documents

Transcript of Chapter 2 the Mass Balances