Mixing Problems-I - Stewart, Calculus: Early Transcendentals, 6th Edition. PDF

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Stewart, Calculus: Early Transcendentals, 6th Edition....

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Mixing Problems – Part I

In these notes, we discuss a physical situation for which it is relatively simple to write the differential equation which serves as the mathematical model of the process. In what are often referred to as “mixing problems”, a large tank of some sort contains a starting volume of a liquid solution of some substance (for example, brine, a solution of ordinary salt in water). The tank is having liquid added to it and/or drained from it. The liquid that is being added to the tank is poured in at a specified rate and may have a concentration of the substance of interest which is different from the concentration currently in the tank. The liquid is being drained from the tank at a rate which may be the same or different from the rate liquid is being added; the concentration of the substance in the draining liquid is related in some way to the concentration presently contained in the tank. (The process can involve any sort of fluid – liquid or gas – being added to and drawn from a container.) We will look at mixing problems which have the simplest conditions for which it is possible to work out a solution function “by hand”. The liquid being added to the tank will have a constant concentration and a constant rate of flow into the tank. The solution in the tank is kept stirred up in a way that all of the substance already present or just added is immediately mixed evenly through the entire volume of the tank (the “instantaneous mixing” assumption*). Thus, the liquid being drained, also at some constant rate of flow, has a concentration which is exactly the same as the concentration of substance within the tank. * This is somewhat unrealistic in practice, but convenient for the purpose of calculation. In this first part, we will investigate the situation in which the rate at which liquid is added is the same as the rate at which liquid is drained out of the tank. This condition makes the differential equation for the model particularly easy to write and to solve. The second (and longer) section will discuss what happens when these two flow rates differ from one another.

Part I -- Flow into and out of the tank are equal There are two differential equations, referred to as “rate equations”, which we can set up in connection with the mixing problem. Each of these is used to describe what is happening within the tank of liquid. The first of these is the “volume flow rate” equation, which tells us how the amount of liquid in the tank is changing. We write the net rate of change of the volume in the tank as

฀฀dV ฀฀ ฀฀dV ฀฀ ฀฀dV ฀฀ d V ( t ) = ฀฀ ฀฀ = ฀฀ ฀฀ − ฀฀ ฀฀ dt ฀฀dt ฀฀net ฀฀ dt ฀฀in ฀฀dt ฀฀out

, [A]

where V ( t ) is the volume of liquid present in the tank at time t measured from some chosen moment, and the two rates in the difference represent the constant rate at which liquid is being added to (“input flow rate”) and at which it is being drained from (“output flow rate”) the tank.

For the case at hand, these two rates are taken to be equal, so we will write

฀฀d V ฀฀ ฀฀d V ฀฀ ฀฀d V ฀฀ ฀฀ ฀฀ = ฀฀ ฀฀ = ฀฀ ฀฀ !!, which thus tells us the rate at which liquid is passing ฀฀dt ฀฀flow ฀฀dt ฀฀in ฀฀ dt ฀฀out ฀฀dV ฀฀ ฀฀dV ฀฀ ฀฀dV ฀฀ d V ( t ) = ฀฀ ฀฀ = ฀฀ ฀฀ − ฀฀ ฀฀ = 0 , through the tank. This means that dt ฀฀ dt ฀฀net ฀฀dt ฀฀flow ฀฀ dt ฀฀flow (that is, there is zero net change in the volume of liquid in the tank), which implies that V ( t ) is a constant. Since the volume will thus always remain the same as it was at time t = 0 , we will express the volume in the tank as V ( t ) = V ( 0 ) = V0 . The second of the rate equations is used to describe the net rate of change of the mass of dissolved substance in the tank (salt, for example); this “mass flow rate equation” resembles the one for volume flow rate:

฀฀dm ฀฀ ฀฀dm ฀฀ ฀฀dm ฀฀ d m( t ) = ฀฀ ฀฀ = ฀฀ ฀฀ − ฀฀ ฀฀ ฀฀dt ฀฀net ฀฀dt ฀฀in ฀฀dt ฀฀out dt

. [B]

The mass of the substance in a particular portion of liquid is equal to the volume of that liquid times the concentration of the substance in it (indeed, this statement provides the definition of “concentration”), so m = V · c . We can now relate the mass flow of the substance to the volume flow of the liquid. For the solution being added to the tank, the concentration is constant: we will

฀฀d V ฀฀ ฀฀dm ฀฀ ฀฀ = ฀฀ ฀฀ ⋅ c in !. The solution found in the ฀฀ dt ฀฀in ฀฀ dt ฀฀in

call it cin ; hence, we can say that ฀฀

tank at time t has a mass of the substance given by m ( t ) = V ( t ) · c ( t ) . We are assuming that the concentration is the same everywhere in the tank because of “instantaneous mixing”; this means that the liquid being drained out has the same concentration of the substance as is found in the tank; the outflow of substance mass is

฀฀d V ฀฀ ฀฀d V ฀฀ ฀฀dm ฀฀ ฀฀ = ฀฀ ฀฀ ⋅ c out = ฀฀ ฀฀ ⋅ c( t ) !!. For any of the mixing problems ฀฀ dt ฀฀out ฀฀ dt ฀฀out ฀฀dt ฀฀out

therefore ฀฀

we will discuss in these notes, the mass flow rate equation is then

฀฀฀฀dV ฀฀ ฀฀ ฀฀฀฀ dV ฀฀ ฀฀ d d m( t ) = [ V ( t ) ⋅ c ( t ) ] = ฀฀฀฀ ฀฀ ⋅ c in ฀฀ − ฀฀฀฀ ฀฀ ⋅ c( t )฀฀ dt dt ฀฀฀฀ dt ฀฀in ฀฀ ฀฀฀฀dt ฀฀out ฀฀

[C]

For the case in Part I, the condition of equal rates of filling and draining permits us to write this equation as

฀฀ ฀฀฀฀ d V ฀฀ ฀฀฀฀d V ฀฀ ฀฀ m( t ) ฀฀ m( t ) ฀฀ ฀฀ d V ฀฀ d ฀฀ = ฀฀ ฀฀ ⋅ ฀฀c in − m( t ) = ฀฀฀฀ ฀฀ ⋅ c in ฀฀ − ฀฀฀฀ ฀฀ ⋅ ฀฀ V0 ฀฀ dt ฀฀ ฀฀฀฀ dt ฀฀flow V ( t ) ฀฀ ฀฀ dt ฀฀flow ฀฀ ฀฀฀฀ dt ฀฀flow

[D]

with the concentration at time t expressed in terms of the substance mass at that time.

We can take another view of this model of the solution in the tank by writing a rate of concentration change equation. If we return to the general mass flow rate equation (equation C above), we can use the condition of constant volume in the tank to write instead

฀฀ ฀฀ ฀฀฀฀ d V ฀฀ ฀฀฀฀d V ฀฀ d [ V0 ⋅ c( t ) ] = ฀฀฀฀ ฀฀ ⋅ c in ฀฀ − ฀฀฀฀ ฀฀ ⋅ c ( t )฀฀ dt ฀฀ ฀฀ ฀฀ ฀฀ ฀฀ ฀฀฀฀ dt ฀฀flow ฀฀ ฀฀฀฀ dt ฀฀flow

⇒ V0 ⋅

฀฀d V ฀฀ d c( t ) = ฀฀ ฀฀ ⋅ [c in − c( t ) ] . ฀฀ dt ฀฀flow dt !

[E]

!

We see that either of the mass flow or the concentration change differential equations we have come up with is a separable equation, so we can use the method we learned about in Section 9.3 of Stewart (6 th edition) or early on in any course on differential equations to solve for the appropriate function. We will do so for the abstract equation first and then look at a problem using specific values of the involved quantities. Equation D for the mass flow rate can be separated as

฀฀dV ฀฀ dm = ฀฀ ฀฀ dt ⇒ m c in − ฀฀dt ฀฀flow V0

⇒ − V0 ln c in −

m V0

∫

dm = c in − m V0

∫

฀฀dV ฀฀ ฀฀ ฀฀ dt ฀฀dt ฀฀flow

฀฀d V ฀฀ = ฀฀ ฀฀ ⋅ t + C . ฀฀ dt ฀฀flow

It is the most convenient at this point to evaluate the arbitrary constant C for the significant initial condition of the problem, that m ( 0 ) = m0 = V ( 0 ) · c ( 0 ) = V0 · c0 :

−V0 ln c in −

m0 m ฀฀d V ฀฀ = ฀฀ ฀฀ ⋅ 0 + C ⇒ C = − V0 ln c in − 0 V0 V0 ฀฀ dt ฀฀flow

⇒ − V0 ln c in − ! !

⇒ ln c in −

m V0

m V0

m ฀฀dV ฀฀ = ฀฀ ฀฀ ⋅ t − V0 ln c in − 0 ! V ฀฀dt ฀฀flow 0

฀฀dV ฀฀ ฀฀ ฀฀ ฀฀ dt ฀฀flow m ⋅ t + ln c in − 0 = − V0 V0

where, in the last expression, we will use the symbol k = writing.

= − kt + ln c in −

฀฀d V ฀฀ ฀฀ ฀฀ ฀฀dt ฀฀flow V0

m0 V0

to save a bit of

,

This would be a good time to discuss the meaning of the absolute value function in an equation such as this last one. When we see the absolute value of a term appear, it is generally a sign that we must consider two cases, one for which the expression within the absolute value brackets is positive, and the other for which it is negative. If we return to the mass flow rate equation for a moment, we see that the rate is zero for

฀฀ ฀฀dV ฀฀ m m ฀฀ dm or m = V0 ⋅ c in = ฀฀ ฀฀ ⋅ ฀฀c in − ฀฀ = 0 ⇒ c in = ฀฀ dt ฀฀flow ฀฀ dt V0 V0 ฀฀

.

This means that the mass of substance in the tank does not change if it already equal to the volume of solution in the tank times the concentration of solution being added. Another way of saying this is that the mass in the tank will be unchanged if the concentration of solution being added to the tank is exactly the same as that of the solution already in there. Such a result from the differential equation is called a “constant solution” (or “stationary solution” or “steady-state solution”). This condition is the “dividing line” between the two cases we must consider in dealing with the absolute value in the result from the differential equation. Thus we have: “Case 0” --

c in =

m ⇒ m( t ) = m0 = V0 ⋅ c in V0

Case I --

c in >

m V0

⇒

c in −

m V0

= c in −

m V0

฀฀ ฀฀ m ฀ m ฀฀ ⇒ ln ฀฀c in − ฀฀ = − kt + ln฀฀c in − 0฀฀! V0 ฀฀ V0 ฀฀ ฀฀ ฀฀ ! !

!

⇒ c in −

m = e −k t ⋅ ฀฀c − m0 ฀฀ ⇒ m = c − ฀฀c − m0 ฀฀ −k t ⋅e ! ฀฀ in ฀฀ in in V0 ฀฀ V0 ฀฀ V0 V0 ฀฀ ฀฀ ฀฀ ฀฀

exponentiating both sides of the equation

!

!

⇒ m( t ) = V0 c in − ( V0 c in − m0) ⋅ e −k t

! Case II --

c in <

m V0

⇒

c in −

m V0

฀฀ ฀฀dV ฀฀฀฀ ฀฀dt ฀฀flow , with k = ! V0

฀฀ m ฀฀ m = = − ฀฀c in − − c in V0 ฀฀ ฀฀ ฀฀ V0

฀฀m ฀฀ m ฀฀ ฀ ⇒ ln ฀฀ − c in ฀฀ = − k t + ln฀฀ 0 − c in ฀฀! ฀฀ ฀฀ ฀฀V 0 ฀฀V 0

!

฀฀ ฀฀ ฀฀m ฀฀m m − c in = e −k t ⋅ ฀฀ 0 − c in ฀฀ ⇒ m = c in + ฀฀ 0 − c in ฀฀ ⋅ e −k t! V0 V0 ฀฀ ฀฀ ฀฀V0 ฀฀V0

!

!

⇒

! !

!

⇒ m( t ) = V0 c in + (m0 − V0 c in ) ⋅ e −k t

We can verify readily that the mass functions for the two cases do meet the initial condition: I --

m( 0 ) = V0 c in − ( V0 c in − m 0) ⋅ e −k ⋅ 0 = V0 c in − (V0 c in − m 0 ) ⋅1 = m 0 ;

II --

m( 0 ) = V0 c in + ( m 0 − V0 c in ) ⋅ e −k ⋅ 0 = V0 c in + ( m 0 − V0 c in ) = m 0

.

We also find that “in the long run”, after a great deal of time has passed, that I --

! !

lim m( t ) = lim [ V0 c in − (V0 c in − m0 ) ⋅ e −k t ] !

t→∞

t→∞

= V0 c in − (V0 c in − m 0 ) ⋅ lim e −k t = V0 c in − ( V0 c in − m0) ⋅ 0 = V0 c in

!

;

t→∞

II --

lim m( t ) = lim [ V0 c in + (m0 − V0 c in ) ⋅ e −k t ] = V0 c in

t→∞

t→∞

.!

! In other words, the two mass functions which arise from the cases where the initial mass of substance in the tank is either m0 < V0 ⋅ c in or m0 > V0 ⋅ c in both asymptotically approach the constant solution function

m( t ) = V0 ⋅ c in

.

We could begin instead with Equation E for the rate of change of concentration in the tank, which we found to be

V0 ⋅

฀฀d V ฀฀ d c( t ) = ฀฀ ฀฀ ⋅ [ c in − c( t ) ] ⇒ dt ฀฀ dt ฀฀flow

฀฀d V ฀฀ ฀฀ ฀฀ ฀฀dt ฀฀flow dc ⋅ (c in − c ) . = V0 dt [F]

As this too is a separable differential equation, we can follow a procedure analogous to the one we used for the mass function:

฀฀dV ฀฀ ฀฀dt ฀฀ ฀฀ ฀฀flow dc = dt = k dt ⇒ V0 (c in − c )

∫

dc = (c in − c )

∫k

dt !

k is the same as it is for m ( t )

! !

!

⇒ − ln c in − c = k t + C ;

since c ( 0 ) = c0 , we can evaluate the arbitrary constant and continue

− ln c in − c( 0 ) = k ⋅ 0 + C ⇒ − ln c in − c 0 = C ! ! ! !

⇒ − ln c in − c = k t − ln c in − c 0

⇒

c in − c = e −k t ⋅ c in − c 0

.

!

Upon examining Equation F for the concentration rate of change, we can see that

dc = k ⋅ (c in − c ) = 0 ⇒ [ c in − c( t ) ] = 0 ⇒ c( t ) = c in dt

,

which is consistent with what we said earlier: the concentration of the substance in the tank does not change if the concentration of solution being added is identical. We again have three cases to consider: “Case 0” --

c in = c ⇒ c( t ) = c 0

Case I --

c in > c ⇒

c in − c = c in − c

⇒ (c in − c ) = (c in − c 0 ) ⋅ e −k t

⇒ c( t ) = c in − (c in − c 0 ) ⋅ e −k t !

! c in − c = − (c in − c ) = c − c in !

Case II --

c in < c ⇒

! !

⇒ (c − c in ) = (c 0 − c in ) ⋅ e −k t

!

⇒ c( t ) = c in +

(c 0

− c in ) ⋅ e −k t !

We will again find that these concentration functions satisfy the initial condition c ( 0 ) = c0 and that the functions for which c0 < cin or c0 > cin both approach the constant solution function asymptotically.

To make all this somewhat less abstract, let’s illustrate the discussion by working out results for a specific problem. Suppose we have a tank which contains 1000 liters of water with salt dissolved in it (brine). Initially ( t = 0 ), there are 6 kg. of dissolved salt in the tank. Through a pipe, a solution with a concentration of 4 grams/liter is being poured into the tank at a rate of 8 liters/minute. The brine in the tank is kept continually stirred up and is being allowed to drain out, also at a rate of 8 liters/minute. We are to find the mass of dissolved salt in the tank after one hour ( t = 60 minutes ). We will solve this problem first for the mass function m ( t ) , and then for the concentration function c ( t ) , to demonstrate the two approaches to this problem. From the statement of the problem, we learn that

฀฀d V ฀฀ ฀฀d V ฀฀ ฀฀d V ฀฀ liters ฀฀ ฀฀ = ฀฀ ฀฀ = ฀฀ ฀฀ = 8 minute !!, so the volume in the tank will remain ฀฀dt ฀฀in ฀฀ dt ฀฀out ฀฀ dt ฀฀flow constant at V ( t ) = V0 = 1000 liters. The concentration of brine being added to the tank is cin = 4 grams/liter, while the concentration in the tank at time t = 0 is

c( 0 ) =

6000 grams m( 0 ) 6 kg. m( 0 ) grams = = = . The mass flow rate = 6 liter 1000 liters V (0) V0 1000 liters

equation thus becomes

฀฀ grams m( t ) ฀฀ dm liters ⋅ ฀฀4 − = 8 ฀฀ minute ฀฀ liter dt 1000 liters ฀฀ (continued)

⇒

dm = 8 dt ⇒ 4 − m 1000

⇒ −1000 ln 4 −

∫

dm = 4 − m 1000

∫8

dt

m = 8t + C 1000

At time t = 0 , the mass of dissolved salt in the tank is m ( 0 ) = 6000 grams , so we must express this last equation as

−1000 ln 4 −

6000 = −1000 ln − 2 = − 1000 ln 2 = 8 ⋅ 0 + C 1000

⇒ C = −1000 ln 2 . This is as we would expect for a situation in which m0 > V0 · cin (Case II), since 6 kg. = 6000 grams > 1000 liters · ( 4 grams/liter ) , so we must write

c in −

฀฀ m m ฀ = − ฀฀c in − ฀฀ V0 V ฀฀ 0 ฀฀ . From here, we have

−1000 ln 4 −

m m = 8t − 1000 ln 2 ⇒ − 1000 ln ( − 4 ) = 8t − 1000 ln 2 1000 1000

⇒ ln ( ! ! !

m m 8 t + ln 2 ⇒ − 4 = 2 ⋅ e −0.008 t! − 4) = − 1000 1000 1000

⇒ m( t ) = 4000 + 2000 ⋅ e −0.008 t grams . !

To answer the question in the problem, we find that the mass of dissolved salt in the

m( 60 ) = 4000 + 2000 ⋅ e −0.008⋅ 60 = 4000 + 2000 ⋅ e −0.48 ≈ 4000 + 2000 ⋅ 0.6188 ≈ 5238 grams .

tank after one hour is

Had the initial mass of dissolved salt in the tank been, say, m ( 0 ) = 1000 grams, we would have m0 < V0 · cin (Case I) , for which the mass function is

m( t ) = 4000 − ( 4000 − 1000) ⋅ e −0.008 t

= 4000 − 3000 ⋅ e −0.008 t grams . We find that

the mass function m ( t ) for the mixing problem where the filling and draining rates are equal is made up of a constant term, which is the stationary solution value V0 · cin , and an exponential decay term in which the difference between the initial mass and the “long-term” mass of salt dies away to zero. A graph illustrating the solutions for “Case 0”, Case I, and Case II is shown below.

If we use the concentration rate of change differential equation for this problem, we have

grams 8 liters/min. dc − c( t ) ] . Following the procedure we’ve ⋅[4 = liter 1000 liters dt

discussed above, the initial concentration of dissolved salt c0 = 6 grams/liter , with cin < c0 (Case II) , leads to the concentration function c( t ) = 4 + ( 6 − 4 ) ⋅ e

−0.008 t

grams

4 + 2 ⋅ e −0.008 t liter . After one hour ( t = 60 minutes ), the concentration of dissolved grams −0.008 ⋅ 60 ≈ 5.238 . We would then salt in the tank falls to c( 60 ) = 4 + 2 ⋅ e

liter multiply this concentration by the 1000 liters of solution in the tank to find the mass of dissolved salt as m ( 60 ) ≈ 1000 liters · 5.238 grams/liter ≈ 5238 grams , as we found previously. For an initial concentration of c0 = 1 gram/liter , we have a situation with

cin > c0 (Case I) and the concentration function is then grams = 4 − 3 ⋅ e −0.008 t liter .

c( t ) = 4 − ( 4 − 1) ⋅ e −0.008 t

The concentration function is given by the concentration of salt in the solution that is being added to the tank, which is a constant, and an exponential decay term for which the difference between that input concentration and the initial concentration of salt in the tank falls off to zero in the long run. We can understand the type of mixing problem we have been discussing in Part I as a process in which the solution that is being added to the tank is “flushing out” the brine solution already present there. After much time has passed, the original contents of the tank have been increasingly replaced so that what is found in the tank comes to resemble the solution that is being poured in. Below is a graph showing the concentration functions for “Case 0”, Case I, and Case II.

In Part II of these notes, we will discuss what happens when the rate at which solution is being added to the tank differs from the rate at which the contents of the tank are being drained.

-- G. Ruffa 14 – 28 June 2010...