Mixing two liquids- can you solve it?

Question

A storage tank is full of liquid A, volume V. Another liquid B is started to pump in at the rate of v vols/unit time and assumed perfectly mixed. An overfow pipe at the top maintains constant vol. in the tank Find the concentration of B at any time t from start.

This is an actual case from industry.
I was solving problems like that by calculus, in first year university, but now in industry we use ready-made tables and graphs and soon forget what we know. They should be banned.

Answer 1

This is an interesting problem. At first, I was unsure whether or not it could be solved without further assumptions, but then I read that the liquid is "assumed perfectly mixed." That's a hefty assumption, but it allows one to perform the work that is necessary to solve the problem.

First, let V(A) and V(B) represent the volume of liquids A and B present in the tank.

dV(A)/dt = -v*(V(A) / V) That is, the concentration of V(A) times the rate of pumping gives the rate of change of V(A). This is because liquid A's volume changes based on what amount of liquid A is pumped out and the percentage of liquid A in the mixture at a given time times the rate of pumping liquid out gives the rate at which liquid A is being pumped out.

dV(B)/dt = v*(V(A) / V) Clearly, the rates have to sum together to 0 because the overall volume does not change. Hence this is the rate of change for V(B).

Finding V(A) is the easier task since both rates of change are in terms of V(A) and one can thus find V(A) by solving a simple differential equation. Thus, we shall find V(A) first and use it to find V(B).

So, to find V(A), set up the equation
dV(A)/dt = -v*[V(A) / V]
V'(A) + (v / V)*V(A) = 0

use the auxiliary equation: r + v/V = 0
r = -v / V

Thus, V(A) = Ce^(-vt / V)
Notice, that C is equal to the initial value of V(A) because at time zero V(A) = C and V(A) = V, so C = V.

Putting these together, our equation for V(A) is:
V(A) = V*e^[(-vt) / V]

To find V(B), we simply use the relation V(B) = V - V(A)

V(B) = V - V*e^[(-vt) / V]

Then, to find the concentration of B (denoted [B]), we use

[B] = V(B) / V

[B] = {V - V*e^[(-vt) / V]} / V

[B] = 1 - e^[(-vt) / V]

This is your final solution. To check its accuracy somewhat superficially, we will check the values of [B] at time zero and the limit of its value as time approaches infinity. Here we notice that the value of [B] is zero at time zero as it should be and it approaches one as time approaches infinity as it should.

Thus, the final answer (giving the concentration of B as a fraction) is:

[B] = 1 - e^[(-vt) / V]

Answer 2

The concentration of B is:
1 - exp(-vt/V)
where "exp" means "e to the power of."

Here's how to derive that answer:
Let c be the concentration of B in the tank.
The rate of change in c is the net of the rate that B is entering the tank, less the rate at which B is flowing out (where both rates are expressed as a proportion of the volume of the tank per unit time).

B flows in at rate v. Expressed as a fraction of the volume of the tank, this is v/V. So this causes c to change at a rate of
v/V.

Meanwhile, B is flowing out of the tank at the rate of cv/V.
That is, liquid is flowing out at rate v/V, and a proportion c of that liquid consists of B.
This causes c to change at a rate of -cv/V.

So c is changing at a net rate of v/V - cv/V, or (v/V)(1-c).
Expressed using calculus, this becomes:
dc/dt = (v/V)(1-c)

Applying algebra:
dc/(1-c) = (v/V)dt

Integrating:
-ln(1-c) = vt/V

Multiplying both sides by -1 and exponentiating:
1-c = exp(-vt/V)

Again applying algebra:
c = 1-exp(-vt/V)

Note:
When solving differential equations, we have to apply initial conditions to find the value of the constant of integration. In this case, the constant turns out to be zero, so it is not shown in the above derivation. When t=0, c is 0, and as t increases without bound c approaches 1. This is consistent with the problem as described (i.e., the initial value of c is 0), so the constant of integration can be ignored.

Answer 3

before everything, there is no longer something you're able to be able to provide a cat to make it end laying off. It basically would not artwork that way. each and every cat sheds. Heck, even human beings shed. yet, in case you have something you're able to be able to desire to offer your cat that it would not desire, you may get a dosing syringe (like they use to offer young toddlers medicine) and basically positioned the liquid interior the back of the cat's throat. this is how I provide all my animals their medicine.

Answer 4

For responder itsverystrange: Nicely done!

Answer 5

Px/V2=r

TFTP

Answer 6

jljll