Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

A tank has $1000 m^3$ of salt solution. The salt concentration is $10\frac{kg}{m^3}$. At time zero, salt-free water starts to flow into the tank at a rate of $10\frac{m^3}{min}$. Simultaneously salt solution flows out of the tank at $10\frac{m^3}{min}$, so that the volume of the solution in the tank is always $1000 m^3$. A mixer in the tank keeps the concentration of of salt in the entire tank constant; the concentration in the effluent is the same at the concentration in the tank. What is the concentration in the effluent as a function of time?

share|improve this question

3 Answers 3

Suppose the concentration at time $t$ is $c(t)$ so $c(0)=10 \; kg/m^3$.

You have $$c'(t) = - \frac{10 \;m^3/min}{1000 \; m^3} c(t)$$

so $c(t) = k \exp(-t/100)$ for some constant $k$ and from the starting condition $k = 10 \; kg/m^3$ so

$$c(t) = 10 \, \exp\left(\frac{-t}{100}\right) \; kg/m^3.$$

share|improve this answer
    
thanks henry for your kind help. –  user8490 Mar 20 '11 at 19:04

Every minute we remove 1% of the salt. So the concentration at t minutes is (0.99^t)*(10 kg/m^3).

share|improve this answer
    
thanks dan can you show me some calculations –  user8490 Mar 20 '11 at 14:36
    
@wali: Dan's calculations are $\frac{10 \;m^3/min}{1000 m^3}=0.01 min^{-1}$ and $1-0.01=0.99$ –  Henry Mar 20 '11 at 17:01
    
Henry, those are indeed my calculations. But as your answer shows they are incorrect, although perhaps a useful approximation! =) –  Dan Brumleve Mar 20 '11 at 19:34
    
Close enough. Yours gives a concentration of about 5.47 after an hour and mine about 5.49. –  Henry Mar 20 '11 at 23:23
    
Works because log(x+1) ~x for small x. –  Dan Brumleve Mar 20 '11 at 23:36

It is a disservice to you to accept this question as-is. I’m re-stating it in terms of parameters. (The specific values given, rather than making the problem simpler, only cloud the issue. And, as Francis Bacon said, the great barrier to progress is not error, but confusion.) In particular, you should not be left with the impression that the fact that it is pure water entering the tank has any mathematical relevance. At time 0 the tank has a certain salt concentration, and then a flow at some possibly different concentration begins to enter the tank, with uniform mixing being instantaneously and perfectly done, and flow from out of the tank occurs at an equal rate. It is obvious that the asymptotic concentration is that of the input flow. The only question is, what is the exact quantitative value of the concentration at time t?

This is a classic textbook problem in DiffyQ, and, I believe, its solution is widely available (e.g., given, if not in the textbook, in study guides). Anyway, it is known as the CSTR problem, with “CSTR” standing for “Continuous Stirred Tank-Reactor”, which in turn is an abbreviation for “Continuous-Flow Stirred Tank-Reactor” (i.e., it is the flow that is being referred to as “continuous”, not the stirring.)

Anyway, the problem should be stated, and solved (I’m hope I’m not being inappropriately dogmatic here) as follows:

A tank contains a saline solution of volume V. At time 0, a saline solution of (a possibly different) constant concentration w begins flowing into the tank at a constant rate q. Mixing is done instantaneously and perfectly, such that the concentration of salt in the tank is uniform at all times. Flow from the tank also starts at time 0 at the same rate q. Let f(t) be the concentration in the tank at time t. Find an expression for f(t), for t ≥ 0.

Solution: The hardest step is the first step, that is knowing to do mass-balance on the tank. That is, salt is accumulating (or disappearing, as the case may be – mathematically it makes no difference) at the rate qw – qf(t). So:

f’(t) = (qw – qf(t))/V (mass divided by volume giving concentration, of course)

So,

f’ = (qw – qf)/V (dropping the argument, expression everything in terms of functions)

So,

f’/(f – w) = -q/V

So,

(log..(f – w))’ = -q/V (by the well-know application of the Chain Rule to the logarithm)

So,

log..(f – w) = -qi/V) + c (with “i” denoting the identity function)

So,

f – w = exp(-qi/V + c)

So,

f = w + Bexp(-qi/V) (where B = exp(c))

So,

f = w + (f(0) – w)exp(-qi/V) (having solved for B by setting t = 0)

Done. Now, plug in whatever numbers you wish.

Here is the link to the Wikipedia article on the CSTR:

http://en.wikipedia.org/wiki/Continuous_stirred-tank_reactor

And here is a link to an article that gives an in-depth discussion of the mathematics involved in multi-faceted related issues:

http://jbrwww.che.wisc.edu/home/jbraw/chemreacfun/ch4/slides-matbal.pdf

Also, the CSTR problem can be regarded as a generalization of Newton’s Law of Cooling, as I mentioned here at MSE, in connection with a question I was asking. Here is the link:

Is Newton’s Law of Cooling as special case of the CSTR problem?

Regards,

Mike Jones

29.May.2011 (Beijing time)

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.