# Mixing Problem

• February 22nd 2010, 05:48 PM
calculuskid1
Mixing Problem
A cistern (water tank) contains 200 litres of pure water. Brine (salty water) that contains 0.02 kg/L of salt is added at a rate of 3 L/min. Brine from a second source with 0.04 kg/L of salt is added at a rate of 1 L/min. Assume that the cistern is instantaneously well-mixed. The cistern is drained at a rate
of 4 L/min. If Q(t) is the amount of salt (in kg) at time t (in min), how much salt is in the cistern after t minutes?
• February 22nd 2010, 10:49 PM
CaptainBlack
Quote:

Originally Posted by calculuskid1
A cistern (water tank) contains 200 litres of pure water. Brine (salty water) that contains 0.02 kg/L of salt is added at a rate of 3 L/min. Brine from a second source with 0.04 kg/L of salt is added at a rate of 1 L/min. Assume that the cistern is instantaneously well-mixed. The cistern is drained at a rate
of 4 L/min. If Q(t) is the amount of salt (in kg) at time t (in min), how much salt is in the cistern after t minutes?

CB
• February 23rd 2010, 01:27 AM
bobey
In our problem, the level of water remains the same, as we have $3\frac{\mbox{L}}{\mbox{min}}$ and $1\frac{\mbox{L}}{\mbox{min}}$ flowing in and $4\frac{\mbox{L}}{\mbox{min}}$ flowing out.

If we call the salt level $S(t)$, then we have

$S(0)=0\,\mbox{kg}.$

To find $\frac{dS}{dt}$, we subtract the rate of salt flowing out from the rate flowing in:

$\frac{dS}{dt}=\frac{dS_{\mbox{\scriptsize{in}}}}{d t}-\frac{dS_{\mbox{\scriptsize{out}}}}{dt}.$

The rate flowing in is


\begin{aligned}
\frac{dS_{\scriptsize{\mbox{in}}}}{dt}&=0.02\frac{ \mbox{kg}}{\mbox{L}}\cdot 3\frac{\mbox{L}}{\mbox{min}}+0.04\frac{\mbox{kg}}{ \mbox{L}}\cdot 1\frac{\mbox{L}}{\mbox{min}} \\
&=0.06\frac{\mbox{kg}}{\mbox{min}}+0.04\frac{\mbox {kg}}{\mbox{min}}\\
&=0.1\frac{\mbox{kg}}{\mbox{min}}.
\end{aligned}

Since the fluid is well-mixed, the rate flowing out will be in proportion to $4\,\mbox{L}$ as the current salt level is to the entire resevoir. Therefore,


\begin{aligned}
\frac{dS_{\scriptsize{\mbox{out}}}}{dt}&=\frac{S}{ 200\,\mbox{L}}\cdot 4\frac{\mbox{L}}{\mbox{min}} \\
&=\frac{S}{50}\,\mbox{min}^{-1}.\\
\end{aligned}

For $\frac{dS}{dt}$, we obtain


\begin{aligned}
\frac{dS}{dt}&=\frac{dS_{\mbox{\scriptsize{in}}}}{ dt}-\frac{dS_{\mbox{\scriptsize{out}}}}{dt}\\
&=0.1\frac{\mbox{kg}}{\mbox{min}}-\frac{S}{50}\,\mbox{min}^{-1}.
\end{aligned}

Now, all that remains is to solve the differential equation using the initial conditions given.
• February 23rd 2010, 02:18 AM
bobey
from the above, we get the equation : ds/dt = 0.1 - s/50

thus, ds/dt + 1/50 s = 0.1 ==> this is linear first degree equation .....(1)

find the integration factor, miu(t) = exp(integration of 1/50 dt)
= exp(1/50 t)

miu(t) x (1) : exp(1/50 t) x (ds/dt + 1/50 s) = 0.1 exp(1/50 t)

d/dt[exp(1/50 t).s] = 0.1 exp(1/50 t)

d[exp(1/50 t).s] = integration of [0.1 exp(1/50 t)] dt

= 0.1 integration of [exp(1/50 t)] dt

exp(1/50 t).s = 0.1/(1/50) exp(1/50 t) + c

thus s(t) = 5 + c exp-(1/50 t) ...(2)

applying initial condition : s(0) = 0 kg in (2) :

0 = 5 + c exp-(1/50 (0))
c= -5 in (2)

thus, the solution : s(t) = 5 - 5 exp-(1/50 t)