How to Solve a Variation of the Concentrating Salt Tank Problem?

[Smed]

I have a variation on the concentrating tank problem that I'm having a bit of trouble solving. I have a tank of 10 kg of pure water at time 0. I add a time dependent concentration of salt and remove the same volume of pure water so that the tank volume never changes. Once the tank has 1 kg of salt, I stop the problem. So unlike the typical problem where I have a known initial concentration, I instead have a target final concentration, but the time of that final concentration is unknown in that it depends on the flow rate.

$$ \frac{dm_{salt}}{dt} = in - out $$
There's no salt exiting, so it becomes,
$$ \frac{dm_{salt}}{dt} = \dot{m}_{in} \frac{m_{salt}(t)}{m_{tank}} $$
Separate the variables and integrate to get
$$ln(m_{salt}) = \frac{\dot{m}_{in} t}{m_{tank}} + C$$
$$m_{salt}(t) = Ae^{\frac{\dot{m}_{in} t}{m_{tank}}}$$

Now I need an initial condition to get the particular solution, but I can't use
$$m_{salt}(0) = 0$$
and I'm not sure how to use my knowledge of the final concentration, since I don't know what time it will occur.

Your help is appreciated.

 

[vela]

Are you sure you're removing only pure water? If so, the problem is pretty trivial. You would just calculate what volume of salt solution added that contains 1 kg of salt. It has nothing to do with the tank, etc. It only depends on the concentration of the incoming solution.

 

[Smed]

Your response made me realize I was missing part of the picture. There's also the rest of the loop (including a pool) which contains all of the salt at time zero. I tried solving the problem from the perspective of the pool concentration depleting rather than the tank concentration accumulating. I then took the inverse as the rate of accumulation in the tank, which seems to give the right answer. I've attached some notes on my thought process.

$$m_{salt}(t) = m_0 \left(1-e^{\frac{\dot{m}_{in}t}{m_{system}}}\right)$$

Is this how it's supposed to be done?

Thanks.

 

[Ray Vickson]

I have a variation on the concentrating tank problem that I'm having a bit of trouble solving. I have a tank of 10 kg of pure water at time 0. I add a time dependent concentration of salt and remove the same volume of pure water so that the tank volume never changes. Once the tank has 1 kg of salt, I stop the problem. So unlike the typical problem where I have a known initial concentration, I instead have a target final concentration, but the time of that final concentration is unknown in that it depends on the flow rate.

$$ \frac{dm_{salt}}{dt} = in - out $$
There's no salt exiting, so it becomes,
$$ \frac{dm_{salt}}{dt} = \dot{m}_{in} \frac{m_{salt}(t)}{m_{tank}} $$
Separate the variables and integrate to get
$$ln(m_{salt}) = \frac{\dot{m}_{in} t}{m_{tank}} + C$$
$$m_{salt}(t) = Ae^{\frac{\dot{m}_{in} t}{m_{tank}}}$$

Now I need an initial condition to get the particular solution, but I can't use
$$m_{salt}(0) = 0$$
and I'm not sure how to use my knowledge of the final concentration, since I don't know what time it will occur.

Your help is appreciated.

I cannot make any sense of your solution, in part because you never define your terms. Exactly what are $m$, $m_{salt}(t)$, $m_{tank}$, etc.? Also, if $\dot{m}_{in}$ is a constant, your DE solution is correct, but not if $\dot{m}_{in}$ depends on $t$. Depending on exactly what your $m$-variables mean, the DE itself may not even be correct.

 

[Chestermiller]

Let C be the concentration of salt in the tank and C_in represent the concentration of salt in the inlet stream. Let V be the volume of the tank and f be the volumetric flow rate into and out of the tank. Then,
$$V\frac{dC}{dt}=fC_{in}$$
Chet