Question

A vat of volume 10001000 gallons initially contains 44 lbs of salt. For t > 0 pure water is pumped into the vat at the rate of 66 gallons per minute; the perfectly stirred mixture is pumped out at the same flow rate. Derive a formula for the concentration of salt in the tank at any time t.

Answer 1

`c(t) = 0.044\cdot e ^(-0.066\cdot t)`

Explanation:

The quantity of salt inside the tank is modelled after the Principle of Mass Conservation:

Salt

`\dot m_(in, salt) - \dot m_(out, salt) = (dm_(tank,salt))/(dt)`

Water

`\dot m_(in,water) - \dot m_(out,water) = (dm_(tank,water))/(dt)`

Given that water is an incompressible fluid, the expression can be simplified into the following expression:

`\dot V_(in, water) - \dot V_(out,water) = (dV_(tank, water))/(dt)`

Both flows have the same rate and tank can be modelled as a steady state system.

`\dot V_(in, water) - \dot V_(out,water) = 0`

The expression for salt concentration in the tank is:

-
`-\dot V_(tank)\cdot c = V_(tank) \cdot (dc)/(dt)`

After some handling, the following homogeneous first-order linear differential equation is found:

`(V_(tank))/(\dot V_(tank)) \cdot (dc)/(dt) + c = 0`

Where
`c (0) = 0.044\,(lbm)/(gal)`. The solution is obtained by using Laplace transforms:

`(V_(tank))/(\dot V_(tank)) \cdot \left[s\cdot C(s) - c(0)\right] + C(s) = 0`

`\left((V_(tank))/(\dot V_(tank))\cdot s + 1\right)\cdot C(s) = (V_(tank))/(\dot V_(tank))\cdot c(0)`

`C(s) = ((V_(tank))/(\dot V_(tank))\cdot c(0) )/(\left((V_(tank))/(\dot V_(tank)) \right)\cdot \left(s + (\dot V_(tank))/(V_(tank)) \right))`

`c(t) = c(0) \cdot e^{-(\dot V_(tank))/(V_(tank))\cdot t }`

Where
`(\dot V_(tank))/(V_(tank)) = 0.066\,min^(-1)`.

The formula for the concentration of salt in the tank is:

`c(t) = 0.044\cdot e ^(-0.066\cdot t)`