Question

ZILLDIFFEQMODAP11 3.1.021. My Notes Ask Your Teacher A tank contains 150 liters of fluid in which 10 grams of salt is dissolved. Brine containing 1 gram of salt per liter is then pumped into the tank at a rate of 5 L/min; the well-mixed solution is pumped out at the same rate. Find the number A(t) of grams of salt in the tank at time t.

Answer 1

Therefore, the number of grams of salt in the tank at time t is
`A(t) = 150-140 e^{-(t)/(30) }`

Step-by-step explanation:

Given:

Tank A contain
`V_(1) = 150` lit

Rate
`\alpha = 5 (L)/(min)`

Dissolved salt
`A = 10` gm

Salt pumped in one minute is
`4 (L)/(min)`

Salt pumped out is
`(5L)/(150L) = (1)/(30)` of initial amount added salt.

To find
`A(t)`

`(dA)/(dt) = Rate _(in) - Rate _(out)`

`A' = 5 - (A)/(30)`

`A' + (A)/(30) = 5`

Solving above equation,

`I .F = e^{\int\limits {p} \, dt }`

`y = e^{\int\limits {(1)/(30) } \, dt }`

`y = e^{(t)/(30) }`

`(Ae^{(t)/(30) } )' = 5 e^{(t)/(30) } + c`

Integrating on both side,

`Ae^{(t)/(30) } = 5 * 30 e^{(t)/(30) } +c`

Add
`e^{-(t)/(30) }` on above equation,

`A = 150 + ce^{-(t)/(30) }`

Here given in question,
`A(t=0) = 10`

`10 =150 +c`

`c = -140`

Put value of constant in above equation, and find the number of grams of salt in the tank at time t.

`A(t) = 150-140 e^{-(t)/(30) }`

Therefore, the number of grams of salt in the tank at time t is
`A(t) = 150-140 e^{-(t)/(30) }`