Thread: Differiential Equation, Mixing problem

1. Differiential Equation, Mixing problem

A 50 litre tank is initially filled with 10 litres pf brine solution containing 20 kg of salt. Starting from time t=0, distilled water is poured into the tank at a constant rate of 4 litres per minute. At the same time, the mixture leaves the tank at a constant rate of k^(1/2) litre per minute, where k^(1/2) >0. The time taken for overflow to occur is 20 minutes.

(a) Let Q be the amount of salt in the tank at time t minutes. Show that the rate of change of Q is given by:

dQ/dt= (-Qk^(1/2))/(10+(4-k^(1/2))t)

* k^(1/2) means square root of k,

Hence, express Q in term of t,

(b) Show that k = 4, and calculate the amount of salt in the tank at the instant outflow occurs.

(c) Sketch the graph of Q against t for 0 < t < 20

Please help me to solve. Thank you....

2. a.)

$\frac{dQ}{dt} = C(t) \times (- \sqrt{k})$

where $C(t)$ is the concentration of salt in the water at time $t$. Hence this states that the rate of salt loss is the concentration at time $t$ multiplied by the rate of loss of salty water volume. Now the concentration is simply the amount of salt $Q$ at time $t$ divided by the total volume of water in the tank at time $t$ so:
$
C(t) = \frac{Q}{V(t)} = \frac{Q}{10 +(4-\sqrt{k}) t}$

hence

$\frac{dQ}{dt} = -\frac{Q \sqrt{k}}{10 +(4-\sqrt{k}) t}$ .

b.)

Here you know the time taken for overflow is 20 minutes and you know the tank has total volume 50 litres so use the relationship for the volume that:

$V(t) = 10 +(4-\sqrt{k}) t$

put $V = 50$ and $t = 20$ and then solve for $k$.

Then substituting back into your original differential equation you should find

$\frac{dQ}{dt} = -\frac{Q}{5 + t}$

so

$\int_{Q_0}^{Q_f} \frac{1}{Q} \, \mathrm{d}Q = - \int_{0}^{20} \frac{1}{5+t} \, \mathrm{d}t$

where $Q_0$ is the initial amount of salt (20 kg) and $Q_f$ is the amount of salt when overflow occurs.

I think you can take it from there, right?

3. See attachment