Question

A tank contains 1000 L of pure water. Brine that contains 0.05 kg of salt per liter of water enters the tank at a rate of 5 L/min. Brine that contains 0.04 kg of salt per liter of water enters the tank at a rate of 10 L/min. The solution is kept thoroughly mixed and drains from the tank at a rate of 15 L/min. How much salt is in the tank (a) after minutes and (b) after one hour?

Answer 1

(a) after t minutes

If $y(t)$ is the amount of salt (in kg) after $t$ minutes, then $y(0) = 0 \ (pure\ water)$ and the total amount of liquid in the tank remains constant at 1000 L.

$\displaystyle \frac{dy}{dt} \\ = \left( 0.05 \frac{\text{kg}}{\text{L}} \right) \left( 5 \frac{\text{L}}{\text{min}} \right) + \left( 0.04 \frac{\text{kg}}{\text{L}} \right)\left( 10 \frac{\text{L}}{\text{min}} \right) - \left( \frac{y(t)}{1000} \frac{\text{kg}}{\text{L}} \right)\left( 15\frac{\text{L}}{\text{min}} \right) \\ \\ = 0.25 + 0.40 - 0.015y = 0.65 - 0.015 y = \frac{130 - 3y}{200} \frac{\text{kg}}{\text{min}}$

$\text{Hence, } \displaystyle \int \frac{dy}{130 - 3y} = \int\frac{dt}{200}$

and $-\frac{1}{3}\ln|130 - 3y| = \frac{1}{200}t + C.$

Because $y(0) = 0$, we have $-\frac{1}{3} \ln 130 = C$

so

$-\frac{1}{3}\ln|130 - 3y| = \frac{1}{200}t -\frac{1}{3} \ln 130\ \Rightarrow \\ \\\ \ln|130 - 3y| = -\frac{3}{200}t + \ln 130 \ \Rightarrow \\ \\ |130 - 3y| = e^{-3t/200 + \ln 130} \ \Rightarrow \\ \\ |130 - 3y| = 130e^{-3t/200}$

Since $y$ is continuous, $y(0) = 0$, and the right-hand side is never zero, we deduce that $130 - 3y$ is always positive.

Thus, $130 - y = 130e^{-3t/200}$ and
$y = \frac{130}{3}\left( 1 - e^{-3t/200}\right) \text{ kg}$

(b)

After one hour,

$y = \frac{130}{3}\left( 1 - e^{-3 \cdot 60/200}\right) = \frac{130}{3}(1 - e^{-0.9} ) \approx 25.7 \text{ kg}$

Answer 2

A tank contains 1000 L of pure water. Brine that contains 0.05 kg of salt per liter of water enters the tank at a rate of 5 L/min. The amount of salt in

the tank after 1 minute and after 1 hour is $\mathbf{ \dfrac{130}{3} (1- e^{\dfrac{-3t}{200}} ) \ kg}$ and 25.715 kg respectively.

Let make an assumption that:

• x should represent the amount of salt after (t) minutes
• Now, we can posit that at x(0) = 0, the tank only contains only pure water.

Now, the rate of change of the salt amount can be expressed as:

$\mathbf{\dfrac{dx}{dt}= \ rate_{(in)} -\ rate_{( out )}}$

The rate of flow of water in the inlet:

= (mass × rate of the salt in the tank)  + (mass of the brine × rate)

= (0.05 kg × 5 L/min ) + (0.04 kg  × 10 L/min)

= 0.25 kg/min + 0.4 kg/min

= 0.65 kg/min

The rate flow of water at the outlet can be computed as:

$\mathbf{= \dfrac{(x)}{1000} \times 15 \ kg/min}$

$\mathbf{= \dfrac{(x)}{200} \times 3 \ kg/min}$

$\mathbf{= \dfrac{(3x)}{200} \ kg/min}$

Recall that: the rate of change of the salt amount can be expressed as: $\mathbf{\dfrac{dx}{dt}= \ rate_{(in)} -\ rate_{( out )}}$

∴

$\mathbf{\dfrac{dx}{dt}=0.65 - \dfrac{3x}{200} }$

By differentiation;

$\mathbf{\dfrac{dx}{dt}= \dfrac{1}{200} }$

$\mathbf{\to 200\dfrac{dx}{dt}}$

∴

$\mathbf{\dfrac{200}{(130-3x)}\ dx = dt}$

By using integration on two both sides, we have:

$\mathbf{\int \dfrac{200}{(130-3x)}\ dx =\int dt}$

$\mathbf{\dfrac{-200}{(3)} \ In \ \Big| 130-3x \Big| =t+C}$

By cross multiply

$\mathbf{ (200)\ In \ \Big| 130-3x \Big| = 3t+3C}$

$\mathbf{ \ In \ \Big| 130-3x \Big| = \dfrac{3t + 3C}{(200)}}$------------ (1)

At the time when t = 0, we know that x is also = 0

∴

$\mathbf{ \ In \ \Big| 130-3(0)\Big| = \dfrac{3t + 3C}{(200)}}$

$\mathbf{ -In \ 130 = \dfrac{ 3C}{(200)}}$

$\mathbf{ In \ 130 = -\dfrac{ 3C}{(200)}}$

$\mathbf{C= \dfrac{- 200}{(3)} In \ 130 }$

Replacing the value of C into equation (1);

$\mathbf{ \ In \ \Big| 130-3x \Big| = \dfrac{3t }{(200)}-In(130)}$

$\mathbf{ \ In \ \Big| 130-3x \Big| = In(130)-\dfrac{3t }{(200)}}$

Dividing through by ㏑

$\mathbf{ \Big| 130-3x \Big| = 130 e^{-\dfrac{3t }{(200)}}}$

Making (x) the subject of the formula, we have:

$\mathbf{ x= \dfrac{130}{3} (1- e^{\dfrac{-3t}{200}} ) \ kg}$

After one hour, the amount of salt in the tank is:

$\mathbf{ x= \dfrac{130}{3} (1- e^{\dfrac{-3(60)}{200}} ) \ kg}$

$\mathbf{ x= \dfrac{130}{3} (1- e^{\dfrac{-180}{200}} ) \ kg}$

$\mathbf{ x= \dfrac{130}{3} (1- e^{-0.9} ) \ kg}$

$\mathbf{ x= \dfrac{130}{3} (0.59344 ) \ kg}$

$\mathbf{ x=25.715 \ kg}$

Therefore, we can conclude that the amount of salt in the tank after t minute and 1 hour is $\mathbf{ \dfrac{130}{3} (1- e^{\dfrac{-3t}{200}} ) \ kg}$ and 25.715 kg respectively.

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