Q35E

Question

Mixing. Suppose a brine containing 0.2 kg of salt per liter runs into a tank initially filled with 500 L of water containing 5 kg of salt. The brine enters the tank at a rate of 5 L/min. The mixture, kept uniform by stirring, is flowing out at the rate of 5 L/min (see Figure 2.6).

(a)Find the concentration, in kilograms per liter, of salt in the tank after 10 min. [Hint: Let A denote the number of kilograms of salt in the tank at t minutes after the process begins and use the fact that

rate of increase in A =rate of input - rate of exit.

A further discussion of mixing problems is given in Section 3.2.]

(b) After 10 min, a leak develops in the tank and an additional liter per minute of mixture flows out of the tank (see Figure 2.7). What will be the concentration, in kilograms per liter, of salt in the tank 20 min after the leak develops? [Hint: Use the method discussed in Problems 31 and 32.]

Step-by-Step Solution

Verified

Answer

(a). The concentration is 0.0281 kg/L.

(b). The concentration is 0.0598 kg/L.

1Step 1(a): Find the concentration at initial conditions

The input and output rate of the flow=5 L/min

Let A be the number of kg present in the tank at t min.

we know that the rate of increase in A = rate of input - rate of exit.

The rate of input of salt is $0.2 kg / L \times 5 L / \min = 1 kg / \min$ .

The rate of output of salt is $\frac{A (t)}{500} kg / L \times 5 L / \min = \frac{A (t)}{100} kg / \min$

Therefore the equation becomes

$\frac{dA (t)}{dt} = 1 - \frac{A (t)}{100} \frac{dA (t)}{dt} = \frac{100 - A (t)}{100} \int \frac{100 - A (t)}{100} dA (t) = \int dt - 100 \ln (100 - A (t)) = t + c \ln (100 - A (t)) = \frac{- t}{100} + C A (t) = 100 - k e^{\frac{- t}{100}}$

At time t=0 ,the salt in the tank is 5 kg. Then k.

$A (0) = 100 - k 5 = 100 - k k = 95$

At t=10

The concentration is $\frac{14.04 kg}{500 L} = 0.0281 kg / L$ .

Hence, the concentration is $0.0281 k g / L$

2Step 2(b): Evaluate the concentration

If there is a leak, we losing our salt at a faster rate and the volume changes. the tank

develops at t=10 and V(10)=500.

The differential equation is

$\frac{dV}{dt} = 5 - (5 + 1) l / \min \frac{dV}{dt} = - 1 l / \min$

By solving this V(t)=-t+c. the initial conditions are V(0)=500 then, V(t)=510-t

Now,

The rate of input of salt=1 kg/min. the rate of output =6l/min.

The concentration of salt in the tank is at time $t = \frac{A (t)}{V (t)} = \frac{A (t)}{510 - t}$ .

The rate of output of salt is $\frac{A (t)}{510 - t} kg / L \times 6 L / \min = \frac{6 A (t)}{510 - t} kg / \min$ .

Now, the differential equation is by part (a)

$\frac{dA (t)}{dt} = 1 - \frac{6 A (t)}{510 - t} A (t) = 100 - 95 e^{- 0.1} \frac{dA}{dt} + \frac{6 A}{510 - t} = 1$

The integrating factor is $e^{\int \frac{6}{510 - t} dt} = e^{- 6 \ln (510 - t)}$ .then

$A (t) (510 - t)^{- 6} = \int {(510 - t)}^{- 6} dt A (t) = \frac{510 - t}{5} + c (510 - t)^{6}$

The initial conditions

$A (t) = \frac{510 - t}{5} + c (510 - t)^{6} A (10) = 100 + c (500)^{6} c = \frac{- 95 e^{- 0.1}}{{(500)}^{6^{}}} A (t) = \frac{510 - t}{5} - \frac{95 e^{- 0.1}}{{(500)}^{6^{}}} {(510 - t)}^{6}$