Differential-Equations · Unit 8 · Video 4 · Interactive Practice

The Mixing Tank: Salt Balance and k = r/V

IKey Formulas

Formula	Name	Where it comes from
$\frac{dx}{dt} = rC_e - r\,\frac{x}{V}$	Salt balance (amount $x$ )	Rate in minus rate out; outflow concentration $= x/V$ (well mixed)
$\frac{dC}{dt} + \frac{r}{V}\,C = \frac{r}{V}\,C_e(t)$	Concentration form	$x = CV$ with $V$ constant, then divide by $V$
$k = \frac{r}{V}$ , units $\frac{1}{\text{time}}$	Analog of the conductivity	Fraction of the tank's volume that flows out per unit time
$A = \frac{1}{\sqrt{1+(\omega/k)^2}}, \quad \phi = \tan^{-1}\frac{\omega}{k}$	Response to $C_e = \cos\omega t$	$C_1 = A\cos(\omega t - \phi)$ , from Unit 7

Key Insight: The tank equation is form 1, $y' + ky = kq_e(t)$ , term for term: $y = C$ , $q_e = C_e$ (the external input) and $k = r/V$ . Only the ratio $r/V$ matters, not $r$ or $V$ separately.

IIVisualization 1 — The Salt Balance

Salt enters at $rC_e$ and leaves at $r\,x/V$ , so $x$ climbs toward $x = VC_e$ , where the two rates balance.

t (min) 10.0 r (L/min) V (L)

IIIVisualization 2 — From Amounts to Concentration

Rewriting the salt balance in terms of $C = x/V$ turns it into form 1 with $k = r/V$ .

Step 1

Step 1 — Balance the amount $x$

\frac{dx}{dt} = \underbrace{rC_e}_{\text{rate in}} - \underbrace{r\cdot\frac{x}{V}}_{\text{rate out}}

\frac{dx}{dt} + \frac{r}{V}\,x = rC_e(t)

Step 2 — Switch to concentration

C = \frac{x}{V}, \quad x = CV \;\Rightarrow\; \frac{dx}{dt} = V\,\frac{dC}{dt}\quad(V \text{ constant})

Step 3 — Substitute, then divide by $V$

V\,\frac{dC}{dt} + \frac{r}{V}\,CV = rC_e \;\Rightarrow\; V\,\frac{dC}{dt} + rC = rC_e

\frac{dC}{dt} + \frac{r}{V}\,C = \frac{r}{V}\,C_e(t)

Step 4 — Match form 1

y' + ky = kq_e(t): \quad y = C,\quad q_e = C_e,\quad k = \frac{r}{V}

With the video's numbers, $r = 1\ \text{L/min}$ and $V = 10\ \text{L}$ give $k = \frac{1}{10}\ \text{min}^{-1}$ .

IVVisualization 3 — How Closely the Lake Follows Its Input

A daily cycle $\cos\omega t$ with $\omega = 2\pi/24$ per hour: raising $k = r/V$ shrinks $\omega/k$ , lifting $A$ and shrinking the lag.

Flow rate r (10³ m³/h) 4 Volume V (10³ m³) 30

💡 The plotted curve is the steady periodic response; the transient from the starting concentration dies out like $e^{-kt}$ .

VQuiz Questions

Problem 1 · Reading Off k and the Inflow

Given: A well-mixed tank holds $V = 50$ L. Liquid flows in and out at $r = 2$ L/min, and the incoming liquid carries $C_e = 4$ g/L of salt. Find $k$ and the rate at which salt enters.

What is k?

What is the salt inflow rate?

✅ Correct!

k = r/V = 2/50 = 0.04

per minute, and salt enters at

rC_e = 2 \cdot 4 = 8

g/min.

❌ That is $V/r$ , upside down.

V/r = 25

min is a time;

k

is the fraction of the tank flowing out per minute,

r/V

❌ Right number, wrong units.

r/V

is (L/min)/L, so

k

has units of one over time:

\text{min}^{-1}

❌ Check $k$ .

k

is the ratio

r/V

, volume per time divided by volume.

❌ That is $VC_e$ , an amount in grams. It is the salt the tank holds once its concentration reaches

C_e

, not a rate.

❌ Check the inflow. Rate in = flow rate × incoming concentration: (L/min)·(g/L) = g/min.

Show solution

The constant is the ratio of flow rate to volume:

k = \frac{r}{V} = \frac{2\ \text{L/min}}{50\ \text{L}} = 0.04\ \text{min}^{-1}.

The salt inflow rate is the liquid's flow rate times its concentration; the liters cancel:

rC_e = 2\ \tfrac{\text{L}}{\text{min}} \cdot 4\ \tfrac{\text{g}}{\text{L}} = 8\ \tfrac{\text{g}}{\text{min}}.

So $\frac{dx}{dt} + 0.04\,x = 8$ : one twenty-fifth of the tank flows out each minute.

Problem 2 · The Outflow Rate

Given: A well-mixed tank of volume $V = 20$ L has flow rate $r = 3$ L/min in and out. Right now it contains $x = 40$ g of salt. Find the rate at which salt is leaving.

✅ Correct! Rate out

= r \cdot \frac{x}{V} = 3 \cdot \frac{40}{20} = 6

g/min.

❌ That multiplies $r$ by the amount $x$ . The units are L·g/min, not a rate of salt;

r

must multiply the outflow concentration

x/V

❌ That is the concentration $x/V$ . It is in g/L; multiply by the flow rate

r

to get grams per minute.

❌ That is $k = r/V$ . It is the fraction of the tank leaving per minute; multiply it by

x

to get the salt leaving.

❌ Not quite. The well-mixed outflow has the tank's concentration

x/V

Show solution

Because the tank is well mixed, the outflow concentration is the tank concentration:

\frac{x}{V} = \frac{40\ \text{g}}{20\ \text{L}} = 2\ \text{g/L}.

The salt outflow rate is the flow rate times that concentration:

r\cdot\frac{x}{V} = 3\ \tfrac{\text{L}}{\text{min}} \cdot 2\ \tfrac{\text{g}}{\text{L}} = 6\ \tfrac{\text{g}}{\text{min}}.

Equivalently $kx$ with $k = 3/20 = 0.15\ \text{min}^{-1}$ : $0.15 \cdot 40 = 6$ g/min.

Problem 3 · Filling a Fresh-Water Tank with Brine

Given: The video's tank: $V = 10$ L of fresh water ( $x(0) = 0$ ), flow rate $r = 1$ L/min, incoming concentration $C_e = 3$ g/L. Find the concentration equation and the amount of salt after $10$ minutes.

Which equation does C satisfy?

How much salt is in the tank at t = 10 min?

✅ Correct!

C' + \tfrac{1}{10}C = \tfrac{1}{10}\cdot 3

, and

x(10) = 30(1 - e^{-1}) \approx 19.0

❌ The right side was not divided by $V$ . Dividing

V C' + rC = rC_e

V

gives

\frac{r}{V}C_e = \frac{3}{10}

on the right: the same

k

multiplies both sides.

❌ Check $k$ . Here

k = r/V = 1/10

per minute, and form 1 has the same

k

on both sides.

❌ That is the long-run amount $VC_e$ . The tank only approaches

30

g; after one time constant

1/k = 10

min it has covered

1 - e^{-1}

of the way.

❌ That is $C(10)$ in g/L. Multiply by

V = 10

L to get the amount of salt.

❌ Check $x(10)$ . Starting from

0

x

rises toward

VC_e = 30

g as

30(1 - e^{-kt})

Show solution

Here $k = \frac{r}{V} = \frac{1}{10}\ \text{min}^{-1}$ , so form 1 reads

C' + \tfrac{1}{10}C = \tfrac{1}{10}\cdot 3 = \tfrac{3}{10}.

The constant input $C_e = 3$ is a steady state, and $C(0) = 0$ , so

C(t) = 3\big(1 - e^{-t/10}\big), \qquad x(t) = VC(t) = 30\big(1 - e^{-t/10}\big).

At $t = 10$ min: $x(10) = 30(1 - e^{-1}) \approx 30(0.632) \approx 19.0$ g.

Check in amounts: $\frac{dx}{dt} + \frac{1}{10}x = 3$ gives the same $x(t)$ , and $x'(0) = 3$ g/min is the inflow rate $rC_e$ .

Problem 4 · The Polluted Lake

Given: The pollutant concentration in a lake's inflow varies about its average as $\cos\omega t$ with a $24$ -hour period, $\omega = \frac{2\pi}{24} = \frac{\pi}{12}$ per hour. The lake's flow rate and volume give $k = r/V = \frac{\pi}{12}$ per hour. Find the amplitude of the lake's response and how many hours its peaks trail the input's.

What is the amplitude A?

How long after an input peak does the response peak?

✅ Correct!

\omega/k = 1

gives

A = 1/\sqrt{2}

and

\phi = \pi/4

, a time lag of

\phi/\omega = 3

hours.

❌ The square root is missing.

A = 1/\sqrt{1 + (\omega/k)^2} = 1/\sqrt{2}

, not

1/(1 + 1)

❌ The amplitude can never exceed $1$ .

1

is the amplitude of the incoming signal;

\sqrt{2}

is the modulus of

1 + i(\omega/k)

before taking the reciprocal.

❌ Check $A$ . Compute

\omega/k

first, then

A = 1/\sqrt{1 + (\omega/k)^2}

❌ That is the phase lag $\phi$ in radians, not a time. Divide by

\omega

\phi/\omega

hours.

❌ Check the lag.

\phi = \tan^{-1}(\omega/k)

is an angle; the time lag is

\phi/\omega

Show solution

The ratio is $\frac{\omega}{k} = \frac{\pi/12}{\pi/12} = 1$ . Then

A = \frac{1}{\sqrt{1 + 1^2}} = \frac{1}{\sqrt{2}} \approx 0.71, \qquad \phi = \tan^{-1}(1) = \frac{\pi}{4}.

The response is $\frac{1}{\sqrt{2}}\cos\big(\omega t - \frac{\pi}{4}\big) = \frac{1}{\sqrt{2}}\cos\big(\omega(t - 3)\big)$ , since

\frac{\phi}{\omega} = \frac{\pi/4}{\pi/12} = 3\ \text{hours}.

To follow the input more closely, the lake needs a larger $k = r/V$ : a faster flow or a smaller volume.

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