riaology
Unit 3 · Lesson 6 of 6

A tank can keep its volume while changing its contents.

Clean water can become salty while equal inlet and outlet flows keep the level fixed. The mass balance must track solute separately from total liquid.

Change the input. Follow the mathematics.

What you'll be able to do
  • Set up the model and calculate its solute mass.
  • Connect the equation to the changing visual.
  • Explain the assumptions and identify a misleading interpretation.
Predict before exploring

What will change if you increase time? Predict the direction, then explain the response.

A model you can changeMove one control at a time.
Well-mixed 100 liter tank. 5 L/min enters and leaves; solute mass 77.68698398515703 g.5 L/min5 L/min100 L · well mixedSolute: 77.687 g

Concentration: 0.7769 g/L. The liquid volume stays fixed while solute accumulates.

Time (min) versus Solute mass (g). Solute mass. Selected solute mass: 77.68698 g. Outflow concentration is the instantaneous tank concentration M/100. Equal volume flows hold volume steady while solute accumulates.0049.6635099.326100Solute mass (g)Time (min)
Solute mass

solute massM(30)=100(1)(1−e−5(30)/100)=77.687 gM(30)=100(1)\left(1-e^{-5(30)/100}\right)=77.687\,\mathrm{g}

Outflow concentration is the instantaneous tank concentration M/100. Equal volume flows hold volume steady while solute accumulates.

Assumptions, units, and limits

Well-mixed 100 L tank, equal inlet/outlet volume flow, constant density/volume, initially no solute.

Slope field for mixing, with current initial-value trajectory. Segment angles account for axis scale.State (same units as main plot)Time · 0 to 100 min

Each segment gives the local rate at that time and state. The red trajectory follows the chosen initial condition; Euler uses finite tangent steps.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

01

From the situation to an equation

Solute enters at flow times inlet concentration. Well-mixed outflow carries current tank concentration, which depends on stored solute mass. Subtract these rates to obtain a linear ODE; constant volume follows from equal volume flows.

M′=qcin−qM100,M(0)=0,M=100cin(1−e−qt/100)M'=q c_{in}-q\frac M{100},\quad M(0)=0,\quad M=100c_{in}(1-e^{-qt/100})
02

Keep the reasoning visible

solute accumulation=solute in−solute out\text{solute accumulation}=\text{solute in}-\text{solute out}

Outflow concentration is the instantaneous tank concentration M/100. Equal volume flows hold volume steady while solute accumulates.

03

What makes this model useful?

Well-mixed 100 L tank, equal inlet/outlet volume flow, constant density/volume, initially no solute.

A constant inlet concentration does not mean constant tank concentration. The outlet concentration follows the evolving mixed contents.

A worked example

Calculate the starting scenario

  1. Use time = 30 min; inlet concentration = 1 g/L; flow rate = 5 L/min.
  2. M(30)=100(1)(1−e−5(30)/100)=77.687 gM(30)=100(1)\left(1-e^{-5(30)/100}\right)=77.687\,\mathrm{g}
  3. Outflow concentration is the instantaneous tank concentration M/100. Equal volume flows hold volume steady while solute accumulates. The checked quantity is solute mass.
Make the idea your own

Calculate. Compare. Explain.

Numeric checks allow 1% rounding error. Values near zero use a small absolute tolerance. Work through the setup before opening a solution.

Calculate and explain

Rebuild the starting calculation

Calculate the solute mass using time 30 min, inlet concentration 1 g/L, flow rate 5 L/min. State the meaning of its sign and units.

Checked quantity: solute mass · g. Follow the definition for finite-cutoff and numerical-model questions.

Hint

Solute enters at flow times inlet concentration. Well-mixed outflow carries current tank concentration, which depends on stored solute mass. Subtract these rates to obtain a linear ODE; constant volume follows from equal volume flows.

Explore the visual solution
Well-mixed 100 liter tank. 5 L/min enters and leaves; solute mass 77.68698398515703 g.5 L/min5 L/min100 L · well mixedSolute: 77.687 g

Concentration: 0.7769 g/L. The liquid volume stays fixed while solute accumulates.

Time (min) versus Solute mass (g). Solute mass. Selected solute mass: 77.68698 g. Outflow concentration is the instantaneous tank concentration M/100. Equal volume flows hold volume steady while solute accumulates.0049.6635099.326100Solute mass (g)Time (min)
Solute mass

solute massM(30)=100(1)(1−e−5(30)/100)=77.687 gM(30)=100(1)\left(1-e^{-5(30)/100}\right)=77.687\,\mathrm{g}

Outflow concentration is the instantaneous tank concentration M/100. Equal volume flows hold volume steady while solute accumulates.

Assumptions, units, and limits

Well-mixed 100 L tank, equal inlet/outlet volume flow, constant density/volume, initially no solute.

Slope field for mixing, with current initial-value trajectory. Segment angles account for axis scale.State (same units as main plot)Time · 0 to 100 min

Each segment gives the local rate at that time and state. The red trajectory follows the chosen initial condition; Euler uses finite tangent steps.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The controls explore variations. These written steps retain the original question's inputs.

Solute enters at flow times inlet concentration. Well-mixed outflow carries current tank concentration, which depends on stored solute mass. Subtract these rates to obtain a linear ODE; constant volume follows from equal volume flows.

M′=qcin−qM100,M(0)=0,M=100cin(1−e−qt/100)M'=q c_{in}-q\frac M{100},\quad M(0)=0,\quad M=100c_{in}(1-e^{-qt/100})
Calculate and explain

Apply the idea to a new scenario

For inlet concentration 0.5 g/L and flow 10 L/min, find solute mass after 20 min in a 100 L tank.

Checked quantity: solute mass · g. Follow the definition for finite-cutoff and numerical-model questions.

Hint

Solute enters at flow times inlet concentration. Well-mixed outflow carries current tank concentration, which depends on stored solute mass. Subtract these rates to obtain a linear ODE; constant volume follows from equal volume flows.

Explore the visual solution
Well-mixed 100 liter tank. 10 L/min enters and leaves; solute mass 43.233235838169364 g.10 L/min10 L/min100 L · well mixedSolute: 43.233 g

Concentration: 0.4323 g/L. The liquid volume stays fixed while solute accumulates.

Time (min) versus Solute mass (g). Solute mass. Selected solute mass: 43.23324 g. Outflow concentration is the instantaneous tank concentration M/100. Equal volume flows hold volume steady while solute accumulates.0024.9995049.998100Solute mass (g)Time (min)
Solute mass

solute massM(20)=100(0.5)(1−e−10(20)/100)=43.2332 gM(20)=100(0.5)\left(1-e^{-10(20)/100}\right)=43.2332\,\mathrm{g}

Outflow concentration is the instantaneous tank concentration M/100. Equal volume flows hold volume steady while solute accumulates.

Assumptions, units, and limits

Well-mixed 100 L tank, equal inlet/outlet volume flow, constant density/volume, initially no solute.

Slope field for mixing, with current initial-value trajectory. Segment angles account for axis scale.State (same units as main plot)Time · 0 to 100 min

Each segment gives the local rate at that time and state. The red trajectory follows the chosen initial condition; Euler uses finite tangent steps.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The controls explore variations. These written steps retain the original question's inputs.

Solute enters at flow times inlet concentration. Well-mixed outflow carries current tank concentration, which depends on stored solute mass. Subtract these rates to obtain a linear ODE; constant volume follows from equal volume flows.

M′=qcin−qM100,M(0)=0,M=100cin(1−e−qt/100)M'=q c_{in}-q\frac M{100},\quad M(0)=0,\quad M=100c_{in}(1-e^{-qt/100})
Challenge the interpretation

Which explanation survives the model?

Choose the statement you can justify, then use the visual to explain why the others fail.

Test it with the visual
Well-mixed 100 liter tank. 5 L/min enters and leaves; solute mass 77.68698398515703 g.5 L/min5 L/min100 L · well mixedSolute: 77.687 g

Concentration: 0.7769 g/L. The liquid volume stays fixed while solute accumulates.

Time (min) versus Solute mass (g). Solute mass. Selected solute mass: 77.68698 g. Outflow concentration is the instantaneous tank concentration M/100. Equal volume flows hold volume steady while solute accumulates.0049.6635099.326100Solute mass (g)Time (min)
Solute mass

solute massM(30)=100(1)(1−e−5(30)/100)=77.687 gM(30)=100(1)\left(1-e^{-5(30)/100}\right)=77.687\,\mathrm{g}

Outflow concentration is the instantaneous tank concentration M/100. Equal volume flows hold volume steady while solute accumulates.

Assumptions, units, and limits

Well-mixed 100 L tank, equal inlet/outlet volume flow, constant density/volume, initially no solute.

Slope field for mixing, with current initial-value trajectory. Segment angles account for axis scale.State (same units as main plot)Time · 0 to 100 min

Each segment gives the local rate at that time and state. The red trajectory follows the chosen initial condition; Euler uses finite tangent steps.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The reasoning. Solute enters at flow times inlet concentration. Well-mixed outflow carries current tank concentration, which depends on stored solute mass. Subtract these rates to obtain a linear ODE; constant volume follows from equal volume flows.

Your learning path

Can you explain the connection?

Mark complete when you can set up the calculation and explain what the visual is showing. Opening a page never completes it automatically.

PreviousNext lesson

Completion is saved on this browser. This is separate from account practice progress.