Unit 01 · Lesson 02 of 08

A useful model starts with a boundary.

Before choosing an equation, choose the question. A tank can be a container, a region of space, or a collection of water particles. Each description makes some questions easier to answer.

Choose what you are studying

What you’ll be able to do

Choose a system that matches a question.
State assumptions before calculating.
Translate a mass balance into a changing liquid level.

Before you touch the controls

If the inlet and outlet stay the same, will doubling the tank area change mass accumulation, level rise, or both?

Explore the modelChange a control. Follow the response.

Predicted liquid height0.43 m

1 mm/s level change.

Inlet2 L/sOutlet1 L/sTank area1 m²Time since starting at 0.40 m30 s

The outer dashed boundary stays fixed while water crosses it. Water density is 1,000 kg/m³; tank height is 1 m. Flow rates are prescribed independently. Stop the model when the tank empties or overflows.

Name the question first.

To predict a tank level, choose the region inside the tank as a control volume. To track a particular parcel of water, follow that material instead. A boundary can be real or imaginary; it is the surface across which you account for transfers.

Keep the useful details.

For this tank, assume constant-density water, vertical walls, a fixed horizontal area, no evaporation or leaks, and prescribed inlet and outlet rates. This is a short-time model before the tank empties or overflows. A gravity outlet would need an additional relation connecting outlet flow to water depth.

Mass accumulation becomes a level change.

Constant density cancels from the mass balance. For a vertical-sided tank, $V=Ah$ . Divide the net volume flow by area to find the rate of level change. A larger area spreads the same accumulating volume over a wider surface.

\frac{dm}{dt}=\rho(\dot V_{in}-\dot V_{out}),\qquad\frac{dh}{dt}=\frac{\dot V_{in}-\dot V_{out}}{A}

A worked example

A small imbalance, a measurable level rise.

Water enters at 2 L/s and leaves at 1 L/s. The net rate is 1 $\mathrm{L}/\mathrm{s} = 0.001$ m³/s.
For $A = 1$ m², $\frac{dh}{dt} = 0.001 \mathrm{m}/\mathrm{s}$ . At ρ = 1,000 kg/m³, $\frac{dm}{dt} = 1 \mathrm{kg}$ /s.
In 10 s at these fixed rates, the level rises 0.010 m. Mass $gained = 10 \mathrm{kg}$ .

Make the idea your own

Predict. Calculate. Question the model.

Try each problem before opening its solution. Then change a value to test whether the reasoning still holds.

01 · Calculate

Does a wider tank fill more slowly?

Water enters at 2 L/s and leaves at 1 L/s. A tank has area 2 m². Find its level rise after 60 s and its mass gain. Use ρ = 1,000 kg/m³.

A small hint

Volume gain is independent of area; height gain is not.

Explore the visual solution

Level rise after 60 s30 mm

Accumulated $volume = 60$ L; mass $gain = 60 \mathrm{kg}$ . The picture starts from a height of 0.40 m.

What if you change it?Tank area2 m²

The visual responds to your control. The worked steps below use the original question.

Original question · Step 1 of 3

Net gain in 60 s = ( $2 - 1$ ) × $60 = 60$ $\mathrm{L} = 0.060$ m³.

Carry this forward. Geometry changes how accumulation appears, not how much mass enters.

02 · Reason

Choose the boundary for a pump.

You want the pressure rise produced by a pump. Compare following a fixed parcel of water with placing a fixed boundary around the pump inlet and outlet. Which is usually simpler for steady operation?

A small hint

Ask whether you want to follow material or measure a device continuously.

Explore the visual solution

DescriptionControl volume

Useful for steady device measurements. Shaft work enters through the device boundary.

What if you change it?

The visual responds to your control. The worked steps below use the original question.

Original question · Step 1 of 3

A material boundary follows the same water; by definition no mass crosses that boundary.

Carry this forward. A system choice is useful because it simplifies the question, not because one apparatus has only one valid boundary.

03 · Challenge

How long until the tank overflows?

A tank has area 0.50 m² and 0.20 m of freeboard. Inflow is 4 L/s; outflow is 1 L/s. Estimate time to overflow. Then explain why the model ends there.

A small hint

Available volume divided by net inflow gives the time.

Explore the visual solution

Time to reach the rim33.33 s

Remaining $capacity = 100$ L. Divide by 3 L/s; the picture shows the tank at the overflow threshold.

What if you change it?Tank area; 0.20 m freeboard0.5 m²

The visual responds to your control. The worked steps below use the original question.

Original question · Step 1 of 3

Available $volume = 0.50 \times 0.20 = 0.100$ $\mathrm{m}^3 = 100$ L.

Carry this forward. A correct equation still has a domain of validity. Identify the physical event that ends it.

Your learning path

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