Conserve mass through a changing pipe area

Follow a volume through the pipe

A supply delivers 4 litres of liquid every second. The pipe narrows from cross-sectional area 40 cm² to 20 cm². Predict whether the narrower section carries fewer litres each second or carries the same volume at a greater speed.

This investigation specifies steady incompressible flow with no branches, leakage or accumulation between the two sections. The supply maintains the chosen volume flow rate. Liquid density is 1000 kg/m³ and both sections have the same elevation. These conditions make mass conservation a useful starting point; they do not describe every real nozzle or pump.

Distinguish three different quantities

Speed v tells how far fluid moves per time, in m/s. Volume flow Q tells how much volume passes a cross-section per time, in m³/s. Mass flow tells how much mass passes per time, in kg/s. They are related but cannot be substituted for one another.

If fluid travels distance vΔt during a short time Δt through area A, the passing volume is AvΔt. Dividing by time gives Q = Av. The model uses uniform section speeds; for a nonuniform profile, v in this volume relation is the area-average normal speed. Multiplying Q by density ρ gives mass flow ρQ.

For a steady stream with no storage or leakage, inlet mass flow equals outlet mass flow. When density is unchanged, this becomes A₁v₁ = A₂v₂. Subscripts 1 and 2 identify the inlet and outlet sections. The OpenStax fluid-dynamics reference describes volume flow and its connection to conservation of mass.

Convert the units before dividing

One litre is 0.001 m³, so 4 L/s is 0.004 m³/s. One square centimetre is 0.0001 m² because both length dimensions must be converted. Thus 40 cm² is 0.004 m² and 20 cm² is 0.002 m².

At the inlet, v₁ = 0.004/0.004 = 1 m/s. At the outlet, v₂ = 0.004/0.002 = 2 m/s. Both carry 4 L/s. Their mass flow is 1000 × 0.004 = 4 kg/s. In one second, the faster narrow-section flow occupies a longer length, but its smaller cross-section keeps the passing volume equal.

The cross-section drawing scales diameter with the square root of area. Halving area does not halve diameter. For circular sections, A = πD²/4, where D is diameter. Quartering area halves diameter and, at fixed Q, multiplies speed by four.

Test the fixed-flow comparison

Keep flow at 4 L/s and compare outlet areas 40, 20 and 10 cm². The outlet speeds are 1, 2 and 4 m/s. Capture two cases and multiply each displayed area in m² by its speed. The products should agree with the prescribed Q.

The speed bars use one 0–8 m/s scale throughout. The diagram changes cross-section while the ledger retains equal inlet and outlet flow. Pipe transition length is schematic, so the drawing cannot be used to infer a travel time through the narrowing.

At a fixed 20 cm² outlet, increase flow from 4 to 8 L/s. Inlet and outlet speeds double to 2 and 4 m/s; mass flow doubles to 8 kg/s. This change is caused by increasing the prescribed supply flow, not by changing the area alone.

Check the zero-flow limit

Set the flow control to zero. Both speeds and both mass-flow values become zero at every area. There is no moving stream to accelerate. The pressure ledger still shows 120 kPa absolute at both sections: a stationary pressurized liquid is possible. Zero flow does not imply zero pressure.

The same geometry can therefore support very different speeds depending on the boundary conditions. The control is a specified volume-flow condition, not a model of how much flow a particular household supply or pump can deliver.

Recognize when the simple comparison changes

If a real pump instead holds a pressure difference fixed, narrowing the pipe can change the total flow rate. The current widget does not predict that pump response. Likewise, flow through branches must be added across branches; it need not be equal in each branch separately.

For a compressible fluid, density may differ between sections, so mass conservation uses ρ₁A₁v₁ = ρ₂A₂v₂. In unsteady flow, material can accumulate between boundaries. These cases require a more general account. The present liquid model deliberately excludes them.

Transfer to a new flow and diameter

A steady incompressible stream carries 6 L/s through a section of area 30 cm². Find its mean speed. A downstream circular section has half the first section’s diameter. Predict its area, speed and mass flow if density is 1000 kg/m³ and no liquid is lost.

Check the continuity transfer

Q = 0.006 m³/s and A₁ = 0.003 m², so v₁ = 2 m/s. Half the diameter gives one quarter of the area: 7.5 cm² or 0.00075 m². The downstream speed is 0.006/0.00075 = 8 m/s. Both sections carry 6 kg/s. Speed increases fourfold; volume and mass flow do not.