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Results

Velocity at Section 2 (v₂)

m/s

Volumetric Flow Rate Q	0.2 m³/s
A₁ · v₁	0.05 × 4
A₂	0.02 m²

What Is the Continuity Equation?

The continuity equation describes the conservation of mass for an incompressible fluid flowing steadily through a pipe or duct. It states that the volumetric flow rate entering a section must equal the flow rate leaving it: $A_1 v_1 = A_2 v_2$. When a pipe narrows, the fluid must speed up to carry the same volume; when it widens, the fluid slows down. This calculator solves for the downstream velocity $v_2$ and the volumetric flow rate $Q$.

Pipe narrowing from a wide section to a narrow section with flow arrows speeding up — As a pipe narrows, fluid speeds up so that $A_1 v_1$ equals $A_2 v_2$.

How to Use This Calculator

Enter the cross-sectional area at the first point ($A_1$) in square metres, the fluid velocity there ($v_1$) in metres per second, and the cross-sectional area at the second point ($A_2$). The tool computes the flow rate $Q = A_1 \cdot v_1$ and the velocity at the second section $$v_2 = \frac{Q}{A_2}$$ All inputs use consistent SI units, so the result is in metres per second.

The Formula Explained

Starting from $A_1 v_1 = A_2 v_2$, we first compute the volumetric flow rate $Q = A_1 \cdot v_1$ (units $\text{m}^3/\text{s}$). Because mass is conserved and the fluid is incompressible, this same $Q$ passes through section 2, so $v_2 = Q / A_2$. A smaller $A_2$ yields a larger $v_2$, which explains why water speeds up at the nozzle of a hose.

Circular pipe cross-section showing area equals pi times radius squared — Volumetric flow rate $Q$ is the cross-sectional area $A$ multiplied by velocity $v$.

Worked Example

Suppose $A_1 = 0.05\ \text{m}^2$, $v_1 = 4\ \text{m/s}$, and $A_2 = 0.02\ \text{m}^2$. The flow rate is $$Q = 0.05 \times 4 = 0.2\ \text{m}^3/\text{s}$$ The downstream velocity is $$v_2 = \frac{0.2}{0.02} = 10\ \text{m/s}$$ As the pipe area shrinks to 40% of its original size, the velocity rises by 2.5×.

FAQ

Does this work for gases? The simple form $A_1 v_1 = A_2 v_2$ assumes incompressible flow, which is a good approximation for liquids and low-speed gases. For high-speed compressible flow, density changes must be included.

What units should I use? Use consistent SI units: areas in $\text{m}^2$ and velocities in $\text{m/s}$, giving $Q$ in $\text{m}^3/\text{s}$. You can use any unit system as long as it is consistent.

Can I find $A_2$ instead? This version solves for $v_2$. To find an area, rearrange $A_2 = A_1 v_1 / v_2$.

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