Physics & Mechanics

Continuity Equation Calculator

Calculate fluid velocity when passing through a pipe with changing cross-sectional areas. Based on the principle of mass conservation.

m/s
Output Velocity (v₂)
4
Volumetric Flow Rate (Q)1 m³/s
Area Ratio (A₁/A₂)2
Velocity Change100.00%

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Continuity Equation for Pipe Flow

The continuity equation is the conservation of mass applied to fluid flow. For a steady incompressible fluid, the amount of volume passing one section of a pipe each second must equal the amount passing every other section. If the pipe narrows, velocity rises; if the pipe widens, velocity falls.

This calculator solves the common two-section form of the equation. Enter the upstream area, upstream velocity, and downstream area, and it returns the downstream velocity plus the conserved volumetric flow rate.

What the Inputs Represent

  • Input Area (A1): the cross-sectional area at the first pipe section.
  • Input Velocity (v1): the average fluid velocity at the first pipe section.
  • Output Area (A2): the cross-sectional area at the second pipe section.

The calculator first computes flow rate with $Q = A_1v_1$, then divides that same flow rate by the output area to find $v_2$.

The Formula

A1v1=A2v2v2=A1v1A2\begin{aligned} A_1 v_1 = A_2 v_2 \quad \Rightarrow \quad v_2 = \frac{A_1 v_1}{A_2} \end{aligned}

Where:
A1A_1=
Cross-sectional area at point 1
v1v_1=
Fluid velocity at point 1
A2A_2=
Cross-sectional area at point 2
v2v_2=
Fluid velocity at point 2

Example Calculation

Water flows at $2,\text{m/s}$ through a pipe with a cross-sectional area of $0.5,\text{m}^2$. The pipe narrows to $0.25,\text{m}^2$.

  1. Calculate flow rate: $Q = 0.5 \cdot 2 = 1.0,\text{m}^3/\text{s}$.
  2. Divide by the new area: $v_2 = 1.0 / 0.25 = 4,\text{m/s}$.
  3. Compare velocities: the output velocity is twice the input velocity because the output area is half as large.

The velocity in the narrow section becomes $4,\text{m/s}$. The flow rate stays $1.0,\text{m}^3/\text{s}$ throughout the pipe.

Assumptions and Limits

This simplified form assumes steady flow, a single inlet and outlet, and an incompressible fluid such as water or hydraulic oil. It ignores density changes, leakage, storage, and branch pipes. For compressible gases, use the mass-flow version $\rho_1 A_1 v_1 = \rho_2 A_2 v_2$ so density changes are included.

Frequently Asked Questions

Not in this simple incompressible form. Gases can change density as pressure changes, so the mass-flow form should include density: $\rho_1 A_1 v_1 = \rho_2 A_2 v_2$.

Yes! By placing your thumb over the hose opening, you are dramatically decreasing the exit area ($A_2$). To maintain the same volumetric flow rate being supplied by the house faucet, the exit water velocity ($v_2$) must drastically increase, spraying the water much further.

The total flow entering the split must equal the sum of the flows leaving it. In simple terms: $Q_{main} = Q_1 + Q_2 + ...$, where each branch flow is area times velocity.

Use consistent units. If area is in square meters and velocity is in meters per second, flow rate is cubic meters per second. If you use square feet and feet per second, the ratio still works, but the flow rate is in cubic feet per second.