How to find velocity after a pipe area change
When the same incompressible flow passes from a wide section to a narrower section, it must move faster through the narrower area to carry the same volume each second.
What the formula is saying
With no branches, leaks or storage, A₁v₁ = A₂v₂. First find the shared flow Q = A₁v₁, then divide Q by A₂.
Read the symbols in plain language
- A₁
- Area 1m²
- v₁
- Velocity 1m/s
- A₂
- Area 2m²
Sort out the units first
Use m² for both areas and m/s for velocity. If given diameters, square their ratio when converting an area ratio; halving a diameter quarters its area.
Let’s solve one together
Read the given values, follow each operation, then check what the result means.
- A₁ · Area 1
- 0.04 m²
- v₁ · Velocity 1
- 1 m/s
- A₂ · Area 2
- 0.01 m²
Find the flow entering
The first section tells us the volume per second.
(0.04) × (1) = 0.04 m³/sSpread that same flow through the new area
Dividing by a smaller area produces a larger velocity.
(0.04) ÷ (0.01) = 4 m/s
Does this worked answer make sense?
Check the answer by multiplying A₂v₂. In the example 0.01 × 4 = 0.04 m³/s, equal to the inlet flow.
Now try your own values
Change a value or its unit. The same method will show your calculation, step by step.
Results update only when you calculate. The lesson example above stays unchanged.
Your turn — check your understanding
Use these new values. Work it out first, then check your answer.
- A₁ · Area 1
- 0.03 m²
- v₁ · Velocity 1
- 2 m/s
- A₂ · Area 2
- 0.02 m²
Find: velocity after a pipe area change
A hint, not the answer
With no branches, leaks or storage, A₁v₁ = A₂v₂. First find the shared flow Q = A₁v₁, then divide Q by A₂.
Use m² for both areas and m/s for velocity. If given diameters, square their ratio when converting an area ratio; halving a diameter quarters its area.
Show the full practice solution
Compare the steps with your work; revealing a solution does not mark the lesson complete.
Find the flow entering
The first section tells us the volume per second.
(0.03) × (2) = 0.06 m³/sSpread that same flow through the new area
Dividing by a smaller area produces a larger velocity.
(0.06) ÷ (0.02) = 3 m/s
Avoid the common trap
Do not keep velocity unchanged when area changes and flow is fixed. Compressible flows require density in the mass-continuity equation.
When this method applies — and when it does not
Steady incompressible flow with no branches, leakage or accumulation between the two sections. Average velocities are used.
For study and understanding, not approval of a real structure, site operation or design. Apply the correct standard, National Annex and professional review to actual engineering work.
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Sources & further reading
References open in a new tab and explain the underlying principles. The teaching text and examples here are SimpleFlick’s own; the source organisations have not endorsed this calculator.
Lesson updated: · Worked examples checked against the implemented formula; not an independent engineering certification.