- Upstream area A₁
- 0.1m²
- Upstream velocity v₁
- 2m/s
- Downstream area A₂
- 0.025m²
8.00m/s
Open with these values8.00m/s
Result: 8.00 m/sThe same volume passes every cross-section each second, so a narrower pipe means a faster fluid: v₂ = A₁ × v₁ ÷ A₂. Quarter the area and the velocity quadruples. Both areas must use the same unit; the answer comes back in the unit of the upstream velocity.
8.00m/s
Open with these values20.00m/s
Open with these values1.00m/s
Open with these valuesv₂ = (A₁ × v₁) ÷ A₂
| A₁, v₁, A₂ | Pipe | v₂ (m/s) |
|---|---|---|
| 0.1, 2, 0.025 | Narrows to a quarter | 8 |
| 1, 1, 1 | Unchanged | 1 |
| 0.05, 4, 0.01 | Narrows to a fifth | 20 |
| 0.2, 3, 0.6 | Widens threefold | 1 |
| 0.0314, 1.5, 0.00785 | Diameter halved | 6 |
Multiply the upstream area by the upstream velocity, then divide by the downstream area: v₂ = (A₁ × v₁) ÷ A₂. A pipe narrowing from 0.1 m² to 0.025 m² with fluid entering at 2 m/s gives 8 m/s.
The continuity equation, A₁v₁ = A₂v₂, expresses conservation of mass for an incompressible fluid: the same volume passes every cross-section each second. Because area times velocity stays constant, a smaller area has to be matched by a higher velocity.
Because the volume passing each point per second is fixed. Halve the area and the velocity doubles; cut it to a quarter and the velocity quadruples. It is the same effect that makes a partly covered hose shoot water farther.
Square metres for both areas and metres per second for the upstream velocity, which returns the downstream velocity in metres per second. The area units cancel, so any consistent pair works — but mixing m² with cm² gives a wrong answer.
For an incompressible fluid in steady flow with no leaks and no branches, so that mass is conserved between the two sections. For a compressible gas at high speed the density changes and the mass form ρ₁A₁v₁ = ρ₂A₂v₂ is needed instead.
Information, not professional advice.
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