Archimedes' buoyancy: a quick recap
A fluid at rest pushes up on any object put in it. This buoyant force equals the weight of the fluid pushed aside: F = ρ × V × g, where ρ is the fluid's density, V the volume under the surface and g = 9.8 m/s².
If the buoyant force is bigger than the object's weight, it floats. This is the same pressure idea as in moving fluids: pressure grows with depth. See Archimedes' principle for the full lesson.
Flow rate and the continuity equation
The volume flow rate Q is the volume of fluid passing a section each second: Q = A × v. A is the area of the cross-section (m²) and v the speed (m/s). Q is in m³/s (1 m³/s = 1000 L/s).
For a liquid that cannot be squeezed, nothing is created or lost along the pipe, so Q is the same everywhere:
A₁v₁ = A₂v₂ (the continuity equation).
So if the area becomes 4 times smaller, the speed becomes 4 times bigger. Remember that area depends on the radius squared: halving the diameter makes the speed 4 times bigger.
Bernoulli's equation
Bernoulli's equation is energy conservation for a flowing fluid. Along a streamline of a smooth, steady, non-sticky (ideal) fluid:
p + ½ρv² + ρgh = constant
p is pressure (pressure energy per volume), ½ρv² is the motion energy per volume, and ρgh is the height energy per volume. For a horizontal pipe h does not change, so p₁ + ½ρv₁² = p₂ + ½ρv₂²: where v is bigger, p is smaller.
Real liquids have some friction (viscosity), so the pressure also falls slowly along the pipe. Bernoulli is a very good model when the liquid is thin and the pipe is short.
The Venturi effect
Join the two ideas. In a pipe with a narrow throat, the continuity equation gives a high speed in the throat, and Bernoulli gives a low pressure there. This pressure drop is the Venturi effect.
A Venturi meter measures the pressure difference Δp between the wide part and the throat (with a U-tube or two gauge tubes). Then
v₁ = √[ 2Δp / ( ρ ( (A₁/A₂)² − 1 ) ) ], and Q = A₁v₁.
Uses: measuring flow in water pipes, carburettors that suck petrol into the air stream, sprayers and paint guns, and the jet pump.
Key formulas and definitions
- Q = A × v (m³/s; 1 L = 0.001 m³)
- A₁v₁ = A₂v₂ (continuity)
- p + ½ρv² + ρgh = constant (Bernoulli)
- Horizontal pipe: Δp = p₁ − p₂ = ½ρ(v₂² − v₁²)
- Venturi: v₁ = √[2Δp / (ρ((A₁/A₂)² − 1))]
- Buoyant force F = ρ V g (ρ water = 1000 kg/m³)
Worked examples
1. Water moves at 2 m/s through a pipe of area 20 cm². It enters a section of area 5 cm². Find the speed there.
A₁v₁ = A₂v₂. 20 × 2 = 5 × v₂, so v₂ = 40 / 5 = 8 m/s.
2. A pipe has radius 2 cm. Water flows at 3 m/s. Find the flow rate in L/s.
A = πr² = 3.14 × (0.02)² = 1.257 × 10⁻³ m². Q = A v = 1.257 × 10⁻³ × 3 = 3.77 × 10⁻³ m³/s = 3.77 L/s.
3. A tap gives 3 L/s. How long does it take to fill a 60 L drum?
Time = volume / Q = 60 / 3 = 20 s.
4. In a horizontal pipe, water (ρ = 1000 kg/m³) speeds up from 2 m/s to 8 m/s. Find the pressure drop.
Δp = ½ρ(v₂² − v₁²) = 0.5 × 1000 × (64 − 4) = 30 000 Pa = 30 kPa.
5. A Venturi meter has A₁ = 20 cm², A₂ = 5 cm² and Δp = 15 000 Pa. Water ρ = 1000 kg/m³. Find v₁ and Q.
A₁/A₂ = 4, so (A₁/A₂)² − 1 = 15. v₁ = √[2 × 15 000 / (1000 × 15)] = √2 = 1.41 m/s. Q = A₁ v₁ = 20 × 10⁻⁴ × 1.41 = 2.83 × 10⁻³ m³/s = 2.83 L/s.
6. A wooden block of volume 0.002 m³ and weight 15 N is pushed fully under water. Find the buoyant force. Will it float up?
F = ρVg = 1000 × 0.002 × 9.8 = 19.6 N. Since 19.6 N > 15 N, the net force is 4.6 N upward, so it rises and floats with about 15 / 19.6 = 77% of its volume under water.
Common mistakes
- Thinking the speed is the same everywhere in a pipe. Only the flow rate Q is the same; speed is bigger where area is smaller.
- Using diameter instead of radius, or forgetting that area depends on the square. Halving the diameter makes v four times bigger.
- Mixing units: 20 cm² = 20 × 10⁻⁴ m², not 20 × 10⁻² m².
- Saying "fast fluid has high pressure". Along a streamline, faster means lower pressure (horizontal flow).