Answer: ΔP = ½ρ v 2 ²(1 - (A 2 /A 1 )²).
- A ΔP = ½ρ v<sub>2</sub>²(1 - (A<sub>2</sub>/A<sub>1</sub>)²)
- B ΔP = ρ v<sub>2</sub>² as frequently described
- C ΔP = ρg(h<sub>1</sub>-h<sub>2</sub>) in most textbook accounts
- D ΔP = ½ρ(v<sub>2</sub>-v<sub>1</sub>)² during normal conditions
Correct answer: A. ΔP = ½ρ v<sub>2</sub>²(1 - (A<sub>2</sub>/A<sub>1</sub>)²)
Explanation: By continuity and Bernoulli: ΔP = ½ρ(v<sub>2</sub>² - v<sub>1</sub>²). Using continuity A<sub>1v</sub>_1 = A<sub>2v</sub>_2: ΔP = ½ρv<sub>2</sub>²(1-(A<sub>2</sub>/A<sub>1</sub>)²). This pressure drop can be measured to find flow rate.
Since the same volume of fluid must pass every cross-section per second (continuity), the fluid speeds up where the pipe narrows; Bernoulli's equation then says this faster-moving fluid has lower pressure - the principle behind a venturi meter and an aircraft wing's lift.
Concept context
Pressure, buoyancy, Bernoulli equation, viscosity, and surface tension.