Fluid dynamics — continuity equation and Bernoulli's principle applications

Question

Water flows through a pipe that narrows from cross-section $A_1 = 4$ cm $^2$ to $A_2 = 1$ cm $^2$ . The speed at the wider section is $v_1 = 2$ m/s and both sections are at the same height. Find the speed at the narrow section and the pressure difference between the two sections. Take $\rho = 1000$ kg/m $^3$ .

(JEE Main & NEET standard problem)

Solution — Step by Step

For an incompressible fluid in steady flow, the volume flow rate is constant:

A_1 v_1 = A_2 v_2

4 \times 2 = 1 \times v_2 \implies v_2 = \mathbf{8 \text{ m/s}}

The fluid speeds up where the pipe narrows — this is intuitive if you’ve ever put your thumb over a garden hose.

For two points at the same height ( $h_1 = h_2$ ), Bernoulli’s equation reduces to:

P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2

P_1 - P_2 = \frac{1}{2}\rho(v_2^2 - v_1^2) = \frac{1}{2}(1000)(64 - 4) = \mathbf{30{,}000 \text{ Pa}}

Pressure is higher in the wider section and lower in the narrow section. This seems counterintuitive but makes sense: the fluid accelerates from wide to narrow, and this acceleration requires a net force — provided by the pressure difference.

Why This Works

Bernoulli’s equation is energy conservation for fluids. Each term represents energy per unit volume:

$P$ = pressure energy (work done by surrounding fluid)
$\frac{1}{2}\rho v^2$ = kinetic energy per unit volume
$\rho gh$ = potential energy per unit volume

When speed increases (KE goes up), pressure must decrease to keep the total constant.

graph TD
    A["Fluid Problem"] --> B{"What's given?"}
    B -->|"Two cross-sections,<br/>one velocity"| C["Continuity: A₁v₁ = A₂v₂"]
    B -->|"Two points with<br/>speed/height/pressure"| D["Bernoulli: P + ½ρv² + ρgh = const"]
    C --> E["Find v₂"]
    E --> D
    D --> F{"Same height?"}
    F -->|"Yes"| G["P₁ + ½ρv₁² = P₂ + ½ρv₂²"]
    F -->|"No"| H["Use full Bernoulli<br/>with ρgh terms"]
    B -->|"Tank with hole"| I["Torricelli: v = √(2gh)"]

Alternative Method — Venturi Meter Application

This exact setup is a Venturi meter — used to measure flow speed. If a manometer is attached showing height difference $h_m$ of mercury:

P_1 - P_2 = (\rho_{\text{Hg}} - \rho_{\text{water}})g h_m

Equating with Bernoulli gives $v_1$ directly. JEE has asked Venturi meter problems multiple times.

For Torricelli’s theorem (speed of efflux from a tank): it’s just Bernoulli with $P_1 = P_2 = P_{\text{atm}}$ and $v_1 \approx 0$ at the tank surface. Result: $v = \sqrt{2gh}$ , same as free-fall speed. This appears in NEET nearly every year.

Common Mistake

Students apply Bernoulli’s equation to viscous or turbulent flow. Bernoulli’s equation assumes the fluid is ideal — incompressible, non-viscous, and in steady (streamline) flow. If the problem mentions viscosity or turbulence, you need the modified Bernoulli equation with energy loss terms. In JEE, always check the problem statement for the word “ideal” or “non-viscous.”

Question

Solution — Step by Step

Why This Works

Alternative Method — Venturi Meter Application

Common Mistake

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