FluidsMedium

Bernoulli equation — tube with constriction

Water flows in a horizontal pipe. Section 1: A₁ = 0.04 m², v₁ = 2 m/s, P₁ = 1.5 × 10⁵ Pa.\nSection 2: A₂ = 0.01 m². Calculate v₂ and P₂.

Given data

A₁ = 0.04 m²v₁ = 2 m/sP₁ = 1.5×10⁵ PaA₂ = 0.01 m²ρ_H₂O = 1000 kg/m³

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What is the water velocity in the narrow section A₂ = 0.01 m²?
Write your working — the numerical result must appear in the text m/s
What is the dynamic pressure term ½ρv₁² in section 1?
What is the dynamic pressure term ½ρv₂² in section 2?
What is the pressure P₂ in the narrow section of the pipe?

Full worked solution

What is the water velocity in the narrow section A₂ = 0.01 m²?
$v_2 = \dfrac{A_1 v_1}{A_2} = \dfrac{0.04 \cdot 2}{0.01}$
$v_2 = A_1 v_1 / A_2 = 0.04 \times 2 / 0.01 = \mathbf{8\,m/s}$ . The area is reduced by a factor of 4, so velocity increases by a factor of 4 (continuity equation $A_1 v_1 = A_2 v_2$ ).
What is the dynamic pressure term ½ρv₁² in section 1?
$\tfrac{1}{2}\rho v_1^2 = \tfrac{1}{2} \cdot 1000 \cdot 4$
$\frac{1}{2}\rho v_1^2 = \frac{1}{2} \times 1000 \times 2^2 = \frac{1}{2} \times 1000 \times 4 = \mathbf{2000\,Pa}$ . This is the dynamic pressure in the wider section.
What is the dynamic pressure term ½ρv₂² in section 2?
$\tfrac{1}{2}\rho v_2^2 = \tfrac{1}{2} \cdot 1000 \cdot 64$
$\frac{1}{2}\rho v_2^2 = \frac{1}{2} \times 1000 \times 8^2 = \frac{1}{2} \times 1000 \times 64 = \mathbf{32000\,Pa}$ . The dynamic pressure is 16 times larger in the narrow section due to $v_2 = 4v_1$ .
What is the pressure P₂ in the narrow section of the pipe?
$P_2 = 150000 + 2000 - 32000$
$P_2 = P_1 + \frac{1}{2}\rho(v_1^2 - v_2^2) = 1.5\times10^5 + (2000 - 32000) = \mathbf{1.2\times10^5\,Pa}$ . Pressure drops where velocity increases — this is the Venturi effect described by Bernoulli's equation.

Result:v₂ = 8 m/s, P₂ = 1.2×10⁵ Pa. Bernoulli: where the pipe narrows, velocity increases and pressure decreases.

Bernoulli equation — tube with constriction

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