Question

A fluid of density \( 800 \, \text{kg/m}^3 \) is flowing through a pipe of varying cross-sectional area. The velocity of the fluid at point A is \( 2 \, \text{m/s} \), and the velocity at point B is \( 4 \, \text{m/s} \). If the cross-sectional area at point A is \( 1 \, \text{m}^2 \), find the cross-sectional area at point B.

• A. \( 0.5 \, \text{m}^2 \)
• B. \( 1.5 \, \text{m}^2 \)
• C. \( 2.0 \, \text{m}^2 \)
• D. \( 4.0 \, \text{m}^2 \)

Hint:

In fluid dynamics, the principle of continuity ensures that the mass flow rate remains constant in an incompressible fluid. This means that if the velocity of the fluid increases, the cross-sectional area must decrease.

Answer 1

The Correct Option is A

To determine the cross-sectional area at point B, the principle of conservation of mass for an incompressible fluid, represented by the continuity equation \( A_1 V_1 = A_2 V_2 \), is applied.

In this equation:

• \( A_1 \) denotes the cross-sectional area at point A.
• \( V_1 \) represents the fluid velocity at point A.
• \( A_2 \) signifies the cross-sectional area at point B.
• \( V_2 \) indicates the fluid velocity at point B.

The provided values are:

• \( A_1 = 1 \, \text{m}^2 \)
• \( V_1 = 2 \, \text{m/s} \)
• \( V_2 = 4 \, \text{m/s} \)

The objective is to calculate \( A_2 \).

Substituting the given values into the continuity equation yields:

\( 1 \cdot 2 = A_2 \cdot 4 \)

\( 2 = 4A_2 \)

Solving for \( A_2 \):

\( A_2 = \frac{2}{4} \)

\( A_2 = 0.5 \, \text{m}^2 \)

Consequently, the cross-sectional area at point B is \( 0.5 \, \text{m}^2 \).