Water flows at 3 m/s through a horizontal pipe where the pressure is 2.3 × 10⁵ Pa. Where the pipe constricts to a quarter of its original area, what is the pressure? A)1.6 × 10⁵ Pa B)2.2 × 10⁵ Pa C)2.3 × 10⁵ Pa D)3.1 × 10⁵ Pa

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Quizplus · Accepted Answer

This problem can be solved using Bernoulli's equation, which in one of its forms states that for an incompressible fluid flowing through a pipe, the sum of the pressure energy, kinetic energy per unit volume, and potential energy per unit volume remains constant. Since the pipe is horizontal, changes in potential energy can be ignored. The equation simplifies to $P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2$, where $P_1$ and $P_2$ are the pressures at points 1 and 2, $v_1$ and $v_2$ are the velocities at points 1 and 2, and $\rho$ is the density of the fluid. Given that the area decreases to a quarter of its original size, the velocity at point 2 ($v_2$) will be four times $v_1$ due to the continuity equation $A_1v_1 = A_2v_2$, leading to $v_2 = 4v_1 = 12 \, \text{m/s}$. Substituting the given values and solving for $P_2$ will show that the pressure decreases due to the increase in kinetic energy, resulting in the correct answer being 1.6 × 10^5 Pa.

Water Flows at 3 M/s Through a Horizontal Pipe Where