Question

Water is flowing through a horizontal pipe in stream line flow. At the narrowest part of the pipe

• A. velocity is maximum and pressure minimum.
• B. pressure is maximum and velocity minimum.
• C. both pressure and velocity are minimum.
• D. both pressure and velocity are maximum.

Hint:

Logic Tip: This phenomenon is known as the Venturi effect. It explains how atomizers, perfume sprays, and carburetor jets work: creating a narrow constriction forces the fluid to speed up, which creates a low-pressure zone that can suck in other fluids.

Answer 1

The Correct Option is A

Concept:
The flow of an ideal fluid through a pipe is governed by two fundamental principles: 1. Equation of Continuity: For an incompressible fluid, the volume flow rate is constant. $A_1v_1 = A_2v_2$. This implies that velocity is inversely proportional to the cross-sectional area ($v \propto 1/A$). 2. Bernoulli's Principle: For horizontal flow, the sum of pressure energy and kinetic energy per unit volume is constant. $P + \frac{1}{2}\rho v^2 = \text{Constant}$.
Step 1: Determine the velocity at the narrowest part.
At the narrowest part of the pipe, the cross-sectional area ($A$) is at its minimum. According to the equation of continuity ($A \cdot v = \text{Constant}$), a minimum area must correspond to a maximum velocity.
Step 2: Determine the pressure at the narrowest part.
According to Bernoulli's principle for a horizontal pipe (where the potential energy term $\rho gh$ is constant and can be ignored): $$P + \frac{1}{2}\rho v^2 = \text{Constant}$$ Since the velocity ($v$) is maximum at the narrowest part, the kinetic energy term ($\frac{1}{2}\rho v^2$) is also at its maximum. For the sum to remain constant, the pressure ($P$) must drop to its minimum value. Thus, at the narrowest part, velocity is maximum and pressure is minimum.