Question

A fluid has stream line flow through a horizontal pipe of variable cross-sectional area. Then:

• A. its velocity is minimum at the narrowest part of the tube and the pressure is minimum at the widest point
• B. its velocity and pressure both are maximum at the widest point
• C. its velocity and pressure both are minimum at the narrowest point
• D. its velocity is maximum at the narrowest point and the pressure is maximum at the widest part
• E. its velocity is maximum and pressure is minimum at the narrowest point

Hint:

When dealing with streamline flow, use the continuity equation and Bernoulli's principle to analyze velocity and pressure at different points. Velocity increases when the area decreases, and pressure decreases when the velocity increases.

Answer 1

Answer: (E)

Step 1: We need to apply Bernoulli’s principle and the continuity equation to solve this problem. The continuity equation states that the mass flow rate is constant for an incompressible fluid: \[ A_1 v_1 = A_2 v_2 \] where: - \( A_1, A_2 \) are the cross-sectional areas of the pipe at two points,
- \( v_1, v_2 \) are the velocities of the fluid at the corresponding points.
This means that when the area decreases, the velocity increases. 
Hence, at the narrowest part of the pipe, the fluid's velocity will be the highest. Step 2: According to Bernoulli’s principle for an incompressible fluid: \[ P + \frac{1}{2} \rho v^2 + \rho gh = {constant} \] where: - \( P \) is the pressure,
- \( \rho \) is the fluid density,
- \( v \) is the velocity,
- \( h \) is the height (which is constant in this case, as the pipe is horizontal).
At the narrowest part of the pipe, the velocity is highest. Since the sum of the pressure and the kinetic energy (term \( \frac{1}{2} \rho v^2 \)) is constant, the pressure will be the lowest at the narrowest part of the pipe. 
Step 3: Thus, the velocity is maximum and the pressure is minimum at the narrowest part of the pipe.