Question

Water is flowing through the non-uniform horizontal pipe. The speed and pressure respectively at the extreme narrow cross section of the pipe would be

• A. maximum and maximum
• B. maximum and minimum
• C. minimum and minimum
• D. minimum and maximum

Hint:

For incompressible flow in a non-uniform pipe, apply the Continuity Equation (\( A_1v_1 = A_2v_2 \)) to relate velocity to cross-sectional area. Apply Bernoulli's Principle \( (P + \frac{1}{2}\rho v^2 + \rho gh = \text{constant}) \) to relate pressure and velocity. In a horizontal pipe, where velocity is maximum, pressure is minimum (and vice versa).

Answer 1

The Correct Option is B

Step 1: Apply the principle of conservation of mass (Continuity Equation).
For an incompressible fluid flowing through a pipe, the volumetric flow rate (\( Q \)) remains constant.
The continuity equation is given by: \[ A_1 v_1 = A_2 v_2 \] where \( A \) is the cross-sectional area and \( v \) is the average fluid velocity.
If the pipe has an extreme narrow cross-section, it means the area \( A \) is minimum at that point.
From the continuity equation, if \( A \) decreases, then \( v \) must increase to keep \( Av \) constant.
Therefore, at the extreme narrow cross-section, the speed of the water will be maximum.
Step 2: Apply Bernoulli's Principle for a horizontal pipe.
Bernoulli's principle for steady, incompressible, inviscid flow along a streamline is given by: \[ \frac{P}{\rho g} + \frac{v^2}{2g} + z = \text{constant} \] For a horizontal pipe, the elevation term \( z \) is constant and can be ignored (or cancels out).
So, Bernoulli's equation simplifies to: \[ \frac{P}{\rho g} + \frac{v^2}{2g} = \text{constant} \] This means that \( P + \frac{1}{2}\rho v^2 = \text{constant} \). This equation shows an inverse relationship between dynamic pressure \( (\frac{1}{2}\rho v^2) \) and static pressure \( P \). If one increases, the other must decrease to maintain the constant total pressure.
Step 3: Combine findings from continuity and Bernoulli's equation.
From Step 1, we found that at the extreme narrow cross-section, the speed (\( v \)) is maximum.
According to Bernoulli's principle (Step 2), if the speed (\( v \)) is maximum, then the kinetic energy term \( \frac{v^2}{2g} \) is maximum. For the sum \( \frac{P}{\rho g} + \frac{v^2}{2g} \) to remain constant, the pressure term \( \frac{P}{\rho g} \) must be minimum.
Therefore, at the extreme narrow cross-section, the pressure will be minimum. Combining both conclusions, at the extreme narrow cross-section, the speed will be maximum and the pressure will be minimum. The final answer is $\boxed{\text{2}}$.