Question

An incompressible liquid flows through a horizontal pipe having cross-sectional areas \( A \) at one end and \( 2A \) at the other end. If the pressure and velocity of the liquid at the lower cross-section are \( P \) and \( v \), then these values at the other end are

• A. \( \frac{v}{2},\, P + \frac{3}{8}\rho v^2 \)
• B. \( v,\, P + \frac{1}{8}\rho v^2 \)
• C. \( \frac{v}{4},\, P + \frac{1}{4}\rho v^2 \)
• D. \( v,\, P + \frac{1}{2}\rho v^2 \)
• E. \( 2P + \rho v^2 \)

Hint:

Always apply continuity first, then Bernoulli.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This problem applies two fundamental principles of fluid dynamics for an incompressible, non-viscous fluid: the equation of continuity and Bernoulli's principle.
Step 2: Key Formula or Approach:
1. Equation of Continuity: For an incompressible fluid, the product of the cross-sectional area (A) and the velocity (v) is constant along a pipe. \(A_1v_1 = A_2v_2\). 2. Bernoulli's Principle: For a horizontal pipe (where the height `h` is constant), the sum of pressure and kinetic energy per unit volume is constant. \(P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2\).
Step 3: Detailed Explanation:
Let's denote the "lower cross-sectional end" as end 1 and the "other end" as end 2. We are given:
At end 1 (lower cross-section): \(A_1 = A\), \(v_1 = v\), \(P_1 = P\).
At end 2 (other end): \(A_2 = 2A\).
We need to find \(v_2\) and \(P_2\). 1. Find the velocity \(v_2\) using the Equation of Continuity. \[ A_1v_1 = A_2v_2 \] \[ A \cdot v = (2A) \cdot v_2 \] \[ v_2 = \frac{A \cdot v}{2A} = \frac{v}{2} \] 2. Find the pressure \(P_2\) using Bernoulli's Principle. The pipe is horizontal, so the height terms cancel out. \[ P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2 \] Substitute the known values: \[ P + \frac{1}{2}\rho v^2 = P_2 + \frac{1}{2}\rho \left(\frac{v}{2}\right)^2 \] \[ P + \frac{1}{2}\rho v^2 = P_2 + \frac{1}{2}\rho \frac{v^2}{4} \] \[ P + \frac{1}{2}\rho v^2 = P_2 + \frac{1}{8}\rho v^2 \] Now, solve for \(P_2\): \[ P_2 = P + \frac{1}{2}\rho v^2 - \frac{1}{8}\rho v^2 \] \[ P_2 = P + \left(\frac{1}{2} - \frac{1}{8}\right)\rho v^2 = P + \left(\frac{4-1}{8}\right)\rho v^2 = P + \frac{3}{8}\rho v^2 \] So, the velocity and pressure at the other end are \(\frac{v}{2}\) and \(P + \frac{3}{8}\rho v^2\).
Step 4: Final Answer:
The values at the other end are velocity \(\frac{v}{2}\) and pressure \(P + \frac{3}{8}\rho v^2\).