Question

Water flows in a horizontal pipe. At a point where speed is $2$ m/s, pressure is $2000$ Pa. At another point, speed becomes $4$ m/s. Find pressure at second point. (Density $= 1000$ kg/m$^3$)

• A. $8000$ Pa
• B. $14000$ Pa
• C. $2000$ Pa
• D. $-4000$ Pa

Hint:

In a horizontal pipe, as the speed of the fluid increases, the pressure must decrease to satisfy Bernoulli's principle. Since the speed doubled, the kinetic energy term increased fourfold, leading to a significant drop in pressure.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
For fluid flowing through a horizontal pipe (where potential energy \(h\) is constant), we apply Bernoulli’s Principle.
It states that the sum of pressure energy and kinetic energy per unit volume remains constant along a streamline for an incompressible, non-viscous fluid.
Algebraically: \(P + \frac{1}{2}\rho v^2 = \text{constant}\).
This means that if the velocity of the fluid increases (as it enters a narrower section), the pressure must decrease accordingly.
Step 2: Key Formula or Approach:
\[ P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2 \]
Step 3: Detailed Explanation:
Given parameters:
- Initial pressure (\(P_1\)) = 2000 Pa.
- Initial speed (\(v_1\)) = 2 m/s.
- Final speed (\(v_2\)) = 4 m/s.
- Density of water (\(\rho\)) = 1000 \(\text{kg/m}^3\).
Rearrange Bernoulli's equation to solve for \(P_2\):
\[ P_2 = P_1 + \frac{1}{2}\rho(v_1^2 - v_2^2) \]
Substituting the values:
\[ P_2 = 2000 + \frac{1}{2}(1000)(2^2 - 4^2) \]
\[ P_2 = 2000 + 500(4 - 16) \]
\[ P_2 = 2000 + 500(-12) \]
\[ P_2 = 2000 - 6000 \]
\[ P_2 = -4000 \text{ Pa} \]
Note: While negative pressure is unusual in elementary problems, it indicates that the pressure at the second point is 4000 Pa below the reference pressure (often atmospheric).
Step 4: Final Answer:
The pressure at the second point is \(-4000\) Pa.