Question

A closed pipe containing liquid showed a pressure \(P_1\) by gauge. When the valve is opened, pressure was reduced to \(P_2\). The speed of water flowing out of the pipe is ( \(\rho\) = density of water )

• A. \( \left[\dfrac{2(P_1 + P_2)}{\rho}\right]^{1/2} \)
• B. \( \left[\dfrac{2(P_1 - P_2)}{\rho}\right]^{1/2} \)
• C. \( \left[\dfrac{\rho}{2(P_1 - P_2)}\right]^{1/2} \)
• D. \( \left[\dfrac{\rho}{2(P_1 + P_2)}\right]^{1/2} \)

Hint:

In fluid flow problems, pressure difference is directly converted into kinetic energy of the fluid.

Answer 1

The Correct Option is B

Step 1: Applying Bernoulli’s principle.
When the valve is opened, the pressure energy of the liquid is converted into kinetic energy of flow.
Step 2: Pressure difference and velocity relation.
The pressure drop is \[ \Delta P = P_1 - P_2. \] According to Bernoulli’s equation, \[ \frac{1}{2}\rho v^2 = P_1 - P_2. \]
Step 3: Solving for velocity.
\[ v = \sqrt{\frac{2(P_1 - P_2)}{\rho}}. \]
Step 4: Conclusion.
The speed of water flowing out is \[ \left[\dfrac{2(P_1 - P_2)}{\rho}\right]^{1/2}. \]