Question

A tube of uniform diameter $D$ is immersed in a steady flowing inviscid liquid stream of velocity $V$, as shown in the figure. Gravitational acceleration is $g$. The volume flow rate through the tube is ________________.

• A. $\dfrac{\pi}{4}D^{2}V$
• B. $\dfrac{\pi}{4}D^{2}\sqrt{2gh_{2}}$
• C. $\dfrac{\pi}{4}D^{2}\sqrt{2g(h_{1}+h_{2})}$
• D. $\dfrac{\pi}{4}D^{2}\sqrt{V^{2}-2gh_{2}}$

Hint:

When a tube stagnates a moving fluid, Bernoulli’s equation is applied between the free stream and stagnation point. Pressure differences convert into measurable heights.

Answer 1

The Correct Option is D

A Pitot-type tube is placed inside a moving inviscid liquid stream. At the tube entrance, the fluid velocity reduces to zero (stagnation point). We apply Bernoulli’s equation between the free stream and the stagnation point at the tube mouth. Step 1: Apply Bernoulli between free-stream and stagnation point.
Free-stream total pressure: \[ P_{\text{free}} + \frac{1}{2}\rho V^{2}. \] Stagnation point pressure inside tube: \[ P_{\text{stagn}} = P_{\text{atm}} + \rho g h_{2}. \] Both free-stream and tube discharge are open to atmosphere, so atmospheric pressure cancels out. Thus: \[ \frac{1}{2}\rho V^{2} = \rho g h_{2}. \] Step 2: Determine velocity inside the tube.
Now apply Bernoulli between stagnation point and exit of the tube (vertical outlet at height \(h_{2}\)): \[ \frac{1}{2}\rho V_{\text{tube}}^{2} = \frac{1}{2}\rho V^{2} - \rho g h_{2}. \] Hence: \[ V_{\text{tube}} = \sqrt{V^{2} - 2gh_{2}}. \] Step 3: Find volume flow rate.
The tube has uniform diameter \(D\), so: \[ Q = A V_{\text{tube}} = \frac{\pi D^{2}}{4}\sqrt{V^{2}-2gh_{2}}. \] Step 4: Final result.
\[ Q = \frac{\pi}{4}D^{2}\sqrt{V^{2}-2gh_{2}}. \] Final Answer: (D)