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GATE 2022 ME (ME2) – Question 43

Fluid Mechanics · Bernoulli's equation and fluid acceleration · 2 marks · Multiple choice

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 represented by g. The volume flow rate through the tube is ______.

upstream-facing inlet h₁ below the free surface, atmospheric outlet h₂ above the free surface.
  1. $\frac\pi4D^2V$
  2. $\frac\pi4D^2\sqrt{2gh_2}$
  3. $\frac\pi4D^2\sqrt{2g(h_1+h_2)}$
  4. $\frac\pi4D^2\sqrt{V^2-2gh_2}$

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Correct answer: (D) $\frac\pi4D^2\sqrt{V^2-2gh_2}$

Explanation

Apply Bernoulli's equation along a streamline from the undisturbed stream to the outlet, which is open to the atmosphere at height $h_2$ above the free surface.

**Step 1: reference state.** In the free stream the fluid moves at speed $V$ at the free-surface level, with atmospheric pressure at the surface. So the total head there is
$$\frac{p_{atm}}{\rho g}+\frac{V^2}{2g}+0 .$$

**Step 2: outlet.** The jet leaves the tube at height $h_2$ with speed $v_o$ and atmospheric pressure:
$$\frac{p_{atm}}{\rho g}+\frac{v_o^2}{2g}+h_2 .$$

**Step 3: equate.**
$$\frac{V^2}{2g}=\frac{v_o^2}{2g}+h_2\;\Rightarrow\;v_o=\sqrt{V^2-2gh_2}.$$

The inlet depth $h_1$ cancels: the extra pressure $\rho gh_1$ at the inlet is balanced by the elevation it must be lifted to.

**Step 4: flow rate.**
$$Q=\frac\pi4D^2\sqrt{V^2-2gh_2}\quad\text{(option D)}.$$