Problem 1: Fluid Mechanics – Inverted U-Tube Siphon
1. Analysis of the Physical Situation
When water is poured very slowly into the rubber tube, it fills the lower sections first. As the water level rises, air gets trapped in the inverted U-sections. For water to flow from one section to the next, it must overcome the hydrostatic pressure head created by the height of the inverted U-tube.
The critical condition occurs when the rising limb of every inverted U-section is filled with water up to height $h$, and the falling limb is filled with air. The water column in the main reservoir (at height $H$) must provide enough pressure to balance the sum of the pressure heads in all the individual sections.
Figure 1: Schematic showing the critical equilibrium state where water fills the rising arm of each section.
2. Mathematical Formulation
We apply the pressure balance equation. The pressure at the bottom of the main tube (point B) is due to the water column of height $H$. This pressure must support the back-pressure created by the $n$ columns of water, each of height $h$.
Let $P_0$ be the atmospheric pressure.
The pressure at the bottom of the rubber tube (Point B) is:
$$ P_B = P_0 + \rho g H $$
Moving along the tube from B to C, we encounter $n$ distinct U-sections. In the critical limiting case, water rises to height $h$ in the left arm of a section and falls to zero in the right arm (air filled). Each section contributes a pressure drop of $\rho g h$.
The pressure at the end C (open to atmosphere) is: $$ P_C = P_B – n(\rho g h) $$
Since the end C is open to the atmosphere, $P_C = P_0$. Substituting this back: $$ P_0 = (P_0 + \rho g H) – n(\rho g h) $$
3. Calculation
Simplifying the pressure balance equation: $$ \rho g H = n \rho g h $$ $$ H = n h $$
Given Values:
- $n = 5$ (Number of U-sections)
- $h = 10 \text{ cm}$ (Height of each section)
Substituting the values: $$ H = 5 \times 10 \text{ cm} $$ $$ H = 50 \text{ cm} $$