1. Initial State Analysis

Before opening the valve, the system is in equilibrium via the bottom Liquid C connection. Since Liquid C levels are initially equal:

$$ P_A = P_B \implies \rho_A g h = \rho_B g h_B $$ $$ h_B = \frac{\rho_A}{\rho_B} h $$

Since $\rho_A > \rho_B$, the column height $h_B$ is greater than $h$.

2. Determining Flow Direction

The connecting tube is at height $0.5h$ from the bottom of Liquid A. Let’s compare pressures at this valve level.

• Left Side ($P_L$): Depth is $0.5h$ below the surface of A.
  $P_L = P_{atm} + \rho_A g (0.5h)$
• Right Side ($P_R$): The valve enters Liquid B. The depth below the surface of B is $(h_B – 0.5h)$.
  $P_R = P_{atm} + \rho_B g (h_B – 0.5h)$

Comparing $P_R$ and $P_L$ using $h_B = (\rho_A/\rho_B)h$:

$$ P_R – P_L = \rho_B g \left( \frac{\rho_A}{\rho_B}h – 0.5h \right) – 0.5\rho_A g h $$ $$ P_R – P_L = \rho_A g h – 0.5\rho_B g h – 0.5\rho_A g h = 0.5 g h (\rho_A – \rho_B) $$

Since $\rho_A > \rho_B$, $P_R > P_L$. Thus, Liquid B flows from Right to Left.

3. Final State Variables

Let the interface of Liquid C shift by distance $y$: it moves down by $y$ in the Left Arm and up by $y$ in the Right Arm.

Let the top level of Liquid B in the Right Arm drop by $x$.
By volume conservation, the amount of Liquid B lost from the Right Arm must appear on top of Liquid A in the Left Arm.

• Total Length of B Lost (Right): The column bottom rises by $y$ and top drops by $x$.
  Initial Length $L_{B,initial} = h_B$.
  Final Length Right $L_{B,right} = (h_B – x) – y$. (Top at $h_B-x$, Bottom at $y$).
  Length Transferred to Left = $L_{B,initial} – L_{B,right} = h_B – (h_B – x – y) = x + y$.

So, a layer of Liquid B with height $(x+y)$ now sits on top of Liquid A in the left arm.

4. Equations of Equilibrium

We need two equations to solve for $x$ and $y$.

Equation 1: Pressure Balance at Bottom Interface ($z = -y$)
The pressure at the new lower C-interface (Left Arm) must equal the pressure at the same horizontal level in the Right Arm.

Left: Weight of B layer + Weight of A column.
Right: Weight of remaining B column + Weight of C column (height difference $2y$).

$$ \rho_B(x+y) + \rho_A h = \rho_B(h_B – x – y) + \rho_C(2y) $$

Substitute $\rho_A h = \rho_B h_B$:

$$ \rho_B(x+y) + \rho_B h_B = \rho_B h_B – \rho_B(x+y) + 2\rho_C y $$ $$ 2\rho_B(x+y) = 2\rho_C y $$ $$ \rho_B x = (\rho_C – \rho_B) y \implies y = \frac{\rho_B}{\rho_C – \rho_B}x \quad \dots(1) $$

Equation 2: Pressure Balance at Valve ($z = 0.5h$)
Flow stops when pressures at the connecting tube are equal.

Left: Pressure due to B layer + Depth in A ($0.5h – y$).
Right: Pressure due to depth in B [$(h_B – x) – 0.5h$].

$$ \rho_B(x+y) + \rho_A(0.5h – y) = \rho_B(h_B – x – 0.5h) $$

Rearranging and using $\rho_B h_B = \rho_A h$:

$$ \rho_B x + \rho_B y + 0.5\rho_A h – \rho_A y = \rho_A h – \rho_B x – 0.5\rho_B h $$ $$ 2\rho_B x – y(\rho_A – \rho_B) = 0.5h(\rho_A – \rho_B) \quad \dots(2) $$

5. Solving for x

Substitute $y$ from Eq(1) into Eq(2):

$$ 2\rho_B x – (\rho_A – \rho_B)\left[ \frac{\rho_B}{\rho_C – \rho_B}x \right] = \frac{h}{2}(\rho_A – \rho_B) $$

Divide by $\rho_B$ and isolate $x$:

$$ x \left[ 2 – \frac{\rho_A – \rho_B}{\rho_C – \rho_B} \right] = \frac{h(\rho_A – \rho_B)}{2\rho_B} $$ $$ x \left[ \frac{2\rho_C – \rho_A – \rho_B}{\rho_C – \rho_B} \right] = \frac{h(\rho_A – \rho_B)}{2\rho_B} $$ $$ x = \frac{h(\rho_A – \rho_B)(\rho_C – \rho_B)}{2\rho_B (2\rho_C – \rho_A – \rho_B)} $$

6. Solving for y

Substitute the expression for $x$ back into Equation 1:

$$ y = \frac{\rho_B}{\rho_C – \rho_B} \cdot \left[ \frac{h(\rho_A – \rho_B)(\rho_C – \rho_B)}{2\rho_B (2\rho_C – \rho_A – \rho_B)} \right] $$

Cancel $\rho_B$ and the term $(\rho_C – \rho_B)$:

$$ y = \frac{h(\rho_A – \rho_B)}{2(2\rho_C – \rho_A – \rho_B)} $$