Kvant Physics Problem 19

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Problem

Fig. 1

The shape of the communicating vessels is shown in Fig. 1. In which direction will water flow through the tube connecting the vessels if the water in one of the vessels is heated?

Setup and Assumptions

The system consists of two open vessels connected at the bottom by a narrow horizontal tube filled with water. One vessel is heated while the other remains at the initial temperature. The liquid is initially at rest and both free surfaces are at atmospheric pressure $p_0$.

Let the initial density of water be $\rho_0$ at temperature $T_0$. After heating, the density in the heated vessel becomes $\rho_1 < \rho_0$, while the unheated vessel retains density $\rho_0$. The free surface levels are allowed to adjust, and the tube allows quasistatic flow.

The water is treated as incompressible on mechanical timescales except for temperature dependence of density. The flow is slow enough that hydrostatic equilibrium is maintained at each instant. Surface tension and viscous pressure losses in the tube are neglected.

Physical Principles

Hydrostatic pressure in a static liquid column is described by

$p = p_0 + \rho g h,$

where $h$ is the depth below the free surface.

Pressure equality at the connection level requires equal pressure at the same geometric point in the tube:

$p_{\text{left}} = p_{\text{right}}.$

Mass conservation relates volume changes in each vessel to density changes when temperature varies locally:

$\Delta m = 0, \quad m = \rho V.$

Thermal expansion of the liquid is described by the linear approximation

$\rho_1 = \rho_0 (1 - \beta \Delta T),$

where $\beta$ is the volumetric expansion coefficient.

Derivation

Let the initial equilibrium have both vessels at the same height $H_0$. The pressure at the bottom connection is initially

$p_b^{(0)} = p_0 + \rho_0 g H_0.$

After heating one vessel, its density becomes $\rho_1 < \rho_0$. Let the new heights be $H_1$ in the heated vessel and $H_2$ in the unheated vessel.

The pressure at the connection point from each side is

$p_b^{(h)} = p_0 + \rho_1 g H_1,$

$p_b^{(c)} = p_0 + \rho_0 g H_2.$

Hydrostatic equilibrium in the tube requires

$p_0 + \rho_1 g H_1 = p_0 + \rho_0 g H_2,$

which simplifies to

$\rho_1 H_1 = \rho_0 H_2.$

Since $\rho_1 < \rho_0$, it follows that

$H_1 > H_2.$

The total volume of liquid is conserved. If the heated vessel gains height, it must receive additional liquid volume from the unheated vessel. The flow therefore proceeds until the pressure balance condition above is satisfied, which requires liquid to move from the colder region into the heated region.

Result

The pressure balance condition leads to

$\rho_1 H_1 = \rho_0 H_2, \quad \rho_1 < \rho_0 \Rightarrow H_1 > H_2.$

Liquid flows from the unheated vessel into the heated vessel until equilibrium is reached.

$\boxed{\text{Water flows toward the heated vessel.}}$

Sanity Checks

Dimensional consistency holds since each term in $p = \rho g h$ has units of pressure $\mathrm{Pa} = \mathrm{kg,m^{-1},s^{-2}}$.

In the limit $\Delta T \to 0$, one has $\rho_1 \to \rho_0$, which yields $H_1 \to H_2$, corresponding to no net flow direction, consistent with symmetry.

In the extreme limit of strong heating where $\rho_1 \ll \rho_0$, the relation $\rho_1 H_1 = \rho_0 H_2$ implies $H_1 \gg H_2$, meaning a large volume shift into the heated side, matching the expectation that the lighter fluid column requires greater height to balance pressure.

The sign of the flow follows directly from comparing $\rho_1$ and $\rho_0$ in the hydrostatic balance equation; reversing this relation would incorrectly predict flow from hot to cold, which would violate pressure continuity at the connecting tube.