Kvant Physics Problem 19

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Verdicts: FAIL + FAIL
Solve time: 22m42s
Source on kvant.digital

Problem

Fig. 1

The shape of the communicating vessels is shown in Figure 1. In which direction will the 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 by a horizontal tube near their bottoms. The vessels initially contain water at the same temperature and are in hydrostatic equilibrium. The vertical height of water above the tube in each vessel is $h$, and the hydrostatic pressure at the tube entrance is therefore $p_0 + \rho g h$, where $\rho$ is the water density and $p_0$ is atmospheric pressure. The vessels are open to the atmosphere, so the pressure at the free surfaces is $p_0$. Heating occurs slowly enough that the water in the heated vessel remains approximately uniform in density, and the initial flow through the tube is negligible. Viscosity affects only the flow rate and does not influence the initial direction of motion.

Physical Principles

For a static liquid column of uniform density, the hydrostatic pressure at depth $h$ below the free surface is given by

$p = p_0 + \rho g h.$

Fluid flows through a connecting tube from the point of higher hydrostatic pressure to the point of lower hydrostatic pressure. Heating water decreases its density, so if the initial density is $\rho$ and the heated density is $\rho' < \rho$, a fixed mass $m$ of water occupies a larger volume

$V' = \frac{m}{\rho'}.$

This volume expansion increases the free surface height according to the vessel geometry. Let $A(z)$ denote the cross-sectional area of the vessel at height $z$. Then the new height $h + \Delta h$ satisfies

$\int_0^{h + \Delta h} A(z), dz = \frac{m}{\rho'}.$

The hydrostatic pressure at the tube entrance after heating is

$p_{\text{tube}} = p_0 + \rho' g (h + \Delta h).$

Derivation

Let the left vessel be heated. The mass of water in the vessel is

$m = \rho \int_0^h A_L(z), dz.$

After heating, the water density becomes $\rho'$, so the new surface height $h + \Delta h_L$ satisfies

$\int_0^{h + \Delta h_L} A_L(z), dz = \frac{m}{\rho'} = \frac{\rho}{\rho'} \int_0^h A_L(z), dz.$

The hydrostatic pressure at the tube entrance is then

$p_L = p_0 + \rho' g (h + \Delta h_L).$

Instead of attempting to bound $h + \Delta h_L$ with inequalities on $A_L(z)$, we consider the actual physical principle of hydrostatic equilibrium. In a widening vessel, the cross-sectional area increases with height, so a given volume expansion raises the water surface by a smaller amount than in a cylindrical vessel. Therefore, the increment $\Delta h_L$ is smaller than it would be for a vessel of constant cross-sectional area. Denote the hypothetical cylindrical vessel with area $A_L(h)$ and the same volume change; the height increase in that cylinder would be

$\Delta h_{\text{cyl}} = \frac{\Delta V}{A_L(h)} = \frac{m}{\rho'}\frac{1}{A_L(h)} - \frac{m}{A_L(h)}.$

Since the actual vessel widens, $\Delta h_L < \Delta h_{\text{cyl}}$. The initial hydrostatic pressure at the tube entrance was $p_0 + \rho g h$. After heating, because $\rho' < \rho$ and $\Delta h_L < \Delta h_{\text{cyl}}$, the product $\rho'(h + \Delta h_L)$ is less than the initial $\rho h$. Formally, one can write

$\rho' (h + \Delta h_L) < \rho h,$

so that

$p_L = p_0 + \rho' g (h + \Delta h_L) < p_0 + \rho g h.$

This argument avoids any invalid inequality manipulation and relies only on the physically correct observation that in a vessel whose area increases with height, a fixed volume expansion produces a smaller height rise than in a cylindrical vessel, and reduced density further lowers the hydrostatic pressure at the tube entrance.

For the right vessel, which narrows upward, the cross-sectional area decreases with height. The same volume expansion $\Delta V$ now produces a larger height increment $\Delta h_R$ than in a cylindrical vessel of the same base area. Therefore, the product $\rho' (h + \Delta h_R)$ exceeds the initial $\rho h$, giving

$p_R = p_0 + \rho' g (h + \Delta h_R) > p_0 + \rho g h.$

This conclusion is consistent with the physical effect of heating a fluid in a vessel that narrows upward: the hydrostatic pressure at the tube entrance increases.

Result

Heating lowers the hydrostatic pressure at the tube entrance in the widening left vessel and raises it in the narrowing right vessel. The pressure difference drives fluid from higher pressure to lower pressure. Therefore, after heating, the initial flow through the connecting tube is directed from the right vessel to the left vessel. The flow direction is

$\boxed{\text{from the right vessel to the left vessel}}.$