Kvant Physics Problem 594
A heated bathhouse contains air saturated with water vapor at temperature $T_{\text{in}}$ and pressure $p_{\text{in}}$, while the outdoor air is at temperature $T_{\text{out}}$ with $T_{\text{out}} <…
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Problem
Imagine that you are in a strongly heated bathhouse (banya), while outside there is frost. Which way will the steam flow when you open the small window?
A. Varlamov
Setup and Assumptions
A heated bathhouse contains air saturated with water vapor at temperature $T_{\text{in}}$ and pressure $p_{\text{in}}$, while the outdoor air is at temperature $T_{\text{out}}$ with $T_{\text{out}} < T_{\text{in}}$ and pressure $p_{\text{out}}$ close to $p_{\text{in}}$. A small window is opened, creating a narrow connection between the two regions.
The gas inside is a mixture of dry air and water vapor; the outside air is colder and drier. The flow is treated as quasi-stationary over the short initial interval after opening. Viscous losses are neglected in the global pressure balance but not in the qualitative direction of motion. Gravity acts vertically with acceleration $g$. The gas in both regions is modeled as ideal.
The unknown is the direction of the visible steam motion through the opening immediately after the window is opened.
Physical Principles
The density of an ideal gas mixture satisfies
$$\rho = \frac{pM}{RT},$$
where $p$ is pressure, $M$ is the effective molar mass, $R$ is the universal gas constant, and $T$ is temperature.
Hydrostatic balance implies that in a gravitational field, denser fluid tends to occupy lower regions while lighter fluid occupies higher regions, producing pressure-driven circulation when two regions of different densities are connected.
Mass continuity requires that any volume flux through the opening must be balanced by an equal and opposite flux elsewhere in the opening cross-section, producing simultaneous inflow and outflow layers when density stratification is present.
The direction of net flow across the opening is determined by the sign of the pressure difference at the same height, which is governed primarily by the density difference between the two air masses.
Derivation
The density inside the bathhouse is
$$\rho_{\text{in}} = \frac{p_{\text{in}} M_{\text{in}}}{R T_{\text{in}}},$$
and the density outside is
$$\rho_{\text{out}} = \frac{p_{\text{out}} M_{\text{out}}}{R T_{\text{out}}}.$$
For typical conditions, the pressure difference between inside and outside is small compared with the temperature contrast, so $p_{\text{in}} \approx p_{\text{out}} = p$. The molar mass difference between humid and dry air is minor compared with the effect of temperature, so $M_{\text{in}} \approx M_{\text{out}} = M$.
Under these approximations,
$$\rho_{\text{in}} \approx \frac{pM}{R T_{\text{in}}}, \quad \rho_{\text{out}} \approx \frac{pM}{R T_{\text{out}}}.$$
Since $T_{\text{in}} > T_{\text{out}}$, it follows that
$$\rho_{\text{in}} < \rho_{\text{out}}.$$
At equal height in the opening, the hydrostatic pressure gradient implies that the heavier external air exerts a larger static pressure than the internal air at the same level. This produces an inward flow of cold air along the lower part of the opening.
To satisfy continuity, an outward compensating flow develops in the upper part of the opening, where the lighter warm humid air leaves the bathhouse. The visible “steam” corresponds to water vapor that condenses into aerosol upon mixing with cold air, so it traces the outward-moving warm jet.
Result
The net circulation through the window is bidirectional, with cold outside air entering at lower levels and warm humid air leaving at upper levels. The visible steam corresponds to the warm humid stream and therefore moves outward from the bathhouse.
In symbolic form, the outward mass flux of humid air dominates the visible phase transport,
$$\dot{m}_{\text{steam}} > 0 \quad \text{from inside to outside}.$$
Substituting typical values such as $T_{\text{in}} \approx 340,\text{K}$ and $T_{\text{out}} \approx 250,\text{K}$ confirms $\rho_{\text{in}}/\rho_{\text{out}} \approx T_{\text{out}}/T_{\text{in}} \approx 0.74$, consistent with outward motion of the warm humid plume.
$$\boxed{\text{Steam flows outward from the bathhouse through the upper part of the opening.}}$$
Sanity Checks
Dimensional consistency follows directly from $\rho = pM/(RT)$, giving units $\mathrm{kg,m^{-3}}$ on both sides since $[p]=\mathrm{Pa}$, $[M]=\mathrm{kg,mol^{-1}}$, and $[R]=\mathrm{J,mol^{-1},K^{-1}}$.
In the limiting case $T_{\text{in}} \to T_{\text{out}}$, the densities become equal and no preferred direction of buoyancy-driven exchange exists, reducing the flow to weak pressure equilibration with no persistent visible jet direction.
If $T_{\text{in}} \gg T_{\text{out}}$, the inequality $\rho_{\text{in}} \ll \rho_{\text{out}}$ strengthens, increasing the intensity of the outward warm jet and the inward cold compensating flow, consistent with stronger steam visibility in colder outdoor conditions.
The dominant sign of the density difference $\rho_{\text{out}} - \rho_{\text{in}}$ determines the circulation structure; a sign error in the temperature dependence of density would reverse the predicted direction of the visible plume.