Why Laundry Dries at 5 °C — Diffusion-Limited Evaporation vs. Heat-Driven Boiling

Far-field humidity sets the droplet lifetime when diffusion limits the rate. Once S crosses 2, the Lee coefficient turns the temperature trace into a sawtooth.

Laundry hung on a balcony dries. The air is at 5 °C. Water boils at 100 °C. Ninety-five degrees are missing, and the water leaves anyway.

The same water inside a kettle waits until 100 °C. Both cases are phase change: liquid becoming vapor. But different things set the rate. The same is true in CFD. Use one source term for both situations and one of them will be wrong.

The two rates do not share a source term. The equations below cover evaporation limited by diffusion and evaporation limited by heat, then both move into the Navier–Stokes equations. The last part is how to pick the Lee model's relaxation coefficient $r$ — and why published values scatter across seven orders of magnitude.

Raise RH from 30% to 95% and the lifetime line lies down; raise temperature and the same 1 mm drop shrinks to a few seconds.

RH30 %T25 °Cd₀1.00 mm

What to watch: the disk radius is d(t)/2 from the same d(t) the cyan trace draws, so the plot and the drop are one object. Push RH from 30 % to 95 % — B_M collapses, the amber line flattens, and the lifetime jumps from minutes to the better part of an hour. Raise T instead and Y_s runs away from Y_∞, so the same drop empties in seconds.

The slope on the right is $B_M$ , and the disk on the left uses that same $d(t)$ . When humidity climbs toward $Y_s$ , the lifetime stretches toward an hour.

The two evaporations have different bottlenecks#

Evaporation needs two things: enough energy for a molecule to leave the interface, and somewhere for that molecule to go once it has left. Which one is in short supply decides the regime.

Diffusion-limited. The gas just above the interface is already saturated. Further evaporation requires that vapor to be carried away. The bottleneck is diffusion on the gas side. Liquid temperature barely moves. Puddles, wet laundry, and room-temperature fuel sprays live here.

Thermally driven. The liquid near the interface has already passed saturation temperature. There is plenty of room for the vapor to go. The bottleneck is the heat supply that pays for the latent heat (the hidden energy cost of the phase change). Boiling, cavitation, and wall condensation live here. The stretch of the boiling curve where a hotter wall carries less heat sits in this regime.

The two regimes are even measured by different dimensionless groups. Diffusion-limited evaporation is scaled by the Spalding mass transfer number $B_M$ (the vapor concentration drop between interface and far field); heat-driven evaporation by the Jakob number $\mathrm{Ja} = c_p \Delta T / L$ (sensible over latent heat). The first pairs with the Schmidt number $\mathrm{Sc} = \nu/D$ , the second with the Prandtl number.

Diffusion-limited — the concentration gradient sets the rate#

Write the vapor mass flux leaving a spherical droplet. Fick's law (diffusion proportional to the concentration gradient) alone is not enough, because the escaping vapor also pushes the whole gas mixture outward. That convective piece is called Stefan flow.

\dot{m}'' = -\rho_g D \frac{\partial Y_v}{\partial r}\bigg|_{s} + \dot{m}'' Y_{v,s}

$\dot{m}''$ is evaporation rate per unit area, $\rho_g$ gas density, $D$ the vapor diffusivity, $Y_v$ the vapor mass fraction, and subscript $s$ the interface. The second term on the right is the share Stefan flow carries back.

Integrate radially under the quasi-steady assumption and a logarithm falls out.

\dot{m}'' = \frac{\rho_g D}{r_s}\ln(1 + B_M), \qquad B_M = \frac{Y_s - Y_\infty}{1 - Y_s}

$r_s$ is droplet radius, $Y_s$ the saturated mass fraction at the interface, $Y_\infty$ the far-field value. $B_M$ is the Spalding mass transfer number — the whole driving concentration drop compressed into one number.

Attach droplet mass conservation, $\dot{m} = -\frac{d}{dt}(\rho_l \pi d^3/6)$ , and the square of the diameter shrinks linearly.

\frac{d(d^2)}{dt} = -K, \qquad K = \frac{8\rho_g D}{\rho_l}\ln(1 + B_M)

That is the d²-law. It is $d^2$ , not $d$ , that plots as a straight line. The lifetime is $t_{life} = d_0^2 / K$ .

$Y_s$ depends only on temperature. Water at 25 °C has a saturation pressure of 3.17 kPa — 3 % of atmospheric — giving $Y_s \approx 0.019$ . Not zero. That is why things dry. At 5 °C it drops to $Y_s \approx 0.0054$ , still not zero. The boiling point never enters the story.

Thermally driven — the heat that arrives pays the latent bill#

Now the interface is pinned at saturation temperature. What sets the rate is not concentration but the imbalance of heat arriving at the interface.

\dot{m}'' L = k_l \frac{\partial T}{\partial n}\bigg|_l - k_v \frac{\partial T}{\partial n}\bigg|_v

$L$ is latent heat, $k_l$ and $k_v$ the liquid and vapor conductivities, $n$ the interface normal. Whatever heat comes in from the liquid side and does not leave on the vapor side goes entirely into phase change.

This is the Stefan condition, and it is awkward for a specific reason: the interface carries two conditions at once. Temperature is fixed at $T = T_{sat}$ (Dirichlet), and simultaneously the flux jump sets the interface velocity. Two conditions, and the boundary location is itself unknown. It is a free-boundary problem.

Solving it honestly with a sharp interface means tracking that interface and satisfying the equation above every step. Most commercial and open-source solvers avoid that road. They let the interface smear over one or two cells and insert a volumetric source term designed so the condition above is satisfied on its own.

Where the source term actually goes#

Define one volumetric evaporation rate $\dot{m}'''$ in kg/m³/s, and it appears in four places.

It enters the phasic continuity equations as a sign-flipped pair.

\frac{\partial (\alpha_l \rho_l)}{\partial t} + \nabla\cdot(\alpha_l \rho_l \mathbf{u}) = -\dot{m}''' , \qquad \frac{\partial (\alpha_v \rho_v)}{\partial t} + \nabla\cdot(\alpha_v \rho_v \mathbf{u}) = +\dot{m}'''

$\alpha$ is volume fraction. Add the two and the sources cancel, so mixture mass is conserved.

In the energy equation it appears as a sink worth the latent heat. If your solver also transports a vapor mass fraction (the diffusion-limited path), the same value shows up there as a source.

\frac{\partial (\rho e)}{\partial t} + \nabla\cdot(\rho \mathbf{u} h) = \nabla\cdot(k\nabla T) - \dot{m}''' L

The fourth place is the one people forget. Phase change changes the divergence constraint on the velocity field.

\nabla\cdot\mathbf{u} = \dot{m}''' \left( \frac{1}{\rho_v} - \frac{1}{\rho_l} \right)

An incompressible solver's pressure Poisson equation usually has zero on the right. With phase change it does not. Water turning to steam at atmospheric pressure expands roughly 1600-fold. Insert the evaporation rate but drop this term and mass disappears while nothing gets pushed aside: the bubble never grows and the pressure field goes strange.

Choosing $r$ — the trap in the Lee model#

The most widely used source term is the Lee model.

\dot{m}''' = r\,\alpha_l \rho_l \frac{T - T_{sat}}{T_{sat}}

$r$ is the relaxation coefficient, in 1/s, and this is where the misunderstanding starts. $r$ is not a material property. It is not something you can measure. Its only job is to hold the interface cell's temperature at $T_{sat}$ . Once the temperature is pinned there, that cell's energy balance becomes the Stefan condition. In other words, $r$ is a numerical device that enforces the free-boundary condition of the previous section by penalty.

So in theory, bigger is better. The trouble appears when you integrate explicitly. Isolate the source term and it is linear decay in $T - T_{sat}$ with decay rate $r L/(T_{sat} c_p)$ . The explicit Euler stability condition is then

S = r\,\frac{L}{T_{sat}\,c_p}\,\Delta t < 2

For water and steam, $L/(T_{sat} c_p) \approx 1.43$ . At $\Delta t = 10^{-5}$ s, $r$ cannot exceed $1.4\times10^{5}$ . Refine the mesh, shrink $\Delta t$ , and the admissible $r$ grows. That is why published values of $r$ range from 0.1 to $10^7$ : every author used a different $\Delta t$ .

Property-based alternatives do exist. The Hertz–Knudsen–Schrage relation starts from kinetic theory and expresses the interfacial rate through a single accommodation coefficient $\sigma$ . The Tanasawa model linearizes it into $\dot{m}'' \propto \sigma (T - T_{sat})$ . Since $\sigma$ is a property, it has measured values — though reported values for water spread from 0.01 to 1, so the uncertainty moves rather than disappears.

Both regimes in one Python script#

On one side humidity decides the answer; on the other, the time step does. Check both in the same script.

import math
 
M_V, M_A, P_ATM = 18.015, 28.96, 101325.0
RHO_L, RHO_G, D_AB = 997.0, 1.18, 2.5e-5     # water, humid air, vapor diffusivity
L_VAP, CP_L, T_SAT = 2.26e6, 4220.0, 373.15
 
def p_sat(t_c):
    """Antoine equation (water, mmHg) -> Pa"""
    return 10 ** (8.07131 - 1730.63 / (233.426 + t_c)) * 133.322
 
def mass_fraction(x):
    """vapor mole fraction -> mass fraction"""
    return x * M_V / (x * M_V + (1 - x) * M_A)
 
def spalding_number(t_c, rh):
    """Spalding mass transfer number B_M = (Y_s - Y_inf) / (1 - Y_s)"""
    x_s = p_sat(t_c) / P_ATM
    y_s = mass_fraction(x_s)
    y_inf = mass_fraction(x_s * rh)
    return (y_s - y_inf) / (1 - y_s)
 
def d2_lifetime(d0_mm, t_c, rh):
    """d^2-law lifetime t = d0^2 / K,  K = 8 rho_g D / rho_l * ln(1 + B_M)"""
    b_m = spalding_number(t_c, rh)
    k = 8 * RHO_G * D_AB / RHO_L * math.log(1 + b_m)     # m^2/s
    return (d0_mm * 1e-3) ** 2 / k, k * 1e6              # s, mm^2/s
 
def lee_source(r, alpha_l, temp):
    """Lee model volumetric evaporation rate [kg/m^3/s]"""
    if temp > T_SAT:
        return r * alpha_l * RHO_L * (temp - T_SAT) / T_SAT
    return 0.0
 
print("diffusion-limited - 1 mm water droplet, 25 degC")
for rh in (0.0, 0.3, 0.6, 0.9, 0.97):
    life, k = d2_lifetime(1.0, 25.0, rh)
    print(f"  RH {rh*100:4.0f} %   B_M = {spalding_number(25.0, rh):.5f}"
          f"   K = {k:.2e} mm^2/s   lifetime = {life/60:7.2f} min")
 
print("\nthermally driven - Lee coefficient vs explicit stability limit")
dt = 1e-5
for r in (1e2, 1e3, 1e4, 1e5, 1e6):
    s = r * L_VAP / (T_SAT * CP_L) * dt
    rate = lee_source(r, 1.0, T_SAT + 1.0)
    print(f"  r = {r:8.0e}   m''' (1 K superheat) = {rate:9.2e} kg/m^3/s"
          f"   S = {s:8.3f}  {'diverges' if s > 2 else 'stable'}")

The output:

diffusion-limited - 1 mm water droplet, 25 degC
  RH    0 %   B_M = 0.02001   K = 4.69e-03 mm^2/s   lifetime =    3.55 min
  RH   30 %   B_M = 0.01406   K = 3.30e-03 mm^2/s   lifetime =    5.04 min
  RH   60 %   B_M = 0.00806   K = 1.90e-03 mm^2/s   lifetime =    8.77 min
  RH   90 %   B_M = 0.00202   K = 4.78e-04 mm^2/s   lifetime =   34.85 min
  RH   97 %   B_M = 0.00061   K = 1.44e-04 mm^2/s   lifetime =  115.98 min
 
thermally driven - Lee coefficient vs explicit stability limit
  r =    1e+02   m''' (1 K superheat) =  2.67e+02 kg/m^3/s   S =    0.001  stable
  r =    1e+03   m''' (1 K superheat) =  2.67e+03 kg/m^3/s   S =    0.014  stable
  r =    1e+04   m''' (1 K superheat) =  2.67e+04 kg/m^3/s   S =    0.144  stable
  r =    1e+05   m''' (1 K superheat) =  2.67e+05 kg/m^3/s   S =    1.435  stable
  r =    1e+06   m''' (1 K superheat) =  2.67e+06 kg/m^3/s   S =   14.352  diverges

Sweep $r$ on the heated column and the stability column in the table moves with it.

log₁₀ r2.0e+3ΔT_wall10 K

What to watch: the amber shaded area is superheat the model failed to convert into vapor. Drag r down to 10² and the area swells while the interface stalls — the wall keeps pouring in heat and nothing boils. Push past 10⁵ and S crosses 2: the profile starts ringing and the history trace turns into a sawtooth. Green sits in between — roughly 10³ to 10⁵ here — and that window moves whenever you change Δt or the mesh.

Drop $r$ to $10^2$ and the amber band — superheat that never became vapor — swells while the interface stalls. The wall keeps adding heat and nothing boils. Past $10^5$ , $S$ crosses 2, the temperature rings, and the history trace turns into a sawtooth. Green covers about two decades, and that window shifts bodily when $\Delta t$ or the mesh changes.

The dry row of the table, 1 mm at 25 °C and RH 0 %, prints $B_M=0.02001$ , $K=4.69\times10^{-3}$ mm²/s, lifetime 3.55 min. With $d_0^2=1$ mm² the lifetime is $1/K$ seconds: $1/0.00469\approx213$ s, and 213/60 is 3.55 min. RH 30 % is 5.04 min with $K=3.30\times10^{-3}$ . RH 97 % is 115.98 min with $K=1.44\times10^{-4}$ . The lifetime ratio 115.98/5.04 is about 23.0, and the inverse $K$ ratio $3.30\times10^{-3}/1.44\times10^{-4}$ is about 22.9. Temperature stays at 25 °C. Between 30 % and 97 % humidity the lifetime spreads by a factor of 23. When a diffusion-limited case misses the experiment, the far-field humidity boundary is usually why.

The $S$ column is the other regime. At $r=10^5$ , $S=1.435$ ; at $r=10^6$ , $S=14.352$ . The limit of 2 sits between them. With $L/(T_{sat} c_p)\approx1.43$ and $\Delta t=10^{-5}$ , $2/(1.43\times10^{-5})\approx1.4\times10^{5}$ . The code's $r=10^5$ is inside that cap and $10^6$ is ten times past it. At 1 K of superheat, $\dot{m}'''$ scales with $r$ from $2.67\times10^{2}$ to $2.67\times10^{6}$ . The $\rho_g=1.18$ inside the d²-law is humid air. The factor of 1600 in the divergence constraint is the volume jump against boiling vapor. Mix those two densities and the laundry lifetime and the kettle expansion become the same number.

Miss the humidity boundary, or let S cross 2#

When a diffusion-limited lifetime misses the lab, suspect the far-field humidity before the model. In the table, RH 30 % and 97 % share 25 °C and still live 5.04 min and 115.98 min. Copy an $r$ from a paper into the heat-limited regime and a different $\Delta t$ either pushes $S$ past 2 or leaves the superheat alive.

Put in an evaporation rate and leave the right-hand side of $\nabla\cdot\mathbf{u}$ at zero, and mass leaves while volume stays. For water and steam that mismatch is about 1600. A drop that flies while it dries has to be read together with the drag correction that vaporization adds. Drop Lee's $r$ into the slot where $B_M$ belongs and the laundry waits for 100 °C.