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How long does it take for a cup of water to evaporate?

To answer this question, I'll assume some basic parameters, and that the water is blown on by a fan, to arrive at an estimate:

  • Water volume: $V=200\ \mathrm{mL}$
  • Top surface area of water: $A_\mathrm s = 0.05\ \mathrm{m^2}$
  • Room temperature: $T_{\infty}=25\ \mathrm{^\circ C}$
  • Water temperature: $T_\mathrm w = 25\ \mathrm{^\circ C}$
  • Relative humidity of water in room air: $50\ \%$
  • Heat transfer convection coefficient from a fan/wind: $h=100\ \ \mathrm{W/(m^2\ K)}$

Let's assume the water is in thermal equilibrium with the surrounding room (a large heat reservoir) so there is no buoyant convection.


I start with the evaporative mass flux given by

$$ n = h_m(\rho_s - \rho_{\infty})$$

and $h_m$ is the mass transfer coefficient, which is found from the heat and mass transfer analogy:

$$ h_m = \frac{h}{\rho c_p Le^{2/3}}$$

where $Le=\frac{\alpha}{D_{\mathrm{H2O},\text{air}}}$ is the Lewis number. So the evaporative mass flow rate is

$$ \dot{m} = n A_\mathrm s = A_\mathrm s \frac{h(\rho_\mathrm s - \rho_{\infty})}{\rho c_p Le^{2/3}}$$

We can estimate the density difference by using the relative humidity of air at ~$50\ \%$ for a normal room:

$$ \rho_\mathrm s - \rho_{\infty} = \rho_\mathrm{sat}(T) - 0.5 \rho_\mathrm{sat}(T) = 0.5\frac{Mp_\mathrm{sat}(T)}{RT} = 0.5\frac{18\ \mathrm{g\ mol^{-1}} \times 3171\ \mathrm{Pa}}{8.315\ \mathrm{m^3\ Pa\ K^{-1}\ mol^{-1}} \times 298\ \mathrm K} = 0.012\ \mathrm{kg/m^3}$$

The Lewis number is calculated from air thermal diffusivity $\alpha=2.2 \times 10^{-5}$ and the binary diffusion coefficient $D_{\mathrm{H2O},\text{air}}$ for diffusion of water vapor through air is given by an experimental correlation (with $p$ in $\mathrm{atm}$):

$$ D_{\mathrm{H2O},\text{air}} = 1.87 \times 10^{-10} \frac{T^{2.072}}{p} = 1.87 \times 10^{-10} \frac{298^{2.072}}{1} = 2.5 \times 10^{-5}$$

The Lewis number is therefore $ Le = \frac{2.2}{2.5} = 0.88$. The mass flow rate from the surface is

$$ \dot{m} = A_s \frac{h(\rho_\mathrm s - \rho_{\infty})}{\rho c_p Le^{2/3}} = 0.05\frac{100 \times 0.012}{1.2 \times 1000 \times 0.88^{2/3}} = 5.4 \times 10^{-5}\ \mathrm{kg/s}$$

Now, I assume that this mass flux remains constant with time since the water is in thermal quasi-equilibrium with the room (a large temperature reservoir), and therefore remains at constant temperature, thus not changing the properties of water.

Mass conservation on the water yields

$$ \frac{\mathrm dm}{\mathrm dt} = -\dot{m}$$

Integrating, we find that the time rate of mass change is linear:

$$ m(t) = m_0 - \dot{m}t$$

To fully evaporate, $m(t)=0$ and

$$t = \frac{m_0}{\dot{m}} = \frac{\rho V}{\dot{m}} = \frac{1.2 \times 0.2}{5.5 \times 10^{-5}} = 4360\ \mathrm s = 1.2\ \mathrm h$$

The water takes 1.2 hours to fully evaporate.


1 hour for evaporation seems pretty fast, but I did use a large convection coefficient from the beginning. Some thoughts/questions:

  1. What if there was no forced convection from a fan? We don't have buoyant natural convection or radiation since the water is in thermal equilibrium with the room. What is the nature of evaporation in this case and how can we calculate the mass loss?
  2. I assumed that the evaporation mass loss is constant throughout time, since the water is in thermal equilibrium with the room (a large reservoir) and not changing temperature. Is this a good assumption?
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  • $\begingroup$ I haven't checked your arithmetic, but your approach is correct. Regarding the question, if there is absolutely no convection, then, as a worst case, you would have a straight diffusion problem. That would mean that you would have concentration buildup in the air surrounding the surface of the cup, and the extent of this region would be increasing with time, with 100% humidity at the surface and 50% humidity far from the surface. $\endgroup$ – Chet Miller Mar 27 '19 at 22:03
  • $\begingroup$ @ChetMiller So that would be like a semi-infinite mass diffusion problem, with similar governing equations and solutions to the heat transfer semi-infinite problem? The mass flux would then be time dependent, correct? $\endgroup$ – Drew Mar 27 '19 at 22:15
  • $\begingroup$ As a practical matter, I think that trying to accurately calculate the rate of evaporation is pretty difficult. There is generally a thin, stagnant layer of air just above the water surface that has a much higher relative humidity than the RH of the room, and that thin layer is an important evaporation rate limiting factor. Don't think it's an easy matter to accurately calculate the RH or thickness of the layer, or how those two parameters may change as a function of the amount of air flow over the surface. The evaporation rate can also be sensitive to tiny oil or other films on the surface. $\endgroup$ – user93237 Mar 27 '19 at 22:16
  • $\begingroup$ Sure. It would probably need to be solved numerically unless you would be willing to approximate the surface of the water as a small circular area embedded in an infinite plane below the semi-infinite half-space. I'm sure Carslaw and Jaeger have the solution to this analogous heat transfer problem. $\endgroup$ – Chet Miller Mar 27 '19 at 22:20
  • $\begingroup$ @SamuelWeir Drew's solution takes into account the concentration boundary layer above the surface. His mass transfer coefficient is equal to the diffusion coefficient divided by the boundary layer thickness. $\endgroup$ – Chet Miller Mar 27 '19 at 22:22

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