Ice sublimation¶
Ice sublimation is modeled with the SublimationLTE class, which balances absorbed sunlight with the energy losses from re-radiated thermal energy and the latent heat of sublimation of the ice.
The grains2 water ice material includes the optical constants of water ice from Warren & Brandt (2008). The latent heat of sublimation follows Delsemme & Miller (1971), and the vapor pressure equation of Lichtenegger & Komle (1991) is also used.
Temperature and mass-loss rate¶
Calculate the temperature and sublimation rate of a 1.0 μm pure water ice grain at 1.0 au from the Sun:
>>> from grains2 import SublimationLTE, waterice
>>>
>>> a = 1.0 # radius, μm
>>> ice = waterice()
>>> rh = 1.0 # heliocentric distance, au
>>>
>>> sublimation = SublimationLTE(a, ice, rh)
>>> print(sublimation.T[0], "K")
160.52729461 K
>>> print(sublimation.phi()[0], "kg/m2/s")
1.35242694e-07 kg/m2/s
Note
The mass loss assumes ice can sublimate over 100% of the surface. This is appropriate for pure ice grains, but is an approximation/simplification for mixtures of ice and dust or vacuum.
Grain sublimation lifetime¶
Grain lifetimes may be calculated with the SublimationLTE.lifetime() method. This requires a list of grain radii, so that it can integrate \(da/dt\) from \(a_i\) to \(a_0\). Formally, the time for a grain to sublimate to \(a=0\) approaches infinity. In order to avoid this non-physical scenario, lifetime will sublimate the grain from \(a_0\) to 0 using a constant radius loss rate (\(da/dt|a_0\)). Users must decide for themselves what to use for \(a_0\).
Calculate the lifetime of a 1.0 μm water ice grain at 1.0 au, with and without considerations for solar wind sputtering:
>>> import numpy as np
>>>
>>> a = np.logspace(-2, 0)
>>> sublimation = SublimationLTE(a, ice, rh)
>>> tau = sublimation.lifetime()
>>> print(tau[-1])
8622.61 s