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A rain droplet with a radius of 2 mm falls down with a velocity of 7 m s 1 . The droplet evaporates with a

A rain droplet with a radius of 2 mm falls down with a velocity of 7 m s1
. The droplet
evaporates with a mass transfer coefficient of 0.14 m s1
. The droplet is initially at 15
C. The air has a temperature of 15C and a relative humidity of 50%.
Calculate the temperature profile in the droplet at different times resulting from
evaporative cooling as the droplet falls down.
Use the following assumptions:
The heat transfer flux at the droplet surface equals the mass transfer flux of
water vapor around the droplet multiplied by the heat of evaporation (45,000 J
mol1
).
The water vapor concentration at the droplet surface can be calculated from
the ideal-gas law, assuming that the water partial pressure is the vapor
pressure at the temperature of the droplet surface. The water partial pressure
in the bulk air equals the water vapor pressure at 15C multiplied by the
relative humidity (0.5). The water mass transfer is driven by this concentration
difference.
The vapor pressure of water is given by the following equation:
v ()()
6757.426 p T Pa exp 58.593654.891124ln
T
=
, where T is the
temperature in K. The equation is valid in the 273 K 303 K range.
Once the evaporation heat flux is known, it can be used as a heat flux boundary at
the outside radius of the droplet, with a virtual point boundary condition similar to the
convection heat flux boundary condition used in the steel ball example.
You can use an existing code (e.g., the steel ball code) as a basis for your code.
Submit a brief (2 pages) report with your results, as well as the Matlab/Octave code of your simulation

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