Evaporation of liquid fuel droplets is often studied in the laboratory by using
ID: 473951 • Letter: E
Question
Evaporation of liquid fuel droplets is often studied in the laboratory by using a porous sphere technique in which the fuel is supplied at the rate just sufficient to maintained a completely wetted surface on the sphere. Consider the use of kerosene at 300 K with a porous sphere of 1 mm diameter. At this temperature the kerosene has a saturated vapour density of 0.015 kg m3 and a latent heat of vaporization of 300 kJ/kg. the mass diffusivity for the vapour-air mixture is 10-5 m2/s. if dry, atmospheric air at V= 15m/s and T = 300 K flows over the sphere, what is the minimum mass rate at which kerosene must be supplied to maintain a wetted surface?Explanation / Answer
We know
the dia of kerosene sphere (D) = 1mm
temperature (T)= 300K
v = 15 m/s
let's refer kerosene as A and air as B
In order to solve the given problem we need to make following assumptions as
1. given system is at steady state condition
2. sphere mount has negligible effect on flow field and hence h
3. negligible kerosene vapor concentration in free stream
refering the standard values we have
for Air (at 300K)
v = 15.89*10-6 m2/s
For Kerosene
A,sat = 0.015 kg/m3
hfg = 300kJ/kg
Kerosene vapor-air DAB = 10-5 m2/s
The kerosene flow rate is
nA = hmA (A,sat - A, )
Using mass transfer analog and neglecting the viscocity ratio, we write sherwood no as
Where reynold's no. (Re) = V*D/v = 15*0.001/15.89*10-6 = 944
and Schimdt no. (Sc) = v/DAB = 15.89*10-6 / 10*10-6 = 1.59
ShD = 2 + (0.4 Re0.5 + 0.06 Re0.67 )Sc0.4
ShD = 2 + (0.4 * 9940.5 + 0.06 * 9940.67 )* 1.590.4 = 23.7
hm = ShD * DAB / D =23.7 * 10*10-6 /0.001 = 0.237 m/s
nA = hm * D2 * A,sat
= 0.0237 * 0.0012 * 0.015
= 1.12*10-8 kg/s
The minimum mass of 1.12*10-8 kg/s of kerosene must be supplied to maintain a wetted surface
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