The permeability coefficient for a 0.1-mm polyethylene film is being measured by maintaining a moisture vapor gradient across the film in a sealed test apparatus. The high moisture vapor side of the film is maintained at 90% RH and a salt (ZnCl2,0.5 H2O) maintains the opposite side at 10% RH .The area of film exposed to vapor transfer is 10cm by 10cm. When the test is conducted at 30°C, a weight gain of 50g in the desiccant salt is recorded after 24 h. From these given data, calculate the permeability coefficient of the film. Assume Pws=4.246 kPa at 30°C.(Example15.1 Introduction-to-Food-Engineering-R-paul-Singh)
Given
Film thickness 0.1 mm 1 10 –4 m
High relative humidity 90%
Low relative humidity 10%
Temperature 30°C
Film area 10 cm*10 cm =100 cm^2=0.01 m^2
Moisture rate of flow= 50 g/24 h =5.787*10 ^-4 g water/s
vapor pressure,
φ=Pw*100/Pws
At 10% relative humidity,
Pw=Pws* φ/100 (φ=% relative humidity)
=4.246*10/100=0.4246kPa
At 90% relative humidity,
Pw=Pws* φ/100 (φ=% relative humidity)
=4.246*90/100=3.821kPa
The permeability coefficient,
PB =(amount of gas vapor)*(thinkness of the film)/(area)(time)(pressure difference across the film)
=5.787*10^-4 *1*10^-4/(0.01)(3.821-0.4246)*1000=1.7*10^-9(g water m)/(m^2 pa s)
The permeability coeffi cient of the film is calculated to be 1.7*10^–9 [(g water m)/ (m^2 Pa s)]; the units may be converted to any other form .
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