1- Consider wool under drying conditions. If the wool contains 0.36 kg water/ kg
ID: 704051 • Letter: 1
Question
1- Consider wool under drying conditions. If the wool contains 0.36 kg water/ kg solid in conatact with air of relative humiditiy Hr 60% and 25 C calculate :
* amount of bound water ..........
* amount of unbound water ......
* amount of free water .........
* What is the state of the free water (bound / unbound and amount)?
28 10 1 Paper, newsprint 2 Wool, worsted 3 Nitrocellulose 4 Silk 5 Leather, tanned 6 Kaolin 7 Tobacco leaf 8 Soap 9 Glue, hide 24 E20 7 4 10 Wood 11 Glass wool 12 Cotton 10//8 12 12 4 0 0 20 40 60 80 100 Relative humidity (%)Explanation / Answer
Read the curve 2 for wool given in question
Amount of unbound moisture = when vapor pressure is equal to pure liquid = amount of water in excess of 100 % relative humidity
Amount of bound moisture = when vapor pressure is less than vapor pressure of pure liquid = amount of water in eless than 100 % relative humidity =
from curve 2 When 100% relative humidity reached amount of water = 27 Kg H2O/100 Kg dry solid
amount of water in bound moisture = 0.27 Kg H2O/Kg dry solid
Bound moisture = 0.27 Kg H2O/Kg dry solid
wool sample contain 0.36 Kg H2O /Kg dry solid
Unbound moisture = excess water = 0.36 Kg H2O /Kg dry solid-0.27 Kg H2O/Kg dry solid =0.09 Kg H2O/Kg dry solid
Unbound moisture =0.09 Kg H2O/Kg dry
from curve 2 at 60 % relative humidity equilibrium water content = 15 Kg H2O/100 Kg dry solid =
0.15 Kg H2O/Kg dry solid
free moisture content = moisture which can be removed = moisture in sample -equilibrium moisture =
0.36 Kg H2O /Kg dry solid-0.15 Kg H2O/Kg dry solid=0.21 Kg H2O/Kg dry solid
free moisture content=0.21 Kg H2O/Kg dry solid
State of free water is both bound and unbound and amount is given above
** if you want to calculate in 100 Kg dry solid then multiply by 100
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