1 Mg of dry mass of a non-porous solid is dried under constant drying conditions in an air stream flowing at 0.75 m/s. The area of surface drying is 55 m^{2}. If the initial rate of drying is 0.3 g/m^{2}s, how long will it take to dry the material from 0.15 to 0.025 kg water/kg dry solid? The critical moisture content of the material may be taken as 0.125 kg water/kg dry solid. If the air velocity were increased to 4.0 m/s, what would be the anticipated saving in time if the process were surface-evaporation controlled?
Question : 1 Mg of dry mass of a non-porous solid is dried under consta...
During the constant rate period, that is whilst the moisture content falls from 0.15 to 0.125 kg/kg, the rate of drying is:
(dw/dt)/A = (0.3/1000) = 0.0003 kg/m^{2}s
At the start of the falling rate period, w = w_{c} = 0.125 kg/kg
and: (dw/dt)/A = m(w_{c}-w_{e})
or: 0.0003 = m(0.125 – 0.025)
and: m = 0.003 kg/m^{2}s kg dry solid
= (0.003/1000) = 3.0 × 10^{-6} kg/m^{2}s Mg dry solid
The total drying time is given by equation 16.14:
t = (1/mA)[\ln (f_{c}/f)+(f_{1}-f_{c})/f_{c}]
where: f = (0.025 – 0) = 0.025 kg/kg (taking w_{e} as zero)
f_{c} = (0.125 – 0) = 0.125 kg/kg
f_{1} = (0.15 – 0) = 0.150 kg/kg
Thus: t = [1/(3.0 × 10^{-6} × 55)][ln(0.125/0.025) + (0.150 – 0.125)/0.125]
= 10,960 s or 10.96 ks (3 h)
As a first approximation it may be assumed that the rate of evaporation is proportional to the air velocity raised to the power of 0.8. For the second case, m may then be calculated as:
m=(3.0\times 10^{-6})(4.0/0.75)^{0.8}=(1.15\times 10^{-5})kg water/m^{2}s Mg dry solid
The time of drying is then:
t = [1/(1.15 × 10^{-5} × 55)](1.609 + 0.20) = 2860 s or 2.86 ks
and the time saved is therefore: (10.96 – 2.86) = 8.10 ks (2.25 h)
equation 16.14:t=\frac{(w_{1}-w_{c})}{mAf_{c}}+\frac{1}{mA}\ln \left ( \frac{f_{c}}{f} \right )=\frac{1}{mA}\left [ \frac{(f_{1}-f_{c})}{f_{c}}+\ln \left ( \frac{f_{c}}{f} \right ) \right ]