Chemical Engineering - Mass Transfer - Discussion
Discussion Forum : Mass Transfer - Section 2 (Q.No. 10)
10.
It takes 6 hours to dry a wet solid from 50% moisture content to the critical moisture content of 15%. How much longer it will take to dry the solid to 10% moisture content, under the same drying conditions? (The equilibrium moisture content of the solid is 5%).
Discussion:
7 comments Page 1 of 1.
Isa Rabiu said:
1 year ago
We have to assume a constant drying rate during the falling rate period.
So, the answer is 51 minutes, therefore B is the correct answer.
So, the answer is 51 minutes, therefore B is the correct answer.
(1)
ALM said:
2 years ago
Tc = Ss/ ANc (Xi-Xc)
6 = Ss/ANc (0.5-0.15)
Ss/ANc = 17.1428.
For case 2:-
Tc = SS/ANc[(Xi-Xc) + (Xc-X*) ln (Xc-X*)/ ( Xf/X*)].
=>Therefore Tc = 17.1428[(0.50-0.15)+(0.15-0.05) ln (0.15-0.05)/(0.10-0.05)].
So, Tc => 2.47 hr => 146min.
6 = Ss/ANc (0.5-0.15)
Ss/ANc = 17.1428.
For case 2:-
Tc = SS/ANc[(Xi-Xc) + (Xc-X*) ln (Xc-X*)/ ( Xf/X*)].
=>Therefore Tc = 17.1428[(0.50-0.15)+(0.15-0.05) ln (0.15-0.05)/(0.10-0.05)].
So, Tc => 2.47 hr => 146min.
(2)
Muhammad Anus said:
5 years ago
50% = 6 hrs = 360 mins.
10% = 360 mins * 10% /50%.
10% = 72 mins.
10% = 360 mins * 10% /50%.
10% = 72 mins.
(8)
Vaibhav said:
7 years ago
Here;
Ss-mass of solid on a dry basis.
A-surface area per unit dry solid.
Nc-critical drying rate.
Ss-mass of solid on a dry basis.
A-surface area per unit dry solid.
Nc-critical drying rate.
Kishor said:
7 years ago
Can you please explain what is SS and ANc?
Chemical engineer said:
9 years ago
tc = (Ss/ANc)*(X1 - Xc)(1).
tc = time for constant drying.
We required moisture % 10. Which is < Xc.
Further drying will done by linear falling rate period.
(Ss/ANc) = tc /(X1 - Xc) =(6/(0.50 -0.15)) = 17.14.
tF = (Ss/ANc)*(Xc " X*)*ln((X1-X*)/(X2-X*))
where tF = falling rate period.
X* = equilibrium moisture content.
X1 = INITIAL MOISTURE CONTENT =0.15.
X2 = FINAL MOISTURE CONTENT = 0.10.
tF = 1.1881 hr =71.28 min.
tc = time for constant drying.
We required moisture % 10. Which is < Xc.
Further drying will done by linear falling rate period.
(Ss/ANc) = tc /(X1 - Xc) =(6/(0.50 -0.15)) = 17.14.
tF = (Ss/ANc)*(Xc " X*)*ln((X1-X*)/(X2-X*))
where tF = falling rate period.
X* = equilibrium moisture content.
X1 = INITIAL MOISTURE CONTENT =0.15.
X2 = FINAL MOISTURE CONTENT = 0.10.
tF = 1.1881 hr =71.28 min.
(7)
Babu said:
9 years ago
Someone solve it.
(2)
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