Chemical Engineering - Mass Transfer - Discussion
Discussion Forum : Mass Transfer - Section 4 (Q.No. 28)
28.
A 25 cm x 25 cm x 1 cm flat sheet weighing 1.2 kg initially was dried from both sides under constant drying rate conditions. It took 1500 seconds for the weight of the sheet to reduce to 1.05 kg. Another 1m x 1m x 1cm flat sheet of the same material is to be dried from one side only.Under the same constant drying rate conditions, the time required for drying (in seconds) from its initial weight of 19.2 kg to 17.6 kg is
Discussion:
6 comments Page 1 of 1.
Divyesh said:
3 years ago
I think the right Answer is 2000.
Venkatesh said:
3 years ago
The right Answer is 2000.
Siddhant ju said:
5 years ago
I think Option C is correct.
Utkarsha said:
6 years ago
1000 sec is the correct answer.
Moisture removed per unit area is directly proportional to the time.
0.15÷0.0625=1500---(1).
1.6÷1 = ? ------------ (2).
So, time required =(1.6x1500)÷(0.15/0.0625).
Moisture removed per unit area is directly proportional to the time.
0.15÷0.0625=1500---(1).
1.6÷1 = ? ------------ (2).
So, time required =(1.6x1500)÷(0.15/0.0625).
Satya said:
6 years ago
I think the answer is 2000 seconds.
In constant drying process t = (Ws/(A*Nc))(Xi-Xc).
1500 = (1.2/(2*625*10^-4)*Nc)*(0.15/1.2) from this Nc = 1/1250.
In second case t=(19.2/(1*1*Nc))*(1.6/19.2) = 2000 seconds.
In constant drying process t = (Ws/(A*Nc))(Xi-Xc).
1500 = (1.2/(2*625*10^-4)*Nc)*(0.15/1.2) from this Nc = 1/1250.
In second case t=(19.2/(1*1*Nc))*(1.6/19.2) = 2000 seconds.
Kalpesh said:
9 years ago
Obviously less then 1500. So 1000 is correct.
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