Chemical Engineering - Heat Transfer - Discussion
Discussion Forum : Heat Transfer - Section 3 (Q.No. 7)
7.
200 kg of solids (on dry basis) is subjected to a drying process for a period of 5000 seconds. The drying occurs in the constant rate period with the drying rate as, Nc = 0.5 x 10-3 kg/m2.s. The initial moisture content of the solid is 0.2 kg moisture/kg dry solid. The interfacial area available for drying is 4 m2/1000 kg of dry solid. The moisture content at the end of the drying period is (in kg moisture/kg dry solid)
Discussion:
10 comments Page 1 of 1.
AnR said:
4 years ago
The interfacial area should be 4m^2/100kg of dry solid.
So for 200 kg dry solid, we can write,
200(0.2-X) = (0.5*10^-3) * (5000)*(8),
X= 0.1.
So for 200 kg dry solid, we can write,
200(0.2-X) = (0.5*10^-3) * (5000)*(8),
X= 0.1.
(3)
Zeeshan Ulhaq said:
5 years ago
Nc = m * (xi-xf)/A * t.
(0.5*10-3) * (4/1000) * 5000 = xi - xf.
Xf = 0.2 - 0.1 = 0.1.
(0.5*10-3) * (4/1000) * 5000 = xi - xf.
Xf = 0.2 - 0.1 = 0.1.
(1)
Shivam said:
5 years ago
Here;
TC = ss / a * NC (X1 -X2).
TC = ss / a * NC (X1 -X2).
Om one8 said:
6 years ago
Interfacial must be 4m2/100 kg of dry solid.
(2)
Sami said:
6 years ago
@Temidayo.
(0.5E-3) how it came?
(0.5E-3) how it came?
Temidayo Gideon said:
6 years ago
Drying time (T) = M(X1 - X2)/AN.
5000 = 200(0.2 - X2) / 4(0.5E-3).
X2 = 0.2 - 0.1 = 0.1.
5000 = 200(0.2 - X2) / 4(0.5E-3).
X2 = 0.2 - 0.1 = 0.1.
Akash said:
7 years ago
Can someone please answer to this?
(1)
Philip Darwin said:
8 years ago
Is it that the interfacial drying area should be "per 100kg of dry solid" and not 1000 kg?
I'm using the equation time (t) = (final moisture - initial moisture)/(constant drying rate x interfacial Area)
I'm using the equation time (t) = (final moisture - initial moisture)/(constant drying rate x interfacial Area)
Astha said:
8 years ago
Please explain the answer of the question.
MUNJESH said:
9 years ago
If the Time must be 500 sec then what will be the answer?
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