Ch4 - Numericals
Ch4 - Numericals
Ch4 - Numericals
Chapter 4: Drying
QB – 1: A porous solid is dried in a batch dryer under constant drying conditions. Six hours
are required to reduce the moisture content from 30 to 10%. The critical moisture content was
found to be 16% and the equilibrium moisture 2%. All moisture contents are on the dry basis.
Assuming that the rate of drying during falling rate period is proportional to the free moisture
content, how long should it take to dry a sample of same solid x from 35 to 6% under the same
drying conditions?
ANS:
Case-1:
Ss Xc − X∗
= [(X1 – XC) + (XC – X*) ln ]
A∗Nc X2 − X∗
Ss 0.16 − 0.02
6= * [(0.3 – 0.16) + (0.16 – 0.02) ln ]
A∗Nc 0.1 − 0.02
Ss
= 27.48
A∗Nc
Case-2:
➢ Now, X1 = 0.35, X2 = 0.06
➢ So, time for drying
Ss Xc − X∗
θ= [(X1 – XC) + (XC – X*) ln ]
A∗Nc X2 − X∗
0.16 − 0.02
= 27.48 * [(0.35 – 0.16) + (0.16 – 0.02) ln ]
0.06 − 0.02
= 10.041 hours
ANS:
Case-1:
Ss Xc − X∗
= [(X1 – XC) + (XC – X*) ln ]
A∗Nc X2 − X∗
Ss 0.14 − 0.04
5= * [(0.35 – 0.14) + (0.14 – 0.04) ln ]
A∗Nc 0.1 − 0.04
Ss
= 19.151
A∗Nc
Case-2:
➢ Now, X1 = 0.35, X2 = 0.06
➢ So, time for drying
Ss Xc − X∗
θ= [(X1 – XC) + (XC – X*) ln ]
A∗Nc X2 − X∗
0.14 − 0.04
= 19.151 * [(0.35 – 0.14) + (0.14 – 0.04) ln ]
0.06 − 0.04
= 7.104 hours
ANS:
Case-1:
Ss Xc − X∗
= [(X1 – XC) + (XC – X*) ln ]
A∗Nc X2 − X∗
Ss 0.14 − 0.04
5= * [(0.25 – 0.14) + (0.14 – 0.04) ln ]
A∗Nc 0.06 − 0.04
Ss
= 18.454
A∗Nc
Case-2:
➢ Now, X1 = 0.20, X2 = 0.06
➢ So, time for drying
Ss Xc − X∗
θ= [(X1 – XC) + (XC – X*) ln ]
A∗Nc X2 − X∗
0.14 − 0.04
= 18.454 * [(0.20 – 0.14) + (0.14 – 0.04) ln ]
0.06 − 0.04
= 4.0773 hours
ANS:
➢ Ss = 150 kg
➢ x1 = 0.32, x2 = 0.17, xC = 0.15
➢ Drying surface = 0.03 m2 /kg dry weight
➢ Nc = 0.75 x 10-3 kg/m2s
➢ Now,
𝑥1 0.32
X1 = = = 0.471
1 − 𝑥1 1 − 0.32
𝑥2 0.17
X2 = = = 0.205
1 − 𝑥2 1 − 0.17
𝑥𝑐 0.15
XC = = = 0.1765
1 − 𝑥𝑐 1 − 0.15
A = 0.3*150 = 4.5 m2
Ss∗(X1 − X2)
θ=
A∗Nc
150∗(0.471 − 0.205)
=
4.5∗0.75∗10−3
= 11822.222 sec
= 3.284 hours
ANS:
Case-1:
Ss Xc − X∗
= [(X1 – XC) + (XC – X*) ln ]
A∗Nc X2 − X∗
Ss 0.20 − 0.04
8= * [(0.40 – 0.20) + (0.20 – 0.04) ln ]
A∗Nc 0.10 − 0.04
Ss
= 22.4132
A∗Nc
Case-2:
➢ Now, X1 = 0.40, X2 = 0.05
➢ So, time for drying
Ss Xc − X∗
θ= [(X1 – XC) + (XC – X*) ln ]
A∗Nc X2 − X∗
0.20 − 0.04
= 22.4132 * [(0.40 – 0.20) + (0.20 – 0.04) ln ]
0.05 − 0.04
= 14.425 hours
ANS:
➢ Ss = 3000 kg dry
➢ A = 60 m2
➢ X1 = 0.15, X2 = 0.025, XC = 0.125, X* = 0
➢ Nc = 3 x 10-4 kg/m2 s
➢ Now, time for drying
Ss∗(X1 − Xc) Ss∗(Xc − X∗) Xc − X∗
θ= + ln
A∗Nc A∗Nc X2 − X∗
Ss Xc
= [(X1 – XC) + (XC) ln ]
A∗Nc X2
3000 0.125
= [(0.15 – 0.125) + (0.125) ln ]
60∗3∗10−4 0.025
= 37696.623 sec
= 10.4713 hours
ANS:
(2∗1.2)∗(X1 − Xc)
1500 =
A∗Nc
(X1 − Xc)
= 625
A∗Nc
➢ Now, for second case
Ss2∗(X1 − Xc)
θ2 =
A∗Nc
= 1000 sec