Drying Worked Problems

Worked Examples:

1) Air containing 0.005 kg of water vapour per kg of dry air is preheated to 52°C in a dryer and passed to the lower shelves. It leaves these shelves at 0! relative hu"idity and is reheated to 52°C and passed over another set of shelves# again leaving at 0! relative hu"idity. $his is again repeated for the third and fourth sets of shelves# after which the air leaves the dryer. %n the assu"ption that the "aterial in each shelf has reached the wet &ul& te"perature and heat loss is neglig negligi&l i&le# e# esti"at esti"ate' e' (i) the te"peratu te"perature re of the "ateria "ateriall on each each tray tray (ii) (ii) the a"ount of water re"oved# in kg*hr# if +00" +*"in of "oist air leaves the dryer. (i)

Air leaves the pre,heater of the dryer at +25°- 

u"idity of inco"ing inco"ing air / 0.005 kg kg water*kg dry air  It enters the first shelf.  o# wet &ul& te"perature / 25°C oistu oisture re is re"ove re"oved d along along constan constantt wet &ul& &ul& te"per te"peratu ature re line line till till 0! .. is reached $his gives the e3it condition of air fro" first shelf. 4ro" the chart# u"idity of air leaving first shelf / 0.01 kg water*kg dry air. ry &ul& te"perature of e3it air is at 26°C and is at hu"idity of 0.01 kg water*kg dry air. $his air is again heated heated to 52°C 52°C dry &ul& &ul& te"perat te"perature ure in second heater heater.. o# air leaves leaves heater at 52°C and hu"idity of 0.01 kg water*kg dry air. 7hen it leaves the econd shelf# the corresponding dry &ul& te"perature is +8°C and hu"idity is 0.02+ kg water* kg dry air. $his air enters the third shelf after preheating to 52 °C. i"ilarly for third shelf# e3it air has a hu"idity of 0.029 kg water* kg dry air and a dry &ul& te"perature is +:°C. $he air leaving the fourth shelf has a hu"idity of 0.0+2 kg water* kg dry air  and a dry &ul& te"perature of 82°C $he solid te"peratures corres respond

to

7;$ 7;$ and and they they are 2+ °C# 26°C# +2 °C and +8 °C respectively. 111 111

Fig. : P 6.1 Humidity vs Temperature

(ii) 4inal "oist air conditions' conditions' ((0.5 "oisture) ? (0.5 dry solid) / 0.5* (1 E 0.5) / "oisture*dry solid ∴ G1 /

1

4or 5 kg wet solid# "oisture / 5 D 0.5 / 2.5 kg.

118

Fig. : P 6.# $a% &ryi"g rate 'urve

∴ 7eight

3K / 0.05#

of dry solid / 5 E 2.5 / 2.5 kg. GK / 0.05* (1 E 0.05) / 0.052

oisture content in ry &asis

/ weight of wet solid E weight of dry solid  weight of dry solid

G# (kg "oisture)*(kg drysolid)  H# (kg*hr"2)

1 5

0. 5

0.88 8.5

0.8 8

0.+ +.5

0.228 0.18 2.0 1.0

 (i) GCr  / 0. kg "oisture*kg dry solid. (ii)4ro" G/ 0. to 0.88 the falling rate curve is non,linear and fro" G / 0.88 to 0.18# falling rate period is linear. G2 / 0.15* (1 E 0.15) / 0.165. o# we can find ti"e for drying fro" 0. to 0.88 graphically and then for G/0.88 to 0.165# we can go in for analytical solution.

115

Fig. : P6.# $(% 1/N vs 

$i"e taken for constant rate drying period. (4ro" G / 1 to G / 0.) tI / >*AHC >G1 E GCr  / >2.5* (5 D 0.) >1 E 0. / 0.+++ hr. tII

⇒  In

falling rate period fro" G / 0.88 to 0.165

tII / *AHC D L(G Cr   E GK) D ln >(G1 E GK)* (G2@ E GK)M (GK / 0.05* (1 E 0.05) / 0.052) tII / >2.5* (5 D 0.) D L(0. E 0.052) D ln >(0.88E 0.052)* (0.16 E 0.052)M /

0.522 hr 

4ro" graph# t III (4ro" G / 0. to G / 0.88) / (0.0++ D 2.5)*0. / 0.18 hr. $otal ti"e / tI ? tII ? tIII / 0.+++ ? 0.522 ? 0.18 / 0.::5 hr or 5:.59 "in.

8) ata on drying rate curve of a particular solid is given &elow. $he weight of the dry "aterial in the solid is 89.0 kg*" 2. Calculate the ti"e reFuired to dry the "aterial fro" 25! to 9! "oisture (dry &asis). ata' G

0.+0

0.20

0.19

0.15

0.18

0.11

0.06

0.05

 H

1.22

1.22

1.18

0.:0

0.90

0.5

0.22

0.05

where G is the "oisture content in kg water*kg dry solid and H is the drying rate in kg*(hr)("2).

11

Fig. : P 6.) 1/N vs  !or !alli"g rate

HC / 1.22 kg*" 2hr# GCr  / 0.2#

G1 / 0.25# G2 / 0.09#

*A / 89 kg*"2

116

$i"e taken for constant rate drying period# tI / >*AHC >G1 E GCr  tI / 89 >0.25E 0.2*1.22 / 1.:6 hr. G

0.19

0.15

1*H

0.9662 1.111

0.18

0.11

0.06

0.05

1.25

1.6956

8.585

20

Area under the curve / 18 D 0.025 D1 /0.+5 tII / 0.+5 D 89 / 1.9 hr  $otal ti"e taken / tI ? tII / 1.:6 ? 1.9 / 19.66 hr

5) In a drying e3peri"ent# a tray drier# containing a single tray of 1 sF." area is used# to dry crystalline solids. $he following data has &een collected' (a) Calculate and plot drying rates. 4ind the critical "oisture content. (&) If dry air is availa&le at 80°C with an a&solute hu"idity of 0.01 kg*kg dry air  and the drier is "aintained at :0°C# calculate the a"ount of air reFuired in first 2 hours. Assu"e the air is heated up to :0°C and the dry air leaves the drier at :0°C with 5! saturation. (c)$est the consistency of the falling rate period. (Choose critical "oisture content and any one point in falling rate period).

l.Ho

$i"e (hr)

7eight of wet "aterial. -g

1. 2. +. 8. 5. . 6. 9. :. 10. 11. 12. 1+. 18. 15.

0.0 0.8 0.9 1.0 1.8 1.9 2.2 2. +.0 +.8 8.2 8. 5.0 .0 Infinite

5.+18 5.2+9 5.12 5.128 5.089 8.:62 8.9:5 8.91: 8.68+ 8.6 8.528 8.89 8.82 8.+80 8.120

119

.Ho.

$i"e# hr

7eight of 7et "aterial# kg

1 2 + 8 5  6 9 : 10 11 12 1+ 18 15

0 0.8 0.9 1.0 1.8 1.9 2.2 2. +.0 +.8 8.2 8. 5  Infinity

5.+18 5.2+9 5.12 5.128 5.089 8.:62 8.9:5 8.91: 8.68+ 8.6 8.528 8.89 8.82 8.+80 8.120

oisture content (dry  &asis) 0.2: 0.261 0.25+ 0.288 0.225 0.20 0.199 0.1: 0.151 0.1++ 0.0:9 0.098 0.068 0.05+ 0.0

rying rate# kg*hr . "2 ,,, 0.1: 0.1: 0.1: 0.1: 0.1: 0.1:25 0.1: 0.1: 0.1: 0.16: 0.18 0.105 0.09 ,,,

4ro" the a&ove data after getting the rate curve it is clear that Gcr / 0.11 $he loss in weight is due to "oisture evaporated. After two hours the weight is / (8.:62?8.9:5) *2 / 8.:+8 kg $he water evaporated in 2 hours is / 5.+18 E 8.:+8 / 0.+9 kg u"idity of inco"ing air / 0.01 kg*kg u"idity of leaving air / 0.0+ kg*kg (4or :0°C with 5! saturation)

11:

Fig.: P6.* &ryi"g rate +urve

7ater carried away &y air / Bs ((G1 E GK)* (G2 E GK)M et us choose readings (11) and (1+) to check the consistency /> 8.12*(1) D (0.1:) D (0.11,0) Dln>0.0:9*0.068 / 0.6 hours ere# GK is taken as / 0 Actual ti"e is 0.9 hours.

) 7oolen cloth is dried in a hot air dryer fro" initial "oisture content of 100! to a final content of 10!. If the critical "oisture content is 55! and the eFuili&riu" "oisture content is ! (at dryer condition)# calculate the saving in drying ti"e if the "aterial is dried to 1! instead of 10!. All other drying conditions re"ain the sa"e. All "oisture contents are on the dry &asis. G1 / 100!#

GCr  / 55!#

G2 / 10!#

GK / ! all are in dry &asis.

120

G2@# / 1! t$1 / *AHC >G1 E GCr  ? L*A HC (GCr  E GK) ln >(G Cr  E GK)* (G 2 E GK)M t$ / *AHC >G1 E GCr  ? L*A HC (GCr  E GK) ln >(G Cr  E GK)* (G 2 E GK)MN... (1) t$@ / *AHC >G1 E GCr  ? L*A HC (GCr  E GK) ln >(G Cr  E GK)* (G 2@ E GK)MN. (2) (i.e.) (1)* (2) is# t$ /

>G1 E GCr  ? L(GCr  E GK) ln >(G Cr  E GK)* (G 2 E GK)M

t$@

>G1 E GCr  ? L(GCr  E GK) ln >(G Cr  E GK)* (G 2@ E GK)M

t$* t$@ / >1 E 0.55 ? L(0.55E 0.0) ln >(0.55 E 0.0)* (0.1 E 0.0)M  >1 E 0.55 ? L(0.55E 0.0) ln >(0.55 E 0.0)* (0.1 E 0.0)M t$* t$@ / 1.69*1.2296

⇒ t$@

/ 0.6+2 t $

t$* t$@ / 1.+56 t$@*t$ / 0.6+22 $he reduction in drying ti"e is# (t$ E t$@)*t$ / 0.269 (i.e.) the ti"e reduces &y 2.69!.

6) A filter cake is dried for 5 hours fro" an initial "oisture content of +0! to 10! (wet  &asis). Calculate the ti"e reFuired to dry the filter cake fro" +0! to ! (wet  &asis) OFuili&riu" "oisture content / 8! on dry &asis Critical "oisture content

/ 18! on dry &asis

Assu"e that the rate of drying in the falling rate period is directly proportional to the free "oisture content.

3i / 0.+#

3f  / 0.10#

G2 / 0.1* (1 E 0.1) / 0.111#

Gi / 0.+* (1 E 0.+) / 0.829# G Cr  / 0.18 GK / 0.08

G2@/ 0.0*(1,0.0) /0.08 t$ / *AHC >(Gi E GCr )? L(GCr  E GK) ln >(G Cr  E GK)* (G 2 E GK)M 5 / *AHC> (0.829 E 0.18) ? L(0.18 E 0.08) ln >(0.18 E 0.08)* (0.111 E 0.08)M *AHC / 15.896 t$@ / *AHC >(Gi E GCr ) ? L(GCr  E GK) ln >(G Cr  E GK)* (G 2@ E GK)M t$@ / 15.896 >(0.829 E 0.18) ? L(0.18 E 0.08) ln >(0.18 E 0.08)* (0.08 E 0.08)M t$@ / .9 hrs.

121

9) 1000 kg dry weight of non,porous solid is dried under constant drying conditions with an air velocity of 0.65 "*s# so that the area of surface drying is 55 " 2. $he critical "oisture content of the "aterial "ay &e taken as 0.125 kg water * kg dry solidsJ (i) If the initial rate of drying is 0.+ g * " 2.s. ow long will it take to dry the "aterial fro" 0.15 to 0.025 kg water per kg dry solidJ (ii) If the air velocity were increased to 8.0 "*s. what would &e the anticipated saving in ti"e if surface evaporation is controlling.

 / 1000 kg# G1 / 0.15#

Air velocity / 0.65 "*s# G2 / 0.025#

assu"e GK / 0#

A / 55" 2#

GCr  / 0.125#

HC / 0.+ g*" 2s or 0.+D10

kg*"2s t$ / (*AHC) D >(G1 E GCr ) ? L(GCr  E GK) ln >(G Cr  E GK)* (G 2 E GK)M t$ / >1000* (55 D0.+D10 ,+) D >(0.15E 0.125) ? L(0.125E 0) ln >(0.125 E 0)* (0.025 E 0)M t$ / +.9066 hr 

(ii) Assu"ing only surface evaporation# and let air "ove parallel to surface  HC P B0.61

B/=DQ

i.e. HC P =0.61 ∴ HC1*

 

HC2 / (=1)0.61* (=2)0.61

(0.+D10 ,+)* HC2 / (0.65)0.61* (8)0.61  HC2 / 0.:95 D10 ,+ kg*"2s. t$ / >1000* (55 D0.:95D10 ,+) D>(0.15E 0.125) ? L(0.125E 0) ln >(0.125 E 0)*(0.025 E 0)M / 1.15:6 hrs.  o# ti"e saved / +.9066 E 1.15:6 / 2.89 hr.

:) A plant wishes to dry a certain type of fi&er&oard. $o deter"ine drying characteristics# a sa"ple of 0.+ 3 0.+ " siRe with edges sealed was suspended fro" a  &alance and e3posed to a current of hot dry air. Initial "oisture content was 65!. $he sheet lost weight at the rate of 1 3 10  E 8 kg*s until the "oisture content fell to 0!. It was esta&lished that the eFuili&riu" "oisture content was 10!. $he dry "ass of the sa"ple was 0.:0 kg. All "oisture contents were on wet &asis. eter"ine the ti"e for  drying the sheets fro" 65! to 20! "oisture under the sa"e drying conditions. 122

,+

31 / 0.65# 3K / 0.1# 3Cr  / 0.  / 0.:0 kg#

A / (0.+ D 0.+) D2 (&oth upper and lower surfaces are e3posed) / 0.19 " 2. A D HC / 10, 8 kg*s# 31 / 0.65# 32 / 0. 2#

GCr  / 0.*0.8 /1.5#

G1 / +#

GK / 0.1*0.: / 0.111

G2 / 0.25#

t$ / (*AHC) D >(G1 E GCr ) ? L(GCr  E GK) ln >(G Cr  E GK)* (G 2 E GK)M t$ / 0.:0*10 , 8 >(+E 1.5) ? L(1.5 E 0.111) ln >(1.5 E 0.111)* (0.25 E 0.111)M / 11.68 hr.

10) A co""ercial drier needed 6 hours to dry a "oist "aterial fro" ++! "oisture content to :! on &one dry &asis. $he critical and eFuili&riu" "oisture content were 1! and 5! on &one dry &asis respectively. eter"ine the ti"e needed to dry the "aterial fro" a "oisture content of +6! to 6! on &one dry &asis if the drying conditions re"ain unchanged.

G1 / 0.++#

GK / 0.05#

G2@ / 0.06#

t$ / 6 hrs

G Cr  / 0.1#

G2 / 0.0:#

G1@ / 0.+6#

t$ / (*AHC) D >(G1 E GCr ) ? L(GCr  E GK) ln >(G Cr  E GK)* (G 2 E GK)M 6 / *AHC D >(0.++E 0.1) ? L(0.1 E 0.05) ln >(0.1 E 0.05)* (0.0: E 0.05)M *AHC / 28.99  How# G1 / 0.+6#

G2 / 0.06

t$ / 28.99 D >(0.+6E 0.1) ? L(0.1 E 0.05) ln >(0.1 E 0.05)* (0.06 E 0.05)M / :.9:+ hr.

11)

A sla& of paper pulp 1.5 "eter 3 1.5 "eter 3 5 ""# is to &e dried under constant drying conditions fro" 5! to +0! "oisture (wet &asis) and the critical "oisture is 1.6 kg free water per kg dry pulp. $he drying rate at the critical point has &een esti"ated to &e 1.80 kg*(" 2)(hr). $he dry weight of each sla& is 2.5 kg. Assu"ing drying to take place fro" the two large faces only# calculate the drying ti"e to &e  provided.

31 / 0.5#

 / 2.5 kg# 12+

A / (1.5 D 1.5) D2 (drying takes place fro" &oth the larger surface only) / 8.5 " 2.  HC / 1.8kg*" 2hr#

32 / 0.+#

G1 / 0.5* (1 E 0.5) / 1.956#

GCr  / 1.6# G 2 / 0.+* (1 E 0.+) / 0.829#

Assu"ing GK / 0 t$ / (*AHC) D >(G1 E GCr ) ? L(GCr  E GK) ln >(G Cr  E GK)* (G 2 E GK)M t$ / (2.5*8.5 D 1.8) D >1.956E 1.6 ? L(1.6 E 0) ln >(1.6 E 0)* (0.829 E 0)M t$ / 0.:6 hr.

12)

A sla& of paper pulp 1.5 "eter 3 1.5 "eter 3 5 ""# thick is to &e dried under  constant drying conditions fro" 15! to 9.5! "oisture (dry &asis). $he eFuili&riu" "oisture is 2.5! (dry &asis) and the critical "oisture is 0.8 kg free water per kg dry pulp. $he drying rate at the critical point has &een esti"ated to &e 1.80 kg*("2)(hr). ensity of dry pulp is 0.22 g"*cc. Assu"ing drying to take place fro" the two large faces only# calculate the drying ti"e to &e provided.

GK / 0.025#

HC / 1.8 kg*" 2hr#

ensity of dry pulp / 0.22 g*cc#

GCr  / 0.8#

G1 / 0.15#

G2 / 0.095#

A / (1.5 D 1.5) D2 / 8.5 " 2#

=olu"e of "aterial / 1.5 D 1.5 D0.5 / 1.125 D10 ,2 "+# ∴  /

1.125 D10 , 2 D 0.22 D 10 + / 2.865 kg

t$ / (*AHC) D >(G1 E GCr ) ? L(GCr  E GK) ln >(G Cr  E GK)* (G 2 E GK)M ;ut here initial "oisture is "ore than the G Cr . o there is no constant rate drying  period and only falling rate period is o&served. ∴  t$ /

*AHC >GCr  E GK ln >(G1 E GK)* (G 2 E GK)M

t$ / >2.865*8.5 D 1.8 D >0.8E 0.025 ln >(0.15 E 0.025)* (0.095 E 0.025)M / 0.125 hr or 6.52 "in

1+) Snder constant drying conditions# a filter cake takes 5 hours to reduce its "oisture content fro" +0! to 10! on wet &asis. $he critical "oisture is 18! and the eFuili&riu" "oisture 8!# &oth on dry &asis. Assu"ing the rate of drying in the falling rate period to &e directly proportional to the free "oisture content# esti"ate the ti"e reFuired to dry the cake fro" +0! to ! "oisture on wet &asis. 31 / 0.+#

32 / 0.1#

G1 / 0.+*0.6/ 0.829# GK / 0.08#

G Cr  / 0.18#

G2 / 0.1*0.: / 0.111# t $ / 5 hrs# 128

31@ / 0.+ and hence# G1@ / 0.829.

32@ / 0.0 and G2@ / 0.0+9#

5 / (*AHC) D >(G1 E GCr ) ? L(GCr  E GK) ln >(G Cr  E GK)* (G 2 E GK)MNN..(1) t$@ / (*AHC) D >(G1@E GCr ) ? L(GCr  E GK) ln >(G Cr  E GK)* (G 2@E GK)M NN(2) ividing (1)* (2) 5 / >0.829E 0.18 ? L(0.18 E 0.08) ln >(0.18 E 0.08)* (0.111 E 0.08)M t$@ >0.829E 0.18 ? L(0.18 E 0.08) ln >(0.18 E 0.08)* (0.0+9E 0.08)M 5*t$@ / 0.+229*0.8+21 ence# t$@ / .: hr.

18) heet "aterial# "easuring 1"2 and 5 c" thick# is to &e dried fro" 85! to 5! "oisture under constant drying conditions. $he dry density of the "aterial is 850 kg per cu&ic "eter and its eFuili&riu" "oisture content is 2!. $he availa&le drying surface is 1 "2. O3peri"ents showed that the rate of drying was constant at 8.9 kg*(hr)("2) &etween "oisture contents of 85! and 20! and thereafter the rate decreased linearly. Calculate the total ti"e reFuired to dry the "aterial fro" 85! to 5!. All "oisture contents are on wet &asis.

A / 1"2#

5c" thick#

3i / 0.85#

3K / 0.02#

3 Cr  / 0.2#

32 /

0.05# GK / 0.02* (1 E 0.02) / 0.02081#

H C / 8.9 kg*" 2hr#

GCr  / 0.25#

G1 / 0.85* (1 , 0.85) / 0.919# G 2 / 0.052.  ensity of dry pulp / 850 kg*" + =olu"e of "aterial / 1 D 5 D10, 2 / 0.05 " + ∴  /

850 D 0.05 / 22.5 kg

t$ / (*AHC) D >(G1 E GCr ) ? L(GCr  E GK) ln >(G Cr  E GK)* (G 2 E GK)M t$ / >22.5* (1 D 8.9)D>(0.919E 0.25) ?L(0.25 E 0.02081) ln >(0.25 E0.02081)*(0.052 E 0.02081)M

/ 8.69 hr.

15) 7et solids containing 120 kg per hour of dry stuff are dried continuously in a specially designed drier# cross circulated with 2#000 kg per hour of dry air under the following conditions' A"&ient air te"perature

/ 20°C

O3haust air te"perature

/ 60°C

125

Ovaporation of water

/ 150 kg per hour  

%utlet solids "oisture content/ 0.25 kg per hour  Inlet solids te"perature

/ 15°C

%utlet solids te"perature

/ 5°C

ower de"and

/5k7

eat loss

/ 19 k 7

Osti"ate heater load per unit "ass of dry air and fraction of this heat used in evaporation of "oisture. ata' ean specific heat of dry air / 1 kT kg, 1 - , 1 Onthalpy of saturated water vapour

/ 2#2 kT per kg

 ean specific heat of dry "aterials / 1.25 kT kg, 1 - , 1 ean specific heat of "oisture

/ 8.19 kT kg, 1 - , 1.

$otal Fuantity of solid 120 kg dry stuff  Air used is 2000 kg*hr dry air 

;asis' 1 hr. eat reFuired for heating 150 kg water fro" 15°C to 5°C / 150 D 8.19 D (5 E 15)/ +1+50 kT eat reFuired for 150 kg water evaporation / 150 D 22 / +:+:00 kT eat reFuired for heating air fro" 20°C to 60°C / 2000 D 1 D (60 U20) / 100000 kT eat reFuired for heating "oisture in solid fro" 15°C to 5°C / 0.25 D 8.19 D (5 E 15) / 52.25 kT eat reFuired for heating dry solid fro" 15°C to 5°C / 120 D1.25 D (5 U15) /6500 kT eat lost / 19 D +00 / 8900 kT

12

o total heat reFuired*hr / +1+50 ?+:+:00 ? 100000 ?52.25 ? 6500 ? 8900 / 5:602.25 kT*hr./ 1 kw (i.e.) 1 k7 of heat is needed for 2000 kg*hr of dry air  ∴ eat

reFuired*"ass dry air / 1* (2000*+00) / 2:9.9 k7

(i) eat needed for evaporation / >+:+:00? +1+50*+00 / 119.1+ (ii) 4raction of this heat needed for evaporation / 119.1+*2:9.9 / 0.+:5+ or +:.5+!.

1) A dru" drier is &eing designed for drying of a product fro" an initial total solid content of 12! to final "oisture content of 8!. An overall heat transfer  coefficient (S) 1600 7*"2 C is &eing esti"ated for the product. An average te"perature difference &etween the roller surface and the product of 95°C will &e used for design purpose. eter"ine the surface area of the roller reFuired to  provide a production rate of 20 kg product per hour. Initial "oisture content / 12! 4inal "oisture content / 8! roduction rate / 20 kg final product*hr  ∴  8

kg "oisture in 100 kg product

In 20 kg product weight of "oisture / (8 D 20)*100 / 0.9 kg ry solid weight / 20 E 0.9 / 1:.2 kg $otal initial "oisture content / 1:.2 D >0.12* (1 E 0.12) / 2.192 kg ∴ 7ater

evaporated / 2.192 E 0.9 / 1.9192 kg*hr 

V  at 95°C / 22:.1 kT*kg eat reFuired / 7DV  (Assu"ing the solid "i3ture enters at 95°C and

∴

only "oisture re"oval &y evaporation is alone considered ) / 1.9192 D 22:.1 / 8168.6+ kT*hr  SD A D W$ / 7DV  1600 D A D 95 / 81686+0*+00 ∴  A

/ 9.025 D 10 , + "2 or 90.25 c" 2.

Exer+ise:

126

1) heet "aterial# "easuring 1"2 and 5 c" thick# is to &e dried fro" 50! to 2! "oisture under constant drying conditions. $he dry density of the "aterial is 800 kg* "+ and its eFuili&riu" "oisture content is negligi&le. $he availa&le drying surface is 1 "2. O3peri"ents showed that the rate of drying was constant at 8.9 kg* (hr)("2) &etween "oisture contents of 50! and 25! and thereafter the rate decreased linearly. Calculate the total ti"e reFuired to dry the "aterial fro" 50! to 2!. All "oisture contents are on wet &asis. (Ans ' .5+ hrs)

2) Calculate the critical "oisture content and the drying rate during the constant rate  period for drying a wet sla& of siRe 20 c" D 65 c" D 5 c"# whose dry weight is 1 kg. ;oth the sides are used for drying. $he stea" used was at + at". pressure and was consu"ed at the rate of 0.1+5 g*s. c"2  of the contact surface. $he following drying data is availa&le for the sa"ple. Assu"e eFuili&riu" "oisture content sis negligi&le. rying $i"e# hrs a"ple weight# kg

0

0.25

1.0

1.5

2.0

2.5

+.0

8.0

.0

9.0

10.0

12.0

1:.:

1:.6

1:.2

19.95

19.

19.+

19.1

16.5

1.:2

1.8

1.15

1.05

+) $he following data is availa&le for drying a su&stance. Osti"ate the drying ti"e needed to dry a si"ilar sa"ple under si"ilar drying conditions fro" 80! to 12! "oisture content# on wet &asis. $he drying surface is 1 " 2*8 kg of dry weight and the initial weight of the wet sa"ple is 90 kgs. G (dry &asis)  H# kg*hr. "2

0.+5 0.25 0.+ 0.+

0.2 0.+

0.19 0.1 0.2 0.28

0.18 0.21

0.10 0.15

0.09 0.06

0.05 0.05

8) 165 kg of wet "aterial with 25! "oisture is to dried to 10! "oisture. Air enters at 5 XC ;$ and a 7;$ of 25 XC. $he velocity of air is 150 c"*s. rying area eFuals 1 "2*80 kg dry weight. G(dry &asis) 0.2 0.22  H# kg*hr. "2 1.5 1.5 (Ans ' .96 hrs)

0.20 1.5

0.19 0.1 1.+ 1.2

0.18 1.08

0.12 0.:

0.1 0.65

0.09 0.

5) A wet solid is dried fro" +5 to 9! "oisture in 5 hrs under constant drying condition. $he critical "oisture content is 15! and eFuili&riu" "oisture content

129

is 5!. All the "oisture contents are reported as percentage on wet &asis. Calculate how "uch longer it would take place under the si"ilar drying conditions to dry fro" 9! to ! "oisture on wet &asis.

) A certain "aterial was dried under constant drying conditions and it was found that 2 hours are reFuired to reduce the free "oisture fro" 20! to 10!. ow "uch longer would it reFuire to reduce the free "oisture to 8!J Assu"e that no constant rate period is encountered.

6) It is desired to dry sheets of "aterial fro" 6+! to 8! "oisture content (wet  &asis). $he sheets are 2 " D+ " D 5 "". $he drying rate during constant rate  period is esti"ated to &e 0.1 kg*hr."2. $he &one,dry density of the "aterial is +0 kg*"+. $he "aterial is dried fro" &oth the sides. $he critical "oisture content is +0! on wet &asis and eFuili&riu" "oisture content is negligi&le. $he falling rate  period is linear. eter"ine the ti"e needed for drying. (Ans ' 2.6:9 hours)

9)

A sla& of paper pulp 1 "eter D 1 "eter D5 ""# is to &e dried under constant drying conditions fro" 0! to 20! "oisture (wet &asis) and the critical "oisture is 1.5 kg water per kg dry pulp. $he drying rate at the critical point has &een esti"ated to &e 1.80 kg*(" 2)(hr). $he dry weight of each sla& is 2.5 kg. Assu"ing that drying rate is linear in falling rate period and drying takes place fro" the two large faces only# calculate the drying ti"e needed. (Ans ' 8.9 hrs)

:) 50 kg of &atch of granular solids containing 25! "oisture is to &e dried in a tray dryer to 12! "oisture &y passing a strea" of air at :2°C tangentially across its surface at a velocity of 1.9 "*s. If the constant rate of drying under these conditions is 0.0009 kg "oisture*"2.s and critical "oisture content is 10!. Calculate the drying ti"e if the surface availa&le is 1.0 "2.(All "oisture contents are on wet &asis).

(Ans ' 2.55 hours)

10) A plant wishes to dry a certain type of fi&er &oard in sheets 1.2 "D 2 " D12 "" . $o deter"ine the drying characteristics a 0.+" D0.+ " &oard with the edge sealed# so that drying takes place only fro" two large faces only# was suspended fro" a  &alance in a la&oratory dryer and e3posed to a current of hot dry air. Initial 12:

"oisture content is 65!# critical "oisture content 0! and eFuili&riu" "oisture content 10!. ry "ass of the sa"ple weighs 0.: kg. Constant drying rate 0.0001 kg*"2.s. eter"ine the ti"e for drying large sheets fro" 65! to 20! "oisture under the sa"e drying conditions. All "oisture contents a re on wet &asis

11) A &atch of wet solid was dried on a tray drier using constant drying conditions and the thickness of "aterial on the tray was 25 "". %nly the top surface was e3posed for drying. $he drying rate was 2.05 kg*" 2.hr during constant rate period. $he weight of dry solid was 28 kg*" 2 e3posed surface. $he initial free "oisture content was 0.55 and the critical "oisture content was 0.22. Calculate the ti"e needed to dry a &atch of this "aterial fro" a "oisture content of 0.85 to 0.+0 using the sa"e drying condition &ut the thickness of 50 "" with drying fro" fro" the top and &otto" surfaces. (Ans ' 1.65 hrs)

12) A sa"ple of porous sheet "aterial of "ineral origin is dried fro" &oth sides &y cross circulation of air in a la&oratory drier. $he sa"ple was + " 2 and "" thick  and edges were sealed. $he air velocity is + "*s. ;$ and 7;$ of air were 52°C and 21°C respectively. $here was no radiation effect. Constant rate drying was 6.5 D 10,5 kg*s until critical "oisture content of 15! (on wet &asis) was o&tained. In the falling rate period# rate of drying fell linearly with "oisture content until the sa"ple was dry. $he dry weight of the sheet was 1.9 kg. Osti"ate the ti"e needed for drying si"ilar sheets 1.2 "D1.2 " D12 "" thick fro" &oth sides fro" 25! to 2! "oisture on wet &asis using air at a ;$ of °C &ut of the sa"e a&solute hu"idity and a linear velocity of 5 "*s. Assu"e the critical "oisture content re"ains the sa"e.

1+) A pig"ent "aterial# which has &een re"oved wet fro" a filter press# is to &e dried  &y e3tending it into s"all cylinders and su&Yecting the" to through circulation drying. $he e3trusions are  "" in dia"eter# 50 "" long and are to &e placed in screens to a depth of 5 "". $he surface of the particles is esti"ated to &e 2:5 "2*"+ of &ed and the apparent density is 1080 kg dry solid*" +. Air at a "ass velocity of 0.:5 kg dry air*" 2s will flow through the &ed entering at 120 °C and a hu"idity of 0.05 kg water*kg dry air# esti"ate the constant drying rate to &e e3pected. 1+0

18) It is necessary to dry a &atch of 10 kg of wet solid fro" +0! to 5! "oisture content under constant rate and falling rate period. $he falling rate is assu"ed to  &e linear. Calculate the total drying ti"e considering an availa&le drying surface of 1"2*80 kg of dry solid.J $he flu3 during constant rate period is 0.000+ kg*" 2s. $he critical and eFuili&riu" "oisture contents are 0.2 and 0.05 respectively. If the air flow rate is dou&led# what is the drying ti"e neededJ $he critical and eFuili&riu" "oisture contents do not change with velocity of air &ut Hc varies as B0.61 where B is the "ass flow rate of air..

15) A rotary dryer using counter current flow is to &e used to dry 12000 kg*hr of wet salt containing 5! water (wet &asis) to 10! water (wet &asis) . eated air at 186 °C with a 7;$ of 50 °C is availa&le. $he specific heat of the salt is 0.21 -cal*kg °C. $he out let te"peratures of air and salt are 62 °C and :+ °C respectively. Calculate the dia"eter of the dryer reFuired.

1) uring the &atch drying test of a wet sla& of "aterial 0.+5 " 2 and 6 "" thick# the falling rate Z H @ was e3pressed as 0.:5 (G E 0.01) where ZH @ is the drying rate in kg* " 2.s and G is the "oisture content in kg "oisture*kg dry solid. $he constant drying rate was 0.+9 kg* " 2.s and sla& was dried fro" one side only with the edges sealed. ensity of the dry "aterial is 1200 kg*" +. It is desired to reduce the "oisture content fro" +5! to 5! on wet &asis. 7hat is the ti"e needed for dryingJ.

1+1