DRYERS AND DRYING
1.
Tobac obacco co in in a war warehou ehouse se,, held held at at 30 C and and 40% 40% rela relati tive ve hum humid idit ity, y, is plac placed ed in in a room at 32 C and 70% relative humidity preparatory to bein wor!ed on. "or each #0 ! o$ tobacco moved $rom the warehouse, what is the bonedry weiht& 'hat is the actual actual weiht o$ this (uantity (uantity o$ tobacco a$ter stayin stayin in the wor!in wor!in room& )olution* m1 = 50 kg "or tobacco at 40% + Regain1 = 13.30% at 70% + Regain2 = 25.00%
-a
Bdw=
Regain 1 0.1330Bdw= 50 − Bdw Bdw
=
44 13 kg
.
m2 / actual weiht
-b
m2 2.
m1 − Bdw
=
( Regain ) ( Bdw Bdw) + Bdw Bdw= ( 0.25) ( 44.13) + 44.13 = 55.16 kg 2
ir ir ente enters rs an an adia adiaba bati tic c drie drierr at ms ms thr throu ouh h a 2m 2m dia diame mete terr duct duct at at 2 C dry dry bulb and 22 C wet bulb temperatures. t is heated to 50 C be$ore reachin the material to be dried and leaves the drier at 44 C and 50% +. The material enters the drier with a moisture content o$ 24%, and leaves with a moisture content o$ 5%. 6etermine -a the mass o$ water removed per ! o$ dry air, -b the volume ow o$ air enterin the reheater, -c the ! o$ water evaporated per second, -d the mass ow rate o$ material leavin the drier, and -e the heat re(uirement o$ drier per ! o$ water evaporated. )olution*
1
DRYERS AND DRYING
at 1, t db
1
= 29 C ,
= 22 C
t wb
1
h1 = 64 2 kJ kg W 1 = 0 0138 kg kg .
.
v 1
= 0 874 m
3
at 2, t db
2
hg2
h2
=
=
3
3
W 3
80
C , W 2
= W = 0 0138 kg kg 1
.
. kJ kg
c pt db +W 2 hg2 2
3
pd
=
2643 7
at 3, t db
p s
kg
.
= 44 C ,
=
φ 3
( 1.0062) ( 80) + ( 0.0138) ( 2643.7 ) = 116.98 kJ kg = 80%
+
= 9 151 kPa = φ p d = ( 0.80)( 9.151) = 7.321 kPa .
3
=
3
0.622 ps
3
pt − ps
3
(
)
0.622 7.321 =
101.325 − 7.321
=
0.0484 kg
kg
-a 8ass o$ water removed per ! dry air / W 3 − W 2 = 0 0484 − 0 0138 = 0 0346 kg kg .
.
.
1 = π ( 2) 2 ( 6) -b 9olume ow rate o$ air enterin the reheater / V 4
-c 8ass o$ water evaporated / V 18 85 a ( W 3 − W 2 ) = 1 (W 3 − W 2 ) = . ( 0.0346 ) m 0.875 v 1
2
=
0.746 kg
s
= 18.85
m3 s
DRYERS AND DRYING
-d 8ass ow rate o$ material leavin the dryer / 5 ( 1 − 0.08) = m 4 ( 1 − 0.24 ) m
4 m
=
m
5
1 21m5
.
4 m 5 m a (W 3 W 2 ) but m 5 m 5 0.746 1.21m 5 = 3 552 kg s m −
=
−
−
=
.
-e
eat re(uirement per ! o$ water evaporated.
a ( h2 h1 ) m a (W 3 W 2 ) m −
=
−
3.
h2 h1 W 3 W 2
116 98
−
=
.
=
−
−
64 2 .
0 0346
=
1525 kJ
kgwate
.
drier is to be desined to reduce the water content o$ a certain material $rom ##% to 10%. ir at 2 C dry bulb temperature and with a humidity ratio o$ 0.00# !! is heated to #0 C in a reheater be$ore enterin the drier. The air leaves the drier at 35 C with 70% relative humidity. :n the basis o$ 1000 ! o$ product per hour, calculate -a the volume ow rate o$ air enterin the reheater, and -b the heat supplied in the reheater. )olution*
t 1, t db
1
=
29
C , W 1
= 0 005 kg kg .
h1 = 42 kJ kg v 1
= 0 862 m
3
.
at 2, t db
2
kg
= 50 C ,
W 2
= W = 0 005 kg kg 1
.
3
DRYERS AND DRYING
h2
= 63 5 kJ kg .
= 38 C ,
at 3, t db
3
W 3
φ 3
= 70%
+
= 0 0298 kg kg .
5 ( 1 − 0.10) = m 4 ( 1 − 0.55) m 5= m 1000 kg hr
(
4 m
=
1000 1 − 0.10 1 − 0.55
a m
=
4 m W 3
)
=
2000 kg hr
1 hr −m 2000 − 1000 = = ( 40 323 kg hr ) = 11 2 kg s − W 0 0298 − 0 005 s 3600 5
,
2
.
.
.
-a 9olume ow rate o$ air enterin the reheater / 1 = m av 1 = (11 2 )( 0 862 ) = 9 65 m3 s V .
.
.
-b eat supplied in the reheater /
4.
=
a ( h2 m
−
h1 ) = 11.2( 63.5 − 42)
=
240.8 kW
dryer is to deliver 1000 !hr o$ palay with ;nal moisture content in the $eed is 1#% at atmospheric condition with 32 C dry bulb and 21 C wet bulb. The dryer is maintained at 4# C while the relative humidity o$ the hot humid air $rom the dryer is 50%. $ the steam pressure supplied to the heater is 2 8
4
DRYERS AND DRYING
= 32 C ,
at 1, t db
1
t wb
1
= 21 C
h1 = 60 6 kJ kg W 1 = 0 0112 kg kg .
.
= 45 C ,
at 2, t db
2
W 2
= W = 0 0112 kg kg 1
.
= 74 9 kJ kg = 0 917 m kg
h2
.
3
v 2
.
= 45 C ,
at 3, t db
3
pd
= 9 593 kPa
ps
=
W 3
=
3
3
h3 -a
= 80%
+
.
φ 3 pd
3
( 0.80) ( 9.593) = 7.674 kPa
=
0.622 ps
3
pt − ps
3
hg3
φ 2
=
=
(
)
0.622 7.674 =
101.325 − 7.674
=
0.0510 kg kg
. kJ kg
2583 2
c pt db
3
+
W 3hg3
=
(1.0062) ( 45) + ( 0.0510) ( 2583.2 ) = 177 kJ kg
4 / m
=
5 (1 − 0.10) m 1 − 0.15
=
(
)
1000 1 − 0.10 1 − 0.15
= 1058.8
kg hr
-b Temperature o$ the humid air $rom the dryer / t db
3
-c
2 ir supplied to dryer / V
av =m
2
5
=
t dryer = 45 C .
DRYERS AND DRYING
a m 2 V -d
=
4 m
−
5 m
W 3 − W 2
=
1058.8 − 1000 0.0510 − 0.0112
= 1477.4
av = (1477 4 )( 0 917 ) = 1354 8 m =m
3
2
.
.
.
kg hr
h
eat supplied to heater in !' 1477 4 a ( h − h ) = =m ( 74 9 − 60 6 ) = 5 87 kW 3600 .
2
1
.
.
.
shfg 5.87 kW -e m s (1890.7) = ( 5.87) ( 3600) m s = 11 18 kg hr m =
.
6
