Mass Transfer Solved problems
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9.6-2) Prediction of Effect of Process Variables on Drying Rate .Using the conditions in example 9.6-3 for the constant rate drying period, do as follows:a Predict the effect on Rc if the air velocity is only 3.! m"s
# Predict the effect if the gas temperat$re is raised tom %6.%&' and it remains the same.
()*+:
rom example 9.6-3
/ .0!% x .0!% m
# / 1!.0 mm
v / 6.2 m"s
d# / 6!.6&'
4 / .2 5g 41"5g d.a
R+7U)R+8:
a Rc if v / 3.! m"s
# Rc with / %6.6&'
U):
a; rom ex. 9.6-3
3037.1
m
kg = ρ
=
=
3037.13600
s
m3.05
m
kg
hr
s
vG ρ
G = 11386.26 kg m2 /h
( )
( )
( ) ( ) ( )36009.286.65
10002433
8703.35
8703.35
11386.260.0204
0204.0
2
8.0
8.0
−=
−=
−
=
=
=
W T T
W
h Rc
K m
W h
Gh
λ
Rc = 1.9479 g! "2 -#
#; < / %6.%&'
4 / .2 5g 41"5g d.a
( )( ) ( )( )[ ]( )7.76273010.01056.42.83x10
1056.41083.2
33-
33
++=
+=
−
−−
x
T H x xV H
VH = 1.0056 m3 /kg da
30044.1
0056.1
010.00.1
m
kg =
+= ρ
=
=
30044.13600
s
m6.1
m
kg
hr
s
vG ρ
G = 22056.1447 kg/ h-m2
( )
K m
W h
Gh
−
=
=
=
2
8.0
8.0
8771.60
22056.14470.0204
0204.0
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( )
( ) ( ) ( )36009.287.76
10002433
8771.60 −=
−= W T T W
h Rc
λ Rc = 4.21 g!#-"2
9.3-1 Humidity from Vaor !r"##ur". $G"a%ko&i#' he air in a room is at 3%.=o' and a total Press$re of 22.3 >pa a#s. containing water vapor with a partialpress$re p / 3.!9 >Pa. 'alc$late:
a; 4$midity#; at$ration h$midity and percentage 4$midityc; Percentage Relative 4$midity
()*+:room / 3%.=o'P / 22.3 >pa a#s
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Pa / 3.!9 >Pa
R+7U)R+8:a; 4#; 4sc; ?R4
U):
rom team a#le @4and#oo5: a#le 1-=;< 3%.= o' / 32.= >Po
/ 6!9.6%660 PaPo
/ 6.!9% >Pa
a; 4$midity, 4
Humidity , H = MW water
MW air [ P A
PT − P0
A ]
H =18
29 [ 3.59
101.3−6.65097 ]=.0235kg H
2O
Kgdryair
#; at$ration 4$midity, 4s
Hs= MW water
MW air [ P0
A
PT − P0
A ]
H =18
29 [ 6.65097
101.3−6.65097 ]=.0426kg H
2O
Kg dry air
c; Percent R4
%RH = H
Hs (100 )
%RH =.0235
.0426(100 )=55.1643
9.3-6 (dia)ati* +aturatio% of (ir
ir enters an adia#atic sat$rator having a temperat$re of %6.% o' and a dew-point temperat$re of 0.6 o'
)t leaves the sat$rator 9 ? sat$rated. Ahat are the final val$es of H and o'B
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90 % RH
Tdew = 40.6 oC
100 % RH
Tdb = 76.7 oC
H2 = .065 Kg water/ dry air
= 46 oC
()*+:
R+7U)R+8:
inal val$es for 4 and o'
U):
$sing ig$re 9.3-1 @(ean5oplis;
)t is proposed to install a #atch drier large eno$gh to handle 3! l# dry solids containing 11l#water. rom the following data, calc$late the total drying time reC$ired.
90% RH
ADIABATIC
SATURATOR
Td = 76.7 oC
D = 40.6 oC
90 % RH90 % RH
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'ritical free moist$re content / .!lbH 2O
lbdry solid
+C$ili#ri$m moist$re content / .0lbH 2O
lbdry solid
Doist$re content of prod$ct / .=lbH 2O
lbdry solid
he c$rve for the falling rate period is a straight line. he rate of drying at a constant rate
period is .6lb
min .
(iven:eed,
3! l# dry solids E1/ .=lb H 2O
lb dry solid
11 l# water
Rc/ .6lb
min F Ec/ .!lbH 2O
lbdry solid
ReCGd: t / B
olGn:
ss$me: / 2ft1
ree Doist$re:
EH / .0
X 1=320
350−¿
.0 / .!=66
E1 / .= I .0 / .0Ec / .! I .0 / .06
t / Ms
ARc [ ( X 1− Xc )+ Xc ln
Xc
X 2]
t /
350
0.6(1)[0.5866−0.46 )+0.46 ln
0.46
0.04¿
t / %19.12%2 min
8R)+R
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poro$s solid is dried in a #atch dryer $nder constant drying conditions. even ho$rs arereC$ired to red$ce the moist$re content from 3!? to 2?. he critical moist$re content was fo$nd to#e 1? and the eC$ili#ri$m moist$re is 0?. ll the moist$re contents are on the dry #asis
ss$ming that the rate of drying d$ring the falling rate period is proportional to the free-moist$recontent, how longs sho$ld it ta5e to dry a sample of the same solid from 3!? to !? $nder the samedrying conditionsB
()*+:
irst 'ondition econd 'ondition
t / % hr s
E2 / .3! - .0 / .32
E1 / .2 - .0 / .6
Ec / .1 - .0 / .26
EH
/ .0
t / B
E2 / .3! - .0 / .32
E1 / .! - .0 / .2
Ec / .1 - .0 / .26
EH/ .0
R+7U)R+8: t for the 1nd condition
U):
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t T = ms
ARc [( x1− xc)+ xcln
Xc
X 2]
7 Hrs= ms
Arc[(0.31−0.16)+0.16 ln(
0.16
0.06)]
lb H 2O
lb dry solution
m/ 11.=63
Rc
t T , 2nd=22.8063[(0.31−0.16)+0.16 ln(0.16
0.01)]
t T , 2nd=13.5381rs
wet solid is dried form 36? to =? in ! ho$rs $nder constant drying conditions. 'riticalmoist$re is 20? and the eC$ili#ri$m moist$re is 0?. ll moist$re contents are on wet #asis.
@a; 4ow m$ch longer @in ho$rs; wo$ld it ta5e, $nder the same dryingconditions, to dry from =? to !? moist$reB
@#; he solid is a 1-in thic5 sla#, 2 ft1 and dried from #oth sides. )t has a density
of21 l# dry solid"ft3 wet solid. Ahat is the drying rate at the instant the moist$re
content is =?
()*+:
irst 'ondition econd 'ondition
t3 / ! hr s
x2 / .36
x1 / .=
xc / .20
xH
/ .0
t3 / B
x2 / .36
x1 / .!
xc / .20
xH
/ .0
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R+7U)R+8:
a; t<=? to !?
#; R < x1/ .=
U):
t T = ms
ARc [( x1− xc)+ xcln
Xc
X 2]
x2/0.36
0.64−
0.04
0.96=0.5208
x2/
0.08
0.92−0.04
0.96=0.0453
x2/0.34
0.86−
0.04
0.96=0.1211
5rs= ms
ARc [(0.5208−0.1211)+0.12 ln
0.1211
0.0453]
ms
ARc=9.6365
1nd 'ondition:
5rs=9.6365[(0.5208−0.1211)+0.12 ln 0.1211
0.011 ]
t/ 6.6!66 hrs
t<=? to !?/@6.6!!!-!; hrst<=? to !?/ 2.6!!! hrs
@#;
< t/.=
'ondition: alling Rate Period
R/ axJ #
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R/ Rc− R
!
Xc− X ! ( x )
Rc /
ms
A (9.6365)
ms / 21 lbdry solid"t 3wet solid (
1 "t 2)( 212
"t )=20lb
Rc/
2(1 "t 2)(9.6365)¿¿
20 lb dry solid¿
R/ Rc
Xc X
at the origin RK/,EK/
R=1.0377
lb
"t 2r (0.0453 )
0.1211
R/ .3==lb
"t 2.r
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2 cfm of air are to #e cooled fr0om 9L to %1L, #y the $se of a horiMontal spray type h$midifier,
employing a co$nterflow of air and water. he air has an initial h$midity of .22 l# water vapor per l# of
dry air. he $nevaporated water collects inside the apparat$s to #e recirc$lated to the spray noMMles,
and ma5e$p water at %L is fed to the p$mp. 'alc$late the following #ased on the data given #elow:
a; cross section of spray cham#er in ft1
#; l#s of water sprayed per ho$r
c;l#s of ma5e $p water reC$ired per ho$r
d;length of spray cham#er in ft
d; h$midity of air leaving the cham#er as l# water per l# of dry air
8ata: ss$me the spray cham#er operates adia#atically with normal #arometric press$re. Ahen
spraying 21 l# water per ho$r per ft1 of cross section and employing an air rate of 10 l# of dry air
per ho$r per ft1 of cross section, the overall coefficient of heat transfer is 9 NU" hr L t3 of spray
cham#er.
()*+:
(x / 21 l#"hr.ft1
(y/ 10 l# d.a" hr.ft1
U / 9 Nt$"hr.L.ft3
R+7U)R+8:
a;
#; a
c; l#s of ma5e$p water
d; Ot
U):
a; m d.a /#a
/m d $ a
y
< pt #: # / .22
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*4 / Q
0.7302(90+460)1
1
29+0.011
18 ¿ ¿
/20. 90 ft3"l# d.a
md.a/10000 "t 3/min
14.094 "t 3/ l b d $ a / %9. !3 l# d.a"min
/
709.53 l b d $ a /min (60min
r )
2400 l b d $a /r$"t 2/ 2%.%0 ft1
#; a /1200 lb
r $ " t 2 @2%.%0 ft1; / 121=!.6! l#"hr
c; + / m d.aQ a- #
a QTdb−72& ' Tw=70& ' / .2!! l# 41"l# d.a
+/ %9.!1@60
1¿
Q .2!!-.22 / .22
+/ 292. !% l# 41"hr
d; m d.a 'ave@ y# - ya; / U (Ty−Tx)l M
'a / .10 J [email protected]!!;/ .10%
'# / .10 J [email protected]; / .10!
'ave/ .106
(Ty−Tx)l=(90−70)−(72−70)
ln 90−70
72−70/ %.=1L
@.106;@%9.!1;@6"2;@9-%1;/ 9(tu
r & '$ "t 3 @%.=1;@2%.%0;@Ot;
Ot / 2!.2 ft
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ir entering an adia#atic cooling cham#er has a temperat$re of 31.1 ℃ and a
percentage relative h$midity of 6!?. )t is cooled #y a cold water spray and sat$rated
with water in the cham#er. fter leaving, it is heated to 13.9 ℃ . he final air has a ?
relative h$midity of 0?. @a; what is the initial h$midity of the airB @#; what is the finalh$midity after heatingB @c; what is the temperat$re #efore heatingB
(iven:
ReC$ired: a; 2
#; 3
c; 3
ol$tion:
a;
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#;
c;
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