mass transfer solved problems

7/25/2019 Mass Transfer Solved problems

http://slidepdf.com/reader/full/mass-transfer-solved-problems 1/14

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



( )

( ) ( ) ( )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



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



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



'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



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):



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



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 #



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



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



*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



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;



#;

c;

mass transfer solved problems

Documents