1
a t S t s r i F
By extracti extracting ng kerosene, kerosene, 2 tons of waxed waxed paper is to be dewaxed dewaxed in a continuous continuous countercurrent extraction system that contains a number of ideal stages. The waxed paper contains, by weight, 25% paraffin wax and 5% paper pulp. The extracted pulp is put through a dryer to e!aporate the kerosene. The pulp, which retains the unextracted wax after e!aporation, must not contain o!er ".2 kg of wax per 1"" kg of wax free pulp. The kerosene used for the extraction contains "."5 kg of wax paper per 1"" kg of wax free kerosene. #xperiments show that the pulp retains 2." kg of kerosene per kg of kerosene and wax free pulp as it is transferred from cell to cell. The extract from the battery is to contain 5 kg of wax per 1"" kg of wax free kerosene. a $ind $ind the the o!e o!erf rflo low w str strea eam. m. b $ind the underflow stream c g wax wax & kg kg ker keros osen enee in in the the und under erfl flow ow d 'o. of stages
(i!en) by b
1)!erflow
5 kgwax aya 100 k kerosen kerosen
) ol!ent
2y2
bx b
$eed ) 3 Tons 0ax paper 25% wax 5% pulp
5 kgwax 1 osen 100 k1xkerose ker n
*e+uired) !erflow and -nderflow streams' stages. olutions) /B )
F + S =V 1 + L1
/aterial balance for the wax 4 Ton x
1000 kg =4000 kg 1 Ton
4000 kg x .25=1000 kg 4000 kg x .75=3000 kg
0ax balance
0.05 kgwax 100 k kerosen kerosenee
2 kg kerosen kerosenee 1 kg pulp pulp
1) -nderflow
0.2 kgwax
1000 kg +
0.05 kgwax 5 kg wax 0.2 kgwax wax ( 0.75 x 4000 kg ) V b= V a + 100 kgkerosene 100 kg kerosen 100 kg pulp kerosenee
ol!ent Balance
V b=V a+
2 kg kerosene kerosene ( 0.75 x 4000 kg ) =V a +6000 kg 1 kg pulp
ol!ing simultaneously we get the streams
V a=20141 . 4141 kg Kerose Kerosene ne
V b=V 2 =20081 . 4141 kg kerosen kerosenee
undeflow =6000 kg kerosene kerosene 0.2 kg wax wax x 0.75 x 4000 kg pulp 100 kg pulp kgwax x b= = 0 . 001 6000 kg kerosen kerosenee kg kerosene kerosene
0ax balance) 20081.4141 y 2 + 1000 =
5 ( 20141.4141) + 6000 x 1 100
equilib equilibrium rium conditio condition n x 1 = y a= y 2= 0.05
kg wax wax kg kerosen kerosenee
'o. of ideal stages ln
N =
(
(
0.001 −0.0005 0.05 −0.01529
) + = )
0.01529 −0.0005 ln 0.05 −0.001
1 4.5398 ≈ 5
y 2=0.01529
1000 kg +
0.05 kgwax 5 kg wax 0.2 kgwax wax ( 0.75 x 4000 kg ) V b= V a + 100 kgkerosene 100 kg kerosen 100 kg pulp kerosenee
ol!ent Balance
V b=V a+
2 kg kerosene kerosene ( 0.75 x 4000 kg ) =V a +6000 kg 1 kg pulp
ol!ing simultaneously we get the streams
V a=20141 . 4141 kg Kerose Kerosene ne
V b=V 2 =20081 . 4141 kg kerosen kerosenee
undeflow =6000 kg kerosene kerosene 0.2 kg wax wax x 0.75 x 4000 kg pulp 100 kg pulp kgwax x b= = 0 . 001 6000 kg kerosen kerosenee kg kerosene kerosene
0ax balance) 20081.4141 y 2 + 1000 =
5 ( 20141.4141) + 6000 x 1 100
equilib equilibrium rium conditio condition n x 1 = y a= y 2= 0.05
kg wax wax kg kerosen kerosenee
'o. of ideal stages ln
N =
(
(
0.001 −0.0005 0.05 −0.01529
) + = )
0.01529 −0.0005 ln 0.05 −0.001
1 4.5398 ≈ 5
y 2=0.01529
2
4n a sing single le step step soli solidl dli+ i+ui uid d extr extract actio ion n soybea soybean n oil oil has to be extract extracted ed from soybea soybean n flakes using hexane as sol!ent. 1"" kg of the flakes with an oil content of 2" wt% are contacted with 1"" kg fresh hexane. 1.5 kg of inert material hold back a constant !alue of 1 kg solution.
sol!ent
extract 6o!erflow7
1
2
extraction
step
21
"
feed
Total Total balance) b alance)
" 8 2 9 M 9 1 8 1 9 1"" 8 1"" 9 200 kg
underflow
Balance for compound :) " w:," 8 2 w:,2 9 / w:,/ with the feed concentration w:," 9 ".; and the suggestion, that no solid particles are included in the o!erflow, so w:,2 9 " follows) 1"" < ".; 8 1"" < " 9 2"" < w:,/ wA,M = 0.4 Balance for compound B) " wB," 8 2 wB,2 9 / wB,/ with the feed concentration wB," 9 ".2 and with the knowledge, that pure hexane is used as sol!ent, wB,2 9 ", follows 1"" < ".2 8 1"" < " 9 2"" < wB,/ wB,M = 0.1
The concentration of compound = 6sol!ent7 in the mixing point / can be determined either by a mass balance for compound = " w=," 8 2 w=,2 9 / w=,/ with w=," 9 ", because no sol!ent is included in the feed, and with w=,2 9 1, pure hexane, follows 1"" < " 8 1"" < 1 9 2"" < w=,/ wC,M = 0.5
or by the rule, that the sum of the mass percent of each compound in the point / has to be 1. w:,/ 8 wB,/ 8 w=./ 9 1 ".3 8 ".1 8 w=./ 9 1 wC.M = 0.5
0ith these concentrations the mixing point / can be drawn in the diagram, which has to be on the connection line of feed point $ and sol!ent =.
4t is gi!en, that 1 kg inert material retains 1.5 kg solution 6extractable substance 8 sol!ent 9 miscella 9 o!erflow7. Therefore the concentration of the underflow is
inert
w
:,-nderflow 9
inert material8extractable subst
1.5
w
:,-nderflow 9w A,L1
9
1.5 8 1
The amount of the lea!ing flows 1 and 1 can be calculated from the mass balance for compound : / w:,/ 9 1 w:,1 8 1 w:,1 with w:,1 9 " 6no solid material in the o!erflow7 and w:,1 9 ".> 6underflow7
L1=
! " # ! " # L 1
=200
0.4 .0.6
L1=133.333 kg 0ith the total balance / 9 181
follows 1 9 / 1 9 2"" 1??.??? V1 = 66.666 kg
The concentrations of B and = in the o!erflow 1 are calculated with the suggestion that no inert material : is included in the o!erflow.
! $ #V 1 =
20 $ = " + $ + % 0 + 20 + 100
100 % = ! % # V 1= " + $ + % 0 + 20 + 100
$ #V 1=0.1667
¿
! ¿ ! % # V 1 =0.8333 The composition of the underflow can be calculated by mass balances for compound B and =. 1 wB,1 8 1, wB,1 9 " wB," 8 2 wB,2
! $ # L 1=
L& ' ! $ #Lo−V 1 '! $ #V 1 L1
=
100 ' 0.2 −66.666 ' 0.1667 133.333
! $ # L 1=0.067 ! " #L 1+ ! $ # L 1+ ! % # L 1=1 ! % # L 1 =1−0.6 −0.67 ! % # L 1 =0.333
Total mass 6kg7
0t% :
0t% B
0t%=
$eed
1""
;"
2"
"
ol!ent 2
1""
"
"
1""
!erflow 1
>>.>>>
"
1>.>>
;?.???
-nderflow 1
1??.???
>"
>.
??.?
Situation for problems no. 23-28
By extraction with kerosene with "."5 lb wax per 1"" lb kerosene, 2 tons of waxed paper per day is to be dewaxed in a continuous countercurrent extraction system that contains a number of ideal stages. The waxed paper contains, by weight, 25 percent paraffin wax and 5 percent paper pulp. The extracted pulp is put through a dryer to e!aporate the kerosene. The pulp, which retains the unextracted wax after e!aporation, must not contain o!er ".2 lbs of wax per 1"" lbs of wax free kerosenefree pulp. #xperiment show that the pulp retains 2." lb of kerosene per lb of kerosene and waxfree pulp as it is transferred from cell to cell. The extract from the battery is to contain 5 lb of wax paper per 1"" lb of waxfree kerosene. @er 1"" lb of waxfree kerosenefree pulp,
2?. The kerosene in the exhausted pulp is e+ual to a. 15" lb c. 11 lb d. 212 lb b. 200 lb 23. The kerosene in the strong solution is e+u al to a. 5>1 lb c. >1 b. >51 lb d. 671 lb 25. The wax in the strong solution is e+ual to a. ?5.?5 lb c. 55.?? lb d. 5?.5? lb b. 33.55 lb 2>. The wax in the underflow to unit 2 is e+ual to a. ; lb c. 12 lb d. 13 lb b. 10 lb 2. The wax in the o!erflow from the second cell to the first is a. 10.22 lb c. 12.11 lb b. 11.12 lb d. 1?.1A lb 2;. The total number of ideal stages is e+ual to a. ? c. 5 d. > b. 4
(i!en)
olute 9 0ax ol!ent 9 erosene 4nert 9 @ulp
Y1
Y2
1
YN+1
2 X1
N XN
F= 2 Tons 25% Solute 75% Inert
olution) 4n $eed) 4nert 9 1"" lb 100 lbinert =133.3333 lb $eed 9 0.75 olute 9 6".25761??.???? lb7 9 ??.???? lb
4n $inal -nderflow) 4nert 9 1"" lb 0.2 lb solute x 100 lbinert =0.2 lb olute 9 100 lbinert ol!ent 9
2 lb sol(ent x 100 lbinert = 200 lb lbinert
!erall ol!ent Balance " 8 '81 9 1 8 2""
#+uation 1
!erall olute Balance 0.05 lb solute 5 lb solute ??.???? 8 6 100 lb sol(ent x '817 9 6 100 lb sol(et x 17 8 ".2
#+uation 2
$rom #+uation 1) '81 1 9 2"" $rom #+uation 2) −4
5 x 10
'81 "."51 9 ??.1???
#+uate #+uation 1 and #+uation 2, sol!e for '81 and 1) '81 9 ;1.?A; lb V1 = 671.3798 lb olute in 1) 5 lb solute 100 lb sol(ent x >1.?A; 9 33.5690 lb
ol!ent Balance in tage 1) " 8 2 9 >1.?A; 82"" 2 9 ;1.?A; lb olute Balance in tage 2)
??.????
(
8
solute
in
2
9
)
5 lb solute x 671.3798 lb sol(ent 100 lb sol(ent
Solut !" V2 = 10.2357 lb
olute in C2 9
5 lb solute x 200 lb sol(ent =10 lb 00 lb sol(ent
(
5 lb solute x 200 lb sol(ent ) 100 lb sol(ent
8
ol!ing for 'umber of tages)
ln [
N =1 + ln [
* N +1 − + N * 2− + 1 * N +1−* 2 + N − + 1
]
]
where) 0.05 lb solute =5 x 10−4 100 lb sol(ent
C '81 9
0.2 lb solute =1 x 10−3 200 lb sol(ent
D ' 9
10.2357 lb solute = 0.0117 C29 871.3798 lb sol(ent
10 lb solute = 0.05 200 lb sol(ent
D1 9
−4
−3
5 x 10 −1 x 10 ln [ 0.0117−0.05
' 9 1 8
−4
ln
[
]
−0.0117 ] 1 x 10 −0.05
5 x 10
−3
' 9 ?.A?A> 9 4 #tag#
@roblems no. 2A?1. 1"" kg of solid containing 5"% of a soluble material were treated with 2"" kg of a sol!ent containing the same solute at ?% concentration in a !essel under the constant agitation. :fter a long time, pressing separated in the solution and the solid. The solid analyEed ".5 kg of sol!ent per kg of inert solid. 2A. The amount of solute in the final underflow is approximately e+ual to a. 1".;2 kg
c. 2.; kg
b. ;.53 kg
d. .1> kg
?". The amount of sol!ent in the extract is approximately e+ual to a. 1">.2 kg
c. 1;.? kg
b. 21>." kg
d. 15>.5 kg
?1. Fow much extract was collectedG a. 2"1.>; kg
c. 21>."; kg
b. 1">.21 kg
d. 1A2.;> kg
$!%"&
!erflow, V1
Vo 9 2"" kg
6#xtract7
?% solute A% sol!ent
$eed, ' 9 1"" kg
-nderflow, L1 0.75 kgsol(ent
5"% solute
solid 9
kginert solid
5"% solid
()u!*d& 2A.7 amount of solute in final underflow
?".7 amount of sol!ent in the extract ?1.7 1
Solut!o"&
4n the $eed, $)
solute =( 0.5 ) ( 100 kg )=50 kg
inert solid =( 0.5 ) (100 kg )=50 kg
4n o)
solute =( 0.03 ) ( 200 kg )=6 kg
sol(ent =( 0.97 ) ( 200 kg )=194 kg
4n o!erflow, 1.)
solute = a sol(ent =Vi −a
olute balance)
Continuation... 4nerts balance)
inerts ∈ F =inerts ∈ L1=50 kg
olution balance)
solution ∈ F + Vo =solution ∈ L1+ V 1
( 50 + 0 ) +200 =93.5 −a + V 1 156.5 + a =Vi
:t #+uilibrium)
(
solute solution
) =(
(
a
) (
156.5+ a
V 1
solute solution
=
V 1
)
L1
56− a 37.5 + 56 −a
)
L1
a =45.1753 kg
2A.7 :mount of solute in underflow, 1)
solute =56 −a =56 −45.1753 =10.8247 kg ?".7 :mount of sol!ent in #xtract, i)
sol(ent =V 1− a=156.5 + a−a =156.5 kg ?1.7 1) V 1= sol(ent V + soluteV =156.5 + 45.1753 =201.6753 kg 1
1
S!tuat!o" +o* *obl-# "o. 3234
: solid B, contains a soluble component, :, of mass fractions
x "=0.25
,
x $= 0.75
and is
to be reco!er : by a sol!ent extraction with =. olid B and sol!ent = are mutually totally insoluble. The extracted solid is to be screw passed to a ".5 kg of solution&kg of B underflow. The entrainment of B in the o!erflow can be neglected. @er kg of feed and to obtain ;5% of : in the extract o!erflow. 32. The composition of the solution in the underflow is a. "."3 c. "."1 d. ".1" b. 0.07 33. The amount of sol!ent in the underflow is a. ".33 b. 0.53
c. ".;; d. 1.??
34. Fow much sol!ent = 6: free7 must be fedG a. 3.5000 kg b. 2.512 kg
c. 1.""" kg d. 5.2?11 kg
$/V&
(/(&
?2. x1 in 1 ??. sol!ent in 1 ?3. sol!ent =
SL/&
4n $eed, $) $ 9 1 kg olute 6:7 ) 6".257617 9 ".25 kg 4nerts 6B7 ) 6".57617 9 ".5 kg
4n -nderflow , 1 ) 4nerts 6B7 ) ".5 kg olution 6: 8 =7 ) 6".576".57 9 ".5>25 kg olute ) 61 ".;576".257 9 "."?5 kg ol!ent ) 6".5>25 "."?57 9 0.5250 kg
x 1=
0.0375 solute = =0.0667 solution 0.5625
olution Balance) 6".25 8 "7 8 = 9 1 8 ".5>25 1 9 = ".?125
H e+uilibrium)
( (
solute solution
) =( V 1
0.2125 % −0.3125
solute solution
) (
C = 3.5000 kg
=
V 1
)
0.0375 0.5625
L L1
)
L L1
S!tuat!o" +o* *obl-# "o. 3538
eeds containing ?"% weight oil are extracted in a countercurrent plant and ;;% of the oil is reco!ered in a solution containing 55% by weight of oil. The seeds are extracted with fresh sol!ent and 1 kg of solution is remo!ed in the underflow in association with e!ery 1.5 kg of insoluble material. ?5. The amount of sol!ent in final extract is approximately e+ual to a. 2>.3 kg b. 21.6 kg
c. 3>.> kg d. 3?." kg
?>. The amount of sol!ent in final underflow is approximately e+ual to a. 2>.3 kg b. 21.> kg
c. 3>.> kg d. 43.07 kg
?. The concentration of oil in the sol!ent stream for stage 1 is approximately e+ual to a. ".55 b. ".";
c. ".1; d. ".?3
?;. Fow many ideal stages are needed to attain the desired separationG a. 4 b. >
c. ; d. 1"
$/V&
(/(&
?5. ol!ent in 1 ?>. ol!ent in ' ?.=oncentration of oil 2 ?;.' SL/&
Basis) 1"" kg of $eed 4n $eed, $) 4nsoluble 9 "."61"" kg7 9 " kg il 9 ".?"61"" kg7 9 ?" kg 4n final !erflow, 1) il 9 ".;; 6?" kg7 9 2>.3 kg ol!ent 9 2>.3 kg
( ) 45 55
9 21.6 kg
y19x19".55
4n final -nderflow, '9193>.>>> kg il9 ".126?" kg7 9 ?.> kg 4nsoluble 9 " kg
olution 9 " kg
(
1 k g solution 1.5 kginsoluble
)
9 3>.>>> kg
ol!ent 9 3>.>>> kg ?.> kg9 43.0667 kg x '9
3.6 =0.0771 46.6667
4n $resh ol!ent, '81) ol!ent 9 21.> kg 8 3?. ">> kg 9 >3.>>> kg olute 9 " y '81 9 " olute Balance around tage 1) ?" kg 8 >3.>>> kg 6y 27 9 2>.3 kg 8 3>.>>> kg6".557 y29 0.3412 9 x2 ln
'9
1+
( (
0− 0.0771 0.3412−0.55
0− 0.3412 ln 0.0771−0.55
) )
9 4.05
ituation for problems ?A32 =alciumcarbonate precipitate can be produced by the reaction of an a+ueous solution of sodium carbonate and calcium oxide. The byproduct is a+ueous sodium hydroxide. $ollowing decantation, the slurry lea!ing the precipitation tank is 5 wt% calcium carbonate, ".1 wt% sodium hydroxide, and the balance water. ne hundred thousand lb&h of this slurry is fed to a twostage, continuous, countercurrent washing system to be washed with 2",""" lb&h of fresh water. The underflow from each thickener will contain 2" wt% solids. ?A. The amount of extract 3". The amount of sodium hydroxide in final extract 31. The amount of sodium hydroxide in final underflow 32. The percent reco!ery of sodium hydroxide in the extract $!%") 20,000 lb/h
(V1)
1
(V2)
(V3)
2 (F) 100,000 lb/h
(L1)
5 t% !al"iu# !arbonate 0$1 t% Soiu# &'ro(ie
Solut!o"& /" 'd&
%alcium %arbonate= 0.05 ( 100,000 )=5000 lb / , Sodium -ydroxide= 0.001 ( 100,000)= 100 lb / ,
!ater = 0.949 ( 100,000 )=94900 lb / ,
(L2 ) 20t% soli
Sol!d Bala"&
Solid ∈ F =Solid ∈ L 2
5000
lb = 0.20 ( L 2 ) ,
L 2 =25000
lb ,
/B) F + V 3 =V 1 + L 2 100,000+ 20,000=V 1 + 25,000
V 1 =95000
lb ,
Stag 1 )u!l!b*!u-:
(
Solute Solute ) =( ) Solution ( V 1 ) Solution ( L 1)
(
Solute Solute ) =( ) 95,000 (V 1 ) 20,000 ( L 1)
Solute ∈V 1 = 4.75 Solute ∈ L 1−−−eqn ( 1 ) Stag 2 )u!l!b*!u-:
(
Solute Solute ) =( ) Solution ( V 2 ) Solution ( L 2)
(
Solute Solute ) =( ) 20,000 ( V 2) 20,000 ( L 2 )
Solute ∈V 2 = Solute ∈ L 2−−−eqn ( 2 ) Solut Bala" !" Stag 2
Solute L 1+ SoluteV 3 =Solute V 2 + Solute L 2 Solute L 1+ 0= Solute V 2 + Solute L 2 Solute L 1=Solute L 2+ Solute L 2 Solute L 1=2 Solute L 2−−−eqn ( 3) %*all Solut Bala"
Solute F + Solute V 3 = SoluteV 1 + Solute L2 100 + 0 =Solute V 1 + Solute L2 −−−eqn ( 4 ) ubstitute e+n 617 to e+n 637 100= 4.75 Solute L 1 + Solute L 2−−−eqn ( 5 )
ubstitute e+n 6?7 to e+n 657 100= 4.75 ( 2 ) Solute L 2 + Solute L 2
Solute L 2=9.52
lb ,
-sing e+n 3) 100 + 0 =Solute V 1 + 9.52
lb Solute V 1=90.48 ,
ercent /eco(ery ( / )=
SoluteV 1− Solute V 3 Solute F
/ =
( 90.48 −0 ) 100
x 100
/ = 90.48
;(BLM 4346
(round roasted coffee contains ;% soluble solids, 2% water, and A"% inert insoluble solids. 4n order to obtain an extract with high soluble solids content without ha!ing to concentrate it for spray drying, a countercurrent extraction process is to be used to prepare the extract. 4t is desired that the final extract contain ".15kg soluble&kg water and that the soluble of the spent coffee grounds not to exceed ".""; kg&kg dry inert solids. The coffee grounds carry 1 kg water&kg of solublefree inert solids and this +uantity is constant with the solute concentration in the extract.
*#IJ) 3?7 The amount of final extract is approximately e+ual to a. 55.;1 kg
b. 3;.53 kg
c. 2.; kg
d. 2;.1 kg
337 The concentration of the solution adhering to the extracted solids is approximately e+ual to a. "."A?>
b. ".""A
c. ".1?"3
d. ".""?2
357 The water&coffee ratio to be used in the extraction is a. 1.?
b. 2.;;
c. ".A;
d. 1.;
3>7 The number of extraction stages needed for this process is a. 5
b. >
c.
d. ;
-T4')
ero, 31 0$15 *g solute
32
3-
3
Y2
Y-
Y
*g &2 1
Fee, F % Solute
4n
2% &2
olute)
2
N
)2
)-
.1
.2
.-
2kg 4nerts) ".A61""79A"kg
4n final underflow) 4nerts9 inerts in $9A"kg olute9"."";6A"79".2kg
D '9 "."";
-
)1
ol!ent)
ol!ent9A"6179A" kg
olution)
the
".";61""79 )= 1 *g &2/*g Inerts
Solent, 3n+1
feed)
Final nero, .n
basis61""
Xn
kg7
Solute/Inerts = 0$00
;kg "."261""79
(
Solute Solution
)
0.72 =0.0079 <44: in ' 9 90 + 0.72
4n final o!erflow, 1) olute balance) ;8"9".28olute in 1
Solute =0.15 C19 D19 Sol(ent
$inal !erflow6extract79solute8sol!ent 9 .2;83;.5???kg9 1 9 55. 8133 kg<43:
4n sol!ent stream, n81 Cn819"6pure water7 ol!ent9 n819G !erall sol!ent bal) 28n819A"83;.5??? n8191?>.5???
*atio)
Vn + 1 136.5333 = =1.3653 100 F
<45:
olute for C2 using olute balance around stage 1 ;8 2C291D18.2; 29 n8191?>.5??? C29G 9 9A"
D19C19".15 ;81?>.5???9A"6".1578.2; C29"."A?>
ol!e for ')
ln '918
( (
ln
) )
0.008 0.0926− 0.15 0− 0.0936 0.008− 0.15
95.69=6 #tag#<46:
;*obl- 47
(i!en) 1
2 C'81
C1
$eed9 5" tons&hr
1
C2
2
N
1
'
D1K
D'
3;% F2 3"% @ulp
*9
3 tons - 20 tons.ulp
12%ugar
*e+uired) ' 9 G olutions)
tons 4n $eed) F2 9 ".3;65"7 9 23 ,r tons @ulp 9 ".3"65"7 9 2" ,r tons ugar 9 ".1265"7 9 > ,r 4n $inal !erflow)
tons ugar 9 ".A6>7 9 5.;2 ,r
olution 9
5.82 tons 0.15 9 ?;.; ,r
F2 9 1 9 ?;.; 5.;2 9 ?2.A;
C1 9
tons ,r
5.82 32.98 9 ".1>5
D1 9 C1 9 ".1>5
at e+uilibrium
4n $inal -nderflow)
tons ugar 9 "."?6>7 9 ".1; ,r tons F2 9 ' 9 2"6?7 9 >" ,r
0.18 60 9 ".""?
D' 9
4n $resh ol!ent) /B 6ol!ent7) ' 8 1 o '81 9 >" 8 ?2.A2 23
tons F2" 9 '81 9 >;.A; ,r C'81 9 " 6pure sol!ent7 ugar Balance :round tage 1) ugar in $ 8 2C2 9 1D1 8 1C1
tons 2 9 ? 9 3 9 L.. 9 '81 9 >;.A; ,r tons 1 9 2 9 ? 9 3 9 L.. 9 ' 9 >" ,r > 8 >;.A;C2 9 >"6".1>57 8 5.;2 C2 9 ".15"A ol!ing for ') 0−0.003 0.1509 − 0.1765 0 −0.1509 ln 0.003 −0.1765
ln ' 9
= 16.36 = 17
81
;*obl- 48
=onstant olution *etention) ; solution flowrates ) D;C solute&solution 4n $eed) F2 9 ".3;65"7 9 23 tons&hr @ulp 9 ".3"65"7 9 2" tons&hr ugar 9 ".1265"7 9 > tons&hr 4n $inal !erflow) ugar 9 ".A6>7 9 5.;2 tons&hr C1 9 ".15
olution 9 1 9
5.82 0.15 9 ?;.; tons&hr
D1 9 C1 9 ".15
6H e+uilibrium7
4n $inal -nderflow) 3 tonssolution
*9
ton dry pulp
ugar 9 "."?6>7 9 ".1; tons&hr olution 9 ' 9 ?62"7 9 >" tons&hr
D* 9
0,18 60 9 ".""?
4n $resh ol!ent F2 9 '81 9 ' 8 1 "
6o!erall solution balance7
F2 9 '81 9 >" 8 ?;.; 623 8 >7 '81 9 >;.; tons&hr
C '81 9 "
6pure sol!ent7
ugar Balance :round tage 1) ugar in $ 8 2C2 9 1D1 8 1C1 2 9 ? 9 3 9 . . . 9 '81 9 >;.; tons&hr 1 8 2 8 ? 9 . . . 9 ' 9 >" tons&hr > 8 >;.;C 2 9 >"6".157 8 ?;.;6".157 C2 9 ".12;2 ol!ing for ') 0 −0.003 ] 0.1282 −0.15 0 −0.1282 ] ln [ 0.003 −0.15
ln [ ' 9
= 15.49 = 16
Situation for problems no. 49-52 : seashore sand contains ;5% insoluble sand, 12% salt and ?% water. 1""" lb&hr of this mixture is to be extracted in a countercurrent washing system with 2""" lb&hr of pure water so that after drying it will contain only ".2% salt. The sand retains ".5 lb of water per pound of insoluble sand. 3A. The mass of salt in the final underflow is e+ual to c. 2.? lb&hr a. 1.7 lb>* b. 1.2 lb&hr d. 2.5 lb&hr 5". The concentration of salt in the final o!erflow is e+ual to a. "."? . 0.07 b. "."5 d. "."A 51. The concentration of salt in the sol!ent stream for stage 1 is approximately e+ual to a. "."2? c. "." b. 0.015 d. ".1A 52. The number of washing is approximately e+ual to a. 3 c. 5 b. 3 d. > (i!en)
2000 lb/hr Y1
Y2
1
YN+1
2 .1
F= 1000 lb/hr 12% Solute 5% Inert -% Solent
olution) 4n $eed ) 4nert 9 1""" lb&hr 6".;57 9 ;5" lb&hr olute 9 1""" lb&hr 6".127 9 12" lb&hr ol!ent 9 1""" lb&hr 6"."?7 9 ?" lb&hr 4n $inal -nderflow ) 4nert 9 ;5" lb&hr 0.5 lb sol(ent ' 850 lbinert = 425 lb sol(ent ol!ent 9 lbinert
olute )
solute ∈ L N =
0.2 ( inert + solute ∈ L N ) 100
solute ∈ L N =
0.2 0.2 inert + solute ∈ L N 100 100
solute ∈ L N =
0.2 lb 0.2 (850 )+ solute ∈ L N 100 ,r 100
solute ∈ L N =1.7034 lb / ,r
N .N
a4ter r'ing = 0$2% s
$inal underflow 9 inert 8sol!ent 8 solute
¿ ( 850 + 425 + 1.7034 )
lb ,r
Final0nderflow =1276.7034 lb / ,r
!erall /aterial Balance 6/B7 )
F + V N +1= L N +V 1
1000
lb lb lb + 2000 = 1276.7034 + V 1 ,r ,r ,r
V 1=1723.2966
lb ,r
!erall olute Balance )
F solute + V N + 1solute =V 1Solute + L Nsolute lb lb + 0 =V 1solute + 1.7034 ,r ,r
120
V 1solute = 118.2966
lb ,r
concentration of solute ∈ V 1=
118.2966 1723.2966
concentration of solute ∈ V 1= 0.06865 ≈ 0.07
sol(ent ∈V 1 =( 1723.2966 −118.2966 )
lb lb =1605 ,r ,r
4n tage 4 )
31=105
32 = 2000 lb/hr
.N= 25
F= -0
solute sol(ent
¿ ¿ ¿
lb ,r solute ∈ L1 = lb lb 1605 425 ,r ,r
118.2966
solute ∈ L1 =31.3246
lb ,r
olute Balance in tage 4 )
F solute + V 2solute =V 1solute + L1solute
lb lb lb + V 2solute =118.2966 + 31.3246 ,r ,r ,r
120
lb solute ∈ V 2=29.6212 ,r
concentration of solute ∈ V 2=
29.6212 2000
concentration of solute ∈ V 1= 0.01481≈ 0.015
ol!ing for 'umber of tages) ln [
N =1 + ln
[
* N +1 − + N * 2− + 1 * N +1−* 2 + N − + 1
]
]
where) 0
C '81 9
1.7034 lb solute =4.008 x 10−3 425 lb sol(ent
D ' 9
29.6212 lb solute = 0.01481 C29 2000 lb sol(ent
D1 9
31.3246 lb solute =0.07370 425 lb sol(ent −3
' 9 1 8
0− 4.008 x 10 ] ln [ 0.01481−0.0737 0.01481−0
ln [
−3
0.07370− 4.008 x 10
' 9 2.?52 9 3 #tag#
]
55. : slurry of flaked soybeans weighing 1"" kg contains 5 kg inert solids and 25 kg of solution 1" weight % oil and A" weight % sol!ent hexane. This slurry is contacted with 1"" kg pure hexane in a single stage so that the !alue of retention for the outlet underflow is 1.5 kg on insoluble solid per kg sol!ent in the adhering solution. The composition of underflow lea!ing the extraction stage in percent by weight oil is (4#') 31
30 = 100 *g he(ane
'1
'0
F = 100 *g
.1 (1
Inert = 75 *g Sol6n = 25 *g
*#I-4*#J) The composition of underflow lea!ing the extraction -T4') 4n $eed) $ 9 1"" kg 4nert 9 5 kg olMn 9 25 kg il 6solute7 9 .1"625 kg7 9 2.5 kg
4nert balance) 4nert in feed 9 4nert in 1 4nert in 19 5 kg 4n -nderfeed 617) 4nert 9 5 kg ol!ent 9 G 9 5" kg 75 kginert
olute 9 G
olute balance) olute in $ 8 olute in " 9 olute in 1 8 olute in 1 2.5 kg 8 " 9 olute in 1 8 olute in 1 #+. 1 solute ∈ V 1=2.5− solute ∈ L1
ol!ent balance) ol!ent in $ 8 ol!ent in " 9 ol!ent in 1 8 ol!ent in 1 22.25 kg 8 1"" kg 9 sol!ent in 1 8 5" kg ol!ent in 1 9 2.5 kg :t #+uilibrium)
solute ∈ V 1 solution∈ V 1
=
solute ∈ L1 solution ∈ L1
#+. 2
solute ∈V 1 solute ∈V 1 + sol(ent ∈ V 1
=
solute ∈ L2+ sol(ent ∈ L2
ubs. #+. 1 to #+. 2) 2.5 −solute ∈ L1
( 2.5− solute∈ L ) +sol(ent ∈V 1
solute ∈ L2
1
=
solute ∈ L1 + sol(ent ∈ L1
olute in 1 9 1."2"3 kg ubs to #+. 1 olute in 1 9 1.3A5 =omposition on underflow lea!ing) V 1
¿
L1
solute ∈ L1
¿
1.4795 kg 1.0204 kg
9 1.45
5>. Tung meal containing 55% oil is to be extracted at a rate of 3""" kg per hour using n hexane containing 5% wt oil as sol!ent. : counter current multiple stage extraction system is to be used. The meal retains 2 kg of sol!ent per kg of oil free meal while the residual charge contains ".11 kg oil per kg oil free meal while the product is composed of 15 weight percent oil. The theoretical number of ideal stages is 6:7 ?
6=7 5
6B7 3
6J7 >
$!%"&
V 1 V n+1 15% oil 5% oil
F =4000
1
2
kg ,r
Ln 55% oil 0.11 kg oil kgoil freemeal
/= ()u!*d& Theoretical number of ideal stages Solut!o"& Basis) 1 hr
2 kgsol(ent
kg freemeal
-
n
4n the $eed, kgoil : 0.55 x 4000 =2200 kg
kgmeal :0.45 x 4000=1800 kg 4n the final underflow, 0.11 kgoil kgoil : x 1800 kg =198 kg kgoil freemeal
kg sol(ent :
2 kg sol(ent x 1800 kg =3600 kg kgfree meal
kgmeal :1800 kg 2200 + V n+ 1= V 1+ 3600 + 198
!erall olution Balance)
V n+1 =V 1+ 1598
e+. 1
!erall olute Balance)
2200 + 0.05 V n+1 = 198 + 0.15 V 1
e+ 2
V 1=20819 kg V n+1 =22417 kg :t e+uilibrium condition,
(
solute solute V 1= L solution solution 1
) (
)
(
0.15 x 20819 20819
)(
∈ = kgoil L 1 3600 + kg oil ∈ L 1
)
g oil in 19 >?5.2A kg olute balance in stage 1)
2200 + kg oil ∈ V 2=0.15 x 20819 + 635.29
g oil in 29 155;.13 kg ol!ent balance in stage 1) 0 + kg sol(ent ∈ V 2=3600 + 0.85 x 20819 g sol!ent in 29 212A>.15 kg
y 2 :
1558.14 = 0.0682 1558.14 + 21296.15
x n :
198 =0.0521 3600 + 198
x 1 :
635.29 =0.15 635.29 + 3600
:t constant underflow, y n+1 − x log y 2− x 1 N −1 = y n+1− y2 log x n− x 1 n
0.05− 0.0521 0.0682 −0.15 N −1 = 0.05− 0.0682 log 0.0521 −0.15 log
= 3.1665 ? 4 #tag#
5. =oconut oil is to be produced from dry copra in two stages. $irst, through expellers to s+ueeEe out part of the coconut oil and then through a counter current multi stage sol!ent extraction process. :fter expelling, the dry copra cake contains 2"% residual oil. 4n the sol!ent extraction operation, A"% of the residual oil in the expeller cake is extracted as a solution containing 5"% by weight oil. 4f fresh sol!ent is used and on kg of solution with e!ery 2 kg of insoluble cake is remo!ed with the underflow, the number of ideal stages is 6:7 3
6=7 >
6B7 5
6J7
$!%"&
V 1 V n+1 A"% reco!ery
1
2
-
n
V n+1
5"% oil
Ln
$ =opra
/=
2"% oil
()u!*d& 'umber of 4deal tages Solut!o"& Basis) 1"" kg =opra /" t> 'd, $9 1"" kg g oil) ".2" x 1""9 2" kg g inert) ".;" x 1""9 ;" kg /" t> Sol%"t, V"@1:
y n+1=
solute =0 solution
4n the $inal -nderflow, n, g inert9 ;" kg 1 kg solution x 80 kginert =40 kg g solution) 2 kginert
Ln :80 kg +40 kg=120 kg solute :0.10 x 20= 2 kg 2 solute : =0.05 solution 40 4n the $inal !erflow, 1 y 1=0.50
kg solute :0.90 x 20 =18 kg
1 kgsolution 2 kgcake
V 1 :
18 kg = 36 kg 0.50
4n the first undeflow, 1 solute = y1=0.50 solution g inert) ;" kg g solution) 3" kg L1 : 80 kg + 40 kg =120 kg kg solute:0.50 x 40 kg =20 kg 4n the !reflow 2, 2, /B on stage 1, F + V 2 = L1+V 1 100 + V 2=120 + 36
V 2=56 kg olute balance on stage 1, solute ∈ F + solute ∈ V 2= Solute ∈ L1+ solute ∈V 1 20 + y 2 x 56= 20 + 18
y 2=0.3214 $or constant underflow, y n+1 − x log y 2− x 1 N −1 = y n+1− y2 log x n− x 1 n
0 −0.05 0.3214 − 0.50 0− 0.3214 log 0.05 −0.50
log
N −1 =
N= 5 stages
5;. *oasted copper ore containing the copper as =u3 is to be extracted in countercurrent stage extractor. #ach hour, a charge consisting of 1" tons gangue, 1.2 tons =u3 and ".5 ton water is to be treated. The strong solution produced is to consist of A"% wt. water and 1"% wt. =u3. The reco!ery of =u3 is to be A;% of that in the ore. @ure water is to be used as fresh sol!ent. :fter each stage, one ton inert gangue retained 2 tons of water plus the copper sulfate dissol!e in that water. #+uilibrium is attained in each stage. The number of stages re+uired is.
(i!en) !er$low water7 A"% water, 1"% =u3
ol!ent6@ure
$eed 1" tons gangue 1.2 tons =u3 ".5 tons water olution) 4nert in feed 9 inert in underflow :mount of solution in underflow 2 ton s olution =20 ton solution 10 toninert x 1 toninert
-nderflow
/=
1 ton gangue 2 tonsolution
1.2 tons x 0.98 =11.76 tons 0.10
!erall balance of solute and sol!ent 11.> 8 2" 9 ".5 8 sol!ent stream ol!ent stream 9 ?"."> tons =omposition of final underflow olute in underflow 9 1.2 1.2x".A; 9"."23 tons 0.024 −3 = 1.2 x 1 0 %wt. 9 20
$or stage 1 11.> tons
?"."> tons
1.2 tons =u3 ".5 tons 0ater :t e+uilibrium solute s olute ( ) =( ) =0.10 solution o(erflow solution underflow
2" tons
olute balance N let x 9 fraction of solute at sol!ent stream 1.2 9 ?"."> x 91.1> 8 2 D 9 ".">5?5 −3
'umber of stages
918
0−1.2 x 1 0 ] ln [ 0.065735 −.10 0 − 0.065735 ] ln [ −3 1.2 x 1 0 −.10
9 A.22> 9 1" stages
ituation for @roblems 5A>? il is to be extracted from meal by means of benEene using a continuous countercurrent extractor. The unit is to be treat 1""" kg of meal 6based on completely exhausted solid7 per hour. The untreated meal contains 3"" kg of oil and is contaminated with 25 kg of benEene. The fresh sol!ent mixture contains 1" kg of oil and >55 kg of benEene. The exhausted solids are to contain >" kg of unextracted oil. #xperiments carried out under conditions identical with those of the proOected battery show that the solution retained depends on the concentration of the solution, as shown in table below. :ll +uantities are gi!en in an hourly basis. =oncentration, kg oil&kg solution "." ".1 ".2 ".?
olution retained, kg&kg solid ".5"" ".5"5 ".515 ".5?"
=oncentration, kg oil&kg solution ".3 ".5 ".> ".
olution retained, kg&kg solid ".55" ".51 ".5A5 ".>2"
5A. The concentration of the strong solution or extract is approximately e+ual to a. ".5> b. ".5;
c. ".>" d. ".>2
>". The concentration of the solution adhering to the extracted solids is approximately e+ual to a. ".1A? b. ".21;
c. "."21 d. ".11;
>1. The mass of the solution lea!ing with the extracted meal is approximately e+ual to
a. 5" kg&h b. ?"> kg&h
c. 31; kg&h d. >21 kg&h
>2. The mass of the extract is approximately e+ual to a. 5;? kg&h b. 512 kg&h
c. 5?> kg&h d. 51 kg&h
>?. The number of stages re+uired is a. ? b. 3
c. > d.
(i!en)
$inal !erflow, i
2, 1
y2
', 2
y '
ol!ent, '81 '
1" kg oil >55 kg benEene
$eed, $ 9 1""" kg meal&hr
1,
2,
3"" kg oil
x1
x2
' >" kg unextracted oil
25 kg benEene 55 kg solid
olution) 4n the feed) $ 9 1""" kg meal&hr olute) 3"" kg oil
4n the ol!ent) '81 9 1"8>55 9 >>5 kg
ol!ent) 25 kg benEene
olute) 1" kg oil
4nert olid) 1""" 63""8257 9 55 kg
ol!ent) >55 kg benEene
olution) 3"" 8 25 9 325 kg&h solution af 9
400 425 9 ".A31
4n the $inal -nderflow) ' olute) >" kg unextracted oil
BenEene) n >"
/B olute) $eed 8 ol!ent 9 $inal 6-nderflow 8 !erflow7 il) 3"" 8 1" 9 >" 8 $inal !erflow
et) a 9 mass fraction of oil in final underflow b 9 mass fraction of oil in final o!erflow bn81 9
10 665 9 "."15
$inal !erflow olute) ?5" kg&h /B ol!ent) $eed 8 ol!ent 9 $inal 6-nderflow 8 !erflow7 BenEene) 25 8 >55 9 33 8 $inal !erflow $inal !erflow ol!ent) 2?? kg&h i 9 ?5" 8 2?? 9 5;? kg&h extracted
By trial and error, b 9
:ssume an81 9 ".1, from table, olution in n 9 ".5"5 kg kgsolid ' 9 ".5"5 61"""7 9 5"5 kg&hr a!n81 9
a 9 b!i 9 ".>",
kg kgsolid
:t stage 1) /B) $eed 8 2 9 1 8 1
H a!n81 9 ".11A, olution
n
n 9 ".5"61"""7 9 5" kg&h a!n81 9
:t e+uilibrium)
from table, olution 9 ".5A5
60 505 9 ".11A
from table, kg kgsolid
350 583 9 ".>"
60 507 9 ".11;
9
".5"
325 8 2 9 5;? 8 5A5 2 9 A5? kg il Balance) 5A5 6".>"7 8 5;? 6".>7 9 325 6".7 8 5?C 2 y2 9 ".3"; 0.015 − 0.118 ] 0.408−0.6 0.015 −0.408 9 3."5 9 3 ] ln [ 0.118 −0.6 ln [
H $inal -nderflow, n) BenEene) 5" >" 9 33 kg&h H $inal !erflow
' 9 1 8
>3. :n oilsand mixture that is 25% 6by mass7 oil and 5% 6by mass7 sand is to be extracted or leached with 5 tons&day of naphtha in a countercurrent extractor. The feed consists of 1"" tons&day of mixture. The final extract 6o!erflow7 produced contains ?5% 6by mass7 oil and >5% 6by mass7 naphtha, and the underflow from each unit consists of ?2% 6by mass7 oil and >;% 6by mass7 sand. The o!erall efficiency of the extraction is ;"% 6by mass7. :ssume the sol!ent is miscible with the oil in all portions and the extractor has reached e+uilibrium conditions in each stage. :ssume there is no sand in the o!erflow. The number stages re+uired to effect the desired separation of oil from sand is a. ?
c. 5
b. 3
d. >
(i!en)
75 tons/a'
erFlo Yoil = 0$-5 Ynahtha = 0$5
Fee, F 100 tons/a'
Xsolution = 0$-2
Xoil =0$25
erall e8"ien" = 0%
*e+uired) 'umber of stages olution) :ssume) 1 day
!" t> +d
61""
tons day of mixture761 day7 9 1"" tons of mixture
a-ou"t o+ *a++!"at 75 tons of sand 0.68
11".2A and 9 1""6".57 9 5 il 9 1""6".257 9 25 MB
$eed 8 naphtha 9 raffinate 8 extract 1"" 8 5 9 11".2A 8 extract #xtract 9 >3.1 a>t>a bala"
:mount of naphtha entering 9 amount of naphtha extract 8 amount of naphtha raffinate et D 9 mass fraction of naphtha in raffinate 5 9 6>3.17 6".>57 8 611".2A7 6D7 = 0.2986
/ass fraction of oil in raffinate 9 1 ".2A;> ".> ; 9 0.0214
> +u*t>* #olut!o" w!ll b #ubtd to a g*a>!al -t>od
9
>5. : copper ore containing 1".?% by mass copper sulfate, ;5.3% by mass inert and 3.? % by mass water is to be extracted with pure water in a counter current extractor. The daily feed consist of 2;1 tons. The final extract produced co ntains 1"% by mass copper sulfate and A"% by mass water. The underflow from each stage consist of >>.% by mass solution and ??.?% by mass inert. The process is to reco!er A2% of the co pper sulfate from the ore. :ssume the extractor has reached e+uilibrium conditions in each stage the minimum number of stages re+uired to effect the desired separation of copper sulfate from the inert. (i!en) !erflow 1"% =u3, A"% water ol!ent %reco!ery 9 A2
$eed 2;1 tons 1".? % =u3 solution ;5.3 % inert 3.? % water olution) Basis) 2;1 tons feed .1"?62;17 9 2;.A3? kg =u3 .;5362;17 9 2?A.A3 kg inert ."3?62;1" 9 12.";? kg water 4nert in feed 9 inert in underflow :mount of solution in underflow 239.974 ( .667 ) =480.6686 tons solution .333
>>. % ??.? % inerts
:mount of o!erflow 0.92 ( 28.943) 0.10
=266.2756
%solute in underflow 28.943 −(28.943 x 0.92) 480.6636
= 4.817 x 1 0−3
ol!ent balance 12.";? 8 sol!ent stream 9 ".A"62>>.25>7 8 3;".>>?> 2.?1533 ol!ent stream 9"5.A1;2
:t stage 1
2>>.25> tons solution
2;.A3? tons=u3 12.";? tons water
"5.A1;2 tons solution
3;".>>;> tons solution
:t e+uilibrium
(
solute solute ) =( ) =0.10 solution o(erflow solution underflow
olute balanceN let x be fraction of solute at sol!ent stream .1"62>>.25>7 8 .1"63;".>>;>7 9 2;.A3? 8 "5.A1;2x D 9 ".">3;
−3
'umber of stages
918
0− 4.817 x 1 0 ln [ ] 0.0648−.10 0− 0.0648 ln [ ] −3 4.817 x 1 0 − .10
9 >.12 9 stages
1. >" tons per day oil sand 625 wt% oil and 5 wt % sand7 is to be extracted with 3" tons per day of naphthalene in a counter current extraction battery. The final extract from the battery is to contain 3" wt% oil and >" wt% naphthalene and the underflow from each unit is expected to consists of ?5 wt% solution and >5 wt% sand. 4f the o!erall efficiency of the battery is 5"%, how many stages will be re+uiredG
(4#') $inal o
n81
$
$inal n
'!"al Vo '!"al %*+low:
'!"al L" '!"al "d*+low:
D oil9 ".3"
D naphthalene8 D oil9 ".?5
D naphthalene9 ".>"
D sand9 ".>5
'd= 60 to"#daD
V"@1= 40 to"#daD
D oil9 ".25 D sand9 ".5
*e+uired) ' 6'umber of tages7 9G
D naphthalene9 1
Jetailed olution) et :9 il 6olute7 B9 and 64nsoluble olid7 =9 'aphthalene 6ol!ent7
4n the $eed $9 >" tons&day :9 ".256>"7 9 15 tons&day B9 ".56>"7 9 35 tons&day
!erall 4nsoluble olid Balance) 6B7$##J96B7-'J#*$0 6B7-'J#*$09 35 tons&day
4n the underflow 6B7-'J#*$09 35 tons&day9".>5 6underflow7 -nderflow9 >A.5? tons&day n96>A.2?76".?579 23.2? tons&day
i+uid Balance) 6 olute8ol!ent7 1583"923.2?8o o9 ?".> tons&day
ol!ent Balance) 3"9 6=7-'J#*$0 8 6".>"76?".>7 6=7-'J#*$09 21.5A; tons&day
4n the -nderflow) Dc 9 66=7-'J#*$0& n79 21.5A;&23.2?9 ".;A Dn9 Da9 1".;A9 ".11
'o. of tages) ' TF#9 18 ln6 6Cn81 D '7& 6C2D177&ln666Cn81 C27& 6D 'D1777
Balance at tage 1) ".256>"78 C263"79 ".3"6 ?".78".3"623.2?7 C29 ".15
ubstitute) Cn819 " D '9 ".11
C29 ".15 D19 ".3" ' TF# 9 18 ln6 6Cn81 D '7& 6C2D177&ln666Cn81 C27& 6D 'D1777 ' TF#9 2.32 stages ' :=T-:9 ' TF#&#$$#=4#'=C9 2.32& ".5" ' :=T-:9 3.;39 5 T:(#
6@rinciples of /ass Transfer and eparation @rocesses by Binay . Jutta7 : solid feed containing 22% of solute, ?% water and 5% inerts 6insoluble7 is to be leached a rate of 1 ton per hour with water in a countercurrent leaching cascad e. The strong leachate lea!ing the unit should ha!e 1>% of the solute in it. Jesired reco!ery of the solute in the feed is AA%. The o!erflow does not ha!e any entrained inert in it, and the amount of solution retained in the sludge is ".35 kg solution per kg inert. :nalytically determine the number of stages re+uired for the separation.
(i!en)
$inal 1
$
olution)
Ba#!#) 1 hour operation
1 ton 9 1""" kg
n81
$inal n
/" 'd
olute) 1"""6".227 9 22" kg 0ater) 1"""6"."?7 9 ?" kg 4nert) 1"""6".57 9 5" kg
/" "d*+low& 0.45 kgsolution
kginert
x 750 kginert =337.5 kg
' 9 ??.5 kg solution
/ass of solute lea!ing with the sludge 6AA% reco!er y7 9 622"76"."17 9 2.2 kg olute 9 2.2 kg ol!ent 9 ??5.? kg
x N =
2.2 = 0.00652 337.5
/" %*+low& Sol%"t bala"&
ol!ent in $ 8 '81 9 ' 8 1 ?" 8 '81 9 ??5.? 8 1 1 9 '81 ?"5.? Solut bala" &
olute in $ 8 '81 9 ' 8 1 22" 8 " 9 2.2 8 6 '81 ?"5.?7 6".1>&".;37
'81 9 133;.5 kg 1 9 113?.35 kg olute in 1 9 1;2.A5 kg ol!ent in 1 9 A>".5 kg
Solut Bala" at Stag 1&
'81 9 133;.5 9 2 D1 9 C1 9 ".1> C '81 9 " @ure sol!ent 22" 8 2y2 9 1x1 8 1;2.A5 22" 8 133;.5y2 9 ??.5 6".1>7 8 1;2.A5 C2 9 "."11
x N =0.00652
-sing the e+uation) ln (
y N +1− x N
N −1 = ln (
y 2− x 1 y N +1− y 2 x N − x1
)
)
0 −0.00652 ) 0.0117 −0.16 N −1 = 0 −0.0117 ln ( ) 0.00652−0.16 ln (
'9 2.2
N= - ST9:;S
