Evaporation Problem
Short Description
A problem on Evaporation Chemical Engineering...
Description
Example 1: A single efect evaporator is to be used to concentrate a ood solution containing 15% (by mass)dissolved solids to 50% solids. The eed stream enters the evaporator at !1 " #ith a eed rate o 1.0 $g s 1. &team is available at a pressure o .' bar and an absolute pressure o 0.0 bar is maintained in the evaporator. Assuming that the properties o the solution are ar e th the e sa sam me as th thos ose e o #a #ate ter r an and d ta ta$i $ing ng th the e ov over eral alll he heat at tr tran ans ser er coe*cient to be +00 , m " 1 calculate the rate o steam consumption and the necessary heat transer surace area. ,or$ing in units o $g s 1 the overall material balance becomes 1.0 - / &ubstituting into the component material balance or 2 - 0.15 and - 0.50 gives 0.15 3 1.0 - 0.50
-4
- 0.+$g s1 - 0.$g s1
rom steam tables (6 the steam and condensate remain saturated at .'0 bar) h& - 15$7 $g1 and h8 - 5+0$7 $g1 The eed enthalpy is determined determined by its temperature. temperature. Assuming the eed to be pure #ater h 2 is e9ual to h at at !1 " and thereore h2 - 5.5$7 $g1. The enthalpies o the vapour and li9uor streams are a unction o the pressure pressure #ithin the evaporator h - 5 $7 $g1 (hg at 0.0 bar) and h 1:+$7 $g1 (h at at 0.0 bar). The enthalpy balance &(15 5+0) - (0.0 3 5) / (0.+0 3 1:+) (1.0 3 5.5) & - 0.;1 $g s1 and
< - 0.;1(15 0.;1(15 5+0)$, - 1'$,
The temperature o steam at .' bar is T & - 1:.1=8 and the temperature o saturated li9uid #ater at the evaporator pressure o 0.0 bar is T > - +!.0=8. Thus rom the rate e9uation e9uation the heat transer area area is
Example 2: An a9ueous solution at 15.5=8 and containing '% solids is concentrated to 0% solids. A single efect evaporator #ith a heat transer surace area o +. m and an overall heat transer coe*cient o 000 , 1
m " 1 is to be used. The calandria contains dry saturated steam at a pressure o 00 $?a and the evaporator operates under a vacuum o ;1.+ $?a. 6 the boiling point rise is 5 " calculate the evaporator capacity.
At 00 $?a the steam and condensate enthalpies are h & - 0 $7 $g 1 h 8 505 $7 $g1 Ts-10.=8. The pressure #ithin the evaporator is 101.+ ;1.+ - 0.0 $?a at #hich the boiling point o #ater is :0.1=8. The evaporator temperature is no# :0.1=8 plus the boiling point elevation and thereore T > - :5.1=8.
-4 & - 1.;: $g s1 2rom steam tables the eed enthalpy at 15.5=8 is h2 - :5 $7 $g1. apour enthalpy h - :0! $7 $g1 (hg at 0.0 bar) / (1.!1 3 5) $7 $g 1 :1;.: $7 $g1 8p (#ater vapour at :0.1=8)- 1.!1 $7 $g1 " 1 The enthalpy o the concentrated li9uor stream at the evaporator temperature is h - $7 $g1 (h at :5.1=8). The component balance becomes 0.0'F - 0.0L => F - 5L , V - 'L S(h& h8) - 'Lh / Lh 5Lh2 1.;:(0 505) - ('L 3 :1;.55) / L +5L L - 0.+!+ $g s1 and the evaporator capacity is F - 1.! $g s1
EXAMPLE 3. Single efect evaporator: steam usage and eat trans!er sur!ace A single efect evaporator is re9uired to concentrate a solution rom 10% solids to +0% solids at the rate o 50 $g o eed per hour. 6 the pressure in the evaporator is $?a absolute and i steam is available at 00 $?a gauge calculate the 9uantity o steam re9uired per hour and the area o heat transer surace i the overall heat transer coe*cient is 100 7 m @ s@1 8@1.
Assume that the temperature o the eed is 1;8 and that the boiling point o the solution under the pressure o $?a absolute is !18. Assume also that the speciBc heat o the solution is the same as or #ater that is '.1;: 10+ 7 $g@18@1 and the latent heat o vaporiCation o the solution is the same as that or #ater under the same conditions. 2rom steam tables the condensing temperature o steam at 00 $?a (gauge) D+00 $?a absoluteE is 1+'8 and latent heat 1:' $7 $g @1F the condensing temperature at $?a (abs.) is !18 and latent heat is ;1 $7 $g @1.
Mass "alance ($g h@1)
&olids
i9uids
Total
2eed
5
5
50
?roduct
5
5;
;+
>vaporation
1:
#eat "alance Geat available per $g o steam - latent heat / sensible heat in cooling to !18 - .1:' 10: / '.1;: 10+(1+' @ !1) - .1:' 10: / 1.; 10 5 - .+' 10: 7 Geat re9uired by the solution - latent heat / sensible heat in heating rom 1;8 to !18 - ;1 10+ 1: / 50 '.1;: 10 + (!1 @ 1;) - +.;1 10; / .: 10 - '.5 10; "g 7 h@1 Ho# heat rom steam - heat re9uired by the solution Thereore 9uantity o steam re9uired per hour - ('.5 10;)I(.+' 10 :) - 1!5 $g h@1 vaporation
5
#eat "alance Area o evaporator tube pDHL - p 0.0' + - 0.+; m 8ondensing steam temperature at 10 $?a (abs) - 1158 rom &team Tables. La$ing a heat balance across the evaporator q = UA KT - :000 0.+; (115 @ 5) - 1.+ 105 7 s@1 Geat re9uired per $g o eed or evaporation - 0.5 +:: 10 + - 1.+' 10: 7
'
Mate o evaporation - (1.+ 10 5)I 1.+' 10 :) - 0.1 $g s@1 Mate o evaporation - +:0 $g h @1 stimate the rate o ra# Juice eed that is re9uired to supply the evaporator. D 5+: $gh@1 E
Pro"lem ,. >stimate (a) the evaporating temperature in each efect (b) the reirements o steam and (c) the area o heat transer surace or a t#o efect evaporator. &team is available at 100 $?a gauge pressure and the pressure in the second efect is 0 $?a absolute. Assume an overall heat@transer :
coe*cient o :00 and '50 7 m @ s@1 8@1 in the Brst and second efects respectively. The evaporator is to concentrate 15000 $g h @1 o ra# mil$ rom !.5 % solids to +5% solids.Assume the sensible heat efects can be ignored and that there is no boiling@point elevation. D (a) 1st. efect !'8 nd. efect :08 (b) 5': $gh @1 0.5+ $g steamI$g #ater (c) '50 m E
STEAM TABLE - SATURATED STEAM
Temperature
Pressure(Absolute)
Enthalpy (sat. vap.)
Latent heat
Spe!"! volume
(#$)
(%Pa)
(%& %'-)
(%& %'-)
(m %'-)
Temperature Table *
*.+
,*
,*
,*+
*.++
,*
,//
/
,
*.0
,*
,/0
1*
*.1
,*/
,/,
0
+
*./
,,
,10
1
1
.*0
,+
,1
,
*
.,
,,*
,01
*+
,
.*
,,
,0
/./
.+*
,,0
,+1
1,.1
+
.1,
,
,+
0.
1
,.*+
,
,/
+.*
,*
,.
,1
,
0.1
,,
,.+
,,
,/
.
,
,.//
,
,
./
,+
.+
,/
,*
*.*
,1
.01
,
,
+.+
*
.,
,+
,
,./
*
0.1
,0
,*0
/.
*
,.
,/,
,1
,.*
+*
/./
,+*
,/
0.+0
0*
.,
,+,0
,
.*
1*
0.
,+
,*/
.
/*
0*.
,++*
,,1
,.+
**
*.
,+0+
,,0
.+0
*
,*.1
,+1
,,
.,
*
.
,+/,
,,*
.,
+/.
,+//
,,0
.*
,*
/1.
,0*+
,,*
*.1/,
,
,,.
,0
,1/
*.00
*
,0*.
,0,
,0
*.++/
.*
,0,0
,+*
*.1,
*
+.
,0
,
*.*/
*
0.1
,00
,
*./
+*
+0.1
,01
,*1
*.*0
1*
**,
,001
,*
*./
,**
,0/
/
*.,0
Pressure Table 0.*
.*
,
,1
,/
/.0
.,
,/
,0/
*/
,.*
.
,,
,0
/./
.*
.+
,,0
,+1
1,.1
.1
.1
,
,+
0.*
0.
,.*
,
,+*
+0.*
,.
,.
,*
,,
.
;
,.
.*
,+
,
.0
,/.*
.*
,
,
.1
,./
.*
,+,
,,
,1.,
*.
0.
,0
,*+
/.,
.1
*.*
,1
,/
.0
+*.
,*.*
,+*
,1
0.+
0./
*.*
,+0
,/
.//
/.
1*.*
,+++
,,0
,.*/
//.+
**
,+0+
,,1
.+/
*,.
/
,+1*
,,
.
*.1
,*
,+1
,,
.
*0.
*
,+10
,,1
.
*/.
*
,+/*
,,,
.,
.
*
,+/
,,,0
.+
.
+*
,+/+
,,,
.*/
.,
0*
,+//
,,+
.*
+./
1*
,0*,
,,
*./01
1.+
/*
,0*
,,*0
*./,/
,*.,
,**
,0*0
,,*,
*.11+
,0.
,*
,00
,1,
*.0/
.+
**
,0,
,+
*.+*+
1./
*
,0,
,1
*.,
.+
**
,0/
,
*.+
0./
*
,0
,,
*.
.+
**
,0/
,*/
*.0
+0.1
0*
,0++
,*0
*.,+
0/./
***
,001
,*
*./
!
View more...
Comments
