Example 1:
Example 1: A single efect evaporator is to be used to concentrate a oodA single efect evaporator is to be used to concentrate a ood
solution containing 15% (by mass)dissolved solids to 50% solids. The eed
solution containing 15% (by mass)dissolved solids to 50% solids. The eed
stream enters the evaporator at !1 " #ith a eed
stream enters the evaporator at !1 " #ith a eed rate o 1.0 $g rate o 1.0 $g ss11. &team is. &team is
available at a pressure o .' bar and an absolute pressure o 0.0 bar is
available at a pressure o .' bar and an absolute pressure o 0.0 bar is
maintained in the evaporator. Assuming that the properties o the solution
maintained in the evaporator. Assuming that the properties o the solution
ar
are e ththe e sasamme e as as ththosose e o o #a#ateter r anand d tata$i$ing ng ththe e ovovereralall l heheat at trtranansserer
coe*cient to be +00 , m
coe*cient to be +00 , m" " 11 calculate the rate o steam consumption calculate the rate o steam consumption
and the necessary heat transer surace area. ,or$ing in units o $g s
and the necessary heat transer surace area. ,or$ing in units o $g s11 the the
overall material balance becomes
overall material balance becomes
1.0 - / 1.0 - /
&ubstituting into the component material balance or
&ubstituting into the component material balance or 22 - 0.15 and - 0.15 and - 0.50 - 0.50 gives
gives 0.15
0.15 3 3 1.0 1.0 - - 0.50 0.50 -4 -4 - - 0.+$g 0.+$g ss11 - - 0.$g 0.$g ss11
rom steam tables (6 the steam and condensate remain saturated at .'0 rom steam tables (6 the steam and condensate remain saturated at .'0 bar)
bar) h
h&& - 15$7 $g - 15$7 $g11 and h and h88 - 5+0$7 $g - 5+0$7 $g11 The eed enthalpy is determ
The eed enthalpy is determined by its temperaturined by its temperature. Assuming the eed to bee. Assuming the eed to be pure #ater h
pure #ater h22 is e9ual to h is e9ual to h at !1 " at !1 " and thereore hand thereore h22 - 5.5$7 $g - 5.5$7 $g11.. The
The enthalpies enthalpies o o the the vapour vapour and and li9uor li9uor streams streams are are a a unction unction o o thethe pressur
pressure e #ithin #ithin the the evaporator evaporator hh - 5 $7 $g - 5 $7 $g11 (h (hgg at 0.0 bar) and h at 0.0 bar) and h - -1:+$7 $g
1:+$7 $g11 (h (h
at 0.0 bar). The enthalpy balance at 0.0 bar). The enthalpy balance
&(15 5+0) - (0.0 3 5) / (0.+0 3 1:+) (1.0 3 5.5) &(15 5+0) - (0.0 3 5) / (0.+0 3 1:+) (1.0 3 5.5) & - 0.;1 $g s
& - 0.;1 $g s11 and and < < - - 0.;1(10.;1(15 5 5+0)$, 5+0)$, - - 1'$,1'$, The temperature o steam at
The temperature o steam at .' bar is .' bar is TT&& - 1:.1=8 and the temperature o - 1:.1=8 and the temperature o saturated li9uid #ater at the evaporator pressure o 0.0 bar is T
saturated li9uid #ater at the evaporator pressure o 0.0 bar is T>> - +!.0=8. - +!.0=8. Thus rom the rate e
Thus rom the rate e9uation the heat transer ar9uation the heat transer area isea is
Example 2:
Example 2: An a9ueous solution at 15.5=8 and containing '% solids is An a9ueous solution at 15.5=8 and containing '% solids is concentrated to 0% solids. A single efect evaporator #ith a heat transer concentrated to 0% solids. A single efect evaporator #ith a heat transer surace area o +. m
surace area o +. m and an overall heat transer coe*cient o 000 , 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 $g1 h8 -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.; 105 - .+' 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 $g h@1
<uantity o steamI$g o #ater evaporated - 1!5I1: - 1.1 $g steamI$g #ater. #eat$trans!er area
Temperature o condensing steam - 1+'8.
Temperature diference across the evaporator - (1+' @ !1) - '+8. ,riting the heat transer e9uation or q in JoulesIsec
q = UA KT
('.5 10;)I+:00 - 100 A '+ A - 1.' m
Area o heat transer surace - 1.' m
(6t has been assumed that the sensible heat in the condensed (cooling rom 1+'8 to !18) steam is recovered and this might in practice be done in a eed heater. 6 it is not recovered useully then the sensible heat component about ;% should be omitted rom the heat available and the remainder o the #or$ing adJusted accordingly).
EXAMPLE %. &oncentration o! tomato 'uice in a clim"ing (lm evaporator Tomato Juice is to be concentrated rom 1% solids to ;% solids in a climbing
Blm evaporator + m high and ' cm diameter. The maimum allo#able
temperature or tomato Juice is 58. The Juice is ed to the evaporator at 58 and at this temperature the latent heat o vaporiCation is +:: $7 $g@1.
&team is used in the Jac$et o the evaporator at a pressure o 10 $?a (abs). 6 the overall heat@transer coe*cient is :000 7 m@ s@18@1 estimate the 9uantity o tomato Juice eed per hour. Ta$e heating surace as + m long 0.0' m diameter. Mass "alance: basis 100@$g eed
&olids i9uids Total
2eed 1 ;; 100
?roduct 1 +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+
Mate o evaporation - (1.+ 105)I 1.+' 10:) - 0.1 $g s@1
Mate o evaporation - +:0 $g h@1
<uantity o tomato Juice eed per hour - +:0 $g
Pro"lem 1. A single@efect evaporator is used to concentrate $gIs o a solution rom 10 to 50 per cent o solids. &team is available at 05 $HIm and evaporation ta$es place at 1+.5 $HIm. 6 the overall heat transer coe*cient is + $,Im " calculate the heating surace re9uired and the amount o steam used i the eed to the evaporator is at !' " and the condensate leaves the heating space at +5. ". The speciBc heat capacity o a 10 per cent solution is +.: $7I$g" the speciBc heat capacity o a 50 per cent solution is +.1' $7I$g ".
Pro"lem 2. 1.! $gIs o a li9uid containing 10 per cent by mass o dissolved solids is ed at ++; " to a or#ard@eed double@efect evaporator. The product consists o 5 per cent by mass o solids and a mother li9uor containing 5 per cent by mass o dissolved solids. The steam ed to the Brst efect is dry and saturated at '0 $HIm and the pressure in the second efect is 0 $HIm. The speciBc heat capacity o the solid may be ta$en as .5 $7I$g " both in solid orm and in solution and the heat o solution may be neglected. The mother li9uor ehibits a boiling point rise o : deg ". 6 the t#o efects are identical #hat area is re9uired i the heat transer coe*cients in the Brst and second efects are 1. and 1.1 $,Im " respectivelyN
Pro"lem 3. An evaporator #or$ing at atmospheric pressure is used to concentrate a solution rom 5 per cent to 0 per cent solids at the rate o 1.5 $gIs. The solution #hich has a speciBc heat capacity o '.1; $7I$g " is ed to the evaporator at !5 " and boils at +;0 ". Kry saturated steam at '0 $HIm is ed to the calandria and the condensate leaves at the temperature o the condensing stream. 6 the heat transer coe*cient is .+ $,Im " #hat is the re9uired area o heat transer surace and ho# much steam is re9uiredN The latent heat o vaporisation o the solution may be ta$en as being the same as that o #ater.
Pro"lem %. A li9uid #ith no appreciable elevation o boiling@point is concentrated in a triple@efect evaporator. 6 the temperature o the steam to the Brst efect is +!5 " and vacuum is applied to the third efect so that the
boiling@point is +5 " #hat are the approimate boiling@points in the three efectsN The overall transer coe*cients may be ta$en as +.1 .+ and 1.1 $,Im " in the three efects respectively.
Pro"lem ). A single@efect evaporator #ith a heating surace area o 10 m is used to concentrate a HaOG solution Po#ing at 0.+; $gIs rom 10 per cent to ++.+ per cent. The eed enters at ++; " and its speciBc heat capacity is +. $7I$g ". The pressure in the vapour space is 1+.5 $HIm and 0.+ $gIs o steam is used rom a supply at +5 ". 8alculate
(a) The apparent overall heat transer coe*cient.
(b) The coe*cient corrected or boiling point rise o dissolved solids. (c) The corrected coe*cient i the depth o li9uid is 1.5 m.
Pro"lem *. A triple@efect bac$#ard@eed evaporator concentrates 5 $gIs o li9uor rom 10 per cent to 50 per cent solids. &team is available at +5 $HIm and the condenser operates at 1+.5 $HIm. ,hat is the area re9uired in each efect assumed identical and the economy o the unitN
The speciBc heat capacity is '.1; $7I$g" at all concentrations and that there is no boiling@point rise. The overall heat transer coe*cients are .+ .0 and 1. $,Im " respectively in the three efects and the eed enters the third efect at +00 ".
Pro"lem +. A single@efect evaporator is to produce a +5% solids tomato concentrate rom a :% solids ra# Juice entering at 1;8. The pressure in the evaporator is 0 $?a(absolute) and steam is available at 100 $?a gauge. The overall heat@transer coe*cient is ''0 7 m@ s@18@1 the boiling temperature o the tomato Juice under the conditions in the evaporator is :08 and the area o the heat@transer surace o the evaporator is 1 m. >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 ,* *./ !
