EVAPORATION. Robert evaporator. Balance equations Material balance (total) Component balance. Heat balance
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1 EAPORATION Roert eorator Balance equation Material alance (total) S S=Solution =aor =Steam K=Steam condenate S Comonent alance S S Heat alance S S v Merkel lot can e ued for otaining entaly data Heat ower conumtion Cae of large eorator (eat lo in te or ace) S S v Cae of mall eorator (eat lo in te team ace) Heat tranort S U v v S A T corr T corr T T T Tcorr T TS U A Falling film eorator work a a eat excanger and logaritmic aroac temerature i ued.
2 Prolem / oiling aquou NaOH olution of 3% i eorated to 4% at.5 ar wit 43 C team. Heat lo i 5 /. Wat i te team conumtion? Solution Notation S = / =.3 =.5 ar T = 43 C =.4 v = 5 / =? Outut tream S S S S S S S S 75 Entaly value,3 Merkel lot.5ar.4 Merkel lot.5ar Heat ower S S v aor entaly Inerolation i alied ecaue no.5 ar data i own in te. 4, 4.93 Pa , 5.33 Pa
3 3 (i mean ),,,,,, Pa 4.93 Pa 5.33 Pa 4.93 Pa Heat ower conumtion S S v 6585 Steam orization eat 36.5 C 43 T Steam flow rate
4 Prolem An atmoeric eorator a 5 m eating area, and i eated wit 65 C aturated team. Te eat lo i 4 %, and te corrected overall eat tranfer coefficient i U corr =. kw/m K. C NaOH i eorated from a) 37% ) 4% to 55% concentration. c) Wat i te team conumtion if te eoration eat of te team i 65.7 /? Solution Notation A = 5 m = ar T = C =.55 T = 65 C a) =.37 v =.4 U corr =. kw/m K = 65.7 / S =? =? Heat ower Dene olution temerature.55 Merkel lot T 55C ar Corrected aroac temerature Tcorr T T 65C 55C C Heat ower U corr AT corr kw. 5m m K C 6kW Entaly value.37 Merkel lot 7 T C.55 Merkel lot 8 ar ar
5 Stream flow rate (alance) rearranged (S in te rigt and ide) S S S S S S S S S S S Heat alance S S v S S S.96 S S S ) = Only te canged value ave to e re-calculated. Entaly.4 Merkel lot T C 7.966kW From te eat alance S S kW c) Wat i te team conumtion if te eoration eat of te team i 65.7 /? 6kW
6 Prolem 3 t/ 5% NaOH olution i to e concentrated to 5% under ar reure. Calculate team conumtion and eat tranfer area if te dilute feed i fed to a Roert eorator in a tate a a) C, ) ule oint, c) uereated to ar Te eating team i at 33 C and containing 5% wetne. Heat lo i 3 kw. irtual overall eat tranfer coefficient i U = kw/m K. Solution Notation a) T = C S = t/ =.5 = ar =.5 T = 33 C x =.95 v = 3 kw U = kw/m K =? A =? Outlet tream flow rate S S S S t.5.5 S S t t S S Entaly value.5 Merkel lot T C.5 Merkel lot 3 ar teamtale ar t t 6
7 Heat ower S S v Steam orization eat T 33C kW 36 Steam conumtion Net team conumtion net net Wet team conumtion 94.6 net t x.95 Heat tranfer area aor temerature ar T C irtual aroac temerature T Area U A U T T A T T 33C C 33C 7.58 kw 33C 36 m K 7m 7
8 ) Boiling feed Entaly.5 Merkel lot ar Heat ower S S v kW 36 Steam conumtion Net team conumtion net Wet team conumtion 8959 net t x.95 Heat tranfer area U A U A T T 7.94 kw 33C 36 m K 63.5m 8
9 c) = ar Entaly.5 Merkel lot ar Heat ower S 3 S v kW 36 Steam conumtion Net team conumtion net Wet team conumtion 8 net t x.95 Heat tranfer area U A U A T T 7.78 kw 33C 36 m K 49.8m 9
10 Prolem 4 One a falling film eorator i ued to concentrate 5 t/ % C NaOH olution to 35%, under.5 ar. Heat lo i 3%. Determine: a) Heat conumtion ) Saturation temerature of te team to e alied if minimum aroac temerature i 6 C c) Required eat tranfer area if average overall eat tranfer coefficient i.5 kw/m K Solution Notation S = 5 t/ =. =.35 =.5 ar T = C lo a) Heat conumtion Outut tream S S S S =.3 t S S t t S S 5.86 t.86 t.4 Entaly value. Merkel lot 4 T C.35 Merkel lot 35.5ar.5ar Heat ower S S
11 ) Saturation temerature of te team to e alied if minimum aroac temerature i 6 C Maximum temerature amongt te roce tream elong to te dene olution..35 Merkel lot T 5C.5ar Steam temerature: T T T 5C 6C C Steam min c) Required eat tranfer area if average overall eat tranfer coefficient i.5 kw/m K Altoug te feed i of C, te dilute olution reace it ule oint in te eorator almot at once. Terefore te oiling oint i to e alied at calculating te logaritmic mean aroac temerature., Merkel lot T.5ar, 88C Aroac temerature at te two endoint T T T C 88C 3C a Steam, T TSteam T C 5C 6C Logaritmic aroac temerature Ta T 3C 6C Tav.65C Ta 3C ln ln T 6C Heat tranfer area UA A U T av T av kw.5.65c 36 m K.7m
12 Prolem 5 C NaOH olution i concentrated to 4 % under. ar reure, reulting in.6 t dene olution and.5 t or in eac our.. ar aturated team, otained y mixing.48 ar live team wit a art of te or. Te oter art of te or i condened in a mixing arometric condener uing 8 m 3 / C cooling water. Calculate: a) Live team conumtion ) Fraction of te or ued in te team c) Temerature of te arometric condener outlet tream d) Flow rate of te fla team otained from te team condenate Solution Notation H: cooling water K: or condenate É: live team : team K: team condenate S: fla team F: exaut =.4 =. ar T = C S =.6 t/ =.5 t/ =. ar É =.48 ar H = 8 m 3 / T H = C
13 a) Live team conumtion Fead tream flow rate t t S S.6.5 Feed tream concentration S S t. t.6 S.4. S t. Entaly value. Merkel lot T C.4 Merkel lot.ar.ar Heat ower S S Steam conumtion Steam orization eat.ar Steam ma flow rate 6 4,
14 ) Fraction of te or utilized in te team Material alance around te comreor É Heat alance around te comreor É É (Te or looe it uereated value a it leave te eorator; it aturated tate i taken into account) Entaly value.ar ar É É Ma flow rate of live team and of utilized or (from alance) É É É É É É É É É É É É É c) Temerature of te arometric condener outlet tream Remaining or Cooling water ma flow rate 3 m H HH 8 3 m 8 aor mixed condenate ma flow rate K H Cooling water entaly value TH C H
15 Heat alance of te arometric mixing condener H K H K Mixed or condenate entaly H H K K Mixed or condenate temerature K 4 TK 5C d) Flow rate of te fla team otained from te team condenate Ma alance K S F Entaly value K.ar K 44. S ar S F ar F (Exaut i conidered a liquid) Steam condenate ma flow rate K Fla team ma flow rate Heat alance K S K K K S F K S S K S F S K K K S S F F F S F F
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