Multiple Effect Evaporator.pptx

8/11/2019 Multiple Effect Evaporator.pptx

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MULTIPLE EFFECT

EVAPORATOR

Prof.Dr.Ir.Achmad Roesyadi,M.Eng



Step By Step

1. Assume values for the temperatures of the first and second

effects.

2. By means of heat-balance equations across each effect,

determine the evaporation in each effect.

3. By means of the rate equations, calculate the heating surface

needed for each effect.

4. If the heating surfaces so determined are not essentially equal

for the three effects, redistribute the temperature drops and

repeat items 2 and 3 till the heating surfaces are equal.



The second trial can usually be made to give the desired

result, so the method is not unreasonably tedious.

If the liquids has an appreciable boiling-point elevation

(with consequent effects on enthalpies), no rigorous solution is

possible because the relation between concentration, pressure,

boiling-point elevations, and enthalpies cannot be simply

formulated mathematically. A series of approximations makes

the trial-and-error method fairly direct.



1. From the known terminal conditions, find the boiling point and enthalpies

for the last effect

2. Assume the amount of evaporation in the first and second effects. Since

for moderately dilute solutions the slope of the Dühring lines is nearly

unity (the boiling-point elevation is nearly independent of pressure), this

gives approximate compositions and approximate boiling-pointelevations.

3. Since all elevations are now known, the working effective temperature

drop may be determined and distributed among the effects.

4. By means of heat-balance equations, calculate the evaporation in the firstand second effects. If these differ appreciably from those assumed in step

2, repeat steps 2 and 3 with the amounts of evaporation just calculated.



5. By means of rate equations, calculate the surface required

for each effect.

6. If the heating surfaces are not essentially equal for the three

effects, revise the temperature distribution of step 3. Unless boiling-point elevations are very large, this will not alter

appreciably the elevations assumed in step 2.

7. Repeat these adjustments till the heating surfaces are equal.

The second revision will usually give the required answer.





F =100,000

X F = 0.10

T F = 100°

h F = 60 t c 1 = 238° C

hc 1 = 204

t c 2

hc 2 t c 3

hc 3

I II III

L1 = F – V 1

t L 1

x 1

h 1

L2 = F – V 2

t L 2

x 2

h 2

L2 = 20,000

t L 3

x 3 = 0.50

h 3 = 202

V 1

Y 1 = 0

t V 1

H 1

V 2

Y 2 = 0

t V 2

H 2

V 3

Y 3 = 0

t V 3

H 3

H S =1159

t S = 236



Solution. The data given directly are

t S = 236°F xF = 0.10

t F = 100°F x1 = 0.50

From the steam tables h F = 60 Btu/lb

H s = 1159 Btu/lb (latent heat = 955 Btu/lb)

h3= 202 Btu/b

TBW III = Saturation temperature (for water) in III = 101°F

Weight of feed, thick liquor, and evaporation :

Evaporation = 80,000 lb/hr = V 1 + V 2 + V 3

To determine the temperature drop available, it is necessary to know the elevations in

boiling point ; but to determine these he concentrations in I and II must be known



Neraca massa :

Asumsi awal : < <

anggap :



TBW Tlarutan KTD

DISTRIBUSI

∆T

F : 100.000 Xf = 0.1 Ts = 236

I V1 : 25.500 F . Xf = L1 . X1 200° 209° 9° 12°

L1 : 74.500 X1 = 0.134 assumed assumed

II V2 : 26.700 F . Xf = L2 . X2 150° 165° 15° 12°

L2 : 47.800 X2 = 0.218 assumed

V3 : 27.800 F . Xf = L3 . X3 101° 171° 70° 17°

III L3 : 20.000 X3 = 0.5 assumed

TOTAL ∆T 135 - 94 41°

∆T TOTAL – KTD = Efektif



Neraca Panas

1st Effect

2nd Effect

3rd Effect

=

=

=

; ;

didapat :

dan

(25.500)

(26.700)

(27.800)



CATATAN : 1) λ ,Hv dicari dari Steam Table

2) hl dicari dengan grafik Enthalpy Concentration Chart

3) Superheat dihitung : C p x KTD

Tλ

Super Heat

λ + sp hl HvCp ∆T

°F sp.

S ke I 236 955 0 955 - 1159

∆T1 12

T larutan I 224 - - - 171 -

KTD I 9

V1 ke II215

TBW1 968 4 972 - 1156

∆T2 12

T larutan II 203 - - - 147 -

KTD II 15

V2 ke III188

TBW2 985 8 993 - 1149

∆T3 17

T larutan III 171 - - - 202 -

KTD III 70V3 ke Kondenser

101TBW3 1037 31 1069 - 1138



Menghitung Luas

dari

=

=

=

=



Karena maka prosedur distribusi ∆T harus diulang

Catatan : Perubahan ∆T ini tidak banyak mempengaruhi KTD

Kondisi data ditabelkan kembali



CATATAN : 1) λ ,Hv dicari dari Steam Table

2) hl dicari dengan grafik Enthalpy Concentration Chart

3) Superheat dihitung : C p x KTD

Tλ

Super Heat

λ + sp hl HvCp ∆T

°F sp.

S ke I 236 955 0 955 - 1159

∆T1 10

T larutan I 226 - - - 171 -

KTD I 9

V1 ke II217

TBW1 967 4 971 - 1156

∆T2 12

T larutan II 205 - - - 149 -

KTD II 15

V2 ke III190

TBW2 984 9 993 - 1151

∆T3 19

T larutan III 171 - - - 202 -

KTD III 70V3 ke Kondenser

101TBW3 1037 31 1069 - 1138



Neraca Panas

=1).

2). =

3). =

didapat :

dan



Menghitung A kembali :

(error 3,5% < 5%)

HASIL DAPAT DITERIMA





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Multiple Effect Evaporator.pptx