EVAPORATOR-01

Tugas : menguapkan H2O

jenis : double pipe evaporator

Umpan masuk evaporator

Tf = 30 oC =
P= 1 atm
Jumlah
Komponen BM
kg/jam kmol/jam
H2O 18.0153 12.4987 0.6938
R-COOCH3 310.6250 1237.3737 3.9835
Total 1249.8725 4.6773

Input
Komponen BM densitas (kg/L)
kg/jam

H2O 18.0153 1.0000 12.4987

R-COOCH3 310.6250 0.8800 1237.3737
Total 1249.8725

Umpan masuk evaporator = 1249.8725 kg/jam

2755.4993518265 lb/hr

hasil atas
Menentukan suhu Evaporator,sehingga produk yang keluar dapat teruapkan
Digunakan steam jenuh (Tabel 7, Kern)
 T = 284 F
P = 52.47 psi 3.569
Viscositas campuran umpan :
1 xi

 i
=
µ=
=

data komponen masuk evaporator:

komponen massa x cp (Btu/lb oF)
C3H5OCl #REF! #REF! 0.000372
H2O #REF! #REF! 0.001001
C3H8O3 #REF! #REF! 0.000705
NaCl 12.4987 #REF! 0.000343
CO2 1237.3737 #REF! 0.005794
total #REF! #REF!

Neraca Panas Evaporator

Kebutuhan Pemanas
diambil T ref = 25 C
T Feed = 30 C 86

Qs = Err:509 Kkal/Jam
Err:509 Btu/Jam

Beban panas yang harus diberikan, Q didapat dari persamaan 9.5

Mc.Cabe and Smith.
 Q  m s  s

dimana : ms Jumlah Steam

λs Kalor Penguapan Uap =
P operasi = 14.7000 psia
P design = 17.6400 psia

ms Err:509 lb/jam

Beda suhu rata-rata

t 2  t1 T1
∆T LMTD =  t  T2
ln  2 
 t1 
t1
t2
= 75.29610968 oF

Perhitungan suhu rata-rata masing-masing fluida

fluida panas (steam) T= 272.26 F

fluida dingin (larutan gliserol) t = 173.26 F

Luas Transfer Panas

Luas Transfer Panas : Q (Pers. 9.1 Mc.Cabe)

A
U D .TLMTD

dimana : A = Luas area transfer panas yang diperlukan (ft2)

 UD = koefisien panas (Btu/ft2.hr.F)
Q = beban panas yang harus diberikan (Btu/hr)

Harga UD diperoleh dari tabel 8 Kern.

Untuk fluida panas berupa steam dan fluida dingin berupa medium organik akan diperoleh UD = 50

diambil UD = 50 Btu/ft2.hr.F

A = Err:509 ft2 (dipilih jenis double pipe evaporator karena

Dimensi Alat
dipilih evaporator double pipe dengan :
hairpins lenght = 12 ft 144
Inner Pipe, IPS = 3 in
Shell ID = 150 in

a. Shell : larutan umpan (gliserol)

Menghitung flow area
D= 155.375 in 12.9479

 .D 2
a s 
4

as = 131.6041 ft2
 W= 1249.872 kg/jam 2755.50

Menghitung mass velocity

W
G s 
as

Gs = 20.9378 lb/jam.ft2

Menghitung Re
µ larutan adalah = #REF! lb/ft.jam

ID .G s
Re s 


Res = #REF!

dengan Re tersebut didapat nilai jH (Fig.28, Kern, 1950) =

Menghitung Pr
k larutan = #REF! btu/ft.jam.F

Cp larutan = #REF! btu/lb.F

c.
Pr 
k
1
1
 c.  3
(Pr)   
3

 k 
 1
1
 c.  3
(Pr)   
3

 k 
Pr^(1/3) = #REF!

Menghitung ho

1
k  c .  3
ho  j H . .  . p
ID  k 

ho = #REF! btu/jam.ft2.oF

b. Inner pipe : steam

Menghitung flow area
De '.3.068
G a in
Re' Da = 0.2557

 .D
2
ap 
4

ap = 0.0513 ft2

W= Err:509 kg/jam Err:509

Menghitung mass velocity

W
G p 
ap

Gp = Err:509 lb/jam.ft2

Menghitung Re
µ steam pada 272,26 oF = 0.019 cp
 0.04598 lb/ft.jam
D .G
Re p  a

Rep = Err:509

denganRe tersebut didapat nilai jH (Fig.24, Kern, 1950) =

Menghitung Pr

dari Tabel 5 (Kern, 1950) didapat k untuk steam 272,26 oF adalah =

dari Fig.2 (Kern, 1950) didapat C untuk steam 272,26 oF adalah =

c.
Pr 
k
1
1
 c.  3
(Pr)   
3

 k 
Pr^(1/3) = 0.8135

Menghitung hi

1
k  c .  3
hi  j H . .  . p
De  k 

hi = 2.4459 btu/jam.ft2.oF

Menghitung hio

ID
hio untuk steam = hio  hi.
2 btu/jam.ft2.oF OD
Menghitung Uc

h . h
Uc  io o
h io  h o

Uc = #REF! btu/jam.ft2.oF

Menghitung UD

1 1
  Rd
U D UC asumsi untuk Rd =

Rd = #REF!

Q  U D . A . T
Q
A 
U D . T

A= Err:509 ft2

dari tabel 11, external surface per lin feet untuk 3-in IPS =

panjang total = Err:509 lin ft

maka jumlah hairpin yang digunakan =

h= L total
n.L
Err:509 ≈ 4

Menghitung Pressure Drop (∆P)

shell
D= 12.9 ft

Re's = #REF!

Fig. 19 (Kern,1950)
f= 0.0035 s=
ρ= 0.1033 lb/ft3

2
∆Ps =
f .G S .ID S .L
5 , 22 .10 10 .D E .sg . S

∆Ps = 2.5603E-10 ft

inner pipe
untuk Re = Err:509

0,264
f  0,0035 
( DG /  ) 0 , 42
f= Err:509
ρ= 0.1033 lb/ft3

2
4 . f .G p L p
Fp 
2 . g . 2 .D

∆Fp = Err:509 ft

F p .
Pp 
144
 F p .
Pp 
144

∆Pp = Err:509 psia

allowable pressure drop = 10 psia

Kesimpulan:
a. Jumlah steam yang dibutuhkan = Err:509 lb/jam
b. Dimensi evaporator :
Dimensi annulus:
OD = 3.5 in 0.0889
ID = 3.068 in 0.0779
panjang hairpin = 20 ft 6.0960
jumlah hairpin = 29 buah
Dimensi inner pipe:
OD = 2.38 in 0.0605
ID = 2.067 in 0.0525
c. Faktor pengotoran evaporator = #REF!
d. Pressure drop (∆P)
annulus #REF! psia
inner pipe Err:509 psia

Menentukan Tebal Dinding Evaporator

P = 1 atm
= 14.7 psia

P perancangan = 1,20 x P
= 17.64 psi
Tebal Shell

P.ri
ts  C
f .E  0.6P
Bahan yang digunakan carbon Stell SA-285 Grade C

IDs = 150 in
ri = 75 in
f = 12650
E = 0.8
C = 0.125 in
ts = 0.2559 in

diambil tebal standar = 0.3125 in

OD = ID + 2.ts
= 150.625 in

Tebal Head

0.885.P.ri
th 
f .E  0.1.P

th = 0.2407 in

diambil tebal standar= 0.25 in

dari tabel 5,7 Brownell didapat

 icr = 9.375
r = 144

a = ID/2
= 77.6875 in

AB  a  icr

AB = 68.3125

BC  r  icr
BC = 134.625

  2

AC  BC  AB  2

AC = 116.005573007

b  r  AC
b = 27.9944269927

OA  t h  b  sf
sf untuk tebal 3/16 in adalah 1,5 - 2
maka diambil 2

OA = 30.2351 in
L awal = 3m
118.1103 in

L total = 178.5806 in
4.5359 m
 86 oF

Input Output
uap cair
kmol/jam Liter/jam
kg/jam kmol/jam Liter/jam kg/jam
0.6938 12.4987 12.3750 0.6869 12.3750 0.1237
3.9835 1406.1065 1237.3737
4.6773 1418.6052 12.3750 12.3750 1237.4975
1249.8725 kg/jam

pat teruapkan
diambil suhu hasil atas (Keluar) = 126.957 C
 = 260.5 F
atm
itas campuran umpan :
1 xi

 i
#REF! cp-1
#REF! cp
#REF! lb/ft.hr

x.cp k (Btu/ft. Jam. 0F) k.x ρ (kg/L) x.ρ

#REF! 0.0155 #REF! #REF! #REF!
#REF! -0.6162 #REF! #REF! #REF!
#REF! 0.2348 #REF! #REF! #REF!
#REF! 37.9419 #REF! 1 #REF!
#REF! 157.7775 #REF! 0.88 #REF!
#REF! #REF! #REF!

Tsteam = 140 C 284 F

F P steam = 52.467 psia 3.569 atm
 921.7 Btu/lb (tabel 7 Kern)

Err:509 kg/jam

= 284.0 F Hot fluid Cold Fluid

= 260.5 F 284.00 Higher T 260.5
= 86.0 F 260.52 Lower T 86.0
= 260.5 F

9.1 Mc.Cabe)

ng diperlukan (ft2)
berikan (Btu/hr)

dium organik akan diperoleh UD = 50-100 Btu/j ft3 F

h jenis double pipe evaporator karena luas transfer panas < 200ft2)

in 3.6576 m

ft (Tabel 11, Kern, 1950, p.844)

lb/jam

16

data steam pada suhu 272,26 oF

Suhu (oF) k (btu/ft.jam.F)
212 0.0137
392 0.0187

P (psia)
392

272.26
 0.0187 x

119.74 =
180
0.5987 =
-2.7673 =
x =

ft (Tabel 11, Kern, 1950, p.844)

lb/jam
 50

0.0154 btu/ft.jam.F

0.18 btu/lb.F

ID
hio  hi.
OD ID 3.068 in 0.2557 ft
 OD 3.5 in 0.2917 ft

o
h o

0.003

0.9170 ft2
 1.1300

asumsi tinggi shell 3m

L= 9.84252 ft
 Err:509 kg/jam

m
m
m

m
OD standar = 156 in 3.9624 m
ID = (OD-2ts) = 155.375 in

(Brownell,258)
in

in

in

in
t Komponen m, kg/jam x
cair C3H5OCl #REF! #REF!
kmol/jam Liter/jam H2O #REF! #REF!
0.0069 0.1237 TOTAL #REF! #REF!
3.9835 1406.1065
1406.2303

asumsi : berdasarkan titik didih air dalam larutan data steam pada suhu 284 oF (Kern, 1950)
 Suhu (oF) P (psia)
281.01 50
287.07 55

P (psia)
287.07

284

55 x

3.07 =
6.06
15.35 =
-317.95 =
x =
 23.48 ∆t2
174.52 ∆t1
-151.0452
u 272,26 oF data steam pada suhu 272,26 oF
Suhu (oF) ρ
267.25 0.0952562
281 0.1174398

P (psia)
281

272.26
 212
0.0137 0.1174398 x

0.0187 - x 8.74 =
0.005 13.75
3.366 .- 180x 0.1938844 =
.- 180x -1.420913 =
0.01537389 btu/ft.jam.F x =

k= 0.015374 btu/ft.jam.F
 µ (cP) x/cP Arus Masuk Arus Keluar
0.2861 #REF! Komponen kg/jam
kg/jam
0.1826 #REF! uap cair
0.4687 #REF! NaCl 12.499 12 0.124
CO2 1237.374 0.000 1237
12.375 1237.497
Total 1249.872
1249.872

a suhu 284 oF (Kern, 1950)

λ (Btu/lb)
924.6
921

281.01
50

55 - x
5
333.3 .- 6,06x
.- 6,06x
52.467 psia

P= 52.467 psia
a suhu 272,26 oF
 267.25
0.095256

0.11744 - x
0.0221835728
1.614797 .- 13,75x
.- 13,75x
0.103339 btu/ft.jam.F

ρ= 0.103339 lb/ft3
Mencari titik didih larutan (Boiling Point)

Ttrial = 148.90098 oC 421.90098269 K

P= 1 atm 760 mmHg

Koefisien Antoine
Komponen
A B C D
C3H5OCl 24.764 -2.8846E+03 -5.6252E+100 -1.1011E-10
H2O 29.8805 -3.1522E+03 -7.3037E+00 2.4247E-09

jadi titik didik larutan umpan = 148.901 oC

Mencari titik didih tiap solution dalam larutan

dengan menggunakan rumus Boiling Point Rise (BPR) (Perry, 7th edition, p.11-111)
BPR = 104.9 N2^ 1.14

Jumlah BP
Komponen N2 BPR ( o F)
kmol/jam o
C
C3H5OCl 0.6938 0.1483 11.9118 116.11
H2O 3.9835 0.8517 87.3543 100
4.6773
 TF TB TC Jumlah jumlah
xi
E K K K kg/jam kmol/jam
5.3331E-07 215.95 389.26 610 12.49872462 0.69378387 0.01
1.8090E-06 273.15 373.15 647.13 1237.3737374 3.98349694 0.99
1249.872462 1

edition, p.11-111)
BP water = 100 oC 212 oF

BP BP solution BP solution
o
F o
F o
C
241 252.910 122.728
212 299.354 148.530
 Ki
Pi yi
Pi/P
0 0 0
3608.55 4.74809 4.700606
4.700606
NERACA PANAS EVAPORATOR

F .Cp f (TF  Tref )  m.  V .Hv  L.Cp L (TL  Tref )

H 1  Q  H 2  H 3
H 1  F .Cp F (TF  Tref )
Q  m.
H 2  V .Hv
H 3  L.Cp L (TL  Tref )

Menghitung Panas yang dibawa umpan masuk

T= 100 C 373 K
Tref = 25 C 298 K

jumlah ∫Cp.dT ∆H1

Komponen BM
(kg/jam) (kcal/kg) (kcal/jam)
C3H5OCl 92.525 12.499 0.0268 0.3348
H2O 18.016 1237.374 0.0748 92.5928
C3H8O3 92.094 Err:509 0.0519 Err:509
NaCl 58.44 Err:509 0.0260 Err:509
CO2 44.01 Err:509 0.2184 Err:509
TOTAL 305.0850 Err:509 0.3979 Err:509

ΔH1 = Err:509 kcal/jam

Menghitung Panas yang dikandung steam

suhu steam = 140 oC 284 oF
tekanan steam pada suhu 284 oF adalah = 52.432 psia
panas laten (λ) steam pada suhu 284 oF = 921.72 Btu/lb
512.067 kcal/kg
Qs = ms.λ
Qs = 512.067 .ms
Menghitung panas yang dibawa uap keluar
T= 126.957 oC Latent heat=Hvb [(Tc-T)/(Tc-Tb)]^0.38

jumlah TB TC Hvb
Komponen BM
(kg/jam) K K (kJ/kmol)
C3H5OCl 92.525 Err:509 389.26 610 4.569
H2O 18.016 Err:509 373.15 647.13 40.063
CO2 44.01 0.000 216.58 304.19 97.225
TOTAL 154.5510 Err:509

Menghitung Hv ∆H2 = V.Hv

Hv = Latent heat + (Cp (T-Tair))
Hv V ∆H2
(kcal/kmol) kmol/jam kcal/jam
2.033 Err:509 Err:509
9.695 Err:509 Err:509
35.141 0.000 0.000
∆H2 = Err:509

Menghitung panas yang dibawa larutan keluar evaporator

T= 126.957 C 399.957 K
Tref = 25 C 298 K
jumlah ∫Cp.dT ∆H3
Komponen BM
(kg/jam) (kcal/kg.K) (kcal/jam)
C3H5OCl 92.525 Err:509 0.0370 Err:509
H2O 18.016 Err:509 0.1020 Err:509
C3H8O3 92.094 Err:509 0.0710 Err:509
NaCl 58.44 0.124 0.0352 0.0044
TOTAL 261.0750 Err:509 0.2452 Err:509

ΔH3 = Err:509 kcal/jam

Menghitung panas yang dibawa kondensat

suhu kondensat 126.957 oC 260.523 oF
panas laten (λ) steam pada suhu 260,523 oF = 938.244 Btu/lb
521.247 kcal/kg
Qc = ms.λ
Qc = 521.247 .ms

jumlah panas keluar = ∆H2 + ∆H3 + Qc

Err:509 + 521.247 .ms
jumlah panas masuk = ΔH1 + Qs
Err:509 + 512.067 .ms

asumsi : tidak ada panas yang hilang ke sekeliling, maka dari neraca panas diperoleh
Panas masuk = Panas keluar

Err:509 + 512.067 .ms = Err:509

-9.180 .ms = Err:509
ms = Err:509

Jadi :
Qs = Err:509 kcal/jam Err:509
Qc = Err:509 kcal/jam

Masuk kcal/jam Keluar kcal/jam

ΔH1 Err:509 ∆H2 + ∆H3 Err:509
Qs Err:509 Qc Err:509
TOTAL Err:509 Total Err:509

out
in
Komponen kcal/jam
kcal/jam uap cair
C3H5OCl 0.335 Err:509 Err:509
H2O 92.593 Err:509 Err:509
C3H8O3 Err:509 0 Err:509
NaCl Err:509 0 0.004
CO2 Err:509 0.000 0
Panas steam Err:509
Panas kondensat Err:509
Total Err:509 Err:509
Ket :
F : jumlah umpan masuk dalam evaporatpr, kg/jam
V : jumlah uap di puncak evaporator, kg/jam
L : jumlah larutan hasil pemekatan di dasar evaporator, kg/jam
m : jumlah steam, kg/jam
CPF : spesifik heat umpan, kkal/kg oC
CPL : panas spesifik larutan, kkal/kg oC
TF : suhu umpan masuk, oC
TR : suhu reference, oC =25 oC
TL : suhu larutan keluar, oC
HV : panas laten penguapan pada suhu didih, kkal/kg
λ : panas laten steam, kkal/kg
DH1 : panas yang dibawa umpan masuk, kkal
∆H2 : panas yang dibawa uap keluar, kkal
∆H3 : panas yang dibawa larutan kaluar, kkal
Q : panas yang dibawa steam, kkal

Cp (kcal/jam) (liquid)
T= 25 C 298 K
T= 100 C 373 K
Komponen A B C D cp (j/mol)
C3H5OCl 52.634 5.7412E-01 -1.5783E-03 1.8687E-06 10373.1567
H2O 92.053 -3.9953E-02 -2.1103E-04 5.3469E-07 5643.0403
C3H8O3 132.145 8.6007E-01 -1.9745E-03 1.8068E-06 19995.5323
NaCl 95.016 -3.1081E-02 9.6789E-07 5.5116E-09 6368.1348
CO2 -3981.02 5.2511E+01 -2.2708E-01 3.2866E-04 40227.0231
Hvb [(Tc-T)/(Tc-Tb)]^0.38----> Coulson p.252 TB air = 100 oC
373.15 K
T= 400.107 K
Hvb Latent heat Cp ∆T Cp.∆T
kcal/kmol (kJ/kmol) kcal/kmol kcal/kmol.K K kcal/kmol
1.092 4.482 1.071 0.0357 26.957 0.962
9.571 38.517 9.202 0.0183 26.957 0.493
23.227 100.630 24.041 0.4118 26.957 11.100

data steam pada suhu 260,523 oF

Suhu (oF) P (psia) λ (Btu/lb)
260 35.43 938.6
265 38.53 935.2

P (psia)
265
 260.523

38.53 x

4.477 = 38.53
5 3.1
+ 521.247 .ms 13.8787 = 192.65
-178.7713 = .- 5x
kg/jam x = 35.75426
kcal/kmol BM kcal/kg
2.4782 92.525 0.0268
1.3481 18.016 0.0748
4.7770 92.094 0.0519
1.5214 58.44 0.0260
9.6104 44.01 0.2184

data steam pada suhu 284 oF

Suhu (oF) P (psia) λ (Btu/lb)
280 49.2 924.6
285 53.24 921

P (psia)
285
 284

280
53.24 x 49.2

1 = 53.24 - x
5 4.04
4.04 = 266.2 .- 5x
-262.16 = .- 5x
x = 52.432 psia

P= 52.432

λ (Btu/lb)
265
 260.523

260 260
35.43 935.2 x 938.6

-x 4.477 = 935.2 - x
3.1 5 -3.4
.- 5x -15.2218 = 4676 .- 5x
-4691.22 = .- 5x
psia x = 938.2444 Btu/lb

P= 35.75426 psia λ= 938.24436 Btu/lb

λ (Btu/lb)
285
 284

280
921 x 924.6

1 = 921 - x
5 -3.6
-3.6 = 4605 .- 5x
-4608.6 = .- 5x
x = 921.72 Btu/lb

psia λ= 921.72 Btu/lb

 T= 126.957 C 400.107 K
P= 1 atm 760 mmHg
Koefisien Antoine
Komponen
A B C D
C3H5OCl 24.764 -2.8846E+03 -5.6252E+00 -1.1011E-10
H2O 29.8805 -3.1522E+03 -7.3037E+00 2.4247E-09
CO2 35.0187 -1.5119E+03 -1.1335E+01 9.3383E-03

Hvb=(8.32*B*Tb^2*Dz)/(Tb+C)^2----> Coulson p.249

∆z=[1-(Pr/(Tr)^3)]^0.5 ; Pr=P/Pc , Tr=Tb/Tc ---> Coulson p.252

Koefisien Antoine
Komponen BM
A B C
(mmHg,C) C3H5OCl 92.525 7.10839 1408.87 217.147
(mmHg,K) H2O 18.016 18.3036 3816.44 -46.13
(mmHg,K) CO2 44.01 22.5898 3103.39 -0.16

µ steam pada 272,26 oF = 0.019 cp

0.04598 lb/ft.jam

dari Tabel 5 (Kern, 1950) didapat k untuk steam 272,26 oF adalah = 0.0154

dari Fig.2 (Kern, 1950) didapat C untuk steam 272,26 oF adalah = 0.18
 Pi
E mmHg
5.3331E-07 1004.7141
1.8090E-06 1932.810907
7.7626E-10 302508.5289

TB TC Pc Hvb
∆z
K K (mmHg) cal/mol kcal/mol
389.26 610 36753.022 0.945938192 1091.5014142395 1.091501
373.15 647.13 165426.1 0.9690505654 9571.11683841 9.571117
216.58 304.19 55369.553 3.7599505352 23227.361400407 23.22736

btu/ft.jam.F data steam pada suhu 272,26 oF

Suhu (oF) k (btu/ft.jam.F)
btu/lb.F 212 0.0137
392 0.0187

P (psia)
392

272.26

212
0.0187 x 0.0137

119.74 = 0.0187 - x
180 0.005
0.5987 = 3.366 .- 180x
-2.7673 = .- 180x
x = 0.015374 btu/ft.jam.F
k= 0.015374
data steam pada suhu 272,26 oF
Suhu (oF) ρ
267.25 0.095256
281 0.11744

P (psia)
281

272.26

267.25
0.117439812 x 0.095256

8.74 = 0.11744 - x
13.75 0.0221835728
0.193884426 = 1.614797 .- 13,75x
-1.42091299 = .- 13,75x
x = 0.103339 btu/ft.jam.F
btu/ft.jam.F ρ= 0.103339 lb/ft3