absorber

Tekanan absorber Tekanan absorber CO2 CO2 C COO BBeennffiieelldd H2 H2 H H22O O GG C CHH4 4 yy1 1 00..1100%% O2 O2 komposisi larutan komposisi larutan K2CO3 K2CO3 DEA DEA V2O5 V2O5 Air Air rich solution rich solution Benfield + CO2 Benfield + CO2 C COO2 2 G G 99440011..66552 2 kkmmooll//hhr r LL C CO O yyo o 1111..88664400332 2 % % xxoo H2 H2 H2O H2O CH4 CH4 O2 O2 B BM M bbeennffiieelld d 5566..00661166 kkoommppoossiissi i ggaas s iin n yy1 1 1 1 %%

kkmmooll//hhr r kkaaddaarr((%%) ) BBM M CCOO2 2 yyaanng g tteerrsseerraap p 112211..11779900442233 C

CHH4 4 11..11887 7 00..00112266225544441 1 116 6 bbeennffiieelld d yyaanng g ddiibbuuttuuhhkkaan n 11559999..113377447777 H H2 2 22883333..77333 3 3300..114400779988666 6 2 2 8899665500..2200556611 H H220 0 33221166..55772 2 3344..221122883388334 4 1188 C COO2 2 11111155..44115 5 1111..886644003311998 8 444 4 G G 4499..0033663344334499 C CO O 11333344..77445 5 1144..119966991199886 6 228 8 L L 2244..9900228833448899 O O2 2 99000 0 99..55772277885577119 9 32 32 rrhho o G G 1188..8800885588660033 T Toottaal l 99440011..66552 2 11000 0 rrhho o L L 11225599..0011882211 A

Avveerraagge e mmool l wwt t 1199..002211775500001 1 kkgg//kkmmooll d

deennssiittaas s ggaas s 1188..880088558866003 3 kkgg//mm33

(L

(L/G)/G)*(r*(rhohoG/rG/rhohoL)L)^0.^0.5 5 0.0.062062071071577577 p

prreessssuurre e ddrroop p 11..55002288773355991 1 iinncchh//fft t 336600..66889966662 2 ccmm//m m ppaacckkiinngg G'^2*F*

G'^2*F*ψ*miu^0.2/(rhoG*rhoL*g)ψ*miu^0.2/(rhoG*rhoL*g) 0.1 0.1 u/ delta P u/ delta P 1.51.5

sangat berpengaruh pada D absoeber sangat berpengaruh pada D absoeber P

Paacckkiinngg==rraasscchhiig g rriinngg, , 3 3 iinncch h ((cceerraammiicc) ) aap p 669 9 mm22//mm33 F

F== 3377

ψ=ratio of the density of water to the density of liquid

ψ=ratio of the density of water to the density of liquid 0.7942696870.794269687 m

g g== 3322..22 rrhhooGGrrhhooLLg g 22997711..883377778833 F F bblla a bblla a 4411..4477669999881199 G G''^^2 2 77..116655002266119911 G G' ' 22..66776677556666555 5 llbb//fftt22sseecc A A 4400..338866775577885 5 fftt2 2 G G aaccttuuaal l 22..667766775566665555 D D 77..11772277333300333 3 fft t 2..12 18866227755661 1 m m D D aattaass G G aaccttuuaal l 1133..0066888888669933 d

deelltta a P P ffllooooddiinngg= = 00..1122FF^^00..7 7 G G aaccttuuaal l 00..668877004499666633 de

deltlta a P P flfloooodidingng= = 1.1.50502828737359591 1 ininchch/f/ft t papackckiningg

L L 00..111188338866994488 a a 44..5544999999777777 b b 33..4455223311227799 c c 33..6655228866555577 h h 5577..33779922669 9 mm

MENENTUKAN DIAMETER BAGIAN BAWAH

MENENTUKAN DIAMETER BAGIAN BAWAH KOLOM ABSORBERKOLOM ABSORBER

L = massa cairan yang masuk kolom per satuan waktu + massa gas yg terserap per satuan w L = massa cairan yang masuk kolom per satuan waktu + massa gas yg terserap per satuan w G = massa gas yang masuk kolom per satuan waktu

G = massa gas yang masuk kolom per satuan waktu

LL= = 11772200..331166552 2 kkmmooll//hhr r rrhho o G G 11..117744221122445544 9 966444433..6699666 6 kkgg//hhr r rrhho o L L 7788..66 2 266..7788999911557 7 kkgg//s s 5599..0066110044882 2 llbb//ss G G 99440011..66552 2 kkmmooll//hhrr 176831.78 kg/hr 176831.78 kg/hr 49.119939 kg/s 49.119939 kg/s 108.28981 108.289818 8 lb/slb/s (L

(L/G)/G)*(r*(rhohoG/rG/rhohoL)L)^0.^0.5 5 0.0.066066661661579579 p

prreessssuurre e ddrroop p 00..6 6 iinncchh//fftt G'^2*F*

G'^2*F*ψ*miu^0.2/(rhoG*rhoL*g)ψ*miu^0.2/(rhoG*rhoL*g) 0.090.09 Packing=raschig ring, 3 inch

Packing=raschig ring, 3 inch F

F== 3377

ψ=ratio of the density of water to the density of liquid

ψ=ratio of the density of water to the density of liquid 0.7942696870.794269687 m

miiuulliiqquuiidd== 55..66 ccpp

g

g== 3322..22

rho

rhoGrhGrhoLg oLg 29712971.83.83778778 F

F blbla a bbla la 4141..47476969989822 G

G''^^2 2 66..444488552233557 7 llbb//fftt22sseecc G

G' ' 22..553399339944333 3 llbb//fftt22sseec c G G aaccttuuaal l 11..77777755776600332 2 llbb//fftt22sseecc

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h h o o a a  yh yh HxHx  Hx  Hx  yo  yo  L  L  HG  HG Ky Ky G G  H   H  llnn )) 1 1 (( 1 1 '' '' ''

g g== 3322..22 rrhhooGGrrhhooLLg g 22997711..883377778833 F F bblla a bblla a 4411..4477669999881199 G G''^^2 2 77..116655002266119911 G G' ' 22..66776677556666555 5 llbb//fftt22sseecc A A 4400..338866775577885 5 fftt2 2 G G aaccttuuaal l 22..667766775566665555 D D 77..11772277333300333 3 fft t 2..12 18866227755661 1 m m D D aattaass G G aaccttuuaal l 1133..0066888888669933 d

deelltta a P P ffllooooddiinngg= = 00..1122FF^^00..7 7 G G aaccttuuaal l 00..668877004499666633 de

deltlta a P P flfloooodidingng= = 1.1.50502828737359591 1 ininchch/f/ft t papackckiningg

L L 00..111188338866994488 a a 44..5544999999777777 b b 33..4455223311227799 c c 33..6655228866555577 h h 5577..33779922669 9 mm

MENENTUKAN DIAMETER BAGIAN BAWAH

MENENTUKAN DIAMETER BAGIAN BAWAH KOLOM ABSORBERKOLOM ABSORBER

L = massa cairan yang masuk kolom per satuan waktu + massa gas yg terserap per satuan w L = massa cairan yang masuk kolom per satuan waktu + massa gas yg terserap per satuan w G = massa gas yang masuk kolom per satuan waktu

G = massa gas yang masuk kolom per satuan waktu

LL= = 11772200..331166552 2 kkmmooll//hhr r rrhho o G G 11..117744221122445544 9 966444433..6699666 6 kkgg//hhr r rrhho o L L 7788..66 2 266..7788999911557 7 kkgg//s s 5599..0066110044882 2 llbb//ss G G 99440011..66552 2 kkmmooll//hhrr 176831.78 kg/hr 176831.78 kg/hr 49.119939 kg/s 49.119939 kg/s 108.28981 108.289818 8 lb/slb/s (L

(L/G)/G)*(r*(rhohoG/rG/rhohoL)L)^0.^0.5 5 0.0.066066661661579579 p

prreessssuurre e ddrroop p 00..6 6 iinncchh//fftt G'^2*F*

G'^2*F*ψ*miu^0.2/(rhoG*rhoL*g)ψ*miu^0.2/(rhoG*rhoL*g) 0.090.09 Packing=raschig ring, 3 inch

Packing=raschig ring, 3 inch F

F== 3377

ψ=ratio of the density of water to the density of liquid

ψ=ratio of the density of water to the density of liquid 0.7942696870.794269687 m

miiuulliiqquuiidd== 55..66 ccpp

g

g== 3322..22

rho

rhoGrhGrhoLg oLg 29712971.83.83778778 F

F blbla a bbla la 4141..47476969989822 G

G''^^2 2 66..444488552233557 7 llbb//fftt22sseecc G

G' ' 22..553399339944333 3 llbb//fftt22sseec c G G aaccttuuaal l 11..77777755776600332 2 llbb//fftt22sseecc

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h h o o a a  yh yh HxHx  Hx  Hx  yo  yo  L  L  HG  HG Ky Ky G G  H   H  llnn )) 1 1 (( 1 1 '' '' ''

A

A 6600..9911999933557 7 fftt22 D

0.1183869 0.11838695 5 kmol/m2.kmol/m2.ss LL''==LLK K 66..6633669966117 7 kkgg//mm22s s DDL L 55..6666000011EE--005 5 mm22//ss σc σc 661 1 mmNN//m m kkl l 0..00 0000011441133225 5 mm//ss σL σL 83.8 mN/m83.8 mN/m a ap p 669 9 mm22//mm3 3 kky y 00..00990088779977885 5 kkmmooll//mm22..ss m miiu u L L 55..6 6 ccpp rrhho o L L 11225599..001188221 1 kkgg//mm3 3 CCtt==rrhhoo//MMRR g g 99..8 8 mm//ss2 2 CCt t 2222..4455777766445 5 kkggmmooll//mm33 a aww//aap p 00..110000111166999 9 kkx x 00..00003311773388335 5 kkmmooll//mm22..ss a aw w 66..9900880077223 3 mm22//mm33 0

0..668877004499666 6 kkmmooll//mm22..s s HH* * = = 00..44553 3 aattmmmm33//kkmmooll G

G''==GGK K 1133..0066888888669 9 kkgg//mm22s s H H = = HH**CCtt//P P 00..337766779911338822 m

miiu u G G 11..55005533EE--005 5 PPaa..s s y y = = 00..33776677991133882 2 XX rrhho o G G 1188..88008855886 6 kkgg//mm3 3 Y Y = = 00..442244 D

DG G 33..44555533EE--006 6 mm22//s s KKyya a 00..00221166440033222 2 kkmmooll//mm33..ss d

dp p 00..00336 6 mm R

R 00..008822005 5 LLaattmm//mmoollKK T

T 333333 KK

kkg g 00..000033336655992 2 kkmmooll//mm22..ss..aattm m 55..66ccp p 00..005566 0.056 0.056 T

Tiinnggggi i AAbbssoorrbbeer r 00..00005566

integrasi dengan simpson rule integrasi dengan simpson rule Y Yo o = = 00..111188664400332 2 XX1 1 = = 00..00001155 Y Y1 1 = = 0..0001 1 XXo o = = 00..6699226633111188 m m = = 00..442244 d dY Y = = 00..0055443322001166 Y Y X X YY* * 11//ff((xx) ) ffaaccttoor r ddAA 0 0..1111886644003 3 00..0000115 5 00..00000066336 6 88..447744226666004 4 1 1 88..447744226666004411 0 0..0066443322002 2 --00..3311337744229 9 --00..11333300227 7 55..006677221122552 2 4 4 2200..2266888855000099 0 0..001 1 --00..6622889988558 8 --00..226666669 9 33..661144115533002 2 2 2 7..2722288330066004477 --00..004444332 2 --00..9944442222888 8 --00..44000033553 3 22..880088772299667 7 4 4 1111..2233449911886677 --00..009988664 4 --11..2255994477117 7 --00..55334400116 6 22..229966886666993 3 2 2 44..559933773333885522

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G G  X  X Y Y Yo Yo XoXo  L  L

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G G dX dX   Z   Z  K Kyya a Y Y HHX  X  

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--00..115522996 6 --11..5577447711446 6 --00..66667766779 9 11..994422880099444 4 4 4 77..7777112233777766 --00..22007722881 1 --11..8888999955775 5 --00..88001133442 2 11..6688333322778 8 2 2 33..336666665555559911 --00..22661166001 1 --22..2200552200005 5 --00..99335500005 5 11..448844999922222 2 4 4 55..993399996688888899 --00..33115599221 1 --22..5522004444334 4 --11..00668866668 8 11..332288446677553 3 2 2 22..665566993355005566 --00..33770022441 1 --22..8833556688663 3 --11..22002233331 1 11..220011779933225 5 1 1 11..220011779933224488 N NooG G 00..0022117766005533 H HooG G 55..4477006666447799 Z Z 00..111199004444559 9 mm

= 27 atm

MENENTUKAN DIAMETER BAGIAN ATAS KOLOM ABSORBER L

x1 0.10% L = massa cairan yang masuk kolom per satuan waktu

G = massa gas yang masuk kolom per satuan waktu - massa gas FCH4out = 1.187 kmol/hr FH2out = 2833.733 kmol/hr

benfield FH2Oout = 3216.572 kmol/hr

BM BM average FCO2out = 1115.415 kmol/hr

30% 138 56.0616 g/mol FCOout = -1334.745 kmol/hr

4% 45 O2 = 16.000 kg/s

0.60% 181.6 FO2out = 900.000 kmol/hr

65.400% 18 Tgas out = 1200.457 C

design reaktor

Diameter Reaktor = 3.500 m packing raschig ring 3 inc Pressure Drop = 0.501 bar

ap 69 m2/m3

rho 561 kg/m3

50.8

G=Gas yg masuk-gas yg terserap 9280.472958 kmol/hr 49.03634349 kg/s g/mol kmol/hr 1727068.476 kmol/hr L=1.4Lmin kg/hr 1kg= 2.2046 lb 1m= 3.2808 ft kg/s 108.105523 lb/s kg/s 54.9007898 lb/s kg/m3 1.17421245 lb/ft3 ket: kg/m3 78.6 lb/ft3 cek lagi BM 19.02175 kg/kmol P 27 atm R 0.082 atmm3/kmolK T 333 K densitas gas 18.808586 kg/m3 1ft=

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h h  x  y  y  y G  L 0 0 min

xh= 0.10% G= 9401.652 kmol/hr yo= 0.1186 yh= 0.001 lb/ft2sec H= 0.1224 Kya= 0.151 kmol/m3dtk kg/m2s Lmin 1142.24106 kmol/hr kmol/m2s kmol/m2s xo 0.69263118

Ca* 0.01660986 kmol CO2/m3 solution

P 0.0053 atm

H* 0.31908764

H=H*

ktu

Menghitung Diffusivitas Liquid

φ 1

MB=berat molekul solven 56.0616

R 0.08205 Latm/molK Tb 194.5 K P 27 atm miu 0.0056 kg/(ms)  H 



_{h o}



h o

 y

y

 L

G

 x

 x







4 2 1 8 10 . . . ) .( 10 . 4 , 7  



 A  B  B  B  AB V   μ T  .M  φ  D P Tb  R V  _A



.

VA 0.034 m3/kgmol

DAB 5.66E-05 m2/s

Menghitung Diffusivitas Gas

Diffusivitas Biner dapat dihitung dengan persamaan Chapman-Enskog

dengan :

T

:

suhu,

K

P

: tekanan, atm

M

ij

: berat molekul gabungan

s

_ij

: diameter tumbukan gabungan

ε

_ij

: karakteristik Lennard-Jones gabungan, K

(Perry)

Hasil perhitungan koefisien difusivitas antara CO₂melalui masing-masing komponen p

Tabel 3. Koefisien difusivitas gas CO

2

melalui campuran gas umpan

Komponen sij Mij WD CO2– H2 3.0955 84.9706 0.5227 0.8845 CO2- CH4 3.8495 170.3136 0.0851 1.1153 CO2 – CO 3.793 144.568 0.0584 1.013 CO2- H2O 3.291 397.412 0.0782 1.6467 DCO2-gas 3.4553E-06 m2/s

Menghitung Koefisien Trnasfer Massa

4 2 5 , 0 5 , 1 10 . . . 001858 , 0 _

W



 D ij ij ij P  M  T   D s  

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 j i ij  BM   BM   M  1 1 ) ( 2 1  j i ij σ  σ  σ 

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1 , 0 575 1 909 4 . 911 , 1 . 54 44

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W

  ij  ,  , ij  D ε kT  ε kT   , 5 , 0 .

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k  ε k  ε k  ε_{ij i  j} k  ε ij

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 0,2 2 05 , 0 2 2 1 , 0 75 , 0 45 , 1 exp 1  p  L  L  L  p  L  p  L c  p w a  ρ  LK  g  ρ a  LK   μ a  LK  σ  σ  a a s  

dengan :

Indeks L menunjukkan cairan

Indeks G menunjukkan gas

a

_w

:

interfacial area

efektif, m

2

 /m

3

s

_C

: tegangan muka kritis untuk bahan

 packing

, untuk metal s

_c

s

_L

: tegangan muka cairan pada suhu 70°C

D

: koefisien difusivitas, m

2

 /s

k 

g

: koefisien transfer massa pada film gas, kmol/m

2

.s.atm

k 

y

: koefisien transfer massa pada film gas, kmol/m

2

.s

k 

_l

: koefisien transfer massa pada film cairan, m/s

k 

_x

: koefisien transfer massa pada film cairan, kmol/m

2

.s

K

y

a

: koefisien transfer massa overall volumetric, kmol/m

3

.s

p

P

o

: tekanan uap, kg/cm

2

g/cm.s miu gas = 15.0531 mP.s 1.50531E-05

kg/m.s ap 38 in2/in3 128

tekanan upa dapat dihitung dengan persamaan Antoine  RT   D a d  a  D  ρ  μ  μ a GK  k   _{p  p}  p G G G G G  p g 2 3 1 7 , 0 ) ( . . 23 , 5

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3 1 4 0 2 1 3 2 . ) ( . 0051 , 0  

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g  μ  ρ d  a  D  ρ  μ  μ a  LK  k   L  L  ,  p  p  L  L  L  L w l P k  k  _y



_g  x l

k



k Ct  

1 1 1  x w  x y K a a k mk  

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16

,

0

39

,

3103

5898

,

22

exp

2

_T 

P

_COo

g terserap per satuan waktu

0.36664089

30.48 cm

333 K 27 atm

enyusun gas umpan, dapat dilihat pada tabel berikut :

Dij 3.5671E-06 7.3810E-07 6.9339E-07 6.5566E-07 4 2 5 , 0 5 , 1

10

.

.

.

001858

,

0

_

W



 D ij ij ij P  M  T   D s  

= 61 mN/m

83.800 mN/m for water chemicalogic

Pa.s

(Coulson and Richardson, 1983)

function secref  %clc,clear

global FH2OO FCH40 FCOO FCO2O FH2o Di Dp VOID RHOB TR Do DReaktor Dekv RI FO20

DReaktor=3.5;%m nilai Re sekitar 4700an dan dP cukup baik Di=DReaktor;

L=15;%panjang reformer m

RHOB=1070;%densitas bulk kg/m3 taken from topsoe Dp=0.0118;%diameter partikel m taken from topsoe

VOID=0.38+0.073*(1+(Di/Dp-2)^2/(Di/Dp)^2);%void fraction froment bischoff 

FH2OO=0.3860;%kgmol/s FCH40=0.3860;%kgmol/s 1389.6 FCOO=0.0224;%kgmol/s FCO2O=0.453;%kgmol/s FH2o=0.0233;%kgmol/s FO20=0.5;%kgmol/s FO=FH2OO+FCH40+FCOO+FCO2O+FH2o+FO20;%kgmol/s %harga awal

To=600+273.15;%suhu umpan masuk K TR=298.15;%suhu referensi K

Po=30;%tekanan atm y1o=0;

y2o=0; y3o=0;

y0=[y1o y2o To Po y3o 0]; Lspan=[0:L/100:L]; [z,y]=ode15s(@persamaan,Lspan,y0); y(end,end) y1=y(:,1); y2=y(:,2); y3=y(:,5); FH2O=FH2OO*(1-y2-2*y3)+2*FCH40*y1;%kgmol/s FCH4=FCH40-FH2OO*((y2+y3));%kgmol/s FCO=FCOO+FH2OO*y3;%kgmol/s FCO2=FCO2O+FH2OO*y3+FCH40*y1;%kgmol/s FH2=FH2o+FH2OO*(3*y2+4*y3);%kgmol/s FO2=FO20-FCH40*y1;%kgmol/s Fsisa=FCH4(end)*3600; Tgout=y(end,3)-273; fprintf('FCH4out = %6.3f kmol/hr\n',Fsisa) fprintf('FH2out = %6.3f kmol/hr\n',FH2(end)*3600) fprintf('FH2Oout = %6.3f kmol/hr\n',FH2O(end)*3600)

fprintf('FCO2out = %6.3f kmol/hr\n',FCO2(end)*3600) fprintf('FCOout = %6.3f kmol/hr\n',FCO(end)*3600) fprintf('O2 = %6.3f kg/s\n',FO20*32)

fprintf('FO2out = %6.3f kmol/hr\n',FO2(end)*3600) fprintf('Tgas out = %6.3f C\n',Tgout)

disp('design reaktor')

fprintf('Diameter Reaktor = %6.3f m\n',Di)

fprintf('Pressure Drop = %6.3f bar\n',Po-y(end,4))

%plotting hasil figure(1) hold on

plot(z,FH2O*3600,'k-',z,FCH4*3600,'b-',z,FCO*3600,'r-',z,FCO2*3600,'c-',z,FH2*3600,'g',z,FO2*3600,'m' xlabel('panjang, z')

ylabel('flow komponen, kgmol/hr')

legend('flow H2O','flow CH4','flow FCO','Flow CO2','Flow H2','FLow O2')

figure(2) hold on plot(z,y(:,3)) xlabel('panjang, z') ylabel('suhu,K') figure(3) hold on plot(z,y(:,4)) xlabel('panjang, z') ylabel('pressure,atm') function dydz=persamaan(z,y)

global FH2OO FCH40 FCOO FCO2O FH2o Di Dp VOID RHOB TR Do DReaktor Dekv RI FO20

%reaksi total combution

%CH4+2O2-->CO2+2H2O dH=-802700 kJ/kmol y1=y(1); y2=y(2); y3=y(5); T=y(3); P=y(4); FH2O=FH2OO*(1-y2-2*y3)+2*FCH40*y1;%kgmol/s FCH4=FCH40-FH2OO*((y2+y3));%kgmol/s FCO=FCOO+FH2OO*y2;%kgmol/s FCO2=FCO2O+FH2OO*y3+FCH40*y1;%kgmol/s FH2=FH2o+FH2OO*(3*y2+4*y3);%kgmol/s FO2=FO20-2*FCH40*y1;%kgmol/s FTOT=FH2O+FCH4+FCO+FCO2+FH2+FO2;%kgmol/s

%fraksi yH2O=FH2O/FTOT; yCH4=FCH4/FTOT; yCO=FCO/FTOT; yCO2=FCO2/FTOT; yH2=FH2/FTOT; yO2=FO2/FTOT; y=[yH2O,yCH4,yCO,yCO2,yH2,yO2]; %pressure parsial PH2O=yH2O*P;%bar PCH4=yCH4*P;%bar PCO=yCO*P;%bar PCO2=yCO2*P;%bar PH2=yH2*P;%bar PO2=yO2*P;%bar %property viscosity

miuCH4=[3.844 4.0112e-1 -1.4303e-4]; miuCO=[2.3811e1 5.3944e-1 -1.5411e-4]; miuCO2=[1.1811e1 4.9838e-1 -1.0851e-4]; miuH2=[27.758 2.1200e-1 -3.2800e-5]; miuH2O=[-3.6826e1 4.2900e-1 -1.6200e-5]; miuO2=[44.224 5.62e-1 -1.13e-4];

konsmiu=[miuH2O;miuCH4;miuCO;miuCO2;miuH2;miuO2]; miunya=konsmiu(:,1)+T.*konsmiu(:,2)+(T^2).*konsmiu(:,3); miuavg=y*miunya*1e-7;%kg/m/s

%data kinetika reaksi

A=[1.17E15;5.43E5;2.83E14];%konstanta tumbukan Er=[240100;67130;243900];%energi aktivasi

M=[18.015 16.043 28.01 44.01 2.016 31.999];%BM masing2 zat R=8.3145;%kj/kmol/K

%perhitungan konstanta laju reaksi

AK1=A(1)*exp(-Er(1)/R/T)*1e-3;%kgmol/kgcat/s AK2=A(2)*exp(-Er(2)/R/T)*1e-3;%kgmol/kgcat/s AK3=A(3)*exp(-Er(3)/R/T)*1e-3;%kgmol/kgcat/s %perhitungan konstanta adsorpsi CO

AKCO=8.23E-5*exp(8497.71/T); AKH2=6.12E-9*exp(9971.13/T); AKH2O=1.77E5*exp(-10666.35/T); AKCH4=6.65E-4*exp(4604.28/T);

%perhitungan konstanta keseimbangan AKP1=exp(-26830/T+30.114); AKP2=exp(4400/T-4.063); AKP3=AKP1*AKP2; %term adsorption DEN=1+AKCO*PCO+AKH2*PH2+AKCH4*PCH4+AKH2O*PH2O/PH2; DEN=DEN^2;

r2=AK1*((PCH4*PH2O-PH2^3*PCO/AKP1)/DEN/PH2^2.5); r3=AK2*((PCO*PH2O-PH2*PCO2/AKP2)/DEN/PH2); r4=AK3*((PCH4*PH2O^2-PH2^4*PCO2/AKP3)/DEN/PH2^3.5); RJ=8.3145;%kJ/kgmol/K k1a=8.11e5*exp(-86000/RJ/T)*1e-3;%kgmol/kgcat/s k1b=6.82e5*exp(-86000/RJ/T)*1e-3;%kgmol/kgcat/s KCCH4=1.26e-1*exp(27300/RJ/T);%bar-1 KCO2=7.78e-7*exp(92800/RJ/T);%bar-1 ohm=1+KCCH4*PCH4+KCO2*PO2; r1=k1a*PCH4*PO2/ohm^2+k1b*PCH4*PO2/ohm; % k1a=3.287e2*exp(-30800/RJ/T)*1e-3%kgmol/kgcat/s % KCCH4=2.02e-3*exp(36330/RJ/T)%bar-1 % KCO2=7.4e-5*exp(57970/RJ/T)%bar-1 % ohm=1+KCCH4*PCH4+KCO2*abs(PO2)^0.5 % r1=k1a*PCH4*abs(PO2)^0.5/ohm^2

%CPdT tiap zat chem prop handbook, cp dalam kJ/kmol/

CPCH4=34.942*(T-TR)-3.9957E-2/2*(T^2-TR^2)+1.9184E-04/3*(T^3-TR^3)-1.5303E-07/4*(T^4-TR^4)+3 CPCO=29.556*(T-TR)-6.5807E-03/2*(T^2-TR^2)+2.0130E-05/3*(T^3-TR^3)+1.2227E-08/4*(T^4-TR^4)+ CPCO2=27.437*(T-TR)+4.2315E-02/2*(T^2-TR^2)+1.9555E-05/3*(T^3-TR^3)+3.9968E-09/4*(T^4-TR^4) CPH2=25.399*(T-TR)+2.0178E-02/2*(T^2-TR^2)-3.8549E-05/3*(T^3-TR^3)+3.1880E-08/4*(T^4-TR^4)-8 CPH2O=33.933*(T-TR)-8.4186E-03/2*(T^2-TR^2)+2.9906E-05/3*(T^3-TR^3)-1.7825E-08/4*(T^4-TR^4)+ CPO2=29.526*(T-TR)-8.8999e-03/2*(T^2-TR^2)+3.8083E-05/3*(T^3-TR^3)-3.2629E-08/4*(T^4-TR^4)+8 %panas reaksi chem prop handbook, kJ/kmol

DHR2=206200+(CPCO+3*CPH2-CPCH4-CPH2O); DHR3=-41100+(CPCO2+CPH2-CPCO-CPH2O); DHR4=164900+(CPCO2+4*CPH2-CPCH4-2*CPH2O); DHR1=-802700+(CPCO2+2*CPH2O-CPCH4-2*CPO2); %CP kJ/kgmol/K CPaCH4=34.942-3.9957E-2*T+1.9184E-04*T^2-1.5303E-07*T^3+3.9321E-11*T^4; CPaCO=29.556-6.5807E-03*T+2.0130E-05*T^2+1.2227E-08*T^3+2.2617E-12*T^4; CPaCO2=27.437+4.2315E-02*T+1.9555E-05*T^2+3.9968E-09*T^3-2.9872E-13*T^4; CPaH2=25.399+2.0178E-02*T-3.8549E-05*T^2+3.1880E-08*T^3-8.7585E-12*T^4; CPaH2O=33.933-8.4186E-03*T+2.9906E-05*T^2-1.7825E-08*T+3.6934E-12*T^4; CPaO2=29.526-8.8999e-03*T+3.8083E-05*T^2-3.2629E-08*T^3+8.8607E-12*T^4; %menghitung Cp campuran

CPa=[FH2O FCH4 FCO FCO2 FH2 FO2]*[CPaH2O;CPaCH4;CPaCO;CPaCO2;CPaH2;CPaO2]; CPa=CPa/FTOT;%kj/kmol/K

%data perhitungan pressure drop+koef perpan overall MAVG=M*[FH2O;FCH4;FCO;FCO2;FH2;FO2];%kg/kgmol MAVG=MAVG/FTOT;%berat molekul rata-rata

mtot=FTOT*MAVG;%kg/s G=mtot/(pi*Di^2/4);%kg/m2/s

RI=0.082;%m3atm/kgmol/K RHOG=P*MAVG/T/RI/1.01325;%kg/m3 REP=Dp*G/miuavg;%bil Re partikel %persamaan differensial dy1=r1*pi*Di^2*RHOB/4/2/FCH40;%/m dy2=(r2-r3)*pi*Di^2*RHOB/4/FH2OO;%/m dy3=((r1+r3+r4)*pi*Di^2*RHOB/4-FCH40*dy1)/FH2OO;%/m dT=(-FH2OO*(dy2*DHR2+dy3*(DHR3+DHR4))-FCH40*dy1*DHR1)/(FTOT*CPa);%K/m dP=-((150*(1-VOID)/REP+1.75)*G.^2/RHOG/Dp*(1-VOID)/VOID.^3)/1e5;%bar/m DHR=(-FH2OO*(dy2*DHR2+dy3*DHR3)-FCH40*dy1*DHR1); dydz=[dy1;dy2;dT;dP;dy3;DHR];

.9321E-11/5*(T^5-TR^5); .2617E-12/5*(T^5-TR^5); -2.9872E-13/5*(T^5-TR^5); .7585E-12/5*(T^5-TR^5); 3.6934E-12/5*(T^5-TR^5); .8607E-12/5*(T^5-TR^5);

steam+CO2 benfield+CO2 (rich benfield) yh G CO2 yang t lean benfie steam L yo 0 xo G

komposisi liquid in komposisi gas in

BM kmol/hr kadar BM

benfield 56.0616 491.524992 30.7369% steam 18

CO2 44 1107.612485 69.2631% CO2 44

total 47.7073597 1599.137477

MENENTUKAN DIAMETER BAGIAN BAWAH KOLOM STRIPPER L = massa cairan yang masuk kolom per satuan waktu

G = massa gas yang masuk kolom per satuan waktu - massa gas yg terserap per satuan waktu

L 1599.137477 kmol/hr xo 50.00% BM liquid 47.7073597 xh 69.26%                      H   y  x  H   y  x  HG  L Kx  L h o o h h a ln ) 1 ( 1 ' ' '

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L 76290.62685 kg/hr yo 0 21.19184079 kg/s H 0.234354 46.71953221 lb/s Gmin 55.4435786 lb/s G actual=1.3Gmin 1897.74812 kmol/hr G actual 83.1653679 lb/s 2846.62218 kmol/hr rho G 1.9867 kg/m3 0.124028881 lb/ft3 rho L 1259.01821 kg/m3 78.6 lb/ft3 (L/G)*(rhoG/rhoL)^0.5 0.154588089 delta P=0.12F^0.7 1.502873591 ich/ft

packing=raschig ring ring, 3 inch

F 37

G'^2*F*ψ*miu^0.2/(rhoG*rhoL*g) 0.2

ψ=ratio of the density of water to the density of liquid 0.794269687

miuliquid= 5.6 cp g= 32.2 rhoGrhoLg 313.9071759 F bla bla 41.47699819 G'^2 1.513644621 G' 1.230302654 lb/ft2sec A 67.59748719 ft2 D bawah 9.279625459 ft 2.828464234 m

MENENTUKAN DIAMETER BAGIAN ATAS KOLOM ABSORBER

L = massa cairan yang masuk kolom per satuan waktu + massa gas yg terserap per satuan waktu G = massa gas yang masuk kolom per satuan waktu

L 1812.498183 kmol/hr 86469.50276 kg/hr 24.01930632 kg/s 52.95296272 lb/s rho L 78.6 lb/ft3 Gmin 62.84099202 lb/s G actual 81.69328962 lb/s rho G 0.124028881 lb/ft3 (L/G)*(rhoG/rhoL)^0.2 0.178370872 delta P=0.12F^0.7 1.502873591 packing=raschig ring, 3 inch

F 37 37

G'^2*F*ψ*miu^0.2/(rhoG*rhoL*g) 0.03

ψ=ratio of the density of water to the density of liquid 0.794269687

miuliquid= 5.6 cp               h o h  Hx  x  x  L Gmin

g= 32.2 rhoGrhoLg 313.9071759 F bla bla 41.47699819 G'^2 0.227046693 G' 0.476494169 A 171.4465673 D 14.77847321 ft 4.50453341 m

L 1599.137 kmol/hr

xh 69.26%

rserap steam 213.3607052 kmol/hr

ld 50.00% 1kg= 2.2046 lb 1m= 3.2808 ft P 4 atm T 175 C 448 K

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H 0.0000736 kmol/m3KPa 0.00745752 mol/Latm

R 0.08205 Latm/molK

T 383 K

H 0.234353685

PERHITUNGAN DIAMETER DAN TINGGI KOLOM ABSORBER

Kondisi Operasi Aborber P tinggi, T rendah

P 27 atm

T 60 C

333 K

Packing=Raschig ring, 3 inch

F 37

ap 102 m2/m3

rho 561 kg/m3

Komposisi Gas Masuk

BM flow,kmol/hr y BM average CO2 44 274.146 0.04817456 2.11968 0.05061281 Yo CO2 CO 28 936.794 0.16461899 4.60933 H2O 18 2426.313 0.42636609 7.67459 CH4 16 0.099 1.7397E-05 0.00028 H2 2 2053.328 0.36082296 0.72165 O2 32 0 0 0.00000 total 5416.534 1 15.12553 BM Gas 15.12553 kg/kmol 5690.68

Larutan penyerap=MDEA yang diaktivasi dengan piperazine Komposisi MDEA

komponen x,wt mol BM x,mol

CO2 0.0000 0 44 MDEA 0.3700 0.003078946 120.171 0.083820053 piperazine 0.0300 0.000348282 86.137 0.009481508 H2O 0.6000 0.033305579 18.015 0.906698439 L,X1 G,Y1 G,Yo L,Xo

0.036732807 BM DMEA 27.266 g/mol Henry, H 0.4399373 Ca* = 0.01661 P = 0.0053 x y X Y H= 0.319088 0.0000 0.0000 0.0000 0.0000 H= H*Ct/P 0.439937 0.1000 0.0440 0.1111 0.0460 y= 0.439937 0.2000 0.0880 0.2500 0.0965 Y= 0.278 0.3000 0.1320 0.4286 0.1520 Ct=rho/BM(liquid) 0.4000 0.1760 0.6667 0.2136 rho L 1015 0.5000 0.2200 1.0000 0.2820 BM L 27.266 0.6000 0.2640 1.5000 0.3586 Ct 37.22585 slope 0.267 xo 0.18956108

yo 0.0506128 slope itungan 0.26602607 rho G 12

y1 0.0001846 0.749155 x1 0 5403.49178 23101.7886 1kg= 1m= Diameter atas

L = massa cairan yang masuk kolom per satuan waktu

G = massa gas yang masuk kolom per satuan waktu - massa gas yg terserap per satuan waktu

G 5416.534 kmol/hr 0.39296293 G 81927.9275 kg/hr

Lmin 20360.915 kmol/hr kmol/m2.s 22.7577576 kg/s

L=2*Lmin 50.1717525 lb/s L 28505.28 kmol/hr L 28505.2803 kmol/hr rho L 63.366041 lb/ft3 777224.974 kg/hr rho G 0.7491552 lb/ft3 215.895826 kg/s 475.963938 lb/s (L/G)*(rhoG/rhoL)^0.5 1.031507476 G'^2*F*ψ*miu^0.2/(rhoG*rhoL*g) 0.023 delta P 1.1739851 inc/ft F= 26 pallring,2i

ψ=ratio of the density of water to the density of liquid 0.79426969

miuliquid= 2 cp g= 32.2 rhoGrhoLg 1528.56612 Fblabla 23.7217834 G'^2 1.48205639 lb/ft2.s G' 1.21739739 lb/ft2.s A 41.2123051 ft2 D 7.24567124 ft 2.20850745 m

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h h  x  H   y  y  y G  L 0 0 min

1.1739851 delta P

Tinggi absorber

xo xo

kalo pake x

integrasi dengan simpson rule Kya 0.02164032

Yo 0.0506128 Kya harusnya segini 0.085

Y1 0.0001846 0.3 Xo 0.0095823 X1 0 m 0.267 dY 0.0050428 Y X Y* 1/f(x) factor dA 0.0506128 0.0095823 0.002558473 20.8097766 1 20.8097766 0.04557 0.0086241 0.002302625 23.112108 4 92.4484319 0.0405272 0.0076658 0.002046778 25.9872606 2 51.9745213 0.0354844 0.0067076 0.001790931 29.6793844 4 118.717537 0.0304415 0.0057494 0.001535084 34.5943557 2 69.1887113 0.0253987 0.0047911 0.001279236 41.4602671 4 165.841068 0.0203559 0.0038329 0.001023389 51.7263504 2 103.452701 0.0153131 0.0028747 0.000767542 68.7496208 4 274.998483 0.0102703 0.0019165 0.000511695 102.474102 2 204.948205 0.0052274 0.0009582 0.000255847 201.142832 4 804.571328 0.0001846 0 0 5416.534 1 5416.534  jumlah 7323.48476 NoG = dX/3*dA NoG 12.310336 HoG=G/Kya HoG 2.3979341 Z 29.519374 m tinggi packing 35.42324929 m ln H=20.267-1.383*10^4/T +0.069*10^5/T2-0.016*10^11/T3+0.012*10^13/T4 ln H -54.834792 H 1.533E-24 komposisi mdea

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1 1

komponen Mr X kmol/hr x X

CO2 44.010 0.0002 273.15 0.0095 0.0096

MDEA 120.171 2390.86 0.0830

piperazine 86.137 270.45 0.0094

500ppm=500mg/L 0.0005

Komposisi Gas Keluar

flow,kmol/hr y CO2 1 0.000184586 0.00018462 y1 CO2 CO 936.794 0.172918896 H2O 2426.313 0.447862994 CH4 0.099 1.8274E-05 H2 2053.328 0.379015249 O2 0 0 5417.534 1

kmolCo2/m3 solution

atm(dari data kelarutan CO2 di udara) atm.m3/kmol x X kg/m3 63.36604 lb/ft3 kg/kmol kmol/m3 kg/m3 lb/ft3 2.2046 lb 3.2808 ft Diameter bawah

L = massa cairan yang masuk kolom per satuan waktu + massa gas yg terserap per G = massa gas yang masuk kolom per satuan waktu

G 5416.534 kmol/hr G actual=0.8*Gflooding 81927.92747 kg/hr G actual

22.75775763 kg/s

2.068023 50.17175247 lb/s

kmol/m2.s L 28523.6991 kmol/hr 20374.07 kmol/hr 777727.1797 kg/hr 28523.7 kmol/hr 216.0353277 kg/s 216.0353 kg/s 476.2714834 lb/s 476.2715 lb/s

(L/G)*(rhoG/rhoL)^0.5 1.032174

G'^2*F*ψ*miu^0.2/(rhoG*rhoL*g) 0.021

inch (metal) delta P 1.1739851

F= 26

ψ=ratio of the density of water to the density of liquid 0.7942697

miuliquid= 2 g= 32.2 rhoGrhoLg 1528.5661 F bla bla 23.721783 G'^2 1.3531819 G' 1.1632635 A 43.130171 D 7.4123473 y = 0.2673x 0.0000 0.0500 0.1000 0.1500 0.2000 0.2500 0.3000 0.3500 0.4000 0.4500 0.0000 0.2000 0.4000 0.6000 0.8000 1.0000 1.2000 1.4000 1.6000

2.2593109 0.013415 0.189561 0=yo/slope L' 56.38672256 kg/m2.s 56.38672 kmol/hr σc 75 mN/m σL 83.8 mN/m ap 102 m2/m3 miu L 0.001 kg/m.s rho L 1015 kg/m3 g 9.8 m/s2 a -1.33423072 b 1.880432151 c 1.187572724 d 0.205498518 e -0.61229049 aw/ap 0.457892245 0.457892245 aw 46.70500901 m2/m3 Ct=rho/MR Ct 37.22584904 kmol/m3 kx 0.021427584 kmol/m2.s Y 0.267 X Kya 0.163875613 kmol/m3.s

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 0,2 2 05 , 0 2 2 1 , 0 75 , 0 45 , 1 exp 1  p  L  L  L  p  L  p  L c  p w a  ρ  LK  g  ρ a  LK   μ a  LK  σ  σ  a a s   P k  k  _y

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satuan waktu

ft2 ft

Series1

m

G' 5.943771074 kg/m2.s miu G 1.50531E-05 Pa.s

rho G 12 kg/m3 DG 3.45527E-06 m2/s dp 0.0508 m R 0.08205 Latm/molK T 333 K a 1698.192263 b 0.713380235 c 0.037245365 d 1.28991E-05 0.00058202 kg 0.000582021 kmol/m2.s.atm DL 5.77416E-10 m2/s a 0.578245734 b 0.024209023 c 1.931014583 d 0.021293809 kl 0.00057561 m/s 0.00057561 ky 0.015714579 kmol/m2.s φ 2.26

MB=berat molekul solven 27.266 kg/kmol

R 0.08205 L.atm/molK Tb 194.5 K P 27 atm miu 0.001 kg/m.s VA 0.034 m3/kmol DAB 5.7742E-10 m2/s  RT   D a d  a  D  ρ  μ  μ a GK  k   _{p  p}  p G G G G G  p g 2 3 1 7 , 0 ) ( . . 23 , 5

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3 1 4 0 2 1 3 2 . ) ( . 0051 , 0  

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g  μ  ρ d  a  D  ρ  μ  μ a  LK  k   L  L  ,  p  p  L  L  L  L w l 1 18 ₂ 0,6

117,3.10 .(

) .

.

 B B  AB  B A

φ .M

T  

 D

 μ V 





Kondisi Operasi Stripper

Stripper beroperasi pada tekanan rendah dan suhu tinggi

P 2 atm

T 115 C

388 K

Larutan MDEA yang sudah jenuh dengan CO2, distripper dengan steam superheated

P steam 4 atm

T steam 175 C

Bahan Isian Packing

Pall ring ceramic dengan diameter 2 inch (0.0508 m)

F 20

ap 102 m2/m3 CO2 yang terb

rho 353 kg/m3 2.553

Komponen Rich Solution

komponen Mr x, wt kmol x,mol rho

CO2 44.010 0.088 2.553 0.00014 x MDEA 120.171 0.35 2390.86 0.08382 0.084 piperazine 86.137 0.029 270.45 0.00948 0.009 H2O 18.015 0.533 25862.39 0.90670 0.907 massa 777224.974 kg/hr 28523.7 1.00000 27.22389845 MR solven 27.2238984 kg/kmol

Tekanan uap dapat dihitung dengan persamaan Antoine

P01=(exp(140.54-(4735/T1)-(21.2 dengan : EXP(94,4914-6789,04/A4-11,4519*LN(A4) P° : tekanan uap, mmHg Rich solution :

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0 0 0 0 0 0 0 0 0 0 0 0

1

1

1

1

1

1

1

 x

 x

 x

 x

G

 L

 y

 y

 x

 x

 x

 x

G

 L

 y

 y

 y

0 0 _{0 0} 0 0 2 0 0 0 0 ( ) ( ) 1 1 1 1 (1 ) ( 1 [( ) ( )] 1 1 1 l  x x  Z l s l  Z x x _s s s s  L dx dZ  K a _{y L  x x}  y G x x  x x  H  y L  x x  y G x x    

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16

,

0

39

,

3103

5898

,

22

exp

2

T 

P

_COo

T : suhu, K P01=(exp(14 EXP(94,4914-6 Po 2849.08751 366.6228899 y 1424.54376 183.3114449 y x X Y 0 0 0 0.0000 0.1 0.0005455 0.000545817 0.1111 0.2 0.001091 0.001092231 0.2500 0.3 0.0016366 0.001639242 0.4286 0.4 0.0021821 0.002186851 0.6667 0.5 0.0027276 0.002735059 1.0000 0.6 0.0032731 0.003283867 1.5000 0.7 0.0038186 0.003833276 2.3333 0.8 0.0043642 0.004383287 4.0000 0.9 0.0049097 0.004933901 9.0000

 jumlah steam stripping: mGm/Lm = 0.7 sd 0.8

ambil m.Gm/Lm = 0.7 (Colburn)

berdasar pengalaman :

Gm/Lm = 0.03347171 taken from stripper Kaltim 3 Lm = 28523.7 kmol/hr

Gm = 954.737137 kmol/hr

massa Gm = 17199.8569 kg/hr 4.777738014 kg/s

P0=(exp(140.54-(4735/T)-(21.268*log(T))+(4.0909E-2*T)))*9.869233E-6; dalam atm po = 366.62289 atm m = 183.311445 mGm/Lm = 0.7 Lm = 28523.7 kmol/hr Gm = 108.921677 kmol/hr massa Gm = 1962.25451 kg/hr 0.545070697 kg/s

menghitung diameter menara

rho L = 1283 kg/m3 rho steam = 1.988 kg/m3 FLV 1.17602595

dari grafik didapat

0.0000 1.0000 2.0000 3.0000 4.0000 5.0000 6.0000 7.0000 8.0000 9.0000 10.0000 0 0.001 cair   ρ gas  ρ G  L F  TOTAL TOTAL  LV 



K4 0.7 diambil percentage flooding 80% Diamter atas s

K4 baru 0.448 L = massa caira

G = massa gas Lm

Gm

massa Gm rho L

miu L = 0.001 kg/m.s rho steam

rho gas(rho cair-rho gas) 2546.6519 FLV

42.9 bla 210.21545 dari grafik dida

GK^2 5.4272893 K4 GK, kg/m2/s = 2.3296543 kg/m2.s 0.129315466 kmol/m2.s K4 baru A = G/GK 2.0508356 m2 GK^2 D = 1.6163321 m D bawah GK LK,kg/m2/s = 96.00818 kg/m2.s 3.526613954 kmol/m2.s A=G/GK D atas Perhitungan Tinggi Menara

Diffusivitas Liquid

Indeks A menunjukkan

solute

Indeks B menunjukkan

solvent 

M

B

: berat molekul solven, gr/gmol

27.224 kg/kmol

V

A

:

volume molar solute

pada titik didihnya, cm

3

 /gmol

Tb

: suhu didih solute, K =

194.5

f

B

: konstanta, untuk cairan (air) f

B

= 2,26

R = 0.08206 L.atm/gmol/K VCO2 = 0.034 m3/kgmol miuB = 0.001 kg/m/s DAB = 2.7150E-09 m2/s

Diffusivitas CO2 dalam gas

Difusivitas biner dapat dihitung dengan menggunakan persamaan Chapman-Enskog sebagai berikut :

0 1 2 4 42, 9. . . ( )  ,  μ cair  GK Fp cair  K 

 ρ gas ρ cair ρ gas

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P Tb  R V  _A



. 1 18 ₂ 0,6

117, 3.10 .(

) .

.

 B B  AB  B A

φ .M

T  

 D

 μ V 





4 5 , 0 5 , 1 . 001858 , 0 T   M _{i }

dengan :

T

:

suhu,

K

P

: tekanan, atm

M

ij

: berat molekul gabungan

s

_ij

: diameter tumbukan gabungan

ε

_ij

: karakteristik Lennard-Jones gabungan, K

(Perry, 1997)

Hasil perhitungan koefisien difusivitas antara CO₂melalui masing-masing komponen penyusun gas um

Tabel 3. Koefisien difusivitas gas CO

₂

melalui campuran gas umpan

Komponen

s

_ij

M

_ij

W

_D

D

_ij

CO2- H2O 3.291 397.412 0.0782 1.6467 1.1133E-05

*---Koefisien Transfer

Massa---Koefisien transfer massa fasa gas dan cair ditentukan menggunakan

metode Onda, yang dinyatakan dalam bentuk persamaan sebagai berikut :

Ct=rho/MR 2 . . 

W

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 D ij ij Ps  

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 j i ij  BM   BM   M  1 1 ) ( 2 1  j i ij σ  σ  σ 





1 , 0 575 1 909 4 . 911 , 1 . 54 44

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W

  ij  ,  , ij  D ε kT  ε kT   , 5 , 0 .

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k  ε k  ε k  ε_{ij i  j} k  ε ij

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 _0,2 2 05 , 0 2 2 1 , 0 75 , 0 45 , 1 exp 1  p  L  L  L  p  L  p  L c  p w a  ρ  LK  g  ρ a  LK   μ a  LK  σ  σ  a a s    RT   D a d  a  D  ρ  μ  μ a GK  k   _{p  p}  p G G G G G  p g 2 3 1 7 , 0 ) ( . . 23 , 5

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3 1 4 0 2 1 3 2 . ) ( . 0051 , 0  

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g  μ  ρ d  a  D  ρ  μ  μ a  LK  k   L  L  ,  p  p  L  L  L  L w l P k  k  _y



_g  x l

k



k Ct  

dengan :

Indeks L menunjukkan cairan

Indeks G menunjukkan gas

a

_w

:

interfacial area

efektif, m

2

 /m

3

s

_C

: tegangan muka kritis untuk bahan

 packing

, untuk metal s

_c

= 75 mN/m

s

L

: tegangan muka cairan pada suhu 70°C

64.009 mN/n

D

: koefisien difusivitas, m

2

 /s

k 

_g

: koefisien transfer massa pada film gas, kmol/m

2

.s.atm

k 

y

: koefisien transfer massa pada film gas, kmol/m

2

.s

k 

_l

: koefisien transfer massa pada film cairan, m/s

k 

_x

: koefisien transfer massa pada film cairan, kmol/m

2

.s

K

_y

a

: koefisien transfer massa overall volumetric, kmol/m

3

.s

P

o

: tekanan uap, kg/cm

2

(Coulson and

miu steam = 15.0531 mP.s 1.50531E-05 kg/m/s

Tekanan uap dapat dihitung dengan persamaan Antoine

..…….(55)

dengan :

P°

: tekanan uap, mmHg

T

:

suhu,

K

σc 75 mN/m GK 2.329654325 σL 64.009 mN/m miu G 1.50531E-05 LK 96.0081802 kg/m2.s rho G 18.015 miu L 0.001 kg/m.s DG 1.11326E-05 ap 102 m2/m3 R 0.08205 rhoL 1283 kg/m3 T 388 g 9.8 m/s2 dp 0.0508 a 881.551006 a -1.6329979 b 0.421822266 b 1.9832196 c 0.037245365 c 0.86751474 d 3.56686E-05 d 0.25604151 kg 0.00049401 aw/ap 0.51293402 aw 52.3192704 m2/m3 w  x  y  y a k  k  k  a K  1 1 

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P P k  o

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         16 , 0 39 , 3103 5898 , 22 exp 2 _T  P_COo

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 _0,2 2 05 , 0 2 2 1 , 0 75 , 0 45 , 1 exp 1  p  L  L  L  p  L  p  L c  p w a  ρ  LK  g  ρ a  LK   μ a  LK  σ  σ  a a s   a μ GK  k   p g . . 23 , 5

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ky 0.000988019 DL 2.7150E-09 m2/s a 16.02778 b 0.05902 Ct=rho/mr c 1.9310146 Ct 47.12771032 d 0.019694 kx 0.08085908 kL 0.0017157 m/s Kxa 0.510684463 kmol/m3.s

Integrasi dengan simpson rule

xo = 0.00014 yo = 0 H 1424.543755 x1 = 0.000 Xo = 0.000139019 dx = 0.0000139 Tinggi Stripper x X f(x) 1/f(x) factor dA 0.000 0 2.661E-06 375743.0487 1 375743.0487 0.000 1.39E-05 1.629E-05 61369.14704 4 245476.5882 0.000 2.7801E-05 2.993E-05 33414.0756 2 66828.15121 0.000 4.1702E-05 4.356E-05 22957.14839 4 91828.59356 0.000 5.5603E-05 5.719E-05 17485.40278 2 34970.80555 0.000 6.9505E-05 7.082E-05 14120.11747 4 56480.46987 0.000 8.3407E-05 8.445E-05 11841.24845 2 23682.4969 0.000 9.7309E-05 9.808E-05 10195.82054 4 40783.28218 0.000 0.00011121 0.0001117 8951.951386 2 17903.90277 0.000 0.00012512 0.0001253 7978.629608 4 31914.51843 0.000 0.00013902 0.000139 7196.245021 1 7196.245021 992808.1024 Luas = dx/3*(Total) luas 4.6000109 HoG 6.9056613 Z 31.766117 m 3 1 4 0 2 1 3 2 . ) ( . 0051 , 0  

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g  μ  ρ d  a  D  ρ  μ  μ a  LK  k   L  L  ,  p  p  L  L  L  L w l k  y k gP



 x l

k



k Ct  

w  x  y  y a k  k  k  a K  1 1 

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0 0 _{0 0} 0 0 2 0 0 0 0 ( ) ( ) 1 1 1 1 (1 ) ( 1 [( ) ( )] 1 1 1 l  x x  Z l s l  Z x x _s s s s  L dx dZ  K a _{y L  x x}  y G x x  x x  H  y L  x x  y G x x    

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wa steam kmol/hr

Lean Solution

massa x,wt htng kmol x, mol x,wt mol x, mol

0 0.000 0.000 0.000 0 0 0.00000

272028.741 0.384 2263.678 0.089 0.37 0.003078942 0.08382 22539.5242 0.032 261.671 0.010 0.03 0.000348283 0.00948 414260.911 0.584 22994.975 0.901 0.6 0.033305061 0.90670

708829.176 25520.323 assumsi hanya CO2 yang terstripping 0.036732286 8*log(T1))+(4.0909E-2*T1)))*9.869233E-6; -0,010454*A4) Lean Solution

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 

0.54-(4735/T1)-(21.268*log(T1))+(4.0909E-2*T1)))*9.869233E-6;

789,04/A4-11,4519*LN(A4)-0,010454*A4)

0.002 0.003 0.004 0.005 0.006

tripper

n yang masuk kolom per satuan waktu + massa gas yg terserap per satuan waktu yang masuk kolom per satuan waktu

28526.253 kmol/hr 954.737137 kmol/hr 17199.8569 kg/hr 4.777738014 kg/s 1283 kg/m3 1.988 kg/m3 1.17613121 pat

0.4 diambil percentage flooding 80% 0.32

4.84579399

2.20131642 kg/m2.s 2.17040039 m2 1.66278118 m

for water

Richardson, 1983)

kg/m2.s kg/m.s kg/m3 Latm/molK K m kmol/m2.s.atm  RT   D a d  a  D  ρ  μ  p G  p  p G G G 3 2 1 7 , 0 ) ( 

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