Reaktor Fixed Bed X

1

REAKTOR

Tugas = Mereaksikan Toluene dengan udara dengan kecepatann umpan masuk sebesar 16749,61871 kg/jam Tipe Alat : Reaktor Fix Bed Multitulbular T = 500 oC

Kondisi operasi :

P = 5 atm

NERACA MASSA 1. Umpan Masuk : C7H8

= 13,238

kmol/ jam = 1217,918 kg/jam

= 0,142

kmol/jam = 11,085

kg/jam

C6H5CHO = 0,0115

kmol/jam = 1,2169

kg/jam

O2

= 113,01

kmol/jam = 3616,171 kg/jam

N2

= 425,12

kmol/jam = 11903,23 kg/jam +

Total

= 551,5125 kmol/jam = 16749,61871 kg/jam

C6H6

2 Reaksi : I. C7H8 + O2 ----------> C6H5CHO + H2O II. C7H8 + 6 O2

S1 = 0,5

---------> C4H2O3 + 3H2O + 3 CO2

S2 = 0,5

III. C6H6 + 7,5 O2 ---------> 6 CO2 + 3 H2O

Katalis : molybdenum oxide dan uranium oxide Konversi : 0,9 (90 %) terhadap C7H8 Reaksi bisa ditulis : I.

A + B

------->

C +D

II. A + 6 B

------->

E +3D +3F

III. A + 7,5 B

------->

6F +3D

Maka pada saat konversi = X A REAKSI 1 : n A ( C7H8 )

= nAo (1 - XA)

nb (O2)

= nBo - nAo X

nC (C6H5CHO) = nCo + nAo X

n D (H2O)

* (S1 + 6 S2) - 7,5 nGo

A

A

= nDo + nAo X

A

S1 ( S1 + 3 S2)

n E (C4H2O3) = nEo + nAo X

A

S2

n F (CO2)

A

3 S2

= nFo + nAo X

REAKSI 3 :

n F (CO2)

= nF + 6 nGo

n D (H2O)

= nD + 3 nGo

Maka pada korversi X A = 0,9 Maka diperoleh hasil reaksi : C7H8

=

1,324

kmol/ jam = 121,792

kg/jam

C6H5CHO

=

5,969

kmol/jam

632,681

kg/jam

O2

=

70,239

kmol/jam =

2247,648

kg/jam

N2

=

425,115

kmol/jam

=

11903,228

kg/jam

H2O

=

24,255

kmol/jam

=

436,593

kg/jam

CO2

=

18,724

kmol/jam

=

823,871

kg/jam

C4H2O3

=

5,957

kmol/jam

=

583,806

kg/jam

Total

=

551,584 kmol/jam

=

16749,619

kg/jam

=

PENENTUAN KONSTANTA KECEPATAN REAKSI Reaksi:

3 I. C7H8 + O2

----------> C6H5CHO + H2O

II. C7H8 + 6 O2

S1 = 0,5

---------> C4H2O3 + 3H2O + 3 CO2

S2 = 0,5

III. C6H6 + 7,5 O2 ---------> 6 CO2 + 3 H2O

Reaksi diatas dapat disederhanakan menjadi: I. A + B

------->

C +D

II. A + 6 B

------->

E +3D +3F

III. G + 7,5 B

------->

F +3D

Persamaan kece patan reaksinya dapat ditulis sebagai berikut:  rA  

1 dN A  k .C AC B V dT

Kinetika Reaksi Reaksi pembentukan Benzene berdasarkan reaksi : C7H8

+

O2

+

H2O

1. Waktu reaksi

=

5 dtk

2. Suhu operasi

=

500oC

3. Perbandingan mol reaktan C7H8 : O2

= 1 : 9.473373

4. Konversi yang dapat dicapai

=



C6H5CHO

Dari Patent diperoleh data sebagai berikut :

0.2

Neraca massa C7H8

+

A

O2



C6H5CHO

+

H2O

B



C

+

D

Mula-mula

:

100

947.3374

Reaksi

:

100 . A

100 . XA

Hasil

: 100 (1-XA) 947.3374 – 100 XA

Komposisi awal : A

=

100

gmol

100 . XA

100 . XA

100 . XA

100 . XA

4 B

=

947.3374

gmol

1047.3374 Tekanan

=

5

atm

Suhu

=

500

o

C

Untuk reactor RAP : xa

V dXa  CAo  Fv  ra 0

k

dXA CA CB

k

dXA CAo(1  XA) CAo ( M  XA)

  CAo 

  CAo 

  CAo  k

k

k

dXA

k

dXA CAo (1  XA) ( M  XA) 2



1 CAo

 (1  XA)



1 CAo

 (1  XA)



1 CAo

 Y dXA

=

dXA ( M  XA)

1 dXA ( M  XA)

1

0.02

XA

1/Y

l

0.000

0.20000

1

0.200

0.020

0.20490

4

0.820

0.040

0.21001

2

0.420

0.060

0.21535

4

0.861

0.080

0.22093

2

0.442

0.100

0.22676

4

0.907

0.120

0.23286

2

0.466

0.140

0.23926

4

0.957

0.160

0.24597

2

0.492

0.180

0.25301

4

1.012

0.200

0.26042

1

0.260 6.837

5 Hasil integrasi

=

0.045580391

Konsentrasi Awal A (S2) Cao

= =

nAo

Pt

nT

RT

100

5

1047.3374 =

(82.06 . 773) gmol/cm3

7.52614E-06

Perbandingan reaktan : M

=

nBo nAo

=

9.473374

Konversi : XA

= k

0.9

=

1 7.52614E-06

k

=

0.045580391 5

1211.255209 cm3/gmol.dt

0

Setiap kenaikan suhu 10oC maka kecepatan reaksi menjadi 2 kali lipatnya : T1

=

773 oK



k1 =

1211.255209 (cm3/gmol)/dt

T2

=

783 oK



k2 =

1514.1 (cm3/gmol)/dt

Persamaan empiris Arhenius : k  Ae (  E .RT )

Atau ln k  ln A 

ln

B T

1211.255209

=

ln A

+

B 773

ln

1211.255209

=

ln A

+

B 783

6 -0.223143551

=

-0.223143551

=

B

=

-13505.96427

=

ln A

B ln

1211.255209

1.65219E-05

+

-13505.96427 773

7.099412463

=

ln A

ln A

=

24.57155253

A

=

46912618686

A

=

46912618686

B

=

-13505.96427

kR1

=

46912618686

+

exp (-13505.96427 /T)

-17.47214007

(cm3/gmol) / dt

PENYUSUNAN MODEL MATEMATIS PADA ELEMEN VOLUME 1. NERACA MASSA PADA ELEMEN VOLUME FA

z

Z

FA

z+∆z

∆Z

∆Z

Masuk – keluar = akumulasi FA Z –[ FA Z +  Z+(-rA) dv ] =Acc dV = A.∆Z dimana A =

 .Di 2 4

Neraca massa elemen volume juga meninjau ruang kosong diantara tumpukan katalis sehingga porositas (ε) berpengaruh. Porositas (ε) didapat dari Brown, fig.219 & 220. Maka : dV =

 .Di 2

FA Z – FA

4 Z+

 .z πDi ε Z=0 4 πDi 2 ε = (-rA) 4

 Z-(-rA)

7

FA Z – FA Z +  Z Z FA  rA πDi 2 ε   Δz 4 dimana FA = F A0(1-XA) FA = -FA0. XA X A (rA ). .Di 2   z 4 ΔX A (r ) . .D i 2  A ε Δz 4FA0

FA0

lim z  0 dX A ( rA ). .Di 2  ε dz 4FA0

(-rA) = kecepatan reaksi = k. CA. CB dX A (kC A C B ). .Di 2  ε dz 4FA0

CA 

=

CB 

=

n A Pt n t RT

n A0 (1  X A )Pt n 1 .RT n B Pt n t RT

n A0 (

n BO  X A )Pt n Ao n1.RT

Maka : 2

 n AO Pt  n   .k(1  X A )( Bo  X A ) .Di 2ε . n Ao dX A  n t RT   dz 4FA0

Dimana : dX A  Perubahan konversi persatuan panjang dz Di = Diameter dalam

...…………(1)

8



= porositas tumpukan katalis

F AO = Kecepatan molar A mula-mula

2. NERACA PANAS PADA ELEMEN VOLUME T z ∆z

z

QR

Qp ∆z T z+∆z Masuk – keluar = akumulasi ∑m.cp (Tz -To ) – (

 m.c (T p

QR

=

 HR.nAO.  XA

QP

=

U.A.  T

Z+

 Z - To) + QR + QP ) = 0

= U.  .DO.  z.(Ts - T)

 m.cp ( T z -T Z+∆Z ) - ∆HR..nAO.  XA - U.  .DO..∆z. (Ts - T) = 0  m.cp ( T Z - TZ+∆Z ) = ∆HR..nAO..  XA + U.  .DO. ∆z. (Ts - T)  HR nAO  XA +U  DO (TS- T ) T Z – T Z+  Z = ∑ m.Cp T

Z T

Z  Z

z

T  z



ΔH R n A0

ΔH R n A0

: z

ΔX A  U.π.π 0 . (Ts  T) Δz  m.Cp

ΔX A  U. .D 0 (TS  T) Δz  m.Cp

Lim  z  0 ……………..….. 2 dimana : dT/dZ

= perubahan suhu persatuan panjang katalis

 HR

= panas reaksi

U

= over all heat transfer coefficient

Do

= diameter luar

9 T

= suhu gas

Ts

= suhu penelitian

m.Cp = kapasitas panas 3. NERACA PANAS UNTUK PENDINGIN PADA ELEMEN VOLUME Tinjauan : elemen panas

Masuk – keluar = akumulasi mp.Cpp. ( Ts

Z+∆Z

- To ) + Qp – mp.Cpp. (( Ts

Z

- To ) = 0

Qp = U.A. ∆T ; dimana : A = π.Do.∆z. dan ∆T = (T – Ts) Sehingga Qp = U.π. Do.∆z. (T – Ts) mp. Cpp. (Ts

Z+∆Z

- Ts Z ) = - U.π. Do.∆z. (T – Ts) : mp. Cpp. ∆z

Ts

Z+∆Z

- Ts

Z

U.π. Do. (T – Ts) = -

∆z

m. Cpp

Ts U. .Do.( T - Ts )  z (m.Cp) p lim z  0 dTs U. .Do. ( T - Ts )  dz ( m.Cp ) p

.……………….(3)

PENURUNAN TEKANAN ( PRESSURE DROP ) Penurunan tekanan dalam pipa yang berisi katalisator (fixed bed) menggunakan rumus 11.6 (Chapter 11, Rase) hal 492, Chemical Reactor Design for Process Plants.

10 gc.dP (1   ) s  1   G  150  1,75 3 2 3  s.dz  Dp    Dp 2

Persamaan di atas dapat ditulis : f k .G 2  1    dP    dz D p . f .gc   3 

……………….(4)

dimana :  1    f k  1,75  150  D .G/  p   dimana : dP  perubahan tekanan per satuan panjang dz

fk = faktor friksi gc = konstanta gravitasi G = kecepatan aliran massa gas dalam pipa, g/cm3 ρf = densitas gas, g/cm3 Dp = diameter partikel katalisator, cm ε

= porositas tumpukan katalisator

μ = viskositas gas, g/cm.jam Sehingga diperoleh 4 persamaan differensial simultan sebagai berikut : 2

1)

2)

dX A  dz

dT  dz

 n AO Pt  n   .k(1  X A )( Bo  X A ) .Di 2ε . n Ao  n t RT 

- ΔH R .n A0

4FA0

ΔX A  U.π.π 0 (TS  T) Δz Σm.Cp

3)

dTs U. .Do. ( T - Ts )  dz ( m.Cp ) p

4)

dP f .G 2  1     k   dz D p . f .gc   3 

Selanjutnya persamaan differensial simultan tersebut diatas diselesaikan dengan program computer dengan Metode Numeris Runge Kutta. OVERALL HEAT TRANSFER 1.

Koefisien transfer panas pipa (hio) Dari pers. 6-2, Kern diperoleh :

11 hio =

 Dp.Gt   μ  

0,8

0,027

 Cp.μ    k  

1/3



k    Di  

…………….(5)

Persamaan diatas berlaku untuk organic liquid, larutan aqueous, dan gas pada Re > 10.000 dimana : Dp

= diameter partikel katalis

Di

= diameter dalam pipa

k

= konduktivitas thermal

μ

= viskositas gas

Cp

= panas jenis gas

Gt

= kecepatan massa per satuan luas

hi

= koefisien transfer panas pipa dalam

hio

= hi.

2.

ID ...................(Kern,1983) OD

Koefisien transfer panas dinding pipa dalam shell ( ho) Dari persamaan

, Kern :

 De.Gp   ho = 0,36 μ p  

0,55

 Cp p . p  kp 

0 , 33

  

 kp    ...........(Kern,1983, p137)  De 

Persamaan diatas berlaku untuk Re antara 2000 – 1.000.000 dimana : ho

= koefisien transfer panas

De

= diameter equivalent

Gp = kecepatan massa pendingin per satuan luas p

= viskositas pendingin

kp

= konduktivitas thermal pendingin

Cpp = panas spesifik pendingin 3. Overall heat transfer coefisient Ud = overall transfer coefisient pada saat kotor Uc = overall transfer coefisient pada saat bersih Rd = faktor tahanan panas pengotor Uc 

hio.ho (hio  ho)

Rd .Uc.Ud  Uc  Ud Rd .Uc.Ud  Ud  Uc ( Rd .Uc  1).Ud  Uc

Rd 

Uc  Ud Uc.Ud

…….(6)

12 Maka : Ud 

Uc ( Rd .Uc  1)

LAY OUT PIPA DALAM REAKTOR ( Kern, 1983, P. 139 ) Pipa dalam reaktor disusun secara square pitch, dimana luas penampang 1 pipa menempati luasan sebesar Pt2. 1 pipa menempati luasan = Pt2 maka luas total penampang reaktor ( over design 10%) As = 1,1. Nt.Pt2 dimana : As = Luas penampang shell Nt = jumlah pipa Pt = pitch Alasan penyusunan pipa secara square pitch : 1. mudah pembersihannya. 2. pressure drop kecil.

FLOW AREA DALAM SHELL B As =

C’

dimana : B

= Jarak buffle, in

C’

= Clearance, in

Pt

= Pitch, in

IDs = Diameter dalam shell, in As

= Flow area shell, in2

DIAMETER EQUIVALEN (De)

IDs .B.C' Pt

………….(7)

13 Diameter equivalen dapat dipahami sebagai diameter dari area dalam shell, bila dipandang sebagai pipa ( Kern, 1983) p.139  μ.OD2 4 Pt 2  4 De =   .OD

De =

 

4 x free area

 = wetted perimeter

4Pt 2  μ.OD 2 4. .OD

………….(8)

DIAMETER SHELL Diameter shell yang dipakai untuk Nt pipa Luas shell = As = 1,1.Nt.Pt2 =

π.(IDs ) 2 4

Diameter shell : IDs =

4. As

…..……..…(9)



KATALISATOR ( Rase, 1977 ) Katalisator yang digunakan berupa molybdenum oxide dan uranium oxide dengan : -

Bentuk = pellet

-

Ukuran D = 0,3175 cm H = 0,3175 cm

-

Bulk density = 1106,13 kg/m3

-

Formula = PbO2 – Mg (SA, Miller Ernest, 1965)

DIAMETER PARTIKEL ( Dp ) Yaitu diameter partikel katalis yang ekuivalen dengan diameter bola dengan volume yang sama dengan volume katalis ( Rase, 1977, p.493 ) V kat = =

 .D 2 4

.H

 .0,31752 .0,3175 4

= 0,025125 cm3 V bola = V kat

14  .Dp

V Bola =

3

4

Maka : Dp

=3 =

3

VB 6



0,025125.6



= 0,36345 cm PEMILIHAN PIPA Dalam pemilihan pipa harus diperhatikan faktor perpindahan panas. Pengaruh bahan isian di dalam pipa terhadap koefisien transfer panas konveksi didelik oleh Colburn ( Smith,JM., p.571) dan diperoleh hubungan pengaruh rasio (Dp/Dt) atau perbandingan diameter katalis dengan diameter pipa dengan koefisien transfer panas pipa berisi katalis disbanding transfer panas konveksi pada pipa kosong. Dp/Dt

0.05

0.01

0.15

0.12

0.25

Hio/h

5.5

7

7

7.5

7.0

Dimana : (Dp/Dt)

= rasio diameter katalis per diameter pipa

(hio/h)

= rasio koefisien transfer panas pipa berisi katalis disbanding koefisien transfer panas pada pipa kosong.

Dari data diatas diperoleh (hio/h)max terjadi pada 7,8 pada (Dp/Dt) = 0,15 Dp  0,15 Dt Dp Dt  0,15 0,36345 Dt  cm 0,15

= 2,42298 cm Dipilih pipa dengan ukuran standar (Kern, table 11) : NPs = 1 in OD

= 1,32 in

ID

= 1,049 in

Sch

= 40

15 JUMLAH PIPA ( Brown, 1950 ) Jumlah pipa ditentukan berdasarkan turbulensi gas dalam pipa berkatalis. Dalam suatu reaksi khusus terjadi tumbukan molekul yang optimum ( well mixed). Keadaan di atas terjadi bila pada keadaan turbulen yaitu bilangan Reynold diatas 3100. Spherecity ( )=

Luas area bola luas area katalis

Luas area bola = . Dp2 = 3,14. 0,36345 = 0,4148 cm2 Luas area katalis

 π.DH 

2. .D 4

= 0,4748 cm2 0,4148

maka  = 0,4748  0,8736 Dari fig. 223 Brown diperoleh  = 0,35

KECEPATAN MASSA MASUK REAKTOR Komponen O2 C7H8 N2

Kgmol/jam 13.1443 124.5205 468.4342

C6H5CHO C6H6 Total

0.0123 0.1399 606.2511

Kg/jam 1209.2717 3984.6553 13116.157 2 1.2990 10.9158 18322.298 8

Kecepatan massa = 18322.2988 Kg/j = 5089.5278 g/dt BM rata-rata = 30.22 Suhu Umpan (T) = 773 oK Tekanan Umpan (P) = 2 atm Densitas gas(rho) = 0.000953 g/cm3 Viskositas gas = 0.000973 g/cm dt Digunakan pipa Standard ID pipa = 1.049 in = 2.664 cm OD pipa = 1.320 in = 3.353 cm BWG = 16 in A. Jumlah pipa maximum :

16 1. Menghitung Gt :

= 18866.13 Gt Gt = 0.217321 g/cm2 dt

2. Menghitung Luas penampang pipa :

= 2.0063 cm2 3. Menghitung Luas penampang total :

= 23419.4434 cm2 Menghitung Jumlah pipa maximum :

= 11673.0938 pipa

B. Jumlah pipa minimum : 1. Menghitung Kecepatan maximum :

17 =

771.5457 cm/dt

2. Menghitung Kecepatan Volume Umpan :

= 5341104.5000 cm3/dt 3. Menghitung Luas penampang total :

= 6922.6025 cm2 Menghitung Jumlah pipa minimum :

= 3450.4744 pipa C. Jumlah pipa : 1. Menghitung Gt : diambil bilangan Reynold (Re) =

7000

= 18866.13 Gt Gt = 7000 / 18866.13 g/cm2 dt = 0.371035 g/cm2 dt 2. Menghitung Luas penampang pipa :

= 2.0063 cm2

18 3. Menghitung Luas penampang total :

= 13717.1025 cm2 Menghitung Jumlah pipa :

= 6837 pipa SIFAT FISIS a) Spesifik Heat Cp = A + BT + CT2 +DT 3 Komponen A 6,713 7,440 4,728 7,701 -3,123 -5,817 -8,107 -2,9

O2 N2 CO2 H2O C4H2O3 C7H8 C6H6 C6H5CHO

Cp ( joule/mol.K ) B C -6 0,879.10 4,170.10-6 -2 0,324.10 6.400.10-6 1,754.10-2 -1,338.10-5 4,595.10-4 2,521.10-6 -2 8,32.10 -5,22.10-5 1,22.10-1 -6,61.10-5 1,13.10-1 -7,21.10-5 1,19.10-1 -6,79.10-5

Cp = Σ Cpi . yi b) Viskositas µ = A + BT + CT2 Komponen

µ ( micropoise.K )

O2

Tc 154,6

Pc 49,8

N2

126,2

33,5

CO2

304,2

72,8

H2O

647,3

217,6

C4H2O3 C7H8 C6H6 C6H5CHO

591,7 562,1 695

406 483 46

D -2,544.10-9 -2,790.10-9 4,097.10-9 -0,859.10-9 1,16.10-8 1,17.10-8 1,70.10-8 1,23.10-8

19 *Sumber: Robert C. Reid Sifat gas dan zat cair Gramedia Pustaka 1991

c) Konduktivitas Thermal

k = ((14,54 . T/Tc ) – 5,14 )2/3 . Cp/τi . 106 Komponen O2

Tc 154,6

Pc 49,8

N2

126,2

33,5

CO2

304,2

72,8

H2O

647,3

217,6

C4H2O3 C7H8 C6H6 C6H5CHO

591,7 562,1 695

406 483 46

d) Sifat Pendingin Cp = 0,509 Btu/lboK µ = 0,40 micropoise.K k = 0,68 Btu/jam.ft.oK PANAS REAKSI T ΔHT

To

ΔH298

ΔHT = ΔH298 + ∫ ΔCp dT Komponen O2 N2 CO2 H2O C4H2O3 C7H8 C6H6 C6H5CHO

 Hf

0 0 -94,05 -68,315 -112,08 11,95 19,82 -8,79

20 *sumber: Dimana : ΔH298 = ΔHfp − ΔHfR = -205,02 − -103,6 = -101,42 kjoule/mol = -24239,38 kkal/kmol ΔCp = α + βT + γT2 α = AP – AR β = B P – BR γ = C P – CR Komponen O2 N2 CO2 H2O C4H2O3 C7H8 C6H6 C6H5CHO

α

A 6,713 7,440 4,728 7,701 -3,123 -5,817 -8,107 -2,9

Cp ( joule/mol.K ) B C 0,879.10-6 4,170.10-6 0,324.10-2 6.400.10-6 -2 1,754.10 -1,338.10-5 4,595.10-4 2,521.10-6 -2 8,32.10 -5,22.10-5 1,22.10-1 -6,61.10-5 1,13.10-1 -7,21.10-5 1,19.10-1 -6,79.10-5

= 64,374 – ( 11,591 + 45,780 ) = 7,003 joule/mol.K

β

= 6,4776.10-2 – ( 3,2301.10-1 + 2,1034.10-2 ) = 2,2286.10-2 joule/mol.K

γ = 3,5143.10-4 – (-1,3067.10-4 + 1,2484.10-4 ) = 3,5726.10-4 joule/mol.K Sehingga :

= =

D -2,544.10-9 -2,790.10-9 4,097.10-9 -0,859.10-9 1,16.10-8 1,17.10-8 1,70.10-8 1,23.10-8

21 = -101,42 kjoule/mol + 7,003 ( 298 - 513 ) +

+

joule/mol

Penyelesaian persamaan defferensial secara numeris dengan metode Runge Kutta : CLS N = 6837 'jml pipa MS = 15 Ptek = 2 'massa pendingin C = 500 'suhu operasi TC1 = C TcC = C D = 490 'suhu keluar td0 = D DOU = 1.32 * 2.54 'diameter luar pipa DI = 1.049 * 2.54 'diameter dalam pipa Pt = 1.25 * DOU 'pitch CL = Pt - DOU 'clearence DE = (4 * (Pt ^ 2 - (3.14 * DOU ^ 2 / 4))) / (3.14 * DOU) 'diameter aquivalen Ass = N * Pt ^ 2 * 1.15 'luas penampang shell ID = (4 * Ass / 3.14) ^ .5 'diameter dalam shell BS = ID / 5 'jarak buffle AT = 3.14 / 4 * DI ^ 2: 'luas penampang pipa ASi = ID * CL * BS / Pt 'flow area dlm shell PRINT PRINT " ┌─────────────────────────────────────────────────────────┐" PRINT " │ R E A K T O R F I X E D B E D M U L T I T U B E │" PRINT " └─────────────────────────────────────────────────────────┘" PRINT PRINT USING " Jumlah pipa = #### pipa"; N PRINT USING " Diameter luar pipa = #.### cm"; DOU PRINT USING " Diameter dalam pipa = #.### cm"; DI PRINT USING " Pitch = #.### cm"; Pt PRINT USING " Diameter Shell = #.### m"; ID / 100 PRINT USING " Jumlah pendingin = ######.### Kg/j"; MS * N * 3.6 PRINT 'MS * N * .373 * (536.6 - 504.8) PRINT : PRINT : PRINT Dp = .5723 BMB = 32 'O2 BMA = 92 'C7H8 BMC = 28 'N2 BMD = 18 'H20 BME = 106 'C6H5CHO BMF = 78 'C6H6 BMG = 98 'C4H2O3 BMH = 44 'CO2

'KECEPATAN MASUK MASING-MASING GAS (KGMOL/JAM) FAIO = 1209.2717# / BMA 'C7H8 FBIO = 3984.6553# / BMB 'O2 FCIO = 13116.1569# / BMC 'N2 FDIO = 0 / BMD 'H2O

22 FEIO FFIO FGIO FHIO FAo FBo FCo FDo FEo FFo FGo FHo

= = = = = = = = = = = =

1.299 / BME 10.9158 / BMF 0 / BMG 0 / BMH FAIO FBIO FCIO FDIO FEIO FFIO FGIO FHIO

/ / / / / / / /

'C6H5CHO 'C6H6 'C4H2O3 'CO2

3.6: 3.6: 3.6: 3.6: 3.6: 3.6: 3.6: 3.6:

FTO = FAo + FBo + FCo + FDo + FEo + FFo + FGo + FHo bmrt = (FAo / FTO) * BMA + (FBo / FTO) * BMB + (FCo / FTO) * BMC + (FDo / FTO) * BMD + (FEo / FTO) * BME + (FFo / FTO) * BMF + (FGo / FTO) * BMG + (FHo / FTO) * BMH GT = FTO * bmrt / AT: GS = MS * N / ASi PRINT " KECEPATAN MASSA MASUK REAKTOR" PRINT PRINT " ┌──────────┬─────────────┬───────────────┐ PRINT " │ Komponen │ Kgmol /jam │ Kg / jam │ PRINT " ├──────────┼─────────────┼───────────────┤ PRINT USING " │ O2 │ #####.#### │ ######.#### │ FBIO * BMB PRINT USING " │ C7H8 │ #####.#### │ ######.#### │ FAIO * BMA PRINT USING " │ N2 │ #####.#### │ ######.#### │ FCIO * BMC PRINT USING " │ H2O │ #####.#### │ ######.#### │ FDIO * BMD PRINT USING " │ C6H5CHO │ #####.#### │ ######.#### │ FEIO * BME PRINT USING " │ C6H6 │ #####.#### │ ######.#### │ FFIO * BMF PRINT USING " │ C4H203 │ #####.#### │ ######.#### │ FGIO * BMG PRINT USING " │ CO2 │ #####.#### │ ######.#### │ FHIO * BMH PRINT " └──────────┴─────────────┴───────────────┘ PRINT USING " Total #####.#### ######.#### 3.6; FTO * bmrt * 3.6 INPUT "", A$ PRINT PRINT " KECEPATAN MASSA GAS KELUAR REAKTOR" PRINT

" " " "; FBIO; "; FAIO; "; FCIO; "; FDIO; "; FEIO; "; FFIO; "; FGIO; "; FHIO; " "; FTO *

XA = .9 XA1 = 1 / .9 * XA S1 = .5 S2 = .5 FA FB FC FD FE FF FG FH

= = = = = = = =

FAo FBo FCo FDo FEo FFo FGo FHo

* (1 - XA) 'C7H8 - FAo * XA * (S1 + 6 * S2) - 7.5 * FFo * XA1 'O2 'N2 + FAo * XA * (S1 + 3 * S2) + 3 * FFo * XA1 'H2O + FAo * XA * S1 'C6H5CHO - FFo * XA1 'C6H6 + FAo * XA * S2 'C4H2O3 + 3 * FAo * XA * S2 + 6 * FFo * XA1 'CO2

FT = FA + FB + FC + FD + FE + FF + FG + FH + FI YA = FA / FT: YB = FB / FT: YC = FC / FT: YD = FD / FT: YE = FE / FT: YF = FF / FT: YG = FG / FT: YH = FH / FT: YI = FI / FT

23 m = BMA * YA + BMB * YB + BMC * YC + BMD * YD + BME * YE + BMF * YF + BMG * YG + BMH * YH + BMI * YI PRINT PRINT PRINT PRINT USING FB * 3.6 * BMB PRINT USING FA * 3.6 * BMA PRINT USING FC * 3.6 * BMC PRINT USING FD * 3.6 * BMD PRINT USING FE * 3.6 * BME PRINT USING FF * 3.6 * BMF PRINT USING FG * 3.6 * BMG PRINT USING FH * 3.6 * BMH PRINT PRINT USING FT * 3.6 * m INPUT "", A$

" " " "

┌──────────┬─────────────┬───────────────┐ │ Komponen │ Kgmol /jam │ Kg / jam │ ├──────────┼─────────────┼───────────────┤ │ O2 │ #####.#### │ ######.#### │

"

│

C7H8

│ #####.####

│

######.####

│ "; FA * 3.6;

"

│

N2

│ #####.####

│

######.####

│ "; FC * 3.6;

"

│

H2O

│ #####.####

│

######.####

│ "; FD * 3.6;

"

│

C6H5CHO │ #####.####

│

######.####

│ "; FE * 3.6;

"

│

C6H6

│ #####.####

│

######.####

│ "; FF * 3.6;

"

│

C4H2O3

│ #####.####

│

######.####

│ "; FG * 3.6;

"

│

CO2

│ #####.####

│

######.####

│ "; FH * 3.6;

" "

└──────────┴─────────────┴───────────────┘ " Total #######.#### ######.#### "; FT * 3.6;

PRINT PRINT " m = bmrt FA = FAo FB = FBo FC = FCo FD = FDo FE = FEo FF = FFo FG = FGo FH = FHo

CO2 "

" " " "; FB * 3.6;

Enthalpi Umpan Masuk Reaktor :"

GOSUB 7000 Qo1 = QTOT INPUT "", P$ PRINT " PRINT PRINT " PRINT PRINT "

PRINT PRINT " PRINT INPUT "", A$ PRINT : PRINT A = 0: B = 0: E1 = E PRINT " PRINT USING " PRINT USING " PRINT USING " PRINT USING " PRINT " PRINT : INPUT "",

Reaksi yang terjadi :" C7H8

+ O2

----------->

C6H5CHO

+ H2O"

C7H8

+ 6 O2

----------->

C4H2O3

+ 3 H2O + 3

C6H6

+ 7.5 O2

----------->

6 CO2

+ 3 H2O"

" KONDISI AWAL" E = Ptek: F = 1 ┌──────────────────────────────────────────┐" │ Suhu gas masuk = ###.# °C │"; │ Suhu pendingin keluar = ###.# °C │"; │ Tekanan awal = ##.# atm │"; │ Increment tebal katalis = #.## cm │"; └──────────────────────────────────────────┘" P$

' Perhitungan RUNGA KUTTA PRINT " ┌──────┬────────┬────────┬───────┬────────┐"

C D E F

24 PRINT " PRINT " PRINT USING "

│ L(cm)│ Xa │ T(c) │ Td (c)│ P(atm) │" ├──────┼────────┼────────┼───────┼────────┤" │ #### │ #.#### │ ###.## │ ###.# │ ##.### │"; A; B; C;

D; E NO = 0 620 GA = A: GB = B: GC = C: GD = D: GE = E: GF = F: GG = G: GH = H GOSUB 910

GB GC GD GE

= = = =

B C D E

K1 = DX * F L1 = T * F EM1 = S * F N1 = P * F K1 L1 EM1 N1

+ + + +

GOSUB 910

GB GC GD GE

= = = =

B C D E

K2 = DX * F L2 = T * F EM2 = S * F N2 = P * F (K2 / 2) (L2 / 2) (EM2 / 2) (N2 / 2)

+ + + +

GOSUB 910

GB GC GD GE

= = = =

B C D E

K3 = DX * F L3 = T * F EM3 = S * F N3 = P * F (K3 / 2) (L3 / 2) (EM3 / 2) (N3 / 2)

+ + + +

GOSUB 910

GB2 GC2 GD2 GE2

= = = =

B C D E

+ + + +

K4 = DX * F L4 = T * F EM4 = S * F N4 = P * F 1 1 1 1

/ / / /

6 6 6 6

* * * *

(K1 + 2 * K2 + (L1 + 2 * L2 + (EM1 + 2 * EM2 (N1 + 2 * N2 +

2 2 + 2

* * 2 *

K3 + K4) L3 + L4) * EM3 + EM4) N3 + N4)

710 A = A + F C = GC2 B = GB2 D = GD2 E = GE2 NO = NO + 1 Qre = Q1 * F * N + Qre Ql = Q2 * F * N + Ql 790 IF B >= .9 THEN 870 IF NO = 10 THEN 800 GOTO 620 800 PRINT USING " │ #### │ #.#### │ ###.## │ ###.# │ ##.### │"; A; B; C; D; E: NO = 0 860 GOTO 620 870 PRINT USING " │ #### │ #.#### │ ###.## │ ###.# │ ##.### │"; A; B; C; D; E PRINT " └──────┴────────┴────────┴───────┴────────┘"

25 PRINT PRINT " PRINT

KECEPATAN MASSA GAS KELUAR REAKTOR"

XA = .9 XA1 = 1 / .9 * XA S1 = .5 S2 = .5 FA FB FC FD FE FF FG FH

= = = = = = = =

FAo FBo FCo FDo FEo FFo FGo FHo

* (1 - XA) 'C7H8 - FAo * XA * (S1 + 6 * S2) - 7.5 * FFo * XA1 'O2 'N2 + FAo * XA * (S1 + 3 * S2) + 3 * FFo * XA1 'H2O + FAo * XA * S1 'C6H5CHO - FFo * XA1 'C6H6 + FAo * XA * S2 'C4H2O3 + 3 * FAo * XA * S2 + 6 * FFo * XA1 'CO2

FT = FA + FB + FC + FD + FE + FF + FG + FH YA = FA / FT: YB = FB / FT: YC = FC / FT: YF = FF / FT: YG = FG / FT: YH = FH / FT: m = BMA * YA + BMB * YB + BMC * YC + BMD * * YG + BMH * YH + BMI * YI PRINT PRINT PRINT PRINT USING FB * 3.6 * BMB PRINT USING FA * 3.6 * BMA PRINT USING FC * 3.6 * BMC PRINT USING FD * 3.6 * BMD PRINT USING FE * 3.6 * BME PRINT USING FF * 3.6 * BMF PRINT USING FG * 3.6 * BMG PRINT USING FH * 3.6 * BMH PRINT PRINT USING FT * 3.6 * m INPUT "", A$

+ FI YD = FD / FT: YE = FE / FT: YI = FI / FT YD + BME * YE + BMF * YF + BMG

" " " "

┌──────────┬─────────────┬───────────────┐ │ Komponen │ Kgmol /jam │ Kg / jam │ ├──────────┼─────────────┼───────────────┤ │ O2 │ #####.#### │ ######.#### │

"

│

C7H8

│ #####.####

│

######.####

│ "; FA * 3.6;

"

│

N2

│ #####.####

│

######.####

│ "; FC * 3.6;

"

│

H2O

│ #####.####

│

######.####

│ "; FD * 3.6;

"

│

C6H5CHO │ #####.####

│

######.####

│ "; FE * 3.6;

"

│

C6H6

│ #####.####

│

######.####

│ "; FF * 3.6;

"

│

C4H2O3

│ #####.####

│

######.####

│ "; FG * 3.6;

"

│

CO2

│ #####.####

│

######.####

│ "; FH * 3.6;

" "

└──────────┴─────────────┴───────────────┘ " Total #######.#### ######.#### "; FT * 3.6;

PRINT PRINT " Enthalpi Hasil reaksi :" PRINT GOSUB 7000 Qo2 = QTOT INPUT "", A$ PRINT PRINT PRINT " NERACA PANAS :" PRINT PRINT " MASUK : PRINT PRINT " 1. Enthalpi Umpan Masuk Reaktor reaksi:" PRINT USING " Qs1 = #########.#### Kcal/jam #########.#### Kcal/jam"; Qo1; Qo2 Qll = Qo1 + (Qre * 3.6) - Qo2 Qp = MS * N * 3.6 * .373 * (td0 - D) Qloss = Qll - Qp

" " " "; FB * 3.6;

KELUAR : " 1. Enthalpi hasil Qs2

=

26 PRINT "

2. Panas Reaksi

2. Panas dibawa

pendingin" PRINT USING " Qr = #########.#### Kcal/jam Qp = #########.#### Kcal/jam"; Qre * 3.6; Qp PRINT " 3. Panas Hilang" PRINT USING " Qloss = #########.#### Kcal/jam"; Qloss tpIN = td0 - (Ql - Qloss) / (MS * 3.6 * N * .373) PRINT " ---------------------------------------------------------------- " PRINT USING " #########.#### Kcal/jam #########.#### Kcal/jam"; Qo1 + Qre * 3.6; (Qo2 + Qp + Qloss) PRINT PRINT INPUT "", A$ PRINT PRINT PRINT USING PRINT USING PRINT USING PRINT USING PRINT USING HH = 40 PRINT USING PRINT USING PRINT PRINT USING PRINT PRINT USING 3048; ID / 100 / PRINT USING / 2 PRINT

"

Dari hasil perhitungan Reaktor diperoleh :"

" " " " "

Jumlah pipa Diameter Shell Jumlah pendingin Panjang terhitung

= = = = =

" "

Panjang Pipa

= ###.# = ###.#

"

Tinggi Head reaktor = #.###

" 2 "

Tinggi reaktor

#### #.### ###### ###.# ###.#

pipa"; N m"; ID / 100 Kg/j"; MS * N * 3.6 m"; A / 100 ft"; A / 100 / .3048 m"; HH * .3048 ft"; HH m"; ID / 100 / 2

= ##.### + 2 . #.### m"; HH * . = ###.#

m"; HH * .3048 + ID / 100

CLOSE 900 END 910 'KOMPOSISI GAS (GMOL/JAM) XA = GB XA1 = 1 / .2 * XA S1 = .5 S2 = .5 FA FB FC FD FE FF FG FH

= = = = = = = =

FAo FBo FCo FDo FEo FFo FGo FHo

* (1 - XA) 'C7H8 - FAo * XA * (S1 + 6 * S2) - 7.5 * FFo * XA1 'O2 'N2 + FAo * XA * (S1 + 3 * S2) + 3 * FFo * XA1 'H2O + FAo * XA * S1 'C6H5CHO - FFo * XA1 'C6H6 + FAo * XA * S2 'C4H2O3 + 3 * FAo * XA * S2 + 6 * FFo * XA1 'CO2

FT = FA + FB + FC + FD + FE + FF + FG + FH YA = FA / FT: YB = FB / FT: YC = FC / FT: YF = FF / FT: YG = FG / FT: YH = FH / FT: m = BMA * YA + BMB * YB + BMC * YC + BMD * * YG + BMH * YH + BMI * YI

+ FI YD = FD / FT: YE = FE / FT: YI = FI / FT YD + BME * YE + BMF * YF + BMG

'KAPASITAS PANAS GAS (CAL/GMOL.K) (Reid,1979) CPA = 6.713 + -8.79E-07 * (GC + 273) + 4.17E-06 * (GC + 273) ^ 2 + -2.54E-09 * (GC + 273) ^ 3 CPB = -5.817 + .122 * (GC + 273) + -.0000661 * (GC + 273) ^ 2 + 1.17E-08 * (GC + 273) ^ 3

27 CPC = 7.44 + -.00324 * (GC + 273) + .0000064 * (GC + 273) ^ 2 + -2.79E09 * (GC + 273) ^ 3 CPD = 7.701 + .00046 * (GC + 273) + 2.52E-06 * (GC + 273) ^ 2 + -8.59E10 * (GC + 273) ^ 3 CPE = 2.9 + .119 * (GC + 273) + -.0000679 * (GC + 273) ^ 2 + 1.23E-08 * (GC + 273) ^ 3 CPF = 8.107 + .113 * (GC + 273) + -.0000721 * (GC + 273) ^ 2 + 1.7E-08 * (GC + 273) ^ 3 CPG = -3.123 + .0832 * (GC + 273) + -.0000522 * (GC + 273) ^ 2 + 1.16E08 * (GC + 273) ^ 3 CPH = 4.728 + .0175 * (GC + 273) + -.0000134 * (GC + 273) ^ 2 + 4.1E-09 * (GC + 273) ^ 3 CPM = (YA * CPA + YB * CPB + YC * CPC + YD * CPD + YE * CPE + YF * CPF + YG * CPG + YH * CPH + YI * CPI) / m 'KAPASITAS pendingin (CAL/GMOL.K) CPP = .373 'RAPAT MASSA CAMPURAN GAS RM = E * m / (GC + 273) / 82.06 'VISKOSITAS GAS (gr/dt.cm) TcA TcB TcC TcD TcE TcF TcG TcH

= = = = = = = =

154.6: 591.7: 126.2: 647.3: 695: 562.1: 500: 304.2:

PcA PcB PcC PcD PcE PcF PcG PcH

= = = = = = = =

49.8 40.6 33.5 217.6 46 483 20 72.8

VA = (.00036 * (4.61 * ((GC + 273) / TcA) ((GC + 273) / TcA)) + 1.94 * EXP(-4.058 * ((GC (1 / 6) * BMA ^ (-.5) * PcA ^ (-2 / 3))) / 100 VB = (.00036 * (4.61 * ((GC + 273) / TcB) ((GC + 273) / TcB)) + 1.94 * EXP(-4.058 * ((GC (1 / 6) * BMB ^ (-.5) * PcB ^ (-2 / 3))) / 100 VC = (.00036 * (4.61 * ((GC + 273) / TcC) ((GC + 273) / TcC)) + 1.94 * EXP(-4.058 * ((GC (1 / 6) * BMC ^ (-.5) * PcC ^ (-2 / 3))) / 100 VD = (.00036 * (4.61 * ((GC + 273) / TcD) ((GC + 273) / TcD)) + 1.94 * EXP(-4.058 * ((GC (1 / 6) * BMD ^ (-.5) * PcD ^ (-2 / 3))) / 100 VE = (.00036 * (4.61 * ((GC + 273) / TcE) ((GC + 273) / TcE)) + 1.94 * EXP(-4.058 * ((GC (1 / 6) * BME ^ (-.5) * PcE ^ (-2 / 3))) / 100 VF = (.00036 * (4.61 * ((GC + 273) / TcF) ((GC + 273) / TcF)) + 1.94 * EXP(-4.058 * ((GC (1 / 6) * BMF ^ (-.5) * PcF ^ (-2 / 3))) / 100 VG = (.00036 * (4.61 * ((GC + 273) / TcG) ((GC + 273) / TcG)) + 1.94 * EXP(-4.058 * ((GC (1 / 6) * BMG ^ (-.5) * PcG ^ (-2 / 3))) / 100 VH = (.00036 * (4.61 * ((GC + 273) / TcG) ((GC + 273) / TcG)) + 1.94 * EXP(-4.058 * ((GC (1 / 6) * BMG ^ (-.5) * PcG ^ (-2 / 3))) / 100

^ .618 - 2.04 * EXP(-.449 * + 273) / TcA)) + .1) / (TcA ^ ^ .618 - 2.04 * EXP(-.449 * + 273) / TcB)) + .1) / (TcB ^ ^ .618 - 2.04 * EXP(-.449 * + 273) / TcC)) + .1) / (TcC ^ ^ .618 - 2.04 * EXP(-.449 * + 273) / TcD)) + .1) / (TcD ^ ^ .618 - 2.04 * EXP(-.449 * + 273) / TcE)) + .1) / (TcE ^ ^ .618 - 2.04 * EXP(-.449 * + 273) / TcF)) + .1) / (TcF ^ ^ .618 - 2.04 * EXP(-.449 * + 273) / TcG)) + .1) / (TcG ^ ^ .618 - 2.04 * EXP(-.449 * + 273) / TcG)) + .1) / (TcG ^

VM = YA * VA * SQR(BMA) + YB * VB * SQR(BMB) + YC * VC * SQR(BMC) VM = VM + YD * VD * SQR(BMD) + YE * VE * SQR(BME) + YF * VF * SQR(BMF) VM = VM + YG * VG * SQR(BMG) + YH * VH * SQR(BMH) + YI * VI * SQR(BMI) VBAH = YA * SQR(BMA) + YB * SQR(BMB) + YC * SQR(BMC) VBAH = VBAH + YD * SQR(BMD) + YE * SQR(BME) + YF * SQR(BMF) VBAH = VBAH + YG * SQR(BMG) + YH * SQR(BMH) + YI * SQR(BMI)

28 VR = VM / VBAH 'VISKOSITAS pendingin (gr/dt.cm) VP = (35.5898 - .04212 * D) * .01 'THERMAL KONDUKTIVITAS (Cal/dt.cm.K) TIA = TcA ^ (1 / 6) * BMA ^ (1 / 2) / TIB = TcB ^ (1 / 6) * BMB ^ (1 / 2) / TIC = TcC ^ (1 / 6) * BMC ^ (1 / 2) / TID = TcD ^ (1 / 6) * BMD ^ (1 / 2) / TIE = TcE ^ (1 / 6) * BME ^ (1 / 2) / TIF = TcF ^ (1 / 6) * BMF ^ (1 / 2) / TIG = TcG ^ (1 / 6) * BMG ^ (1 / 2) / TIH = TcG ^ (1 / 6) * BMG ^ (1 / 2) / jp KA KB KC KD KE KF KG KH

= = = = = = = = =

.000001 ((14.52 ((14.52 ((14.52 ((14.52 ((14.52 ((14.52 ((14.52 ((14.52

KM = YA ^ .333) KM = KM (BME ^ .333) KM = KM (BMI ^ .333) BAWAH = BAWAH = 333) BAWAH = 333)

* * * * * * * *

(GC (GC (GC (GC (GC (GC (GC (GC

+ + + + + + + +

273) 273) 273) 273) 273) 273) 273) 273)

/ / / / / / / /

TcA) TcB) TcC) TcD) TcE) TcF) TcG) TcG)

-

PcA PcB PcC PcD PcE PcF PcG PcG

5.14) 5.14) 5.14) 5.14) 5.14) 5.14) 5.14) 5.14)

^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^

(2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2

/ / / / / / / / / / / / / / / /

3) 3) 3) 3) 3) 3) 3) 3) 3) 3) 3) 3) 3) 3) 3) 3)

* * * * * * * *

(jp (jp (jp (jp (jp (jp (jp (jp

/ / / / / / / /

TIA) TIB) TIC) TID) TIE) TIF) TIG) TIG)

+ YF * KF * (BMF ^ .333) + YD * KD * (BMD ^ .333) + YE * KE * + YG * KG * (BMG ^ .333) + YH * KH * (BMH ^ .333) + YI * KI * YA * (BMA ^ .333) + YB * (BMB ^ .333) + YC * (BMC ^ .333) BAWAH + YD * (BMD ^ .333) + YE * (BME ^ .333) + YF * (BMF ^ . BAWAH + YG * (BMG ^ .333) + YH * (BMH ^ .333) + YI * (BMI ^ .

'KONDUKTIVITAS pendingin (CAL/JAM.M.K) KP = (.84333 - .0005807 * (GD + 273)) 'PERHITUNGAN KOEFISIEN TRANSFER PANAS RE = 50 * GT / N * Dp / VR HI = (.27 * KM * (RE) ^ .8 * (CPM * VR / KM) ^ (1 / 3)) / DI HI = 7.8 * HI HIO = HI * DI / DOU Rs = DE * GS / VP PR = CPP * VP / KP HO = .46 * KP / DE * Rs ^ .55 * PR ^ .333 UC = (HIO * HO) / (HIO + HO) UD = UC / (11.06557 * UC + 1) / 100 'KECEPATAN REAKSI DAN PANAS REAKS kx = 4691261868600# * EXP(-13505.96427# / (GC + 273)) 'PANAS REAKSI = = = =

CPA CPB CPC CPD CPE CPF CPG CPG

* KA * (BMA ^ .333) + YB * KB * (BMB ^ .333) + YC * KC * (BMC

KM = KM / BAWAH

HR1 HR1 HR1 HR1

* * * * * * * *

-79310 + HR1 + (1 HR1 + (1 HR1 + (1

3.905 * ((GC + 273) - 298) / 2) * (-.00344) * ((GC + 273) ^ 2 - 298 ^ 2) / 3) * (-3.539E-06) * ((GC + 273) ^ 3 - 298 ^ 3) / 4) * (2.295E-09) * ((GC + 273) ^ 4 - 298 ^ 4)

HR2 = -579550 + -.297 * ((GC + 273) - 298)

29 HR2 = HR2 + (1 / 2) * (.0148) * ((GC + 273) ^ 2 - 298 ^ 2) HR2 = HR2 + (1 / 3) * (-4.372E-05) * ((GC + 273) ^ 3 - 298 ^ 3) HR2 = HR2 + (1 / 4) * (2.481E-08) * ((GC + 273) ^ 4 - 298 ^ 4) HR3 HR3 HR3 HR3

= = = =

-757490 + 9.2305 * ((GC + 273) - 298) HR3 + (1 / 2) * (-.00667) * ((GC + 273) ^ 2 - 298 ^ 2) HR3 + (1 / 3) * (-3.193E-05) * ((GC + 273) ^ 3 - 298 ^ 3) HR3 + (1 / 4) * (2.406E-08) * ((GC + 273) ^ 4 - 298 ^ 4)

HR = .75 * HR1 + .2 * HR2 + .05 * HR3 'PERSAMAAN DIFFERENSIAL ' (dx/dz) RR = 82.06 TT = GC + 273 MM = FBo / FAo CA0 = FAo / FTO * E1 / RR / TT RC = kx * CA0 ^ 2 * (1 - GB) * (MM - GB) DX = (3.14 * DI ^ 2 * RC * .36) / (4 * (FAo / N)) ' (dT/dz) Q1 = (-HR) * DX cal/dt cm Q2 = (UD * 3.14 cal/dt cm MCPR = FA * CPA + FG * CPG + FH * CPH T = (Q1 - Q2) /

* FAo / N * DOU * (GC - GD))

' (cal/gmol) (1/cm) (gmol/dt) = ' (cal/cm2 oC dt) (cm) (oC)

=

FB * CPB + FC * CPC + FD * CPD + FE * CPE + FF * CPF + (MCPR / N)

' (dTs/dz) S = -((UD * 3.14 * DOU * (GC - GD)) / (MS * CPP))

' (dP/dz) fk = (150 * (1 - .36) / RE + 1.75) / 10000 P = (GT / N) ^ 2 * (1 - .36) * fk P = -.00001 '((P / ((Dp) * (RM) * 981 * .36 ^ 3))) RETURN

7000

FT = FA + FB + FC + FD + FE + FF + FG + FH TC = C GOSUB 8000 RETURN 8000

XX YY ZZ WW

CPA 4 * WW CPB WW CPC WW CPD WW CPE CPF CPG WW CPH WW

= = = =

(TC (TC (TC (TC

+ + + +

273) 273) 273) 273)

^ ^ ^

298 2 - 298 ^ 2 3 - 298 ^ 3 4 - 298 ^ 4

= 6.713 * XX + -8.79E-07 / 2 * YY + 4.17E-06 / 3 * ZZ + -2.54E-09 / = -5.817 * XX + .122 / 2 * YY + -.0000661 / 3 * ZZ + 1.17E-08 / 4 * = 7.44 * XX + -.00324 / 2 * YY + .0000064 / 3 * ZZ + -2.79E-09 / 4 * = 7.701 * XX + .00046 / 2 * YY + 2.52E-06 / 3 * ZZ + -8.59E-10 / 4 * = 2.9 * XX + .119 / 2 * YY + -.0000679 / 3 * ZZ + 1.23E-08 / 4 * WW = 8.107 * XX + .113 / 2 * YY + -.0000721 / 3 * ZZ + 1.7E-08 / 4 * WW = -3.123 * XX + .0832 / 2 * YY + -.0000522 / 3 * ZZ + 1.16E-08 / 4 * = 4.728 * XX + .0175 / 2 * YY + -.0000134 / 3 * ZZ + 4.1E-09 / 4 *

30 QS1 QS2 QS3 QS4 QS5 QS6 QS7 QS7

FA FB FC FD FE FF FG FH FT

= = = = = = = =

FA FB FC FD FE FF FG FH

* * * * * * * *

CPA CPB CPC CPD CPE CPF CPG CPG

* * * * * * * *

3.6 3.6 3.6 3.6 3.6 3.6 3.6 3.6

QTOT = QS1 + QS2 + QS3 + QS4 + QS5 + QS6 + QS7 + QS8 PRINT PRINT USING " Suhu operasi = ###.## C"; TC PRINT " Suhu refferensi = 25 C " PRINT PRINT " ┌──────────┬────────────┬──────────┬───────────────┐" PRINT " │ Komponen │ Kgmol /jam │ Cp dT │ Qs = m Cp dT │ " PRINT " ├──────────┼────────────┼──────────┼───────────────┤" PRINT USING " │ O2 │ #####.#### │ #####.## │ ########.#### │ "; * 3.6; CPA; QS1 PRINT USING " │ C7H8 │ #####.#### │ #####.## │ ########.#### │ "; * 3.6; CPB; QS2 PRINT USING " │ N2 │ #####.#### │ #####.## │ ########.#### │ "; * 3.6; CPC; QS3 PRINT USING " │ H2O │ #####.#### │ #####.## │ ########.#### │ "; * 3.6; CPD; QS4 PRINT USING " │ C6H5CHO │ #####.#### │ #####.## │ ########.#### │ "; * 3.6; CPE; QS5 PRINT USING " │ C6H6 │ #####.#### │ #####.## │ ########.#### │ "; * 3.6; CPF; QS6 PRINT USING " │ C4H2O3 │ #####.#### │ #####.## │ ########.#### │ "; * 3.6; CPG; QS7 PRINT USING " │ CO2 │ #####.#### │ #####.## │ ########.#### │ "; * 3.6; CPH; QS8 PRINT " └──────────┴────────────┴──────────┴───────────────┘" PRINT USING " Total #######.#### ########.#### "; * 3.6; QTOT RETURN

Hasil Running Program computer ┌─────────────────────────────────────────────────────────┐ │REAKTOR FIXED BED MULTI TUBE│ └─────────────────────────────────────────────────────────┘ Jumlah pipa = 6837 pipa Diameter luar pipa = 3.353 cm Diameter dalam pipa = 2.664 cm Pitch = 4.191 cm Diameter Shell = 4.194 m Jumlah pendingin = 369198.000 Kg/j

KECEPATAN MASSA MASUK REAKTOR ┌──────────┬─────────────┬───────────────┐ │ Komponen │ Kgmol /jam │ Kg / jam │ ├──────────┼─────────────┼───────────────┤ │ O2 │ 124.5205 │ 3984.6553 │ │ C7H8 │ 13.1443 │ 1209.2717 │ │ N2 │ 468.4342 │ 13116.1572 │ │ H2O │ 0.0000 │ 0.0000 │

31 │ C6H5CHO │ 0.0123 │ 1.2990 │ │ C6H6 │ 0.1399 │ 10.9158 │ │ C4H203 │ 0.0000 │ 0.0000 │ │ CO2 │ 0.0000 │ 0.0000 │ └──────────┴─────────────┴───────────────┘ Total 606.2511 18322.2988 KECEPATAN MASSA GAS KELUAR REAKTOR ┌──────────┬─────────────┬───────────────┐ │ Komponen │ Kgmol /jam │ Kg / jam │ ├──────────┼─────────────┼───────────────┤ │ O2 │ 82.0665 │ 2626.1272 │ │ C7H8 │ 1.3144 │ 120.9272 │ │ N2 │ 468.4342 │ 13116.1563 │ │ H2O │ 24.0795 │ 433.4310 │ │ C6H5CHO │ 5.9272 │ 628.2800 │ │ C6H6 │ 0.0000 │ 0.0000 │ │ C4H2O3 │ 5.9149 │ 579.6617 │ │ CO2 │ 18.5844 │ 817.7147 │ └──────────┴─────────────┴───────────────┘ Total 606.3211 18322.2988 Enthalpi Umpan Masuk Reaktor : Suhu operasi = 500.00 C Suhu refferensi = 25 C ┌──────────┬────────────┬──────────┬───────────────┐ │ Komponen │ Kgmol /jam │ Cp dT │ Qs = m Cp dT │ ├──────────┼────────────┼──────────┼───────────────┤ │ O2 │ 13.1443 │ 3571.98 │ 46951.0313 │ │ C7H8 │ 124.5205 │ 19696.54 │ 2452622.0000 │ │ N2 │ 468.4342 │ 3395.24 │ 1590446.2500 │ │ H2O │ 0.0000 │ 4065.76 │ 0.0000 │ │ C6H5CHO │ 0.0123 │ 22865.14 │ 280.2058 │ │ C6H6 │ 0.1399 │ 23612.95 │ 3304.5408 │ │ C4H2O3 │ 0.0000 │ 13115.67 │ 0.0000 │ │ CO2 │ 0.0000 │ 5110.12 │ 0.0000 │ └──────────┴────────────┴──────────┴───────────────┘ Total 606.2511 4093604.0000 Reaksi yang terjadi : C7H8 + O2 C7H8 + 6 O2

-----------> ----------->

C6H6 + 7.5 O2 ----------->

C6H5CHO + H2O C4H2O3 + 3 H2O + 3 CO2 6 CO2

+ 3 H2O

KONDISI AWAL ┌──────────────────────────────────────────┐ │ Suhu gas masuk = 500.0 °C │ │ Suhu pendingin keluar = 490.0 °C │ │ Tekanan awal = 2.0 atm │ │ Increment tebal katalis = 1.00 cm │ └──────────────────────────────────────────┘ ┌──────┬────────┬────────┬───────┬────────┐ │ L(cm)│ Xa │ T(c) │ Td (c)│ P(atm) │ ├──────┼────────┼────────┼───────┼────────┤ │ 0 │ 0.0000 │ 500.00 │ 490.0 │ 2.000 │ │ 10 │ 0.0208 │ 504.25 │ 489.9 │ 2.000 │ │ 20 │ 0.0428 │ 508.35 │ 489.8 │ 2.000 │ │ 30 │ 0.0661 │ 512.33 │ 489.6 │ 2.000 │

32 │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │ │

40 │ 0.0905 │ 516.15 │ 489.4 │ 50 │ 0.1159 │ 519.82 │ 489.1 │ 60 │ 0.1422 │ 523.32 │ 488.9 │ 70 │ 0.1693 │ 526.63 │ 488.6 │ 80 │ 0.1971 │ 529.73 │ 488.2 │ 90 │ 0.2254 │ 532.60 │ 487.9 │ 100 │ 0.2540 │ 535.20 │ 487.5 │ 110 │ 0.2826 │ 537.52 │ 487.1 │ 120 │ 0.3111 │ 539.52 │ 486.6 │ 130 │ 0.3394 │ 541.18 │ 486.2 │ 140 │ 0.3671 │ 542.49 │ 485.7 │ 150 │ 0.3940 │ 543.41 │ 485.2 │ 160 │ 0.4201 │ 543.95 │ 484.7 │ 170 │ 0.4451 │ 544.11 │ 484.2 │ 180 │ 0.4690 │ 543.89 │ 483.7 │ 190 │ 0.4916 │ 543.31 │ 483.2 │ 200 │ 0.5130 │ 542.39 │ 482.7 │ 210 │ 0.5330 │ 541.16 │ 482.2 │ 220 │ 0.5517 │ 539.65 │ 481.8 │ 230 │ 0.5691 │ 537.91 │ 481.3 │ 240 │ 0.5852 │ 535.97 │ 480.8 │ 250 │ 0.6002 │ 533.87 │ 480.4 │ 260 │ 0.6141 │ 531.65 │ 479.9 │ 270 │ 0.6269 │ 529.34 │ 479.5 │ 280 │ 0.6388 │ 526.98 │ 479.1 │ 290 │ 0.6498 │ 524.59 │ 478.7 │ 300 │ 0.6599 │ 522.21 │ 478.4 │ 310 │ 0.6694 │ 519.85 │ 478.0 │ 320 │ 0.6782 │ 517.53 │ 477.7 │ 330 │ 0.6864 │ 515.27 │ 477.4 │ 340 │ 0.6940 │ 513.07 │ 477.1 │ 350 │ 0.7011 │ 510.95 │ 476.8 │ 360 │ 0.7078 │ 508.92 │ 476.5 │ 370 │ 0.7141 │ 506.97 │ 476.2 │ 380 │ 0.7201 │ 505.12 │ 476.0 │ 390 │ 0.7257 │ 503.35 │ 475.8 │ 400 │ 0.7310 │ 501.68 │ 475.5 │ 410 │ 0.7360 │ 500.10 │ 475.3 │ 420 │ 0.7408 │ 498.60 │ 475.1 │ 430 │ 0.7454 │ 497.19 │ 475.0 │ 440 │ 0.7498 │ 495.87 │ 474.8 │ 450 │ 0.7539 │ 494.62 │ 474.6 │ 460 │ 0.7579 │ 493.45 │ 474.5 │ 470 │ 0.7618 │ 492.36 │ 474.3 │ 480 │ 0.7655 │ 491.33 │ 474.2 │ 490 │ 0.7691 │ 490.36 │ 474.0 │ 500 │ 0.7725 │ 489.46 │ 473.9 │ 510 │ 0.7759 │ 488.61 │ 473.8 │ 520 │ 0.7791 │ 487.82 │ 473.7 │ 530 │ 0.7823 │ 487.08 │ 473.5 │ 540 │ 0.7853 │ 486.39 │ 473.4 │ 550 │ 0.7883 │ 485.74 │ 473.3 │ 560 │ 0.7912 │ 485.13 │ 473.2 │ 570 │ 0.7940 │ 484.56 │ 473.1 │ 580 │ 0.7967 │ 484.02 │ 473.0 │ 590 │ 0.7994 │ 483.52 │ 473.0 │ 600 │ 0.8020 │ 483.04 │ 472.9 │ 610 │ 0.8045 │ 482.60 │ 472.8 │ 620 │ 0.8070 │ 482.18 │ 472.7 │ 630 │ 0.8095 │ 481.79 │ 472.6 │ 640 │ 0.8119 │ 481.42 │ 472.6 │ 650 │ 0.8142 │ 481.07 │ 472.5 │ 660 │ 0.8165 │ 480.74 │ 472.4 │ 670 │ 0.8187 │ 480.42 │ 472.4 │ 680 │ 0.8209 │ 480.13 │ 472.3 │

2.000 │ 1.999 │ 1.999 │ 1.999 │ 1.999 │ 1.999 │ 1.999 │ 1.999 │ 1.999 │ 1.999 │ 1.999 │ 1.998 │ 1.998 │ 1.998 │ 1.998 │ 1.998 │ 1.998 │ 1.998 │ 1.998 │ 1.998 │ 1.998 │ 1.997 │ 1.997 │ 1.997 │ 1.997 │ 1.997 │ 1.997 │ 1.997 │ 1.997 │ 1.997 │ 1.997 │ 1.996 │ 1.996 │ 1.996 │ 1.996 │ 1.996 │ 1.996 │ 1.996 │ 1.996 │ 1.996 │ 1.996 │ 1.995 │ 1.995 │ 1.995 │ 1.995 │ 1.995 │ 1.995 │ 1.995 │ 1.995 │ 1.995 │ 1.995 │ 1.994 │ 1.994 │ 1.994 │ 1.994 │ 1.994 │ 1.994 │ 1.994 │ 1.994 │ 1.994 │ 1.994 │ 1.993 │ 1.993 │ 1.993 │ 1.993 │

33 │ 690 │ 0.8231 │ 479.85 │ 472.2 │ 1.993 │ │ 700 │ 0.8252 │ 479.58 │ 472.2 │ 1.993 │ │ 710 │ 0.8273 │ 479.33 │ 472.1 │ 1.993 │ │ 720 │ 0.8294 │ 479.09 │ 472.1 │ 1.993 │ │ 730 │ 0.8314 │ 478.86 │ 472.0 │ 1.993 │ │ 740 │ 0.8334 │ 478.64 │ 472.0 │ 1.993 │ │ 750 │ 0.8354 │ 478.43 │ 471.9 │ 1.992 │ │ 760 │ 0.8373 │ 478.23 │ 471.9 │ 1.992 │ │ 770 │ 0.8392 │ 478.04 │ 471.8 │ 1.992 │ │ 780 │ 0.8410 │ 477.85 │ 471.8 │ 1.992 │ │ 790 │ 0.8429 │ 477.68 │ 471.7 │ 1.992 │ │ 800 │ 0.8447 │ 477.51 │ 471.7 │ 1.992 │ │ 810 │ 0.8464 │ 477.34 │ 471.6 │ 1.992 │ │ 820 │ 0.8482 │ 477.18 │ 471.6 │ 1.992 │ │ 830 │ 0.8499 │ 477.03 │ 471.5 │ 1.992 │ │ 840 │ 0.8516 │ 476.89 │ 471.5 │ 1.992 │ │ 850 │ 0.8533 │ 476.74 │ 471.4 │ 1.991 │ │ 860 │ 0.8549 │ 476.61 │ 471.4 │ 1.991 │ │ 870 │ 0.8566 │ 476.47 │ 471.3 │ 1.991 │ │ 880 │ 0.8582 │ 476.34 │ 471.3 │ 1.991 │ │ 890 │ 0.8598 │ 476.22 │ 471.3 │ 1.991 │ │ 900 │ 0.8613 │ 476.10 │ 471.2 │ 1.991 │ │ 910 │ 0.8629 │ 475.98 │ 471.2 │ 1.991 │ │ 920 │ 0.8644 │ 475.86 │ 471.2 │ 1.991 │ │ 930 │ 0.8659 │ 475.75 │ 471.1 │ 1.991 │ │ 940 │ 0.8673 │ 475.64 │ 471.1 │ 1.991 │ │ 950 │ 0.8688 │ 475.54 │ 471.0 │ 1.990 │ │ 960 │ 0.8702 │ 475.43 │ 471.0 │ 1.990 │ │ 970 │ 0.8717 │ 475.33 │ 471.0 │ 1.990 │ │ 980 │ 0.8731 │ 475.23 │ 470.9 │ 1.990 │ │ 990 │ 0.8744 │ 475.13 │ 470.9 │ 1.990 │ │ 1000 │ 0.8758 │ 475.04 │ 470.9 │ 1.990 │ │ 1010 │ 0.8771 │ 474.95 │ 470.8 │ 1.990 │ │ 1020 │ 0.8785 │ 474.85 │ 470.8 │ 1.990 │ │ 1030 │ 0.8798 │ 474.76 │ 470.8 │ 1.990 │ │ 1040 │ 0.8811 │ 474.68 │ 470.7 │ 1.990 │ │ 1050 │ 0.8823 │ 474.59 │ 470.7 │ 1.989 │ │ 1060 │ 0.8836 │ 474.51 │ 470.7 │ 1.989 │ │ 1070 │ 0.8849 │ 474.42 │ 470.6 │ 1.989 │ │ 1080 │ 0.8861 │ 474.34 │ 470.6 │ 1.989 │ │ 1090 │ 0.8873 │ 474.26 │ 470.6 │ 1.989 │ │ 1100 │ 0.8885 │ 474.19 │ 470.6 │ 1.989 │ │ 1110 │ 0.8897 │ 474.11 │ 470.5 │ 1.989 │ │ 1120 │ 0.8908 │ 474.03 │ 470.5 │ 1.989 │ │ 1130 │ 0.8920 │ 473.96 │ 470.5 │ 1.989 │ │ 1140 │ 0.8931 │ 473.89 │ 470.4 │ 1.989 │ │ 1150 │ 0.8943 │ 473.81 │ 470.4 │ 1.988 │ │ 1160 │ 0.8954 │ 473.74 │ 470.4 │ 1.988 │ │ 1170 │ 0.8965 │ 473.67 │ 470.4 │ 1.988 │ │ 1180 │ 0.8976 │ 473.61 │ 470.3 │ 1.988 │ │ 1190 │ 0.8986 │ 473.54 │ 470.3 │ 1.988 │ │ 1200 │ 0.8997 │ 473.47 │ 470.3 │ 1.988 │ │ 1203 │ 0.9000 │ 473.45 │ 470.3 │ 1.988 │ └──────┴────────┴────────┴───────┴────────┘ KECEPATAN MASSA GAS KELUAR REAKTOR ┌──────────┬─────────────┬───────────────┐ │ Komponen │ Kgmol /jam │ Kg / jam │ ├──────────┼─────────────┼───────────────┤ │ O2 │ 82.0665 │ 2626.1272 │ │ C7H8 │ 1.3144 │ 120.9272 │ │ N2 │ 468.4342 │ 13116.1563 │ │ H2O │ 24.0795 │ 433.4310 │ │ C6H5CHO │ 5.9272 │ 628.2800 │

34 │ C6H6 │ 0.0000 │ 0.0000 │ │ C4H2O3 │ 5.9149 │ 579.6617 │ │ CO2 │ 18.5844 │ 817.7147 │ └──────────┴─────────────┴───────────────┘ Total 606.3211 18322.2988 Enthalpi Hasil reaksi : Suhu operasi = 473.45 C Suhu refferensi = 25 C ┌──────────┬────────────┬──────────┬───────────────┐ │ Komponen │ Kgmol /jam │ Cp dT │ Qs = m Cp dT │ ├──────────┼────────────┼──────────┼───────────────┤ │ O2 │ 1.3144 │ 3359.46 │ 4415.7666 │ │ C7H8 │ 82.0665 │ 18267.09 │ 1499115.5000 │ │ N2 │ 468.4342 │ 3197.49 │ 1497813.5000 │ │ H2O │ 24.0795 │ 3823.43 │ 92066.2578 │ │ C6H5CHO │ 5.9272 │ 21285.38 │ 126162.1016 │ │ C6H6 │ 0.0000 │ 22025.61 │ 0.0000 │ │ C4H2O3 │ 5.9149 │ 12185.40 │ 226458.5938 │ │ CO2 │ 18.5844 │ 4789.26 │ 0.0000 │ └──────────┴────────────┴──────────┴───────────────┘ Total 606.3211 3446031.7500 NERACA PANAS : MASUK :

KELUAR :

1. Enthalpi Umpan Masuk Reaktor 1. Enthalpi hasil reaksi: Qs1 = 4093604.0000 Kcal/jam Qs2 = 3446031.7500 Kcal/jam 2. Panas Reaksi 2. Panas dibawa pendingin Qr = 2518927.2500 Kcal/jam Qp = 2716473.5000 Kcal/jam 3. Panas Hilang Qloss = 450026.0000 Kcal/jam -------------------------------- --------------------------------6612531.5000 Kcal/jam 6612531.0000 Kcal/jam Dari hasil perhitungan Reaktor diperoleh : Jumlah pipa = 6837 pipa Diameter Shell = 4.194 m Jumlah pendingin = 369198 Kg/j Panjang terhitung = 12.0 m = 39.5 ft Panjang Pipa = 12.2 m = 40.0 ft Tinggi Head reaktor = 2.097 m Tinggi reaktor = 12.192 + 2 . 2.097 m = 14.3 m

1. Menghitung tebal shell ────────────────────── Digunakan bahan Carbon steel SA 178 grade C Tekanan design (p) = 17.64 psi

35 Allowable stress = 18750 psi efisiensi sambungan = 0.85 faktor korosi = 0.125 in Jari-jari tangki = 82.56 in Tebal Shell : p .ri t shell = ─────────────── + c S . e - 0.4 . p 17.64 . 82.56 = ──────────────────────────────── + 0.125 18750.00 . 0.85 - 0.4 . 17.64 = 0.216 in Dipakai tebal shell 1/4 in 2. Menghitung tebal head ────────────────────── Bentuk head : Elliptical Dished Head Digunakan bahan Carbon steel SA 178 grade C Tekanan design (p) = 32.24 psi Allowable stress = 18750 psi efisiensi sambungan = 0.85 faktor korosi = 0.125 in Jari-jari tangki = 82.56 in Tebal Head : 0.885 . p .d t head = ──────────────────── + c 2 . S . e - 0.2 . p 0.885 . 32.24 . 165.12 = ──────────────────────────────────── + 0.125 2 . 18750.00 . 0.85 - 0.2 . 32.24 = 0.273 in Dipilih tebal head 1/4 in 3. Menghitung ukuran pipa ────────────────────── Diameter Optimum pipa berdasarkan Pers. 15 Peters, hal.525 a. Pipa pemasukan Umpan Reaktor : Kecepatan Umpan Densitas Umpan

= 40309.059 lb/j = 0.1684 lb/ft^3

Di = 2.2 . ( G/1000 )^ 0.45 . den ^ (-0.31) = 2.2 . ( 40309.059 / 1000 )^ 0.45 . 0.1684 ^ (-0.31) = 20.171 in Dipakai pipa dengan ukuran : 20.00 in b. Pipa pengeluaran hasil Reaktor :

36 Kecepatan hasil = 40309.059 lb/j Densitas hasil = 0.1673 lb/ft^3 Di = 2.2 . ( G/1000 )^ 0.45 . den ^ (-0.31) = 2.2 . ( 40309.059 / 1000 )^ 0.45 . 0.1673 ^ (-0.31) = 20.212 in Dipakai pipa dengan ukuran : 20.00 in c. Pipa pemasukan dan pengeluaran pemanas: Kecepatan HITEC = 812235.6250 lb/j Densitas HITEC = 54.6624 lb/ft^3 Di = 2.2 . ( L/1000 )^ 0.45 . den ^ (-0.31) = 2.2 . ( 812235.625 / 1000 )^ 0.45 . 54.6624 ^ (-0.31) = 12.975 in Dipakai pipa dengan ukuran : 12.00 in 4. Menghitung tebal Isolasi Diameter shell = 13.76 ft Tinggi shell = 40.00 ft Tebal shell = 0.018 ft Luas permukaan head = 356.70 ft² Luas permukaan shell = 1728.24 ft² Total luas permukaan = 2084.94 ft² Suhu permukaan isolasi = 140.00 °F = 600.00 °R Suhu dalam reaktor = 956.39 °F = 1416.39 °R Suhu udara lingkungan = 86.00 °F = 546.00 °R Konduktifitas thermal diding shell = 28.0000 Btu ft/(j ft² F) Digunakan Isolasi Fine Diatomaceous earth powder Konduktifitas thermal isolasi = 0.1250 Btu ft/(j ft² F) Koeffisien transfer panas konveksi (hc) : ┌ ┐0.25 hc = 0.3 │ Tw - Tu │ └ ┘ ┌ ┐0.25 = 0.3 │ 140.00 - 86.00 │ └ ┘ = 0.8132418 Btu/j ft² F Koeffisien transfer panas Radiasi (hr) : ┌ 4 4┐ k e │ (Tw/100) - (Tu/100) │ └ ┘ hr = --------------------------------[ Tw - Tu ] ┌ 4 4┐ 0.178 . 0.8 │ 6.00 - 5.46 │ └ ┘ = ---------------------------------

37 [ 600.00 - 546.00 ] = 5.8827672 Btu/j ft² F A . (T1 - Tu) Q loss = ----------------------------┌ t1 t2 1 ┐ │ ---- + ---- + --------- │ └ k1 k2 (hr + hc) ┘ 2084.940 . (956.39 - 86.00) = -------------------------------------------┌ 0.018 T is 1 ┐ │ ------- + -------- + ------------------ │ └ 28.00 0.125 0.8132 + 5.8828 ┘ ┌ 0.018 T is 1 ┐ 2084.940 (956.39 - 86.00) │ ------- + -------- + -------- │= --------------------------└ 28.00 0.125 6.6960 ┘ 450026.000 ┌ │ └ ┌ │ └

0.018 T is 1 ┐ ------- + -------- + -------- │= 4.032 28.00 0.125 6.6960 ┘ T is ┐ 0.00064 + -------- + 0.1493 │= 4.032 0.1250 ┘

┌ T is ┐ │ -------- │= └ 0.1250 ┘

3.882

T isolasi =

0.485 ft

=

5.824 in

Kesimpulan : 1. Tugas 2. Type alat = Reaktor Fixed Bed Multitube, karena reaksinya sangat eksotermis 3. Kondisi operasi 4. Katalis 5. Pipa reactor 6. Ukuran reaktor Jumlah pipa Diameter Shell Jumlah pendingin Panjang terhitung Panjang Pipa

= 6837 = 4.194 = 369198 = 12.0 = 39.5 = 12.2 = 40.0

Tinggi Head reaktor = 2.097 Tinggi reaktor

pipa m Kg/j m ft m ft m

= 12.192 + 2 . 2.097 m = 14.3 m

38 7.