Lampiran Iii Spesifikasi Pralatan.docx

LAMPIRAN III PERHITUNGAN SPESIFIKASI ALAT

1.

ACCUMULATOR - 01 (ACC-01)

Fungsi : Tempat menampung kondensat yang berasal dari condensor-01 Tipe

: Silinder horizontal dengan penutup ellipsoidal

Gambar

:

input

ACC - 01

output

Kondisi Operasi: Tekanan

= 0,6 atm

Temperatur

= 82,767 oC

Laju alir

= 93,448.729 kg/jam

Densitas

= 304, 711 kg/m3

Residence Time

= 5 menit

Perhitungan Desain Accumulator - 01 a. Kapasitas Accumulator, Vt Volume liquid

= Laju alir/Densitas x holding time = (93,448.729 kg/jam)/(304, 711 kg/m3) x 0,083

jam = 25,557 m3 Faktor keamanan

= 10%

Kapasitas acc.

= 1,1 x 25,557 m3 = 36,025 m3

b. Desain Ukuran Accumulator 

Volume Silinder, Vs Vs

=

 4

 D 2 .Lsilinder

Lsilinder = 4.D

=  . D3



Volume Ellipsoidal, Ve Ve



=

( .D 3 ) 24

Volume Total Accumulator, VT VT

= Vs + 2.Ve = (  . D 3 ) + (2.( = 3,402.D3



Diameter Accumulator, D D

=

3

=

3

VT acc 3,402 28,112 3,402

= 2,022 m Maka, Vs

= 25,950 m3

Ve

= 1,081 m3

VT

= 28,112 m3

c. Panjang Accumulator 

Panjang Silinder L

= 4.D = 4 . 2,022 m = 8,087 m

( .D 3 ) )) 24



Panjang Ellipsoidal h

=

1 D 4

=

1  2,022 m 4

= 0,505 m 

Panjang Total Accumulator, LT LT

= L + 2h = (8,087 m + (2. 0,505 m)) = 9,098 m

d. Tebal Dinding Accumulator 

Ketebalan Dinding Bagian Head, thead t =

P . Da  Cc 2.S .Ej  0,2.P

Dimana: t

= ketebalan dinding bagian head, m

P = tekanan design

= 8,818 psi

Da = diameter vessel

= 79,598 in

S = working stress yang diizinkan = 13.700 psi C = faktor korosi yang diizinkan

= 0,013 in

Ej = faktor efisiensi pengelasan

= 0,850

Maka didapatkan: t =

(8,818 psi )(79,598 in )  0,013 in (2 x13.700 x0,85)  (0,2  x8,818 psi )

= 0,028 in = 0,001 m 

Ketebalan Dinding Bagian Silinder, tsilinder t =

P . ri  Cc S .Ej  0,6.P

Dimana: ri

= 49,028 in

Maka didapatkan: t =

(8,818 psi )(79,598in )  0,013 in (13.700  0,85)  (0,6  8,818 psi )

= 0,028 in = 0,001 m

OD

= ID + 2.tsilinder = (2,022 + (2. 0,001)) m = 2,023 m

Ringkasan spesifikasi Accumulator-01 (ACC-01) IDENTIFIKASI Nama Alat

Accumulator-01

Kode

ACC-01

Jumlah

1 buah

Fungsi

Menampung kondensat dari condensor-01 DATA DESAIN

Tipe

Silinder horizontal dengan ellipsoidal head

Kapasitas

25,557 m3

Tekanan

0,600 atm

Temperatur

82,767 oC

Diameter

2,022 m

Panjang

9,098 m

Tebal Dinding

0,001 m

Bahan Konstruksi

Carbon steel

Dengan Perhitungan yang sama untuk Accumulator selanjutnya analog dengan perhitungan Accumulator ACC-01. 2.

ACCUMULATOR - 02 (ACC-02) IDENTIFIKASI

Nama Alat

Accumulator-02

Kode

ACC-02

Jumlah

1 buah

Fungsi

Menampung kondensat dari condensor-02 DATA DESAIN

Tipe

Silinder horizontal dengan ellipsoidal head

Kapasitas

4,469 m3

Tekanan

0,370 atm

Temperatur

55,071 oC

Diameter

1,131 m

Panjang

5,087 m

Tebal Dinding

0,000462 m

Bahan Konstruksi

Carbon steel

3.

ADSORBER-01 (AD-01/02)

Fungsi

: Untuk menghilangkan atau menyerap kandungan H2O keluaran dari Kolom Destilasi-01 (KD-01)

Tipe

: Silinder vertikal dengan ujung ellipsoidal

Gambar

:

a. Data Temperatur, T

=82,767 °C

Tekanan, P

= 0,6 atm

Laju alir massa, W

= 42931,198

Densitas campuran

= 0,928

Faktor keamanan

= 10%

Adsorbtion time

= 0,500 jam

b. Kapasitas Kolom, Vk V

=

Laju alir massa × t Densitas

= 23141,008 m3 Vk

= (1 + f) × k = 25455,109 m3

kg m3

kg jam

c. Volume Packing, Vp Dalam adsorber-01/02 ini diharapkan bahwa semua H2O yang terdapat di dalam produk keluaran Kolom destilasi-01 dapat terserap seluruhnya sehingga tidak ada vapor H2O yang akan terbawa karena dikhawatirkan terjadinya reaksi balik. Kemampuan penyerapan silica gel terhadap H2O yang dihasilkan adalah 420,656 kg silica gel/kg H2O. Sedangkan untuk menentukan Volume Packing (Vp) yang dibutuhkan yaitu: Diketahui jumlah uap air yang akan di serap sebesar 2148,416 kg dan faktor penyerapan (Fa) adalah 0,014 maka jumlah adsorben yang digunakan: Wp

= Fa × Wa = 323,582 kg

Excess silica gel

= 97,074 kg

Total silica gel

= Wp + excess zeolit = 323,582 kg + 97,074 kg = 425,656 kg

Sehingga untuk kebutuhan penyerapan 8 jam, dibutuhkan silica gel sebanyak: Wpt

= Total silica gel x 8 jam = 425,656 kg x 8 jam = 3365,248 kg

Volume packing, Vp Vp

=

Wpt ρcampuran

(Pers. 16.4 Perry, 1997)

= 3627,908 m3 Tinggi packing, Tp

=

4 × Vp π × D2

= 31,870 m Pemasangan bed Packing yakni 7,484 meter di atas dasar Adsorber (Muklis, 2012). d. Volume Total Desiccantor Volume total

= Vk + Vp = 398,631 m3

e. Diameter Kolom Adsorber (Tabel 10-64, Volume of Partially Filled Horizontal Cylinder, Perry, 1997) Volume bagian Silinder, Vs Vs

= π Dt2 Hs

Hs

3

= Dt 2

3

= π Dt2 ( Dt) 2

3

= π Dt3 8

Volume bagian Ellipsoidal, VE π

VE = Dt2 He

He

6

π

1

6

4

= Dt2 ( Dt) =

π 12

Dt3

Sehingga, Vt

= Vs + Ve = 1,440 Dt3

Dt

= (Vt/1,440)1/3

Dt

= 12,042 m

Sehingga, jari-jari tangki (R) = 6,021 m f. Tinggi Tangki Tinggi silinder, Hs Hs

3

= Dt 2

= 15,063 m Tinggi ellipsoidal, He He

1

= Dt 4

= 3,011 m Tinggi Total, Ht

(Treyball, Pers. 3.9, 1981)

= Hs + He = 18,074 m

1

= Dt 4

g. Tebal Dinding t

=(

P×D

2 S E - 0,2 P

)+C

(Tabel. 4, Peters and Timmerhaus)

Dimana : Tekanan design, P

= 0,6 atm

Diameter vessel, D

= 12,042 m

Working stress allowable, S

= 932,230 m

Joint efficiency, E

= 0,850 m

Korosi maksimum, C

= 0,0003 m

P×D

t

=(

t

= 0,005 m = 0,489 cm

2 S E - 0,2 P

)+C

h. Outside Diameter OD = D + 2t = 12,052 m IDENTIFIKASI Nama Alat

Adsorber-01 (DS-01/02)

Kode

AD-01

Jumlah

1 buah

Fungsi

Untuk menyerap air keluaran dari kolom destilasi-01 DATA DESAIN

Tipe

Silinder vertikal dengan ujung ellipsoidal

Kapasitas

3627,908 m3

Adsorben

Silica Gel

Tekanan

0,600 atm

Temperatur

82,767 oC

Diameter

12,042 m

Tinggi

18,074 m

Tebal Dinding

0,489 m

Bahan Konstruksi

Carbon steel

4.

CONDENSER - 01 (CD-01)

Fungsi : Mengkondesasikan Top produk KD-01 Tipe

: Shell and Tube Heat Exchanger

Gambar

: Aliran inlet Tube

Shell

Rear End

Head

Aliran outlet

Fluida Panas

Water in

:

Top produk KD-01

W

= 93.448,729 kg/hr

= 206. 018,938 lb/hr

T1

= 82,767 oC

= 180,981 oF

T2

= 82,767 oC

= 180,981 oF

Fluida Dingin

:

Air

w

= 297.253,863 kg/hr = 655.331.811 lb/hr

t1

= 28oC

= 82,4oF

t2

= 50oC

= 122oF

Perhitungan design sesuai dengan literatur pada buku Donald Q. Kern (1965). 1.

Beban Panas CD-01 Q

= 27.374.702,744 kJ/hr = 25.702.434,670 Btu/hr

2.

LMTD

Fluida Panas (oF)

Fluida Dingin (oF)

Selisih

180,981

Suhu tinggi

122,000

58,981

180,981

Suhu rendah

82,400

98,581

Selisih

-39,6

LMTD =

t 2  t1 ln (t 2 / t1 )

= 77.094 oF Ft

=1

t

= 77.094 oF

3.

(Fig.18, Kern)

Temperatur Rata-rata Tavg =

T1  T 2 2

= 180.981 oF tavg

=

t1  t 2 2

= 102,200 oF 4.

Menentukan luas daerah perpindahan panas Asumsi UD = 210 Btu/hr.ft2.oF A

=

Q U D . t

=

25.702.434,670 150  47,697

(Tabel 8, Kern)

= 2,932.691 ft2 Karena A > 200 ft2, maka digunakan Shell & Tube Heat Exchanger 5.

Spesifikasi tube dan shell 

Tube Side

= Aliran top produk KD-01

Panjang tube (L)

= 19 ft

Outside Diameter (OD) = 1 in BWG

= 18

Pass

=2

a”

= 0,262 ft2/lin ft A = L x a" 2,932.691 = 19  0,262

Jumlah tube, Nt

= 589,580

Dari tabel.9 Kern, didapat nilai yang mendekati Nt perhitungan adalah Nt

= 589,580



Corrected Coefficient, UD A

= Nt x L x a'' = 589,580 x 19 ft x 0,262 ft2 = 2,932.691

UD

=

Q U D . t

= 149,319 karena nilai Ud perhitungan mendekati nilai Ud asumsi, maka data untuk shell : Shell

= Air/pendingin

ID

= 39 inch

(Tabel 9, Kern)

Baffle Space (B = ID/5) = 19,5 inch Pass

=2

Pt

= 1,25 in triangular pitch

6.

Perhitungan desain bagian tube 

Flow Area/tube, a’t a’t

= 0,902 in2

(Tabel 10, Kern)

at

=

Nt  a ' t 144  n

(Pers. 7.48, Kern)

=

589,580  0,902 144  2

= 1,882 ft2 

Laju Alir, Gt Gt

= W/at =

206.018,938 1,882

= 109.450,484 lb/hr.ft2



Bilangan Reynold, Ret μ

= 0,010 cp

ID

= 0,902 inch = 0,075 ft

Ret

= ID.Gt/ μ =

= 0,025 lb/ft hr (Tabel 10, Kern)

0,075 109.450,484 / 0,010 0,025

= 322.801,472 

Dengan L/D = 252,782 diperoleh Jh = 600



(Fig.24, Kern)

Nilai hi CP

= 15,678 Btu/lb.oF

k

= 0,012 Btu/hr ft.oF

 c.     k 

= 5,770

 k   Cp   hi  J H      D  k 

1/ 3

      w 

0 ,14

    Koreksi viskositas diabaikan, karena   w  hi

0 ,14

=1

 0,012  1/ 3 = 252,782    3,491  0,075  = 140,728 Btu/hr ft2 oF

hio

= hi x ID/OD

(Pers. 6.5, Kern)

= 126,937Btu/hr.ft2.oF 7.

Perhitungan desain bagian shell ID = Diameter dalam shell

= 39 in

B = Baffle spacing

= 19,5 in

Pt = tube pitch

= 1,25 in

C’ = Clearance

= Pt – OD = 1,25 – 1 = 0,250 in



Flow Area, as as

= ID x C’B/144 PT =

39  0,250  19,5 144 1,25

= 1,056 ft2 

Laju Alir, Gs Gs

= w/as =

655,331.811 1,056

= 620,432.484 lb/hr.ft2 

Bilangan Reynold, Res de

= 0,902 in

(Fiq.28 Kern)

De (Equivalent diameter) = 0,06 ft μ

= 0,011 cp

Res

= =

= 0,025 lb/ft hr

GS De



620,432..484 x0,060 0,282

= 132 ,158.142 Maka: jH 

= 140

Nilai ho Cp

= 176,447 Btu/lb.oF

k

= 0,259 Btu/hr ft.oF

(Fig.28, Kern)

 Cp.     k 

1

3

= 5,770

Koreksi viskositas diabaikan karena tidak significant, maka diperoleh : ho

= jH . (k/De). (Cpμ/k)1/3\

0,259 x14,262 0,06

= 140 x

= 398,871 Btu / hr ft2 oF

8.

Clean Overall Coefficient, UC UC

=

hio  ho hio  ho

=

126,937 x398,871 126,937  398,871

= 96,293 Btu/hr.ft2.oF

9.

Dirt Factor, Rd Rd

=

U C U D U C U D

=

146,319  96,293 149,319 x96,293

= 0,004 hr.ft2.oF/Btu 10. 

Pressure drop Bagian tube

Untuk NRe

= 322,801.472

Faktor friksi

= 0,00012

s

= 0,108

ΔPt

=

f Gt 2 L n 5, 22 x 10 10 x De s f t

(Fig 26, Kern)

=

0,00012 x109,450.484 2 x19 x(2) 5,22  1010 x0,075x0,108x1

= 0,129 psi V2 / 2g

= 0,004

ΔPr

= ( 4n/s ) ( V2/2g )

(Fig 27, Kern)

= 0,149 Psi ΔPT

= ΔPt + ΔPr = 0,129 psi + 0,149Psi = 0,278 psi



Shell Side Re

= 132,158.142

f

= 0,0005

N+1

= 12 L / B

(Fig.29, Kern)

= 12 x (228/19,5) = 140,308 Ds

= 3,250 ft

s

=1

ΔPs

=

fGs2 Ds ( N  1) 5,22 1010 Desf s

0,0005x620,432.484 2 x3,250 x(140,308) 5,22  1010 x0,060 x1 = 0,028 psi

IDENTIFIKASI Nama Alat

Condenser – 01

Kode

CD – 01

Jumlah

1

Fungsi

Untuk mengkondensasikan top KD – 01

Tipe

Shell and Tube Heat Exchanger DATA DESAIN Uc

96,293

Btu/hr ft2 oF

Ud

149,319

Btu/hr ft2 oF

Rd Calculated

0,004

hr ft2 oF/Btu

TUBE SIDE Length

19

Ft

OD

1

In

Passes

2

BWG

18

Pitch

1,250

Nt ΔPT

601 0,278

in Triangular Pitch tubes Psi SHELL SIDE

ID

39

in

B

19,5

in

Passes ΔPs

2 0,028

psi

Dengan Perhitungan yang sama untuk Accumulator selanjutnya analog dengan perhitungan Condenser-01 (CD-01). 5.

CONDENSER - 02 (CD-02) IDENTIFIKASI

Nama Alat

Condensor – 02

Kode

CD – 02

Jumlah

1

Fungsi

Untuk mengkondensasikan top KD – 02

Tipe

Shell and Tube Heat Exchanger DATA DESAIN Uc

241,610

Btu/hr ft2 oF

Ud

126,710

Btu/hr ft2 oF

Rd Calculated

0,004

hr ft2 oF/Btu

TUBE SIDE Length

12,550

Ft

OD

1

In

Passes

2

BWG

18

Pitch

1,250

Nt ΔPT

605 10,440

in Triangular Pitch tubes Psi SHELL SIDE

ID

37

in

B

18,5

in

Passes ΔPs

2 2,407

psi

6.

COOLER - 01 (C-01) Fungsi

: Menurunkan suhu keluaran R-01 menuju PC-01

Tipe

: Double Pipe Heat Exchanger

Gambar

:

Fluida Panas

Fluida Dingin

: Keluaran R-01 W

= 66804,593 kg/jam = 39441,925 lb/jam

T1

= 274,726 oC

= 526,507 oF

T2

= 234,818 oC

= 454,672 oF

: Air W

= 102249,854 kg/jam = 225422,0726 lb/jam

t1

= 28 oC

= 82,4 oF

t2

= 50 oC

= 122 oF

Perhitungan: 1. Beban Panas C-01 Q = 9416393,534 kJ/jam = 8925149,699 Btu/jam 2. LMTD Fluida Panas (oF)

Fluida Dingin (oF)

Selisih

526,507

Suhu tinggi

122

404,507

454,672

Suhu rendah

82,4

372,272

Selisih

32,234

LMTD

t 2  t1 = 388,166 oF ln (t 2 / t1 )

=

(Coulson & Richardson, 2005), Hal. 752 Ft = 1,000

(Coulson & Richardson, 2005), Grafik 12.19

Ft x LMTD

= 388,166 oF

Tc = T avg

= 490,590 oF

tc = t avg

= 102 oF

Asumsi UD = 135 Btu/hr,ft2,oF Q A = U D . t

(Kern, 1957), Tabel 8, Hal 840

A = 170,319 ft2 Karena A > 200 ft2, maka dipilih HE jenis Double Pipe Heat Exchanger, Rencana Klasifikasi Data Pipa

Outer Pipe

Inner Pipe

IPS (in)

10

8

SN

40

40

OD (in)

10,75

8,625

ID (in)

10,02

7,981

a” (ft2/ft)

2,814

2,258

Cold Fluid (Air) : Annulus a.

Flow Area, aa D1 = 10,02 in = 0,835 ft D2 = 8,625 in = 0,719 ft aa = =

 4

 4

(D22 – D12) (0,172 2 – 0,138 2) = 0,008 ft2

Equivalent Diameter, De De =

D

2 2

 D1 D1

2



= 0,2513 ft

(Kern, 1957), Eq. 6.3

b. Kecepatan Massa, Ga Ga = W/aa = 1589837,288 lb/hr,ft2 c.

Reynold number, Re Pada tavg = 102,2 oF μ

= 0,664 lb/hr ft

Rea

= De,Ga/μ =

0,2513 𝑥 1589837,288 0,664

= 248636,841

JH

= 600

k

= 0,3642 Btu/hr.ft2(oF/ft)

cp

= 0,0037 Btu/lb.oF

 c  Pr =    k 

1

(Kern, 1957), Fig. 24

3

= 0,2533 d. Koefisien perpindahan panas k  c  = JH   De  k 

ho

1

3

    w

  

0 ,14

= 220,258 Btu/hr,ft2,oF Hot Fluid (Keluaran RB-02) : Inner Pipe a. Flow Area, ap D = 7,981 in = 0,6651 ft ap =

 4

D2

=

 4

(0,6651)2 = 0,3472 ft2

b. Kecepatan Massa, Gp Gp = w/ap

= 424148,836 lb/hr ft2

c. Reynold number, Re Pada Tavg = 490,6 oF μ

= 0,0181 cP

= 0,0438 lb/hr

Rep = D x Gp/μ =

424148,836 x 7,981 0,0181

= 6436659,383

JH = 1000

(Kern, 1957), Fig. 24

= 0,0778 Btu/hr,ft2(oF/ft)

d. k

cp = 0,0026 Btu/lb,oF  c  Pr =    k 

1

3

= 0,1133 k e. hi = JH De

 c     k 

1

3

    w

  

0 ,14

= 13,2618 Btu/hr,ft2,oF f. Koreksi hi pada permukaan OD hio = hi x ID/OD = 12,2716 Btu/hr,ft2,oF g. Clean Overall Coefficient, UC UC

=

hioxho = 11,624 Btu/hr,ft2,oF hio  ho

h. Design Overall Coefficient, UD 1 1   Rd U D UC

(Kern, 1957), Eq. 6.10

Rd ditentukan 0,002 untuk masa servis 1 tahun 1 𝑈𝐷

=

1 𝑈𝐶

+ 0,002

UD = 11,359 Btu/hr,ft2,oF i. Required Length Q A = U . t D = 2024,057 ft2 Dari tabel 11 Kern, untuk 10-in IPS standard pipe, external surface/foot length = 2,258 ft Required length =

2024,057 ft2 2,258 ft

= 896,394 ft

Diambil panjang 1 harpin = 20 ft Jumlah harpin yang dibutuhkan =

896,394 40

= 22,41

Maka, dipakai 22 harpin 20 ft Actual Length

= 22 x 20 ft x 2 = 896 ft

Actual Surface = L x a” = 2024,0571 ft2 j. Actual Design Coefficient, UD UD =

Q A. t

= 11,36 Btu/hr,ft2,oF k. Dirt Factor, Rd Rd =

U C U D U C U D

= 0,002 hr,ft2,oF/Btu PRESSURE DROP Cold Fluid : Annulus a. De’ = (D2 – D1) = 0,1163 ft NRe = 248636,8407 0,264 ƒ = 0,0035  (Re a) 0, 42 = 0,0049 ρ b. Fa

= 61,94 lb/ft3 =

4 fGa2 L 2 g 2 De

= 8,7453 ft c. V

=

G = 7,1298 ft/s 3600 

Fl

V 2  = 3 x   = 1,5787 ft  2g 

Pa

=

( Fa  Fl )  144

(Eq, 3,47b, (Kern, 1957))

= 4,4408 psi

Hot Fluid: Inner Pipe a. Rep = 6436659 ƒ ρ

0 , 264 (Re p ) 0 , 42 = 3,615 lb/ft3 0 ,0035  = 0,0039

(Eq, 3,47b, (Kern, 1957))

4 fGp 2 L 2 b. ΔFp = 2 g D = 343,0956 ft

Pp

Fp .  = 144

= 8,6131 psi Identifikasi Nama Alat

Cooler-01

Kode Alat

C-01

Uc

11,624 Btu/hr ft2 oF

Ud

11,360 Btu/hr,ft2,oF

Rd

0,002 hr ft2 oF/Btu

Tipe

Double Pipe Heat Exchanger

Fungsi

Menurunkan temperatur keluaran R-01

Jumlah

1 Unit Inner Tube

Length

896 ft

OD

7,981 in

ΔPI

8,6131 psi Outer Tube

ID ΔPO

2,067 in 4,4408 psi

7.

COOLER – 02 (C-02) Fungsi

: Menurunkan suhu keluaran R-01 menuju PC-01

Tipe

: Double Pipe Exchangers

Gambar

:

Fluida Panas

Fluida Dingin

: Keluaran R-01 W

= 66804,593 kg/jam = 147278,7418 lb/jam

T1

= 234,818 oC

= 454,672 oF

T2

= 194,910 oC

= 382,838 oF

: Air W

= 99775,417 kg/jam = 219966,880 lb/jam

t1

= 28 oC

= 82,4 oF

t2

= 50 oC

= 122 oF

Perhitungan: 1. Beban Panas C-02 Q = 9188517,718 kJ/jam = 8709161,937 Btu/jam 2. LMTD Fluida Panas (oF)

Fluida Dingin (oF)

Selisih

454,672

Suhu tinggi

122

322,672

382,838

Suhu rendah

82,4

300,438

Selisih

32,234

LMTD

t 2  t1 = 316,281 oF ln (t 2 / t1 )

=

(Coulson & Richardson, 2005), Hal. 752 Tc = T avg

= 418,755 oF

tc = t avg

= 102 oF

Asumsi UD = 145 Btu/hr.ft2.oF Q A = U D . t

(Kern, 1957), Tabel 8, Hal. 840

= 189,9042 ft2 Karena A < 200 ft2, maka dipilih HE jenis Double Pipe Exchangers Rencana Klasifikasi

Data Pipa

Outer Pipe

Inner Pipe

IPS (in)

12

10

SN

30

40

OD (in)

12,75

10,75

ID (in)

12,09

10,02

a” (ft2/ft)

3,338

2,814

Cold Fluid (Air) : Annulus a. Flow Area, aa D1 = 10,75 in = 0,8958 ft D2 = 12,09 in = 1,0075 ft aa = =

 4

 4

(D22 – D12) (1,0075 2 – 0,8958 2) = 0,1668 ft2

Equivalent Diameter, De

D De =

2 2

 D1 D1

2



= 0,2373 ft

b. Kecepatan Massa, Ga

(Kern, 1957), Eq. 6.3

Ga = W/aa = 1318406,1877 lb/hr,ft2 c.

Reynold number, Re Pada tavg = 102,2 oF μ

= 0,664 lb/hr ft = De,Ga/μ

Rea

=

0,2373 x 1318406,1877 0,664

= 194660,115

JH

= 500

k

= 0,3642 Btu/hr.ft2(oF/ft)

cp

= 0,0037 Btu/lb.oF

 c  Pr =    k 

1

(Kern, 1957), Fig. 24

3

= 0,2533 d. Koefisien perpindahan panas k  c  = JH   De  k 

ho

1

3

    w

  

0 ,14

= 194,4172 Btu/hr,ft2,oF Hot Fluid (Keluaran R-01) : Inner Pipe a. Flow Area, ap D = 10,02 in = 0,835 ft ap =

 4

D2

=

 4

(0,835)2 = 0,5473 ft2

b. Kecepatan Massa, Gp Gp = w/ap

= 269089,9374 lb/hr ft2

c. Reynold number, Re Pada Tavg = 418,8 oF μ

= 0,0167 cP

= 0,0403 lb/hr

Rep = D x Gp/μ =

0,835 x 269089,9374 0,0167

= 5573058,093

JH = 1000

(Kern, 1957), Fig. 24

= 0,071 Btu/hr,ft2(oF/ft)

d. k

cp = 0,0031 Btu/lb,oF

 c  Pr =    k 

1

3

= 0,1201 k e. hi = JH De

 c     k 

1

3

    w

  

0 ,14

= 10,2154 Btu/hr,ft2,oF f. Koreksi hi pada permukaan OD hio = hi x ID/OD = 9,5217 Btu/hr,ft2,oF g. Clean Overall Coefficient, UC UC

=

hioxho = 9,0771 Btu/hr,ft2,oF hio  ho

h. Design Overall Coefficient, UD 1 1   Rd U D UC

(Kern, 1957), Eq. 6.10

Rd ditentukan 0,002 untuk masa servis 1 tahun 1 𝑈𝐷

=

1 9,0771

+ 0,002

UD = 8,9153 Btu/hr,ft2,oF i. Required Length Q A= U D . t = 3088,643 ft2 Dari tabel 11 Kern, untuk 10-in IPS standard pipe, external surface/foot length = 2,814 ft Required length =

3088,643 ft2 2,814 ft

= 1097,5989 ft

Diambil panjang 1 harpin = 20 ft

Jumlah harpin yang dibutuhkan =

1097,5989

40

= 27,44

Maka, dipakai 27 harpin 20 ft Actual Length

= 27 x 20 ft x 2 = 1098 ft

Actual Surface = L x a” = 3088,6433 ft2 j. Actual Design Coefficient, UD UD =

Q A. t

= 8,9153 Btu/hr,ft2,oF k. Dirt Factor, Rd Rd =

U C U D U C U D

= 0,002 hr,ft2,oF/Btu PRESSURE DROP Cold Fluid : Annulus a. De’ = (D2 – D1) = 0,1117 ft NRe = 194660,1147 0,264 ƒ = 0,0035  (Re a) 0, 42 = 0,0051 ρ b. Fa

= 61,94 lb/ft3 =

4 fGa2 L 2 g 2 De

= 7,90703025 ft c. V

Fl

=

G = 5,9126 ft/s 3600 

V 2  = 3 x   = 1,0857 ft  2g 

(Eq, 3,47b, (Kern, 1957))

Pa

=

( Fa  Fl )  144

= 3,8681 psi Hot Fluid: Inner Pipe a. Rep = 5573058 ƒ ρ

0 , 264 (Re p ) 0 , 42 = 2,392 lb/ft3 = 0,0039 0 ,0035 

b. ΔFp =

(Eq, 3,47b, (Kern, 1957))

4 fGp 2 L 2 g 2 D

= 309,422 ft

Pp

Fp .  = 144

= 5,1398 psi

Identifikasi Nama Alat

Cooler-02

Kode Alat

C-02

Uc

9,0771 Btu/hr,ft2,oF

Ud

8,9153 Btu/hr,ft2,oF

Rd

0,002 hr,ft2,oF/Btu

Tipe

Double Pipe Heat Exchanger

Fungsi

Menurunkan temperatur keluaran R-01

Jumlah

1 Unit Inner Tube

Length

1098 ft

OD

10,75 in

ΔPI

5,1398 psi Outer Tube

ID ΔPO

12,09 in 3,8681 psi

8.

COOLER - 03 (C-03) Fungsi

: Menurunkan suhu keluaran R-01 menuju PC-01

Tipe

: Shell and Tube Exchangers

Gambar

:

Fluida Panas

Fluida Dingin

: Keluaran R-01 W

= 66804,593 kg/jam = 147278,7418 lb/jam

T1

= 194,910 oC

= 382,838 oF

T2

= 155,000 oC

= 311,000 oF

: Air W

= 97264,051 kg/jam = 214430,274 lb/jam

t1

= 28 oC

= 82,4 oF

t2

= 50 oC

= 122 oF

Perhitungan: 1. Beban Panas C-03 Q = 9188517,718 kJ/jam = 8709161,937 Btu/jam 2. LMTD Fluida Panas (oF)

Fluida Dingin (oF)

Selisih

382,838

Suhu tinggi

122

260,838

311,000

Suhu rendah

82,4

228,600

Selisih LMTD

=

32,238

t 2  t1 = 244,365 oF ln (t 2 / t1 )

(Coulson & Richardson, 2005), Hal. 752 Ft = 1,00

(Coulson & Richardson, 2005), Grafik 12.19

Ft x LMTD = 244,365 oF Tc = T avg = 346,919 oF tc = t avg = 102 oF Asumsi UD A =

= 80 Btu/hr,ft2,oF Q U D . t

(Kern, 1957), Tabel 8, Hal. 840

A = 445,5002 ft2 Karena A > 200 ft2, maka dipilih HE jenis Sheel and Tube, Rencana Klasifikasi Tube Side : Panjang Tube (L)

= 16 ft

Outside Diameter (OD)

= 1,5 inch

BWG

= 18

Pass

=2

a"

= 0,3925 A = L x a"

Jumlah Tube (Nt) Nt

= 70,9395

Tube sheet

= 1,875 triangular pitch

(Kern, 1957), Tabel 10, Hal 843

dari tabel 9 Kern, didapat Jumlah Tube (Nt) yang mendekati adalah Nt

= 72

Koreksi UD A

= Nt x L x a'' = 452,160 ft2

UD =

Q A .Δt

UD = 78,8217 Btu/hr ft2 oF

(koreksi memenuhi)

Karena nilai Ud perhitungan sama dengan nilai Ud asumsi, maka data untuk Shell yaitu : Shell side : ID

= 21,25 inci

Baffle Space (B )

= ID/2= 10,625 inci

Pass (n)

=2

Pt

= 1,875

Hot Fluid (keluaran FT-01) : Tube Side 1. Flow area per Tube (at') = 1,54 in2 Total flow area (at)

= Nt x a't / 144 x n = 0,385 ft2

2.

Laju alir, Gt

= W / at = 382542,1865 lb/(hr) (ft2)

3.

Bilangan Reynold, Ret Pada Tavg

= 346,919 oF

Viskositas ( μ )

= 0,0152 cp

= 0,0368 lb/ft hr

ID

= 1,40 in

= 0,1167 ft

Re

= D x Gt / μ = 2932320,7466

4.

L/D

= 16 ft / 1,40 ft = 137,143

jH 5.

6.

= 1000

Prandl Number ( Pr )

(Kern, 1957), Fig, 24, Hal. 834

 cp x     k 

= 

k

= 0,0640 Btu/hr ftoF

Cp

= 0,0028 Btu/lboF

Pr

= 0,0016

Dengan (μ/μw) = 1 untuk bahan kimia kecuali untuk hidrokarbon, Koreksi viskositas diabaikan karena tidak signifikan, maka didapat : hi  j H

 k   Cp      D  k 

1

3

    w

  

0 ,14

hi

= 64,5256 Btu/hr ft2oF

hio

= hi (ID / OD) = 60,224 Btu/hr ft2oF

Cold Fluid (Air): Shell Side Suhu rata-rata

= 102,2oF

Baffle spacing (B)

= 10,625 in

Clerance (C')

= pitch – OD = 1,875 – 1,5 = 0,375

1. Luas area laluan (as)

= (ID x C' x B) / (144 Pt) = 0,3136 ft2

2.

Laju alir, (Gs)

= W / as = 683802,5616 lb/hr ft2

3.

Reynold Number (Res) = D x Gs / μ Pada tavg

= 102,2oF

Viskositas (u)

= 0,664 cp = 1,607 lb/ft hr

Diameter ekivalen (De) = 1,08 in = 0,09 ft = D x Gt / μ

Jadi, Re

= 38299,2075 4.

JH

= 140

(Kern, 1957), Fig, 28, Hal. 838

5.

k

= 0,3642 Btu/hr ft0F

Cp

= 0,0037 Btu/lb 0F

Prandl Number ( Pr )

 Cp x   =   k  = 0,0163

6.

Koreksi viskositas diabaikan karena tidak significant, maka didapat : ho  jH

 k   De

  Cp       k 

1

3

    w

  

0 ,14

= 143,5032 Btu/hr ft2 °F

ho 7.

Clean overall coefficient Uc = (hio x ho) / (hio + ho) = 42,421 Btu/hr ft2 0F

8.

Dirt factor, Rd Rd = (UD - Uc ) / (Uc x UD) = 0,0109

PRESSURE DROP Tube side 1. Untuk NRe

= 2932320,7466

Faktor friksi ( f ) = 0,0001

2.

s

= 0,2766

Pt

=

(Kern, 1957), Fig, 26

f Gt 2 L n 5, 22 x 10 10 x De s f t

= 0,139 psi 3.

Gt

= 382542,1865 lb/(hr) (ft2)

V2/ 2g

= 0,054 psi

Pr

= (4n / s) (V2/ 2g)

(Kern, 1957), Fig, 27

= 1,5621 psi 4.

PT

= Pt + Pr = 1,7011 psi

Shell Side 1.

Untuk NRe

= 38299,2075

Faktor friksi ( f )

= 0,0002

(Kern, 1957), Fig, 29

Number of cross, (N+1) N+1

(Kern, 1957), Eq 7.43

= 12 L / B = 216,8471

2.

s

= 1,000

Ps

=

f G s Di ( N  1) 5,22 x 1010 x De S f s 2

= 5,733 psi

Identifikasi Nama Alat

Cooler-03

Kode Alat

C-03

Uc

42,4211 Btu/hr ft2 oF

Ud

78,8217 Btu/hr ft2 oF

Rd

0,0109 hr ft2 oF/Btu

Tipe

Shell and Tube Heat Exchanger

Fungsi

Menurunkan temperatur keluaran R-01

Jumlah

1 Unit Tube Side

Length

16 ft

OD

1,5 in

Passes

2

BWG

18

Pitch

1,875 in (Triangular Pitch)

Nt

72

ΔPT

1,7011 psi Shell Side

ID

21,25 in

B

10,625 in

Passes ΔPs

2 5,733 psi

9.

EKSPANDER - 01 (EP-01) Fungsi

:

Menurunkan tekanan gas Hidrogen keluaran FD-01

Bentuk

:

Turbin

Gambar

: out E-01 10. in

a.

Data

Laju alir massa, W

= 5709,431 = 95,157

Densitas, ρ

BM campuran = 2,000 R

= 8,3145

k

=

jam

kg min

kg

= 0,9997

Cp campuran = 28,780

kg

m3 kJ kmol K kg

kmol

kJ kmol K

Cp campuran Cp campuran - R

= 1,406 b.

Kondisi Operasi

Tekanan masuk, Pin

= 18 atm

Tekanan keluar, Pout = 5 atm Temperatur, T

= 155 °C = 428,150 K

Faktor keamanan

= 10%

.... (Mc. Cabe, Hal. 820)

c.

Laju Alir Volumetrik, Q

Q

=

laju alir massa densitas

= 95,186

m3 min

= 312,289 d.

ft3 min

Kapasitas Ekspander

Faktor keamanan

= 10%

Kapasitas ekspander, Q1

= (100% + 10%) × Q = 343,518

e. Pw

min

Daya Ekspander, Pw =

0,0643 k T Q1 520 (k - 1) η

P1

k-1 k

[(P ) 2

- 1] (Mc. Cabe, Pers. 8.30, 2005)

dimana, k

= 1,406

T

= 107,058 °C

Q1

= 343,518

η

= 0,85

P1

= 18 atm

P2

= 5 atm

ft3 min

sehingga, Pw

ft3

=

0,0643 k T Q1 520 (k - 1) η

P1

k-1 k

[(P ) 2

= 33,166 hp = 33 hp

- 1]

IDENTIFIKASI Nama Alat

Ekspander 01

Kode Alat

EP-01

Jumlah

1 buah

Fungsi

Untuk menurunkan tekanan gas Hidrogen yang keluar dari top Flash Drum-01 DATA DESIGN

Tipe

Turbin

Temperature design

155,000

Tekanan design

o

C

18 Atm 343,518 ft3/min

Kapasitas

DATA MEKANIK Temperatur keluar

155,000

o

C

Tekanan masuk

18 Atm

Tekanan keluar

5 Atm

Power Bahan konstruksi

35,000 Hp Carbon Steel

11.

FLASH DRUM - 01 (FD-01) Fungsi

: Untuk memisahkan antara aliran Crude Etanol dengan Hidrogen

Tipe

: Silinder Vertikal dengan Tutup Elipsoidal

Bahan Konstruksi : Carbon steel SA-285, Cr. C Gambar

:

1. Data Desain : Tekanan

: 18 atm

Temperatur

: 155 oC

Laju Alir Uap, WV

: 5709,431 kg/jam

Laju Alir Liquid, WL

: 61095,141 kg/jam

Densitas Uap

: 709,2 kg/m3

Densitas Liquid

: 987,8 kg/m3

2. Vapor volumetric flowrate, QV QV

laju alir massa densitas =

=

5709,431 kg/jam 709,2 kg/m3

= 8,0505228 m3/jam = 0,1341754 m3/min

3. Kecepatan uap maksimum, UV UV

=

    L g 0,035 g   

𝑘𝑔 𝑚3

987,8

= 0,035 [(

   

0,5

   

–709,2

709,2

𝑘𝑔 𝑚3

𝑘𝑔 𝑚3

0,5

)

]

= 0,02194 m/s 4. Vessel area minimum, A

A

Qv = Uv

1,7974 m3 / s = 0,2810 m / s = 6,3970 m2 5. Diameter vessel minimum, D

D

 4A       =

=(

0,5

4 𝑥 6,3970 𝑚2 3,14

0,5

)

= 2,8547 m 6. Liquid volumetric flowrate, QL QL

laju alir massa densitas =

=

61.095,142 𝑘𝑔/𝑗𝑎𝑚 987,8 𝑘𝑔/𝑗𝑎𝑚

= 61,850 m3/jam = 0,0172 m3/s 7. Tinggi Liquid, HL Holding time, t

= 10 menit

(Walas, Hal. 612)

= 0,1667 jam HL

=

t A

QLx

= 0,0172

𝑚3 𝑗𝑎𝑚

0,1667 𝑗𝑎𝑚

𝑥

6,7281 𝑚2

= 1,6853 m

8. Tinggi vessel, HV Jarak top ke nozzle inlet, Hv

= 4 ft + D

Hal. 461, Coulson

= 1,2192 + 2,9276 m = 4,1470 m 𝐷

Jarak nozzle inlet ke level liquid maksimum, HZ = =

Hal. 461, Coulson

2

2,9276 m 2

= 0,4572 m Ht

= HL + HV + HZ = 1,6853 m + 4,1470 m + 0,4572 m = 6,2893 m

9. Volume vessel, Vt Digunakan vertikal Flash Drum dengan head tipe ellipsoidal head 𝜋

D2 Hs

Volume shell vessel, Vs

=

Tinggi shell, Hs

= 12,304 m

4

Maka : Volume shell vessel , Vs

=

𝜋 4

D2 Hs

= 20,696 m3 Volume head, Vh

=

𝜋 12

D3

= 6,5657 m3 Volume vessel, Vt = 20,696 m3 + (2 x 6,5657 m3) = 33,827 m3 10. Tebal dinding vessel, t

= Vs + 2Vh

Untuk Silinder

:t=

P r C S  E - 0,6  P

Untuk Ellipsoidal Head

:t=

P D C 2S  E - 0,2  P ( (Peter & Timmerhaus, 1991), Tabel 4)

Dimana : P

= Tekanan design

= 1 atm

D

= Diameter vessel

= 56,1939 in

R

= Jari-Jari vessel

= 28,0969 in

S

= Working stress allowable

=

= Joint effisiensi

= 0,85

= Korosi maksimum

= 0,010 in

= 14,9690 psi

13.700 psi

(Table 4, Peter,

hal538) E

(Table 4, Peter,

hal538) C

(Table 6, Peter, hal542)

Maka : Tebal dinding silinder : t

=

Pr  CC S  E - 0,6  P

=

(114,6960 psi  28,0969 in)  0,010 in (13.700 psi  0,85) - (0,6  14,6960 psi)

= 0,00785 m Tebal dinding ellipsoidal head : t

=

PD  CC 2S  E - 0,2  P

=

(14,6960 psi  56,1939 in)  0,010 in (2  13.700 psi  0,85) - (0,2  14,6960 psi)

= 0,0809 in = 0,0021 m 11. Outside diameter, OD

OD

= D + 2t = 2,9276 m + (2 x 0,03408 m) = 2,9957 m

IDENTIFIKASI Nama Alat

Flash Drum-01

Kode Alat

FD-01

Jumlah

1 Unit

Fungsi

Untuk memisahkan gas Hidrogen dari Crude Etanol DATA DESAIN

Tipe

Silinder Vertikal dengan Tutup Elipsoidal

Temperatur

155 oC

Tekanan

18 atm DATA MEKANIK

Laju Alir Uap

8,0505228 kg/jam

Laju Alir Liquid

61,849708 kg/jam

Diameter Vessel

2,9276 m

Tinggi Vessel

6,2893 m

Tebal Dinding

34,0782 mm

Bahan Konstruksi Carbon steel SA-285, Cr. C

12.

HEATER - 01 (H-01) Fungsi

: Memanaskan bahan baku asam asetat dari T-01

Tipe

: Shell and Tube Heat Exchanger

Gambar

:

Aliran inlet Tube

Shell

Rear End

Head

Aliran outlet

Fluida Panas

Water in

: Steam

W

= 33.641,509 kg/hr

= 74.166,745 lb/hr

T1

= 350 oC

= 662oF

T2

= 350 oC

= 662oF

Fluida Dingin

: Output Vaporizer-01 (As. Asetat, Etil Asetat, dan Air)

w

= 85.202,804 kg/hr

= 187.839,805 lb/hr

t1

= 188oC

= 244,580 oF

t2

= 155oC

= 311,000 oF

Perhitungan design sesuai dengan literatur pada buku Donald Q. Kern (1965). 2. Beban Panas H-01A Q

= 30.139.428,492 kJ/hr = 28.567.084,647 Btu/hr

3. LMTD Fluida Panas (oF)

Fluida Dingin (oF)

Selisih

662

Suhu tinggi

311,000

351,000

662

Suhu rendah

244,580

417,420

Selisih

LMTD =

-66,420

t 2  t1 ln (t 2 / t1 )

= 383,251 oF Ft

=1

t

= 383,251 oF

(Fig.18, Kern)

4. Temperatur Rata-rata Tavg =

T1  T 2 2

= 662oF tavg

=

t1  t 2 2

= 277,790 oF 5. Menentukan luas daerah perpindahan panas Asumsi UD = 150 Btu/hr.ft2.oF A

= =

(Tabel 8, Kern)

Q U D . t 30.139.428,492 150 𝑥 383,251

= 496,925 ft2 Karena A > 200 ft2, maka digunakan Shell & Tube Heat Exchanger 6. Spesifikasi tube dan shell 

Tube Side

= Aliran bahan baku asam asetat dari T-01

Panjang tube (L)

= 14 ft

Outside Diameter (OD) = 1,25 in BWG

= 18

Pass

=4

a”

= 0,327 ft2/lin ft A = L x a"

Jumlah tube, Nt

=

496,925 14 𝑥 0,327

= 108,513 Dari tabel.9 Kern, didapat nilai yang mendekati Nt perhitungan adalah Nt

= 105



Corrected Coefficient, UD A

= Nt x L x a'' = 105 x 14 ft x 0,327 ft2 = 480,837 ft2

UD

=

Q U D . t

= 163,551 karena nilai Ud perhitungan mendekati nilai Ud asumsi, maka data untuk shell : Shell

= steam

ID

= 21,25 inch

(Tabel 9, Kern)

Baffle Space (B = ID/2) = 10,625 inch Pass

=4

Pt

= 1,5625

in triangular pitch

7. Perhitungan desain bagian tube 

Flow Area/tube, a’t a’t

= 1,040 in2

(Tabel 10, Kern)

at

=

Nt  a ' t 144  n

(Pers. 7.48, Kern)

105 𝑥 1,040

=

144 𝑥 4

= 0,1896 ft2 

Laju Alir, Gt Gt

= W/at =

187.839,805 0,1896

= 990.803,369 lb/hr.ft2



Bilangan Reynold, Ret μ

= 0,0083 cp

ID

= 1,150 inch = 0,0958 ft

Ret

= ID.Gt/ μ

= 0,0201 lb/ft hr (Tabel 10, Kern)

= 11.404.274,503 

Dengan L/D = 159,652 diperoleh Jh = 1000



(Fig.24, Kern)

Nilai hi CP

= 0,0013 Btu/lb.oF

k

= 0,0109 Btu/hr ft.oF

 c.     k 

= 0,00236

 k   Cp   hi  J H      D  k 

1/ 3

    w

  

0 ,14

    Koreksi viskositas diabaikan, karena   w  hi

= 15,125 Btu/hr ft2 oF

hio

= hi x ID/OD

8. Perhitungan desain bagian shell ID = Diameter dalam shell

= 21,25 in

B = Baffle spacing

= 10,625 in

Pt = tube pitch

= 1,5625 in = Pt – OD = 1,5625 – 1 = 0,3125 in



Flow Area, as as

=1

(Pers. 6.5, Kern)

= 13,915 Btu/hr.ft2.oF

C’ = Clearance

0 ,14

= ID x C’B/144 PT

= =

23,25  0,250  11,625 144 1,25 21,25 x 0,3125 x 10,625 144 𝑥 1,5625

= 0,3136 ft2 

Laju Alir, Gs Gs

= w/as =

74.166,745 𝑙𝑏/ℎ𝑟 0,3136 ft2

= 236.512,360 lb/hr.ft2 

Bilangan Reynold, Res de

= 0,720 in

(Fiq.28 Kern)

De (Equivalent diameter) = 0,06 ft μ

= 0,022 cp

Res

=

= 0,054 lb/ft hr

GS De



= 577.209,671 Maka: jH 

= 500

(Fig.28, Kern)

Nilai ho Cp

= 0,0004 Btu/lb.oF

k

= 0,015 Btu/hr ft.oF

 Cp.     k 

1

3

= 0,0008

Koreksi viskositas diabaikan karena tidak significant, maka diperoleh : ho

= jH . (k/De). (Cpμ/k)1/3 = 9,470 Btu / hr ft2 oF

9. Clean Overall Coefficient, UC UC

=

hio  ho hio  ho

= 5,635 Btu/hr.ft2.oF 10. Dirt Factor, Rd Rd

=

U C U D U C U D

= 0,1713 hr.ft2.oF/Btu 11. Pressure drop 

Bagian tube Untuk NRe

= 11.404.274,503

Faktor friksi

= 0,0001

s

= 5,873

ΔPt

f Gt 2 L n = 5, 22 x 10 10 x De s f t

(Fig 26, Kern)

= 0,047 psi V2 / 2g

= 1,000

ΔPr

= ( 4n/s ) ( V2/2g )

(Fig 27, Kern)

= 2,7244 Psi ΔPT

= ΔPt + ΔPr = 0,047 psi + 2,7244 Psi = 2,7711 psi



Shell Side Re

= 577.209,671

f

= 0,001

N+1

= 12 L / B = 189,741

Ds

= 1,771 ft

s

=1

ΔPs

=

fGs2 Ds ( N  1) 5,22 1010 Desf s

= 4,7481 psi

(Fig.29, Kern)

IDENTIFIKASI Nama Alat

Heater– 01

Kode

H-01

Jumlah

1

Fungsi

Menaikkan suhu bahan baku keluaran Vaporizer

Tipe

Shell and Tube Heat Exchanger DATA DESAIN Uc

5,635

Btu/hr ft2 oF

Ud

163,551

Btu/hr ft2 oF

Rd Calculated

0,171

hr ft2 oF/Btu

TUBE SIDE Length OD

14

Ft

1,250

In

Passes

4

BWG

18

Pitch

1,5625

Nt ΔPT

105 2,771

in Triangular Pitch Tubes Psi

SHELL SIDE ID

21,25

in

B

10,625

in

Passes ΔPs

4 4,748

psi

13.

HEATER - 02 (H-02)

Fungsi

: Menaikkan suhu Feed sebelum masuk R - 01

Tipe

: Double Pipe Heat Exchanger

Gambar

:

Return bend

t1 Gland

Gland

Gland

T2

T1 Tee Return Head

t2

Fluida Panas

:

Saturated steam

Flowrate,

W1 = 12.120,647 Kg/jam

= 26.721,421 lb/hr

T1

= 350 oC

= 662 oF

T2

= 350 oC

= 662 oF

Fluida Dingin

:

Bahan baku untuk masuk reaktor

Flowrate,

W2

= 67.556,947 Kg/jam

= 14214,2607 lb/hr

t1

= 123,757 oC

= 254,763 oF

t2

= 174,171 oC

= 345,508 oF

Perhitungan design sesuai dengan literatur pada Donald Q. Kern (1965). a.

Beban Panas H - 02 Q = 10.858.887,717 kJ/jam = 10.292.390,404 Btu/hr

b.

LMTD Fluida Panas

Fluida Dingin

(oF)

(oF)

Selisih

662 (T1)

Suhu Tinggi (th)

345,508 (t2)

316,492

662 (T2)

Suhu Rendah (tc)

254,763 (t1)

407,237

Selisih

-90,745

t 2  t1 ln T1 / T2 

LMTD 

∆t = 359,960 oF c.

Temperatur rata-rata Tc

= T avg

= 662,000 oF

tc

= t avg

= 300,135 o F

Penentuan tipe Heater : Asumsi UD = 170 Btu/hr.ft2.F A

Q U D  t

A = 168,1948 ft2 Karena A < 200 ft2 , maka dipilih jenis Double Pipe Heat Exchanger Dari Tabel.10 Kern didapat spesifikasi data : Rencana Klasifikasi : Data Pipa

Annulus

Inner Pipe

IPS (in)

12

10

SN

30

40

OD (in)

12,75

10,75

ID (in)

12,09

10,02

a” (ft2/ft)

3,338

2,814

FLUIDA PANAS : Annulus a. Flow Area, aa D2 =

12,09 inch

= 1,0075 ft

D1 =

10,75 inch

= 0,8958 ft

aa =

 (D22 – D12) 4

=

0,1668 ft2

Equivalent Diameter De

=

D

2

2

 D1 D1

2



= 0,2373 ft b. Kecepatan Massa, Ga Ga

= W/aa = 16.159,0510 lb/hr.ft2

Pada T = 662 oF μ =

0,0432 lb/ft.hr

Rea

= De.Ga/μ = 363.718,389

JH = 700 k

(Fig. 28, Shell-side heat transfer curve)

= 0,0283 Btu/hr.ft2(oF/ft)

CP = 0,0021 Btu/lb.oF  c     k 

c.

1

3

= 0,1991 Koefisien perpindahan panas

ho =

JH

k  c    De  k 

1

3

    w

  

0 ,14

= 16,6098 Btu/hr.ft2.oF

FLUIDA DINGIN: Inner Pipe a. Flow Area, ap D =

10,02 inch

ap =

 2 D 4

=

 (0,835 ft)2 4

= 0,5437 ft2 b. Kecepatan Massa, Gp

= 0,835 ft

Gp

= w/ap = 272120,433 lb/hr.ft2

Pada 300,135 oF µ

= 0,0142 lb/hr ft

Rep

= D.Gp/μ = 6.630.847,0291

JH

= 1000

(Fig. 24, Tube-side heat transfer

curve) k

= 0,0156 Btu/hr.ft2(oF/ft)

c

= 82,6785 Btu/lb.oF

 c     k 

c.

1

3

= 0,1042

Koefisien Perpindahan Panas hi

= JH

k  c    De  k 

1

3

    w

  

0 ,14

= 10,645 Btu/hr.ft2.oF Koreksi hi pada permukaan OD hio = hi x ID/OD = 9,922 Btu/jam ft2 oF d. Clean Overall Coefficient, UC UC = =

hio  ho hio  ho

6,2116 Btu/hr.ft2.oF

e. Design Overall Coefficient, UD 1 1   Rd U D UC

Rd diasumsikan 0,001 UD

= 6,1354 Btu/hr.ft2.oF

f.

Required Surface

A =

Q U D  t

= 4.660,3701 ft2 Dari tabel 11 Kern, untuk 10-in IPS standard pipe, external surface/foot length a”= 2,814 ft2. Required length

= 1656,137 ft

Diambil panjang 1 hairpin = 20 ft, maka jumlah hairpin yang dibutuhkan = 42 buah Actually Length

= Jumlah hairpin x panjang hairpin x 2 = 4660,3701 ft

Actually surface

= L x a” = 4660,3701

g. Dirt Factor, Rd UD = = Rd =

Q A  t

6,1354 Btu/hr.ft2.oF

U C U D U C U D

= 0,002 hr.ft2.oF/Btu PRESSURE DROP FLUIDA PANAS : Annulus a. De’ = (D2 – D1) = 0,1117 ft Rea ƒ

= 363718,389

= 0,0035 

0,264 (Re a' ) 0, 42

= 0,0047 ρ

=

7,083 lb/ft3

b. ΔFa =

4 fGa2 L 2 g 2 De

= 171,2243 Va

=

Ga 3600 

= 6,281 ft/s

V 2  = 1 x    2g 

d. Fl

= 1,2252 e. ΔPa =

(Fa  F1)  144

= 8,4824 Psi

FLUIDA DINGIN : Inner Pipe a. Rep = ƒ

6630847,0291

= 0 , 0035 

0 , 264 (Re p ) 0 , 42

= 0,0039 ρ

= 0,3154 lb/ft3

b. ΔFp =

4 fGp 2 L 2 g 2 D

= 2726,9200 ft Fp .  c. ΔPp = 144 = 5,9727 Psi

IDENTIFIKASI Nama Alat

Heater – 02

Kode

H-02

Jumlah

1

Fungsi

Untuk memanaskan bahan baku reaktor

Tipe

Double pipe heat exchanger DATA DESAIN Uc

6,2116

Btu/hr ft2 oF

Ud

6,1354

Btu/hr ft2 oF

Rd Calculated

0,002

hr ft2 oF/Btu

ANNULUS IPS

12

Sch. No

30

In

OD

12,75

in

ID

12,09

in

a”

3,338

Ft2

Pa

8,482

Psi

INNER IPS

10

Sch. No

40

In

OD

10,75

in

ID

10,02

in

a”

2,814

Ft2

Pp

5,973

Psi

14.

HEATER - 03 (H-03)

Fungsi

: Menaikkan suhu Feed sebelum masuk R - 01

Tipe

: Double Pipe Heat Exchanger

Gambar

:

Return bend

t1 Gland

Gland

Gland

T2

T1 Tee Return Head

t2

Fluida Panas

:

Saturated steam

Flowrate,

W1 = 12.521,484 Kg/jam

= 27.605,115 lb/hr

T1

= 350 oC

= 662 oF

T2

= 350 oC

= 662 oF

Fluida Dingin

:

Bahan baku untuk masuk reaktor

Flowrate,

W2

= 67.556,947 Kg/jam

= 14214,2607 lb/hr

t1

= 174,171 oC

= 345,508 oF

t2

= 224,586 oC

= 436,253 oF

Perhitungan design sesuai dengan literatur pada Donald Q. Kern (1965). a.

Beban Panas H - 03 Q = 11.217.997 kJ/jam = 10.632.766,243 Btu/hr

b.

LMTD Fluida Panas

Fluida Dingin

(oF)

(oF)

Selisih

662 (T1)

Suhu Tinggi (th)

436,2530 (t2)

225,7470

662 (T2)

Suhu Rendah (tc)

345,5078 (t1)

316,4922

Selisih

-90,7452

t 2  t1 ln T1 / T2 

LMTD 

∆t = 268,5693 oF c.

Temperatur rata-rata Tc

= T avg

= 662,000 oF

tc

= t avg

= 390,880 o F

Penentuan tipe Heater : Asumsi UD = 199 Btu/hr.ft2.F A

Q U D  t

A = 198,947 ft2 Karena A < 200 ft2 , maka dipilih jenis Double Pipe Heat Exchanger Dari Tabel.10 Kern didapat spesifikasi data : Rencana Klasifikasi : Data Pipa

Annulus

Inner Pipe

IPS (in)

14

10

SN

30

40

OD (in)

14

10,75

ID (in)

13,25

10,02

a” (ft2/ft)

3,665

2,814

FLUIDA PANAS : Annulus c. Flow Area, aa D2 =

13,25 inch

= 1,0075 ft

D1 =

10,75 inch

= 0,8958 ft

aa =

 (D22 – D12) 4

=

0,3271ft2

Equivalent Diameter

De

=

D

2

2

 D1 D1

2



= 0,4651 ft

d. Kecepatan Massa, Ga Ga

= W/aa = 84.397,8044 lb/hr.ft2

Pada T = 662 oF μ =

0,0432 lb/ft.hr

Rea

= De.Ga/μ = 375.746,786

JH = 700 k

(Fig. 28, Shell-side heat transfer curve)

= 0,0283 Btu/hr.ft2(oF/ft)

CP = 0,0021 Btu/lb.oF  c     k 

c.

1

3

= 0,1991 Koefisien perpindahan panas

ho =

JH

k  c    De  k 

1

3

    w

  

0 ,14

= 8,4725 Btu/hr.ft2.oF FLUIDA DINGIN: Inner Pipe b. Flow Area, ap D =

10,02 inch

ap =

 2 D 4

=

 (0,835 ft)2 4

= 0,5437 ft2 b. Kecepatan Massa, Gp Gp

= w/ap

= 0,835 ft

= 272120,433 lb/hr.ft2

Pada 300,135 oF µ

= 0,0142 lb/hr ft

Rep

= D.Gp/μ = 6.034.241,255

JH

= 1000

(Fig. 24, Tube-side heat transfer

curve) k

= 0,0938Btu/hr.ft2(oF/ft)

c

= 0,0029 Btu/lb.oF

 c     k 

c.

1

3

= 0,1055

Koefisien Perpindahan Panas hi

= JH

k  c    De  k 

1

3

    w

  

0 ,14

= 11,8593 Btu/hr.ft2.oF Koreksi hi pada permukaan OD hio = hi x ID/OD = 11,0540 Btu/jam ft2 oF d. Clean Overall Coefficient, UC UC = =

hio  ho hio  ho

4,7963 Btu/hr.ft2.oF

e. Design Overall Coefficient, UD 1 1   Rd U D UC

Rd diasumsikan 0,002 UD

= 4,7507 Btu/hr.ft2.oF

f.

Required Surface

A =

Q U D  t

= 8333,5353 ft2 Dari tabel 11 Kern, untuk 10-in IPS standard pipe, external surface/foot length a”= 2,814 ft2. Required length

= 1656,137 ft

Diambil panjang 1 hairpin = 20 ft, maka jumlah hairpin yang dibutuhkan = 42 buah Actually Length

= Jumlah hairpin x panjang hairpin x 2 = 2961,4553 ft

Actually surface

= L x a” = 2961,4553

g. Dirt Factor, Rd UD = = Rd =

Q A  t

4,7505 Btu/hr.ft2.oF

U C U D U C U D

= 0,002 hr.ft2.oF/Btu

PRESSURE DROP FLUIDA PANAS : Annulus a. De’ = (D2 – D1) = 0,2083 ft Rea ƒ

= 375.746,7858

= 0,0035 

0,264 (Re a' ) 0, 42

= 0,0047 ρ

=

7,083 lb/ft3

d. ΔFa =

4 fGa2 L 2 g 2 De

= 45,4122 ft

Va

=

Ga 3600 

= 3,3099 ft/s

V 2  = 1 x    2g 

d. Fl

= 0,3402 e. ΔPa =

(Fa  F1)  144

= 2,2504 Psi

FLUIDA DINGIN : Inner Pipe a. Rep = ƒ

663421,2553

= 0 , 0035 

0 , 264 (Re p ) 0 , 42

= 0,0039 ρ

= 0,8803 lb/ft3

b. ΔFp =

4 fGp 2 L 2 g 2 D

= 628,3145 ft Fp .  c. ΔPp = 144 = 3,841 Psi

IDENTIFIKASI Nama Alat

Heater – 03

Kode

H-03

Jumlah

1

Fungsi

Untuk memanaskan bahan baku reaktor

Tipe

Double pipe heat exchanger DATA DESAIN Uc

4,796

Btu/hr ft2 oF

Ud

4,750

Btu/hr ft2 oF

Rd Calculated

0,002

hr ft2 oF/Btu

ANNULUS IPS

14

Sch. No

30

In

OD

14,00

in

ID

13,25

in

a”

3,665

Ft2

Pa

2,250

Psi

INNER IPS

10

Sch. No

40

In

OD

10,75

in

ID

10,02

in

a”

2,814

Ft2

Pp

3,841

Psi

15.

HEATER- 04 (H-04) Fungsi

: Memanaskan bahan baku sebelum masuk R-01

Tipe

: Shell and Tube Heat Exchanger

Gambar

:

Aliran inlet Tube

Shell

Rear End

Head

Aliran outlet

Fluida Panas

Water in

: Steam

W

= 33.641,509 kg/hr

= 74.166,745 lb/hr

T1

= 350 oC

= 662oF

T2

= 350 oC

= 662oF

Fluida Dingin

: Output Vaporizer-01 (As. Asetat, Etil Asetat, dan Air)

w

= 85.202,804 kg/hr

= 187.839,805 lb/hr

t1

= 188oC

= 244,580 oF

t2

= 155oC

= 311,000 oF

Perhitungan design sesuai dengan literatur pada buku Donald Q. Kern (1965). 12. Beban Panas H-01A Q

= 30.139.428,492 kJ/hr = 28.567.084,647 Btu/hr

13. LMTD Fluida Panas (oF) 662

Suhu tinggi

Fluida Dingin (oF)

Selisih

311,000

351,000

662

Suhu rendah

244,580

Selisih

LMTD =

417,420 -66,420

t 2  t1 ln (t 2 / t1 )

= 383,251 oF Ft

=1

t

= 383,251 oF

(Fig.18, Kern)

14. Temperatur Rata-rata Tavg =

T1  T 2 2

= 662oF tavg

=

t1  t 2 2

= 277,790 oF 15. Menentukan luas daerah perpindahan panas Asumsi UD = 150 Btu/hr.ft2.oF A

= =

(Tabel 8, Kern)

Q U D . t 30.139.428,492 150 𝑥 383,251

= 496,925 ft2 Karena A > 200 ft2, maka digunakan Shell & Tube Heat Exchanger 16. Spesifikasi tube dan shell 

Tube Side

= Aliran bahan baku asam asetat dari T-01

Panjang tube (L)

= 14 ft

Outside Diameter (OD) = 1,25 in BWG

= 18

Pass

=4

a”

= 0,327 ft2/lin ft A L x a"

Jumlah tube, Nt

= =

496,925 14 𝑥 0,327

= 108,513 Dari tabel.9 Kern, didapat nilai yang mendekati Nt perhitungan adalah Nt

= 105



Corrected Coefficient, UD A

= Nt x L x a'' = 105 x 14 ft x 0,327 ft2 = 480,837 ft2

UD

=

Q U D . t

= 163,551 karena nilai Ud perhitungan mendekati nilai Ud asumsi, maka data untuk shell : Shell

= steam

ID

= 21,25 inch

(Tabel 9, Kern)

Baffle Space (B = ID/2) = 10,625 inch Pass

=4

Pt

= 1,5625

in triangular pitch

17. Perhitungan desain bagian tube 

Flow Area/tube, a’t a’t

= 1,040 in2

(Tabel 10, Kern)

at

=

Nt  a ' t 144  n

(Pers. 7.48, Kern)

105 𝑥 1,040

=

144 𝑥 4

= 0,1896 ft2 

Laju Alir, Gt Gt

= W/at

=

187.839,805 0,1896

= 990.803,369 lb/hr.ft2 

Bilangan Reynold, Ret μ

= 0,0083 cp

ID

= 1,150 inch = 0,0958 ft

Ret

= ID.Gt/ μ

= 0,0201 lb/ft hr (Tabel 10, Kern)

= 11.404.274,503 

Dengan L/D = 159,652 diperoleh Jh = 1000



(Fig.24, Kern)

Nilai hi CP

= 0,0013 Btu/lb.oF

k

= 0,0109 Btu/hr ft.oF

 c.     k 

= 0,00236

 k   Cp   hi  J H      D  k 

1/ 3

    w

  

0 ,14

    Koreksi viskositas diabaikan, karena   w  hi

= 15,125 Btu/hr ft2 oF

hio

= hi x ID/OD

=1

(Pers. 6.5, Kern)

= 13,915 Btu/hr.ft2.oF 18. Perhitungan desain bagian shell ID = Diameter dalam shell

= 21,25 in

B = Baffle spacing

= 10,625 in

Pt = tube pitch

= 1,5625 in

C’ = Clearance

0 ,14

= Pt – OD = 1,5625 – 1 = 0,3125 in



Flow Area, as = ID x C’B/144 PT

as

= =

23,25  0,250  11,625 144 1,25 21,25 x 0,3125 x 10,625 144 𝑥 1,5625

= 0,3136 ft2 

Laju Alir, Gs Gs

= w/as =

74.166,745 𝑙𝑏/ℎ𝑟 0,3136 ft2

= 236.512,360 lb/hr.ft2 

Bilangan Reynold, Res de

= 0,720 in

(Fiq.28 Kern)

De (Equivalent diameter) = 0,06 ft μ

= 0,022 cp

Res

=

= 0,054 lb/ft hr

GS De



= 577.209,671 Maka: jH 

= 500

(Fig.28, Kern)

Nilai ho Cp

= 0,0004 Btu/lb.oF

k

= 0,015 Btu/hr ft.oF

 Cp.     k 

1

3

= 0,0008

Koreksi viskositas diabaikan karena tidak significant, maka diperoleh : ho

= jH . (k/De). (Cpμ/k)1/3\

= 9,470 Btu / hr ft2 oF

19. Clean Overall Coefficient, UC UC

=

hio  ho hio  ho

= 5,635 Btu/hr.ft2.oF 20. Dirt Factor, Rd Rd

=

U C U D U C U D

= 0,1713 hr.ft2.oF/Btu 21. Pressure drop 

Bagian tube Untuk NRe

= 11.404.274,503

Faktor friksi

= 0,0001

s

= 5,873

ΔPt

=

(Fig 26, Kern)

f Gt 2 L n 5, 22 x 10 10 x De s f t

= 0,047 psi V2 / 2g

= 1,000

ΔPr

= ( 4n/s ) ( V2/2g )

(Fig 27, Kern)

= 2,7244 Psi ΔPT

= ΔPt + ΔPr = 0,047 psi + 2,7244 Psi = 2,7711 psi



Shell Side Re

= 577.209,671

f

= 0,001

N+1

= 12 L / B

(Fig.29, Kern)

= 189,741 Ds

= 1,771 ft

s

=1

ΔPs

fGs2 Ds ( N  1) = 5,22 1010 Desf s = 4,7481 psi

IDENTIFIKASI Nama Alat

Heater– 04

Kode

H-04

Jumlah

1

Fungsi

Menaikkan suhu bahan baku keluaran Vaporizer

Tipe

Shell and Tube Heat Exchanger DATA DESAIN Uc

9,761

Btu/hr ft2 oF

Ud

155,760

Btu/hr ft2 oF

Rd Calculated

0,096

hr ft2 oF/Btu

TUBE SIDE Length OD

14

Ft

1,500

In

Passes

1

BWG

18

Pitch

1,875

Nt ΔPT

76 3,138

in Triangular Pitch Tubes Psi

SHELL SIDE ID

21,25

in

B

10,625

in

Passes

1

ΔPs

16.

A.

1,037

psi

KOLOM DESTILASI - 01 (KD-01)

Fungsi

: Memisahkan Etanol dan Air

Tipe

: Sieve Tray Tower

Material

: Carbon Steel

Gambar

:

Menentukan kondisi operasi. 

Feed P = 1 atm

= 760,000 mmHg

T = 82,767 oC

= 355,917 K

Komponen

Pi

Xi=Yi/Ki

Ki = Pi / P

Yi=Xi . Ki

asam asetat

230.462

0.052

0.303

0.016

air

397.047

0.257

0.522

0.134

etanol

903.672

0.519

1.189

0.617

etil acetate

908.285

0.060

1.195

0.072

Total

2,439.466

0.887

3.210

1.000



Top P = 0,6 atm

= 456,000 mmHg

T = 67,096 oC

= 340,246 K

Komponen

Kmol

Yi

Ki = Pi / P

Xi = Yi / Ki

asam asetat

0.009

0.000

0.270

0.000

air

35.954

0.038

0.452

0.084

etanol

913.022

0.959

1.050

0.913

etil acetate

3.233

0.003

1.188

0.003

Total

952.217

1.000

2.959

1.000



Bottom P = 1 atm

= 760,000 mmHg

T = 98,787 oC

= 371,971 K

Komponen

Kmol

Yi

Ki = Pi / P

Xi = Xi*Ki

asam asetat

14.505

0.016

0.540

0.009

air

862.884

0.942

0.957

0.902

etanol

38.043

0.042

2.139

0.089

etil acetate

0.129

0.000

1.922

0.000

Total

915.562

1.000

5.559

1.000

B. Desain Kolom Destilasi a. Menentukan Relatif Volatilitas, α Komponen kunci : Light Key

: Etanol

Heavy Key

: Air

 αD

K LK K HK

= KLK / KHK = 1,050 / 0,452 = 2,324

αB

= KLK / KHK = 2,139 / 0,957 = 2.234

 Avg 



Top

x  Bot 

2,324 x 2,234

=

= 2,279 b. Menentukan Stage Minimum Dengan menggunakan metode Fenske ( R. Van Winkle;eg : 5.118 ; p 236)

SM 

Log  X LK / X HK D x  X HK / X LK B  Log ( Avg )

SM 

Log 25,395D x 22,682B  Log (3.118)

NM = 7.717 c. Menentukan rasio refluks minimum, Rmin n

1 – q = 1

xF (   ) /  n

(L/D)min + 1 =

q=1

xD

 (   ) /  1

Komponen

Xf

Α

(α-θ)/α

Xf/(α-θ)/α

XD/(α-θ)/α

asam asetat

0.014

0.580

-1.031

-0.014

0.000

air

0.265

1.000

-0.179

-1.482

-0.211

etanol

0.716

2.276

0.482

1.485

1.989

etil acetate

0.005

2.288

0.485

0.010

0.007

Total

1.000

0.000

0.000

1.784

(L/D)m = 1,784 – 1 = 0,784

(L/D) = 1,5 x (L/D)m = 1,5 x 0,784 = 1.177 d. Teoritical Tray Pada Actual reflux – Methode Gilliland Diketahui

: Rm

= 0,784

Nm

= 7.717

L / D  ( L / D) m 1,177  0,784  L/ D 1 1,177  1 = 0,466 Dari grafik 5.18 hal hal.243 Van Winkle (Gilland Corelation) diperoleh :

N  Nm = 0,466 N 1 N

= 17,823 stage ~ 18 stage

R akt

= 1,177

N teori

= 17,823

e. Menentukan Feed Location. Feed location ditentukan dengan menggunakan metode Kirkbride: m

 B  X    X   2  0,206 Log   HK   LK B   dengan:  D  X LK  F   X HK D   B = molar flow pada bagian bawah produk

Log

D

p

=

= molar flow pada bagian atas produk

(XHK)f = konsentrasi heavy key pada feed (XLK)f = konsentrasi light key pada feed (XHK)d = konsentrasi heavy key pada bagian atas produk (XlK)b = konsentrasi light key pada bagian bawah produk

 18,163.581 0,495 1.211 0,206 Log   42,931.198 

Log

m

=

p

Log

m p

m

= -0,596 = 0,754 p

Ntheoritical

= m + p

17,823

= 0,754 p + p

17,823

= 0,754 p

p

= 10,163

m

= 7,660

sehingga: m (rectifying section)

= 7,660 tray = 8 tray

p (stripping section)

= 10,163 tray = 10 tray

berdasarkan dari nilai m dan p, dapat ditentukan bahwa feed yang masuk pada tray ke 19 dari bagian atas kolom destilasi. C. Desain kolom bagian atas (Rectifying section) a. Data fisik untuk rectifying section D

= 42,931.198 kg/jam

L

=R.D = 1,177 x (42,931.198 kg/jam) = 50,517.531kg/jam = 14,033 kg/det

V

=L+D = 50,517.531 kg/jam + 42,931.198 kg/jam = 93,448.729 kg/jam = 25.958 kg/det Data Fisik

Vapour

Liquid

Mass Flow rate (kg/det)

25.998

14.033

Density (kg/m3)

4.010

733.744

Volumetric Flow rate (m3/det)

6.474

0.019

Surface tension (N/m)

16.751

b. Diameter kolom  Liquid – Vapour Flow Factor (FLV)

LW VW

V L

FLV

=

FLV

= 0,541 x 0,074 = 0,040

 Ditentukan tray spacing = 0,5 m  Dari figure 11.27 buku Chemical Engineering, vol. 6, . JM. Couldson didapat nilai konstanta K1 = 0,094  Kecepatan Flooding (uf) = K1 *

uf

= 0,094

 L  V V 733,744  4,010 4,010

= 4,873 m/s 

 Desain untuk 85 % flooding pada maksimum flowrate ( u ) 

u

= 0,85 . uf = 0,85 . 4,873 m/s = 4,142 m/s

 Maksimum volumetric flow rate (Uv maks)

Uv maks

=

V

V

=

93,448.729kg / s 4,010kg / m 3

= 16,385 m3/s  Net area yang dibutuhkan (An) An

=

U V maks 

u =

16,385 m 3 / s 7,298 m / s

= 6,474 m2

 Cross section area dengan 12 % downcormer area (Ac) Ac

=

An 1  0,12

=

6,474m 2 1  0,12

= 1,776 m2  Diameter kolom (Dc) Dc

=

4 Ac 3,14

=

4 . (1,776m 2 ) 3,14

= 1,504 m c. Desain plate  Diameter kolom (Dc)  Luas area kolom (Ac) Ac

=

Dc 2 . 3,14 4

= 1,504 m

=

(1,504) 2 . 3,14 4

= 1.776 m2  Downcomer area (Ad) Ad

= Persen downcomer x Ac = 0,12 x 1.776 m2

= 0.213 m2  Net area (An) An

= Ac – Ad = 1,776 m2 – 0.213 m2 = 1,563 m2

 Active area (Aa) Aa

= Ac – 2 Ad = 1,776 m2 – (2 x 0.213 m2) = 1,350 m2

 Hole area (Ah) ditetapkan 10% dari Aa sebagai trial pertama Ah

= 10% . Aa = 0,135 m2

 Nilai weir length (lw) ditentukan dari figure 11.31, JM. Couldson ed 6

0,213 x 100 1,776

Ordinat

=

Ad x 100 Ac

=

Absisca

=

Iw Dc

= 0,760

= 12

Sehingga : lw

= Dc . 0,760 = 1,504 m . 0,760 = 1,143 m



Penentuan nilai weir height (hw) , hole diameter (dh), dan plate thickness, (nilai ini sama untuk kolom bagian atas dan bawah) Weir height (hw)

= 50 mm

Hole diameter (dh)

= 5 mm

Plate thickness

= 5 mm

d. Pengecekan Check weeping  Maximum liquid rate (Lm,max) Lm,max

= =

L 3600

50,517.531 kg / jam 3600

= 14,033 kg/det  Minimum liqiud rate (Lm,min) Minimum liquid rate pada 70 % liquid turn down ratio Lm,min

= 0,7 Lm, max = 0,7 (14,033 kg/det) = 9,823 kg/det

 Weir liquid crest (how) 2

how

 Lm  = 750     l Iw 

how,maks

 Lm, maks  = 750     l Iw 

3

(J.M.Couldson. Eq.11.85) 2

3

  14,033 kg / det = 750   3  733,744kg / m x 1,143m  = 37,614 mm liquid how,min

 Lm, min  = 750     l Iw 

2

3

  9,823kg / det = 750   3  733,744kg / m x 1,143m  = 28.231 mm liquid Pada rate minimum hw + how

= 50 mm + 28.231 mm

2

3

= 78.231 mm Dari figure 11.30 JM. Coulson ed 6 K2

= 30,9

 Minimum design vapour velocity (ŭh) ŭh

=

=

K 2  0,90 25,4  d h  1  V  2 30,9  0,90 25,4  5 1,9941 / 2

= 0,285 m/s  Actual minimum vapour velocity (Uv,min actual) Uv,min actual

=

minimum vapour rate Ah

=

70% x 6,474 m 3 /s 0,135 m 2

= 3,357 m/s  Nilai ini dapat diterima, karena minimum operating rate harus berada diatas nilai weep point. Plate pressure drop  Jumlah maksimum vapour yang melewati holes (Ǚh) Ǚh

=

Uv, maks Ah

6,474 m 3 / s = 0,135 m 2 = 3,357 m/s  Dari figure 11.34 JM. Couldson ed 6, untuk : Plate thickness hole diameter

= 1

Ah Ah = Aa Ap

= 0,1

Ah x 100 Ap

= 10,0

Didapat nilai Orifice coeficient (Co) = 0,750  Dry plate drop (hd) 2

hd

    Uh = 51   V  Co   L   = 5120,034x0,005 = 15,583 mm liquid

 Residual head (hr) hr

=

=

12,5 .10 3

L 12,5 .10 3 733,744kg / m 3

= 17,036 mm liqiud  Total pressure drop (ht) ht

= hd + (hw + how) + hr = 15,583 + (78,231) + 17,036 = 110,850 mm liquid

Nilai ht yang didapat tidak jauh berbeda dari 100 mm air yang merupakan basis asumsi pressure drop. Downcomer liquid backup  Downcomer pressure loss (hap) hap

= hw – (10 mm) = 50 – 10 = 40 mm

 Area under apron (Aap) Aap

= hap . lw = 40 x 10-3 . 1,143 m = 0,046 m2

Karena nilai Aap lebih kecil dari nilai Ad (0,213 m2), maka nilai Aap yang digunakan pada perhitungan head loss di downcomer (hdc)  Head loss in the downcomer (hdc) hdc

 Lm, max  = 166     L Aap 

2

  13,680 kg/s = 166   3  733,744kg / m 0,046 

2

= 0,00012 mm  Back up di downcomer (hb) hb

= (hw+ how) + ht + hdc = (78,231) + 110,850 + 0,00012 = 189,081 mm = 0,189 m

(plate spacing + weir height)/2 = 0,275 m hb harus lebih kecil dari (plate spacing + weir height)/2, Ketentuan bahwa nilai hb harus lebih kecil dari (plate spacing + weir height)/2, telah terpenuhi. (J.M.Coulson..p.474) Check resident time (tr) tr

=

=

Ad hbc  L Lm, maks

0,213 m 2 . 0,189m .733,744kg / m 3 14,033kg / s

= 2.107 s Check Entrainment  Persen flooding actual. uv

=

Uv maks An

6,474m 3 / s = 1,563 m 2

= 4,162 m/s % flooding =

=

uv x100 uf

4,142m / s x100% 4,873 m/s

= 85 %  Untuk nilai FLV = 0,040 dari figure 11.29 JM. Couldson ed 6 Didapat nilai ψ = 0,060 Ketentuan bahwa nilai ψ harus lebih kecil dari 1, telah terpenuhi. e. Trial plate layout Digunakan plate type cartridge, dengan 50 mm unperforted strip mengelilingi pinggir plate dan 50 mm wide calming zones.

 Dari figure 11.32 JM. Couldson ed 6 pada

1,143m lw = = 0,760 Dc 1,504m

Di dapat nilai θC = 100O  Sudut subtended antara pinggir plate dengan unperforated strip (θ) θ

= 180 - θC = 180 – 100 = 80O

 Mean length, unperforated edge strips (Lm) Lm

   = Dc  hw x 3,14    180 

 80  = 1,504m  0,050m  x 3,14    180 

= 2,029 m  Area of unperforated edge strip (Aup) Aup

= hw . Lm = 0,050 m . 2,029 m = 0,101 m2

 Mean length of calming zone (Lcz) Lcz

  = ( Dc  hw) sin  C   2   96  = (1,504 m  0,050 m) sin    2

= 1,081 m  Area of calming zone (Acz) Acz

= 2 ( Lcz . hw) = 2 (1,081 m . 0,050m) = 0,108 m2

 Total area perforated (Ap) Ap

= Aa – (Aup + Acz) = 1,350 m2 – (0,101 + 0,108) m2 = 1,140 m2

Dari figure 11.33 JM. Couldson ed 6 di dapat nilai Ip/dh = 2,75 untuk nilai Ah/Ap = 0,115 . Nilai Ip/dh = 2,750, harus berada dalam range 2,5 – 4.0 .  Jumlah holes Area untuk 1 hole (Aoh) Aoh

dh 2 = 3,14 4 (5 x10 3 m) 2 = 3,14 4 = 0,00001963 m2

Jumlah holes =

Ah Aoh

0,135m 2 = 0,00001963m 2 = 6.878,814 holes = 6.879 holes

f. Ketebalan kolom bagian atas. Ketebalan dinding bagian head, thead t=

P.Da  Cc 2.S.E j  0,2.P

( Peter Tabel.4 Hal 537)

Ketebalan dinding bagian silinder, tsilinder t=

P.ri  Cc S .E j  0,6.P

( Peter Tabel.4 Hal 537)

Keterangan : P = Tekanan Desain

= 0,6 atm

Da = Diameter Kolom

= 1,504 m

ri = Jari-jari Kolom

= 0,752 m

S = Tekanan kerja yang diperbolehkan = 932,226 atm Cc = Korosi maksimum

= 0,003 m

Ej = Efisien pengelasan

= 0,85

thead

=

0,6atm x 1,504 m  0,003 m 2.(932,226 atm x 0,85)  0,2 x 0,6 atm

= 0,004 m tsilinder =

= 0,358 cm

0,6atm x 0,752m  0,003 m (932,226 atm x 0,85)  0,6 x 0,6 atm

= 0,004 m = 0,358 cm Sehingga : OD = ID + 2tsilinder

= 1,504 m + 2 (0,004 m) = 1,511 m

D. Desain kolom bagian bawah (Striping section) a. Data fisik untuk stripping section F

= 41.119,744 kg/jam

L

= 18.163,581 kg/jam

V

= 41.119,789 kg/jam

q

= 1

q

=

V’

= V  ( q 1) F

L’

= F + L

L '  L  F

(RE.Treyball, Eq.9.126) (RE.Treyball, Eq.9.127)

= 41.119,744 kg/jam + 18.163,581 kg/jam = 59.283,325 kg/jam = 16,468 kg/det V’

= V = 41.119,789 kg/jam = 11,422 kg/det

Data Fisik

Vapour

Liquid

Mass Flow rate (kg/det)

11,422

16,468

Density (kg/m3)

2,201

926,494

Volumetric Flow rate (m3/det)

5,190

0,018

b. Diameter kolom  Liquid –Vapour Flow Factor (FLV)

V L

FLV

=

LW VW

FLV

=

59.283,325 kg / jam 2,201kg / m 3 41.119,789 kg / jam 926,494 kg/ m 3

= 0,048  Ditentukan tray spacing = 0,5 m  Dari figure 11.27 buku Chemical Engineering, vol. 6, . JM. Couldson didapat nilai konstanta K1 = 0,082  Kecepatan Flooding (uf)

 L  V V

= K1

uf

= 0,360

926,494 kg / m 3  2,201kg / m 3 2,201kg / m 3

= 7,370 m/s 

 Desain untuk 85% flooding pada maksimum flow rate ( u ) 

u

= 0,85 . uf = 0,85 . 7.370 m/s = 6,264 m/s

 Maksimum volumetric flow rate (Uv maks) Uv maks

=

V  V . 3600

=

41.119,789 kg / jam 2,201kg / m 3 . 3600

= 5,190 m3/s  Net area yang dibutuhkan (An) An

=

U V maks 

u 5,190 m 3 / s = 6,264 m / s

= 0,828 m2  Cross section area dengan 12 % downcomer area (Ac) Ac

=

An 1  0,2

0,828 m 2 = 1  0,12 = 0,941 m2  Diameter kolom (Dc) Dc

=

4 Ac 3,14

=

4 (0,941 m 2 ) 3,14

= 1,095 m c. Desain plate  Diameter kolom (Dc)

= 1,095 m

 Luas area kolom (Ac) Ac

=

Dc 2 . 3,14 4

=

(1,095 m) 2 . 3,14 4

= 0,941 m2  Downcomer area (Ad) Ad

= persen downcomer x Ac = 0,12 (0,941 m2) = 0,113 m2

 Net area (An) An

= Ac – Ad = 0.941 m2– 0,113 m2 = 0,828 m2

 Active area (Aa) Aa

= Ac – 2 Ad = 0,941 m2 – 2 (0,113 m2) = 0,715 m2

 Hole area (Ah) ditetapkan 10% dari Aa sebagai trial pertama Ah

= 10 % . Aa = 10% . 0,715 m2 = 0,072

 Nilai weir length (lw) ditentukan dari figure 11.31, JM. Couldson ed 6

0,113 x 100 0,941

Ordinat

=

Ad x 100 Ac

=

Absisca

=

Iw Dc

= 0,760

= 12.000

Sehingga : lw

= Dc . 0,76 = 1,095 m . 0,76 = 1,795 m

 Penentuan nilai weir height (hw) , hole diameter (dh), dan plate thickness, (nilai ini sama untuk kolom atas dan kolom bawah) Weir height (hw)

= 50 mm

((J M.Couldson. p.571)

Hole diameter (dh)

= 5 mm

((J M.Couldson. p.573)

Plate thickness

= 5 mm

((J M.Couldson. p.573)

d. Pengecekan Check weeping  Maximum liquid rate (Lm,max) Lm,max

=

L 3600

=

59.283,325 kg/jam = 16,468 kg/s 3600

 Minimum liqiud rate (Lm,min) Minimum liquid rate pada 70 % liquid turn down ratio

Lm,min

= 0,7 Lm, max = 0,7 (16,468 kg/s) = 11,527 kg/s

 Weir liquid crest (how) 2

how

 Lm  = 750     l Iw 

how,maks

 Lm, maks  = 750     l Iw 

3

2

3

  16,468 kg / s = 750   3  926,494kg / m . 0,832m 

2

3

= 5,773 mm liquid how,min

 Lm, min  = 750     l Iw 

2

3

  11,527 kg / s = 750   3  926,494 kg / m . 0,832m 

2

3

= 4,551 mm liquid

Pada rate minimum hw + how

= 50 mm + 4,551 mm = 54,551 mm

Dari figure 11.30 JM. Couldson ed 6, diperoleh K2 = 31,00  Minimum design vapour velocity (ŭh) Ŭh

=

=

K 2  0,90 25,4  dh 1  V  2 31  0,90 25,4  5 1,629

1

2

= 0,349 m/s  Actual minimum vapour velocity (Uv,min actual)

Uv,min actual

=

minimum vapour rate Ah

70% x 5,190 m 3 /s = 0,072m 2 = 5,078 m/s  Jadi minimum operating rate berada diatas nilai weep point.

Plate pressure drop  Jumlah maksimum vapour yang melewati holes (Ǚh) Ǚh

=

Uv, maks Ah

=

16,834 m 3 /s 0,333m 2

= 50,575 m/s  Dari figure 11.34 JM. Couldson ed 6, untuk : Plate thicness hole diameter

= 1

Ah Ah = Aa Ap

= 0,1

Ah x 100 Ap

= 10

Sehingga didapat nilai Orifice coeficient (Co) = 0,845  Dry plate drop (hd) 2

hd

    Uh = 51   V  Co   L   2

 5,078 m/s  2,201kg / m 3 = 51   3  0,845  926,494 kg / m = 3,964 mm liquid

 Residual head (hr) hr

=

hr

=

12,5 .10 3

L 12,5 .10 3 1.233,024kg / m 3

= 10,138 mm liqiud  Total pressure drop (ht) ht

= hd + (hw + how) + hr = 3,964 mm+ 54,551mm + 10,138 mm = 68,653 mm liquid

Selisih nilai total pressure drop harus lebih besar dari 100 mm liquid, maka desain dapat diterima. (Coulson p.474)

Downcomer liquid backup  Downcomer pressure loss (hap) hap

= hw – 10 mm = 50 – 10 = 40 mm

 Area under apron (Aap) Aap

= hap . lw = 40 x 10-3m . 0,832 m = 0,05 m2

Karena nilai Aap lebih kecil dari nilai Ad, maka nilai Aap yang digunakan pada perhitungan head loss di downcomer (hdc)  Head loss in the downcomer (hdc) hdc

 Lm, max  = 166     L Aap 

2

  16,468 kg/s = 166  3 2   926,494kg / m . 0,832m 

2

= 0,00013 mm  Back up di downcomer (hb) hb

= (hw + how) + ht + hdc = 54,551 mm+ 68,653 mm+ 0,00013 mm = 123,204 mm = 0,123 m

(plate spacing + weir height)/2 = 0,275 m. hb harus lebih kecil dari (plate spacing + weir height)/2 Ketentuan bahwa nilai hb harus lebih kecil dari (plate spacing + weir height)/2, telah terpenuhi.

Check resident time (tr) tr

=

=

Ad hbc  L Lm, maks

0,113 m 2 . 0,123m .926,494kg / m 3 16,468 kg / s

= 0,783 s Ketentuan bahwa nilai tr harus lebih besar dari 3 s, telah terpenuhi Check Entrainment  Persen flooding actual. uv

=

Uv maks An

5,190 m 3 / s = 0,828 m 2 = 6,264 m/s % flooding =

=

uv x 100 uf

6,264m/s x 100 7,370 m/s

= 85

 Untuk nilai FLV = 0,048 dari figure 11.29 Didapat nilai ψ = 0,8 00 Ketentuan bahwa nilai ψ harus lebih kecil dari 1 telah terpenuhi.

e. Trial plate layout Digunakan plate type cartridge, dengan 50 mm unperforted strip mengelilingi pinggir plate dan 50 mm wide calming zones.

 Dari figure 11.32 JM. Couldson ed 6 pada

Iw = 0,760 Dc

Di dapat nilai θC = 100OC  Sudut subtended antara pinggir plate dengan unperforated strip (θ) θ

= 180 - θC = 180 – 100 = 80OC

 Mean length, unperforated edge strips (Lm) Lm

   = Dc  hw x 3,14    180 

 80  = 1,095 m  50 x 10 3 m  x 3,14    180 

= 1,458 m  Area of unperforated edge strip (Aup) Aup

= hw . Lm = 50 x 10-3 m . 1,458 m = 0,073 m2

 Mean length of calming zone (Lcz) Lcz

  = ( Dc  hw) sin  C   2   96  = (1,095m  50 x 10 3 m) sin    2 

= 0,776 m  Area of calming zone (Acz) Acz

= 2 ( Lcz . hw) = 2 ( 0,776 m . 50 .10-3m) = 0,078 m2

 Total area perforated (Ap) Ap

= Aa – (Aup + Acz) 0,715 m2– (0,073 + 0,078) m2 = 0,565 m2

Dari figure 11.33 JM. Couldson ed 6 di dapat nilai Ip/dh = 2,75 untuk nilai Ah/Ap = 0,127 Nilai Ip/dh = 2,750, harus berada dalam range 2,5 – 4.0 .  Jumlah holes Area untuk 1 hole (Aoh) Aoh

= 3,14

dh 2 4

= 3,14

(5 x10 3 m 2 4

= 0,00001963 m2

Jumlah holes =

Ah Aoh

0,072m 2 = 0,00001963m 2 = 3.645,577 holes = 3.645,000 holes f. Ketebalan minimum kolom bagian bawah. Ketebalan dinding bagian head, thead t=

P.Da  Cc 2.S.E j  0,2.P

( Peter Tabel.4 Hal 537)

Ketebalan dinding bagian silinder, tsilinder t=

P.ri  Cc S .E j  0,6.P

( Peter Tabel.4 Hal 537)

Dimana : P = Tekanan Design

= 1 atm

Da = Diameter Kolom

= 1,095 m

ri = Jari-jari Kolom

= 0,548 m

S = Tekanan kerja maksimum

= 932,226 atm

Cc = Korosi maksimum

= 0,003 m

Ej = Efisien pengelasan

= 0,850

thead

=

1atm x 1,095 m  0,003 m 2.(932,226 atm x 0,850)  0,2 x 1 atm

= 0,004m = 0,4 cm tsilinder =

1atm x 0,548m  0,003 m (932,226 atm x 0,85)  0,6 x 1atm

= 0,004 m = 0,4 cm Sehingga : OD = ID + 2tsilinder = 1,095 m + 2 (0,004m)

= 1,102 m

E. Total Pressure Drop Pressure drop per plate 

Rectifying Section

= 110,850 mm liquid = 110,850 mm x10-3 m x 9,8 m/s2x 994,9345 kg/m3 = 1.080,830 Pa



Stripping Section

= 68,653 mm liquid = 68,653 mm x 10-3 m x 9,8 m/s2 x 994,9345 kg/m3 = 669,338 Pa

Total Pressure Drop

= (N1 x P1) + (N2 x P2) = (1.080,830 Pa x 8000) + (669,338 Pa x 10.000) = 15.340,521 Pa = 0,151 atm

F. Tinggi Kolom Destilasi H

= [(N1+1)Tray spacing1 + (N2+1)Tray spacing2] = [(18,741 . 0,5) + (4,024 . 0,5)] = 11,500 m

Heatas = tinggi tutup ellipsoidal atas = ¼ x ID = ¼ x 1,803m = 0,451 m

Hebawah= tinggi tutup ellipsoidal bawah = ¼ x ID = ¼ x 2,362 m = 0,591 m

Ht

= H + Heatas + Hebawah

= 11,500 m + 0,451 m + 0,591 m = 12,541 m

Nama Alat Alat Kode Jenis Jumlah Operasi Fungsi

Tekanan Temperatur

IDENTIFIKASI Kolom Destilasi KD-01 Sieve Tray Tower 1 buah Kontinyu Memisahkan etanol dan air DATA DESAIN Top 0,600 Atm 67,096 oC

Bottom 1,00 Atm 98,787 oC

KOLOM Tinggi kolom Material Diameter Tray spacing Jumlah tray Tebal silinder Tebal head

Downcomer area Active area Hole Diameter Hole area Tinggi weir Panjang weir Tebal pelat Pressure drop per tray

12,541 m Carbon Steel Top 1,435 M 0,500 M 8,000 Buah 0,004 M 0,004 M PELAT Top 2 0,213 m 2 1,350 m 5,000 Mm 2 0,135 m 50,000 Mm 1,143 M 5,000 Mm 110,850 mm liquid

Bottom 1,095 m 0,500 m 10,000 buah 0,004 m 0,004 m Bottom 2 0,113 m 2 0,715 m 5,000 mm 2 0,072 m 50,000 mm 0,832 m 5,000 mm 68,653 mm liquid

Tipe aliran cairan Desain % flooding Jumlah hole

17.

Single pass 85,000 % 6.879,000 Buah

Single pass 85,000 % 3.646,000 Buah

KOLOM DESTILASI - 02 (KD-02)

Dengan Perhitungan yang sama untuk Kolom Distilasi selanjutnya analog dengan perhitungan Kolom Distilasi KD-01. Nama Alat Alat Kode Jenis Jumlah Operasi Fungsi

Tekanan Temperatur

IDENTIFIKASI Kolom Destilasi KD-02 Sieve Tray Tower 1 buah Kontinyu Memisahkan etanol dan etil asetat DATA DESAIN Top Bottom 0,370 atm 1,00 Atm 55,071 oC 78,335 oC KOLOM

Tinggi kolom Material Diameter Tray spacing Jumlah tray Tebal silinder Tebal head

Downcomer area Active area Hole Diameter Hole area Tinggi weir

20,040 m Carbon Steel Top 1,245 m 0,500 m 6,000 Buah 0,284 m 0,232 m PELAT Top 2 0,146 m 2 0,924 m 5,000 Mm 2 0,092 m 50,000 mm

Bottom 2,916 m 0,500 m 32,000 buah 0,005 m 0,005 m Bottom 2 0,801 m 2 5,073 m 5,000 mm 2 0,507 m 50,000 mm

Panjang weir Tebal pelat Pressure drop per tray Tipe aliran cairan Desain % flooding Jumlah hole 18.

0,946 m 5,000 mm 323,196 mm liquid Single pass 85,000 % 44.415,000 Buah

2,216 m 5,000 mm 481,872 mm liquid Single pass 85,000 % 51.609,000 Buah

KOMPRESOR- 01 (K-01)

Fungsi

: Untuk menaikkan tekanan keluaran MP-01

Tipe

: Centrifugal Compressor

Gambar

:

1. Data : Laju alir massa (w)

=

57407,4074

Densitas ()

=

9,3578 kg/m3

=

1,0004

k =

Cp campuran Cp cam  R

2. Kondisi Operasi : Tekanan, Pin Pout Temperatur, Tin 3. Volumetric Flowrate, Q Q

=

W 

= 1 atm = 5 atm = 118,100 0C

kg/jam

= 3611,3592 ft3/min 4. Power Kompresor (Pw) Pw

0,0643 k T Q1 = 520 (k  1) 

 P  2  P1

  

( k  1) / k

  1 

(McCabe, 2009)

Dimana : K

= 1,0004



= 0,8000

P2

= 5 atm

P1

= 1 atm

Q

= 3611,3592 ft3/min

Pw

0,0643 .(1,2071) .(91). (15102,3040 = 520 (1,2071  1) 0,80

)  2     1 

1, 2071 1 / 1, 2071

  1 

= 93,3970 Hp 5.

Rasio Kompresi, Rc Rc

= (Pout / Pin)

(E. E. Ludwig, 1999)

= (5 atm / 1 atm) = 5,000 Jumlah stage, n

=1

Rc perstage

= (Rc)1/n = (5)1/1 = 5

6.

Pada Stage 1 Pi

= 1 atm

RC1

= (Po* / Pi)

Maka : Po*

= Rc1 x Pi = 5 x 1 atm = 5,000 atm

(E. E. Ludwig, 1999)

Temperatur yang keluar dari Kompressor stage 1 : T2





= T1  P2  p 

1

( K 1) / Kn

(E. E. Ludwig, 1999)



= TinRc(K-1)/Kn = 118,1749 oC 7.

Kapasitas Kompressor Laju alir volumetrik

= 3611,3592 ft3/min

faktor keamanan

= 10%

maka : kapasitas kompressor

= (100% + 10%) x 3611,3592 ft3/min = 3972,4951 ft3/min IDENTIFIKASI

Nama Alat

Kompresor 01

Kode Alat

K-01

Jumlah

1 buah

Fungsi

Untuk mengalirkan & menaikkan tekanan gas dari Vaporizer-01 menuju Mixing Point-02 DATA DESIGN

Tipe Temperature design Tekanan design Kapasitas

Centrifugal Compressor 118,175

o

C

5 Atm 3972,495 ft3/min DATA MEKANIK

Temperatur keluar

118,175

o

C

Tekanan masuk

1 Atm

Tekanan keluar

5 Atm

Power Bahan konstruksi

93,397 Hp Carbon and Steel Alloy

19.

KOMPRESOR - 02 (K-02)

Dari perhitungan yang sama maka untuk kompressor selanjutnya dihitung dengan cara analog dengan perhitungan kompressor K-01. IDENTIFIKASI Nama Alat

Kompresor 02

Kode Alat

K-02

Jumlah

1 buah

Fungsi

untuk mengalirkan dan menaikkan tekanan gas dari mixing point-02 menuju reaktor-01 DATA DESIGN

Tipe Temperature design Tekanan design Kapasitas

Centrifugal Compressor 299,257

o

C

18 Atm 3972,495 ft3/min DATA MEKANIK

Temperatur keluar

118,175

o

C

Tekanan masuk

5 Atm

Tekanan keluar

18 Atm

Power Bahan konstruksi

299,257 Hp Carbon and Steel Alloy

20.

PARTIAL CONDENSOR - 01 (PC-01) Fungsi

: Mengubah sebagian fase crude etanol sebelum masuk FD-01

Tipe

: Shell and Tube Heat Exchanger

Gambar

: Aliran inlet Tube

Head

Fluida Panas

Shell

Rear End Aliran outlet

Water in

: Produk reaktor-01

W

= 66.804,593 kg/hr

= 147.278,740 lb/hr

T1

= 155 oC

= 311,000 oF

T2

= 155 oC

= 311,000 oF

Fluida Dingin

: Air Pendingin

w

= 85.202,804 kg/hr

= 187.839,805 lb/hr

t1

= 28 oC

= 82,400 oF

t2

= 50 oC

= 122,000 oF

Perhitungan design sesuai dengan literatur pada buku Donald Q. Kern (1965). 1. Beban Panas PC-01 Q

= 17.853.473,026 kJ/hr = 16.922.075,190 Btu/hr

2. LMTD

Fluida Panas (oF)

Fluida Dingin (oF)

Selisih

311,000

Suhu tinggi

122,000

189,000

311,000

Suhu rendah

82,400

228,600

Selisih

LMTD =

-39,600

t 2  t1 ln (t 2 / t1 )

= 288,600 oF Ft

=1

t

= 228,600 oF

(Fig.18, Kern)

3. Temperatur Rata-rata Tavg =

T1  T 2 2

= 311,000 oF tavg

=

t1  t 2 2

= 102,200 oF 4. Menentukan luas daerah perpindahan panas Asumsi UD = 80 Btu/hr.ft2.oF A

= =

(Tabel 8, Kern)

Q U D . t 16.922.075,190 80 𝑥 228,600

= 925,310 ft2 Karena A > 200 ft2, maka digunakan Shell & Tube Heat Exchanger 5. Spesifikasi tube dan shell 

Tube Side

= Aliran bahan baku asam asetat dari T-01

Panjang tube (L)

= 16 ft

Outside Diameter (OD) = 1,25 in BWG

= 18

Pass

=4

a”

= 0,2618 ft2/lin ft A L x a"

Jumlah tube, Nt

= =

925,310 16 𝑥 0,2618

= 220,900

Dari tabel. 9 Kern, didapat nilai yang mendekati Nt perhitungan adalah Nt

= 217



Corrected Coefficient, UD A

= Nt x L x a'' = 217 x 16 ft x 0,2618 ft2 = 908,969 ft2

UD

=

Q U D . t

= 81,44 karena nilai Ud perhitungan mendekati nilai Ud asumsi, maka data untuk shell : Shell

= Air pendingin

ID

= 29 inch

(Tabel 9, Kern)

Baffle Space (B = ID/2) = 14,50 inch Pass

=4

Pt

= 1,5625

in triangular pitch

6. Perhitungan desain bagian tube 

Flow Area/tube, a’t a’t

= 1,040 in2

(Tabel 10, Kern)

at

=

Nt  a ' t 144  n

(Pers. 7.48, Kern)

217 𝑥 1,040

=

144 𝑥 4

= 0,3918 ft2 

Laju Alir, Gt Gt

= W/at

=

147.278,740 0,3918

= 375.897,530 lb/hr.ft2 

Bilangan Reynold, Ret μ

= 0,0147 cp

ID

= 1,150 inch = 0,0958 ft

Ret

= ID.Gt/ μ

= 0,0355 lb/ft hr (Tabel 10, Kern)

= 1.014.706,845 

Dengan L/D = 166,956 diperoleh Jh = 1000



(Fig.24, Kern)

Nilai hi CP

= 0,0027 Btu/lb.oF

k

= 0,0610 Btu/hr ft.oF

 c.     k 

= 0,0016

 k   Cp   hi  J H      D  k 

1/ 3

    w

  

0 ,14

    Koreksi viskositas diabaikan, karena   w  hi

= 74,334 Btu/hr ft2 oF

hio

= hi x ID/OD

7. Perhitungan desain bagian shell ID = Diameter dalam shell

= 29 in

B = Baffle spacing

= 14,50 in

Pt = tube pitch

= 1,5625 in = Pt – OD = 1,5625 – 1 = 0,3125 in



Flow Area, as

=1

(Pers. 6.5, Kern)

= 68,387 Btu/hr.ft2.oF

C’ = Clearance

0 ,14

= ID x C’B/144 PT

as

=

29 x 0,3125 x 14,50 144 𝑥 1,5625

= 0,584 ft2 

Laju Alir, Gs Gs

= w/as =

427.400,031 𝑙𝑏/ℎ𝑟 0,584 ft2

= 731.814,560 lb/hr.ft2 

Bilangan Reynold, Res de

= 0,910 in

(Fiq.28 Kern)

De (Equivalent diameter) = 0,076 ft μ

= 0,664 cp

Res

=

= 1,607 lb/ft hr

GS De



= 34.536,454 Maka: jH 

= 120

(Fig.28, Kern)

Nilai ho Cp

= 0,0037 Btu/lb.oF

k

= 0,3642 Btu/hr ft.oF

 Cp.     k 

1

3

= 0,2536

Koreksi viskositas diabaikan karena tidak significant, maka diperoleh : ho

= jH . (k/De). (Cpμ/k)1/3 = 146,181 Btu / hr ft2 oF

8. Clean Overall Coefficient, UC UC

=

hio  ho hio  ho

= 46,591 Btu/hr.ft2.oF 9. Dirt Factor, Rd

Rd

=

U C U D U C U D

= 0,009 hr.ft2.oF/Btu

10. Pressure drop 

Bagian tube Untuk NRe

= 1.014.706,845

Faktor friksi

= 0,0001

s

= 0,1821

ΔPt

f Gt 2 L n = 5, 22 x 10 10 x De s f t

(Fig 26, Kern)

= 0,2482 psi V2 / 2g

= 0,02

ΔPr

= ( 4n/s ) ( V2/2g )

(Fig 27, Kern)

= 1,7572 Psi ΔPT

= ΔPt + ΔPr = 0,2482 psi + 1,7572 Psi = 2,0053 psi



Shell Side Re

= 34.536,455

f

= 0,0015

N+1

= 12 L / B = 158,900

Ds

= 2,4167 ft

s

=1

ΔPs

=

fGs2 Ds ( N  1) 5,22 1010 Desf s

= 7,793 psi

(Fig.29, Kern)

IDENTIFIKASI Nama Alat

Parsial Condensor-01

Kode

PC-01

Jumlah

1

Fungsi

Mengubah fase keluaran R-01

Tipe

Shell and Tube Heat Exchanger DATA DESAIN Uc

46,591

Btu/hr ft2 oF

Ud

81,438

Btu/hr ft2 oF

Rd Calculated

0,009

hr ft2 oF/Btu

TUBE SIDE Length OD

16

Ft

1,25

In

Passes

4

BWG

18

Pitch

1,5625

Nt ΔPT

217 2,005

in Triangular Pitch tubes Psi SHELL SIDE

ID

29

in

B

14,5

in

Passes ΔPs

4 7,793

psi

21.

POMPA- 01 (P-01)

Fungsi

: Mengalirkan larutan asam asetat menuju Vaporizer (VP-01)

Tipe

: Centrifugal pump

Bahan Konstruksi

: Carbon Steel

Gambar

:

Data Desain Temperatur, T

: 30 C

Flowrate, ms

: 2870,3704 kg/jam

Densitas fluida, 

: 884,600 kg/m3

Viskositas,

: 0,3993 cp

Tekanan uap, Puap

: 122,0567 mmHg

Faktor keamanan, f

: 10 %

1. Kapasitas Pompa, Qf mf

= (1+ f ) x ms = (1 + 0,01) x 2870,3704 lb/jam = 120,7106 lb/min

Qf

= =

mf

 120,7106 lb/min 58,553 𝑙𝑏/𝑓𝑡 3

= 2,0616 ft3/min = 0,0344 ft3/sec = 15,4216 gal/min

2. Menentukan Ukuran Pipa Diameter Pipa Untuk aliran turbulent yang mempunyai range viskositas 0,02 – 20 cp maka digunakan rumus diameter dalam optimum pipa Dopt = 3,9 Qf 0,45 x 0,13

(Peter, 1991)

= 3,9 x (0,0344 ft3/sec)0,45 x 58,553 lb/ft3) 0,13 = 1,4523 in Dari tabel 10-18 Properties of steel pipe, Perry's chemical Engineers' Handbook, hal 10-72 – 10-74, dimensi pipa yang digunakan adalah : a) Untuk Suction Pipe IPS

= 2,5 in

SN

= 40

ID

= 2,469 in

OD

= 2,875 in

Ls

=3m

a”

= 1,704 in2

b) Untuk Discharge Pipe IPS

= 2 in

SN

= 40

ID

= 2,067 in

OD

= 2,375 in

Ld

=7m

a”

= 0,023 in2

3. Perhitungan Pada Suction a. Suction friction loss Suction velocity Vs

=

Qf a"

= 1,474 ft/sec = 5308,748 ft/jam

V2 2 gc

= 0,0338 ft. lbf/lb

Reynold Number NRe

=

D .V . 



= 90.086,3962 Material yang digunakan untuk konstruksi pipa adalah “Commercial Steel Pipe” Dari figure 14-1. Fanning friction factors for long straight pipes. Peter, hal 482, diperoleh : Equivalent roughness,  = 0,0002 ft

 D

=

0,0002 𝑓𝑡 0,256 𝑓𝑡

= 0,0009 Pada NRe = 90.086,3962 dan ε/D = 0,0009, dari figure 14-1. Fanning friction factors for long straight pipes. Peter, hal 482, diperoleh : fanning factor, f = 0,0045

b. Skin friction loss, Hfs

H

fs



2  f  L V2 x D gc

(Peter, 1991)

Equivalentlength dari fitting dan valve diperoleh dari Tabel II.1 Alat Industri Kimia, Prof. Dr. Ir. Syarifuddin Ismail, hal 35 : Elbow 90o std

: 32

Gate valve

:7

jadi equivalent length dari fitting dan valve = 1 elbow 90o std + 1 gate valve = 2 (32)+ 1 (7) = 71

L

= Ls + (Lfitting . ID) = 0,0009 ft

Maka :

2  f  L V2 H  x fs D gc = 0,069 ft. lbf/lb c. Sudden Contraction Friction Loss, Hfc

H

fc



Kc V 2  2α gc

(Peter, 1991)

Dimana : Kc

= 0,4 (1 – Sb / Sa)

Sa

= Luas penampang 1

= A>>>

Sb

= Luas penampang 2,

= A <<<



= 1 untuk aliran turbulent

Sb/ Sa diabaikan karena luas Sa sangat besar dibandingkan dengan luas Sb Maka : Kc

= 0,4 (1– A2/A1 ) = 0,4 (1-1) = 0,4

H

fc

 Kc x

1 V2 x 2  gc

= 0,0068 ft. lbf/lb

d. Fitting dan Valve Friction Loss, Hff

H

ff

Kf x

V2 2 gc

(Pers. II.7, Syarifuddin)

Nilai Kf diperoleh dari Tabel II.2 Alat Industri Kimia, Prof. Dr. Ir. Syarifuddin Ismail, hal 35 : Elbow 90o std

: 0,9

Gate valve, wide open : 0,2 Kf = 2 elbow 90o std + 1 gate valve

= 1 (0,9) + 1 (0,2) =2

V2 H Kf x ff 2 gc = 0,0676 ft.lbf/lb

e. Total Suction Friction Loss, Hf suc Hf suc

= Hfs + Hfc + Hff = 0,1177 ft. lbf/lb D

A

VZ-01

T-01

C B

P-01

f. Suction Head, Hsuc

Pa  Pb





2 2 g Za  Zb  Va Vb  Hf gc 2 g

Za

=3m

Zb

=0m

Static suction, Zs

= Za – Zb

g/gc

= 1 lbf/lb

static suction head, Hs

=

= 9,8425 ft

g Za  Zb gc

= 1 lbf/lb x 9,8425 ft = 9,8425 ft. lbf/lb

Pressure head, Hp : Pa

= 1 atm

= 14,6960 psi = 2.116,2240 lbf/ft2

Pa 

= 72,2841 ft. lbf/lb

Velocity head, Hv Va – Vb = 0 Hv

= 0 ft. lbf/lb

Maka :

Pa  Pb



g Va 2 Vb2  Za  Zb   Hf gc 2 g

g Va 2  Vb2 ( Za  Zb) + = + - Hf gc   2 g

Pb Pb



Pa

= 82,0090 ft. lbf/lb

Pb = 4801,8676 lbf/ft2 = 33,3463 psi g. Net Positive Suction Head (NPSH) Vapor Pressure Corection, Hp uap Hp uap

=

Puap



= 67,3031 ft/lbf/lb Total NPSH

= Hsuc - Hp uap = 14,7059 ft/lbf/lb

4. Perhitungan Pada Discharge a. Discharge friction loss Discharge velocity Vd

=

Qf a"

= 2,4299 ft/sec = 8747,7953 ft/jam

V2 2g c

= 0,0918 ft. lbf/lb

Reynold Number, NRe =

D .V . 



= 115624,8347 Material yang digunakan untuk konstruksi pipa adalah “Commercial Steel Pipe” Dari figure 14-1. Fanning friction factors for long straight pipes. Peter, hal 482, diperoleh : Equivalent roughness,  = 0,0002 ft

 D

= 0,0011

Pada NRe = 115.501,3240 dan ε/D = 0,001, dari figure 14-1. Fanning friction factors for long straight pipes. Peter, hal 482, diperoleh : fanning factor, f = 0,0045

b. Skin friction loss, Hfs

2  f  L V2 H  x fs D gc

(Peter, 1991)

Equivalent length dari fitting dan valve diperoleh dari Tabel II.1 Alat Industri Kimia, Prof. Dr. Ir. Syarifuddin Ismail, hal 35 : Elbow 90o std

: 32

Gate valve

:7

jadi equivalent length dari fitting dan valve

= 2 elbow 90o std + 1 gate

valve = 71 L

= Ls + (Lfitting . ID)

= 0,200 ft Maka :

2  f  L V2 H  x fs D gc = 0,200 ft. lbf/lb c. Sudden Contraction Friction Loss, Hfc

H

fc



Kc V 2  2α gc

(Peter, 1991)

Dimana : Kc = 0,4 (1 – Sb / Sa) Sa

= Luas penampang 1

= A>>>

Sb = Luas penampang 2,

= A <<<



= 1 untuk aliran turbulent

Sb/ Sa diabaikan karena luas Sa sangat besar dibandingkan dengan luas Sb Maka : Kc

= 0,4 (1 – A2/A1 ) = 0,500

1 V2 H  Kc x x fc 2  gc = 0,0229 ft. lbf/lb d. Fitting dan Valve Friction Loss, Hff

H

ff

Kf x

V2 2 gc

(Pers. II.7, Syarifuddin)

nilai Kf diperoleh dari Tabel II.2 Alat Industri Kimia, Prof. Dr. Ir. Syarifuddin Ismail, hal 35 : Elbow 90o std

: 0,9

Gate valve

: 1,8

jadi nilai Kf

= 2 elbow 90o std + 1 gate valve =2

H

ff

Kf x

V2 2 gc

= 0,0918 ft . lbf/lb

e. Total Discharge Friction Loss, Hf dis Hf dis = Hfs + Hfc + Hff = 0,3147 ft. lbf/ lb D

A

V-01

T-01

C B

P-01

g. Suction Head, Hsu

Pc  Pd





2 2 g Zc  Zd  Vc Vd  Hf gc 2 g

Zc

=0m

Zd

=4 m

Static suction, Zs

= Zd – Zc = 13,1234 ft

g/gc

= 1 lbf/lb

static suction, head, Hs

=

g Zd  Zc  gc

= 1 lbf/lb x 13,1234 ft = 13,1234 ft. lbf/lb Pressure head, Hp :

Pd

= 1,1 atm

Pd 

=

2327,838 lbf/ft2 64,835 lb/ft 3

= 79,5125 ft. lbf/lb

Velocity head, Hv Vc – Vd = 0 Hv

= 0 ft. lbf/lb

Maka :

Pc  Pd

 Pc

 Pc

 Pc

=



Pd



2 2 g Zc  Zd  Vc Vd  Hf gc 2 g

+

g Vd 2 Vc 2 ( Zd  Zc) + - Hf gc 2 g

= 92,3212 ft. lbf/lb = 92,3212 = 5405,6775 = 37,5394

ft. lbf/lb x lbf/ft2 Psi

64,835 lb/ft3

h. Net Positive Suction Head (NPSH) Vapor Pressure Corection, Hp uap Hp uap

=

=

Puap

 53,619 lbf/ft 2 64,835 lb/ft 3

= 0,827 ft Total NPSH

= Hsuc - Hp uap = (13,1234 ft – 0,827 ft) = 12,2964 ft

5. Differential Pressure (Total Pump), ΔP

Differential pressure = Discharge pressure – Suction pressure = 4,1931 psi 6. Total Head = Discharge head – Suction head

Total head

= 10,3122 ft. lbf/lb 7. Effisiensi Pompa,  Kapasitas pompa, Qf = 15,4216 gal/min Efisiensi pompa diperoleh dari figure 14-37 efficiencies of centrifugal pump, Peters 4th edition hal 520 diperoleh: Effisiensi pompa,  = 61 % 8. Break Horse Power (BHP) Persamaan Bernoulli : Ws

=

P



 Z 

V 2  H f 2 gc

= 10,3122 ft.lbf/lb

m f  Ws p

BHP

=

BHP

= 2894,8665 ft. lbf/min = 0,0877 Hp

9. Requirement Driver (Besarnya Tenaga Pompa) MHP =

BHP Effisiensi motor

Dari gambar 14-38, efficiencies of three-phase motor, Peter (hal 521) diperoleh : Effisiensi motor

= 80%

Jadi : MHP

= 0,1097 Hp

Dipilih pompa

= 1 Hp

IDENTIFIKASI Fungsi

Mengalirkan larutan etil asetat menuju Mixing Point (MP-01)

Tipe

Centrifugal Pump

Temperatur. oC

30,000

Densitas. kg/m3

937,928

Laju alir massa. kg/jam

2986,549

Viskositas. Cp

0,245

Tekanan uap. Psi

1415,262

Safety factor. %

10%

Kapasitas pompa. gal/min

15,421

Volumetric Flowrate. ft3/det

0,034 SUCTION

DISCHARGE

NPS. In

2

1,5

SN

40

40

ID. In

2,067

1,610

OD. In

2,375

1,900

3

7

Velocity.ft/s

1,475

6,835

Total friction loss. ft, lbf/lb

0,118

0,315

Tekanan operasi. Psi

33,346

37,539

L. m

NPSH. ft, lbf/lb

14,706

Required motor driver. Hp

1,000

Jumlah

1 buah

Bahan

Carbon steel

Dengan perhitungan yang sama maka untuk pompa selanjutnya dilakukan perhitungan cara analog dengan perhitungan pompa P-01. 22.

POMPA- 02 (P-02) IDENTIFIKASI

Fungsi

Mengalirkan larutan asam asetat menuju Mixing Point (MP-01)

Tipe

Centrifugal Pump

Temperatur. oC

30,000

Densitas. kg/m3

1.054,000

Laju alir massa. kg/jam

54.537,037

Viskositas. Cp

0,909

Tekanan uap. Psi

19,256

Safety factor. %

10%

Kapasitas pompa. gal/min

250,599

Volumetric Flowrate. ft3/det

0,558 SUCTION

DISCHARGE

NPS. In

3

1,5

SN

40

40

ID. In

3,068

1,610

OD. In

3,500

1,900

4

7

Velocity.ft/s

10,884

6,835

Total friction loss. ft, lbf/lb

3,890

0,315

Tekanan operasi. Psi

17,416

21,049

L. m

NPSH. ft, lbf/lb

37,299

Required motor driver. Hp

1,000

Jumlah

1 buah

Bahan

Carbon steel

23.

POMPA- 03 (P-03) IDENTIFIKASI

Fungsi

Mengalirkan larutan dari Kd-01 yang masuk ke Reboiler menuju Utilitas

Tipe

Centrifugal Pump

Temperatur. oC

98,787

Densitas. kg/m3

166,458

Laju alir massa. kg/jam

18.163,581

Viskositas. Cp

0,987

Tekanan uap. Psi

1.480,708

Safety factor. %

10%

Kapasitas pompa. gal/min

528,476

Volumetric Flowrate. ft3/det

1,177 SUCTION

DISCHARGE

NPS. In

3,5

1,5

SN

40

40

ID. In

3,55

1,610

OD. In

4

1,900

L. m

5

7

Velocity.ft/s

17,14

6,835

Total friction loss. ft, lbf/lb

14,084

0,315

Tekanan operasi. Psi

29,086

31,738

NPSH. ft, lbf/lb

6,288

Required motor driver. Hp

2,000

Jumlah

1 buah

Bahan

Carbon steel

24.

POMPA- 04 (P-04) IDENTIFIKASI

Fungsi

Mengalirkan Produk Etanol menuju tangki penyimanan produk

Tipe

Centrifugal Pump

Temperatur. oC

72,982

Densitas. kg/m3

169,987

Laju alir massa. kg/jam

34.490,947

Viskositas. Cp

0,9988

Tekanan uap. Psi

654,206

Safety factor. %

10%

Kapasitas pompa. gal/min

982,693

Volumetric Flowrate. ft3/det

2,189 SUCTION

DISCHARGE

NPS. In

3,5

3

SN

40

40

ID. In

3,55

3,07

OD. In

4

3,5

L. m

5

5

Velocity.ft/s

31,869

42,679

Total friction loss. ft, lbf/lb

35,432

49,512

Tekanan operasi. Psi

12,810

15,195

NPSH. ft, lbf/lb

2,170

Required motor driver. Hp

2,500

Jumlah

1 buah

Bahan

Carbon steel

25.

POMPA- 05 (P-05) IDENTIFIKASI

Fungsi

Mengalirkan Produk Etanol menuju refluks Kolom destilasi

Tipe

Centrifugal Pump

Temperatur. oC

55,071

Densitas. kg/m3

150,786

Laju alir massa. kg/jam

4.197,151

Viskositas. Cp

0,9988

Tekanan uap. Psi

654,206

Safety factor. %

10%

Kapasitas pompa. gal/min

134,810

Volumetric Flowrate. ft3/det

0,300 SUCTION

DISCHARGE

NPS. In

3,5

3

SN

40

40

ID. In

3,55

3,07

OD. In

4

3,5

L. m

5

5

Velocity.ft/s

4,372

5,855

Total friction loss. ft, lbf/lb

0,667

0,932

Tekanan operasi. Psi

15,296

18,646

NPSH. ft, lbf/lb

40,471

Required motor driver. Hp

1,000

Jumlah

1 buah

Bahan

Carbon steel

26.

POMPA- 06 (P-06) IDENTIFIKASI

Fungsi

Mengalirkan Produk Etanol menuju Tanki

Tipe

Centrifugal Pump

Temperatur. oC

55,071

Densitas. kg/m3

150,786

Laju alir massa. kg/jam

4.197,151

Viskositas. Cp

0,9988

Tekanan uap. Psi

654,206

Safety factor. %

10%

Kapasitas pompa. gal/min

250,310

Volumetric Flowrate. ft3/det

0,557 SUCTION

DISCHARGE

NPS. In

3,5

3

SN

40

40

ID. In

3,55

3,07

OD. In

4

3,5

L. m

5

5

Velocity.ft/s

8,118

10,871

Total friction loss. ft, lbf/lb

2,299

3,212

Tekanan operasi. Psi

15,189

18,497

NPSH. ft, lbf/lb

38,840

Required motor driver. Hp

1,000

Jumlah

1 buah

Bahan

Carbon steel

27.

REAKTOR - 01 (R-01)

Fungsi

: Sebagai tempat terjadinya reaksi antara Asam Asetat, Etil Asetat dan Hidrogen menjadi Etanol dan Air.

Tipe

: Fixed Bed Multi Bed Reactor

Operasi

: Kontinyu

Gambar

:

Kondisi Operasi : Temperatur, T

= 274,726 oC

Tekanan, P

= 18 atm

Konversi Asam Asetat bed 1 = 80 % Konversi Asam Asetat bed 2 = 90 % Konversi Etil Asetat bed 1 = 5 % Konversi Etil Asetat bed 2 = 95 % Laju alir massa, W

= 66804,593 kg/jam

BM rata-rata, BMav

= 38,477 kg/mol

Densitas Campuran, 

= 5,52 kg/m3

Viskositas Campuran, 

= 0,0073 kg/m.s

Data Katalis :

Bed 1 Nama katalis

= Platinum (Pt) –Timah (Sn)

Porositas (Φ)

= 0,42

Diameter katalis (dp)

= 0,003 mm = 3,000 m = 2845,8923 kg/m3

Densitas katalis Bed 2 Nama katalis

= CuZn-Al2O3

Porositas (Φ)

= 0,42

Diameter katalis (dp)

= 0,003 mm = 3,000 m = 2845,8923 kg/m3

Densitas katalis Reaksi yang terjadi

:

Bed 1 CH3COOH + 2H2

CH2H5OH + H2O

CH3COOH

C2H8O2 + H2O

+ C2H8O2

C2H8O2 + 2H2

2CH2H5OH

CH3COOH + 2H2

CH2H5OH + H2O

C2H8O2 + 2H2

2CH2H5OH

Bed 2

Perhitungan Pada Desain Reaktor Reaksi 1 1.

Volumetric Flowrate Umpan, Qf Qf Laju alir Volumetrik (Qf)

=

M



= 12.102,281 m3/jam = 3,3617 m3/s

2.

Konsentrasi mula-mula, C a) Konsentrasi umpan CH3COOH, [CAo]

Umpan masuk, (Fao)

= 907,133 kmol/jam

Konsentrasi [CAo]

=

Fao Qf

= 0,075 kmol/m3

b) Konsentrasi umpan H2, [CBo] Umpan masuk (Fbo)

= 4698,580 kmol/jam

Konsentrasi O2 [CBo]

=

Fbo Qf

= 0,389 kmol/m3 Reaksi 2 1.

Volumetric Flowrate Umpan, Qf Qf Laju alir Volumetrik (Qf)

=

M



= 12.102,281 m3/jam = 3,3617 m3/s

2.

Konsentrasi mula-mula, C a) Konsentrasi umpan CH3COOH, [CAo] Umpan masuk, (Fao)

= 181,426 kmol/jam

Konsentrasi [CAo]

=

Fao Qf

= 0,015 kmol/m3 b) Konsentrasi umpan C2H5OH, [CBo] Umpan masuk (Fbo)

= 725,706 kmol/jam

Konsentrasi O2 [CBo]

=

Fbo Qf

= 0,060 kmol/m3 Reaksi 3

1.

Volumetric Flowrate Umpan, Qf Qf Laju alir Volumetrik (Qf)

=

M



= 12.102,281 m3/jam = 3,3617 m3/s

2.

Konsentrasi mula-mula, C a) Konsentrasi umpan CH4H8O2, [CAo] Umpan masuk, (Fao)

= 68,870 kmol/jam

Konsentrasi [CAo]

=

Fao Qf

= 0,0057 kmol/m3 b) Konsentrasi umpan H2, [CBo] Umpan masuk (Fbo)

= 3247,168 kmol/jam

Konsentrasi O2 [CBo]

=

Fbo Qf

= 0,2683 kmol/m3 Reaksi 4 1.

Volumetric Flowrate Umpan, Qf Qf Laju alir Volumetrik (Qf)

=

M



= 12.102,281 m3/jam = 3,3617 m3/s

2.

Konsentrasi mula-mula, C a) Konsentrasi umpan CH3COOH, [CAo] Umpan masuk, (Fao)

= 145,141 kmol/jam

Konsentrasi [CAo]

=

Fao Qf

= 0,012 kmol/m3 b) Konsentrasi umpan H2, [CBo]

Umpan masuk (Fbo)

= 3240,281 kmol/jam

Konsentrasi O2 [CBo]

=

Fbo Qf

= 0,267 kmol/m3 Reaksi 5 1.

Volumetric Flowrate Umpan, Qf Qf Laju alir Volumetrik (Qf)

=

M



= 12.102,281 m3/jam = 3,3617 m3/s

2.

Konsentrasi mula-mula, C a) Konsentrasi umpan CH4H8O2, [CAo] Umpan masuk, (Fao)

= 65,427 kmol/jam

Konsentrasi [CAo]

=

Fao Qf

= 0,0054 kmol/m3 b) Konsentrasi umpan H2, [CBo] Umpan masuk (Fbo)

= 2979,0267 kmol/jam

Konsentrasi O2 [CBo]

=

Fbo Qf

= 0,2462 kmol/m3 3.

Residence Time

4.

Residence Time, = 0,4 _ 30 detik

(Sumber: US Patent 9,670119 B2)

Maka diambil, = 15 detik 5.

Menentukan Kinetika Reaksi Reaksi 1 : CH3COOH + 2H2

k1

Reaksi merupakan bentuk reaksi orde dua :

C2H5OH + H2O

-r1

= k1 . C A . C B

(Levenspiel, 1999)

Nilai k dapat dihitung menggunakan persamaan Arrhennius : k1

= A . e –E/RT

A

=

(Levenspiel, 1999)

 1 1  A B  N . 8 .R.T .    3 2  10   M A MB  2

Dimana diameter partikel : σA

= 4,3992 A

σB

= 6,0168 A = 6,0168 x 10-8 cm

= 4,3992 x 10-8 cm

(Welty, 2008 )

Berat Molekul MA

= 60,052 kg/kmol

MB

= 2,0160 kg/kmol

N (Bil. Avogadro)

= 6,02 x 1023 A

R

= 8,314

kJ/kmol.K

Menentukan Energi Aktivasi ∆Hf 298 CH3COOH

= -438.150,000 kJ/kmol

∆Hf 298 CH3OH

= 0,000 kJ/kmol

ECH3COOH

=  H f 298  R T = -442.705,041 kJ/kmol =  H f 298  R T

EH2

= -4.555,041 E

kJ/kmol

= (ECH3COOH + ECH3OH) /2 = -223.630,041

- E/RT

kJ/kmol

= -223.630,041 / (8,314 kJ/kmol.K) (423,15 K) = 49,095

Maka:  1 1   E / RT A  B  N =  2  103 8 .R.T . M  M .e   B   A 2

k1

k1 = 3,79 x 1012 cm3/kmol.s k1 = 3,79 x 106 m3/kmol.s

(Smith, J.M, 2001)

Diketahui dari perhitungan: CAo1

= 0,075

kmol/m3

CBo1

= 0,388

kmol/m3

X

= 0,800

k1

= 3,79 x 106 m3/kmol.s

Menghitung laju reaksi: -r1

= k1[CA][CB]2

-r1

= k1 . CAO2 (1 – X) (M – X)

-r1

= 9,592 x 108 kmol/m3.s

Reaksi 2 : k2

CH3COOH + C2H5OH

C4H8O2

Reaksi merupakan bentuk reaksi orde dua : -r2

= k2 . C A . C B

(Levenspiel, 1999)

Nilai k dapat dihitung menggunakan persamaan Arrhennius : k2

= A . e –E/RT

A

= A B  N

(Levenspiel, 1999)

 1 1  . 8 .R.T .   3  10  M A MB  2

 

2

Dimana diameter partikel : σA

= 2,573 A

σB

= 30,247 A = 3,0248 x 10-7 cm

= 2,5728 x 10-8 cm

(Welty, 2008 )

Berat Molekul MA

= 60,052 kg/kmol

MB

= 46,069 kg/kmol

N (Bil. Avogadro)

= 6,02 x 1023 A

R

= 8,314

kJ/kmol.K

Menentukan Energi Aktivasi ∆Hf 298 CH3COOH

= -438.150,000 kJ/kmol

∆Hf 298 C2H5OH

= -235,310 kJ/kmol

=  H f 298  R T

ECH3COOH

= -442.705,041 kJ/kmol =  H f 298  R T

EC2H5OH

= -4.790,351 kJ/kmol E

= (ECH3COOH + EC2H5OH) /2 = -223.747,696

- E/RT

kJ/kmol

= -223.747,696 / (8,314 kJ/kmol.K) (423,15 K) = 49,120

Maka:  1 1   E / RT A  B  N =  2  103 8 .R.T . M  M .e   B   A 2

k2

(Smith, J.M, 2001)

k2 = 2,313 x 1013 cm3/kmol.s k2 = 2,313 x 107 m3/kmol.s Diketahui dari perhitungan: CAo2

= 0,015

kmol/m3

CBo2

= 0,388

kmol/m3

X

= 0,050

k2

= 2,313 x 107 m3/kmol.s

Menghitung laju reaksi: -r2

= k2[CA][CB]2

-r2

= k2 . CAO2 (1 – X) (M – X)

-r2

= 2,6984 kmol/m3.s

Reaksi 3 : C4H8O2

k3

+ H2

C2H5OH

Reaksi merupakan bentuk reaksi orde dua : -r3

= k3 . C A . C B

(Levenspiel, 1999)

Nilai k dapat dihitung menggunakan persamaan Arrhennius : k3

= A . e –E/RT 2 A B  N  

2

 1 1  . 8 .R.T .   3  10  M A MB 

(Levenspiel, 1999)

A

=

Dimana diameter partikel : σA

= 2,2306 A

= 2,2306 x 10-8 cm

σB

= 5,3196 A

= 5,3196 x 10-8 cm

(Welty, 2008 )

Berat Molekul MA

= 88,206 kg/kmol

MB

= 2,0159 kg/kmol

N (Bil. Avogadro)

= 6,02 x 1023 A

R

= 8,314 kJ/kmol.K

Menentukan Energi Aktivasi ∆Hf 298 C4H8O2

= -426,800 kJ/kmol

∆Hf 298 H2

= 0,000 kJ/kmol

EC4H8O2

=  H f 298  R T = -4.981,841 kJ/kmol =  H f 298  R T

EH2

= -4.555,041 kJ/kmol E

= (ECH3COOH + EC2H5OH) /2 = -4.768,441

- E/RT

kJ/kmol

= -4.768,441 / (8,314 kJ/kmol.K) (423,15 K) = 1,0468

Maka: K3

 1 1   E / RT A  B  N .e 8 .R.T .    = 3 2   10  MA MB  2

2001) K3 = 4,467 x 1012 cm3/kmol.s K3 = 4,467 x 106 m3/kmol.s Diketahui dari perhitungan: kmol/m3

CAo1

= 0,0057

CBo1

= 0,2683 kmol/m3

X

= 0,050

(Smith, J.M,

= 4,467 x 106 m3/kmol.s

k3

Menghitung laju reaksi: -r3

= k1[CA][CB]2

-r3

= k1 . CAO2 (1 – X) (M – X)

-r3

= 2,6984 kmol/m3.s

Reaksi 4 : k4

CH3COOH + 2H2

C2H5OH + H2O

Reaksi merupakan bentuk reaksi orde dua : -r4

= k4 . C A . C B

(Levenspiel, 1999)

Nilai k dapat dihitung menggunakan persamaan Arrhennius : k4 A

= A . e –E/RT 2 = A B  N  

(Levenspiel, 1999)

 1 1  . 8 .R.T .   3  10  M A MB 

2

Dimana diameter partikel : σA

= 2,2306 A

= 2,2306 x 10-8 cm

σB

= 5,3196 A

= 5,3196 x 10-8 cm

Berat Molekul MA

= 60,052 kg/kmol

MB

= 2,0159 kg/kmol

N (Bil. Avogadro)

= 6,02 x 1023 A

R

= 8,314 kJ/kmol.K

Menentukan Energi Aktivasi ∆Hf 298 CH3COOH

= -438,150 kJ/kmol

∆Hf 298 H2

= 0,000 kJ/kmol

ECH3COOH

=  H f 298  R T = -4.993,191 kJ/kmol

EH2

=  H f 298  R T = -4.555,041 kJ/kmol

(Welty, 2008 )

E

= (ECH3COOH + EC2H5OH) /2 = -4.774,116

- E/RT

kJ/kmol

= -(-4.774,116 / (8,314 kJ/kmol.K) (423,15 K) = 1,0481

Maka:

 1 1   E / RT A  B  N .e 8 .R.T .   3 2   10  MA MB  2

k4 = 

(Smith, J.M, 2001)

k4 = 2,753 x 10-9 cm3/kmol.s k4 = 2,753 x 10-15 m3/kmol.s Diketahui dari perhitungan: kmol/m3

CAo4

= 0,0120

CBo4

= 0,2677 kmol/m3

X

= 0,900

k4

= 2,753 x 10-15 m3/kmol.s

Menghitung laju reaksi: -r4

= k4[CA][CB]2

-r4

= k4 . CAO2 (1 – X) (M – X)

-r4

= 6,006-13 kmol/m3.s

Reaksi 5 : k5

C4H8O2 + 2H2

C2H5OH + H2O

Reaksi merupakan bentuk reaksi orde dua : -r5

= k5 . C A . C B

(Levenspiel, 1999)

Nilai k dapat dihitung menggunakan persamaan Arrhennius : k5 A

= A . e –E/RT 2 = A B  N  

2

 3  10

(Levenspiel, 1999)  1 1  . 8 .R.T .   M A MB 

Dimana diameter partikel : σA

= 6,4187 A

= 6,4187 x 10-8 cm

σB

= 5,3196 A

= 9,1663 x 10-7 cm

Berat Molekul

(Welty, 2008 )

MA

= 88,206 kg/kmol

MB

= 2,0159 kg/kmol

N (Bil. Avogadro)

= 6,02 x 1023 A

R

= 8,314 kJ/kmol.K

Menentukan Energi Aktivasi ∆Hf 298 C4H8O2

= -426,800 kJ/kmol

∆Hf 298 H2

= 0,000 kJ/kmol

EC4H8O2

=  H f 298  R T = -4.981,841 kJ/kmol =  H f 298  R T

EH2

= -4.555,041 kJ/kmol E

= (ECH3COOH + EC2H5OH) /2 = -4.768,441

- E/RT

kJ/kmol

= -4.768,441 / (8,314 kJ/kmol.K) (423,15 K) = 1,0468

Maka: k5

 1 1   E / RT A  B  N =  2  103 8 .R.T . M  M .e   B   A 2

k5 = 9,976 x 10-7 cm3/kmol.s k5 = 9,976 x 10-13 m3/kmol.s Diketahui dari perhitungan: kmol/m3

CAo5

= 0,0054

CBo5

= 0,2462 kmol/m3

X

= 0,05

K5

= 9,976 x 10-13 m3/kmol.s

Menghitung laju reaksi: -r5

= k5[CA][CB]2

-r5

= k5 . CAO2 (1 – X) (M – X)

-r5

= 6,1034-15 kmol/m3.s

5. Menentukan Volume Reaktor

(Smith, J.M, 2001)

Mass flow rate (W)

= 66804,593 kg/jam

Densitas (ρ)

= 5,520

kg/m3

Volumetric flow rate (Q) = 12.102,281 m3/jam

1) Volume reaktor =.Q

Vf

= 0,00417 jam × 12.102,281 m3/jam = 50,4262 m3 Vr

= (1+f) Vf = 60,5114 m3

2) Menentukan Volume Katalis (Vk) dan Berat Katalis (Wk) Menghitung Volume Katalis Bed 1 f

= 0,420

VTR

= 60,5114 m3

Vk

= (1– Φ). Vr = (1– 0, 420) . 60,5114 m3

Vk

= 35,0966 m3

Menghitung Berat Katalis ρk

= 770 kg/m3

Wk

= ρk. Vk

Wk

= 2845,892 kg/m3 x 35,0966 m3 = 99.881,189 kg

Bed 2 f

= 0,450

VTR

= 60,5114 m3

Vk

= (1– Φ). Vr = (1– 0,450) . 60,5114 m3

Vk

= 33,2813 m3

Menghitung Berat Katalis

ρk

= 1,7462 kg/m3

Wk

= ρk. Vk

Wk

= 1,7462 kg/m3 x 33,2813 m3 = 58.127,6082 kg

3) Volume Total Reaktor (VRt) VRt

= Vr +Vk

VRt

= 60,5114 m3 + (35,0966 + 33,2813 m3) = 128,8893 m3

4) Menentukan Ukuran Kolom Reaktor a)

Diameter Accumulator, D Volume Silinder, Vs Vs

= (π/4) D2 HS

Dimana : HS = ¾ D

= 3/8 (π D3) Volume Ellipsoidal, Ve Ve

= (π/6) D2 He

Ve

= (1/24) π D3

Volume Total, VR VR

= VS + 2Ve

VR

= 3/8. π D3 + 2 (1/24) π .D3

VR

= 3/8 .π D3 + 2/24 .π D3

VR

= 1,4392 D3

Diameter Tangki, DR VR

DR

=(

DR

= 4,4741 m

1,4392

)1/3

b) Tinggi Silinder, Hs Hs

= 3/2 DR

Hs

= 6,7111 m

c) Tinggi Ellipsoidal, He He

= ¼ DR = 1,1185 m

Dimana : He = ¼ D

d) Tinggi Reaktor, HR HR

= Hs + 2He

HR

= 6,7111 m + 1,1185 m = 8,9481 m

10) Menentukan Tebal Dinding Reaktor, t t

(Peters, 1991)

P . Da C .2SE  0,2 . P

Dimana : Tekanan design, P

= 18 atm

Jari-jari kolom, ri

= 2,237 m

Working stress allowable, S

= 932,227 atm

(Peters, 1991)

Welding joint efficiency, E

= 0,85

(Peters, 1991)

Tebal korosi yang diizinkan, C = 0,003 m Maka : t

P . Da C .2SE  0,2 . P

t = 0,0547 m 11) Outside Diameter Reaktor, OD OD = ID + 2.t = 4,4741 m + 2 . (0,0547 m) OD = 4,5823 m 12) Menentukan Ketebalan Jaket Pendingin id

OD

H

Keterangan : Outside Diameter, OD

= 4,582 m

Tinggi Silinder, H

= 8,948 m

Diameter Reaktor beserta Jaket Bagian Dalam, id Flowrate Cooling Water (m)

= 802.808,878 kg/jam

Densitas Cooling Water ()

= 1000 kg/m3

Residence Time

= 0,0042 jam

Volumetric Flowrate

=

m



= 3,345 m3/jam Volume Jaket Cooling System

= Volumetric Flowrate x Residence Time = 3,345 m3/jam x 0,0042 jam = 0,0140 m3

VJaket

= (Volume Reaktor + Jaket) – (Volume Reaktor)

VReaktor + Jaket

=

1 1  (id ) 2 H   (id ) 3 4 24

VReaktor

=

1 1  (OD ) 2 H   (OD ) 3 4 24

Maka : VJaket

1 1 1  1  =   (id ) 2 H   (id ) 3     (OD ) 2 H   (OD ) 3  24 24 4  4 

0,0140 m3

=

Id

= 4,6181 m





Maka, Tebal Jaket



1 1  H id 2  OD 2   id 3  OD 3 4 24

= id – OD = 4,6181 m – 4,582 m = 0,0358 m = 3,5784 cm



IDENTIFIKASI Nama Alat

Reaktor

Alat Kode

R-01

Jenis

Multi Bed Fixed Bed Reactor

Jumlah

1 buah

Operasi

Continue Tempat bereaksi asam asetat, etil asetat dan hidrogen

Fungsi

dengan bantuan katalis Pt-Sn dan CuZn-Al2O3 membentuk crude ethanol KONDISI OPERASI

Tekanan

18 atm

Temperatur

274,726 OC Bed 1

Katalis

Pt-Sn

Volume

35,0996 m2

Material

Carbon Steel Bed 2

Katalis

CuZn- Al2O3

Material

33,2813 m2

Material

Carbon Steel DATA DESAIN

Tekanan

18 atm

Temperatur

274,726 OC

Diameter

4,474 m

OD

4,582 m

Tinggi

8,948 m

Tebal Dinding Reaktor

0,054 m

Tebal Jaket Pendingin

0,036 m

Bahan Konstruksi

Carbon Steel

28.

REBOILER- 01 (RB-01) Fungsi

: Untuk memanaskan kembali residu KD - 01

Tipe

: Shell and Tube Heat Exchanger

Gambar

:

Aliran inlet Shell

Tube

Rear End

Head

Aliran outlet

Fluida Panas

:

Water in

Steam

W

= 28,436.666kg/hr

= 62,692.044 lb/hr

T1

= 350 oC

= 662oF

T2

= 350 oC

= 662oF

Fluida Dingin

:

Residu KD - 01

w

= 59,283.325 kg/hr

= 130,697.205 lb/hr

t1

= 98,787oC

= 209,817oF

t2

= 98,787oC

=209,817oF

Perhitungan design sesuai dengan literatur pada buku Donald Q. Kern (1965). 22. Beban Panas RB-01 Q

= 25,476,409.467 kJ/hr = 24,146,872.086 Btu/hr

23. LMTD

Fluida Panas (oF)

Fluida Dingin (oF)

Selisih

662

Suhu tinggi

209,817

452,183

662

Suhu rendah

209,817

452,183

Selisih

-

LMTD = 452,183 oF Ft

=1

t

= 452,183 oF

(Fig.18, Kern)

24. Temperatur Rata-rata Tavg =

T1  T 2 2

= 662oF tavg

=

t1  t 2 2

= 209,817oF

25. Menentukan luas daerah perpindahan panas Asumsi UD = 156 Btu/hr.ft2.oF A

=

Q U D . t

=

24,146,872.086 60  452,183

(Tabel 8, Kern)

= 890,011 ft2 Karena A > 200 ft2, maka digunakan Shell & Tube Heat Exchanger

26. Spesifikasi tube dan shell 

Tube Side

= Residu KD - 01

Panjang tube (L)

= 18,1 ft

Outside Diameter (OD) = 1 in BWG

= 18

Pass

=2

a”

= 0,302 ft2/lin ft A = L x a" 890,011 = 18,1 0,302

Jumlah tube, Nt

= 162,821 Dari tabel.9 Kern, didapat nilai yang mendekati Nt perhitungan adalah Nt

= 162,821



Corrected Coefficient, UD A

= Nt x L x a'' = 162,821 x 18,1 ft x 0,302 ft2 = 890,011

UD

=

Q U D . t

= 60.000

karena nilai Ud perhitungan mendekati nilai Ud asumsi, maka data untuk shell : Shell

= steam

ID

= 31 inch

(Tabel 9, Kern)

Baffle Space (B = ID/2) = 15,5 inch Pass

=2

Pt

= 1,25 in triangular pitch

27. Perhitungan desain bagian tube 

Flow Area/tube, a’t a’t

= 0,302 in2

(Tabel 10, Kern)

at

=

Nt  a ' t 144  n

(Pers. 7.48, Kern)

=

162,821 0,302 144  2

= 0,170 ft2 

Laju Alir, Gt Gt

= W/at =

130,697.205 0,170

= 220.955,906 lb/hr.ft2 

Bilangan Reynold, Ret μ

= 0,379 cp

ID

= 0,902 inch = 0,075 ft

Ret

= ID.Gt/ μ =

= 0,917 lb/ft hr (Tabel 10, Kern)

0,075 766,764.180 0,917

= 62,890.438 

Dengan L/D = 240,808 diperoleh Jh = 200



(Fig.24, Kern)

Nilai hi CP

= 41,743 Btu/lb.oF

k

= 0,347 Btu/hr ft.oF

 c.     k 

= 4,796

 k   Cp   hi  J H      D  k 

1/ 3

      w  

0 ,14

    Koreksi viskositas diabaikan, karena   w  hi

0 ,14

  0,419  1/ 3 = 200     6,501  0,075 

=1

= 7.256,248 Btu/hr ft2 oF hio

= hi x ID/OD

(Pers. 6.5, Kern)

= 6.545,136 Btu/hr.ft2.oF 28. Perhitungan desain bagian shell ID = Diameter dalam shell

= 31 in

B = Baffle spacing

= 15,5 in

Pt = tube pitch

= 1,25 in

C’ = Clearance

= Pt – OD = 1,25 – 1 = 0,250 in



Flow Area, as as

= ID x C’B/144 PT =

31 0,250  15,5 144 1,25

= 0,667 ft2 

Laju Alir, Gs Gs

= w/as =

147.457,379 0,667

= 220.955,906 lb/hr.ft2 

Bilangan Reynold, Res de

= 0,720 in

= 0,06 ft

μ

= 0,022 cp

= 0,054 lb/ft hr

Res

= =

GS De



93,940.211x0,06 0,054

= 103,873.817

(Fiq.28 Kern)

Maka: jH 

= 350

(Fig.28, Kern)

Nilai ho Cp

= 282,970 Btu/lb.oF

k

= 0,028 Btu/hr ft.oF

 Cp.     k 

1

3

= 8,148

Koreksi viskositas diabaikan karena tidak significant, maka diperoleh : ho

= jH . (k/De). (Cpμ/k)1/3\ = 350 x

0,028 x8,148 0,06

= 1.349,169 Btu / hr ft2 oF

29. Clean Overall Coefficient, UC UC

=

hio  ho hio  ho

=

3,990.751x1,349.169 3,990.751  1.349,169

= 1.008,292 Btu/hr.ft2.oF

30. Dirt Factor, Rd Rd

=

U C U D U C U D

=

1.008,292  60,000 1.008,292 x60,000

= 0,016 hr.ft2.oF/Btu

31. Pressure drop 

Bagian tube Untuk NRe

= 62,890.438

Faktor friksi

= 0,00023

(Fig 26, Kern)

s

= 0,424

ΔPt

=

V2 / 2g

= 0,015

ΔPr

= ( 4n/s ) ( V2/2g )

f Gt 2 L n 5, 22 x 10 10 x De s f t

= 2,943 psi

(Fig 27, Kern)

= 0,283 Psi ΔPT

= ΔPt + ΔPr = 2,943 psi + 0,283 Psi = 3,226 psi



Shell Side Re

= 103,873.817

f

= 0,0003

N+1

= 12 L / B

(Fig.29, Kern) = 12 x (217/15,5)

= 168,155 Ds

= 2,583 ft

s

=1

ΔPs

=

fGs2 Ds ( N  1) 5,22 1010 Desf s

= 0,367 psi

IDENTIFIKASI Nama Alat

Reboiler - 01

Kode

Rb – 01

Jumlah

1

Fungsi

Untuk memanaskan kembali residu KD – 01

Tipe

Shell and Tube Heat Exchanger DATA DESAIN Uc

1.008,292

Btu/hr ft2 oF

Ud

24,803

Btu/hr ft2 oF

Rd Calculated

0,039

hr ft2 oF/Btu

TUBE SIDE Length

18,10

Ft

OD

1

In

Passes

2

BWG

18

Pitch

1,250

Nt ΔPT

454 0,785

in Triangular Pitch Tubes Psi SHELL SIDE

ID

31

In

B

15,5

In

Passes ΔPs

2 2,031

psi

29.

REBOILER- 02 (RB-02) IDENTIFIKASI

Nama Alat

Reboiler - 01

Kode

Rb – 01

Jumlah

1

Fungsi

Untuk memanaskan kembali residu KD – 01

Tipe

Shell and Tube Heat Exchanger DATA DESAIN Uc

1.008,292

Btu/hr ft2 oF

Ud

24,803

Btu/hr ft2 oF

Rd Calculated

0,039

hr ft2 oF/Btu

TUBE SIDE Length

18,10

Ft

OD

1

In

Passes

2

BWG

18

Pitch

1,250

Nt ΔPT

454 0,785

in Triangular Pitch Tubes Psi SHELL SIDE

ID

31

In

B

15,5

In

Passes ΔPs

2 2,031

psi

30.

TANGKI- 01 (T-01) Fungsi

: Menampung bahan baku hidrogen

Tipe

: Spherical tank

BahanKonstruksi

: Carbon steel

Gambar

:

a. Data Temperatur, T

: 30 oC

Tekanan, P

: 5 atm

Lajualir massa, Ws

: 3.687,754 kg

Densitas, 

: 8,586 kg/m3

Faktor keamanan,f

: 10 %

Lama penyimpanan

: 3 hari

Jumlah

: 5 buah

b. KapasitasTanki,

W.t

Vt



Vt

= 30.924,564 m3



Kapasitas dalam satu tanki :

Vt(1 tanki)

=

Vt ' n tanki '

= 1.236,983 m3 Safety factor Vt’

= 10%

= (1+0,1) x Vt = 1,1 x 1.236,983 m3 = 1.484,379 m

C. Diameter Tangki Volume total, Vt = Volume bola = ( 4/3 x π R3 ) R3

= (V x 3)/(4 x 3,14) = (1.484,379 x 3)/(4 x 3,14) = 354,549 m3

R

= 7,078 m

D

= 2 x 14,155 m = 557,299 m

D. Tebal dinding tanki, t t

PxR C 2 S x E  0,2 x P

=

Dimana : P

= Tekanan design

= 220,440 psi

R

= Jari – jari kolom

= 151,496 in

S

= Working stress allowable

= 13.700 atm

Ej

= Welding joint efficiency

= 0,85

C

= Tebalkorosi yang diijinkan

= 0,013

t

= 0,892 in = 0,023 m

Outside diameter (OD)

= ID + t = 14,155 m+ 0,023 m = 14,178 m

IDENTIFIKASI Nama Alat

Tangki-01

Kode Alat

T-01

Jumlah

5 Unit

Fungsi

Tempat menyimpan bahan baku Hidrogen DATA DESAIN

Tipe

Spherical Tank

Kapasitas

1.484,379

m3

Tekanan

5

atm

Temperatur

30,000

0

Diameter

14,155

m

OD

14,178

m

Tebal Dinding

0,023

m

Bahan Konstruksi

Carbon steel

C

392.131,542

gal

31.

TANGKI - 02 (T-02) Fungsi

: Untuk menampung bahan baku Asam Asetat

Tipe

: Silinder vertical dengan head type ellipsoidal

Bahan Konstruksi

: Stainless Steel

Gambar

: He

Hs

Dt

A.

Data-data

:

Temperatur, T

: 30 OC

Tekanan, P

: 1 atm

Laju alir, Ws

: 54.537,037 kg/jam

Densitas, 

: 481,113 kg/m3

Faktor keamanan,f

: 10 %

Lama penyimpanan

: 3 hari

Jumlah

:5

B.

Kapasitas Tangki, Vt Laju alir massa



Vt

=

Vt

= 8.161,634 m3

x lama persediaan

Kapasitas dalam satu tanki :

Vt(1 tanki)

Vt ' ' = n tanki = 1.632,327 m3

Volume tangki,Vt

= (1 + f) x Vt = (100% + 10%) x 1.413,390 m3 = 1.795,559 m3

C.

Diameter Tangki Volume bagian silinder, Vs 2 =r H

Vs

H = 3/2 D 2

D 3      D =  2  2 

3  D3 = 8 = 1,178 D3 Volume bagian head, Vh Vh

𝜋

= 24 x D3

h=¼D

(Tabel 4, Peter, hal 538)

=0,131 D3 Jadi, Vt

= Vs + Vh = 1,178 D3 + 0,131 D3 = 1,308 D3

Dt

𝑉𝑡 1/3

= 1,308

= (1.795,559 m3/ 1,308)1/3 = 11,113 m = 218,758 in D.

Tinggi Tangki, Ht Tinggi Silinder

=H

= 3/2 D

Tinggi Head = h

= ¼ D = 2,778 m

= 16,669 m

Ht

=H+h = 16,669 m + 2,778 m = 19,448 m

E. Tebal dinding tangki,t

  P. r   C t    S . E  0.6 P 

(Table 4, hal 537,Peters and Timmerhaus)

Keterangan: P = Tekanan design

= 1 atm= 14,696 psi

R = Jari-jari vessel

= 208,504 in

S = Working stress allowable = 11.500 psi

(table 4, Peter, hal 538)

E = Joint effisiensi

= 0,85

(table 4, Peter, hal538)

C = Korosi maksimum

= 0,013 in

(table 6, Peter, hal 538)

Maka : t = 0,342 in = 0,009 m F.

Outside diameter, OD

OD

= D + 2t =11,113 m+ 2 (0,017) m = 11,130 m

IDENTIFIKASI Nama Alat

Tangki-02

Kode Alat

T-02

Jumlah

5 Unit

Fungsi

Tempat menyimpan Asam Asetat DATA DESAIN

Tipe

Silinder Vertikal dengan ellipsoidal head

Kapasitas

1.795,559

m3

Tekanan

1

Atm

Temperatur

30,00

0

Diameter

11,113

m

OD

11,130

m

Tinggi

19,448

m

Tebal Dinding

0,009

m

Bahan Konstruksi

Stainless steel

474.336,703

gal

212,078

in

63,804

ft

C

Dengan perhitungan yang sama maka untuk pompa selanjutnya dilakukan perhitungan cara analog dengan perhitungan pompa T-02. 32.

TANGKI - 03 (T-03) IDENTIFIKASI

Nama Alat

Tangki-03

Kode Alat

T-03

Jumlah

1 Unit

Fungsi

Tempat menyimpan Etil Asetat DATA DESAIN

Tipe

Silinder Vertikal dengan ellipsoidal head

Kapasitas

582,975

m3

Tekanan

1

Atm

Temperatur

30,00

0

Diameter

7,638

m

OD

7,643

M

Tinggi

13,366

M

Tebal Dinding

0,003

M

Bahan Konstruksi

Stainless steel

154,006

gal

300,706

in

43,853

ft

C

33.

TANGKI - 04 (T-04) IDENTIFIKASI

Nama Alat

Tangki-04

Kode Alat

T-04

Jumlah

5 Unit

Fungsi

Tempat menyimpan Produk Etanol DATA DESAIN

Tipe

Silinder Vertikal dengan ellipsoidal head

Kapasitas

2.022,180

m3

Tekanan

1

Atm

Temperatur

30,00

0

Diameter

11,562

m

OD

11,577

m

Tinggi

20,234

m

Tebal Dinding

0,008

m

Bahan Konstruksi

Carbon steel

534.203,422

gal

455,196

in

66,383

ft

C

34.

TANGKI - 05 (T-05) IDENTIFIKASI

Nama Alat

Tangki-05

Kode Alat

T-05

Jumlah

5 Unit

Fungsi

Tempat menyimpan Produk Etanol DATA DESAIN

Tipe

Silinder Vertikal dengan ellipsoidal head

Kapasitas

855,555

m3

Tekanan

1

Atm

Temperatur

30,00

0

Diameter

8,680

m

OD

8,691

m

Tinggi

15,190

m

Tebal Dinding

0,006

m

Bahan Konstruksi

Carbon steel

226.013,892

gal

341,724

in

49,835

ft

C

35.

VAPORIZER - 01 (VP-01) Fungsi

:Menguapkan bahan baku asam asetat.

Tipe

: Shell and Tube Heat Exchanger

Gambar

:

Aliran inlet Tube

Head

Shell

Rear End Aliran outlet

Fluida Panas

Fluida Dingin

Water in

: W

= 55.685,454 kg

= 122.765,266 lb

T1

= 350oC

= 662oF

T2

= 350oC

= 662oF

w

= 85.202,804 kg

= 187.839,806 lb

t1

= 118,100 oC

= 244,580 oF

t2

= 118,100 oC

= 244,580 oF

:

Perhitungan design sesuaidenganliteraturpada Donald Q. Kern (1965). a. Beban Panas VP-01 Q = 35.295.784,129 kJ

= 33.454.438,367 Btu

b. LMTD Fluida Panas (oF)

Fluida Dingin (oF)

Selisih

662

Suhu tinggi

244,580

417,420

662

Suhu rendah

244,580

417,420 0,000

Selisih LMTD =  t= 417,420 oF Ft t

= 1,000

(Fig.18, Kern)

= 417,420 oF c. Temperatur Rata-rata Tc = 662 oF

; tc = 244,580 oF



Asumsi UD = 150 Btu/hr.ft2.oF



A =

(Tabel 8, Kern)

𝑄 𝑈𝐷 𝑋

t

A = 534,305 ft2 Karena A > 200 ft2, maka digunakan Shell & Tube Heat Exchanger 

Rencana Klasifikasi -

Tube Side (Cold Fluid)

Panjang tube (L)

= 12 ft

Outside Diameter (OD) = 1 in BWG

= 18

Pass

=4

a”

= 0,2618 ft2/lin ft A L x a"

Jumlah tube, Nt

=

= 170,074 dari tabel.9 Kern, didapat nilai yang mendekati Nt perhitungan adalah Nt = 170  A

Corrected Coefficient, UD =Nt x L x a''

=534,072 ft2 UD =

Q A .Δt

UD = 150,065

(Nilai UD sudah mendekati UD asumsi)

karena nilai Ud perhitungan mendekati dengan nilai Ud asumsi, maka data untuk shell : - Shell side (Hot Fluid) ID

= 21,25 inch

(Tabel 9, Kern)

Baffle Space (B = ID/2)

= 10,625 inch

Pass

=4

Pt

= 1,25 in triangular pitch

Fluida Dingin: a. Flow Area/tube, a’t a’t = 0,639 in2

(Tabel 10, Kern)

at = Nt.a’t/144 x n = 0,189 ft2 b. Laju Alir, Gt Gt = W/at = 996.002,284 lb/hr.ft2 c.

Bilangan Reynold, Ret Pada Tavg = 244,580 oF μ = 0,876 cp

= 2,119 lb/ft hr

ID= 0,902 inch

= 0,075 ft

D = 0,0752 ft Ret = D.Gt/μ = 35.331,698 d. Dengan L/D = 159,645 diperoleh Jh = 110 e. Nilai hi

(Fig.24, Kern)

(Tabel 10, Kern)

`

Pada Tavg

= 244,580 oF

Cp

= 0,002 Btu/lb.oF

k

= 0,099 Btu/hr ft.oF

 c.     k 

= 0,043

 k   Cp   hi  J H      D  k 

1/ 3

      w  

0 ,14

    Koreksi viskositas diabaikan, karena   w  hi

= 282,008 Btu/hr ft2 oF

hio

= hi x ID/OD

0 ,14

= 51,143 Btu/hr.ft2.oF Fluida Panas: a.

Flow Area, as as = ID x C’B/144 PT

(Pers.7.1, Kern)

ID = Diameter dalam shell

= 15,25 in

B = Baffle spacing

= 10,625 in

Pt = tube pitch

= 1,25 in

C’ = Clearance

= Pt – OD

= 1,25 – 1 = ¼ in as =

b.

15,25  0,25  7,6 = 0,161 ft2 144 1,25

Laju Alir, Gs Gs = w/as =

124.929,069 = 773.546,30 lb/hr.ft2 0,161

c. Bilangan Reynold, Res

=1

tavg

= 244oF

Cp

= 0,756Btu/lb.oF

k

= 0,115Btu/hr ft.oF

μ

= 0,325 Cp

 Cp.     k 

1

3

= 1,753

De = 0,990 inch Res=

= 0,787lb/ft hr

= 0,083 ft

(Fig.28, Kern)

GS D



= 81.041,476 jH = 180

(Fig.28, Kern)

d. Nilai ho Koreksi viskositas diabaikan karena tidak significant, maka diperoleh : ho

= jH . (k/De). (Cpμ/k)1/3 = 422,540Btu/ hr ft2oF

e. Clean Overall Coefficient, UC UC =

hio  ho = 158,783 Btu/hr.ft2.oF hio  ho

f. Dirt Factor, Rd Rd =

U C U D = 0,003 hr.ft2.oF/Btu U C U D

PRESSURE DROP Tube Side 1) Untuk NRe Faktor friksi s

= 686.580,738 = 0,00011 = 0,0200

(Fig 26, Kern)

f Gt 2 L n 5, 22 x 10 10 x De s f t

2) ΔPt = = 3,906 psi 3) V2 / 2g

= 0,005

ΔPr

(Fig 27, Kern)

= ( 4n/s ) ( V2/2g ) = 3,999 psi

ΔPT

= 7,906 psi

Shell Side 1) Faktor Friksi Re

= 81.041,476

f

= 0,00017

2) Number of cross, (N + 1) N+1

= 12 L / B = 339,93

Ds

= ID / 12 = 1,270 ft s

= 1,051

ΔPs

=

fGs2 Ds ( N  1) 5,22  1010 Desf s = 9,708 psi

IDENTIFIKASI

(Fig.29, Kern)

Nama Alat

Vaporizer

Kode

VP-01

Jumlah

1

Fungsi

Untuk mengubah fase asam asetat dan etil asetat

Tipe

Shell and Tube Heat Exchanger DATA DESAIN Uc

121,158

Btu/hr ft2 oF

Ud

150,065

Btu/hr ft2 oF

Rd Calculated

0,003

hr ft2 oF/Btu

TUBE SIDE Length

12

Ft

OD

1

In

Passes

4

BWG

18

Pitch

1,250

Nt ΔPT

170 0,001

in Triangular Pitch Tubes Psi SHELL SIDE

ID

21,25

In

B

10,625

In

Passes ΔPs

4 1,762

psi