lampiran perhitungan

Lampiran I

PERHITUNGAN NERACA MASSA, NERACA ENERGI DAN EFFISIENSI PADA

PRIMAY REFORMER

1. Menghitung Neraca Massa Pada Primary Reformer

Diketahui :

Data aktual Natural Gas pada Primary Reformer hasil laboratorium pusat PT. PUSRI

Tanggal 23 Agustus 2012.

Basis perhitungan 1 jam operasi

a. Komposisi Natural Gas

Komponen Hasil

Satuan 11 Juni 2012

CO2 4,50 %mol

CH4 83,17 %mol

C2H6 5,86 %mol

C3H8 3,40 %mol

i-C4H10 0,59 %mol

n-C4H10 0,73 %mol

i-C5H12 0,32 %mol

n-C5H12 0,16 %mol

C6H14 0,21 %mol

Specific Gravity 0,6984

Gross Heating Value 1104,7417 Btu/Cuft

b. Flowrate natural gas masuk Primary reformer

Laporan Kerja Praktek

PT. Pupuk Sriwidjaja Palembang L I - 2

Jurusan Teknik Kimia

Program Studi Teknik Energi

Flow natural gas untuk Primary Reformer berdasarkan data actual tanggal 23

Agustus 2012 adalah 24.912,93 m3/hour

Asumsi: Natural Gas masuk Primary Reformer merupakan gas ideal

Mol natural gas = volume : 22.4 liter / mol

= 24.912,93 m3/hr : 0.0224 m

3/mol

= 1112184,375 mol/hr

c. Komposisi Natural Gas masuk

Basis 1 mol Fuel

Komponen Komposisi

(%) BM Mol (kmol) Massa (kg/hr) volume (m

3/hr)

CO2 4,5 44 4,5 198 112108,2

CH4 83,17 16 83,17 1330,72 2072008

C2H6 5,86 30 5,86 175,8 145989,8

C3H8 3,4 44 3,4 149,6 84703,96

i-C4H10 0,59 58 0,59 34,22 14698,63

n-C4H10 0,73 58 0,73 42,34 18186,44

i-C5H12 0,32 72 0,32 23,04 7972,138

n-C5H12 0,16 72 0,16 13,76 3986,069

C6H14 0,21 86 0,21 18,07 5231,715

TOTAL 98,94 2045,55 2464885

Asumsi : semua Hidrokarbon bereaksi sempurna (dalam kmol/h)

CH4 + 2 O2 CO2 + 2 H2O

83,17 166,34 83,17 166,34





C2H6 + 3,5 O2 2 CO2 + 3 H2O

5,86 29,3 11,72 17,58

C3H8 + 5 O2 3 CO2 + 4 H2O

3,4 8,5 10,2 13,6

iso-C4H10 + 6,5 O2 4 CO2 + 5 H2O

0,59 3,84 2,36 2,95

nC4H10 + 6,5 O2 4 CO2 + 5 H2O

0,73 4,75 2,93 3,65

Iso-C5H12 + 8 O2 5 CO2 + 6 H2O

0,32 2,56 1,6 1,92

nC5H12 + 8 O2 5 CO2 + 6 H2O

0,16 1,28 0,8 0,96

C6H14 + 9,5 O2 6 CO2 + 7 H2O

0,21 1,995 1,26 1,47

O2Stoikiometri

=( 166,34 + 29,3 + 8,5 + 3,84 + 4,75 + 2,56 + 1,28 +1,995 ) kmol/hr

= 218,565 kmol/hr





d. Analisa Exhaust Primary Reformer (dry basis)

Data diambil dari Laboratorium PUSRI III untuk tanggal 23 Agustus 2012

CO2 = 6,0 % = 0,06

O2 = 8,06 % = 0,0806

CO = 0,01 % = 0,0001

CO2 hasil pembakaran = (83,17 + 11,72 + 10,2 + 2,36 + 2,93 + 1,6 + 0,8 + 1,26 )

= 114,04 kmol/ hr

Massa CO2 = mol x BM CO2

= 114,04 kmol/hr x 44.01 kg/mol

= 5018,9004 kg/hr

Mol CO2 dalam NG = 4,5 kmol/hr

Mol total CO2 pada Primary Reformer

= 114,04 kmol/hr + 4,5 kmol/hr

= 118,54 kmol/hr

Massa CO2 total = 118,54 kmol/hr 44,01 kg/kmol

= 5216,9454 kg/hr

Mol Exhaust Primary Reformer (dry basic)

Mol Flue Gas (dry basis) = mol CO2 total : % CO2 outlet ke Primary Reformer





= 5216,9454 kmol/hr : 0,06

= 85449,09 kmol/hr

Mol CO2 = 6,0 % x 85449,09 kmol/hr = 5126,9454 kmol/hr

Massa CO2 = 5126,9454 kmol/hr x 44.01 kg/kmol= 225636,8671 kg/hr

Mol O2 = 8,06 % x 85449,09 kmol/hr = 6887,196654 kmol/hr

Massa O2 = 6887,196654 kmol/hr x 32 kg/kmol= 220663,73 kg/hr

Mol CO = 0,01 % x 85449,09 kmol/hr = 8,544909 kmol/hr

Massa CO = 8,544909 kmol/hr x 28 kg/kmol = 239,257452 kg/hr

e. Massa Balance O2 dan N2 (udara)

Mol O2 Input = O2 Stoikiometri + O2 exhaust Primary Reformer

= 218,565 kmol/h + 13107,89041kmol/h

= 13326,45541 kmol/h

Massa O2 Input = 13326,45541 kmol/h x 32 kg/kmol

= 426446,5731 kg/h

Mol N2 Input = (79/21) x 13326,45541kmol/hr

= 50132,85607 kmol/hr

Massa N2 Input = 50132,85607 kmol/hr x 28 kg/kmol

= 1403719,97 kg/hr

Excess Udara = (100/21) x 13326,45541 kmol/hr

= 63459,31148 kmol/hr





f. H2O hasil pembakaran (yang terbentuk di Primary Reformer)

Mol H2O terbentuk = ( 166,34 + 17,58 + 13,6 + 2,95 + 3,65 + 1,92 + 0,96 + 1,47 )kmol/hr

= 208,47 kmol/hr

Massa H2O terbentuk = 208,47 kmol/hr x 18 kg/kmol

= 3752,46 kg/hr

Mass Balance pada Primary Reformer

Komposisi Input (Kg/hr) Output (Kg/hr)

CO2 198 -

CH4 1330,72 -

C2H6 175,8 -

C3H8 149,6 -

i-C4H10 34,22 -

n-C4H10 42,34 -

i-C5H12 23,04 -

n-C5H12 13,76 -

C6H14 18,07 -

Udara

O2 426446,5731 -

N2 1403719,97 -

flue gas

CO2 - 225636,8671

CO - 239,257452





O2 - 220663,73

H2O - 3752,46

N2 -

1403719,97

TOTAL 1832152,093 1832137,318

2. Menghitung Neraca Massa Tail Gas pada Primary Reformer

Diketahui :

Data aktual Tail Gas pada Primary Reformer hasil laboratorium pusat PT. PUSRI

Tanggal 23 Agustus 2012.

Komponen Komposisi

(%)

BM Mol(kmol) Massa(kg/hr)

Ar 6,57 39,95 6,57 262,4715

N2 61,27 28 61,27 1715,56

CH4 22,69 16 22,69 363,04

H2 9,47 2 9,47 18,94

Total 100 2360,0115

a. Reaksi aktual

CH4 + 2 O2

CO2 + 2 H2O

22,69 45,38 22,69 45,38

H2 + 0,5 O2

H2O

9,47 4,735 9,47





O2 reaksi = ( 45,32 + 4,735 ) kmol/hr

= 50,055 kmol/hr

Massa O2 reaksi = 50,055kmol/hr x 32kg/kmol = 1601,76 kg/hr

b. Analisa pembakaran pada Primary Reformer

Mol O2 udara = 21/79 x 50,055 kmol/hr = 13,30576kmol/hr

Massa O2 udara = 13,30576kmol/hr x 32 kg/kmol=425,784kg/hr

Massa O2 total = ( 1601,76 + 425,784 )kg/hr = 2027,544 kg/hr

Mol N2 udara = 79/21 x 13,30576kmol/hr = 50,055kmol/hr

Massa N2 udara= 50,055 kmol/hr x 28 kg/kmol = 1401,54kg/hr

Mol N2 bahan bakar = 61,27kmol/hr

Massa N2 bahan bakar = 61,27kmol/hr x 28kg/kmol = 1715,56kg/hr

Mol CO2 pembakaran = 0,2269 kmol/hr

Massa CO2 pembakaran = 0,2269kmol/hr x 28kg/kmol = 6,3532kg/hr

c. H2O hasil dari pembakaran di Primary Reformer

Mol H2O pembakaran =( 45,38 + 9,47 )kmol/hr = 54,85 kmol/hr

Massa H2O pembakaran = 54,85kmol/hr x 18kg/kmol= 987,3 kg/hr

Massa Balance pada Primary Reformer

Komponen Input (kg/hr) Outpu (kg/hr)





Ar 262,4715

N2 1715,56 1715,56

CH4 363,04

H2 18,94

Udara 2027,544

N2 1401,54

CO2 6,3532

H2O 987,3

Total 4387,555 4110,753

3. Menghitung Neraca Energi

Udara

Primary

Reformer

Natural Gas

Tail Gas

Gas Proses





INPUT

1.Menghitung Panas Natural Gas

Panas dari natural gas :

T1 = 250

C = 298 K ( temperatur referen)

T2 = 290,30C = 563,3 K ( temperature Natural gas)

Komposisi A B 10-2

C 10-5

D 10-9

kg

CO2 5,316 1,428 -0,8362 1,784 198

CH4 4,75 1,200 0,3030 -2,63 1330,72

C2H6 1,648 4,124 -1,530 1,74 175,8

C3H8 -0,966 7,279 -3,755 7,58 149,6

i-C4H10 -1,89 9,936 -5,495 11,192 34,22

n-C4H10 0,94 8,873 -4,380 8,36 42,34

i-C5H12 1,618 10,85 -5,365 10,1 23,04

n-C5H12 2,464 4,5351 -1,4111 -0,8587 13,76

C6H14 3,025 5,3722 -1,6791 -1,157 18,07

Cp = (A (T2 T1))+(1/2 B (T2 T1)2)+(1/3 C (T2 T1)

3)+(1/4 D (T2 T1)

4)

Satuan Cp diatas yaitu Kkal/kg K.

Maka didapatlah :

Tabel Kapasitas Panas Natural Gas

Komposisi Cp (Kkal/kg K)





CO2 1863,039057

CH4 1698,081966

C2H6 1795,457506

C3H8 2081,014381

i-C4H10 2668,002005

n-C4H10 3109,701519

i-C5H12 3926,167252

n-C5H12 2160,799057

C6H14 2588,934006

Sehingga dapat dihitung Q (panas) dari komposisi Natural Gas, dengan rumus :

Q = m. Cp. Dt

a. CO2

Q = m. Cp. dT

= m x ((aT) + (bT2

x 10-2

/2) - (cT3x10

-5/3) + (dT

4x10

-9/4))

= (198 kg/hr) x (1863,039057 Kkal/kg)

= 368881,7333 Kkal/hr

b. CH4

Q = m. Cp. dT

= m x ((aT) + (bT2

x 10-2

/2) - (cT3x10

-5/3) + (dT

4x10

-9/4))

= (1330,72 kg/hr) x (1698,081966 Kkal/kg)

= 2259671,634Kkal/hr

c. C2H6

Q = m. Cp. dT





= m x ((aT) + (bT2

x 10-2

/2) - (cT3x10

-5/3) + (dT

4x10

-9/4))

= (175,8 kg/hr) x (1795,457506 Kkal/kg)

= 315641,4296 Kkal/hr

d. C3H8

Q = m. Cp. dT

= m x ((aT) + (bT2

x 10-2

/2) - (cT3x10

-5/3) + (dT

4x10

-9/4))

= (149,6 kg/hr) x (2081,014381 Kkal/kg)

= 311319,7514 Kkal/hr

e. i-C4H10

Q = m. Cp. dT

= m x ((aT) + (bT2

x 10-2

/2) - (cT3x10

-5/3) + (dT

4x10

-9/4))

= (34,22 kg/hr) x (2668,002005 Kkal/kg)

= 91299,02861 Kkal/hr

f. n-C4H10

Q = m. Cp. dT

= m x ((aT) + (bT2

x 10-2

/2) - (cT3x10

-5/3) + (dT

4x10

-9/4))

= (42,34 kg/hr) x (3109,701519 Kkal/kg)

=131664,7623 Kkal/hr

g. i-C5H12

Q = m. Cp. dT

= m x ((aT) + (bT2

x 10-2

/2) - (cT3x10

-5/3) + (dT

4x10

-9/4))

= (23,04 kg/hr) x (3926,167252 Kkal/kg)

= 90458,89349 Kkal/hr

h. n-C5H12

Q = m. Cp. dT

= m x ((aT) + (bT2

x 10-2

/2) - (cT3x10

-5/3) + (dT

4x10

-9/4))

= (13,76 kg/hr) x (2160,799057 Kkal/kg)

= 29732,595Kkal/hr

i. C6H14

Q = m. Cp. dT





= m x ((aT) + (bT2

x 10-2

/2) - (cT3x10

-5/3) + (dT

4x10

-9/4))

= (18,07 kg/hr) x (2588,934006 Kkal/kg)

= 46782,0375 Kkal/hr

Tabel Q total Natural Gas

Komposisi Q (Kkal)

CO2 368881,7333

CH4 2259671,634

C2H6 315641,4296

C3H8 311319,7514

i-C4H10 91299,02861

n-C4H10 131664,7623

i-C5H12 90458,89349

n-C5H12 29732,595

C6H14 46782,0375

TOTAL 3645452

2. Menghitung Panas Sensibel Udara

T1 = 250

C = 298 K ( temperatur referen)

T2 = 28 0C = 301 K ( temperatur udara masuk )

komposisi kmol A B 10-3

C 10-6

O2 248,5461 6,117 3,167 -1,005

N2 935,0068 6,457 1,389 -0,069





Cp = A + ½ B (T1+T2) + 1/3 C (T22+T1.T2+T1

2) ( Hougen, 1943 )

Satuan Cp diatas yaitu Kkal/kmol K.

a. O2

Cp udara = 6,117 + ½ (3,167x10-3

)( 301+298) + ⅓(-1,005x10-6

) (3012+ (298x301)

+2982)

= 6,975366995 Kkal/ kmol K

b. N2

Cp udara = 6,457 + ½ (1,389x10-3

)( 301+298) + ⅓(-0,069x10-6

) (3012+ (298x301)

+2982)

= 6,866816131 Kkal/ kmol K

Maka didapatlah :

Tabel Kapasitas Panas Udara

Komposisi Cp (Kkal/kmol K)

O2 6,975366995

N2 6,866816131

Sehingga panas sensibel udara dapat dihitung dengan menggunakan rumus :

Q = n. Cp. dT

a. O2

Q = n. Cp. dt

= 248,5461 kmol x 6,975366995 kkal/kmol K x (301-298)K

= 5201,100788 kkal

b. N2

Q = n. Cp. dt

= 935,0068 kmol x 6,866816131 kkal/kmol K x (301-298)K

= 19261,55933 kkal





Tabel Q total Udara

Komposisi Q (Kkal)

O2 5201,100788

N2 19261,55933

TOTAL 24462,66012

3. Menghitung Panas Pembakaran Tail Gas

Komponen Massa (kg/hr) Cp(Kkal/kg K)

Ar 262,4715 11,02736

N2 1715,56 7,057

CH4 363,04 7,370

H2 18,94 87,01

Sehingga panas pembakaran tail gas yaitu :

Q= n x Cp x dT

Ar

Q = n . Cp .dT

= m x ((aT) + (bT2

x 10-2

/2) - (cT3x10

-5/3) + (dT

4x10

-9/4))

= 262,4715kg/hr x 66127.595 Kkal/kg

= 17356609,05 Kkal

N2

Q = n . Cp .dT





= m x ((aT) + (bT2

x 10-2

/2) - (cT3x10

-5/3) + (dT

4x10

-9/4))

= 1715,56 kg/hr x 5601,199 Kkal/kg

= 9609192,956 Kkal

CH4

Q = n . Cp .dT

= m x ((aT) + (bT2

x 10-2

/2) - (cT3x10

-5/3) + (dT

4x10

-9/4))

=363,04 kg/hr x 20080,271 Kkal/kg

=7289941,584 Kkal

H2

Q = n . Cp .dT

= m x ((aT) + (bT2

x 10-2

/2) - (cT3x10

-5/3) + (dT

4x10

-9/4))

= 18,94kg/hr x 828,964 Kkal/kg

=15700,578Kkal

Tabel Qtotal Tail Gas

Komponen Q (Kkal)

Ar 17356609,05

N2 9609192,956

CH4 7289941,584





H2 15700,578

Total 3487319

Total QInput

Komponen Q (Kkal) (%)

Q Natural Ga 3645452 14,19313214

Q Udara Inlet 24462,66012 0,199204768

Q Tail Gas 3487319 85,60766309

TOTAL 7157233,66 100

OUTPUT

1. Menghitung panas Gas Proses

T1 = 790,5 0C

T2 = 190,5 0C

Komponen Massa(kg/hr) Cp (Kkal/kg K)

Ar 0,3995 87,01

N2 19,32 7,475

CH4 168,64 14,511

H2 137,64 7,087

CO2 572 11,986





CO 194,32 7,568

Sehingga panas gas proses,yaitu :

Q = m . Cp . dT

Ar

Q = m . Cp . dT

= 0,3995Kg/hr x 87,01Kkal/kg K x 600 K

= 20856,297Kkal

N2

Q = m . Cp . dT

= 19,32Kg/hr x 7,475Kkal/kg K x 600 K

= 86650,2Kkal

CH4

Q = m . Cp . dT

= 168,64Kg/hr x 14,511Kkal/kg K x 600 K

= 1468281,024Kkal

H2

Q = m . Cp . dT





= 137,64Kg/hr x 7,087Kkal/kg K x 600 K

= 585272,808Kkal

CO2

Q = m . Cp . dT

= 572Kg/hr x 11,986Kkal/kg K x 600 K

= 4113595,2Kkal

CO

Q = m . Cp . dT

= 194,32Kg/hr x 7,568Kkal/kg K x 600 K

= 882368,256Kkal

Tabek Q total Gas Proses

Komponen Q(Kkal)

Ar 20856,297

N2 86650,2

CH4 1468281,024

H2 585272,808

CO2 4113595,2

CO 882368,256





Total 7017023,488

QInput = QOutput , dimana QOutput = QGas Proses

QNatural Gas + QUdara + QTail Gas = QGas Proses

( 3645452 + 24462,66012 + 3487319 ) Kkal = 7017023,488 Kkal

7157233,66 Kkal = 7017023,488 Kkal

Efisiensi panas pembakaran pada Primary Reformer

η = 100%

= 100 %

= 98,04 %

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