lampiran perhitungan edit lagi pusta

Perhitungan untuk data ke 1

4.4.1. Menghitung panas hasil pembakaran untuk FICI

1) Menghitung panas yang diserap oleh Crude Oil

Feed = 2685

TonHari x

124

HariJam x

10001

KgTon x

2 ,204621

LbKg

= 246.641,86

LbJam

Sg = 0.8540

T-in rate = 167,4 oC = 333,32 oF

T-out rate = 191,24 oC = 376,23 oF

o API=141 ,5

SGR (60 o F /60 o F )−131 ,5

........................(Persamaan 1.4 Kern)

=

141 ,50 ,8540

−131 ,5

=34,19

Untuk mencari nilai kapasitas panas (Cp) dapat dicari dengan grafik Specific Heat of Hydrocarbon

Liquids. (Lampiran 1)

Cp pada 34,19 oAPI

T-in rate = 167,4 oC = 333,32 oF

Cp = 0,61

BtuLb oF

T-out rate = 191,24 oC = 376,23 oF

Cp = 0,64

BtuLb

oF

Maka Cp rata-rata =

0 ,61+0 ,642 = 0,625

BtuLb oF

Mencari Entalpi Feed = Cp (TOutlet – TInlet)

= 0,625

BtuLb oF (376,23 – 333,32) oF

= 26,82

BtuLb

Mencari panas Feed = M Feed x H

Q FICI = 246.641,86

LbJam x 26,82

BtuLb

= 6.614.934,68

BtuJam ................................................(1)

1) Menghitung panas yang dilepaskan oleh Fuel Oil

Fuel Oil = 3,70

TonHari x

124

HariJam x

10001

KgTon x

2 ,204621

LbKg

= 339,87

LbJam

LHV(FO) = 13080 - 3425 Sg2 ..................(Diktat “Fuel and Combution” 2001)

= 13080 – 3425 (0,9146)2

= 10.190,75

KcalKg ⇒ 18.343,22

BtuLb

Mencari panas Fuel = M Fuel Oil x LHV(FO)

Q Fuel oil = 339,87

LbJam x 18.343,22

BtuLb

= 6.234.310,18

BtuJam ................................................(2)

2) Menghitung panas yang diserap steam

Jumlah pemakaian steam yang digunakan untuk steam atomizing

Dimana pemakaian FO untuk FICI adalah: 3,70

TonHari = 154,41

KgFOJam

Tekanan steam atomizing = 7,5

Kg

Cm2 x 14,22

Psia

Kg .Cm2 = 106,65 Psia

Suhu steam atomizing = 310 oC = 590 oF

Maka didapat enthalpy = 1173,7

BtuLb (Lampiran 4)

Enthalpy steam pada suhu 600F adalah 28,06

BtuLb (Lampiran 3)

Tekanan FO adalah = 4,86 bar

Dengan menggunakan grafik atomizing consumtion didapatkan

Pemakaian steam atomizing sebesar 0,285

KgsteamKgFO (Lampiran 2)

Maka Msteam = 0,285

KgsteamKgFO x 154,41

KgFOJam

= 44

KgsteamJam x 2,20462

LbKg

= 97,00

LbsteamJam

Mencari panas steam = M steam x H

Q steam = 97,00

LbJam x (1173,7

BtuLb - 28,06

BtuLb )

= 110.127,08

BtuJam .........................................(3)

4) Menghitung panas yang dilepaskan oleh Fuel Gas

Fuel Gas = 4,11

TonHari x

124

HariJam x

10001

KgTon x

2 ,204621

LbKg

= 377,54

LbJam

LHV(FG)= (155+1425 Rd)x(

6 .8718Rd )....(Diktat ”Fuel and Combution” 2001)

= (155+1425 x0,7061)x(

6 .87180 .7061 )

= 11.300,783

KcalKg ⇒ = 20.341,43

BtuLb

Mencari panas Fuel = M Fuel gas x LHV(FG)

Q Fuel gas = 377,54

LbJam x 20.341,43

BtuLb = 7.679.703,48

BtuJam ................................................(4)

4.4.2. Menghitung panas hasil pembakaran untuk FIC2


Feed = 2626

TonHari x

124

HariJam x

10001

KgTon x

2 ,204621

LbKg

= 241.222,17

LbJam

Sg = 0,8540

T-in rate = 248,52 oC = 479,33 oF

T-out rate = 291,24 oC = 556,23 oF

o API=141 ,5

SGR (60 o F /60 o F )−131 ,5

.................... (Persamaan 1.4 Kern)

=

141 ,50 ,8540

−131 ,5

=34,19



Cp pada 34,19 oAPI

T-in rate = 248,52 oC = 479,33 oF

Cp = 0,685

BtuLb oF

T-out rate = 291,24 oC = 556,23 oF

Cp = 0,7025

BtuLb oF

Maka Cp rata-rata =

0 , 685+0 , 7202 = 0,7025

BtuLb oF


= 0,7025

BtuLb oF (556,23 – 479,33) oF

= 54,02

BtuLb


Q FIC2 = 241.222,17

LbJam x 54,02

BtuLb

= 13.030.821,62

BtuJam ...............................................(5)


Fuel Oil = 3,70

TonHari x

124

HariJam x

10001

KgTon x

2 ,204621

LbKg

= 339,87

LbJam


= 13080 – 3425 (0,9146)2

= 10.190,75

KcalKg ⇒ 18.343,22

BtuLb


Q Fuel oil = 339,87

LbJam x 18.343,22

BtuLb

= 6.234.310,18

BtuJam ................................................(6)




TonHari = 154,41

KgFOJam


Kg

Cm2 x 14,22

Psia

Kg .Cm2 = 106,65 Psia



BtuLb (Lampiran 4)


BtuLb (Lampiran 3)


Dengan menggunakan grafik atomizing consumtion (Lampiran 2) didapatkan


KgsteamKgFO

Maka Msteam = 0,26


KgFOJam

= 40,15


LbKg ⇒ 88,51

LbsteamJam


Q steam = 88,51

LbJam x (1173,7

BtuLb - 28,06

BtuLb )

= 101.400,59

BtuJam .......................................................(7)


Fuel Gas = 4,11

TonHari x

124

HariJam x

10001

KgTon x

2 ,204621

LbKg

= 377,54

LbJam

LHV(FG) = (155+1425 Rd)x(


= (155+1425 x0,7061)x(

6 .87180 .7061 )

= 11.300,783

KcalKg ⇒ = 20.341,43

BtuLb


Q Fuel gas = 377,54

LbJam x 20.341,43

BtuLb

= 7.679.703,48

BtuJam ................................................(8)

5) Menghitung panas yang diserap Udara dari APH

Tin = 32 oC = 305 K

Tout = 294 oC = 523 K

Pada Tin = HIn = 1,00423

Kj

Kg0C = 20,97

BtuLb (Lampiran 5)

Pada Tout = Hout = 1,04460

Kj

Kg0C = 21,82

BtuLb

Menghitung Total Panas yang diserap udara

Mass flow = 41.200 (diambil dari dalam data sheet)

= 41.200

KgJam x

2 ,204621

LbKg

= 90.830,34

LbJam

Mencari panas udara = M Feed x H

= 90.830,34

LbJam x (21,82

BtuLb - 20,97

BtuLb )

= 77.205,789

BtuJam ...................................................(9)

Tabel 5 Neraca Panas Furnace I

No Deskripsi Btu/hr Btu/hr

1 Heat Release absorb

# Fuel Oil

# Fuel Gas

TOTAL

12.468.620,36

15.359.406,96

27.828.027,32

2 Heat Absorb

# Crude Oil

# Steam

# Udara

TOTAL

19.645.756,63

211,527,67

77.205,79

19.934.490,09

3 Heat loss 7.893.637,23

4.4.3 Menghitung Efisiensi

Efisiensi () =

Heat absorb totalHeat release

x100 %

=

19 . 934 . 490,0927 . 828 .027,32

x100 % = 71.63 %

Perhitungan untuk data hari ke 2



Feed = 2685

TonHari x

124

HariJam x

10001

KgTon x

2 ,204621

LbKg

= 246.641,86

LbJam

Sg = 0.8540

T-in rate = 167,4 oC = 333,32 oF

T-out rate = 191,24 oC = 376,23 oF

o API=141 ,5

SGR (60 o F /60 o F )−131 ,5

........................(Persamaan 1.4 Kern)

=

141 ,50 ,8540

−131 ,5

=34,19



Cp pada 34,19 oAPI

T-in rate = 167,4 oC = 333,32 oF

Cp = 0,61

BtuLb oF

T-out rate = 191,24 oC = 376,23 oF

Cp = 0,64

BtuLb

oF

Maka Cp rata-rata =

0 ,61+0 ,642 = 0,625

BtuLb oF


= 0,625

BtuLb oF (376,23 – 333,32) oF

= 26,82

BtuLb


Q FICI = 246.641,86

LbJam x 26,82

BtuLb

= 6.614.934,68

BtuJam ................................................(1)


Fuel Oil = 3,70

TonHari x

124

HariJam x

10001

KgTon x

2 ,204621

LbKg

= 339,87

LbJam


= 13080 – 3425 (0,9146)2

= 10.190,75

KcalKg ⇒ 18.343,22

BtuLb


Q Fuel oil = 339,87

LbJam x 18.343,22

BtuLb

= 6.234.310,18

BtuJam ................................................(2)




TonHari = 154,41

KgFOJam


Kg

Cm2 x 14,22

Psia

Kg .Cm2 = 106,65 Psia



BtuLb (Lampiran 4)


BtuLb (Lampiran 3)





Maka Msteam = 0,285


KgFOJam

= 44


LbKg

= 97,00

LbsteamJam


Q steam = 97,00

LbJam x (1173,7

BtuLb - 28,06

BtuLb )

= 110.127,08

BtuJam .........................................(3)


Fuel Gas = 4,11

TonHari x

124

HariJam x

10001

KgTon x

2 ,204621

LbKg

= 377,54

LbJam

LHV(FG)= (155+1425 Rd)x(


= (155+1425 x0,7061)x(

6 .87180 .7061 )

= 11.300,783

KcalKg ⇒ = 20.341,43

BtuLb


Q Fuel gas = 377,54

LbJam x 20.341,43

BtuLb = 7.679.703,48

BtuJam ................................................(4)



Feed = 2626

TonHari x

124

HariJam x

10001

KgTon x

2 ,204621

LbKg

= 241.222,17

LbJam

Sg = 0,8540

T-in rate = 248,52 oC = 479,33 oF

T-out rate = 291,24 oC = 556,23 oF

o API=141 ,5

SGR (60 o F /60 o F )−131 ,5

.................... (Persamaan 1.4 Kern)

=

141 ,50 ,8540

−131 ,5

=34,19



Cp pada 34,19 oAPI

T-in rate = 248,52 oC = 479,33 oF

Cp = 0,685

BtuLb oF

T-out rate = 291,24 oC = 556,23 oF

Cp = 0,7025

BtuLb oF

Maka Cp rata-rata =

0 ,685+0 , 7202 = 0,7025

BtuLb oF


= 0,7025

BtuLb oF (556,23 – 479,33) oF

= 54,02

BtuLb


Q FIC2 = 241.222,17

LbJam x 54,02

BtuLb

= 13.030.821,62

BtuJam ...............................................(5)


Fuel Oil = 3,70

TonHari x

124

HariJam x

10001

KgTon x

2 ,204621

LbKg

= 339,87

LbJam


= 13080 – 3425 (0,9146)2

= 10.190,75

KcalKg ⇒ 18.343,22

BtuLb


Q Fuel oil = 339,87

LbJam x 18.343,22

BtuLb

= 6.234.310,18

BtuJam ................................................(6)




TonHari = 154,41

KgFOJam


Kg

Cm2 x 14,22

Psia

Kg .Cm2 = 106,65 Psia



BtuLb (Lampiran 4)


BtuLb (Lampiran 3)




KgsteamKgFO

Maka Msteam = 0,26


KgFOJam

= 40,15


LbKg ⇒ 88,51

LbsteamJam


Q steam = 88,51

LbJam x (1173,7

BtuLb - 28,06

BtuLb )

= 101.400,59

BtuJam .......................................................(7)


Fuel Gas = 4,11

TonHari x

124

HariJam x

10001

KgTon x

2 ,204621

LbKg

= 377,54

LbJam

LHV(FG) = (155+1425 Rd)x(


= (155+1425 x0,7061)x(

6 .87180 .7061 )

= 11.300,783

KcalKg ⇒ = 20.341,43

BtuLb


Q Fuel gas = 377,54

LbJam x 20.341,43

BtuLb

= 7.679.703,48

BtuJam ................................................(8)


Tin = 32 oC = 305 K

Tout = 294 oC = 523 K


Kj

Kg0C = 20,97

BtuLb (Lampiran 5)


Kj

Kg0C = 21,82

BtuLb



= 41.200

KgJam x

2 ,204621

LbKg

= 90.830,34

LbJam


= 90.830,34

LbJam x (21,82

BtuLb - 20,97

BtuLb )

= 77.205,789

BtuJam ...................................................(9)




# Fuel Oil

# Fuel Gas

TOTAL

12.468.620,36

15.359.406,96

27.828.027,32

2 Heat Absorb

# Crude Oil

# Steam

# Udara

19.645.756,63

211,527,67

77.205,79

TOTAL 19.934.490,09

3 Heat loss 7.893.637,23


Efisiensi () =


x100 %

=

19 . 934 .490,0927 . 828 .027,32

x100 % = 71.63 %




Feed = 2685

TonHari x

124

HariJam x

10001

KgTon x

2 ,204621

LbKg

= 246.641,86

LbJam

Sg = 0.8540

T-in rate = 167,4 oC = 333,32 oF

T-out rate = 191,24 oC = 376,23 oF

o API=141 ,5

SGR (60 o F /60 o F )−131 ,5

........................(Persamaan 1.4 Kern)

=

141 ,50 ,8540

−131 ,5

=34,19



Cp pada 34,19 oAPI

T-in rate = 167,4 oC = 333,32 oF

Cp = 0,61

BtuLb oF

T-out rate = 191,24 oC = 376,23 oF

Cp = 0,64

BtuLb

oF

Maka Cp rata-rata =

0 ,61+0 ,642 = 0,625

BtuLb oF


= 0,625

BtuLb oF (376,23 – 333,32) oF

= 26,82

BtuLb


Q FICI = 246.641,86

LbJam x 26,82

BtuLb

= 6.614.934,68

BtuJam ................................................(1)


Fuel Oil = 3,70

TonHari x

124

HariJam x

10001

KgTon x

2 ,204621

LbKg

= 339,87

LbJam


= 13080 – 3425 (0,9146)2

= 10.190,75

KcalKg ⇒ 18.343,22

BtuLb


Q Fuel oil = 339,87

LbJam x 18.343,22

BtuLb

= 6.234.310,18

BtuJam ................................................(2)




TonHari = 154,41

KgFOJam


Kg

Cm2 x 14,22

Psia

Kg .Cm2 = 106,65 Psia



BtuLb (Lampiran 4)


BtuLb (Lampiran 3)





Maka Msteam = 0,285


KgFOJam

= 44


LbKg

= 97,00

LbsteamJam


Q steam = 97,00

LbJam x (1173,7

BtuLb - 28,06

BtuLb )

= 110.127,08

BtuJam .........................................(3)


Fuel Gas = 4,11

TonHari x

124

HariJam x

10001

KgTon x

2 ,204621

LbKg

= 377,54

LbJam

LHV(FG)= (155+1425 Rd)x(


= (155+1425 x0,7061)x(

6 .87180 .7061 )

= 11.300,783

KcalKg ⇒ = 20.341,43

BtuLb


Q Fuel gas = 377,54

LbJam x 20.341,43

BtuLb = 7.679.703,48

BtuJam ................................................(4)



Feed = 2626

TonHari x

124

HariJam x

10001

KgTon x

2 ,204621

LbKg

= 241.222,17

LbJam

Sg = 0,8540

T-in rate = 248,52 oC = 479,33 oF

T-out rate = 291,24 oC = 556,23 oF

o API=141 ,5

SGR (60 o F /60 o F )−131 ,5

.................... (Persamaan 1.4 Kern)

=

141 ,50 ,8540

−131 ,5

=34,19



Cp pada 34,19 oAPI

T-in rate = 248,52 oC = 479,33 oF

Cp = 0,685

BtuLb oF

T-out rate = 291,24 oC = 556,23 oF

Cp = 0,7025

BtuLb oF

Maka Cp rata-rata =

0 ,685+0 , 7202 = 0,7025

BtuLb oF


= 0,7025

BtuLb oF (556,23 – 479,33) oF

= 54,02

BtuLb


Q FIC2 = 241.222,17

LbJam x 54,02

BtuLb

= 13.030.821,62

BtuJam ...............................................(5)


Fuel Oil = 3,70

TonHari x

124

HariJam x

10001

KgTon x

2 ,204621

LbKg

= 339,87

LbJam


= 13080 – 3425 (0,9146)2

= 10.190,75

KcalKg ⇒ 18.343,22

BtuLb


Q Fuel oil = 339,87

LbJam x 18.343,22

BtuLb

= 6.234.310,18

BtuJam ................................................(6)




TonHari = 154,41

KgFOJam


Kg

Cm2 x 14,22

Psia

Kg .Cm2 = 106,65 Psia



BtuLb (Lampiran 4)


BtuLb (Lampiran 3)




KgsteamKgFO

Maka Msteam = 0,26


KgFOJam

= 40,15


LbKg ⇒ 88,51

LbsteamJam


Q steam = 88,51

LbJam x (1173,7

BtuLb - 28,06

BtuLb )

= 101.400,59

BtuJam .......................................................(7)


Fuel Gas = 4,11

TonHari x

124

HariJam x

10001

KgTon x

2 ,204621

LbKg

= 377,54

LbJam

LHV(FG) = (155+1425 Rd)x(


= (155+1425 x0,7061)x(

6 .87180 .7061 )

= 11.300,783

KcalKg ⇒ = 20.341,43

BtuLb


Q Fuel gas = 377,54

LbJam x 20.341,43

BtuLb

= 7.679.703,48

BtuJam ................................................(8)


Tin = 32 oC = 305 K

Tout = 294 oC = 523 K


Kj

Kg0C = 20,97

BtuLb (Lampiran 5)


Kj

Kg0C = 21,82

BtuLb



= 41.200

KgJam x

2 ,204621

LbKg

= 90.830,34

LbJam


= 90.830,34

LbJam x (21,82

BtuLb - 20,97

BtuLb )

= 77.205,789

BtuJam ...................................................(9)




# Fuel Oil

# Fuel Gas

TOTAL

12.468.620,36

15.359.406,96

27.828.027,32

2 Heat Absorb

# Crude Oil

# Steam

# Udara

TOTAL

19.645.756,63

211,527,67

77.205,79

19.934.490,09

3 Heat loss 7.893.637,23


Efisiensi () =


x100 %

=

19 . 934 . 490,0927 . 828 .027,32

x100 % = 71.63 %




Feed = 2685

TonHari x

124

HariJam x

10001

KgTon x

2 ,204621

LbKg

= 246.641,86

LbJam

Sg = 0.8540

T-in rate = 167,4 oC = 333,32 oF

T-out rate = 191,24 oC = 376,23 oF

o API=141 ,5

SGR (60 o F /60 o F )−131 ,5

........................(Persamaan 1.4 Kern)

=

141 ,50 ,8540

−131 ,5

=34,19



Cp pada 34,19 oAPI

T-in rate = 167,4 oC = 333,32 oF

Cp = 0,61

BtuLb oF

T-out rate = 191,24 oC = 376,23 oF

Cp = 0,64

BtuLb

oF

Maka Cp rata-rata =

0 ,61+0 ,642 = 0,625

BtuLb oF


= 0,625

BtuLb oF (376,23 – 333,32) oF

= 26,82

BtuLb


Q FICI = 246.641,86

LbJam x 26,82

BtuLb

= 6.614.934,68

BtuJam ................................................(1)


Fuel Oil = 3,70

TonHari x

124

HariJam x

10001

KgTon x

2 ,204621

LbKg

= 339,87

LbJam


= 13080 – 3425 (0,9146)2

= 10.190,75

KcalKg ⇒ 18.343,22

BtuLb


Q Fuel oil = 339,87

LbJam x 18.343,22

BtuLb

= 6.234.310,18

BtuJam ................................................(2)




TonHari = 154,41

KgFOJam


Kg

Cm2 x 14,22

Psia

Kg .Cm2 = 106,65 Psia



BtuLb (Lampiran 4)


BtuLb (Lampiran 3)





Maka Msteam = 0,285


KgFOJam

= 44


LbKg

= 97,00

LbsteamJam


Q steam = 97,00

LbJam x (1173,7

BtuLb - 28,06

BtuLb )

= 110.127,08

BtuJam .........................................(3)


Fuel Gas = 4,11

TonHari x

124

HariJam x

10001

KgTon x

2 ,204621

LbKg

= 377,54

LbJam

LHV(FG)= (155+1425 Rd)x(


= (155+1425 x0,7061)x(

6 .87180 .7061 )

= 11.300,783

KcalKg ⇒ = 20.341,43

BtuLb


Q Fuel gas = 377,54

LbJam x 20.341,43

BtuLb = 7.679.703,48

BtuJam ................................................(4)



Feed = 2626

TonHari x

124

HariJam x

10001

KgTon x

2 ,204621

LbKg

= 241.222,17

LbJam

Sg = 0,8540

T-in rate = 248,52 oC = 479,33 oF

T-out rate = 291,24 oC = 556,23 oF

o API=141 ,5

SGR (60 o F /60 o F )−131 ,5

.................... (Persamaan 1.4 Kern)

=

141 ,50 ,8540

−131 ,5

=34,19



Cp pada 34,19 oAPI

T-in rate = 248,52 oC = 479,33 oF

Cp = 0,685

BtuLb oF

T-out rate = 291,24 oC = 556,23 oF

Cp = 0,7025

BtuLb oF

Maka Cp rata-rata =

0 , 685+0 , 7202 = 0,7025

BtuLb oF


= 0,7025

BtuLb oF (556,23 – 479,33) oF

= 54,02

BtuLb


Q FIC2 = 241.222,17

LbJam x 54,02

BtuLb

= 13.030.821,62

BtuJam ...............................................(5)


Fuel Oil = 3,70

TonHari x

124

HariJam x

10001

KgTon x

2 ,204621

LbKg

= 339,87

LbJam


= 13080 – 3425 (0,9146)2

= 10.190,75

KcalKg ⇒ 18.343,22

BtuLb


Q Fuel oil = 339,87

LbJam x 18.343,22

BtuLb

= 6.234.310,18

BtuJam ................................................(6)




TonHari = 154,41

KgFOJam


Kg

Cm2 x 14,22

Psia

Kg .Cm2 = 106,65 Psia



BtuLb (Lampiran 4)


BtuLb (Lampiran 3)




KgsteamKgFO

Maka Msteam = 0,26


KgFOJam

= 40,15


LbKg ⇒ 88,51

LbsteamJam


Q steam = 88,51

LbJam x (1173,7

BtuLb - 28,06

BtuLb )

= 101.400,59

BtuJam .......................................................(7)


Fuel Gas = 4,11

TonHari x

124

HariJam x

10001

KgTon x

2 ,204621

LbKg

= 377,54

LbJam

LHV(FG) = (155+1425 Rd)x(


= (155+1425 x0,7061)x(

6 .87180 .7061 )

= 11.300,783

KcalKg ⇒ = 20.341,43

BtuLb


Q Fuel gas = 377,54

LbJam x 20.341,43

BtuLb

= 7.679.703,48

BtuJam ................................................(8)


Tin = 32 oC = 305 K

Tout = 294 oC = 523 K


Kj

Kg0C = 20,97

BtuLb (Lampiran 5)


Kj

Kg0C = 21,82

BtuLb



= 41.200

KgJam x

2 ,204621

LbKg

= 90.830,34

LbJam


= 90.830,34

LbJam x (21,82

BtuLb - 20,97

BtuLb )

= 77.205,789

BtuJam ...................................................(9)




# Fuel Oil

# Fuel Gas

TOTAL

12.468.620,36

15.359.406,96

27.828.027,32

2 Heat Absorb

# Crude Oil

# Steam

# Udara

TOTAL

19.645.756,63

211,527,67

77.205,79

19.934.490,09

3 Heat loss 7.893.637,23


Efisiensi () =


x100 %

=

19 . 934 . 490,0927 . 828 .027,32

x100 % = 71.63 %




Feed = 2685

TonHari x

124

HariJam x

10001

KgTon x

2 ,204621

LbKg

= 246.641,86

LbJam

Sg = 0.8540

T-in rate = 167,4 oC = 333,32 oF

T-out rate = 191,24 oC = 376,23 oF

o API=141 ,5

SGR (60 o F /60 o F )−131 ,5

........................(Persamaan 1.4 Kern)

=

141 ,50 ,8540

−131 ,5

=34,19



Cp pada 34,19 oAPI

T-in rate = 167,4 oC = 333,32 oF

Cp = 0,61

BtuLb oF

T-out rate = 191,24 oC = 376,23 oF

Cp = 0,64

BtuLb

oF

Maka Cp rata-rata =

0 , 61+0 ,642 = 0,625

BtuLb oF


= 0,625

BtuLb oF (376,23 – 333,32) oF

= 26,82

BtuLb


Q FICI = 246.641,86

LbJam x 26,82

BtuLb

= 6.614.934,68

BtuJam ................................................(1)


Fuel Oil = 3,70

TonHari x

124

HariJam x

10001

KgTon x

2 ,204621

LbKg

= 339,87

LbJam


= 13080 – 3425 (0,9146)2

= 10.190,75

KcalKg ⇒ 18.343,22

BtuLb


Q Fuel oil = 339,87

LbJam x 18.343,22

BtuLb

= 6.234.310,18

BtuJam ................................................(2)




TonHari = 154,41

KgFOJam


Kg

Cm2 x 14,22

Psia

Kg .Cm2 = 106,65 Psia



BtuLb (Lampiran 4)


BtuLb (Lampiran 3)





Maka Msteam = 0,285


KgFOJam

= 44


LbKg

= 97,00

LbsteamJam


Q steam = 97,00

LbJam x (1173,7

BtuLb - 28,06

BtuLb )

= 110.127,08

BtuJam .........................................(3)


Fuel Gas = 4,11

TonHari x

124

HariJam x

10001

KgTon x

2 ,204621

LbKg

= 377,54

LbJam

LHV(FG)= (155+1425 Rd)x(


= (155+1425 x0,7061)x(

6 .87180 .7061 )

= 11.300,783

KcalKg ⇒ = 20.341,43

BtuLb


Q Fuel gas = 377,54

LbJam x 20.341,43

BtuLb = 7.679.703,48

BtuJam ................................................(4)



Feed = 2626

TonHari x

124

HariJam x

10001

KgTon x

2 ,204621

LbKg

= 241.222,17

LbJam

Sg = 0,8540

T-in rate = 248,52 oC = 479,33 oF

T-out rate = 291,24 oC = 556,23 oF

o API=141 ,5

SGR (60 o F /60 o F )−131 ,5

.................... (Persamaan 1.4 Kern)

=

141 ,50 ,8540

−131 ,5

=34,19



Cp pada 34,19 oAPI

T-in rate = 248,52 oC = 479,33 oF

Cp = 0,685

BtuLb oF

T-out rate = 291,24 oC = 556,23 oF

Cp = 0,7025

BtuLb oF

Maka Cp rata-rata =

0 ,685+0 , 7202 = 0,7025

BtuLb oF


= 0,7025

BtuLb oF (556,23 – 479,33) oF

= 54,02

BtuLb


Q FIC2 = 241.222,17

LbJam x 54,02

BtuLb

= 13.030.821,62

BtuJam ...............................................(5)


Fuel Oil = 3,70

TonHari x

124

HariJam x

10001

KgTon x

2 ,204621

LbKg

= 339,87

LbJam


= 13080 – 3425 (0,9146)2

= 10.190,75

KcalKg ⇒ 18.343,22

BtuLb


Q Fuel oil = 339,87

LbJam x 18.343,22

BtuLb

= 6.234.310,18

BtuJam ................................................(6)




TonHari = 154,41

KgFOJam


Kg

Cm2 x 14,22

Psia

Kg .Cm2 = 106,65 Psia



BtuLb (Lampiran 4)


BtuLb (Lampiran 3)




KgsteamKgFO

Maka Msteam = 0,26


KgFOJam

= 40,15


LbKg ⇒ 88,51

LbsteamJam


Q steam = 88,51

LbJam x (1173,7

BtuLb - 28,06

BtuLb )

= 101.400,59

BtuJam .......................................................(7)


Fuel Gas = 4,11

TonHari x

124

HariJam x

10001

KgTon x

2 ,204621

LbKg

= 377,54

LbJam

LHV(FG) = (155+1425 Rd)x(


= (155+1425 x0,7061)x(

6 .87180 .7061 )

= 11.300,783

KcalKg ⇒ = 20.341,43

BtuLb


Q Fuel gas = 377,54

LbJam x 20.341,43

BtuLb

= 7.679.703,48

BtuJam ................................................(8)


Tin = 32 oC = 305 K

Tout = 294 oC = 523 K


Kj

Kg0C = 20,97

BtuLb (Lampiran 5)


Kj

Kg0C = 21,82

BtuLb



= 41.200

KgJam x

2 ,204621

LbKg

= 90.830,34

LbJam


= 90.830,34

LbJam x (21,82

BtuLb - 20,97

BtuLb )

= 77.205,789

BtuJam ...................................................(9)




# Fuel Oil

# Fuel Gas

TOTAL

12.468.620,36

15.359.406,96

27.828.027,32

2 Heat Absorb

# Crude Oil

# Steam

# Udara

TOTAL

19.645.756,63

211,527,67

77.205,79

19.934.490,09

3 Heat loss 7.893.637,23


Efisiensi () =


x100 %

=

19 . 934 . 490,0927 . 828 .027,32

x100 % = 71.63 %




Feed = 2685

TonHari x

124

HariJam x

10001

KgTon x

2 ,204621

LbKg

= 246.641,86

LbJam

Sg = 0.8540

T-in rate = 167,4 oC = 333,32 oF

T-out rate = 191,24 oC = 376,23 oF

o API=141 ,5

SGR (60 o F /60 o F )−131 ,5

........................(Persamaan 1.4 Kern)

=

141 ,50 ,8540

−131 ,5

=34,19



Cp pada 34,19 oAPI

T-in rate = 167,4 oC = 333,32 oF

Cp = 0,61

BtuLb oF

T-out rate = 191,24 oC = 376,23 oF

Cp = 0,64

BtuLb

oF

Maka Cp rata-rata =

0 ,61+0 ,642 = 0,625

BtuLb oF


= 0,625

BtuLb oF (376,23 – 333,32) oF

= 26,82

BtuLb


Q FICI = 246.641,86

LbJam x 26,82

BtuLb

= 6.614.934,68

BtuJam ................................................(1)


Fuel Oil = 3,70

TonHari x

124

HariJam x

10001

KgTon x

2 ,204621

LbKg

= 339,87

LbJam


= 13080 – 3425 (0,9146)2

= 10.190,75

KcalKg ⇒ 18.343,22

BtuLb


Q Fuel oil = 339,87

LbJam x 18.343,22

BtuLb

= 6.234.310,18

BtuJam ................................................(2)




TonHari = 154,41

KgFOJam


Kg

Cm2 x 14,22

Psia

Kg .Cm2 = 106,65 Psia



BtuLb (Lampiran 4)


BtuLb (Lampiran 3)





Maka Msteam = 0,285


KgFOJam

= 44


LbKg

= 97,00

LbsteamJam


Q steam = 97,00

LbJam x (1173,7

BtuLb - 28,06

BtuLb )

= 110.127,08

BtuJam .........................................(3)


Fuel Gas = 4,11

TonHari x

124

HariJam x

10001

KgTon x

2 ,204621

LbKg

= 377,54

LbJam

LHV(FG)= (155+1425 Rd)x(


= (155+1425 x0,7061)x(

6 .87180 .7061 )

= 11.300,783

KcalKg ⇒ = 20.341,43

BtuLb


Q Fuel gas = 377,54

LbJam x 20.341,43

BtuLb = 7.679.703,48

BtuJam ................................................(4)



Feed = 2626

TonHari x

124

HariJam x

10001

KgTon x

2 ,204621

LbKg

= 241.222,17

LbJam

Sg = 0,8540

T-in rate = 248,52 oC = 479,33 oF

T-out rate = 291,24 oC = 556,23 oF

o API=141 ,5

SGR (60 o F /60 o F )−131 ,5

.................... (Persamaan 1.4 Kern)

=

141 ,50 ,8540

−131 ,5

=34,19



Cp pada 34,19 oAPI

T-in rate = 248,52 oC = 479,33 oF

Cp = 0,685

BtuLb oF

T-out rate = 291,24 oC = 556,23 oF

Cp = 0,7025

BtuLb oF

Maka Cp rata-rata =

0 , 685+0 , 7202 = 0,7025

BtuLb oF


= 0,7025

BtuLb oF (556,23 – 479,33) oF

= 54,02

BtuLb


Q FIC2 = 241.222,17

LbJam x 54,02

BtuLb

= 13.030.821,62

BtuJam ...............................................(5)


Fuel Oil = 3,70

TonHari x

124

HariJam x

10001

KgTon x

2 ,204621

LbKg

= 339,87

LbJam


= 13080 – 3425 (0,9146)2

= 10.190,75

KcalKg ⇒ 18.343,22

BtuLb


Q Fuel oil = 339,87

LbJam x 18.343,22

BtuLb

= 6.234.310,18

BtuJam ................................................(6)




TonHari = 154,41

KgFOJam


Kg

Cm2 x 14,22

Psia

Kg .Cm2 = 106,65 Psia



BtuLb (Lampiran 4)


BtuLb (Lampiran 3)




KgsteamKgFO

Maka Msteam = 0,26


KgFOJam

= 40,15


LbKg ⇒ 88,51

LbsteamJam


Q steam = 88,51

LbJam x (1173,7

BtuLb - 28,06

BtuLb )

= 101.400,59

BtuJam .......................................................(7)


Fuel Gas = 4,11

TonHari x

124

HariJam x

10001

KgTon x

2 ,204621

LbKg

= 377,54

LbJam

LHV(FG) = (155+1425 Rd)x(


= (155+1425 x0,7061)x(

6 .87180 .7061 )

= 11.300,783

KcalKg ⇒ = 20.341,43

BtuLb


Q Fuel gas = 377,54

LbJam x 20.341,43

BtuLb

= 7.679.703,48

BtuJam ................................................(8)


Tin = 32 oC = 305 K

Tout = 294 oC = 523 K


Kj

Kg0C = 20,97

BtuLb (Lampiran 5)


Kj

Kg0C = 21,82

BtuLb



= 41.200

KgJam x

2 ,204621

LbKg

= 90.830,34

LbJam


= 90.830,34

LbJam x (21,82

BtuLb - 20,97

BtuLb )

= 77.205,789

BtuJam ...................................................(9)




# Fuel Oil

# Fuel Gas

12.468.620,36

15.359.406,96

TOTAL 27.828.027,32

2 Heat Absorb

# Crude Oil

# Steam

# Udara

TOTAL

19.645.756,63

211,527,67

77.205,79

19.934.490,09

3 Heat loss 7.893.637,23


Efisiensi () =


x100 %

=

19 . 934 . 490,0927 . 828 .027,32

x100 % = 71.63 %

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