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Udara 300 C

RH 49 % Flue Gas 4000 C

Bbg CO = 0,96 %

Flow rate = 92,2 ton/jam O2 = 5,55 %

t = 200 C CO2 = 22,09 %

P = 20,10 bar N2 = 71,4 %

ρ = 0,809 Kg/m3

CO2 = 5,02 % n C4H10 = 0,21 %

CH4 = 85,34 % i C5H12 = 0,06 %

C2H6 = 6,09 % n C5H12 = 0, 04%

C3H8 = 1,72 % C6H14 = 0,07 %

I C4H10 = 0,23 % N2 = 1,17 %

Penyelesaian :

Basis 1 jam operasi :

m = 92,2 ton/jam x 1 jam = 92,2 ton

Vt = mρ

= 92 ,2 ton x

1000kg1 ton

0,809 kg/m3

= 113.967,862 m3

FURNACE

T ?

nt = PVRT

= 20,10 x̄ 113.967,862m3

0,08205l atmgrmol K

x1,01325 ¿̄atm

x10−3m3

l.293K ¿

= 94.040.653, 92 grmol

Mencari n komponen

% n = % V

nknt

% = V kV k

%

nk = V kV k

. nt

mol masing-masing komponen

CO2 = 5,02 % x 94.040.653, 92 grmol = 4.720.840,827 grmol

CH4 = 85,34 % x 94.040.653, 92 grmol = 80.254.294,055 grmol

C2H6 = 6,09 % x 94.040.653, 92 grmol = 5.727.075,824 grmol

C3H8 = 1,72 % x 94.040.653, 92 grmol = 1.617.499,247 grmol

i C4H10 = 0,23 % x 94.040.653, 92 grmol = 216.293,504 grmol

n C4H10 = 0,21 % x 94.040.653, 92 grmol = 197.485,373 grmol

i C5H12 = 0,06 % x 94.040.653, 92 grmol = 56.424,392 grmol

n C5H12 = 0,04 % x 94.040.653, 92 grmol = 37.616,262 grmol

C6H14 = 0,07 % x 94.040.653, 92 grmol = 65.828,458 grmol

N2 = 1,17 % x 94.047.963, 94 grmol = 1.100.275,651 grmol

Neraca Massa Tanp Reaksi Kimia (N2 dan O2)

79 % N2

Udara 79 % N2

21 % O2 A G

bbg Gas hasil pembakaran

N2 masuk = N2 keluar

N2 bb + N2 udara = N2 keluar

1.100.275,651 grmol + 0,79 A = 0,714 G

1.100.275,651 grmol = 0,714 G – 0,79 A

O2 masuk = O2 keluar

4.720.840,827 grmol + 0,21 A = (12

x 0,0096 + 0,0555 + 0,2209) G

4.721.207,79 grmol + 0,21 A = 0,2812 G

4.721.207,79 grmol = 0,2812 G - 0,21 A

0,714 G – 0,79 A = 1.100.275,651 x 0,21 0,14994 G – 0,1659A = 231.057,8867 grmol

0,2812 G – 0,21 A = 4.721.207,79 x 0,79 0,222148 G – 0,1659A = 3729464,235 grmol

-

FURNACE

- 0,07221 G = - 3498406,367

G = 48449013 grmol

A = 42395243 grmol

Mol gas hasil pembakaran = G = 48449013 grmol

Mol udara = A = 42395243 grmol

N2 dari udara = 0,79 x 42395243 grmol = 33492320 grmol

N2 masuk = N2 bahan bakar + N2 dari udara = 1.100.275,651 + 33.492.320 = 34292595,65

O2 dari udara = 0,21 x 42395243 grmol = 8903021,8 grmol

Analisis orsat

CO = 0,96 % x 48449013 gmol = 465110,525 gmol

O2 = 5,55% x 48449013 gmol = 2688920,222 gmol

CO2 = 22,09 % x 48449013 gmol = 10702786,972 gmol

N2 = 71,4 % x 48449013 gmol = 34592595,282 gmol

Total = 48449013 gmol

O2 gas hasil = (1 x mol O2) + (1 x mol CO2) + (2 x mol CO)

= 2688920,222 gmol + 10702786,972 gmol + 930221,05 gmol

= 14321928,24 gmol

H2 bahan bakar

CH4 = 80.254.294,055 / 2 grmol = 40127147,028 grmol

C2H6 = 5.727.075,824 / 3 grmol = 1909025,275 grmol

C3H8 = 1.617.499,247 / 4 grmol = 404374,812 grmol

i-C4H10 = 216.293,504 / 5 grmol = 43258,701 grmol

n-C4H10 = 197.485, 373 / 5 grmol = 39497,075 grmol

i-C5H12 = 56.424,392 / 6 grmol = 9404,065 grmol

n-C5H12 = 37.616,262 / 6 grmol = 6269,377 grmol

C6H14 = 65.828,458 / 7 grmol = 9404,065 grmol

Total H2 bahan bakar = 42548380,397 grmol

Reaksi yang terjadi

H2 dari bahan bakar + ½ O2 dari udara → H2O yang teruapkan

Mol H2O teruapkan = Mol H2 bahan bakar = 42548380,397 grmol

Neraca massa

KOMPONENInput Output

Mol (gmol) Massa (gr) Mol (gmol) Massa (gr)

CH4 = 16 80.254.294,055 1.284.068.705 - -

C2H6 = 30 5.727.075,824 171.812.274,7 - -

C3H8 = 42 1.617.499,247 71.169.966,87 - -

i-C4H10 = 58 216.293,504 12.545.023,23 - -

n-C4H10 = 58 197.485,373 11.454.151,63 - -

i-C5H12 =72 56.424,392 4.062.556,224 - -

n-C5H12 = 72 37.616,262 2.708.270,864 - -

C6H14 = 86 65.828,458 5.661.247,388 - -

O2 = 32 8.903.021,8 284.896.697,6 14.321.928,24 458.301.703,7

CO2 = 44 4.720.840,827 207.716.996,4 10.702.786,972 470922626,8

CO = 28 - - 465.110,525 13023094,7

H2O = 18 - - 42.548.380,397 765870847,1

N2 = 28 34.292.595,65 960.192.678,2 34.592.595,282 968592667,9

Total 136.088.975,4 3.016.288.568 102.630.801,4 2.676.710.940

A

G

B

Reaksi Pembakaran

1. CO2

2. CH4 + 2O2 CO2 + 2H2O

3. C2H6 + 72

O2 2CO2 + 3H2O

4. C3H8 + 102

O2 3CO2 + 4H2O

5. i C4H10 + 132

O2 4CO2 + 5H2O

6. n C4H10 + 132

O2 4CO2 + 5H2O

7. i C5H12 + 8O2 5CO2 + 6H2O

8. n C5H12 + 8O2 5CO2 + 6H2O

9. C6H14 + 192

O2 6CO2 + 7H2O

10. N2

-∆H0R = (n. ∆H0cCH4 + n. ∆H0cC2H6 + n. ∆H0cC3H8 + n. ∆H0ci-C5H12 + n.

∆H0cn-C5H12 + n. ∆H0Cnn-C4H10 + n. ∆H0ci-C4H10 + n.∆H0cC6H14)

= (80.254.294,055 x -212.800 + 5.727.075,824 x -372.820 + 1.617.499,247 x -491.990 + 216.293,504 x -649.380 + 197.485,373 x –806.700 + 37.616 x -845.160 + 65.828,458 x -1.002.570)

= 2,014202389 x 1013grcal

Misalkan t = 10000 C

Cp O2 = 7, 941 gcal/gmol K

Cp CO = 7,587 gcal/gmol K

Cp CO2 = 11,92 gcal/gmol K

Cp N2 = 7,507 gcal/gmol K

Hp =

{(7,587gcalgmol

. k x 465.110,53 grmol)+( 11,92gcalgmol

. k x 10.702.786,97 grmol)+(7,507gcalgmol

. k x34.592 .595,28grmol )+(7,941grcalgrmol

. k x14.321 .928,24 grmol)} (1000-250 C)

= (3528793,553 grcal +127577220,7 grcal +259686612,8 grcal +113730432,2 grcal ) (975)

= 504.523.059,20 grcal (975)

= 4,9191 x 1011 grcal


Cp O2 = 8,203 gcal/gmol K


Cp CO2 = 12.69 gcal/gmol K


Hp =

{(7,866gcalgmol . k

x 465.110,53 grmol)+(12,69gcalgmol . k

x10.702 .786,97grmol )+(7,786gcalgmol . k

x34.592 .595,28 grmol)+(8,203grcalgrmol . k

x14.321 .928,24 grmol)} (1500-250 C)

=3658559,43 grcal + 135818366,8 grcal + 269337946,9 grcal + 117482777,4 grcal ) (1475)

= 526297650.5 grcal (1475)

= 7.7629 x 1011 grcal




Cp CO2 = 13,10 gcal/gmol K


Hp =

{(8,069gcalgmol . k


x10.702 .786,97grmol )+(7,994gcalgmol . k

x 34.592.595,28 grmol)+(8,423grcalgrmol .k

x14.321 .928,24 grmol)} (2000-250 C)

= 3752976,87 grcal + 140206509,3 grcal + 276533206,7 grcal + 120633601,6 grcal ) (1975)

= 1.0687 x 1012 grcal




Cp CO2 = 13, 24gcal/gmol K


Hp =

{(8,127gcalgmol . k


x 10.702.786,97 grmol)+(8.054gcalgmol . k

x34.592 .595,28 grmol)+(8,491grcalgrmol . k

x 14.321.928,24 grmol )}

(2200-250 C)

= 3779953,28 grcal + 141704899,5 grcal + 278608762,4 grcal + 121607492,7 grcal ) (2175)

= 1.1869 x 1012 grcal

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