teknik pembakaran

Nama: Maya Elvisa NIM: 061340411653Kelas: 5 EG BMata Kuliah: Teknik Pembakaran

Flue Gas

1.

FURNACE

Bahan bakar Batubara

O2 = 21 %

Udara Teoritis

C = 70 %H2= 15 %O2 = 21 %N2= 4 %N2 = 79 %O2 = 3 %S = 2 %H2O= 6 %

Penyelesaian :Basis 100 grmol bahan bakar.

Mol masing-masing komponen :C = 70 % x 100 grmol = 70 grmolH2 = 15 % x 100 grmol = 15 grmolN2 = 4 % x 100 grmol = 4 grmolO2 = 3 % x 100 grmol = 3 grmolS = 2 % x 100 grmol = 2 grmolH2O = 6 % x 100 grmol = 6 grmol

Reaksi Pembakaran

C + O2 CO2 ......(1) M70 grmol - -B70 grmol 70 grmol 70 grmolS - - 70 grmol BM12 gr/grmol 32 gr/grmol 44 gr/grmol

m - - 3080 gr

H2 + O2 H2O ....(2)M15 grmol - -B15 grmol 7,5 grmol 15 grmol

S - - 15 grmolBM 2 gr/grmol 32 gr/grmol 18 gr/grmol

m - - 270 gr

S + O2 SO2 ..(3)M2 grmol - -B2 grmol 2 grmol 2 grmol

S - - 2 grmolBM32 gr/grmol 32 gr/grmol 96 gr/grmol

m - - 128 gr

O2 teoritis= O2 (1) + O2 (2) + O2 (3)= (70 grmol + 7,5 grmol + 2 grmol)= 79,5 grmol

O2 teoritis dari udara= O2 teoritis O2 bahan bakar= 79,5 grmol 3 grmol= 76,5 grmol

N2 dari udara= x 76,5 grmol= 287,79 grmol

N2 keluar= N2 dari udara N2 bahan bakar= 287,79 grmol + 4 grmol= 291,79

H2O keluar= H2O reaksi + H2O bahan bakar= 15 grmol + 6 grmol= 21 grmol

KomponenInputOutput

n (grmol)Massa (gr)n (grmol)Massa (gr)

C 70 840--

H2 15 30--

N2 291,79 8170,12291,79 8170,12

O2 79,5 2544--

S2 64 --

H2O6 108 21 378

CO2-- 70 3080

SO2-- 2 128

Total464,2911756,12 384,79 11756,12

Flue Gas

CO2= 3080 gr = 70 grmol = 18,19 %H2O= 378 gr = 21 grmol = 5,46 %SO2= 128 gr = 2 grmol = 0,52 %FURNACE

N2= 8170,12 gr = 291,79 grmol = 75,83 %BatubaraBasis 100 grmol

Total= 11756,12 gr = 384,79 grmol = 100 %Udara Teoritis

C = 840 grH2= 30 grO2 = 76,5 grmol = 2448 gr N2= 112 grN2 = 287,79 grmol = 8058,12 grO2 = 96 grS = 64 grSub Total= 364,29 grmol = 10506,12 grH2O= 108 gr

Sub Total= 1250 gr

2. Flue Gas

FURNACE

Bahan bakar Batubara

O2 = 21 %

Udara 23 % Exhaust

C = 70 %H2= 15 %O2 = 21 %N2= 4 %N2 = 79 %O2 = 3 %S = 2 %H2O= 6 %

Penyelesaian :Basis 100 grmol bahan bakar.

Mol masing-masing komponen :C = 70 % x 100 grmol = 70 grmolH2 = 15 % x 100 grmol = 15 grmolN2 = 4 % x 100 grmol = 4 grmolO2 = 3 % x 100 grmol = 3 grmolS = 2 % x 100 grmol = 2 grmolH2O = 6 % x 100 grmol = 6 grmol

Reaksi Pembakaran

C + O2 CO2.(1) M70 grmol - -B70 grmol 70 grmol 70 grmolS - - 70 grmol BM12 gr/grmol 32 gr/grmol 44 gr/grmol

m - - 3080 gr

H2 + 1/2O2 H2O .(2)M15 grmol - -B15 grmol 7,5 grmol 15 grmol

S - - 15 grmolBM 2 gr/grmol 32 gr/grmol 18 gr/grmol

m - - 270 gr

S + O2 SO2 ..(3)M2 grmol - -B2 grmol 2 grmol 2 grmol

S - - 2 grmolBM32 gr/grmol 32 gr/grmol 96 gr/grmol

m - - 128 gr

O2 teoritis= O2 (1) + O2 (2) + O2 (3) = (70 grmol + 7,5 grmol + 2 grmol)= 79,5 grmol

O2 teoritis dari udara= O2 teoritis O2 bahan bakar = 79,5 grmol 3 grmol= 76,5 grmol

% excess= x 100 %

23 %= x 100 %

O2 Supply= 94,095 grmol

O2 Berlebih= O2 Supply O2 teoritis dari udara= 94,095 grmol 76,5 grmol= 17,595 grmol


N2 keluar= N2 dari udara + N2 bahan bakar= 353,976 grmol + 4 grmol= 357,976 grmol

H2O keluar= H2O reaksi + H2O bahan bakar= 15 grmol + 6 grmol= 21 grmol

KomponenInputOutput


C70 840--

H215 30--

N2 357,976 10023,328 357,976 10023,328

O2 97,0953107,04 17,595563,04

S2 64--

H2O6 108 21 378

CO2-- 70 3080

SO2-- 2 128

Total 548,07114172,368 468,571 14172,368

Flue Gas

CO2= 3080 gr = 70 grmol =14,94 % H2O = 378 gr = 21 grmol = 4,48 % SO2 = 128 gr = 2 grmol = 0,43 %FURNACE

N2 = 10023,328 gr = 357,976 grmol =76,39 % O2 = 563,04 gr = 17,595grmol = 3,76 %BatubaraBasis 100 grmol

Total= 14172,368 gr = 468,571grmol = 100,00 %Udara Teoritis

C = 840 grH2= 30 grO2 = 94,095 grmol = 3011,04 gr N2= 112 grN2 = 353,976 grmol = 9911,328 grO2 = 96 grS = 64 grSub Total= 448,071 grmol = 12922,368 grH2O= 108 gr

Sub Total= 1250 gr

3.

Flue Gas

FURNACE

Bahan bakar gas

O2 = 21 %

Udara

CH4= 7 grmolC2H6= 3 grmolO2 = 21 %N2 = 79 %Laju alir 12000 L/jamT = 25P = 1 atm

Penyelesaian :

Basis 1 jam operasi

Volume= Laju alir 1 jam operasi = 12000 L/jam 1 jam operasi = 12000 L

P V= n R Tn= =

n udara = 490,7795 grmol

n O2= n udara = 490,7795 grmol= 103,0637 grmol

Reaksi pembakaran

CH4 + 2O2 CO2+ 2H2O .(1) M 7 grmol - - -B 7 grmol 14 grmol 7 grmol14 grmolS - - 7 grmol 14 grmol BM 16 gr/grmol 32 gr/grmol 44 gr/grmol 18 gr/grmol

m - - 308 gr 252 gr

C2H6 + O2 2CO2+ 3H2O (2) M 3 grmol - - -B 3 grmol 10,5 grmol 6 grmol9 grmolS - - 6 grmol 9 grmol BM 30 gr/grmol 32 gr/grmol 44 gr/grmol 18 gr/grmol

m - - 264 gr 162 gr

O2 teoritis= O2 (1) + O2 (2) = 2 (7 grmol) + (3 grmol)= 24,5 grmol

% excess= x 100

= x 100

= 320,6682 %

O2 Berlebih= O2 Supply O2 teoritis = 103,0637 grmol 24,5 grmol= 78,3637 grmol


KomponenInputOutput


CH47112--

C2H6390--

O2103,06373298,038478,56372514,0384

N2387,715810856,0424387,715810856,0424

CO2--13572

H2O--23414

Total500,779514356,0808502,279514356,0808

Flue Gas

CO2 = 572 gr = 13 grmol = 2,59 % H2O = 414 gr = 23 grmol = 4,58 % N2 = 10856,0424 gr = 387,7158 grmol = 77,19 %FURNACE

O2 = 2514,0384 gr = 78,5637 grmol = 15,64 %Bahan Bakar Gas

Total = 14356,0808 gr = 502,2795 grmol= 100,00 %Udara Teoritis

CH4 = 112 grC2H6= 90 grO2 = 21%= 103,0637 grmol = 3298,0384 gr N2 = 79%= 387,7158 grmol = 10856,0424 grSub Total= 202 grSub Total= 490,7795 grmol = 14154,0808 gr

Illustrasi 2Gas kota mempunyai komposisi volumeCO22,6 %C2,73H4,728,4O20,7H239,3CO32,9C1,14H4,2810,1N25,4100 %a. Hitunglah mol O2 teoritis yang harus disuplai untuk pembakaran 1 mol gasb. Hitung nilai kalor gas kal/grmol dan Btu/cuft

JawabBasis:100 grmol gas

KomponenCO2= 2,6 % x 100 grmol = 2,6 grmolC2,73H4,72= 8,4 % x 100 grmol = 8,4 grmolO2= 0,7 % x 100 grmol = 0,7 grmolH2= 39,3 % x 100 grmol = 39,3 grmolCO= 32,9 % x 100 grmol = 32,9 grmolC1,14H4,28= 10,1 % x 100 grmol = 10,1 grmolN2= 5,4 % x 100 grmol = 5,4 grmol

Reaksi pembakaran

C2,73H4,72 + 7,82/2 O22,73 CO2 + 2,36 H2OM8,4 - - -(grmol)B8,4 32,844 22,932 19,824 (grmol)

S - -22,932 19,824 (grmol)BM37,48 32 44 18(gr/grmol)m - -1009,008356,832(gr)

H2+ O2H2OM39,9 - -(grmol)B39,919,9539,9(grmol)

S --39,9(grmol)BM23218(gr/grmol)

m--718,2(gr)CO + O2CO2M392,9 - -(grmol)B32,916,4532,9(grmol)

S --32,9(grmol)BM283244(gr/grmol)

m--718,2(gr)

C1,14H4,28 + 4,42/2 O21,14 CO2 + 2,14 H2OM10,1 - - -(grmol)B10,122,321 11,514 21,614 (grmol)

S - -11,514 21,614 (grmol)BM17,96 32 44 18(gr/grmol)m - -506,616389,052(gr)

O2 teoritis=mol O2(1) + mol O2(2) + mol O2(3) + mol O2(4) =(32,844 + 19,95 + 16,45 + 22,321) grmol=91,565 grmol

a. O2 supply===0,908 grmol

b. Basis=1 grmol gasNilai kalor:Hidrogen= 0,399 grmol x 68317 kal/grmol =27259 kalKarbon monoksida= 0,329 grmol x 67636 kal/grmol =22252 kalUnsaturated= 0,084 grmol { (2,73 x 9820) + (4,72 x 28200) + 28200 }kal/grmol=36120 kalParaffin= 0,101 grmol {(1,14 x 158100) + 54700}kal/grmol =23730 kalTotal = 109361 kal109361 kal = 511 btu/cuft

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