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  • 1. How many kmole of air are needed to burn 1 kmol of carbon?Jawab: Pembakaran sempurna: C + O2 1 CO2

    1 kmole O2 memerlukan CO2 . karna udara memiliki kandungan 21 % O2, iniberarti bahwa didalam melakukan pembakaran Carbon diperlukan udara (O2)sebanyak 4,76 kmol.

    2. Calculate the theoretical air-fuel ratio on a mass and mole basis for the combustion ofethanol, C2H5OH ?

    Jawab: Reaksi : C2H5OH + VO2(O2 + 3.76N2) aCO2+ bH2O + cN2Penyetaraan C : 2 = a Balance H: 6 = 2b b = 3Penyetaraan O : 1 + 2VO2 = 2a + b = 4 + 3 = 7 VO2 = 3(udara/bahan bakar) mol = VO2(1 + 3.76)/1

    = 3 4.76= 14.28

    (udara/bahan bakar) massa = (VO2MO2 + VN2 MN2)/MFuel= (331.999 + 11.2828.013)/46.069= 8.943

    3. A certain fuel oil has the composition C10H22. If this fuel is burned with 150% theoreticalair, what is the composition of the products of combustion?

    Jawab: C10H22 + (1/) VO2 (O2 + 3.76 N2) a H2O + b CO2 + c N2 + d O2Pembakaran sempurna : = 1, d = 0,

    NAMA : JUNIVER HUTASOITNIM : 5133122015PRODI : 2013

    BAHAN BAKAR DANPELUMASAN

    1. How many kmole of air are needed to burn 1 kmol of carbon?Jawab: Pembakaran sempurna: C + O2 1 CO2

    1 kmole O2 memerlukan CO2 . karna udara memiliki kandungan 21 % O2, iniberarti bahwa didalam melakukan pembakaran Carbon diperlukan udara (O2)sebanyak 4,76 kmol.

    2. Calculate the theoretical air-fuel ratio on a mass and mole basis for the combustion ofethanol, C2H5OH ?

    Jawab: Reaksi : C2H5OH + VO2(O2 + 3.76N2) aCO2+ bH2O + cN2Penyetaraan C : 2 = a Balance H: 6 = 2b b = 3Penyetaraan O : 1 + 2VO2 = 2a + b = 4 + 3 = 7 VO2 = 3(udara/bahan bakar) mol = VO2(1 + 3.76)/1

    = 3 4.76= 14.28

    (udara/bahan bakar) massa = (VO2MO2 + VN2 MN2)/MFuel= (331.999 + 11.2828.013)/46.069= 8.943

    3. A certain fuel oil has the composition C10H22. If this fuel is burned with 150% theoreticalair, what is the composition of the products of combustion?

    Jawab: C10H22 + (1/) VO2 (O2 + 3.76 N2) a H2O + b CO2 + c N2 + d O2Pembakaran sempurna : = 1, d = 0,

    NAMA : JUNIVER HUTASOITNIM : 5133122015PRODI : 2013

    BAHAN BAKAR DANPELUMASAN

    1. How many kmole of air are needed to burn 1 kmol of carbon?Jawab: Pembakaran sempurna: C + O2 1 CO2

    1 kmole O2 memerlukan CO2 . karna udara memiliki kandungan 21 % O2, iniberarti bahwa didalam melakukan pembakaran Carbon diperlukan udara (O2)sebanyak 4,76 kmol.

    2. Calculate the theoretical air-fuel ratio on a mass and mole basis for the combustion ofethanol, C2H5OH ?

    Jawab: Reaksi : C2H5OH + VO2(O2 + 3.76N2) aCO2+ bH2O + cN2Penyetaraan C : 2 = a Balance H: 6 = 2b b = 3Penyetaraan O : 1 + 2VO2 = 2a + b = 4 + 3 = 7 VO2 = 3(udara/bahan bakar) mol = VO2(1 + 3.76)/1

    = 3 4.76= 14.28

    (udara/bahan bakar) massa = (VO2MO2 + VN2 MN2)/MFuel= (331.999 + 11.2828.013)/46.069= 8.943

    3. A certain fuel oil has the composition C10H22. If this fuel is burned with 150% theoreticalair, what is the composition of the products of combustion?

    Jawab: C10H22 + (1/) VO2 (O2 + 3.76 N2) a H2O + b CO2 + c N2 + d O2Pembakaran sempurna : = 1, d = 0,

    NAMA : JUNIVER HUTASOITNIM : 5133122015PRODI : 2013

    BAHAN BAKAR DANPELUMASAN

  • C disetarakan : b = 10H disetarakan : a = 22/2 = 11O disetarakan : 2 VO2 = a + 2b = 11 + 20 = 31

    : => VO2 = 15.5Keadaan sebenarnya : 1/ = 1.5 => VO2 = 1.5 15.5 = 23.25

    H disetarakan : a = 11,C disetarakan : b = 10,N disetarakan : c = 23.25 3.76 = 87.42O2 disetarakan : d = 23.25 - 10 - 11/2 = 7.75 (kelebihan oksigen)

    4. Methane is burned with 200% theoretical air. Find the composition and the dew point ofthe products?

    Jawab: Persamaan reaksi untuk campuran stokiometri adalah :CH4 + VO2 (O2 + 3.76 N2) a H2O + b CO2 + c N2

    C disetarakan : 1 = bH disetarakan : 4 = 2aO disetarakan : 2 VO2 = a + 2b = 2 + 2 1 => VO2 = 2N2 disetarakan : 3.76 VO2 = c = 7.52

    200% theoretical air: VO2 = 2 2 = 4 so now more O2 and N2CH4 + VO2 (O2 + 3.76 N2) a H2O + b CO2 + c N2 + d O2

    N2 disetarakan : 3.76 VO2 = c = 15.04Ekstra oksigen : d = 4 1 - 1 = 2Produk : 2 H2O + 1 CO2 + 15.04 N2 + 2 O2

    Fraksi mol uap air : y = , = 0,0998Parsial tekanan uap air : Pv = yv Po = 0.0998 101 = 9.98 kPa Pg(Tdew)

    Pv = 9.98 kPa

    Tdew = 45.8oC

  • 5. Ethane (C2H6) is burned with 20 percent excess air during a combustion process.Determine (a) the air-fuel ratio on mole and massbasis and (b) the product composition?

    6. If I burn 1 kmol of hydrogen H2 with 6 kmol air what is A/F ratio on a mole basis andwhat is the percent theoretical air?

    Jawab: Combustion Eq. stoichiometric:

    H2 + VO2(O2 + 3.76 N2) 1 H2O + 3.76 VO2 N2VO2 = 0.5 ; (A/F)S

    = VO2 (1 + 3.76) / 1= 2.38

    6 kmol dari udara adalah : 1.26 O2 + 4.74 N2 .A/F ratio mol adalah 6, sehingga persen udara teoritis adalah% teoritis udara = ( / )( / ) x 100= , X 100 = %

    7. Acetylene (C2H2) is burned withstoichiometric amount of air during a combustionprocess. Asumming complate combustion, determine (a) the air- fuel ratio on a mass andon a mole basis and (b) composition of product gas ?

    Jawab : Pembakaran sempurna C2H2 yaitu :C H + a (O + 3,76N ) 2CO + H O + 3,76 aNO2 disetarakan : = 2 + 0,5 = 2,5

    Substitusikan:C2H2 + 2,5 (O2 + 3,76N2) 2CO2 + H2O + 9,4N2

    a. AF = ( )( ) = ( , , )( )( ) ( )( / ) = 13,3 kg air / kg fuelb. AF = = ( , , ) = ,

  • 8. Octane (C8H18) is burned with 250 percent theoritical air, which enters the combustionschamber at 250C. Asumming complate combustion and a total presure of 1 atm,determine(a) the air-fuel ratio on mole and mass basis, (b) the ER and (c) the product composition?

    Jawab: C8H18+2,5ath[O2+3,76N2] 8CO2+9H2O+1,5athO2+(2,5x3,76)athN2O2 disetarakan : 2,5ath = 8+4,5+1,5ath ath= 12,5

    Disubstitusikan :C8H18+31,25[O2+3,76 N2] 8 CO2+9 H2O+18,75 O2+117,5 N2AF = = ( , , )( / )( )( / ) ( )( / ) = ,

    (b) P = = , (101,325kPa) = ,(c) Tdp = Tsat 5,951 kPa = ,

    9. Propane (C3H8) is burned with 75 percent excess airduring a combustion process.Assuming complate combustion, determine (a) the air-fuel ratio on mole and mass basis,(b) the ER and (c) the product composition?Jawab: C3H8+1,75ath [O2+3,76 N2] 3 CO2 + 4 H2O + 0,75ath O2 + (1,75X3,76)ath N2

    O2 disetarakan : 1,75ath = 3 + 2 + 0,75ath ath = 5Disubstitusikan: C3H8+8,75 [O2 + 3,76 N2] 3 CO2 + 4 H2 O + 3,75 O2 + 32,9 N2AF = = ( , , )( / )( ) ( )( / ) = ,

    10. One kmol of ethane (C2H6) is burned with an unknown amount of air during acombustion process. An analytsis of the combustion products reveals that the combustionis coplate, and there are 3 kmol of free O2 in the products. Determine (a) the air-fuel ratioand (b) the percentage of theoritical air used during this process?