Question

a) Explain the reason why heat is released when a hydrocarbon fuel is burned in a combustion chamber.

b) Why excess air is necessary to use in an actual combustion? Explain.

c) \(n\) - Dodecane \(\left(C_{12} H_{26}\right)\) in gas form is burned in a combustion chamber with \(150 \%\) excess air.

Determine the following:

i) the chemical reaction equation showing all reactants and products.

ii) the required air to fuel ratio by mass, \(A F_{m}\left(\frac{k g \text { air }}{k g \text { fuel }}\right)\),

iii) the rate of fuel consumed if \(300 \mathrm{~kg}\) of air per minute is supplied to the combustion, \(\left(\frac{k g \text { fuel }}{\min }\right)\)

iv) find if there will be condensation when the products are discharged to the atmosphere at \(45^{\circ} \mathrm{C}, 101 \mathrm{kPa}\), Give your reasoris.

Molar mass: H: 1 amu, \(0: 16\) amu, \(C: 12\) amu, \(N: 14\) amu

Answer 1

(a)

When Hydrocarbon fuel is burned in the combustion chamber, the bonds between the carbon-carbon and carbon-hydrogen elements get broken. Due to that break up of bonds, heat is generated. This is the reason why heat is released when hydrocarbon in burnt in combustion chamber.

(b)

Theoretically we don't need the excess air. But in real life practice, by providing theoretical can, chances of complete combustion is less. Due to that some unburnt fuel may go to the exhaust without being used. That's why we provide access air, so that the complete combustion will take place .

(c)

(i) Chemical Reaction:

As HOMEWORKLIB RULES's policy I have answered first 4 sub questions. Kindly ask remaining ones separately.

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Answer 2

Answer 3

Answer 4

(a)

When Hydrocarbon fuel is burned in the combustion chamber, the bonds between the carbon-carbon and carbon-hydrogen elements get broken. Due to that break up of bonds, heat is generated. This is the reason why heat is released when hydrocarbon in burnt in combustion chamber.

(b)

Theoretically we don't need the excess air. But in real life practice, by providing theoretical can, chances of complete combustion is less. Due to that some unburnt fuel may go to the exhaust without being used. That's why we provide access air, so that the complete combustion will take place .

(c)

(i) Chemical Reaction:

As HOMEWORKLIB RULES's policy I have answered first 4 sub questions. Kindly ask remaining ones separately.

Thanks

Answer 5

(a)

When hydrocarbon fuel is burned in the combustion chamber, the bonds between the carbon-carbon and carbon-hydrogen elements get broken. Due to that break up of bonds, heat is generated.

This is the main reason why heat is released when hydrocarbon is burnt in combustion chamber.

(b)

Theroretically we don't need the excess air, but in real life practice, by providing theoretical air, chances of complete combustion is less.

Due to that some unburnt fuel may go to the exhaust without being used. That's why we provide access air, so that the complete combustion will take place.

(c) (I) and (II)

Answer 6

1. Combustion reactions break the bonds of hydrocarbon molecules, and the resulting water and carbon dioxide bonds always release more energy than was used to break the original hydrocarbon bonds. That's why burning materials mainly made up of hydrocarbons produces energy and is exothermic.

2. The excess air results in oxygen that isn’t consumed during combustion, and this oxygen absorbs otherwise usable heat and carries it out of the stack. The chemically ideal amount of air entering a furnace is just enough for all the oxygen in the air to be consumed. However, this ideal (known as the stoichiometric air-to-fuel ratio) is difficult to reach because fuel and air don’t completely mix, meaning that a certain amount of excess air will always be necessary for complete combustion. In fact, too little excess air results in inefficient burning of fuel, soot buildup, and unnecessary greenhouse gas emissions.

3.
C2H₂ 6 + + (1.s) ath (O2+ 3.76N2 ) d Co2 xH2O + 02 + y d=12 2. 13 1.5 ath- d - a 2 ya 65 )( ath) - dan 124 6.5 = 18.5 Ath= 2 Y = 9:25 E lou:34 115 X 18.5% 3.76 2 = 15x ath X 3.76 Sto; chiometric equation: 12 Colt Ci thot (27.75)/ 02 + 3.76 N2 ) - 13 H2o+ 9.25 Ozt 104:34 N,

d (ii) Ais fuel ratio. | Mair) ( Nair Afm (Mfuel) ( Mfuel) 2611) Mfuel- 212 H26= (12x12) + 170 kg farol Mhina 28-97 lcg/kmol Afm = (2897)| 27.75) (170)11) 4.73 kga kg fuel Ass. (iii) ma= 300 ks/min rif ܟܢܝ 4.73 ma mid Afm rig- nia 1 3oo 473 63.4 ks min Afon amount of fel consumed

Mole fraction of H₂O in products. Total moles, Okottal Notal = 12 +13+ 9.25 + 10434 138.59 kmoles c00938 Thoot NHDO 13858 Ntotal Partial pressure due to H2o: (101 kpa) (0.0938) Hoo & Protal. y Hoo 9.47 kla Phoo. = Gilen, T= u5ec, usingtable (steem). 9,5953 kpa Peat = @ To usec condension oceurs MO PH2O psat at products. ine dow point T=45°C is not

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