Question 3 Pet coke from a refinery plant in Alberta has following ultimate anal
ID: 701165 • Letter: Q
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Question 3 Pet coke from a refinery plant in Alberta has following ultimate analysis (composition) on a dry basis, percent by mass: C 84%, H =6%, O = 2%, S 6%, N = 1%, Ash-190 and the fuel contains 15% moisture a) If the excess air is 30%, calculate the air-fuel ratio on a mass basis. b) Calculate the dew point of the product if the pressure is 0.1 MPa. c) Calculate the amount of energy released per kmol of Pet coke. (Assume 30% excess air and product leaves at 1000 K) Use the following Dulong and Petit formula to calculate the heat of formation (Higher Heating Value) of the pet coke HHV 33823 C + 144249 (H - O/8) 9418 S (kJ/kg) Hint: formula for converting percentage of different elements into kilo moles. kmol orx component in fuel/1 kg of fuel-% of X component (100*Atomic weight of X component)Explanation / Answer
Assume: Weight of fuel on dry basis is 100 kg.
Available moisture is 15%, means 85% is dry weight.
Weight of total fuel is = 100 kg/ 85% = 117.647 kg.
Weight of moisture in the fuel is 117.647-100 = 17.647 kg of moisture.
Weight to mole analysis of petcoke
Weight (kg) = (Weight%*
Total weight on dry basis)
Theriotical oxygen requirement: Amount of oxygen required to combust 100% of fuel
For Carbon (Reaction) ===> C + O2 --> CO2 ------------------ (1)
(7 Kmol) (7 Kmol) (7 Kmol)
For Hydrogen ===> H2 + 0.5O2 --> H2O -------------------------------(2)
(3 Kmol) (1.5 Kmol) (3 Kmol)
(Since 6 Kmol of H equal to 3 Kmol of H2)
For Sulfur ===> S + O2 --> SO2--------------------------(3)
(0.1875 Kmol) (0.1875 Kmol) (0.1875 Kmol)
Theriotical Oxygen requirement = (O2 requirement in Rxn 1 to 3) - (O2 available in feed (petcoke))
= (7+1.5+0.1875)-0.125 = 8.5625 Kmol of O2 required to combust 100 % of feed.
Theriotical air requirement:
In air, mole ratio of O2 and N2 is 21% and 79% respectively. Based on this the 1 Kmol of O2 accompanies with (79/21) = 3.7619 Kmole of N2.
So one mole of O2 is avaialbe in (1+3.7619)= 4.7619 Kmol of air
Theriotical air requirement = (Theriotical oxygen requirement * 4.7619) = 40.7737 Kmoles.
Sub Part A: If the excess air 30%, air-fuel ratio is
If the excess air 30% is supplied means 1.3 times theriotical air is supplied for combustion
Total kmole of air supplied = 40.7737*1.3 = 53.0058 Kmoles.
Molecular Weight of air = 29 kg/kmol
Total weigth of air supplied = 1537.168 Kg
Total fuel combustion = 117.647 kg
So Air to fuel ratio is (AFR) = Weight of air/ weight of fuel = 1537.168/117.647 = 13.06 kg of air/ kg of fuel.
Sub Part B: Dew Point if total pressure 0.1 MPa.
The mole fraction of a component in gas phase is equal to ratio of partial pressure of component to total pressure.
xH2O = No of moles of H2O (NH2O)/ Total no of moles in gas (N) = pH2O/ P .
Total Pressure P = 0.1 Mpa
Moles analysis in product
Assume Considered 100% combsution since excess air supplied
CO2 from reaction1 is 7 kmol
H2O from reaction 2 is 3 Kmol
H2O from mositure in the feed in Kmole = Weight of moisture (H2O) in the feed / M.Wt of H2O = 17.647/18
=0.98 Kmol
Total H2O in product gas is = 3.98 Kmol.
SO2 from reaction 3 is 0.1875 Kmol
N2 available in the feed is = N in elemental form in feed/2 = 0.0714/2 =0.0357 Kmol
N2 supplied by air = (Total air supplied*0.79) = 53.008*0.79 = 41.87 Kmol.
Total N2 in product = 41.9057 Kmol
Total O2 in product = (53.008*0.21) - (O2 used in combustion or Theriotal O2 since 100 % combustion)
= 11.13168- 8.5625 = 2.5619 Kmol.
Total prodcut in Kmoles = 2.5619+41.9057+0.1875+3.98+7 = 55.635 Kmol
Mole fracttion of H2O = 3.98/55.635 = 0.071
Partial pressure of H2O in prodcut = 0.071*0.1 MPa = 0.0071 MPa * 7500 = 53.25 mm of Hg.
The vapor pressure of water at 39.35 Degree Centigrade is 53.25 mm of Hg.
Temperature at dew point is 39.35 oC
Element Weight (%)Weight (kg) = (Weight%*
Total weight on dry basis)
MolecularWeight Kg. Moles C 84 84%*100 = 84 12 7 H 6 6%*100 = 6 1 6 O 2 2%*100 = 2 16 0.125 S 6 6%*100 = 6 32 0.1875 N 1 1%*100 = 1 14 0.0714 Ash 1 1%*100 = 1 -Related Questions
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