Calculations 1 and 2 Final

CALCULATIONS 1 AND 21. The gases from a sulfur burner have the following analysis: 9.86% SO2, 8.54% O2, and 81.60% N2.After passage of the gases through a catalytic converter, the analysis is 0.605% SO2, 4.50% O2, and 94.9% N2. What percentage of the SO2 entering the converter has been oxidized to SO3? a. 80% b. 90% c.95% d. 100% Given: 605%SO2, 4.5%02, 94.9%N2 Required: %SO2 converted Solution: Basis: 100kmol feed nSO2=8.54 kmol N2in = N2out 81.6=P(.949) P=85.98524763 SO2out=.00605(85.98524763)=.5202 kmol 𝟖.𝟓𝟒−.𝟓𝟐𝟎𝟐 %= 𝐱𝟏𝟎𝟎 = 𝟗𝟑. 𝟗𝟏% = 𝟗𝟓% 𝟖.𝟓𝟒 2. When 0.01 mole of a substance consisting of O, H, and C is burned, the following products are obtained: 896 cm3 of CO2 at STP and 0.72 g water. It is found that the ratio of oxygen mass to the mass of hydrogen plus carbon in the substance is 4/7. What is the chemical formula of the substance? One mole of CO2 has a volume of 22,400 cm3 at STP. a. C2H4O b. C5H6O2 c. C4H8O2 d. C8H16O4 Given: 0.01 mole substance (C, H, O) @STP O 4 : H+C 7 896 cm3 CO2 O2 0.72 g H2O 1 mole CO2 = 22,400 cm3 @STP Required: a. Chemical formula of substance  Solution: V1 V2 = n1 n2 22, 400 cm3 896 cm3 = 1 mole n2 n2 = 0.04 mole CO2 1 mole H2O mole H20 = 0.72 g × = 0.04 mole H2O 18 g H2O 1 mole C mole C = 0.04 mole CO2 × = 0.04 mole C 1 mole CO2 2 mole H mole H = 0.04 mole H2O × = 0.08 mole H 1 mole H2O 12 grams C g C = 0.04 mole C × = 0.48 g C 1 mole C 1 gram H g H = 0.08 mole H × = 0.08 g H 1 mole H 4 g O = × (0.48 g + 0.08 g) = 0.32 g O 7 1 mole O mole O = 0.32 g × = 0.02 mole O 16 g O C = 0.04 moles Answer: C2H4O, H= 0.08 moles O = 0.02 moles 0.04 C= =2 0.02 0.08 H= =4 0.02 0.02 O= =1 0.02 3. Determine the mole percent of CO2 in the products of combustion of C8H15 when 200% theoretical air is used. a. 5.5% b. 6.5% c. 7.5% d. 8.5% Given: 200% theoretical air C8H15 + 25/2O2 8CO2 + 9H2O Required: % CO2 in products Solution: Basis: 100 moles C8H15 8 mols CO2 CO2: 100 mols C8H18 ( ) = 800 mols CO2 1 mol C8H18 9 mols H2O H2O: 100 mols C8H18 ( ) = 900 mols H2O 1 mol C8H18 25 mols O2 2 O2 supplied: 100 mols C8H18 (1 mol C8H18 ) (200%) = 2500 mols O2 25/2 mols O2 O2 free = O2 sup-O2 used = 2500 mols O2 − 100 mols C8H18 ( 1 mol C8H18 ) = 1250 mols O2 79 mol N2 N2 sup = N2 in SG = 2500 mols O2 (21 mol O2) = 4909.7619 mols N2 800 % CO2 = × 100 = 6.47% 800 + 900 + 1250 + 4909.7619 4. Coal fired in a furnace has a heating value of 13800 Btu/lb and contains 78.05%C and 1.2%S. The proximate analysis shows 4%M, 24%FC, 8%A and the analysis of the refuse shows 8%VCM, 25%FC and 32% ash. Calculate the % of C lost in the refuse a. 8.0% b. 4.33% c. 9.5% d. 17.0% Given: 78.05%C HV = 13,800 BTU/lb 1.2%S Proximate Analysis Refuse 4%M 8% VCM 24%FC 25% FC 8%A 32 % Ash Required: %C lost in the Refuse: X=25 moles in refuse Solution: FC = C in Refuse Basis: 100 moles of coal C = 0.25(25) = 6.25 moles C 6.25moles C in Refuse Ash Balance: % C Lost = 78.05 moles C in Coal x100 8 mols Ash = 0.32(X) % C lost = 8.007 5. 250 lbs per hour of 98% H2SO4 enters an absorption tower of a contact sulfuric acid plant. If 20% oleum is produced per hour, how many pounds of SO3 are absorbed? a. 70 lb/h b. 80 lb/h c. 90 lb/h d. 100 lb/h Given: 250 lbs 98% H2SO4 Required: lbs SO3 absorbed Solution: Let x = SO3 absorbed SO3 absorbed = SO3 in the converter gas SO3 Balance: SO3 entering = SO3 leaving 1H2SO4 SO3 80lb H2SO4 SO3 80lb x + (250)(. 98) ( 98lb ) (H2SO4) ( SO3 ) = (250 + x) [. 2 + .8 ( 98lb ) (H2SO4) ( SO3 )] lb x = 90.28 hr SO3 absorbed 6. A stoichiometric problem was solved on the basis of 100 moles dry flue gas (DFG). The given condition at the stack outlet are as follows 780 mmHg, 970 K and the partial pressure is 24 mm Hg. The volume of the wet flue gas is: a. 6.10m3 b. 8.00m3 c. 8719 L d. 7754 L Given: 100 Moles DFG P=780 mmHg T= 970 K Required: Total Volume PH20= 24 mmHg Solution: nH2O P = H20 nTotal+nH20 PTOTAL nH2O 24 mmHg = 100 mol+nH20 780 mmHg nH20= 3.1746 mol (PV = nRT)total L atm (100+3.1746)mol x 0.08205 x 970 K 1m3 mol K V= 780 mmHg x 1000 L= 8.0010 m3 ≈ 8 m3 atm 760 mmHg 7. Impure sulfur is burned to SO2 for conversion to SO3 in a sulfuric acid plant. Orsat analysis of the burner shows 9.32% SO2, 6.93% O2. The charge fuel contains 48% sulfur. What percent of the sulfur fired leaves as SO3? a. 39.21% b. 29.91% c. 30.08% d. 35.78% Given: Burner Gas (Orsat) 9.32 % SO2 Burner 6.93 % O2 (83.75 % N2) Feed 48 % S 52 % inerts Required: %S converted to SO3 Solution: Basis: 100 kmol of Burner Gas (Orsat) 3 S+ O → SO3 2 2 21 O2 O2 supplied = 83.75kmol N2 [ ] = 22.26 kmol O2 79 N2 O2 disappearance = 22.26 − 9.32 − 6.93 = 6.01 kmol O2 O2 disappearance = O2 used to produce SO3 1 SO3 SO3 produced = 6.01 kmol O2 [ ] = 4.01 kmol SO3 3⁄ O 2 2 4.01 %S → SO3 = x100% 4.01 + 9.32 %𝐒 → 𝐒𝐎𝟑 = 𝟑𝟎. 𝟎𝟖𝟐𝟓% 8. An automobile uses a gasoline with an octane number of 85. Air is supplied 30% in excess such that the molal ratio of CO2 to CO is 5.2 and H2 to CO is 1:1 in the exhaust gas. What is the percent of O2 free in the stack gas? a. 2.95 b.7.37 c. 7.79 d. 9.51 Given: Octane number = 85 Ratio of CO2 to CO is 5:2 and H2 to CO is 1:1 30% Excess Air Required: O2 free in stack gas Solution: Density of iso-octane is 0.6918 and n-heptane is 0.684 g/ml. Volume Density Wt. % Wt. C8H18 85 0.6918 58.80 85.14 C7H16 15 0.684 10.26 14.86 Total 69.06 100 Basis: 100 kg of gasoline Wt. MW Mole At C At H C8H18 85.14 114 0.7468 5.97 13.44 C7H16 14.86 100 0.1486 1.04 2.38 Total 7.01 15.82 1 mol O2 1 mol O2 O2 theo = 7.01 mol C x 1 mol C + 15.82 mol H x 4 mol H = 10.97 mol O2 O2 supplied = 10.97 mol O2(1.30) = 14.26 mol O2 79 mol N2 N2from air= 14.26 mol O2 x 21 mol O2 = 53.64 mol N2 1 mol CO2 5 mol CO2 produced = 7.01 mol x 1 mol C x 7 mol = 5.01 mol CO2 1 mol CO 2 mol CO produced = 7.01 mol C x x = 2.01 mol CO 1 mol C 7 mol H2= CO= 2.01 mol H2 Free O2= (0.30)(10.97) +1+1 = 5.29 mol H2  Mol % mol CO2 5.01 7.37 CO 2.01 2.95 H2 2.01 2.95 O2 5.29 7.79 N2 53.64 78.94 67.94 100 The O2 Free in the stack gas is 7.79% 9. If moist hydrogen containing 4% water by volume is burned completely in a furnace with 25% excess air, calculate the percent moisture in the flue gas produced from the furnace. a. 11.89% b. 20.90% c. 29.61% d. 41.56% GIVEN: Moist H2 Flue Gas FURNACE 4% H2O O2 N2 H2O 25% excess air N2 REQUIRED: %H2O SOLUTION: H2 + ½ O2 H2O Basis: 1 mole moist H2 ½ mole O2 O2 theo = 0.96 (1 mole H2) = 0.4800 mole O2 1 mole H2 1 mole H2 O2 supplied = 0.4800 mole O2 (1.25) = 0.6 mole O2 79 mole N2 Mole N2 = 0.6000 mole O2 = 2.2571 mole N2 21 mole O2 21 mole O2 O2 excess = O2 free = 0.4800 mole O2 (0.25) = 0.1200 mole O2 1 mole H2O Mole H2O = 0.96 (1 mole H2) + 0.04 mol H2O= 1 mole H2O 1 mole H2 1 mole H2 COMPOUND MOLE %COMPOSITION O2 0.1200 3.5533% N2 2.2571 66.8454% H2O 1 29.6112% TOTAL 3.3771 100% 10. 15% oleum is to be produced using an absorbing acid, 40% H2SO4, and a burner gas containing 884.4 kg SO3. Calculate the mass of product solution that would be obtained if the gas leaving the absorption unit is SO3 free. a. 1222 kg b. 1622 kg c. 1922 kg d. 2222 kg Given: 15% oleum product Input: 40%H2SO4 mSO3 = 884.4 kg Required: m product Solution: TMB: y+884.4=x 80g SO3 80 CMB: 884.4 + (0.4y H2SO4) (98 g H2SO4) = (. 15 + .85 (98)) (X) X = 1621 kg 11. A mixture of pure sulfur and pyrites analyzing 85% FeS2 and 15% gangue is burned in a standard pyrites burner. The burner gas contains 10% SO2, 7% O2 and 83% N2 on an SO3-free basis and contains 1 mol SO3 per 100 mol SO3-free burner gas. The cinder contains 2%S as SO3. Calculate the percentage of FeS2 in the charge. a. 54.89% b. 73.35% c. 80.0% d. 91.23% Required: %FeS2 in charge Solution: Basis: 100 mol SO3-free Burner Gas g 1 mol(119.85 ⁄mol) FeS2 %FeS2 in cinder = 2% [ g ] = 3.7453% 2 mol(32 ⁄mol) S 4 mol FeS2 FeS2 → SO2 = 10 mol SO2 ( ) = 5 mol 8 mol SO2 4 mol FeS2 FeS2 → SO3 = 1 mol SO3 ( ) = 0.5 mol 8 mol SO3 n FeS2 = 5.5 mol 2 mol S . 98 n gangue = 5.5 (0.037453) mol FeS2 ( ) ( ) = 20.1872 mol 1 mol FeS2 . 02 . 85 n FeS2 in feed = 20.1872 ( ) = 114.3941 mol . 15 5.5 + 114.3941 %FeS2 in charge = [ ] (100) = 89.0867% 114.3941 + 20.1872 12. Coal fired in a furnace analyzes 57.1%C, 8%ash, 1.4%N, and 0.77%S. The refuse contains 24.5%C and 75.5%ash. Orsat analysis of the stack gas shows 11.21%CO2, 1.57%CO, 7.45%O2, and 79.77%N2. Complete the ultimate analysis of the coal. a. 4.2%O & 28.53%H c. 4.76%O & 27.97%H b. 5.95%O & 26.78%H d. 26.85%O & 5.88&H Given: Stack Gas: Air 11.21%CO2 1.57%CO 7.45%O2 79.77%N2 Furnace Coal: 57.1%C 8%ash Refuse: 1.4%N 24.5%C 0.77%S 75.5%ash Since there is no VCM in the refuse, the coal type is Coked Coal; BASIS: 100kmol DSG Overall Ash Balance (in kg): 0.08F = 0.755R Carbon Balance (in kg): 153.36 + 0.245R = 0.571F F = 281.3740kg R = 29.8144kg Excess O2 = 6.06kmol Theo O2 =21.2 – 6.06 = 15.14kmol Considering the Modified Analysis of the Coked Coal: netH = 7.01kg CW = 100 - 2.49 - 0.77 - 1.4 - 8 - 57.1 = 30.24% 7.01kg = 281.3740kg(%netH) %netH = 2.49% O = (8/9)(30.24) = 26.85% H = (1/9)(30.24) = 5.88% Using Dulong’s Formula: CV = 0.338C + 1.44netH + 0.094S CV = 0.338(57.1) + 1.44n(2.49) + 0.094(0.77) CV = 22.96MJ/kg 13. Determine the amount of O2 theoretically required for the combustion of 100 kmol of blast furnace gas analyzing 25% CO, 10% CO2, 5% H2, 8% CH4, 48% N2, and 4% O2. a. 37 kmol b. 31 kmol c. 27 kmol d. 20 kmol Given: Burner gas: 100 kmol 25% CO 8% CH4 10% CO2 48% N2 5% H2 4% O2 Required: Theo O2 Solution: 1 CO + O2 CO2 2 2H2 + O2 H2O CH4 + 2O2 CO2 + H2O 1 mole O2 1 moles O2 2 moles O2 Theo O2 = (25 moles CO) ( 2 ) + (5 moles H2) (2 moles H2) + (8 moles CH4) (1 mole CH4) 1 mole CO = 31 moles O2. 14. A well-known reaction to generate hydrogen from steam is the so-called water gas shift reaction: CO + H2 O → CO2 + H2 If the gaseous feed to the reactor consists of 30 moles CO, 12 moles CO2, and 35 moles steam per hour at 800 ˚C, while 18 moles of H2 are produced per hour, the limiting reactant is: a. CO b.steam c. CO2 d. H2 The ratio of CO to steam is 1:1. There are 30 moles of CO and 35 moles of steam, 30<35. The limiting reactant is CO. 15. In the preceding problem, the degree of completion of the reaction is: a. 0.10 b. 0.60 c. 0.45 d. 0.80 mols H2 produced Degree of Completion = mols H2 used in feed 18 mol H2 produced 1 mol H2 = 0.60 30 mol H20 used in feed( ) 1 mol H20 16. To prepare a solution of 50% sulfuric acid, a dilute waste acid containing 28.0% sulfuric acid is fortified with a purchased acid containing 96.0% sulfuric acid. How many kilograms of the purchased acid must be bought for a 100 kg of dilute acid? a. 45 b. 55 c. 65 d. 75 Given: F2 100 kg 28% H2SO4 F1 P 96% H2SO4 50% H2SO4 Required: kilograms of purchased acid, F1 Solution: TMB: F1 + 100=P CMB (H2SO4 balance): F1 (0.96) + (100) (0.28) = P(0.50) F1= 47.8261 kg 17. A solution of specific gravity 1 is 35% by weight A and the rest is B. If the specific gravity of A is 0.7, what is the specific gravity of B? a. 1.16 b. 1.5 c. 1.71 d. 1.8 Given: Liquid solution: SG = 1 ; xA=0.35 ; xB=0.65 SG of A = 0.7 Required: SG of B Solution: Basis: 1000 kg solution mA= 0.35X1000 = 350 kg mB= 1000 – 350 = 650 kg Density of liquid solution = 1x1000 = 1000kg/m3 1000 kg Volume of liquid solution = kg = 1 m3 1000 m3 350 kg 650 kg 1 m3 = + kg ρB 700 3 m ρB = 1300 kg/m3 1300 kg/m3 SG of B = 1000 kg/m3 = 1.3 18. You are asked what size of containers to use to ship 1000 lbs of cottonseed oil (SG = 0.926). What would be the minimum size of drum expressed in gallons? a. 85.6 b. 103.9 c. 129.5 d. 254.2 Given: 1000lbs cottonseed SG=0.926 Required: minimum size of drum in gallon Solution:  lb lb ρ = 0.926 (62.4 3 ) = 57.7824 3 ft ft m 1000lb v= = = 𝟏𝟐𝟗. 𝟒𝟔𝟖𝐠𝐚𝐥 ρ lb 1ft 3 57.7824 3 [ ] ft 7.481gal 19. Coal is mixed with water to form a slurry for transportation by pipelines. If 2 tons/hr of coal are mixed with water to give a slurry containing 50% coal by weight, find the mass (tons/hr) of water. a. 2 b. 4 c. 6 d. 3.5 Given: F = 2 tons/hr X2 = 0.5 Required: W Solution: TMB: P=F+W P = 2+W – eqn 1 CMB: 4 = 2+W 0.5P = 2 W = 4-2; P = 4 tons/hr W = 2 tons/hr 20. Humid air at temperature 600 F and a total pressure of 1 atm passed through a humidifier at the rate of 1000 ft3/min. If the partial pressure of water vapour in air is reduced from 45 mmHg to 10 mmHg. How many pounds of water is removed per hour? a. 1.09 b. 82.6 c. 76.3 d. 65.2 Given: Required: lbs water removed per hour, W Solution: 45 18 lbH2O Y1 = 760−45 (29) = 0.0391 lb da 10 18 lbH2O Y2 = 760−10 (29) = 8.2759x10^ − 3 lb da lbH2O Y1 – Y2 = 0.0308 lb da 1 0.0308 0.7302(600+460) VH = (29 + 18 )( 1 ) ft3 min hr VH = 28.0415 ( )( ) lb da 1000 ft3 60min lb da/hr = 2141.7488 lb H2O/hr = 0.0308 (2141.7488) lb H2O/hr = 65.9659 21. To prepare a solution of 50% sulfuric acid, a dilute waste acid containing 28% sulfuric acid is fortified with a purchased acid containing 96% sulfuric acid. How many kilograms of the purchased acid must be bought for a 100 kg dilute acid. a. 45 b. 55 c. 65 d. 75 Given: 50% sulfuric acid 96% sulfuric acid 28% sulfuric acid at 100 kg Required: kg of 96% sulfuric acid Solution:  Let x = kg of 96% sulfuric acid 0.5(100 + x) = 0.28(100) + 0.96(x) x= 47.8261 kg 22. A radioactive material was originally weighed 10 grams and after 6 months, only 1.7 grams remain. What is the half-life of the radioactive material? a. 1.35 mos b. 2.35 mos c. 0.35 mos d. 4.35 mos Given: NO= 10 grams N= 1.7 grams t= 6 months Required: half-life, t1/2 Solution: N = No e−kt 1.7 = (10)(e(−k)(6) ) k = 0.2953 ln 2 t 1⁄ = 2 k ln 2 t 1⁄ = 2 0.2953 t 1⁄ = 2.3473 months~2.35 months 2 23. A pressure gauge is attached and fixed to a delivery pump. The gauge reads 75.49 kPa. If the atmospheric is assumed to as 10.8 meters of water, determine the absolute pressure of water pumped. a. 1 atm b. 6589 kPa c. 18.48 mH2O d. 36 psi Given: Pgage= 75.49kPa Patm = 10.8 mH2O Required: P abs of water Solution: Pabs = Pgage + Patm  10.33 mH2O Pabs = 75.49 kPa [ ] + 10.8 mH2O = 𝟏𝟖. 𝟒𝟗𝟔𝟏 𝐦𝐇𝟐𝐎 101.325 kPa 24. A high volatile bituminous coal was found to contain 6% moisture, 60% fixed carbon, 6% ash, 2.4% S and 2.6% N with a heating value of 14000 BTU/lb. Calculate the % carbon in the coal. a. 69.81% b. 78.10% c. 78.43% d. 83.89% Given: HHV = 14000 BTU/lb Proximate Analysis: Moisture - 6% Fixed Carbon – 60% VCM – 28 % Ash – 6% Required: % Carbon Solution: Using Calderwood Equation: 1.55 28 C = 5.88 + 0.00512 (14000 − 40.5(2.4)) + 0.0053 (80 − 100 ( )) 60 C = 78.2778% 25. Calculate the % oxygen in the ultimate analysis. (Refer to no 24) a. 2.8% b. 3.8% c. 4.0% d. 5.8% Given: HV= 14000 BTU/lb 6% moisture, 60% FC, 6% ash 2.4%S, 2.6 %N Required: % O in the ultimate analysis Solution: % VCM=100-6-60-6=28 % VCM 100(28/60) = 46.66< 80 , use (-) using Calderwood to solve for %C VCM 1.55 C=5.88+0.00512(HHV-40.5S) + 0.0053[80 − 100 ( FC )] C=78.2778% C  HHV= 145.44C +620.228(net H) +40.50 S net H=4.0599 % CW=100-78.2778-6-4.0599-2.6-2.4-6=0.6623% CW HTOTAL =HNET H +HCW +HM HTOTAL =4.0599+0.6623(2/18) +6(2/18) HTOTAL =4.8% %O2 = 100-4.8-78.2778-2.4-6-2.6 %O2 = 5.9222 % 26. Calculate the % hydrogen in the VCM. (Refer to no. 24) a. 2% b. 4% c. 4.8% d. 6% Given: VCM =28% 6%M N=2.6% S=2.4% C=78.43% CW=0.5461% Net H=4.0239% Required: % H in VCM Solution: 1 %H in VCM = (0.5461) + 4.0239 = 4% 9 27. 100 moles of benzene are burned with 30% excess air. Assuming complete combustion, what is the percentage of water in the flue gas? a. 4.7% b. 6.3% c. 12.5% d. 76.5% Given: 100 mols C6H6 burner Flue gas 30% excess air  15 C6 H6 + 2 2 O → 6CO2 + 3H2 O Required: % water in flue gas Solution: 6 mols CO2 nCO2 = 100 mols C6 H6 [1 mol C ] = 600 mols CO2 6 H6 15 mols O2 nO2 entering = 100 mols C6 H6 [12mol C ] (1.3) = 975 mols O2 6 H6 79 mols N nN2 = 975 mols O2 entering [ 21 mol O 2 ] = 3667.8571 mols N2 2 3 mols H2 O nH2 O = 100 mols C6 H6 [ ] = 300 mols H2 O 1 mol C6 H6 15 mols O2 nO2free = 975 mols O2 entering − 100 mols C6 H6 [12mol C ] = 225 mols O2 6 H6 nflue gas = nCO2 + nN2 + nH2 O + nO2free = 4792.8571 mols flue gas 300 mols H O 2 𝐱𝐇𝟐𝐎 = 4792.8571 mols flue (100) = 𝟔. 𝟐𝟓𝟗𝟑% gas 28. If moist hydrogen containing 4% water by volume is burnt completely in a furnace with 25% excess air. Calculate the percent moisture of the flue gas produced from the furnace. a.11.89% b. 20.90% c. 29.61% d. 41.46% Given: 25% excess air furnace % moisture of 4% 𝐻2 𝑂 flue gas 96%𝐻2 Required: % moisture of flue gas Solution: 1 H2 + O2 → H2 O 2 1 mol O2 nO2 theo = 0.96 molsH2 O × 2 = 0.48 mol O2 1 mol H2 nO2 supplied = 0.48 mol O2 (1.25) = 0.6 mol O2 nO2 free = 0.48 mol O2 (0.25) = 0.12 mol O2 79 nN2 = 0.6 mol O2 ( ) = 2.2571 mol O2 21  1 H2 O nH2 O = 0.6 mols O2 × + 0.04 = 1 mol H2 O 1 2 mol O 2 Composition Kmols % kmol O2 0.12 3.5533 H2 2.2571 66.8354 𝐇𝟐 𝐎 1 29.6112 Total 3.3771 100 29. If a fuel is composed mainly of saturated hydrocarbon, what is the ratio of carbon to hydrogen in the fuel? a. 0.271 b. 0.324 c. 0.613 d. 0.890 Solution: C=1 H=4 C/H = ¼=0.25 30. A high speed diesel engine burns fuel to give an exhaust gas analyzing 7.14% CO2, 4.28% CO, 8.24% O2 and 80.34% N2. The cetane number of the fuel fired is Given: Assume diesel composition to correspond to cetane Density of cetane = 0.7751 g/mL Density of methyl naphthalene = 1.025 g/mL AIR Exhaust Gas 7.14% CO2 Fuel ENGINE 4.28% CO 8.24% O2 Diesel Oil 80.34% N2 Required: cetane # Solution: Basis: 100 moles of dry exhaust gas 21 O2 from air = 80.34 x = 21.356 mols 79  4.28 O2 unaccounted for = 21.356 − 7.14 − − 8.24 = 3.836 mols 2 at net H = 3.836 x 4 = 15.344 mols = at total H at C = 11.42 mols FUEL n At C At H C16H34 x 16x 34x C11H10 y 11y 10y at C bal: 16x + 11y = 11.42 at H bal: 34x + 10y = 15.344 x = 0.255 y = 0.667 226 volume of cetane = 0.255 x = 74.35 m3 0.7751 142 volume of MT = 0.667 x = 92.404 m3 1.025 74.35 % cetane by volume = x 100 = 44.58% 92.404 + 74.35 𝐂𝐞𝐭𝐚𝐧𝐞 𝐍𝐮𝐦𝐛𝐞𝐫 = 𝟒𝟒. 𝟓𝟖