Calculations 1 And 2 Final

CALCULATIONS 1 AND 2 1. 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 𝟖.𝟓𝟒

%=

= 𝟗𝟑. 𝟗𝟏% = 𝟗𝟓%

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

Given: 0.01 mole substance (C, H, O)

@STP

O 4 : H+C 7

896 cm3 CO2

O2 1 mole CO2 = 22,400 cm3 @STP

Required: a. Chemical formula of substance

0.72 g H2O

d. C8H16O4

Solution: V1 V2 = n1 n2 22, 400 cm3 896 cm3 = 1 mole n2 n2 = 0.04 mole CO2 mole H20 = 0.72 g ×

1 mole H2O = 0.04 mole H2O 18 g H2O

1 mole C = 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 mole C = 0.04 mole CO2 ×

C = 0.04 moles H= 0.08 moles O = 0.02 moles 0.04 =2 0.02 0.08 H= =4 0.02 0.02 O= =1 0.02 C=

Answer: C2H4O,

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 ) 1 mol C8H18

= 800 mols CO2

9 mols H2O ) 1 mol C8H18

= 900 mols H2O

CO2: 100 mols C8H18 (

H2O: 100 mols C8H18 (

25

mols O2

2 O2 supplied: 100 mols C8H18 (1 mol ) (200%) = 2500 mols O2 C8H18

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 % CO2 =

800 × 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%

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

c. 9.5%

d. 17.0%

Required: %C lost in the Refuse: Solution:

X=25 moles in refuse FC = C in Refuse

Basis: 100 moles of coal

C = 0.25(25) = 6.25 moles C

Ash Balance:

% C Lost =

8 mols Ash = 0.32(X)

6.25moles C in Refuse x100 78.05 moles C in Coal

% 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 ) 98lb

x + (250)(. 98) (

H2SO4 SO3 80lb ) (H2SO4) ( SO3 )] 98lb

= (250 + x) [. 2 + .8 (

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

Given: 100 Moles DFG P=780 mmHg T= 970 K PH20= 24 mmHg

Required: Total Volume Solution: nH2O P = H20 nTotal+nH20 PTOTAL nH2O 24 mmHg = 100 mol+nH20 780 mmHg

nH20= 3.1746 mol (PV = nRT)total V=

(100+3.1746)mol x 0.08205 780 mmHg atm 760 mmHg

L atm x 970 K mol K

1m3

x 1000 L= 8.0010 m3 ≈ 8 m3

d. 7754 L

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 6.93 % O2 (83.75 % N2)

Burner Feed 48 % S 52 % inerts Required: %S converted to SO3

Solution: Basis: 100 kmol of Burner Gas (Orsat) S+

3 O → SO3 2 2

O2 supplied = 83.75kmol N2 [

21 O2 ] = 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 SO3 produced = 6.01 kmol O2 [

%S → SO3 =

1 SO3 ] = 4.01 kmol SO3 3⁄ O 2 2

4.01 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

69.06

100

Total

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

7.01

15.82

Total

O2 theo = 7.01 mol C x

1 mol O2 + 1 mol C

15.82 mol H x

1 mol O2 4 mol H

= 10.97 mol O2

O2 supplied = 10.97 mol O2(1.30) = 14.26 mol O2 N2from air= 14.26 mol O2 x

79 mol N2 = 21 mol O2

CO2 produced = 7.01 mol x

1 mol CO2 5 mol x 7 mol 1 mol C

CO produced = 7.01 mol C x

1 mol CO 1 mol C

x

53.64 mol N2

2 mol 7 mol

H2= CO= 2.01 mol H2 Free O2= (0.30)(10.97) +1+1 = 5.29 mol H2

= 5.01 mol CO2 = 2.01 mol CO

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 O2 N2 H2O

4% 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)

1 mole H2

= 0.4800 mole O2

1 mole H2 O2 supplied = 0.4800 mole O2 (1.25) = 0.6 mole O2

Mole N2 = 0.6000 mole O2

79 mole N2 21 mole O2 21 mole O2

= 2.2571 mole N2

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%

3.3771

100%

TOTAL

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

CMB:

884.4 + (0.4y H2SO4) (98 g H2SO4) = (. 15 + .85 (98)) (X)

80g SO3

80

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% [ ] = 3.7453% g 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 %FeS2 in charge = [

5.5 + 114.3941 ] (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:

Air

Furnace Coal: 57.1%C 8%ash 1.4%N 0.77%S

Refuse: 24.5%C 75.5%ash

Stack Gas: 11.21%CO2 1.57%CO 7.45%O2 79.77%N2

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

Given: Burner gas: 100 kmol 25% CO

8% CH4

10% CO2

48% N2

5% H2

4% O2

Required: Theo O2 Solution: CO +

1 2

O2

CO2

d. 20 kmol

2H2 + O2

H2O

CH4 + 2O2

CO2 + H2O

1

Theo O2 = (25 moles CO) ( 2

mole O2

1 mole CO

1 moles O2

2 moles O2

) + (5 moles H2) (2 moles H2) + (8 moles CH4) (1 mole CH4)

= 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 30 mol H20 used in feed(

= 0.60

1 mol H2 ) 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

Given: F2 100 kg 28% H2SO4

F1 96% H2SO4 Required: kilograms of purchased acid, F1

P 50% H2SO4

d. 75

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 Volume of liquid solution =

1000 kg 1000

kg m3

= 1 m3 1 m3 =

350 kg 650 kg + kg ρB 700 3 m

ρB = 1300 kg/m3 1300 kg/m3 kg/m3

SG of B = 1000

= 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

Given: 1000lbs cottonseed SG=0.926 Required: minimum size of drum in gallon Solution:

c. 129.5

d. 254.2

lb lb ) = 57.7824 3 3 ft ft m 1000lb v= = = 𝟏𝟐𝟗. 𝟒𝟔𝟖𝐠𝐚𝐥 lb 1ft 3 ρ 57.7824 3 [ ] ft 7.481gal ρ = 0.926 (62.4

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

10

18

Y1 = 760−45 (29) = 0.0391

lbH2O lb da

Y2 = 760−10 (29) = 8.2759x10^ − 3 Y1 – Y2 = 0.0308 1

VH = (29 +

lbH2O lb da

lbH2O lb da

0.0308 0.7302(600+460) )( ) 18 1

VH = 28.0415

ft3 min hr ( )( ) 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

Given: 50% sulfuric acid 96% sulfuric acid 28% sulfuric acid at 100 kg

Required: kg of 96% sulfuric acid Solution:

c. 65

d. 75

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 k ln 2 = 0.2953

t 1⁄ = 2

t 1⁄

2

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 Given:

b. 6589 kPa

Pgage= 75.49kPa Patm = 10.8 mH2O

Required: P abs of water Solution: Pabs = Pgage + Patm

c. 18.48 mH2O

d. 36 psi

Pabs = 75.49 kPa [

10.33 mH2O ] + 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%

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 )] FC

C=5.88+0.00512(HHV-40.5S) + 0.0053[80 − 100 ( C=78.2778% C

d. 5.8%

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: %H in VCM =

1 (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%

Given: 100 mols C6H6

burner

30% excess air

Flue gas

d. 76.5%

C6 H6 +

15 O 2 2

→ 6CO2 + 3H2 O

Required: % water in flue gas Solution: 6 mols CO2 ] 6 H6

nCO2 = 100 mols C6 H6 [1 mol C

15

= 600 mols CO2

mols O2

nO2 entering = 100 mols C6 H6 [12mol C

6 H6

] (1.3) = 975 mols O2

79 mols N

nN2 = 975 mols O2 entering [ 21 mol O 2 ] = 3667.8571 mols N2 2

nH2 O =

3 mols H2 O 100 mols C6 H6 [ ] 1 mol C6 H6

= 300 mols H2 O 15

mols O2

nO2free = 975 mols O2 entering − 100 mols C6 H6 [12mol C

6 H6

] = 225 mols O2

nflue gas = nCO2 + nN2 + nH2 O + nO2free = 4792.8571 mols flue gas 300 mols H O

2 (100) = 𝟔. 𝟐𝟓𝟗𝟑% 𝐱𝐇𝟐𝐎 = 4792.8571 mols flue 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%

Given:

d. 41.46%

25% excess air

furnace 4% 𝐻2 𝑂

% moisture of flue gas

96%𝐻2 Required: % moisture of flue gas Solution: 1 H2 + O2 → H2 O 2 nO2 theo

1 mol O2 = 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

nH2 O = 0.6 mols O2 ×

Composition

1 H2 O + 0.04 = 1 mol H2 O 1 mol O 2 2

Kmols

% kmol

O2

0.12

3.5533

H2

2.2571

66.8354

1

29.6112

3.3771

100

𝐇𝟐 𝐎 Total

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 4.28% CO 8.24% O2 80.34% N2

ENGINE

Fuel Diesel Oil Required: cetane #

Solution: Basis: 100 moles of dry exhaust gas O2 from air = 80.34 x

21 = 21.356 mols 79

O2 unaccounted for = 21.356 − 7.14 −

4.28 − 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 = 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 volume of cetane = 0.255 x

𝐂𝐞𝐭𝐚𝐧𝐞 𝐍𝐮𝐦𝐛𝐞𝐫 = 𝟒𝟒. 𝟓𝟖