1. Crude petroleum oil is generally considered to be formed from animal and vegetable debris accumulating in sea basins or estuaries and decomposed by anaerobic bacteria resulting in a black viscous product. A typical elemental analysis shows 80% C, 13% H, 1% N, 3% O, and 3% S. During a certain combustion, air supplied is less than the theoretical so that all of the is used up. 70% of the C burns to , the rest to CO; the molal ratio of CO to in the exhaust gas is 1:2. Assume that the Sulfur in the fuel burns to and the Nitrogen combines with the nitrogen from air. Calculate: a) Orsat analysis of the exhaust gas b) % of the theoretical air which is supplied for combustion c) Equivalence Ratio
air
Given:
exhaust gas
fuel 80% C 13% H 1% N 3% O 3% S
70 % C 30% C CO
=
Required:
S
a. Orsat analysis of the exhaust gas b. % of the theoretical air which is supplied for combustion c. Equivalence Equivalence Ratio
+
O2
0.09375
SO2
0.09375
0.09375
= 12
H2 = 2 CO H2 (unburned) = 2(2.001) = 4.002 H2O = 6.5 – 6.5 – 4.002 4.002 = 2.498
Solution:
Basis: 100 kg Crude Petroleum
8012 = 6.6767 13 1 = 13 1 14 =0.0714 3 16 =0.1875 3 32 = 0.09375 C
+
4.67
C 2.001
+
H2 2.498 O2 theo =
2.498
2.498
6.67 0.0937 093755 0.1875 1875
= 7.013
4.67
(0.70)(6.67) (0.70)(6. 67) = 4.67
4.002
H2O
O2 air = O2 used
CO2
2
2
= 9.92 moles
O2
O
O
+
CO (0.30)(6.67) (0.30)(6. 67) = 2.001
N2 air = 7.013
N2 total = 26.38 +
= 26.38
. = 26.4517
a.) ORSAT ANALYSIS Gas n
%
CO2
4.67
12.56%
CO
2.001
5.38%
H2
4.002
10.76%
SO2
0.09375
0.25%
N2
26.4517
71.04%
=
37.21845 100.00%
b.) % theo air =
100
. 100% .
theo air = 70.7 %
d)
ER = .. = . ..
ER = 1.4
2. An alcogas mixture made up of 85% gasoline and 15% ethanol is used as fuel for an engine in the presence of 17.05 m 3 air/ kg alcogas supplied essentially dry at 30°C and 740 mmHg. 80% of the C burns to CO 2, the rest to CO; molal ratio of H 2 to CO is 1:2. Assume that gasoline has the same composition as a mixture of iso-octane-heptane with 95% octane number. Use a density of 0.6918 g/mL for iso-octane and 0.684 g/mL for n-heptane. Calculate: a) Orsat analysis of the exhaust gas b) % excess air
air
Given:
30℃
, 740 mmHg
exhaust gas
fuel 85% gasoline 95% octane number 15% ethanol
− =0.6818 − =0.684
80 % C 20% C CO
= 17.05
Required:
= 66.705 kgmol
a) Orsat analysis of the exhaust gas
O2 supplied = (0.21)(66.705) = 14.01435
b) % excess air
N2 supplied = (0.79)(66.705) = 52.42065 C converted to CO 2 = [(0.7083 x 8) + (0.0425 x 7) + (0.3261 x 2)](0.80) = 5.2929 kgmol CO 2
Solution:
C converted to CO = 5.2929 x (20/80) = 1.3232 kgmol CO
Basis: 100 kg alcogas
H2 unburned = 1.3232/2 = 0.6616 kgmol H 2
Gasoline = 95% octane number; 95 kg C8H18, 5 kg C7H16 m = ρV = (0.6918)(95) = 65.721 kg C 8H18(95% w)
= 14.0143510.29955 1.32322 0.6616 = 4.7072 a) ORSAT ANALYSIS
(0.684)(5) = 3.42 kg C 7H16 (5% w) 0.95 x 85 kg = 80.75 kg C 8H18 / 114 kg (MW)
n CO2 5.2929 CO 1.3232 H2 0.6616 O2 4.7072 N2 52.72065
= 0.7083 kgmol C 8H18 0.05 x 85 kg = 4.25 kg C 7H16 / 110 kg (MW) = 0.0425 kgmol C 7H16
% 8.18 2.04 1.02 7.27 81.48
15 kg C2H5OH / 46 kg (MW) b) % excess air
= 0.3261 kgmol C 2H5OH
8+. 7 0.3261 2 ℎ 0. 0 425 = +0..7083 . [ ] 0.3261 740 100 1000 ( ) 17. 0 5 760 1 = = 0.0820630273.151000 = 10.29955
% = ℎ 100 ℎ = ..–. 100 =.%
3. A primary blend of gasoline (90% octane no.), alcohol (75% ethanol, 25% methanol) and benzole (75% benzene, 15%toluene and 10% xylene) is burned completely in 35% excess O 2. Analysis of the blend shows 70% gasoline, 15% benzole and 15% alcohol. Calculate the complete analysis of the exhaust gas.
air
Given:
35% exhaust gas
fuel 70% gasoline 90% octane number 15% alcohol 75% ethanol 25%methanol 15% benzole 75% benzene 15% toluene 15% xylene Required: Complete analysis of the exhaust gas
80 % C 20% C CO
Solution: Basis: 100 kg fuel Gasoline =90% octane no.
CTotal=(0.5526)(8)+(0.7)(7)+(0.1442)(6)+(0.0245)(7)+(0.01415 )(8)+(0.2446)(2)+0.1172= 6.6671 HTotal=(0.5526)(18)+(0.7)(16)+(0.1442)(6)+(0.0245)(8)+(0.014 15)(10)+(0.2446)(6)+(0.1172)(4)= 14.2059
90% C8H18; ρ=0.6918 10% C7H16; ρ=0.684
OTotal=0.2446(0.5) +0.1172(0.5) =0.1809
mC8H18= (0.6918) (90) =62.262 kg= 63kg/114 =0.5526
Theo O2=6.6671+
mC7H16= (0.684) (10) =6.84 kg= 6.84kg/100 =0.7
. -0.1809 = 10.0337
O2 supplied= 10.0337(1.35) =13.5509 N2 supplied= 13.5509(79/21) = 50.9772 Benzole 75% benzene
(0.75) (15) = 11.25 kg C 6H6
15% C7H8
(0.15) (15) = 2.25 kg C 7H8
10% C8H10
(0.10) (15) = 1.5 kg C 8H10
Actual combustion C
+
6.6671 mC6H6= 11.25kg/78 =0.1442 kmol C 6H6 mC7H8= 2.25kg/92=0.0245 kmol C 7H8 mC8H10= 1.5kg/106 =0.01415kmol C 8H10
H2
+
7.10295
O2
CO2
6.6671
6.6671
O
H2O
3.5515
7.10295
2
O2free= 13.5509-10.0377 = 3.5132
Alcohol 75% C2H5OH
(0.75) (15) = 11.25/46= 0.2446 kmol
25% CH3OH
(0.25) (15) = 3.75/32 = 0.1172 kmol
Complete Analysis % CO2-6.6671
9.77
H2O-7.10295
10.41
O2-3.5132
5.15
N2-50.9772
74.67
4. A boiler uses high grade distillate fuel oil with a calorific value of 43.38 MJ/kg. Analysis of the stack gases at 375 torrs shows 9% , 1.21% CO, 0.37% , 7.05% , and 82.37% . Assuming that the fuel oil consists only of hydrocarbons. Calculate: a. % excess air b. Weight % composition of the fuel oil c. Air- fuel ratio d. % CV lost due to: i. Unburnt combustibles ii. Uncondensed water iii. Sensible heat
Given:
air
exhaust gas
℃
375 , 765 torrs
9% 1.21% CO
fuel
0.37%
CV= 43.43 Required:
℃
7.05%
82.37%
a. % excess air b. Weight % composition of the fuel oil
and 765
c. Air- fuel ratio d. Barrels per metric tonne of fuel oil at assume 1 = 6.2898 American barrels e. % CV lost due to: iv. Unburnt combustibles v. Uncondensed water vi. Sensible heat f. Adiabatic flame temperature
30
C API;
+
1.21 H2
+
10.49
½ O2
CO2
0.605
1.21
O
2
H2O
5.245
10.49
O2used= O2supplied- O2unused = 21.90-7.05 O2used= 14.85
Solution: Basis: 100 kg mol SG
14.85-(9+0.605) = 5.245 =(10.49+0.37)(2) =21.72
N2 supplied = 82.37
=21.90 O free= x’s + + . . Excess = 7.05 - Excess = 6.26 O2 supplied= 82.37(
b. Fuel oil Composition
2
C = 10.21 X 12 = 122.52 H = 21.72
a) % excess air
100 % = = ..−. 100 =.% C 9
+
O2
CO2
9
9
c. AFR =
.+. =20.96 .+.
% 84.94 15.06
d. % CV lost due to i) Unburnt combustible
.. 100 =5.4761% ... +. . % = ..+. 100 =1.6914% %CO+% =5. 4 7611. 6 914=7. 1 7% ii) Uncondensed water .. %Uncondensed = ..+. 100 = %CO=
7.38%
Q=n
. ..+. 100 =19.51%
%=
iii) Sensible heat
CO2 CO H2O H2 O2 N2
n
37.11 29.14 33.58 23.03 29.355 29.125
9 1.21 10.49 0.37 7.05 82.37
dT
5. A furnace is fired with fuel oil with a partial analysis of 7.6% S and 2.8%N. Orsat analysis of the stack gas shows 9.44% CO 2, 1.19% CO, 0.4% SO2 , 0.47% H2 ,6.8%O2, and 81.7%N2. Air is supplied is at 23oC, 755 mmHg and 85% RH. Calculate: a.)%excess air b.)Analysis of the fuel oil c.)m3air/kgfuel
℃
air 23 , 755mmHg, 85% RH
Given:
exhaust gas
9.44% 1.19% CO
fuel
0.37%
7.6% S 2.8% N
7.05% 82.37% 0.4% S
Required: a.)%excess air b.)Analysis of the fuel oil c.)m3air/kgfuel
Solution:
Let y= O2 in H2O
Basis: 100 kgmol DSG
Actual combustion
S bal:
.F=0.004(100)
F=168.42kg
C
+
9.44
NF = (0.028) (168.42)=4.7158/14=0.33684 kgmol
C
N2 supplied= 81.7-0.33684(1/2) = 81.53
1.19
+
O2
CO2
9.44
9.44
O
CO
0.595
1.19
O
H2O
y
2y
2
O2 supplied = 81.53(21/79) = 21.67
. . = 5.97 O x’s= 6.8 O2 x’s= O2theo 2
H2 2y
+
2
%x’s air %x’s air=
. .−.x100 =38.03%
H bal:
C bal:
%.= 9.44+1.19
(168.42)(0. 1386-x)= 0.47(2)+4y
1
O bal:
%H+%O= 1-(0.7574+0.076+0.028)
.+21.67(2) = 9.44(2) + 1.19+ 0.4(2) + 6.8(2) + 2y
%H+%O= 0.1386
Solving for x 1
Let x=%O
X=0.0245 x100 = 2.46%
%H=0.1386-x
%H=0.1386-0.0246=0.114 x100 = 11.4%H
%C= 0.7574x100= 75.74%
2
2
Ps=
.−.– . ++.
Ps =20.90mmHg nH2Oair = Analysis of Fuel Oil 7.6%S
...= 2.487 −.∗.
nair=103.2+2.487= 105.687
.. =2584.11 m
3
2.8%N
V=
75.74%C
=. =15.34 .
11.4%H 2.46%O
X
6. Coal tar fuels are liquid fuels obtained by blending coal tar distillation products such as carbolic oil, naphthalene oil, creosote oil, anthracene oil, and medium pitch. A sample elemental analysis shows 85.9% C, 6.3% H, 1.2% S, 5.5% O, and 1.1% N. If this fuel is burned in excess air at 30 and 755 mmHg with 90% RH, it produces a burner gas containing 10.64% , 3.19% CO and 0.64% . Assuming negligible N and S, Calculate: a. % excess air b. Complete Orsat analysis of burner gas c. stack gas (200 , 100 kPa) / kg coal tar
℃
d.
℃
℃
air 30 , 755mmHg, 90% RH
Given:
℃
burner gas 200 , 100 kPa
10.64% 3.19% CO
fuel 85.9% C 6.3% H 1.2% S 5.5% O 1.1% N
0.64%
negligible N & S
. . moles N = (8.56 + x)
Required:
free O2 = x +
a. % excess air b. Complete Orsat analysis of burner gas c. stack gas/ kg coal tar d.
2
by BG bal:
Solution:
. . + (8.56
51.76 = 5.507 + 1.651 + 0.3313 + x +
Basis: 100 kg Coal tar fuel
+ x) O2 theo =
. . . = 8.56 moles
x = 2.326 a.) % excess air =
Let x = x’s air O2 O2 air = 8.56 + x N2 air = (8.56 + x)
by C bal:
. = 0.10640.0319
100 . = . 100 = 27.18 %
b.) ORSAT ANALYSIS Gas
n
%
CO2
5.507
12.56%
moles BG = 51.76 moles
CO
1.651
5.38%
moles CO2 in BG = 0.1064 x 51.76 = 5.507
H2
0.3313 10.76%
moles CO in BG = 0.0319 x 51.76 = 1.651
O2
3.317
0.25%
moles H2 in BG = 0.0064 x 51.76 = 0.3313
N2
40.95
71.04%
51.76
100.00%
= .+.+. . =22.26
c.) Ps =
.− .− .n+ . Ps = 31.59 mm Hg
2.0284
Moles H2O from air =
d.) AFR =
.. = 40.9510.886 −. .
.+.+ .
= 15.32
7. Same as Problem 6, but assume that all the combustibles are converted to fuel to NO at 700
℃
and 120 kPa. Will the amounts of
which are 180 mg/L and 150 mg/L respectively?
SO
CO HO , and
, with Sulfur converted to
SO
and the nitrogen in the
and NO (mg/Ncm) be within the allowable requirements of the Philippine Clean Air Act
℃
air 30 , 755mmHg, 90% RH
Given:
℃
burner gas 700 , 120 kPa
10.64% 3.19% CO
fuel
0.64%
85.9% C 6.3% H 1.2% S 5.5% O 1.1% N
All combustibles are converted to & Sulfur converted to and Nitrogen to NO
Required: Amounts of
and NO (mg/Ncm) be equal to 180
and 150 respectively
Solution: At S in the fuel =
SO
O
. =0.0725 kgmol
NO = 0.512 kgmol
formed = 0.0725 kgmol
At N in the fuel =
SO
. =0.152 kgmol
NO formed = 0.512 kgmol
mg SO = 0.0725 kgmol × k × =4.64 × 10 mg × =4.54 × 10 mg mg NO = 0.512 kgmol × k Moles in the Stack Gas:
CO HO= . =6.085 N : 13.83 kgmol
= 79.12 kgmol
kgmol
= 4.49 kgmol
= 0.0725 kgmol 103.75 kgmol
O =13.830.0725 . . =16.613 =103.75 × .× =2535 m SO = . × =1829.9 . NO= . × =1798.4 . ∴ SO NO Theoretical Ncm =
and
exceeds the allowable requirements
kgmol
8. A German fuel blend called Reichkraftskoff is made up of 50% motor benzole (75% benzene, 15% toluene, 10% xylene), 25% tetralin ( ) and 25% industrial alcohol ( 80% ethanol, 20% methanol). After combustion in excess air, a stack gas containing 9.13% and 1.83% CO is obtained. Calculate: a. %excess air b. Complete orsat analysis of the stack gas
air
Given:
stack gas
9.13% 1.83% CO
fuel 50% motor benzole 75% benzene 15% toluene 15% xylene 25% tetralin ( ) 25% industrial alcohol 80% ethanol 20% methanol
Required:
CBAL 6.7582 = (0.913+0.0183)(DSG)
a. %excess air b. Complete orsat analysis of the stack gas
DSG=61.6131
Solution:
O2 SUPPLIED =9.8361+ X
Basis: 100 kg fuel N2 SUPPLIED= (8.8361+X)( 50 kg benzole
. =0.4808 . 15% =50(0.15) = =0.8815 15% =50(0.15) = =0.0472 25kg / 132 = 0.1894 kmol 25kg Industrial Alcohol = 0.4348 kmol C H OH 80% C H OH = 25(0.80)= 20% CH OH = 25(0.20)= = 0.1563 kmol CH OH = 0.4808 (6)+ 75%
=50(0.75) =
2
5
2
3
5
3
0.0815(7)+0.047258(8)+0.1894(10)+0.4348(2)+0.1563(1) = 6.7528 kmol H TOTAL=
0.4808 (6)+ 0.0815(8)+0.047258(10)+0.1894(12)+0.4348(6)+0.1563(4) = 9.5156 kmol O2 TOTAL = 0.4348(.5) + 0.1563(.5) = 0.2956 kmol O2 THEO = 6.7528 +
. – 0.2956 = 8.8361kmol
CO2=6.6253 CO= 1.1275
. .+ .
O2= X N2=
X=4.4218
(100) . = . 100
a. %excess air =
=50%
b. Orsat analysis Gas CO2 CO O2 N2
n 5.6253 1.1275 4.9856 49.8750
% 9.13% 1.83% 8.09% 80.95%
61.6134
100.00%
9. Biodiesel from a palm oil has an approximate formula of
℃
CHO CO
250 and 750 mmHg with a complete analysis of 12.08%
℃ O
. Excess air is supplied at 30
, 0.25% CO, 0.55%
H
, 2.12%
, 100kPa and 85%RH. The exhaust gas leaves at
, 71.17%
N
and the rest is
H O
. Find
a) formula of biodiesel b) Equivalence Ratio
℃
air 30 , 100 kPa, 85% RH
Given:
℃
exhaust gas 250 , 750 mmHg
12.08% 0.25% CO
fuel
0.55%
Required: a. formula of biodiesel b. Equivalence Ratio
2.12%
71.17% The rest is
Basis: 100 kgmol of exhaust gas
H O N O
in exhaust gas = 100 – 12.08 – 0.25 – 0.55 – 2.12 – 71.17 = 13.83 kgmol
N
in exhaust gas =
in air = 71.17
in air = 71.17 kgmol
= 18.92 kgmol
O
N
℃ =31.589 mmHg [.− .−. n+. × ] × . HO 71.17 18.92 × kPa × .. = 3.345 moles − .. CO H H 11.035×2=22.07 O 18.92 12.08 . 2.12 .−. =0.6475 0.6475×2=1.295 At 30 , T = 30 + 273 = 303 K, Ps is computed as: Ps =
Moles of
in air =
C in fuel = moles of
+ moles of CO = 12.08 + 0.25 = 12.33 moles
in fuel = (13.83 + 3.345) + 0.55 = 11.035 moles
Total
in fuel =
moles
in fuel =
Total O in fuel =
a) formula of biodiesel
. =9.52×2=19 . . =17.04×2=34 H= . . =1×2=2 O= . CHO C=
mole
moles
b) Equivalence Ratio Using mass balance Let FO = wt. of Fuel Oil FO + (71.17 + 18.92)(29) + (3.45)(18) = 12.08(44) + 0.25(28) + 0.55(2) + 2.12(32) + 71.17(28) +13.83(18) FO = 176.34 kg
O O =2.12 . . =1.72 O O O . fffu .. ir ctu ∅= = =0. 9 1 . f fu f ir tichitric . Excess
= free
Theoretical
=
moles
supplies – excess
= 18.92 – 1.72 = 17.2 moles
.
10. Biodiesel made from jatropha was found to have a Gross Calorific Value of 39.63 MJ/kg and an analysis of 14.2% palmitic acid, C16H32O2, 6.9% stearic acid, C 18H36O2, 43.1% oleic acid, C 18H34O2, and 35.8% linoleic acid, C 18H32O2. This fuel is burned in excess air at 32°C and 98 kPa with 75% RH. Partial orsat analysis of the exhaust gas shows 11.63% CO 2, 0.61% CO, and 0.92% H 2. The exhaust gases leave at 300°C and 740 mmHg. Find: a) % excess air b) AFR, Air-Fuel Ratio c) Combustion efficiency
℃
air 32 , 98 kPa, 75% RH
Given:
℃
exhaust gas 300 , 740 mmHg
11.63% 0.61% CO
fuel 14.2% palmitic acid , 6.9% stearic acid, 43.1% oleic acid, 35.8% linoleic acid,
0.92%
Required:
CO = .0061 x 52.1023 = 0.3178;
a) % excess air
H2 = 0.0092 x 52.1023 = 0.4793
b) AFR, Air-Fuel Ratio
Free O2 = x +
c) Combustion efficiency Solution: Basis: 100 kg biodiesel
. + .
52.1023=6.05950.3178 0.31780.2 4793 9.0012 (7921) 0.4793 x = 2.3070 a) %excess air =
C bal:
14.216 6.918 43.118 35.818 256 =2840.11630.0282061 280 EG = 52.1023 kgmol Ctotal = 6.3773 kgmol Htotal =
. . . . = 11.9375 kgmol
O2 = 0.3605 kgmol
ℎ =6.3773 . 0.3605=9.0012; let x = excess O2 O2 supplied = 9.0012 + x N2 supplied = (9.0012 + x)(79/21) CO2 = 0.1163 x 52.1023 = 6.0595;
. 100 =. % .
b) AFR
= . –.−.n+ . . .+. = ..−. . . . =. = ... +
= 35.406
mmHg
= 2.018
c) Combustion Efficiency
1 8 % = 0.039.317963 283. 100 100=2.27% % = 0.439.79463286.10003 100=3.46%
% 11.92375 0.4794 44.04 = 39.63 100 100=6.10% CO2 CO H2O O2 N2 H2
n 6.0595 0.3178 7.5074 2.7056 42.5404 0.4794
Cp 0.04654 0.03025 0.03603 0.03183 0.02995 0.01459
nCp 0.2821 0.009616 0.2711 0.0861 0.2624 0.006994 1.9183 MJ/°C
1.9183 100 % ℎ= 30025 39.63 100 =13.31% =10013. 3 16. 1 03. 4 62. 2 7= .%
