1. A 650 Bhp diesel engine uses fuel oil of 28°API gravity, fuel consumption
is 0.65 lb/Bhp – hr. Cost of fuel is P 7.95 per liter. For continuous operation,
determine the minimum volume of the cubical day tank in cm3, ambient
temperature is 45°C.
The theoretical volume required:
mRT
V=
P
141.5
141.5
SGat 15.6°C =
=
=0.887147
131.5+°API 131.5+28
Solving for m:
F
O
m= ( ) = 11.5C+34.5 (H- ) +4.3S
A t
8
SGat 45°C =SGat 15.6°C [1-0.0007(t2 -t1 )]
SGat 45°C =SGat 15.6°C [1-0.0007(45-15.6)]
SGat 45°C =0.86889
F
.1861
( ) = 11.5(.6565)+34.5 (.0587) +4.3(.0151)
A t
8
m=8.84
ρat 45°C =SGat 45°C ρwater =0.86889(1 kg⁄L)=0.86889 kg⁄L
mf
BPsfc =
BP
mf =650 Bhp(0.65 lb⁄Bhp-hr)
mf =422.5 lb⁄hr
ρat 45°C =
mf
Vat 45°C
lb
1 kg
422.5 hr (2.205 lb)
L
Vat 45°C =
=220.5227
0.86889 kg⁄L
hr
L 1 m3
100 cm 3 24 hr
Volume of day tank=220.5227 (
) (
)
)(
hr 1000 L
1m
day
Volume of day tank=5,292,544.8 cm3 ⁄day
2. It is required to find the theoretical volume of air at 20°C and 100kPa
absolute pressure to burn one kilogram of Franklin County coal. The ultimate
analysis of coal as fixed is as follows.
C = 65.65 % O = 18.61 %
Mois = 3 %
H = 5.87 % S = 1.51 %
S = 1.51%
Solution:
kgair
kgfuel
Then:
(8.84
V=
V=7.43
kgair
kJ
) (.287
) (20+273)K
kgfuel
kg-K
100 kPa
m3 air
kg coal
3. Calculate the minimum volume of day tank of 28° API fuel having a fuel
consumption of 1kg/s.
V=
m
Density
Solving for density:
SGat 15.6°C =
141.5
141.5
=
=0.887
131.5+°API 131.5+28
Then:
V=
kg
1 s
kg
0.887 (1000 3 )
m
V =97.41m3
(
3600s 24hr
)(
)
hr
day
4. Calculate the higher heating value of liquid dodecane fuel. The chemical
formula for DOdecane is 𝐶12 𝐻26.
Theoretical weight of air:
F
O
( ) = 11.5C+34.5 (H- ) +4.3S
A t
8
Qs =13,500C+60,890H Btu/lb
F
.07
( ) = 11.5(.715)+34.5 (.05) +4.3(.36)
A t
8
Where:
Weight of Carbon
%C=
(100)
𝑭
( ) = 𝟗. 𝟑𝟎𝒌𝒈𝒂𝒊𝒓 /𝒌𝒈𝒇𝒖𝒆𝒍
𝑨 𝒕
Weight of Fuel
12(12)
%C=
12(12)+1(26)
6. There are 20 kg of flue gasses formed per kg of fuel oil burned in the
combustion of a fuel oil C12H26. What is the excess air in percent?
(100)
A. 26.67%
B. 18.34%
%C=84.6
Weight of Hydrogen
%H=
SOLUTION:
(100)
Weight of Fuel
C12H24 + 18.5O2 + 18.5(3.76) N2
18.5(3.76) N2
12(26)
%H=
12(12)+1(26)
C. 12.34%
D. 20.45%
12CO2 + 13H2O +
(100)
Theo. A/F =
18.5+ 3.76(18.5)
1
%H=15.29%
Theo. A/F = 88.06 mol/mol
Then:
Theo. A/F in kg/kg =
Qs =13,500(.8461)+60,890(.1529) Btu/lb
88.06 (28.97)
12(12) + 26(1)
Qs =48,225.04 Btu/lb
Theo. A/F in kg/kg = 15 kg air/ kg fuel
5. A bituminous coal has the following compositions:
Mass of air = 20 kg flue gas – 1 kg fuel gas
C = 71.5 %
H = 5.0 %
O = 7.0 %
N = 1.3 %
S = 36 %
Ash = 8.2 %
W = 3.4 %
Mass of air = 19 kg air
Actual A/F = 19 kg air / kg fuel
Calculate for the complete combustion the theoretical weight of air
required in 𝑘𝑔𝑎𝑖𝑟 /𝑘𝑔𝑓𝑢𝑒𝑙
Actual A /F = Theo. A/F (1 + e)
19 = 15(1 + e)
7.108C + 7.05H2 + 0.016S + 0.004N2 + 1.25(10.649)O2 +
1.25(10.649)(3.76)N2→ 7.108CO2 + 7.05H2O + 0.016SO2 + 50.054N2 +
0.25(10.469)O2
e = 0.2667
e = 26.67%
7. A bituminous coal has the following compositions:
C = 71.5%
H = 5.0%
O = 7.0%
S = 3.6%
Ash = 8.2 %
W = 3.4%
N = 1.3%
Determine the theoretical weight of Oxygen in lb/lb of coal
Theo. A/F = 11.5C + 34.5(H-O/8) + 4.3S
Theo. A/F = 11.5(0.715) + 34.5(0.05-0.07/8) + 4.3(0.036)
Theo. A/F = 9.8 lb air/ lb coal
mT = total mols in product
mT = 7.108 + 7.05 + 0.016 + 50.054 + 2.662
mT = 66.89 mols
Partial pressure of H2O = (7.05/66.89)(170)
Partial pressure of H2O = 17.92 kPa
9. A circular fuel tank 45 feet long and 5.5 feet diameter is used for oil
storage. Calculate the number of days the supply tank can hold for
continuous operation at the following conditions:
Steam flow = 2000 lbs/hr
Steam dry and saturated at 200 psia
Feedwater temperature = 230˚F
Boiler efficiency = 75%
Fuel oil = 34˚API
O2 in air by weight = 23.2%
From steam tables:
Therefore:
Theoretical weight of O2 = 0.232(9.8)
At 200 psi (1.380 Mpa), hs = 2789.6 kJ/kg
At 230˚F(110˚C), hf = 461.3 kJ/kg
Theoretical weight of O2 = 2.274 lb/ lb coal
8. A steam generator burns fuel oil that has the following chemical analysis
by mass in percent:
C = 85.3H2 = 14.1
S = 0.5
N2 = 0.1
Converting the given mass analysis to molal analysis:
C
H2
S
N2
85.3/12 =
14.1/2
0.5/32
0.1/28
7.108
=
7.050
=
0.016
=
0.004
Combustion reaction with 125%theoretical air:
h = 41,130 + 139.6(34)
h = 45,876 kJ/kg
ms = 2000/2.205
ms = 907 kg/hr
ms (hs -hf )
m f Qh
907(2789.6-461.3)
0.75=
mf (45,876)
mf = 61.376 kg/hr
141.5
SG@15.6C̊ =
131.5+34
SG@15.6C̊ =0.855
ηb =
O
Qh =33, 820C+144,212 (H- ) +9,340S
8
0.035
Qh =33, 820(0.69)+144,212 (0.025) +9,340(0.07)
8
Qh =26,961.45 kJ/kg
1000kg
Density=0.855 (
)
m3
Density=0.855kg/m3
π 5.5 2 45
Volume of tank= (
) (
)
4 .281
3.281
Volume of tank=30.297m3
12. A diesel power plant uses fuel that has a density of 892.74 kg/m3 at
15.67 ˚C. Find the heating values of fuel.
Total weit of fuel=30.297m3 x855kg/m3
Total weit of fuel=25,904 kg
25,904
Number of days=
61.376(24)
Number of days=17.58 days
10. Determine the minimum volume of day tank in m3 of 28˚API fuel having
a fuel consumption of 200 kg/hr.
SG=
Density of fuel
Density of water
SG=
892.74
1000
SG=0.89274
ÅPI=
141.5
-131.5
0.89274
141.5
131.5+ÅPI
SG@15.6C̊ =0.887
ÅPI=27
Density of fuel = 0.887(1000)
Density of fuel = 887 kg/m3
m
w=
V
200
887=
V
0.22548m3 24 hrs
V=
x
hr
day
V=5.41m3
Q=44, 899.31 kJ/kg
SG@15.6C̊ =
11. A certain coal has the following ultimate analysis:
C = 69%
N2 = 5%
H2 = 2.5%
S = 7%
O2 = 3.5%
Ash = 5%
Determine the heating value of fuel used.
Q=41,130+139.6ÅPI
Q=41,130+139.6(27)
13. A boiler burns fuel oil with 15% excess air. The fuel may be represented
by C14H30. Calculate the molal air fuel ratio.
Fuel+Air→ Product of Combustion
C14 H20 +O2 +3.76N2 →CO2 +H2 O+ 3.76N2
C14 H20 +21.5O2 +21.5(3.76)N2 → 14CO2 + 15H2 O+ 21.5(3.76)N2
Moisture = 8%
A 21.5+21.5(3.76)
Theoretical =
F
1
A
Theoretical =102.34
F
A
Actual = 102.34(1.15)
F
𝑘𝑔𝑎𝑖𝑟
A
mol air
Actual =117.69
F
mol fuel
Actual A/F = 20 kg flue gases – 1 kg fuel = 19 𝑘𝑔𝑓𝑢𝑒𝑙
% Excess air =
14. A diesel power plant consumes 650L of fuel at 26˚C in 24 hours with 28˚
API. Find the fuel rate in kg/hr.
=
𝐴𝑐𝑡𝑢𝑎𝑙 𝐴/𝐹−𝑇ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝐴/𝐹
𝑇ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝐴/𝐹
× 100
19−15
× 100
15
= 26.67%
141.5
131.5+ÅPI
141.5
SG@15.6C̊ =
131.5+28
SG@15.6C̊ =
16. A fuel has the following volumetric analysis:
CH4 = 68%
C2H6 = 32%
SG@15.6C̊ =0.887
SG@15.6C̊ =0.887[1-0.0007(26-15.6)]
SG@15.6C̊ =0.88
Assume complete combustion with 15% excess air at 101.325 kPa, 21℃ wet
bulb and 27℃ dry bulb. What is the partial pressure of the water vapour in
kPa?
Combustion reaction with theoretical air:
Density of fuel= 0.88(1 kg/L)
Density of fuel= 0.88kg/L
CH4 + 0.32 C2H6 + 2.48 O2 + 2.48 (3.76) N2➔ 1.32 CO2 + 2.32 H2O + 2.48
(3.76) N2
𝑚
𝑤=
𝑉
Combustion reaction with 15% excess air:
V = 650/24
V = 27.0833 L/hr
0.88 = m/27.0833
m = 23.83 kg/hr
CH4 + 0.32 C2H6 + 1.15 (2.48) O2 + 1.15 (2.48) (3.76) N2
➔ CO2 + 2.32 H2O + 1.15 (2.48) (3.76) N2+ 0.15 (2.48) O2
Total mols in products of combustion:
15. There are 20 kg of flue gases formed per kg fuel oil burned in the
combustion of a fuel oil C12H26. What is the excess air in percent?
Solving for the theoretical air-fuel ratio:
C12H26+18.5 O2+18.5 (3.76) N2 = 12 CO2+13 H2O+18.5 (3.76) N2
Theo A/F =
18.5+18.5(3.76)
=
1
88.06(28.97)
𝑚𝑜𝑙𝑠𝑎𝑖𝑟
88.06 𝑚𝑜𝑙𝑓𝑢𝑒𝑙
𝑘𝑔𝑎𝑖𝑟
= 12(12)+26(1) = 15 𝑘𝑔𝑓𝑢𝑒𝑙
= 1.32 + 2.32 + 10.723 + 0.372
= 14.735 mols
Partial Pressure of water vapour, Pw =
𝑉𝑤
(𝑃)
𝑉
2.32
= 14.735 (101.325) = 𝟏𝟓. 𝟗𝟓 𝒌𝑷𝒂
17. A diesel electric plant supplies energy for Meralco. During a 24-hour
period, the plant consumed 200 gallons of fuel at 28℃ and produced 3930
kw-hr. Industrial fuel fuel used 28°API and was purchased at P5.50 per liter
at 15.6℃. What should be the cost of fuel to produce one kw-hr?
Solving for density at 15.6℃:
°API =
28 =
141.5
𝑆𝐺15.6
141.5
𝑆𝐺15.6
𝑂
Theo A/F = 11.5 C + 34.5 (𝐻 − 8 ) + 4.3 S
= 11.5 (0.74) + 34.5 (0.006 −
− 131.5
0.08
)+
8
4.3 (0.01)
= 10.278 kg air per kg coal
− 131.5
N2 in product = 100 – 12 – 0.1 – 6.5 = 81.4% by volume
SG15.6 = 0.887
Total kg in products = 0.12 (44) + 0.001 (28) + 0.065 (32) + 0.814 (28)
= 5.28 + 0.028 + 2.08 + 22.792
= 30.18
Density at 15.6℃ = 0.887 (1) = 0.887 kg/li
Solving for density at 28℃:
𝑘𝑔 𝑁2
𝑘𝑔 𝑝𝑟𝑜𝑑
SGt = SG15.6 [ 1 – 0.0007 (t – 15.6) ]
SG28℃= 0.887 [ 1 – 0.0007 (28 – 15.6) ] = 0.879
=
𝐶 𝑏𝑢𝑟𝑛𝑒𝑑
𝑘𝑔 𝑐𝑜𝑎𝑙
22.792
30.18
= 0.74 – 0.008 = 0.732
Density at 28℃ = 0.879 (1) = 0.879 kg/li
Price per kg =
𝑃5.50
0.887
5.28
Note:
1 kg C + 2.67 kg O2 = 3.67 CO2
1 kg C + 1.33 kg O2 = 2.33 kg CO
Cost per kw-hr
200𝑔𝑎𝑙
3.7854𝑙𝑖
𝑔𝑎𝑙
×
0.879𝑘𝑔
𝑃6.20
× 𝑘𝑔 =
𝑙𝑖
₱1.05 per kw-hr
18. The following is an analysis of coal in percent:
C = 74
H2 = 6
O2 = 8
N2 = 1.6
CO = 0.1
𝑘𝑔 𝑔𝑎𝑠
𝑘𝑔 𝑐𝑜𝑎𝑙
S=1
Ash = 9.4
O2 = 6.5
The refuse contains 0.008 kg of carbon per kg of coal burned. Determine the
percentage excess air used.
1
= 0.732 × 0.0481 = 15.218
𝑁2 𝑠𝑢𝑝𝑝𝑙𝑖𝑒𝑑
𝑘𝑔 𝑐𝑜𝑎𝑙
If burned in a boiler, the coal produces the following Orsat analysis in
percent:
CO2 = 12
0.028
C in prod = 30.18 (3.76) + 30.18(2.33) = 0.0481 𝑘𝑔
= 𝑃6.20 𝑝𝑒𝑟 𝑘𝑔
= 3930𝑘𝑤−ℎ𝑟 ×
= 0.7552
= 15.218 (0.7552) = 11.493
𝐴𝑖𝑟 𝑠𝑢𝑝𝑝𝑙𝑖𝑒𝑑
𝑘𝑔 𝑐𝑜𝑎𝑙 𝑏𝑢𝑟𝑛𝑒𝑑
=
11.493
0.768
= 14.965
Note: N2 is 76.8% by weight in air
Percentage of Excess Air =
14.965−10.278
10.278
= 𝟒𝟓. 𝟔%
19. A steam generator burns fuel oil that has the following chemical analysis
by mass in percent:
C = 85.3
N2 = 0.1
H2 = 14.1
𝑚𝑜𝑙𝑠 𝑎𝑖𝑟
S = 0.5
= 132.09 𝑚𝑜𝑙 𝑓𝑢𝑒𝑙
132.09(28.97)
Combustion takes place in 125 percent in theoretical air. The flue gases
leave the air pre-heater at 0.17 Mpa. What is the partial pressure of the
stack gases to avoid condensation in kPa? Take molecular weight of the flue
gases as 28.8.
𝑘𝑔 𝑎𝑖𝑟
= 12(12)+26(1) = 22.51 𝑘𝑔 𝑓𝑢𝑒𝑙
Amount of wet products = 22.51 + 1 = 23.51
𝑘𝑔 𝑎𝑖𝑟
𝑘𝑔 𝑓𝑢𝑒𝑙
= 23.51 (45) = 1058 kg/min
Converting the given mass analysis to molal analysis:
Solving for the gas constant of the wet products:
Component
C
H2
S
N2
Mass Analysis
85.3
14.1
0.5
0.1
Molal Analysis
85.3/12 = 7.108
14.1/2 = 7.050
0.5/32 = 0.016
0.1/28 = 0.004
14.178
Components
CO2
H2O
N2
O2
Combustion reaction with 125% theoretical air:
M=
7.208 C + 7.050 H2 + 0.016 S + 0.004 N2 + 1.25 (10.649) O2 + 1.25 (10.649)
(3.76) N2 -> 7.108 CO2 + 7.050 H2O + 0.016 SO2 + 50.054 N2 + 0.25 (10.649)
O2
Total mols in products = 7.108 + 7.050 + 0.016 + 50.054 + 2.662
= 66.89 mols
7.050
Partial Pressure (H2O) = (66.890) × 170 = 𝟏𝟕. 𝟗𝟐 𝒌𝑷𝒂
20. A fuel oil is burned with 50% excess air. What is the volume rate of flow
in m3/min of the wet products at a pressure of 102 kPa and a temperature
of 350℃ when the fuel is burned at the rate of 45 kg/min? Assume that the
combustion requirements of the fuel oil are similar to those of C12H26.
R=
3,979.52
138.59
8.3143
28.71
No. of Mols
12
13
104.34
9.25
No. of kg
12 × 44 = 528.00
13 × 18 = 234.00
104.34 × 28 = 2,921.52
9.25 × 32 = 296.00
3,979.52
= 28.71
= 0.2896
Solving for the volume flow of the wet products:
PV = mRT ; 102 V = 1058 (0.2896) (350 + 273) ; V = 1,871 m3/min
21. Methyl alcohol (CH3OH) is burned with 20% excess air. How much
unburned oxygen in kg-mol-oxygen/kg-mol-fuel will there be in the products
if the combustion is complete?
Combustion Reaction with Theoretical Air:
CH3OH + (1.5) O2 + 1.5 (3.76) N2-> CO2 + 2 H2O + (1.5) (3.76) N2
Combustion reaction with 50% excess air:
Combustion Reaction with 20% Excess Air:
C12H26 + 1.50 (18.5) O2 + 1.50 (18.5) (3.76) N2➔ 12 CO2 + 13 H2O + 1.50
(18.5) (3.76) N2 + 0.50 (18.5) O2
CH3OH + (1.5) (1.2) O2 + 1.5 (1.2) (3.76) N2➔ CO2 + 2 H2O + (1.5) (1.2) (3.76)
N2 + (0.20) (1.5) O2
𝐴𝑖𝑟
𝐹𝑢𝑒𝑙
=
1.5 (18.5)+1.50 ( 18.5)(3.76)
1
MO2 = 0.20 (1.5) = 0.30 kgmol O2/kgmol fuel
22. What is the specific gravity of an oil which has a Baume reading of 28
degrees?
140
−
𝑆𝐺
°Baume =
28 =
140
−
𝑆𝐺
Percent excess air =
130
130
SG = 0.886
23. A test run using this coal showed a dry products of combustion analysis
by volume of nitrogen equals 82.5%, molecular weight of 30 kg flue gas per
mol dry flue gas, and the weight of this dry flue gas is 15.03 kg per kg of
coal. The actual ash-pit sample was 0.15 kg per kg of coal, of which 20% was
carbon. What is the percentage excess air supplied to the fuel combustion in
percent?
Consider an ultimate analysis the same as that given in Bd. Exam April 1995,
as follows:
C = 74%
O2 = 8%
S = 1%
H2 = 6%
N2 = 1.6%
Ash = 9.4%
𝑂
Theo A/F = 11.5𝐶 + 34.5 (𝐻 − 8 ) + 4.3𝑆
= 11.5 (0.74) + 43.5 (0.06 −
= 10.278 kg air per kg coal
𝑘𝑔 𝑁2
𝑘𝑔 𝑝𝑟𝑜𝑑
𝑘𝑔 𝑔𝑎𝑠
𝑘𝑔 𝑐𝑜𝑎𝑙
=
0.825(28)
30
= 0.77
= 15.03 (given)
𝑁2 𝑠𝑢𝑝𝑝𝑙𝑖𝑒𝑑
𝑘𝑔 𝑐𝑜𝑎𝑙
= 15.03 (0.77) = 11.573
𝐴𝑖𝑟 𝑠𝑢𝑝𝑝𝑙𝑖𝑒𝑑
𝑘𝑔 𝑐𝑜𝑎𝑙
=
(Note: N2 is 76.8% by weight in air)
11.573
0.768
= 15.069
0.08
)+
8
4.3(0.01)
15.69−10.278
10.278
= 𝟒𝟔. 𝟔𝟏%