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A fuel gas consists of 75% butane (C4H10), 10% propane (C3H8) and 15%
butene (C4H8) by volume.
It is to be fed to the combustion chamber in 10% excess air at 25ºC,
where it is completely burnt to carbon dioxide and water. The flue gases
produced are to be used to generate 5 bar steam from water at 90ºC.
Data:
Net calorific value (MJ m–3) at 25ºC of:
Butane (C4H10) = 111.7 MJ m–3
Butene (C4H8) = 105.2 MJ m–3
Propane (C3H8) = 85.8 MJ m–3
Air is 21% oxygen, 79% nitrogen by volume and 23.3% oxygen and
76.7% nitrogen by mass.
Atomic mass of C = 12, O = 16, N=14 and H = 1.
I write the question and after the solution ,i just want to know if what i did is ok or i have to do something else,
thank you in advance for your help.(also i used a table with enphaly for different temperatures but didnt post it here)
Calculate:
(i) the net calorific value (CV) per m3 of the fuel/air mix at 25ºC
(ii) the net calorific value (CV) per kmol of the fuel/air mix at
25ºC.
(i) NCV of fuel in MJ/M3 @25°C= (0.75 x 111.7 + 0.1 x 105.2 + 0.15 x 85.8) MJ/M3= 107.2 MJ/M3(ii) (ii) NCV of the fuel will be 2.401MJ/Kg-mol.
Weight of 1 kg-mol of the fuel mix = 56.3 kg
At atm pressure the volume of 1 kg mol = 22.4m3
And hence the NCV = 107.2 x 22.4
= 2401.28 MJ/Kg-mol.
Determine the actual fuel:air ratio
(i) by volume
(ii) by mass.
(i). Fuel gas composition,
Butane: C4H10
Propane: C3H8
Butane: C4H8
Oxygen sent is 10% excess
Stoichiometric equations (the combustion equation)
C4H10 + 6.5 O2 4CO2 + 5H2O
C3H8 +5O2 3CO2 + 4H2O
C4H8 +6O2 4CO2 + 4H2O
Butane : 75%
Propane : 10%
Butane : 15%
1 Mole of fuel contains
0.75 Moles of Butane
0.1 Moles of Propane
0.15 Moles of Butane
Corresponding oxygen requirement for,
Butane = 0.75 x 6.5 = 4.875 Moles of O2
Propane = 0.10 x 5 = 0.5 Moles of O2
Butane = 0.15 x 6 = 0.9 Moles of O2
Total = 6.275 Moles of O2
Amount of O2 in 1 mole of air =0.2095 moles
Hence, 6.275 moles of O2 Present in 25.95 moles of air
Hence fuel to air ratio is 1 mole of fuel needed for 29.95 moles of air for complete combustion.
Since the excess air supplied is 10%
The actual air intake for 1 mole of fuel
Combustion = 32.945 mole / 1 mole of fuel
By mass, the amount of fuel intake, for mole
Butane = 0.75 x (4 x 12 +10 x 1) = 43.5 Kg
Propane = 0.10 x (3 x 12 + 8 x 1) = 4.4 Kg
Butane = 0.15 x (4 x 12 +8 x 1) = 8.5 Kg
1 mole of fuel molecular mt = 56.3 kg-mol.
Corresponding air intake (with 10% Excess)
= 1.1 x 6.275 x 29 kg/kg-mole
= 200.1725 kg/kg-mole
Hence fuel to air ratio by mass = 56.3/200.1725 = 0.28
Actual air to fuel ratio by mass =3.56
A/F) Mass = 3.56
A/F) Volume = 32.945
Determine the composition of the flue gases by volume (assuming
the inlet air is dry):
(i) on a wet basis
(ii) on a dry basis
Composition of fuel gas by volume: (wet basis)
(i) Butane, Propane, Butane’s combustion products along with excess oxygen and nitrogen.1 mole of fuel yields = 0.75 x (4CO2 + 5H2O) + 0.1 x (3CO2 + 4H2O) + 0.15 x (4CO2 + 4H2O) + Excess O2 + N2
= 39 CO2 + 4.75 H2O + 0.64 O2 + 26.03 N2
Excess O2 = (0.21)*32.95 = 6.9195 - 6.275 = 0.645 moles
Nitrogen present = 0.79 * 32.95 = 26.03 moles(i) On dry basis, the consistent of the three gases are
3.9 CO2 + 0.645 O2 + 26.03 N2.
Determine the ‘furnace efficiency’ if the flue gases leave the boiler at
300ºC.
The furnace efficiency is the measure of the furnace combustion efficiency.
Furnace efficiency
If the flue gases are leaving at 300°C
Heating value of the gas
= (3.9x26.61+.645x18.33+26.03x17.4+4.75x20.89)
= 667.75 MJ/kg-mol
Hence the efficiency of the furnace is given by
= 667.75/2401
= 27.8%
If 5% of the heat available for steam production is lost to the
atmosphere, determine the amount of steam raised per hour when the
total flow of flue gases is 1400 kmol h–1.
Total flow rate of the flue gases = 1400Kmol/hr.
5% of the heat available is lost.
The actual amount of the heat available in the flue gases = 2401.28MJ/kg-mol
Hence considering the loss of 5% of heat energy the actual heat available for making steam,
= 0.95 x 2401.28MJ/kg-mol
= 2281.2MJ/kg-mol
Steam enthalpy at 5 bar = 2748KJ/kg = 49.464MJ/kg-mol
Saturated Water enthalpy at 90 degree Celsius = 385KJ/kg = 6.930MJ/kg-mol
Hence the heat requirement for steam generation = 49.5-6.93 = 42.57MJ/kg-mol
Flue gas flow rate = 1400 kgmol/hr
Hence the steam generation rate = (heat available/heat consumption),
= (1400*2281.2)/42.6
=~ 75 tons/hr.
Determine the dew point temperature assuming that the flue gas
pressure is 1.00 bar and the inlet air:
(i) is dry
(ii) contains 0.8 kg water per kmol of air at the temperature of the
inlet air.
The gas is 1.00 bar,
(i) Inlet air is dry
Flue gas pressure is 1.00 bar
If inlet air is dry, the amount of water vapor in the products = 4.75 moles
Hence % of vapor by volume = 4.75/3.9+4.75+0.645+26.03
= 13.45%
Hence from the charts, the dew points of water vapor in the gas = 53°C(ii) In inlet air contains 0.8kg water/kg-mol of air
Actual air intake = 32.945 kg mol
Hence water present = 26.356 kg
Hence the actual kg-mol of water will be increased by this content proportionately
26.356 kg = 1.46 kg-mol
Hence total H2O vapor = 1.46+4.75 = 6.21 kg mol
=(6.21/ 3.9+4.75+0.645+26.03 ) +1.46
= 16.88% by volume
From charts corresponding dew point = 58°C
If the flue gases exiting the boiler are used to preheat the water fed to
the boiler from a temperature of 28ºC to 90ºC and assuming:
• a mean specific heat capacity for water over this temperature
range to be 4.2 kJ kg–1 K–1
• a mean molar heat capacity for the flue gases up to 300ºC to be
31 kJ kmol–1 K–1
• 10% of the heat required to heat the water is lost in the heat
exchanger
• all water entering the system is converted to steam
determine the final outlet temperature of the flue gas and state if the
dew point will be reached in both of the cases given in previous question
Flue gases used to preheat water
From 28°C to 90°C
Near specific heat of water = 4.2 kj/kg
Water heat capacity of flue gas = 31kj/kg-mol kHeat lost = 10%
Water completely became steam
Mw x 4.2 x (90-28) = 31 x (-Tf + 300) x 1kg mol
8.4 Mw = 300- Tf
Tf = 300-8.4 (Mw)
Amount of water Mw/ in kg/kg-mol of the gas if increased proportionately Tf will come down.
It is possible to get the dew points of both the previous cases.
butene (C4H8) by volume.
It is to be fed to the combustion chamber in 10% excess air at 25ºC,
where it is completely burnt to carbon dioxide and water. The flue gases
produced are to be used to generate 5 bar steam from water at 90ºC.
Data:
Net calorific value (MJ m–3) at 25ºC of:
Butane (C4H10) = 111.7 MJ m–3
Butene (C4H8) = 105.2 MJ m–3
Propane (C3H8) = 85.8 MJ m–3
Air is 21% oxygen, 79% nitrogen by volume and 23.3% oxygen and
76.7% nitrogen by mass.
Atomic mass of C = 12, O = 16, N=14 and H = 1.
I write the question and after the solution ,i just want to know if what i did is ok or i have to do something else,
thank you in advance for your help.(also i used a table with enphaly for different temperatures but didnt post it here)
Calculate:
(i) the net calorific value (CV) per m3 of the fuel/air mix at 25ºC
(ii) the net calorific value (CV) per kmol of the fuel/air mix at
25ºC.
(i) NCV of fuel in MJ/M3 @25°C= (0.75 x 111.7 + 0.1 x 105.2 + 0.15 x 85.8) MJ/M3= 107.2 MJ/M3(ii) (ii) NCV of the fuel will be 2.401MJ/Kg-mol.
Weight of 1 kg-mol of the fuel mix = 56.3 kg
At atm pressure the volume of 1 kg mol = 22.4m3
And hence the NCV = 107.2 x 22.4
= 2401.28 MJ/Kg-mol.
Determine the actual fuel:air ratio
(i) by volume
(ii) by mass.
(i). Fuel gas composition,
Butane: C4H10
Propane: C3H8
Butane: C4H8
Oxygen sent is 10% excess
Stoichiometric equations (the combustion equation)
C4H10 + 6.5 O2 4CO2 + 5H2O
C3H8 +5O2 3CO2 + 4H2O
C4H8 +6O2 4CO2 + 4H2O
Butane : 75%
Propane : 10%
Butane : 15%
1 Mole of fuel contains
0.75 Moles of Butane
0.1 Moles of Propane
0.15 Moles of Butane
Corresponding oxygen requirement for,
Butane = 0.75 x 6.5 = 4.875 Moles of O2
Propane = 0.10 x 5 = 0.5 Moles of O2
Butane = 0.15 x 6 = 0.9 Moles of O2
Total = 6.275 Moles of O2
Amount of O2 in 1 mole of air =0.2095 moles
Hence, 6.275 moles of O2 Present in 25.95 moles of air
Hence fuel to air ratio is 1 mole of fuel needed for 29.95 moles of air for complete combustion.
Since the excess air supplied is 10%
The actual air intake for 1 mole of fuel
Combustion = 32.945 mole / 1 mole of fuel
By mass, the amount of fuel intake, for mole
Butane = 0.75 x (4 x 12 +10 x 1) = 43.5 Kg
Propane = 0.10 x (3 x 12 + 8 x 1) = 4.4 Kg
Butane = 0.15 x (4 x 12 +8 x 1) = 8.5 Kg
1 mole of fuel molecular mt = 56.3 kg-mol.
Corresponding air intake (with 10% Excess)
= 1.1 x 6.275 x 29 kg/kg-mole
= 200.1725 kg/kg-mole
Hence fuel to air ratio by mass = 56.3/200.1725 = 0.28
Actual air to fuel ratio by mass =3.56
A/F) Mass = 3.56
A/F) Volume = 32.945
Determine the composition of the flue gases by volume (assuming
the inlet air is dry):
(i) on a wet basis
(ii) on a dry basis
Composition of fuel gas by volume: (wet basis)
(i) Butane, Propane, Butane’s combustion products along with excess oxygen and nitrogen.1 mole of fuel yields = 0.75 x (4CO2 + 5H2O) + 0.1 x (3CO2 + 4H2O) + 0.15 x (4CO2 + 4H2O) + Excess O2 + N2
= 39 CO2 + 4.75 H2O + 0.64 O2 + 26.03 N2
Excess O2 = (0.21)*32.95 = 6.9195 - 6.275 = 0.645 moles
Nitrogen present = 0.79 * 32.95 = 26.03 moles(i) On dry basis, the consistent of the three gases are
3.9 CO2 + 0.645 O2 + 26.03 N2.
Determine the ‘furnace efficiency’ if the flue gases leave the boiler at
300ºC.
The furnace efficiency is the measure of the furnace combustion efficiency.
Furnace efficiency
If the flue gases are leaving at 300°C
Heating value of the gas
= (3.9x26.61+.645x18.33+26.03x17.4+4.75x20.89)
= 667.75 MJ/kg-mol
Hence the efficiency of the furnace is given by
= 667.75/2401
= 27.8%
If 5% of the heat available for steam production is lost to the
atmosphere, determine the amount of steam raised per hour when the
total flow of flue gases is 1400 kmol h–1.
Total flow rate of the flue gases = 1400Kmol/hr.
5% of the heat available is lost.
The actual amount of the heat available in the flue gases = 2401.28MJ/kg-mol
Hence considering the loss of 5% of heat energy the actual heat available for making steam,
= 0.95 x 2401.28MJ/kg-mol
= 2281.2MJ/kg-mol
Steam enthalpy at 5 bar = 2748KJ/kg = 49.464MJ/kg-mol
Saturated Water enthalpy at 90 degree Celsius = 385KJ/kg = 6.930MJ/kg-mol
Hence the heat requirement for steam generation = 49.5-6.93 = 42.57MJ/kg-mol
Flue gas flow rate = 1400 kgmol/hr
Hence the steam generation rate = (heat available/heat consumption),
= (1400*2281.2)/42.6
=~ 75 tons/hr.
Determine the dew point temperature assuming that the flue gas
pressure is 1.00 bar and the inlet air:
(i) is dry
(ii) contains 0.8 kg water per kmol of air at the temperature of the
inlet air.
The gas is 1.00 bar,
(i) Inlet air is dry
Flue gas pressure is 1.00 bar
If inlet air is dry, the amount of water vapor in the products = 4.75 moles
Hence % of vapor by volume = 4.75/3.9+4.75+0.645+26.03
= 13.45%
Hence from the charts, the dew points of water vapor in the gas = 53°C(ii) In inlet air contains 0.8kg water/kg-mol of air
Actual air intake = 32.945 kg mol
Hence water present = 26.356 kg
Hence the actual kg-mol of water will be increased by this content proportionately
26.356 kg = 1.46 kg-mol
Hence total H2O vapor = 1.46+4.75 = 6.21 kg mol
=(6.21/ 3.9+4.75+0.645+26.03 ) +1.46
= 16.88% by volume
From charts corresponding dew point = 58°C
If the flue gases exiting the boiler are used to preheat the water fed to
the boiler from a temperature of 28ºC to 90ºC and assuming:
• a mean specific heat capacity for water over this temperature
range to be 4.2 kJ kg–1 K–1
• a mean molar heat capacity for the flue gases up to 300ºC to be
31 kJ kmol–1 K–1
• 10% of the heat required to heat the water is lost in the heat
exchanger
• all water entering the system is converted to steam
determine the final outlet temperature of the flue gas and state if the
dew point will be reached in both of the cases given in previous question
Flue gases used to preheat water
From 28°C to 90°C
Near specific heat of water = 4.2 kj/kg
Water heat capacity of flue gas = 31kj/kg-mol kHeat lost = 10%
Water completely became steam
Mw x 4.2 x (90-28) = 31 x (-Tf + 300) x 1kg mol
8.4 Mw = 300- Tf
Tf = 300-8.4 (Mw)
Amount of water Mw/ in kg/kg-mol of the gas if increased proportionately Tf will come down.
It is possible to get the dew points of both the previous cases.