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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.

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.

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

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

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)

= 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

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,

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

Actual air intake = 32.945 kg mol

Hence water present = 26.356 kg

Hence the actual kg-mol of water will be

26.356 kg = 1.46 kg-mol

Hence total H2O vapor = 1.46+4.75 = 6.21 kg mol

=(6.21/

= 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.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 are3.9 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 dryFlue 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 airActual 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 proportionately26.356 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.