CHE Calculations 2

Lecture notes # 2 : Gaseous Fuels

COMBUSTION

- is defined as the burning of a fuel and oxidant to produce heat and/or light. It starts with the mixing of fuel and oxidant, and sometimes in the presence of other species or catalysts. The fuel can be gaseous, liquid, or solid and the mixture may be ignited with a heat source.

Complete Combustion

- complete combustion occurs when the fuel and oxygen are in the perfect combination, or ratio, to completely burn the fuel. This condition also is referred to as stoichiometric or zero excess air combustion.

Incomplete Combustion

- In incomplete combustion, the reaction is inefficient and produces both the products that would be created in complete combustion and other products as well like carbon monoxide.

Products of Combustion

- The products of combustion of a sulfur-free fuel consist of carbon dioxide, water vapor, nitrogen, oxygen. But carbon monoxide and unburned hydrocarbons are produce for incomplete combustion.

Reactions in a Combustion Process:

C + O2 CO2

H + O2 H2O

S + O2 SO2

Theoretical Oxygen and air

- The amount of oxidant (oxygen and air) just sufficient to burn the carbon, hydrogen, and sulfur in a fuel to carbon dioxide, water vapor, and sulfur dioxide.

Excess Air and oxygen

- More than the theoretical amount of oxygen and air necessary to achieve complete combustion.

Flue gas

- Gaseous combustion products from a furnace. Flue gas is also called stack gas.

Orsat Analysis

- Volumetric analysis of a gas excluding any water vapor present, i.e. the composition is reported on a dry basis.

Procedure:

a. Gas sample is placed in a container with water to keep the condition saturated. Temperature and pressure are both constant.

b. Gas is washed in succession with different solvents to absorb specific components. Measured change in volume is proportional to change in moles.

Solvents:

V0 V1 V2 V3 V4

O2 CO2 CnH2n CO

V5 V6 V7 N2

H2 H2S SO2

Calculations: PT constant

Gas VT T constant

Water

PYROGALLOL KOH H2SO4 CuCl2

Pd Black Pd Acetate PbO2

Initial Conditions: Total Volume, Vt

Total Pressure, Pt

Temperature, T

Saturation - fixes the amount of water vapor in the gas sample.

Partial Pressure of water, Pw

Pw = Pwo

Since temperature is constant and vapor pressure is a function of temperature alone, then the partial pressure of water is constant.

=

Total volume of dry gas, Vdg,0

Vdg,o = Vo – Vw

Sample is the washed with the first solvent, e.g pyrogallol

New condition after washing , Vt

=

Vw = Vt

Volume of dry gas , Vdg, t

Vdg, t = Vt - Vw

Volume difference of the dry gas is the volume of gas absorbed

VO2= Vdg, o - Vdg, t

Volume percent O2 in the sample on a dry basis:

Volume % O2 = x 100

Process is repeated for other solvents. For other components,

Volume % component a = x 100

*Note: 1. Volume % = mole % ( for ideal gas only)

2. One of the important determinations in combustion calculations is the DEW POINT of the flue gas. The greater the moisture present in the gas, the higher is its dew point. If the flue gases are cooled below the dew point, H2O consumed and may dissolve in SO3 and SO2 to form acids. These acids are corrosive and affect the tubes.

GASEOUS FUEL

Gaseous fuels of value in commerce include natural gases, gases manufactured purely for use as fuels, and gases obtained as by-product of some industries.

Classification of Gaseous fuel

1. Natural gas – is a combustible gas that occurs in porous rock of the earth’s crust and is found with or near accumulations of crude oil. It consists of hydrocarbons with a very low boiling point. Methane is the main constituent with a boiling point of 119 K.

2. Manufacture Gases

a. wood gas ( from wood)– by distillation or carbonization

b. peat gas (from peat) – by distillation or carbonization

c. coal gas (from coal – by carbonization (coal gas)

* By gasification

- in air (producer gas)

- in air and steam (water gas)

- in Os and steam (Lurgi gas)

* By hydrogenation

- as a by-product of the reduction of ores ( Blast Furnace gas)

d. oil gas – from petroleum cracking and hydrogenation

e. from carbides w/ H2O (acetylene)

f. by electrolysis w/ electricity ( H2)

Common Conditions for Gas Volume measurements are:

Standard temperature and pressure (STP) which is 0o C and 760 mmHg, respectively.

Dry condition - means no H2O vapor is present with the gas.

Wet condition – means the gas is saturated with H2O vapor.

At saturation:

Partial Pressure of H2O in gas ( P) = Vapor pressure @ partial saturation

= vapor pressure (P0) x

The vapor pressure is computed by using Antoine’s Equation for vapor pressure (P0)

Log P0 = A -

Where: P0 is in mmHg and T is in oC

For H2O: A = 7.96681, B = 1668. 21, C= 228

Computation:

-Analysis of the product (Orsat or Complete) is composed of CO2 , CO, SO2, H2, free O2 and H2O. Orsat Analysis of the gas mixture does not include water.

O2 theo = at C + at S + at H / 4 – moles O2

Free O2 = excess O2 + +

x’s O2 = O2 supplied – O2 theo

% x’ s air = x 100

= x 100

% x’s O2 =% x’s air = x 100

*Note: For the computation of gaseous fuel a mole basis is used for the % composition of the fuel or flue gas.

A. Composition on wet and dry basis

1. Wet basis to dry basis

Example 1:A stack gas contains 60 % mole N2, 15 % CO2, 10% O2, and the balance H2O. Calculate the molar composition of the gas on a dry basis.

Given: Air

fuel Stack gas 60% mole N2

15% mole CO2

10% mole O2 H2O

Req’d: molar composition of the gas on a dry basis

Solution:Basis: 100 mol wet gas

nt dry gas = 60 mol N2 + 15 mol CO2 + 10 mol O2

= 85 mol

Burner

mole N2 = = 0.706

mole CO2 = = 0.176

mole O2 = = 0.118

2. Dry basis to wet basis

Example 2:An Orsat Analysis ( a technique for stack gas) yields the following dry basis composition:

N2 65 %CO2 14 %CO 11 %O2 10%

A humidity measurement shows that the mole fraction of H2O in the stack gas is 0.0700. Calculate the stack gas composition on a wet basis.

Given:

air

Fuel Stack gas (0.0700 H2O)

(orsat analysis) 60% mole N2

15% mole CO2

11% mole CO 10% mole O2

Required: Stack gas composition on a wet basis

Solution: Basis:100 mole Dry gas

0.0700 = 0.930

Burner

= 0.0753

Hence the gas in the assumed basis contains:

n H2O in the stack gas = 100 mole dry gas x 0.0753

= 7.53 moles

nt stack gas = 7.53 mole + 100 mole = 107.53

n O2 = x 100 = 9.3 %

n N2 = x 100 = 60.45 %

n CO2 = x 100 = 13.02 %

n CO2 = x 100 = 10.23 %

n H2O = x 100 = 7.00 %

B. Calculation based on fuel analysis

Example 3:

1. Pure ethane is burned completely in 20 % excess air. Air is supplied at 740 torrs and is substantially dry. Calculate:

a. Orsat analysis of the products of combustion.

b. kg dry air supplied / kg of fuel gas

c. cubic meters of air / kg ethane

d. cubic meters of the products of combustion measured at 400o C, 100 kPa/ kg ethane

e. Partial pressure of H2O in the products of combustion

Given: Air (20 % x’s)

Fuel( C2H6) Stack gas ( 400 oC, 100 kPa)

Req’d: a. orsat analysis b. kg air supplied / kg of fuel gas c. m3 of air / kg ethane d. m3 of the combustion products / kg ethane e. P of H2O in the combustion products

Solution:

Basis: 1kgmole ethane or 30 kg ethane

n C = 2

n H = 6

Theo O2 = mole C + mole H/ 4 – mole O2

= 2 + 6/4 = 3.5 kgmoles

Excess O2 = (% x’s) (theo O2)

= (.20) (3.5) = 0.7 kgmoles

O2 supp = theo O2 + x’s O2

= 3.5 kgmoles + 0.7 kgmoles = 4.2 kgmoles

N2 supp = O2 supp (79 /21)

= 4.2 kgmoles (79/21) = 15.8 kgmoles

C + O2 CO2

CO2 produced = 2 kgmoles C x = 2 kgmoles

4H + O2 2H2O

Burner

H2O produced = 6 kgmolesH x = 3kgmoles

Free O2 = excess O2 = 0.7 kgmoles

a. OrsatAnalysis:

Components Moles %

CO2 2 10.81

O2 0.7 4.43

N2 15.8 84.04

Total 18.5 100

b. = = 19.33

c. = nRT/ P

kg ethane

( 4.2 kgmole + 15.8 kgmole) (22.4) x x

= ________________________________________ = 16.74 30 kg

d. nt of combustion products = 18.5 kgmole + 3 kgmoles = 21.5 kgmole

= nRT / P

kg ethane

(21.5 kgmole) (22.4) x x

= ________________________________________ = 40.1 30 kg

e. =

P = 100 kPa x = 13.95 kPa

Example 4:Pure methane is burned with 40 % x’s air and 25 % of its carbon content is converted to CO and the rest to CO2. 90% of its hydrogen burns to water, the rest remains as unburned H2. Air supplied is 230C, 758 mmHg with 80% RH. Calculate:a. Orsat Analysis of the combustion products.b. cubic meter of air supplied per kg methanec. cubic meter of combustion products @ 350 oC and normal barometric pressure per kg methane

Given: air (230C, 758 mmHg) 40% x’s

80% RH Fuel ( CH4) stack gas (350 oC, 1 atm)

25 %C to CO 75%C to CO2

90% H to H2O 10% H to H2

Req’d : a. Orsat analysisb. m3 of air supplied per kg methanec. m3 of combustion products per kg methane

Solution:Basis: 1 kgmole methanen C = 1 kgmolen H = 4 kgmoletheo O2 = 1 kgmole + 4kgmole / 4 = 2 kgmolex’s O2 = 0.4 ( 2 kgmole) = 0.8 kgmoleO2 supp = 2 kgmole + 0.8 kgmole = 2.8 kgmoleN2 supp = (2.8 kgmole) (79/21) = 10.533 kgmoleCO2 produced = 1 kgmole x 0.75 = 0.75 kgmoleCO produced = 1 kgmole – 0.75 kgmole = 0.25 kgmoleunburned H2 = 0.1 x (4/2) = 0.2 kgmoleH2O formed = 0.9 x (4/2) = 1.8 kgmoleFree O2 = 0.8 kgmole + (0.25/2) + (0.2/2) = 1.025 kgmole

a. Orsat AnalysisComponents Moles %

CO2 0.75 5.88CO 0.25 1.96H2 0.20 1.57O2 1.025 8.03

Burner

N2 10.533 82.56Total 12.758 100

b. @ 23 oC

log Po = A - = 7.98861 –

= 20.92 mmHg

n H2O from air = (2.8 kgmole + 10.533) = 0.301 kgmole

=

= = 20.75

c. moles of H2O in combustion products = 0.301 kgmole + 1.8kgmole = 2.101 kgmole

=

= = 47.47

C. Calculation based on flue gas analysis

- If there is no N2 given in the fuel, the N2 in the flue gas may be assumed to be coming from air

- O2 balance will determined O2 unaccounted. This O2 was used to burned H2 to H2O and was not accounted since H2O is not included in the Orsat analysis

- If combustible matter losses (i.e. soot) are small, all the carbon in the fuel are accounted for in the flue gas.

- NET HYDROGEN is the H in the fuel that uses O2 from air for combustion.Net H = Total H in the fuel ( if O2 is not present in the fuel)

Example 5: A pure saturated hydrocarbon (CnH2n +2) is burned with excess. Orsat analysis of the products of combustion shows 9.02 % CO2, 1.63% CO, 5.28% O2 and no free H2.calculate:a. the formula of the hydrocarbonb. % x’s air

c. kg dry air / kg of hydrocarbon

Given: Air

Fuel Stack gas Pure HC 1.63 %CO

9.08%CO2

5.28%O2

Required: a. formula of HC

b. x’s air c. kg dry air / kg HC

Solution: Basis: 100 kgmoles dry stack gas

N2 is flue gas = 100 kgmoles – (1.63+9.08 + 5.28) kgmole = 84.01 kgmoleO2 supp = 84.01 kgmole ( 21/79) = 22.33 kgmole

O2 unaccounted = 22.33 kgmole – (9.08 + +5.28) kgmole

= 7.16 kgmoleNet H = H2 in the fuel = 7.16 kgmole (2) = 14.32 kgmoleTotal H in the fuel = 14.32 kgmole x 2 = 28.64 kgmoleTotal C in the fuel = 9.08 kgmole + 1.63 kgmole = 10.71 kgmole

= =

N = 2.966 ≈ 3a. Formula of HC: C3H8

b. x’s O2 =5.28 kgmole - = 4.465 kgmole

% x’s air = x 100

= 25%

c. = = 19.62

Burner

Example 6:A pure gas consisting of methane and ethane is burnt with air to yield a flue gas where orsat analysis is 10.57 % CO2, 3.79% Os, and 85.64% N2.Calculate:a. analysis of the fuel in mole %.b. excess air

Given: Air

Pure gas Flue gas CH4 (Orsat analysis)

C2H6 10.57 % CO2,3.79% Os, 85.64% N2

Req’d: a. analysis of the fuel in mole% b. x’s airSolution:

Basis: 100 moles dry flue gasGas Moles Atoms C Moles O2

CO2 10.57 10.57 10.57Os 3.79 3.79N2 85.64Total 10.57 14.36

N2 supplied = 85.64 molesO2 supplied = 85.64 moles ( 21/79) = 22.77 molesO2 unaccounted = 22.77 moles – 14.36 moles = 8.4 moles

H2 + 1/2O2 H2O 16.8 8.4 16.8

Let x = moles CH4

Y = moles C2H6

C CH4 + C CsH6 = CT

x + 2y = 10.57

H2 (CH4) + H2 (C2H6) = HT

2x + 3y = 16.8

x = 1.89y = 4.34a. mole fraction

Fuel components Mole Mole %CH4 1.89 30.34

Burner

C2H6 4.34 69.66Total 6.23 100

b. theo O2 = 10.57 + 8.4 = 18.97 moles

% x’s air = x 100

= 20.03%

D. Calculation based on partial analysis of the stack gas

- carbon is used as a tie substance to relate the fuel with the stack gas

Example 7:The burning of pure butane with excess air gives a stack gas with analyzing 11.55% CO2 on a dry basis. Assuming complete combustion, calculate: a. % x’s airb. complete orsat analysis of the stack gas

Given: Air

Pure Butane Stack gas 11. 55 % CO

Required: a. % x’s air b. orsat analysis

Solution:Basis: 100 moles C4H10

Mole C = 400 molesMole H = 1000 moles

Theo O2 = 400 moles + = 650 moles

Let x = x’s O2

O2 supp = x + 650 moles N2 supp = x + 650 moles (79/21) Free O2 in the stack gas = x

Tie substance at C 400 moles = 0.1155 ( moles dry stack gas) Moles DSG = 3,463.2

DSG balance:

Burner

3463.2 moles = 400 moles + x + (650 +x) (79/21) X = 129.77

a. % x’s air = x 100 = 19.96 %

b. Orsat analysis:Component N %CO2 400 11.55O2 129.77 3.75N2 2933.43 84.70Total 3463.2 100

Example 8:The flue gas from a certain furnace burning a gaseous fuel of negligible Nitrogen content is found by analysis to contain 12% CO2, 7.5% O2, and 80.5% N2. Calculate the percent excess air used and the H to C ration in the fuel.

Given: Air

Gaseous Fuel Flue gasNo N 12 % CO2

7.5 % O2 80.5% N2

Required: a. % x’s air b. H/C ratio in fuel

Solution:Basis: 100 lbmol dry flue gas

lbmolCO2 12O2 7.5N2 80.5 Total 100

N2 balance: mol N2 from air = 80.5 lbmol N2

Furnace

mol O2 supp = 80.5 lbmol N2 (

= 21.40 lbmol O2

Assume: no oxygen from fuel

Complete combustion:Mole O2 excess = mol free O2 in the flue gas

= 7.5 lbmol O2

% x’s O2 = x 100

Solve O2 unaccounted for:

mole unaccounted O2 = 21.40 lbmol O2 – 12 lbmol CO2 x - 7.5 lbmol O2

= 1.90 lbmol O2

Mole H2O = 1.90 lbmol O2 x

= 3.80 lbmol

H balance:

mol H in fuel = 3.80 lbmol H2O x

= 7.60 lbmol H

C balance:

mol C in fuel = 12 lbmol CO2 x

= 12 lbmol C

= = 0.63

E. Calculation based on fuel and flue gas analysis

Example 9:A coal containing 80% C is completely burnt in a furnace. The flue gas analysis shows 14.5 % Co2, 3.76% O2, and no CO. What is the % net hydrogen in the coal and % x’s air.

Given: Air

Coal stack gas 80% C 14.5% CO2

3.76% O2

81.24 % N2

Req’d: a. % net hydrogen in the coal b. % x’s air

Solution:Basis: 100 moles Dry flue gas

Component Moles Atoms C Moles O2

CO2 14.5 14.5 14.5O2 3.76 3.76N2 81.24Total 100 14.5 18.26

N2 supp = 81.24 molesO2 supp = 81.24 mole(21/79) =21.59 molesO2 unaccounted = (21.59 – 18.26 ) moles = 3.33 moles HsO produced = 3.33 moles O2 x ( 2 moles HsO/ 1 mole O2) = 6.66 mole C balance:Let x = kg of coal0.80 x = (14.5 mole C ) (12 g/ mole)x = 217.8 g

Furnace

a. % H2 = x 100 = 6.12%

b. the O2 = 14.5 mole + ( 6.66 mole) (2/4) = 17.83

%x’s air = x 100 = 21.1 %

PROBLEMS:1. Ethane is burned with 50% excess air. The percentage of the ethane is 90%; of the ethane burned, 25% reacts to form CO and the balance CO2. Calculate the molar composition of the stack gas on a dry basis and the mole ratio of water to dry stack gas. ANS. 0.113 mol H2O per mole of dry stack gas.

2. A hydrocarbon gas is burned with air. The dry basis product gas composition is 1.5mole% CO, 6.0% CO2, 8.2% O2, and 84.3% N. There is no atomic oxygen in the fuel. Calculate the ratio of the hydrogen to carbon fuel in the gas and speculate on what fuel might be. Then calculate the percent excess air fed to the reactor. ANS. 3.97 mol of H/ mol C, 49.8% excess air, fuel is CH4.

3. One hundred mol/h of butane (C4H10) and 5000 mol/h of air are fed onto a combustion chamber. Calculate the percent excess air. ANS. 61.6%

4. Ammonia is burned to form nitric oxide in the following reaction:

4NH3 +5O2 ====== 4NO + 6H2O

Calculate the ratio lb-mole of O2 react / lb-mole of NO formed and if ammonia is fed to ac continuous reactor rate of 100.0 kmol NH3/h, what oxygen feed rate would respond to 40.0 excess O2? If 50.0 kg of ammonia and 100 kg of oxygen are fed to a batch reactor, determine the limiting reactant, the percentage by which the other reactant is in excess, and the extent of reaction and mass of NO produced if the reaction proceeds to completion. ANS. 17.6 % excess NH3.

5. A mixture if a saturated hydrocarbon and N2 is burned in excess air supplied at 25C, 740 torrs with 90% RH. An orsat analysis of the stack gas shows 7.6% CO2, 2.28% Co, 1.14%H2, 6.03% O2, and 82.95% N2 with dew a point of 53.4C. The stack gases leave at 300C , 755 mmHg with a volume ratio of 2.049 m3 wet stack gas / m3 of air.a. calculate formula for the hydrocarbonb. volume % analysis of the fuelc. % excess airANS. C2H6, 74.70% C2H6 and 25.30% N2, excess air is 24.96%

References: CHE calculation by LauritoPerry’s Chemical Engineering Handbook 7th edition.Elementary Principles of Chemical Processes by Richard Felder and Ronald Rousseau Industrial Stoichiometry by Lewis, Radasch, Lewishttp://eyrie.shef.ac.uk/eee/cpe630/comfun5.htmlhttp://www.eoearth.org/article/Combustion