Combustion Solutions

1)

Calculate for a natural gas with the composition by volume given below:

a) Calorific valueb) Relative densityc) Wobbe numberd) Theoretical air requiremente) The volume of and % composition by volume of the dry combustion products with 10% excess

air at metric standard condition (msc).

COMPONENT GROSS R1 COMP

COMPONENTS VOLUME (%) CV (MJ/M3)

RD CV (MJ/M3)

RD O2 O2 CO2 H2O

Nitrogen (N2) 2.53 0.0253 - 0.9668 - 0.024 - - - -Carbon dioxide

(C02)0.52 0.0052 - 1.5188 - 0.008 - - 0.052 -

Methane (CH4) 91.43 0.9143 37.71 0.5537 34.48 0.506 2.0 1.83 0.914 1.830Ethane (C2H6) 4.10 0.0410 66.35 1.0378 2.78 0.043 3.5 0.144 0.082 0.123

Propane (C3H8) 0.99 0.0099 93.94 1.5219 0.93 0.015 5.0 0.05 0.030 0.040Butane (C4H10) 0.33 0.0033 121.80 2.0059 0.40 0.007 6.5 0.021 0.013 0.017Pentane (C5H12) 0.10 0.0010 149.66 2.4900 0.15 0.0025 8.0 0.008 0.005 0.006

38.68 0.605 2.053 1.05 2.02

CH4 + 2.0 O2 -----------> CO2 + 2 H2OC2H6 + 3.5 O2 -----------> 2CO2 + 3 H2OC3H8 + 5.0 O2 -----------> 3CO2 + 4 H2OC4H10 + 6.5 O2 -----------> 4CO2 + 5 H2OC5H12 + 8.0 O2 -----------> 5CO2 + 6 H2O

a) CV =38.7 MJ/m3

b) RD = 0.605

c) WN = = = 48.85 MJ/m3

d) Theoretical air = = 9.78

e) 10% excess air (Dry)

1.1 9.78 = 10.758

Volume CO2 = 1.05Volume N2 Air + N2 gas – 8.499 + 0.0253 = 8.524

Volume of 10% O2 2.053 0.1 = 0.205

% CO2 = 100 = 10.57%

% O2 = 100 = 2.10%

% N2 = 100 = 87.21%

Flue gasVol % Req

O2/VolO2 Req CO2 H2O N2

Methane (CH4) 88.73 2.0 1.7746 0.8873 1.7746 -Ethane (C2H6) 4.48 3.5 0.1568 0.0896 0.1344 -

Propane (C3H8) 0.85 5.0 0.0425 0.0255 0.0340 -Butane (C4H10) 0.37 6.5 0.02405 0.0148 0.0185 -Pentane (C5H12) 0.12 8.0 0.0096 0.0060 0.0072 -

Hexane plus (C7H16) 0.05 11.0 0.0055 0.0035 0.0040 -N2 4.77 - - - 0.0477

CO2 0.63 - 0.0063 - -100 2.01305 1.0330 1.9727 0.0477

a) Air requirement = 2.01305 = 9.5859

b) Theoretical Products: 1.033 1.973N2 gas 0.048N2 air 7.573 10.627Total vol. wet = 10.624Total vol. dry = 8.651

% Composition dry % Composition Wet

CO2 CO2

N2 N2

100.00 O2

2)

Gases can be made up of a mixture of a number of gases and the CV can be found for each individual gas using a calorimeter. Calculate the gross CV of a substitute natural gas with the following composition

GAS CV MJ/m3 % by Vol.CO 11.97 3.0 0.36CH4 37.69 34.1 12.85C2H6 66.03 12.9 8.52H2 12.10 38.4 4.65

C3H8 93.97 11.6 10.90CV 37.28 MJ/m3

3)

For a natural gas with the % composition by volume given below, calculate

(a) Theoretical air requirement.

(b) The volume and percentage composition by volume at the theoretical combustion products.

(c) The % composition by volume of the dry combustion products with 10% excess air at metric standard condition (msc).

(a) CH4 + 2.0 O2 -----------> CO2 + 2 H2O C2H6 + 3.5 O2 -----------> 2CO2 + 3 H2O C3H8 + 5.0 O2 -----------> 3CO2 + 4 H2O C4H10 + 6.5 O2 -----------> 4CO2 + 5 H2O C5H12 + 8.0 O2 -----------> 5CO2 + 6 H2O C6H14 + 9.5 O2 -----------> 6CO2 + 7 H2O

GAS (%)COMPOSITION OXYGENO2 REQ

ACTUALO2 REQ

PRODUCEDCO2

PROCUCED H2O

CH4 88.73 0.8873 2.0 1.7746 0.8873 1.7746C2H6 4.48 0.448 3.5 0.1568 0.0896 0.1344C3H8 0.85 0.0085 5.0 0.0425 0.0255 0.0340C4H10 0.37 0.0037 6.5 0.0241 0.0148 0.0185C5H12 0.12 0.0012 8.0 0.0096 0.0060 0.0072

C6H14 0.05 0.0005 9.5 0.0048 0.0030 0.0035N2 4.77 0.0477 - - - -

CO2 0.63 0.0063 - - 0.0063 -2.0124 1.0325 1.9722

Theoretical air requirement =

(b) Theoretical combustion products (wet)

Volume

CO2 - 1.0325

H2O - 1.9722

Gas - 0.0477

Air - 7.5682

Total 10.00

Percentage composition of combustion products.

% CO2 - 9.72 -

% H2O - 18.57 -

% N2 - 71.71

100.00

Dry % composition

% CO2 - 11.94 -

% N2 - 88.05 -

99.99

N2 7.6159

(c) % composition by volume of the dry combustion products with 10% excess air.

Total air used with 10% excess air 9.58 × 1.1 = 10.538

Volume

CO2 = 1.0325

O2 = 0.2012

N2 = 8.3266

9.5603

% CO2 = 10.80

% O2 = 2.10

%N2 = 87.10

100.00

(d) Importance of CO2 in combustion products.

1. The combustion products from any fuel burned with theoretical air for complete combustion (no excess), contains a fixed and characteristic quantity of CO2.

2. For pure dry carbon, this is 21% CO2 by volume.3. Any air supplied in excess of theoretical, reduces the % CO2 proportionally.4. The % CO2 can be measured using any suitable type of gas analysis equipment. The result

is usually reported on a dry basis.5. With fuels containing free hydrogen or hydrocarbon gases, the carbon dioxide is lower

because the air supplied to burn the H2 forms water which condenses leaving the extra nitrogen which dilutes the combustion products.

6. If the fuels contain sulphur, this turn to SO2 and is often analysed as CO2. The error is very small with low sulphur fuels.

7. There must be very little CO2 and the ratio is a very important parameter (<0.02).

4) A fuel oil is burned with 50 percent excess air, and the combustion characteristics of the fuel oil are similar to C12H26. Determine the volumetric analysis of the products of combustion and determine the dewpoint for the products of combustion (the last part of the question will be covered in later modules).

Can you also suggest the relevance of the dewpoint temperature with respect to the products of combustion?

€

C12H26 + (1.5)(18.5)O2 + (1.5)(18.5)(3.76)N2 →

12CO2 + 13H2O + 104.3N2 + 9.25O2

The total moles of the products are:

€

12 + 13 + 104.2 + 9.25 = 138.55

∴

CO2 =12

138.55= 0.0866

H20 =13

138.55= 0.0938

N2 =104.2

138.55= 0.7528

O2 =9.25

138.55= 0.0668

The partial pressure of water is (0.0938)(101)=9.47 kPa. The saturation temperature corresponding to this pressure is 45C, which is also known as the dewpoint temperature. If the temperature drops below this point, precipitation of water will occur. Should this occur, the liquid water will contain dissolved gases, which can form a corrosive substance (sulphur in the fuel). To prevent this, the temperature of the products of combustion is kept well above the dewoint in smoke stacks and exhaust piping.