BUDAPEST UNIVERSITY OF TECHNOLOGY

AND ECONOMICS

Mohd Basit Wani

BKI553

COMBUSTION HOMEWORK
2021/20/05
 𝐴𝑡𝑚𝑜𝑠𝑝ℎ𝑒𝑟𝑖𝑐 𝑝𝑟𝑒𝑚𝑖𝑥𝑒𝑑 𝑝𝑟𝑒𝑣𝑎𝑝𝑜𝑟𝑖𝑧𝑒𝑑 𝑡𝑢𝑟𝑏𝑢𝑙𝑒𝑛𝑡 𝑙𝑖𝑞𝑢𝑖𝑑 − 𝑓𝑢𝑒𝑙𝑒𝑑 𝑏𝑢𝑟𝑛𝑒𝑟

𝐷𝑖𝑎𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑚𝑖𝑥𝑖𝑛𝑔 𝑡𝑢𝑏𝑒(𝐷) = 26.8𝑚𝑚

𝐿𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑖𝑥𝑖𝑛𝑔 𝑡𝑢𝑏𝑒 (𝐼) = 75.5𝑚𝑚
𝐼𝑛𝑛𝑒𝑟 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟(𝑎𝑛𝑛𝑢𝑙𝑢𝑠)(𝐷𝑖 ) = 0.8𝑚𝑚;
𝑎𝑛𝑑 𝑜𝑢𝑡𝑒𝑟 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟(𝑎𝑛𝑛𝑢𝑙𝑢𝑠) (𝐷𝑜 ) = 1.4𝑚𝑚
𝐷𝑖𝑎𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑓𝑢𝑒𝑙 𝑗𝑒𝑡 (DF, 𝑖) = 0.4𝑚𝑚

• Determine the inlet and outlet mass flows of the combustion chamber. Give the
combustion power. The O2, NOX, THC and CO content of the flue gas is measured.

• Give the flue gas emissions at the reference oxygen level both in ppm and mg/m3.

• Determine the highest possible turbulent flame speed in order to avoid the flame
flashback in the mixing tube.

• Calculate the flue gas temperature of perfect and complete combustion, considering a
heat loss for the combustion chamber, which is 30% of the combustion power.
 Data Sheet
Fuel Heptane (H₃C(CH₂)₅CH₃)

Fuel-air surface tension 0.02086 N/m

Fuel dynamic viscosity 0.00041 Pa s

Fuel density at ambient pressure and temperature 672.81 kg/m3

Fuel mass flow 0.34 g/s

Atomizing gauge pressure of air 0.75 bar

Atomizing air mass flow 0.4094 g/s

Evaporation constant at droplet heat up 0.0207 mm2/s

Time of droplet heating up 0.0339 ms

Fuel specific heat capacity at the reference temperature 2581.03 J/kgK

Fuel thermal conductivity at the reference temperature 0.0322 W/mK

Density of the fuel at it's vapor pressure and surface temperature 689.19 kg/m3

Heat of vaporization at the droplet surface temperature 352.62 kJ/kg

O2 content of the flue gas 13.4 V/V%

NOX concentration in the flue gas 80 ppm

THC concentration in the flue gas 32 ppm

CO concentration in the flue gas 30 ppm

Reference O2 content of the flue gas 4 V/V%

Ambient air temperature 33 °C

Preheated combustion air temperature 610 °C

 Sauter Mean Diameter (SMD)
In Plain-jet air blast atomizer, the SMD is calculated by the equation:

2 −0,5
𝜌𝐴 𝜔𝑅 𝑑0 𝜇𝐿 1
𝑆𝑀𝐷 = 𝑑0 (0.48 ( ) + 0.35 ) (1 + 𝐴𝐿𝑅)……………………..(1)
𝜎 √𝜌𝐿 𝜎𝑑0

wher (𝑤𝑅) is the relative velocity of air-to-liquid, ALR is the air-to-fuel mass flow ratio, and

𝑑0 is the diameter of fuel jet.

𝑚̇𝑎 0.4034
ALR = = = 1.2041
𝑚̇𝐹 0.34

where, 𝑚̇𝑎 is the Atomizing air mass flow and 𝑚̇𝐹 is the Fuel mass flow.

Air Density (𝑎 ):

𝑃𝑎 + 𝑃𝑎𝑚𝑏𝑖𝑒𝑛𝑡 0.75 × 100000 + 101325 𝑘𝑔
𝑎 = = = 2.007 3
𝑅𝑢 8314.5 𝑚
× 𝑇𝑎𝑚𝑏𝑖𝑒𝑛𝑡 × 306
28.97 28.97
where, 𝑃𝑎 = Atomizing gauge pressure of air

Calculating the velocityof air(𝑉𝐹 ) and relative velocity (𝜔𝑅 ) :

𝑚̇𝐹 ∗10−3 0.34∗10−3

Fuel Velocity (𝑽𝑭 ) = 𝜋 = π = 4.02 𝑚/𝑠
𝐹 ∗ 4 ∗(𝐷𝐹,𝑖 )2 672.81∗ 4 ∗0.4 2

𝛾−1
2𝛾 𝑅𝑢 𝑃𝑎𝑚𝑏𝑖𝑒𝑛𝑡 𝛾
𝜔𝑅 = √ ∗ 𝑇𝑎𝑚𝑏𝑖𝑒𝑛𝑡 ∗ (1 − ( ) )
𝛾 − 1 𝑀𝑎𝑖𝑟 𝑃𝑔𝑢𝑎𝑔𝑒 + 𝑃𝑎𝑚𝑏𝑖𝑒𝑛𝑡

1.4−1
2∗1.4 8314.5 101.325 1.4
𝜔 𝑅 =√ ∗ 306 ∗ (1 − ( ) ) = 300 𝑚/𝑠
1.4−1 28.97 75+101.325
Substituting all the valves in equation (1), we get
1
−
2.007 ∗ 3002 ∗ 0.0004 2 0.00041 1
𝑆𝑀𝐷 = 0.0004 (0.48 ( ) + 0.35 ) (1 + )
0.02086 √672.81 ∗ 0.02086 ∗ 0.0004 1.204

𝑆𝑀𝐷 = 7.37 × 10−6 𝑚

Droplet Diameter:

𝐷ℎ𝑢 = √𝑆𝑀𝐷2 − (ℎ𝑢 × 𝑡ℎ𝑢 )

where, ℎ𝑢 is the evaporation constant at droplet heat up and 𝑡ℎ𝑢 is the Time of droplet heating
up.

𝐷ℎ𝑢 =7.32 × 10−6 𝑚

Fuel vapour pressure(𝑷𝑭𝒔 ) from the Clausius-Clapeyron equation can be given

as:
𝑃𝐹𝑠 𝐿 1 1
𝑙𝑛 ( ) = − 𝑅 ∗ (𝑇 − 𝑇 )
𝑃𝑎𝑚𝑏𝑖𝑒𝑛𝑡 𝐹𝑠 𝑏𝑛

where, 𝐿 is Heat of vaporization, 𝑇𝐹𝑠 is the fuel surface temperature and 𝑇𝑏𝑛 is the boiling point
of the fuel
Fuel mass fraction on the droplet surface:
𝑃𝐹𝑠 ∗𝑀𝐹
𝑤𝐹𝑠 =
𝑃𝐹𝑠 ∗𝑀𝐹 +(𝑃𝑎𝑚𝑏𝑖𝑒𝑛𝑡 −𝑃𝐹𝑠 )∗𝑀𝐴

where, 𝑀𝐹 and 𝑀𝐴 are the molar mass of fuel and air respectively.

Mass transfer number:

𝑤𝐹𝑠
𝐵𝑀 =
1 − 𝑤𝐹𝑠
Under steady state conditions: Droplet surface temperature is constant and the mass and heat
transfer numbers are equal
 𝐵𝑀 =𝐵𝑇
Refrence Temperature:
𝑇𝑎𝑚𝑏𝑖𝑒𝑛𝑡 − 𝑇𝐹𝑠
𝑇𝑟 = 𝑇𝐹𝑠 +
3

Refrence fuel and air mass flow:

𝑊𝐹𝑎𝑚𝑏𝑖𝑒𝑛𝑡 − 𝑊𝐹𝑠 2
𝑊𝐹𝑟 = 𝑊𝐹𝑠 + = 𝑊𝐹𝑠
3 3

𝑊𝐴𝑟 = 1 − 𝑊𝐹𝑠

Specific heat capacity of gas mixture:

𝐶𝑃𝑎𝑖𝑟 (𝑇𝑟,𝑃𝑎𝑚𝑏𝑖𝑒𝑛𝑡 ) = −2 × 10−7 T 3 + 0.0006 T 2 − 0.2159 T + 1025.5

𝐽
𝐶𝑃𝑎𝑖𝑟 = 1048.7
𝑘𝑔𝐾
𝐽
𝐶𝑃𝐹 (𝑇𝑟,𝑇𝐹𝑠 ) = 2581.03
𝑘𝑔𝐾

𝐽
𝐶𝑝𝑔 = 𝐶𝑃𝐹 𝜔𝐹𝑟 + 𝐶𝑃𝑎𝑖𝑟 𝜔𝐴𝑟 = 1796.3
𝑘𝑔𝐾

where, 𝐶𝑃𝐹 is fuel specific heat capacity at the reference temperature and 𝐶𝑝𝑔 is
the specific heat capacity of the the mixture.

Thus the heat transfer number is calculated as:

𝑇𝑎𝑚𝑏𝑖𝑒𝑛𝑡 − 𝑇𝐹𝑠
𝐵𝑇 = 𝐶𝑝𝑔 ( )
𝐿
𝑀𝐹
We can calculate the valve of 𝑇𝐹𝑠 ( fuel surface temperature) using itrations untill both
mass and heat transfer numbers are equal( 𝐵𝑀 =𝐵𝑇 )

𝑇𝐹𝑠 𝐵𝑀 𝐵𝑇
360 7.91 2.4
355 4.96 2.37
350 3.41 2.33
349 3.19 2.33
348 2.98 2.32
347 2.8 2.49
346 2.63 2.30
345 2.47 2.30
344 2.32 2.29
343.75 2. 2.29

At 𝑇𝐹𝑆 = 346.68K 𝐵𝑀 =𝐵𝑇 =2.73

Using the valve of ( 𝑇𝐹𝑠 ) we can calculate the valve of above equations:

𝑃𝐹𝑆 =44.74𝐾𝑃𝑎 , 𝑇𝑟 = 525.45𝐾, 𝑤𝐹𝑠 =0.7319, 𝑊𝐹𝑟 = 0.487, 𝑊𝐴𝑟 = 0.512

𝐽 𝐽
𝐶𝑃𝑎𝑖𝑟 = 1048.7 , 𝐶𝑝𝑔 =1796.3
𝑘𝑔𝐾 𝑘𝑔𝐾
Thermal Conductivity:
𝑊
𝑘𝑎𝑖𝑟 (𝑇𝑟 ) = −3 × 10−8 𝑇 2 + 10−4 𝑇 − 0.0011 = 0.043
𝑚𝐾
𝑊
𝑘𝐹 (𝑇𝑟 ) = 0.0322
𝑚𝐾
𝑊
𝑘𝑔 (𝑇𝑟 ) = 𝑘𝐹 𝜔𝐹𝑟 + 𝑘𝐹 𝜔𝐴𝑟 =0.0378𝑚𝐾

The rate of fuel evaporation:

𝑘𝑔
𝑀̇𝐹 = 2𝜋𝐷𝑑𝑟𝑜𝑝 ln(1 + 𝐵𝑇 )
𝐶𝑝𝑔
𝑚𝑔
𝑀̇𝐹 = 0.00127
𝑠

Fuel Evaporation Constant:

The fuel evaporation constant is given as:

8𝑘𝑔 𝑙𝑛(1 + 𝐵𝑇 ) 8𝑘𝑔 𝑙𝑛(1 + 𝐵𝑇 ) 𝑚2

𝑠𝑡 = = = 3.21 × 10 −7
𝐶𝑝𝑔 𝐹,𝑠𝑡 𝐶𝑝𝑔 𝐹,𝑠𝑡 𝑠

Where, 𝐹,𝑠𝑡 is the Density of the fuel at it's vapor pressure and surface temperature.

Evaporation time of the droplet:

Evaporation time of the droplet can be calculated as:

𝑆𝑀𝐷2 − 𝐷𝑑𝑢 2
𝑡𝑠𝑡 = = 0.00218𝑚𝑠
𝑠𝑡
 stoichiometric calculations (Heptane C7H16)

Calculate the specific stoichiometric air demand:

7 ∗ 𝑀𝐶 7 ∗ 12 𝑘𝑔 𝐶
𝑊𝐶 = = = 0.84
𝑀𝐹 (7 ∗ 12 + 16) 𝑘𝑔 𝑓𝑢𝑒𝑙

16 ∗ 𝑀𝐻 16 ∗ 1 𝑘𝑔 𝐻
𝑊𝐻2 = = = 0.16
𝑀𝐹 (7 ∗ 12 + 16) 𝑘𝑔 𝑓𝑢𝑒𝑙

Mass of air required with respect to per Kg mass of Carbon and Hydrogen:
2 ∗ 𝑀𝑂 79 2𝑀𝑁 2 ∗ 16 79 ∗ 2 ∗ 14 103 𝑘𝑔 𝑎𝑖𝑟
𝑚𝑎𝑖𝑟,𝐶 = + = + =
𝑀𝐶 21 𝑀𝐶 12 21 ∗ 12 9 𝑘𝑔 𝐶

2𝑀𝑂 79 2𝑀𝑁 2∗16 79∗2∗14 103 𝑘𝑔 𝑎𝑖𝑟

𝑚𝑎𝑖𝑟,𝐻 = + = + =
4𝑀𝐻 21 4𝑀𝐻 4∗1 21∗4∗1 3 𝑘𝑔 𝐻

Stoichiometric air demand:

103 103 𝑘𝑔 𝑎𝑖𝑟
𝑚𝑎𝑖𝑟,𝑠𝑡 = 𝑚𝑎𝑖𝑟,𝐶 ∗ 𝑊𝐶 + 𝑚𝑎𝑖𝑟,𝐻 ∗ 𝑊𝐻 = ∗ 0.84 + ∗ 0.16=15.1
9 3 𝑘𝑔 𝑓𝑢𝑒𝑙

𝑣 79 𝑣𝑚𝑛 𝑣 79 𝑣
𝒴𝑎𝑖𝑟,𝑠𝑡 = [ 𝑀𝑚𝑛 + 21 𝑚𝑛
] ∗ 𝑊𝐶 + [4𝑀 𝑚𝑛
+ 21 4𝑀 ] ∗ 𝑊𝐻2
𝐶 𝑀𝐶 𝐻 𝐻

22.4 79 22.4 22.4 79 22.4 𝑁𝑚3 𝑎𝑖𝑟

𝒴𝑎𝑖𝑟,𝑠𝑡 = [ + ] ∗ 𝑊𝐶 + [ + ] ∗ 0.16 = 11.73
12 21 12 4 ∗ 1 21 4 ∗ 1 𝑘𝑔 𝑓𝑢𝑒𝑙

H2O, N2 and CO2 content of the fuel with respect to one kg fuel:
2∗𝑀𝑂 +4∗𝑀𝐻 2∗16+4∗1 𝑘𝑔 𝐻2 𝑂
𝑚𝐻2 𝑂,𝑓𝑢𝑒𝑙 = ∗ 𝑊𝐻 = *0.16= 1.44
4∗𝑀𝐻 4∗1 𝑘𝑔 𝑓𝑢𝑒𝑙

𝑣𝑚𝑛 ∗2 22.4∗2 𝑁𝑚3 𝐻2 𝑂

𝒴𝐻2 𝑂,𝑓𝑢𝑒𝑙 = ∗ 𝑊𝐻 = ∗ 0.16 = 1.792
4∗𝑀𝐻 4∗1 𝑘𝑔 𝑓𝑢𝑒𝑙

2∗𝑀𝑂 +𝑀𝑐 2∗16+12 𝑘𝑔 𝐶𝑂2

𝑚𝐶𝑂2 ,𝑓𝑢𝑒𝑙 = ∗ 𝑊𝐶 = ∗ 0.84 = 3.08
𝑀𝐶 12 𝑘𝑔 𝑓𝑢𝑒𝑙

𝑣𝑚𝑛 22,4 𝑁𝑚3 𝐶𝑂2

𝒴𝐶𝑂 ,𝑓𝑢𝑒𝑙 = ∗ 𝑊𝐶 = ∗ 0.84 = 1.568
2 𝑀𝐶 12 𝑘𝑔 𝑓𝑢𝑒𝑙
 𝑘𝑔 𝑁2
𝑚𝑁2,𝑓𝑢𝑒𝑙 = 0.77 ∗ 𝑚𝑎𝑖𝑟,𝑠𝑡 = 0.77 ∗ 15.1 = 11.627
𝑘𝑔 𝑓𝑢𝑒𝑙

𝑁𝑚3 𝑁2
𝒴𝑁2,𝑓𝑢𝑒𝑙 = 0.79 ∗ 𝒴𝑎𝑖𝑟,𝑠𝑡 = 0.79 ∗ 11.73 = 9.26
𝑘𝑔 𝑓𝑢𝑒𝑙

The stoichiometric specific mass and volume of the flue gas is given as:
𝑘𝑔 𝑤𝑒𝑡 𝑓𝑙𝑢𝑒 𝑔𝑎𝑠
𝑚𝑓𝑔,𝑤,𝑠𝑡 = 𝑚𝐻2𝑂 + 𝑚𝐶𝑂2 +𝑚𝑁2 = 1.44+3.08+11.627=16.147 𝑘𝑔 𝑓𝑢𝑒𝑙

𝑁𝑚3 𝑤𝑒𝑡 𝑓𝑙𝑢𝑒 𝑔𝑎𝑠

𝒴𝑓𝑔,𝑤,𝑠𝑡 = 𝒴𝐻2 𝑂 + 𝒴𝐶𝑂2 +𝒴𝑁2 = 1.792+1.568+9.26=12.62
𝑘𝑔 𝑓𝑢𝑒𝑙

The stoichiometric specific dry mass and volume of flue gas is given as:
𝑘𝑔 𝑑𝑟𝑦 𝑓𝑙𝑢𝑒 𝑔𝑎𝑠
𝑚𝑓𝑔,𝑑,𝑠𝑡 = 𝑚𝑓𝑔,𝑤,𝑠𝑡 − 𝑚𝐻2 𝑂 = 16.147 − 1.44 = 14.707 𝑘𝑔 𝑓𝑢𝑒𝑙

𝑁𝑚3 𝑑𝑟𝑦 𝑓𝑙𝑢𝑒 𝑔𝑎𝑠

𝒴𝑓𝑔,𝑑,𝑠𝑡 = 𝒴𝑓𝑔,𝑤,𝑠𝑡 − 𝒴𝐻2 𝑂 =12.62 − 1.792 = 10.828 𝑘𝑔 ℎ𝑒𝑝

The actual air- to fuel equivalence ratio (based on the measured O2 content of the flue gas)
is given as:

𝒴𝑓𝑔,𝑑,𝑠𝑡 𝒴𝑂2,𝑚𝑒𝑎𝑠𝑢𝑟𝑒𝑑 10.828 ∗ 13.4

𝑚𝑒𝑎𝑠𝑢𝑟𝑒𝑑 = +1 = + 1 = 2.627
𝒴𝑎𝑖𝑟,𝑠𝑡 (21 − 𝒴𝑂2,𝑚𝑒𝑎𝑠𝑢𝑟𝑒𝑑 ) 11.73 ∗ (21 − 13.4)

where, 𝒴𝑂2,𝑚𝑒𝑎𝑠𝑢𝑟𝑒𝑑 is the measured O2 content of the flue gas in volume percent.

Actual air demand:

𝑘𝑔 𝑎𝑖𝑟
𝑚𝑎𝑖𝑟 = 𝑚𝑒𝑎𝑠𝑢𝑟𝑒𝑑 ∗ 𝑚𝑎𝑖𝑟,𝑠𝑡 = 2.627 ∗ 15.1 = 39.6677 𝑘𝑔 𝑓𝑢𝑒𝑙

Therefore, the air mass flow is given as:

ṁ F × 3600 × mair 39.6677 ∗ 3600 ∗ 39.6677

𝑀̇𝑎𝑖𝑟 = = = 48.5532 𝑘𝑔/ℎ
1000 1000

where, ṁ F is the fuel mass flow.

Dry flue gas quantity
𝑘𝑔 𝑑𝑟𝑦 𝑓𝑙𝑢𝑒 𝑔𝑎𝑠
𝑚𝑓𝑔,𝑑 = 𝑚𝑓𝑔,𝑑,𝑠𝑡 + (𝑚𝑒𝑎𝑠𝑢𝑟𝑒𝑑 − 1)𝑚𝑎𝑖𝑟,𝑠𝑡 = 14.707 + (1.627) ∗ 15.1 = 39.275 𝑘𝑔 𝑓𝑢𝑒𝑙

0.34
𝑚̇𝑓𝑔,𝑑 = 𝑚̇𝐹 𝑚̇𝑓𝑔,𝑑 = 1000 *39.275*3600= 48.0726 𝑘𝑔/ℎ

𝒴𝑓𝑔,𝑑 = 𝒴𝑓𝑔,𝑑,𝑠𝑡 +(𝑚𝑒𝑎𝑠𝑢𝑟𝑒𝑑 − 1)∗ 𝒴𝑎𝑖𝑟,𝑠𝑡 = 10.828 + (2.627 − 1) ∗ 11.73

𝑁𝑚3 𝑑𝑟𝑦 𝑓𝑙𝑢𝑒 𝑔𝑎𝑠

𝒴𝑓𝑔,𝑑 = 29.91 𝑘𝑔 𝑓𝑢𝑒𝑙

Wet flue gas quantity

𝑚𝑓𝑔,𝑤 = 𝑚𝑓𝑔,𝑤,𝑠𝑡 + (𝑚𝑒𝑎𝑠𝑢𝑟𝑒𝑑 − 1)𝑚𝑎𝑖𝑟,𝑠𝑡 = 16.147 + (2.627 − 1) ∗ 15.1

𝑘𝑔 𝑑𝑟𝑦 𝑓𝑙𝑢𝑒 𝑔𝑎𝑠

𝑚𝑓𝑔,𝑤 = 40.7147
𝑘𝑔 𝑓𝑢𝑒𝑙

𝑚̇𝑓𝑔,𝑤 = 𝑚̇𝐹 ∗ 𝑚
̇ 𝑓𝑔,𝑤 = 49.83 𝑘𝑔/ℎ

𝒴𝑓𝑔,𝑤 = 𝒴𝑓𝑔,𝑤,𝑠𝑡 +(𝑚𝑒𝑎𝑠𝑢𝑟𝑒𝑑 − 1)∗ 𝒴𝑎𝑖𝑟,𝑠𝑡 = 12.62 + (2.627 − 1) ∗ 11.73

𝑁𝑚3 𝑑𝑟𝑦 𝑓𝑙𝑢𝑒 𝑔𝑎𝑠

𝒴𝑓𝑔,𝑑 = 31.704
𝑘𝑔 𝑓𝑢𝑒𝑙
0.34
𝒴̇𝑓𝑔,𝑤 = 𝑚̇𝐹 ∗ 𝒴𝑓𝑔,𝑤 = ∗ 31.704 ∗ 3600 = 38.805 𝑚3 /ℎ
1000

Combustion Power:

LHV=39.91 ∗ 𝑊𝑐 + 117.83 ∗ 𝑊𝐻2 − 18 ∗ 𝑊𝐻2 ∗ 𝐿𝐻20

𝑀𝐽
LHV = 39.91 ∗ 0.84 + 117.83 ∗ 0.16 − 18 ∗ 0.16 ∗ 2.51 = 45.14
𝐾𝑔

𝑄̇𝑐𝑜𝑚𝑏 = 𝑚̇𝐹 ∗LHV

𝑄̇𝑐𝑜𝑚𝑏 =0.34*10-3*45*106 =15.3 𝐾𝑊

Reference air-to-fuel equivalence ratio:
𝒴𝑓𝑔,𝑑,𝑠𝑡 ∗𝒴𝑂2 ,𝑟𝑒𝑓 10.828∗4
𝑟𝑒𝑓 = 𝒴 + 1= +1=1.217
𝑎𝑖𝑟,𝑠𝑡 ∗(21−𝒴𝑂2 ,𝑟𝑒𝑓 ) 11.73∗(21−4)

where, 𝒴𝑂2,𝑟𝑒𝑓 is the refrence O2 level.

Reference dry flue gas quantity:

𝑁𝑚3 𝑑𝑟𝑦 𝑓𝑙𝑢𝑒 𝑔𝑎𝑠
𝒴𝑓𝑔,𝑑,𝑟𝑒𝑓 = 𝒴𝑓𝑔,𝑑,𝑠𝑡 +(𝑟𝑒𝑓 − 1)∗ 𝒴𝑎𝑖𝑟,𝑠𝑡 = 10.828+(1.217-1) *11.73=13.37 𝑘𝑔 𝑓𝑢𝑒𝑙

NOx emission at refrence O2 level:

𝐶𝑁𝑂𝑋,𝑚𝑒𝑎𝑠𝑢𝑟𝑒𝑑 = 80 𝑝𝑝𝑚:
𝒴𝑓𝑔,𝑑 29.91
𝐶𝑁𝑂𝑋 = 𝐶𝑁𝑂𝑋 ,𝑚𝑒𝑎𝑠𝑢𝑟𝑒𝑑 = 80 ∗ = 178.96 𝑝𝑝𝑚
𝒴𝑓𝑔,𝑑,𝑟𝑒𝑓 13.37

2𝑀𝑁 +𝑀𝑜 2∗14+16 𝑚𝑔

𝑐𝑁𝑂𝑋 = 𝐶𝑁𝑂𝑋 = 178.96 ∗ = 351.52
𝑣𝑚𝑛 22.4 𝑚3

THC emission at reference O2 level:

𝐶𝑇𝐻𝐶,𝑚𝑒𝑎𝑠𝑢𝑟𝑒𝑑 = 32 𝑝𝑝𝑚:
𝒴𝑓𝑔,𝑑 29.91
𝐶𝑇𝐻𝐶 = 𝐶𝑇𝐻𝐶,𝑚𝑒𝑎𝑠𝑢𝑟𝑒𝑑 = 32 ∗ = 71.58 𝑝𝑝𝑚
𝒴𝑓𝑔,𝑑,𝑟𝑒𝑓 13.37

4𝑀𝐶 + 10𝑀𝐻 4 ∗ 12 + 10 ∗ 1 𝑚𝑔
𝑐𝑇𝐻𝐶 = 𝐶𝑇𝐻𝐶 = 71.58 ∗ = 185.34 3
𝑣𝑚𝑛 22.4 𝑚

CO emission at refrence O2 level:

𝐶𝐶𝑂,𝑚𝑒𝑎𝑠𝑢𝑟𝑒𝑑 = 30 𝑝𝑝𝑚:
𝒴𝑓𝑔,𝑑 29.91
𝐶𝐶𝑂 = 𝐶𝐶𝑂,𝑚𝑒𝑎𝑠𝑢𝑟𝑒𝑑 = 30 ∗ = 67.11ppm
𝒴𝑓𝑔,𝑑,𝑟𝑒𝑓 13.37

𝑀𝐶 +𝑀𝑂 12+16 𝑚𝑔
𝑐𝐶𝑂 = 𝐶𝐶𝑂 = 67.11 ∗ =83.88
𝑣𝑚𝑛 22.4 𝑚3
Maximum turbulent flame speed to avoid the flame flashback in the mixing
tube:

Mass flow in mixing tube is given as:

𝐾𝑔
𝑚̇𝑚𝑡 = 𝑚̇𝑓𝑢𝑒𝑙 + 𝑀̇𝑎𝑖𝑟 = 0.0134 + 0.00034 = 0.01374
𝑠

Air and fuel vapour density:

𝑃𝑎𝑚𝑏𝑖𝑒𝑛𝑡 101325 𝑘𝑔
𝑎𝑖𝑟 = 𝑅𝑢 = 8314.5 = 0.399 𝑚3
×𝑇𝑝𝑟𝑒ℎ𝑒𝑎𝑡 ∗883
28.97 28.97

𝑃𝑎𝑚𝑏𝑖𝑒𝑛𝑡 101325 𝑘𝑔
𝑓𝑢𝑒𝑙,𝑣𝑎𝑝𝑜𝑢𝑟 = = = 3.51 3
𝑅𝑢 8314.5 𝑚
× 𝑇𝐹𝑆 ∗ 346.68
𝑀𝑓𝑢𝑒𝑙 100
Therefore, density of the gas in mixing tube:

𝑚̇𝑚𝑡 0.01374 𝑘𝑔
𝑚𝑖𝑥𝑖𝑛𝑔𝑡𝑢𝑏𝑒 = = = 0.397 3
𝑀̇𝑎𝑖𝑟 𝑚̇𝑓𝑢𝑒𝑙 0.0134 0.00034 𝑚
+
𝑎𝑖𝑟 + 𝑓𝑢𝑒𝑙,𝑣𝑎𝑝𝑜𝑢𝑟 0.399 3.51

The maximum turbulent flame speed is given as:

4 ∗ 𝑚̇𝑚𝑡 4 ∗ 0.01374 𝑚
𝑆𝑇,𝑚𝑎𝑥 = = = 61.35
𝑚𝑡 ∗ 𝜋 ∗ 𝑑𝑚𝑡
2 0.397 ∗ 𝜋 ∗ 26.82 ∗ 10−6 𝑠