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Theoretical air and excess air formula

Theoretical Air - an overview ScienceDirect Topic

  1. Stoichiometric air or theoretical air is the exact amount of air required to provide the right amount of oxygen for complete combustion. The amount of air required for stoichiometric combustion is fairly constant on the air/gas weight ratio, with an approximate value of 16
  2. The additional air is termed excess air, but the term theoretical air may also be used. 200% theoretical air is 100% excess air. The chemical equation for methane burned with 25% excess air can be expressed as CH4 + 1.25 x 2 (O2 + 3.76 N2) -> CO2 + 2 H2O + 0.5 O2 + 9.4 N2 Excess Air and O 2 and CO 2 in Flue Ga
  3. Atmospheric Air contains approximately 21% oxygen (O 2) by volume. The other 79% of other gases is mostly nitrogen (N 2), so we will assume air to be composed of 21% oxygen and 79% nitrogen by volume. Thus each mole of oxygen needed to oxidize the hydrocarbon is accompanied by 79/21 = 3.76 moles of nitrogen
  4. imum amount of air which supplies the required amount of oxygen for complete combustion of a fuel is called the stoichiometric or theoretical air. The amount of air in excess of the stoichiometric air is called excess air
  5. Theoretical Air = 100/23 * Net O 2 Net O 2 = (8/3 * C + 8 * H + S) - O 2 =100/23 (8/3 * 0.48 + 8 * 0.051 + 0.003) - 0.1

% theoretical air = 100% + % excess air (9.4) Slightly insufficient air results in CO being formed; some hydrocarbons may result from larger deficiencies. The parameter that relates the amount of air used in a combustion process is the air-fuel ratio (AF), which is the ratio of the mass of air to the mass of fuel The total quantity of air supplied varies with the quantity of the fuel, rate of combustion, system of firing and the draught intensity. The excess air may approach 100 per cent but the modern practice is to use 25 to 50 per cent. The excess air is indicated by CO 2 % in flue gases This is a section of a Class...Check out the complete class HERE: www.ChemicalEngineeringGuy.com/Coursesor Check out the Mass Balance Playlist in YouTube her.. Solutions for Chapter 14 Problem 109P: A fuel is burned with 80 percent theoretical air. This is equivalent to(a) 20% excess air(b) 80% excess air(c) 20% deficiency of air(d) 80% deficiency of air(e) stoichiometric amount of air

Video: Stoichiometric Combustion - Engineering ToolBo

The theoretical air required to burn the coal is = 1.64 / 23.2% = 7.1 kg of Air for 1 kg of Coal. This is the theoretical air required to burn the coal. Quick Calculation. The heating value of coal also depends on the elemental carbon and hydrogen. This means that the air required and the heating value have an almost fixed relationship Theoretical and Excess Air Calculation Example Given the following ultimate analysis of a coal: C = 71.6% H = 4.8% S = 3.4% O = 6.3% Ash = 9.1% If the excess air (EA) is 20% calculate: a) the theoretical air (W ta); b) total or actual air (W a); and c) flue gas (W g) all in lbm/(lbm of fuel). a) Using formula for W ta, enter values for fuel The amount of air in excess of the stoichiometric amount is called excess air. The amount of excess air is usually expressed in terms of the stoichiometric air as percent excess air or percent theoretical air. For example 150% = 1.5 × the theoretical air. Also, it may be referred to as 20% excess air; i.e., 120% stoichiometric air

Chapter 11: Combustion (Updated 5/31/10) - Ohio Universit

  1. ed by combustion calculations and is called stoichiometric air or theoretical air
  2. Excess air is the amount of air that exceeds theoretical air. So if you were to input 400 moles of air, and the theoretical air is found to be 300 moles, your excess air would equate to 100 moles. Percent Excess air is a simple formula which is: n air, fed − n air, theoretical n air, theoretical × 100
  3. With those measurements, the following formulas can be used to calculate excess air: For example, if the oxygen dry reading in flue gas is 2.5%, then the excess-air calculation would be: 0.895 x 0.025 / (0.21-0.025) = 12.1% excess air. Too much excess air leads to lower flame temperature. That means less heat gets into the system
  4. imum air in stoichiometric mixture. The stoichiometric air/fuel ratio (AFR) can be calculated from the reaction equation (g/g). For gas AFR is usuall

• The amount of air in excess of stoichiometric is called excess air • Usually expressed as percent excess air or percent theoretical air • For example, - 50% excess air = 150% theoretical air - 200% excess air = 300% theoretical air • Equivalence ratio is ratio of actual FA ratio to stoichiometric FA ratio: ( ) (). actual stoich FA. Excess air increase the amount of oxygen to the combustion and the combustion of fuel. when fuel and oxygen from the air are in perfect balance - the combustion is said to be stoichiometric The combustion efficiency increases with increased excess air - until the heat loss in the excess air is larger than the heat provided by more efficient. 2. Theoretical or Stoichiometric Air Requirements The quantity of oxygen theoretically required to ensure complete combustion As the composition of air is known, the Theoretical or Stoichiometric Air can easily be calculated by multiplying the stoichiometric oxygen by the molar ratio of 100/21 i.e. 4.76. Please correct in the notes page 3 There is a quick and dirty approach to estimate the actual air excess ratio, based on the measured oxygen content in the exhaust gas, which works quite well: Air Ratio = actual air volume (or mass).. Introduces percent excess air for combustion reactions. Made by faculty at Lafayette College and produced by the University of Colorado Boulder, Department o..

To convert from oxygen level to excess air percentage, the following simple formula can be used: Excess air = 92 O2 / (21 - O2) with O2 expressed in vol% (dry). Using this equation, we see that 3% O2 translates to 15% excess air, and 5% O2 is equal to 35% excess air. Okay, so what is the cost of excess excess air If the air supplied is 20% more than the stoichiometric value, find the analysis of the dry products by mass. SOLUTION If 20% excess air is supplied then the air supplied is: 120% x 13.03 = 15.637 kg Oxygen is also 20% excess so 0.2 x 2.997 = 0.599kg is left over. Nitrogen in the air is 77% x 15.637 = 12.04kg List of products Since 23.2 mass-percent of air is actually oxygen, we need : 3.99 * 100/23.2 = 17.2 kg air for every 1 kg of methane. So the stoichiometric air-fuel ratio of methane is 17.2. When the composition of a fuel is known, this method can be used to derive the stoichiometric air-fuel ratio

The amount of excess air can be tailored as part of the design to control the adiabatic flame temperature. The considerable distance between present temperatures in a gas turbine engine and the maximum adiabatic flame temperature at stoichiometric conditions is shown in Figure 3.24(b) , based on a compressor exit temperature of (922 K) taken as 0.09 pounds per cubic foot. The temperature of the ambient air shall be taken as 94oF. The volume of gases which are emitted by the boiler is a function of the firing rate and associated theoretical air requirements and excess air used in the combustion process. The more excess air used, the larger the volume of exhaust gase The operator of the heater measures excess air indirectly by checking the firebox oxygen level. To convert from oxygen level to excess air percentage, use the following simple formula: with O2 expressed in vol% (dry). Using this equation, we see that 3% O2 translates to 15% excess air, and 5% O2 is equal to 35% excess air. The cost of excess air The constant-pressure adiabatic flame temperature of such substances in air is in a relatively narrow range around 1950 °C. This is because, in terms of stoichiometry , the combustion of an organic compound with n carbons involves breaking roughly 2 n C-H bonds, n C-C bonds, and 1.5 n O 2 bonds to form roughly n CO 2 molecules and n H 2 O. Percent Excess Air: The amount of air in excess of the stoichiometric amount is called excess air. The percent excess air, %EA, is defined as %EA ¼ 100 m a m as m as ¼ 100 m a m as 1 (2.17) For example, a mixture with %EA¼50 contains 150% of the theoretical (stoichio-metric) amount of air

In the problem we are told octane, C8H18, is combusted with 200% theoretical air (or 100% excess air) so a =1. Using the formulas developed for the general case with excess air: C's → 8 2 vCO =x = H2's → 9 2 18 2 2 = = = y vH O O2's → 12.5 4 18 8 2 4 =+ =+ = y vO - 50% excess air = 150% theoretical air - 200% excess air = 300% theoretical air •Equivalence ratiois ratio of actual FA ratio to stoichiometric FA ratio:() ( excess air. [For example, for 300% theoretical air (200% excess air), (1+a)=3 so a=2]. The problem states that octane, C 8H 18, is combusted with 200% theoretical air (or 100% excess air) so a =1. Using the formulas developed for the general case with excess air the stochiometric coefficients can be determined as shown below The amount of air supplied can be expressed alternatively as a percent excess or a percent deficiency of air. Thus, 150% of theoretical air is equivalent to 50% excess air, and 80% of theoretical air is the same as a 20% deficiency of air. For example, consider the complete combustion of methane with 150% theoretical air (50% excess air). The.

amount of air, or 100% theoretical air. Stoichiometric or theoretical combustion: The ideal combustion process during which a fuel is burned completely with theoretical air. Excess air: The amount of air in excess of the stoichiometric amount. Usually expressed in terms of the stoichiometric air as percent excess air or percent theoretical air with 50 percent excess air. The combustion is incomplete with 10 percent of the carbon in the fuel forming carbon monoxide. Calculate the mole fraction of carbon monoxide and the apparent molar mass of the products Percent Excess Air = 10.00 % Percent Available Heat = 0.00 % There is no available heat (heat delivered to the home) because we are looking at the burner which is before the heat exchanger that captures the heat for the home. Chart Value For Flame Temperature = 3310 There is a theoretical ratio of air-to-fuel that provides exactly enough air to completely burn the fuel, converting it to $\ce{CO2}$ and $\ce{H2O}$. This theoretical ratio can be calculated knowing (a) the exact composition of the fuel and (b) how much oxygen is in the air (generally ~21%), and (c) the formula weight(s) of the molecules that. How is excess air calculated? Excess Air = 100 x (20.9%) / (20.9% -O2m%) - 100% Where O2 m% = The measured value of oxygen in the exhaust

For complete combustion to occur, we have to have excess air, or air supplied in excess of what is needed typically because of poor mixing of the fuel and air during the combustion process. If excess air is not provided we will not have the complete conversion of carbon to CO2, and will end up with the formation of partially oxidized compounds. ) Percent excess air ercent excess air, first we have to calculate theoretical air. Theoretical air is the air required for complete combustion of the following reactions: H 60.5O 6H 6O note that both CO and H 6 require 0.5 mole of oxygen. Hence of 7 air retical air / Theoretical air L 78.89% Excess Air. In order to ensure complete combustion, combustion chambers are fired with excess air. Excess air increases the amount of oxygen and nitrogen entering the flame increasing the probability that oxygen will find and react with the fuel. The addition of excess air also increases turbulence, which increases mixing in the combustion chamber Theoretical air demand = air needed for complete conversion of carbon to carbon dioxide and hydrogen to water vapor % excess air = 100 x (actual air used - theoretical air demand) / theoretical air demand = 100 x (22.7 - (10.81 + 8.11))/( 10.81 + 8.11) = 20 excess air or emission rate correction factor determinations, unless approved by the Administrator. If both percent CO 2 and percent O 2 are measured, the analytical results of any of the three procedures given below may also be used for calculating the dry molecular weight (see Method 3). 8.1 Single-Point, Grab Sampling and Analytical Procedure

15-2 15-13 Methane is burned with the stoichiometric amount of air during a combustion process. The AF and FA ratios are to be determined. Assumptions 1 Combustion is complete.2 The combustion products contain CO2, H2O, and N2 only. Properties The molar masses of C, H2, and air are 12 kg/kmol, 2 kg/kmol, and 29 kg/kmol, respectively (Table A-1). Analysis This is a theoretical combustion. Daniel A. Vallero, in Air Pollution Calculations, 2019 9.2 Air and combustion. The theoretical, sufficient concentration of O 2 to achieve complete combustion is that needed to react with the total C in the combustible material, that is, fuel. The air needed to achieve this is known as theoretical air or stoichiometric air, which depends on the chemical makeup of the fuel and the. The adiabatic flame temperature is a function of excess air and it is maximal for 0% excess air, and thus for theoretical air. Hence, the higher is excess air the lower will be the adiabatic flame temperature. As a practical consequence, excess air can be used to reach an adiabatic flame temperature lower than the. 2H2 + O2 2H2O (Moisture) Air quantity is calculated keeping in mind the complete combustion of fuel, so for the complete combustion of fuel excess air is kept to around 20% - 50% depending upon the type of fuel, size of fuel particles and degree of mixing

• Lambda = Volume (Practical Air / Theoretical Air) =14.28/9.52= 1.5 • Excess of Air = ( Lambda - 1 ) * 100 = ( 1,5 - 1 ) * 100 = 50% • Excess of Air measured from O2 ( 7.5% ) = %O2 measured * 100 / ( 20.9 - %O2 measured ) x Coeff KL= 50% • To little excess of air is inefficient because it permits unburned fuel, in the form of. excess air (2.1% O2 in flue gases), but tests show an actual ratio of 25% excess air (4.5% O 2 in flue gases). The chart shows an actual available heat of 22% compared to an ideal of 29%. Fuel Savings = 100 X (( 29 - 22 ) / 29 ) = 24% Note: The graph on the front page is for combustion air at ambient temperature (about 60ºF) usin

For sufficient air, We have already said that mass wise there is 23.2 % O 2 presents in air. Hence the amount of air required to provide 2.67 gm of O 2 is As per ideal combustion theory, after combustion of one gm carbon (C), product of combustion contains only 3.67 gm of CO 2 and of N 2.. Coal Combustion for Insufficient Air The theoretical amount of air is determined from the stoichiometric reaction to be The actual combustion process (1) 50 percent excess air (2) only 90 percent of the carbon burns to CO 2, with the remaining 10 percent forming CO. 1 st Law Analysis of Reacting Systems a th = 3 + 2 = 5 2 2 2 2 2 8 3 76. 3 4 3) 76. 3 () (N a O H CO N O a H C th th       2 2 2 2 2 2 8 3 2. 28 65. 2 4 3. 0 7

Fuels and Combustio

Excess air is the air flow in excess of the stoichiometric air-to-fuel ratio; excess air is expressed as a percentage of 100% theoretical air, i.e., if the air-to-fuel ratio is 1.1 times the stoichiometric air-to-fuel ratio, the excess air is 10% of theoretical air. Why reduce excess air EXCESS AIR AND BOILER EFFICIENCY In theory, to have the most efficient combustion in any combustion process, the quantity of fuel and air would be in a perfect ratio to provide perfect combustion with no unused fuel or air. This type of theoretical perfect combustion is called stoichiometric combustion. In practice, however, for safety and. Quantity of excess air can be determined by measuring the % Air in the flue gas. With the measured volume of % Air, the excess air can be calculated by the following formula % excess air = ((Theoretical Air - Actual Air)/(Actual Air)) X 100 Regards Shivshankar. Edited by Shivshankar, 20 April 2012 - 01:05 PM Knowing the recommended kerosene mass flow rate, the theoretical air quantity necessary for the combustion process and the air/oxidant excess, the air/oxidant mass flow rate and the fuel mass flow rate can be calculated for each case, in order to be further used as input data in numerical simulations of the combustion process High excess air levels (>45 percent) may result in increased NOx formation because the excess nitrogen and oxygen in the combustion air entering the flame will combine to form thermal NOx. Low excess air firing involves limiting the amount of excess air that is entering the combustion process in order to limit the amount of extra nitrogen and.

The calculation of the theoretical flame temperature is based on the assumption that the heat released by the combustion process is completely absorbed by reaction products and excess air. The temperature of the product gases is the flame temperature of interest 10 , 11 , 27 often termed T exit , T 2 or T f Excess air × 100 (Where O 2 is the % of oxygen in flue gas = 12% ) = 12 x 100 / (21 - 12) = 133% excess air Theoretical air required to burn 1 kg of oil = 14 kg (Typical value for all fuel oil) Total air supplied = Theoretical air x (1 + excess air/100) Total air supplied = 14 x 2.33 kg / kg of oil = 32.62 kg / kg of oil Sensible heat loss = m. well, by your definition of equivalence ratio, that would mean that you're inputting twice as much air as is theoretically needed, so your equation would go: C8H18 + 2*12.5 O2 + 2*47N2 --> 8CO2 + 9H2O + 2*47N2 + 12.5 O

it takes 8lbs of oxygen to burn hydrgen.8/.2315=34.56lbs of air or 8 x 4.32=34.56 it takes 1lb of oxygen to burn sulfur.1/.2315=4.32lbs of air oxygen and air always combined at the rate of 8 to 1 note .2315 = the weight of oxygen in 1lb of air To calculate Excess Air, % use the following formula: = (Air supplied - Theoretical air) x 100 Theroretical air. The theoretical air required will vary depending on the fuel that is being fired. I found a table in the ASHRAE 1985 Fundamentals Handbook (pg. 15.8) that lists these values How stoichiometric air-fuel ratio is calculated. In order to understand how the stoichiometric air-fuel ratio is calculated, we need to look at the combustion process of the fuel. Combustion is basically a chemical reaction (called oxidation) in which a fuel is mixed with oxigen and produces carbon dioxide (CO 2), water (H 2 O) and energy (heat). Take into account that, in order for the.

Fuels and combustion (2014)

Excess air is expressed as a percentage increase over the stoichiometric requirement and is defined by: Excess air will always reduce the efficiency of a combustion system. It is sometimes convenient to use term excess air ratio, defined as: Where sub-stoichiometric (fuel-rich) air-to-fuel ratios may be encountered, for instance, in the primary. Excess Air and Air-to-Fuel Ratio. a liquid and gas fuel burner achieves this desired balance by operating at 105% to 120% of the optimum theoretical air. For natural gas fired burners the stoichiometric air required is 9.4-11 ft. 3 / 1.0 ft. 3 of natural gas, or approximately an air-to-gas ratio of approximately 10:1. This results in an.

How can I calculate the theoretical air needed for

1.8mm, air temperature is 553 to 773K and its pressure is one standard atmosphere, and rule is represented in Fig. 4. Results of experiments and simulation are as follows: 4.1 Rule of evaporation rate in high-temperature air and the compare of experiments and theoretical calculation 4.1.1 Rule of evaporation rate and drop diameter with tim One mole of octane is burned with 120% theoretical air. Assuming that the octane (gas state) and air enter the combustion chamber at 25degree C and the excess oxygen and nitrogen in the reaction will not dissociate, calculate the following (put final answer in table below) Excess Air: Theoretical air amount that is exceeded by the amount of air fed. Percent Excess air Formula: [(moles air) fed - (Moles air) theoretical / (moles air) theoretical] x 100% . It is useful to know the fuel feed rate and stoichiometry in the equations to help with calculations like theoretical O ₂ and air feed rates. If the actual.

2H_2(g) + O_2(g) rarr 2H_2O(l). Clearly, there are 4 hydrogen atoms..... Clearly, there are 4 hydrogen atoms, but 2 hydrogen molecules, and 2 oxygen atoms, but 1 oxygen molecule on the left hand side of the equation. And on the RHS, there must necessarily be 4 hydrogen atoms and 2 oxygen atoms, because it is a balanced chemical equation The total amount of excess oxygen in the flue gases is (O - ½CO); hence N - 3.782(O - ½CO) represents the nitrogen content in the air actually required for combustion and N ÷ (N - 3.782[O - ½CO]) is the ratio of the air supplied to that required. This ratio minus one will be the proportion of excess air So that: the amount of excess air used is 85.7% of theoretical air used. 4- Propylene (C 3 H 6) is burned with 50 percent excess air during a combustion process. Assuming complete combustion and a total pressure of 105 kPa, determine (a) the air- fuel ratio and (b) the temperature at which the water vapor in the products will start condensing to find the convective heat transfer coefficient for the air. The heat sink design of the radiator must then be analyzed using the Effectiveness-NTU method to find the theoretical effectiveness, overall heat transfer rate of the radiator, and outlet temperatures of both air and water One mole of octane is burned with 120% theoretical air. Assuming that the octane and air enter the combustion chamber at 25{eq}^o{/eq}C and the excess oxygen and nitrogen in the reaction will not.

Combustion: Combustion Equations ,Enthalpy of Formation

1 Answer to A fuel is burned with 90 percent theoretical air. This is equivalent to (a) 10% excess air (b) 90% excess air (c) 10% deficiency of air (d) 90% deficiency of air (e) stoichiometric amount of air O 2 produces more amount of MgO than Mg (25.2g MgO vs. 3.98 MgO), therefore O 2 is the excess reactant in this reaction. Step 6: Find the amount of remaining excess reactant by subtracting the mass of the excess reactant consumed from the total mass of excess reactant given. Mass of excess reactant calculated using the limiting reactant Theoretical Yields When reactants are not present in stoichiometric quantities, the limiting reactant determines the maximum amount of product that can be formed from the reactants. The amount of product calculated in this way is the theoretical yield, the amount obtained if the reaction occurred perfectly and the purification method were 100%. Consider a fuel that is burned with ( a ) 130 percent theoretical air and (b) 70 percent excess air. In which case is the fuel burned with more air? Ask your homework questions to teachers and professors, meet other students, and be entered to win $600 or an Xbox Series X.

Solved: All Formulas Are Given In The Formula Sheet, If Th

How much Air is Required for Complete Combustion

The amount of air per mole of fuel is 1.5 times the theoretical amount determined by Eq. 13.4 . The excess air supplied appears in the products as O2 and as a greater amount of N2 than in Eq. 13.4 , based on the theoretical amount of air. The air-fuel ratio on a molar basis is, AF = (1.5) (2) (4.76)/1 = 14.28/ GHT = Total wet exhaust gases = Excess air + GHE GST = Total dry exhaust gases = Excess air + GSE CALCULATION 3 : Theoretical emission of SO2 with exhaust gases ST = maximum emission of SO2 = % S x 10,000 x 64 / 32 = % S x 20,000 = mg S / Kg Diesel oil Concentration of maximum theoretical SO2 in gases, relative to % O2 measured

Theoretical Air and Excess Oxygen in Combustion (MB1 2-50

Considering the combustion of methane with 20% excess air, the excess air (0.2×9.52) of 1.9 volumes will appear in the flue gases as (0.21×1.9)=0.4 volumes of oxygen and (1.9-0.4)=1.5 volumes of nitrogen Air contains 21 mol percent O2 and 79 mol percent of N2. The minimum amount of air which supplies the required amount of oxygen for complete combustion of a fuel is called the stoichiometric or theoretical air. The amount of air in excess of the stoichiometric air is called . excess air Theoretical air requirement = [ (11.6 × C) + {34.8 × (H2 - O2/8)} + (4.35 × S)]/100 kg/kg of fuel Excess Air supplied (EA) = O2% / (21-O2%) X 100 Actual mass of air supplied/ kg of fuel (AAS) = {1 + EA/100} × theoretical air Step-3 Calculate different percentage los

Solved: A fuel is burned with 80 percent theoretical air

, 86.85%. Determine (a) the air-fuel ratio on both a molar and a mass basis,(b) the percent theoretical air,(c) the dew point temperature of the products, in C, if the mixture were cooled at 1 atm. SOLUTION Known: Methane is burned with dry air. The molar analysis of the products on a dry basis is provided air is needed. Step 3: • Determine if additional air is needed. • If less than max., no additional air is needed. • If greater than or equal to max., additional air is needed. Table 1 Appliances Table Job: Prepared by: Date: Appliance Input rating (Btu/hr) Additional air needed? (Check one) Yes Room volume = Maximum appliance input

Stoichiometric Air or Theoretical Air Quantity

HVAC Talk: Rise in O2 and Excess Air During Combustion2(PDF) Excess Air Ratio Management in a Diesel Engine withEffect of Excess air on Combustion efficiency | Download

The combination law explains what happens to air when it's compressed into a smaller volume. It tells us that when air is compressed, the pressure and temperature of the air increases, as the volume of the space containing air decreases. By pushing air into a smaller space, we force it to become hotter and more pressurized A fuel is burned with 70 percent theoretical air This is equivalent to (a) 30 \% excess air (b) 70 \% excess air (c) 30 \% deficiency of air (d) 70 \% deficien Our Discord hit 10K members! Meet students and ask top educators your questions 15-45 15-72E Hydrogen is burned with 20 percent excess air during a steady-flow combustion process. The exit temperature of product gases is to be determined. Assumptions 1 Steady operating conditions exist.2 Air and combustion gases are ideal gases.3 Kinetic and potential energies are negligible. 4 There are no work interactions.5 The combustion chamber is adiabatic • When air is 30% excess than theoretical • When 30% excess air is preheated to 227. OC and 327Oc 3) Calculate theoretical maximum adiabatic flame temperature of fuel gas of composition 96 % CH4, 0.8 % CO2 and 3.2 % N2 when burnt with theoretical air. Assume fuel and air are mixed at 25O Air standard diesel engine cycle: In the diesel engine, air is compressed adiabatically with a compression ratio typically between 15 and 20. This compression raises the temperature to the ignition temperature of the fuel mixture which is formed by injecting fuel once the air is compressed Please cite this article as:Sh. Kharazmi, A. H. Benisi, A.Mozafari, An Experimental and Theoretical Study of the Effects of Excess Air Ratio and Waste Gate Opening Pressure Threshold on Nox Emission and Performance in a Turbocharged CNG SI Engine, International Journal o

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