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Adiabatic flame temperature

In the study of combustion, there are two types of adiabatic flame temperature depending on how the process is completed, constant volume and constant pressure, describing the temperature the combustion products theoretically reach if no energy is lost to the outside environment.

The constant volume adiabatic flame temperature is the temperature that results from a complete combustion process that occurs without any work, heat transfer or changes in kinetic or potential energy. The constant pressureadiabatic flame temperature is the temperature that results from a complete combustion process that occurs without any heat transfer or changes in kinetic or potential energy. Its temperature is lower than the constant volume process because some of the energy is utilized to change the volume of the system (i.e., generate work).

It is a commonly misunderstood that the adiabatic flame temperature is the maximum temperature that can be achieved for given reactants because any heat transfer from the reacting substances and/or any incomplete combustion would tend to lower the temperature of the products. However, since the assumptions inherent in the adiabatic flame temperature assume chemical equilibrium, states in thermal equilibrium but not chemical equilibrium are not constrained by this limit. In fact, several fuel rich acetylene and methane flames have been found to exceed their adiabatic flame temperatures by hundreds of degrees.

Common flames

In daily life, the vast majority of flames one encounters are those of organic compounds including wood, wax, fat, common plastics, propane, and gasoline. 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 2n C–H bonds, n C–C bonds, and 1.5n O2 bonds to form roughly n CO2 molecules and n H2O molecules.

Because most combustion processes that happen naturally occur in the open air, there is nothing that confines the gas to a particular volume like the cylinder in an engine. As a result, these substances will burn at a constant pressure allowing the gas to expand during the process.

Common flame temperatures

Assuming initial atmospheric conditions (1 bar and 20°C), the following table list the adiabatic flame temperature for various gases under constant pressure conditions. The temperatures mentioned here are for a stoichiometric fuel-oxidizer mixture (i.e. equivalence ratio \phi = 1).

Note this is a theoretical flame temperature produced by a flame that loses no heat (i.e. closest will be the hottest part of a flame) where the combustion reaction is quickest. And where complete combustion occurs, so the closest flame temperature to this will be a non-smokey, commonly bluish flame

Thermodynamics

From the first law of thermodynamics for a closed reacting system we have,

{}_RQ_P - {}_RW_P = U_P - U_R

where, {}_RQ_P and {}_RW_P are the heat and work transferred during the process respectively, and U_R and U_P are the internal energy of the reactants and products respectively. In the constant volume adiabatic flame temperature case, the volume of the system is held constant hence there is no work occurring,

{}_RW_P = \int\limits_R^P {pdV} = 0

and there is no heat transfer because the process is defined to be adiabatic: {}_RQ_P = 0 . As a result, the internal energy of the products is equal to the internal energy of the reactants: U_P = U_R . Because this is a closed system, the mass of the products and reactants is constant and the first law can be written on a mass basis,

U_P = U_R \Rightarrow m_P u_P = m_R u_R \Rightarrow u_P = u_R .

In the constant pressure adiabatic flame temperature case, the pressure of the system is held constant which results in the following equation for the work,

{}_RW_P = \int\limits_R^P {pdV} = p\left( {V_P - V_R } \right)

Again there is no heat transfer occurring because the process is defined to be adiabatic: {}_RQ_P = 0 . From the first law, we find that,

- p\left( {V_P - V_R } \right) = U_P - U_R \Rightarrow U_P + pV_P = U_R + pV_R

Recalling the definition of enthalpy we recover: H_P = H_R . Because this is a closed system, the mass of the products and reactants is constant and the first law can be written on a mass basis,

H_P = H_R \Rightarrow m_P h_P = m_R h_R \Rightarrow h_P = h_R .

We see that the adiabatic flame temperature of the constant pressure process is lower than that of the constant volume process. This is because some of the energy released during combustion goes into changing the volume of the control system. One analogy that is commonly made between the two processes is through combustion in an internal combustion engine. For the constant volume adiabatic process, combustion is thought to occur instantaneously when the piston reaches the top of its apex (Otto cycle or constant volume cycle). For the constant pressure adiabatic process, while combustion is occurring the piston is moving in order to keep the pressure constant (Diesel cycle or constant pressure cycle).

If we make the assumption that combustion goes to completion (i.e. CO_2 and H_2O), we can calculate the adiabatic flame temperature by hand either at stoichiometric conditions or lean of stoichiometry (excess air). This is because there are enough variables and molar equations to balance the left and right hand sides,

{\rm{C}}_\alpha {\rm{H}}_\beta {\rm{O}}_\gamma {\rm{N}}_\delta + \left( {a{\rm{O}}_{\rm{2}} + b{\rm{N}}_{\rm{2}} } \right) \to \nu _1 {\rm{CO}}_{\rm{2}} + \nu _2 {\rm{H}}_{\rm{2}} {\rm{O}} + \nu _3 {\rm{N}}_{\rm{2}} + \nu _4 {\rm{O}}_{\rm{2}}

Rich of stoichiometry there are not enough variables because combustion cannot go to completion with at least CO and H_2 needed for the molar balance (these are the most common incomplete products of combustion),

{\rm{C}}_\alpha {\rm{H}}_\beta {\rm{O}}_\gamma {\rm{N}}_\delta + \left( {a{\rm{O}}_{\rm{2}} + b{\rm{N}}_{\rm{2}} } \right) \to \nu _1 {\rm{CO}}_{\rm{2}

Air-fuel ratio

Air-fuel ratio (AFR) is the mass ratio of air to fuel present during combustion. If exactly enough air is provided to completely burn all of the fuel, the ratio is known as the stoichiometric mixture (often abbreviated to stoich). AFR is an important measure for anti-pollution and performance tuning reasons. Lambda (λ) is an alternative way to represent AFR.

stoichiometric mixture fraction The relative amounts of oxygen enrichment and fuel dilution can be quantified by the stoichiometric mixture fraction, Zst, defined as Zst=(1+YF,0WOvO/YO,0WFvF)−1, where YF,0 and YO,0 represent the fuel and oxidizer mass fractions at the inlet, WF and WO are the species molecular weights, and vF and vO are the fuel and oxygen stoichiometric coefficients, respectively.

In industrial fired heaters, power plant steam generators, and large gas-fired turbines, the more common term is percent excess combustion air. For example, excess combustion air of 15 percent means that 15 percent more than the required stoichiometric air is being used.

A stoichiometric mixture is the working point that modern engine management systems employing fuel injection attempt to achieve in light load cruise situations. For gasoline fuel, the stoichiometric air/fuel mixture is approximately 14.7; i.e. the approximate mass of air is 14.7 mass of fuel. Any mixture less than 14.7 to 1 is considered to be a rich mixture, any more than 14.7 to 1 is a lean mixture - given perfect (ideal) "test" fuel (gasoline consisting of solely n-heptane and iso-octane). In reality, most fuels consist of a combination of heptane, octane, a handful of other alkanes, plus additives including detergents, and possibly oxygenators such as MTBE (methyl tert-butyl ether) or ethanol/methanol. These compounds all alter the stoichiometric ratio, with most of the additives pushing the ratio downward (oxygenators bring extra oxygen to the combustion event in liquid form that is released at time of combustions; for MTBE-laden fuel, a stoichiometric ratio can be as low as 14.1:1). Vehicles using an oxygen sensor(s) or other feedback-loop to control fuel to air ratios (usually by controlling fuel volume) will usually compensate automatically for this change in the fuel's stoichiometric rate by measuring the exhaust gas composition, while vehicles without such controls (such as most motorcycles until recently, and cars predating the mid-1980s) may have difficulties running certain boutique blends of fuels (esp. winter fuels used in some areas) and may need to be rejetted (or otherwise have the fueling ratios altered) to compensate for special boutique fuel mixes. Vehicles using oxygen sensors enable the air-fuel ratio to be monitored by means of an air fuel ratio meter.

Synopsis

In theory a stoichiometric mixture has just enough air to completely burn the available fuel. In practice this is never quite achieved, due primarily to the very short time available in an internal combustion engine for each combustion cycle. Most of the combustion process completes in approximately 4-5 milliseconds at an engine speed of 6000 rpm. This is the time that elapses from when the spark is fired until the burning of the fuel air mix is essentially complete after some 80 degrees of crankshaft rotation.

Catalytic converters are designed to work best when the exhaust gases passing through them show nearly perfect combustion has taken place.

A stoichiometric mixture unfortunately burns very hot and can damage engine components if the engine is placed under high load at this fuel air mixture. Due to the high temperatures at this mixture, detonation of the fuel air mix shortly after maximum cylinder pressure is possible under high load (referred to as knocking or pinging). Detonation can cause serious engine damage as the uncontrolled burning of the fuel air mix can create very high pressures in the cylinder. As a consequence stoichiometric mixtures are only used under light load conditions. For acceleration and high load conditions, a richer mixture (lower air-fuel ratio) is used to produce cooler combustion products and thereby prevent detonation and overheating of the cylinder head.

In the typical air to natural gas combustion burner, a double cross limit strategy is employed to ensure ratio control. (This method was used in World War 2). The strategy involves adding the opposite flow feedback into the limiting control of the respective gas (air or fuel).This assures ratio control within an acceptable margin.

Other terms used

There are other terms commonly used when discussing the mixture of air and fuel in internal combustion engines.

Mixture

Mixture is the predominant word that appears in training texts, operation manuals and maintenance manuals in the aviation world.

AFR

The Air fuel ratio is the most common reference term used for mixtures in internal combustion engines.

AFR = \frac{m_{air}}{m_{fuel}}

It is the ratio between the mass of air and the mass of fuel in the fuel-air mix at any given moment.

For pure octane the stoichiometric mixture is approximately 14.7:1 or λ of 1.00 exactly.

In naturally aspirated engines powered by octane, maximum power is frequently reached at AFRs ranging from 12.5 - 13.3:1 or λ of 0.850 - 0.901.

FAR

Fuel Air ratio is commonly used in the gas turbine industry as well as in government studies of internal combustion engine and refers to the ratio of fuel to the air, it is 1/AFR.

Lambda

Most practical AFR devices actually measure the amount of residual oxygen (for lean mixes) or unburnt hydrocarbons (for rich mixtures) in the exhaust gas. Lambda (λ) is the ratio of actual AFR to stoichiometry for a given mixture. Lambda of 1.0 is at stoichiometry, rich mixtures are less than 1.0, and lean mixtures are greater than 1.0.

There is a direct relationship between lambda and AFR. To calculate AFR from a given lambda, multiply the measured lambda by the stoichiometric AFR for that fuel. Alternatively, to recover lambda from an AFR, divide AFR by the stoichiometric AFR for that fuel. This last equation is often used as the definition of lambda:

\lambda = \


From Yahoo Answers

Question:What are the balanced chemical equations showing complete and incomplete combustion of alcohol with a formula of C2H6O?

Answers:Complete combustion : C2H6O +3 O2 = 2 CO2 + 3 H2O incomplete combustion : C2H6O + 2 O2 = 2 CO + 3 H2O

Question:I can't seem to find the answer anywhere!! :(

Answers:In the incomplete combustion of acetylene with oxygen, water and carbon monoxide are formed, rather than carbon dioxide which is formed in the complete combustion. So the balanced equation for the incomplete combustion is: 2C2H2 + 3O2 4CO + 2H2O Hope this helps.

Question:We put ethanol (C2H5OH) in a beaker and then set it on fire using a burning splint. We then covered it with a larger beaker and observed. What is the chemical equation of this hydrocarbon combustion? C2H5OH ---> CO2 + H2O Is that correct (unbalaned) Also if it is correct how do I balance that?

Answers:Ethanol burns with a pale blue, non luminous flame to form carbon dioxide and steam. C2H5OH + 3O2 ==> 2CO2 + 3H2O Ethanol

Question:If you were in the business of buying, transporting, and storing alcohols for use as home-heating fuels, which of these alcohols: Ethanol, Propanol, or Butanol would you choose to work with? I'm not sure which one would be the best to use. I'm not sure if this would help answer the question but the accepted values for combustion of the three alcohols are: Ethanol: 1396 kJ/mol Propanol: 2008 kJ/mol Butanol: 3318 kJ/mol

Answers:If i get the jist right basically you want to be able to get the most heat from the least amount of weight. Which would lower your shipping costs. I also think that perhaps volume might be important to know, cause it factors into cost. so first you have to find the mass of each molecule, i used approxiamations. ethanol=44g/mol propanol=60 g/mol butanol=74 g/mol so now you have to convert from kj/mol to kj/g for ethanol 1396kj/mol X mol/46g= ~30.34 kj/g for propanol 2008 kj/mol X mol/60g= ~33.46 kj/g for butanol 3318 kj/mol X mol/74g= ~44.83 kj/g so by weight butanol is your best bet. Now if you take volume int the equation you need to use the density of each substance ethanol= .7893 g /cm^3 propanol= .8657g /cm^3 butanol = .8095g /cm^3 so now we just multiply density by the answer before. for ethanol .7983g /cm^3 x 30.34 kj/g=~24.22 kj/cm^3 for propanol .8657 g/cm^3 x 33.46 kj/g=~28.97 kj/cm^3 for butanol .8095g/cm^3 x 44.83 kj/g= ~36.29 kj/cm^3 so if you had liter of each {(1 ml= 1 cm^3) 1000ml=1liter)} butanol would yield ~36000kj of energy compared to ~29000kj for propanol and ~24000kj for ethanol so butanol is the most economic , and to be honest it all comes down to dollar and cents, oh and they are all liquids at room temperture. so butanol