Unlike Enthalpy, Where We Can Only Ever Know Changes in H, We Can Know Absolute Values of S.

Chapter eleven: Combustion
(Thanks to David Bayless for his aid in writing this section)

Introduction - Upwardly to this bespeak the heat Q in all bug and examples was either a given value or was obtained from the Offset Law relation. However in various heat engines, gas turbines, and steam power plants the estrus is obtained from combustion processes, using either solid fuel (e.g. coal or wood). liquid fuel (e.thousand. gasolene, kerosine, or diesel), or gaseous fuel (e.grand. natural gas or propane).

In this chapter we innovate the chemistry and thermodynamics of combustion of generic hydrocarbon fuels - (C_xH_y), in which the oxydizer is the oxygen independent in atmospheric air. Note that we volition not cover the combustion of solid fuels or the circuitous blends and mixtures of the hydrocarbons which make upward gasolene, kerosene, or diesel fuel fuels.

Atmospheric Air contains approximately 21% oxygen (O _two ) past volume. The other 79% of "other gases" is generally nitrogen (N ₂ ), and then nosotros will presume air to exist composed of 21% oxygen and 79% nitrogen by volume. Thus each mole of oxygen needed to oxidize the hydrocarbon is accompanied past 79/21 = three.76 moles of nitrogen. Using this combination the molecular mass of air becomes 29 [kg/kmol]. Notation that it is assumed that the nitrogen volition not unremarkably undergo whatsoever chemical reaction.

The Combustion Process - The basic combustion process can be described by the fuel (the hydrocarbon) plus oxydizer (air or oxygen) called the Reactants , which undergo a chemic process while releasing heat to grade the Products of combustion such that mass is conserved. In the simplest combustion process, known as Stoichiometric Combustion , all the carbon in the fuel forms carbon dioxide (CO ₂ ) and all the hydrogen forms water (H _ii O) in the products, thus we tin can write the chemical reaction as follows:

where z is known as the stoichiometric coefficient for the oxidizer (air)

Notation that this reaction yields five unknowns: z, a, b, c, d, thus we demand 5 equations to solve. Stoichiometric combustion assumes that no excess oxygen exists in the products, thus d = 0. We obtain the other four equations from balancing the number of atoms of each element in the reactants (carbon, hydrogen, oxygen and nitrogen) with the number of atoms of those elements in the products. This means that no atoms are destroyed or lost in a combustion reaction.

Element	Amount in reactants	=	Amount in Products	Reduced equation
Carbon (C)	x		a	a = x
Hydrogen (H)	y		2b	b = y/2
Oxygen (O)	2z		2a+b	z = a + b/2
Nitrogen (N)	ii(three.76)z		2c	c = 3.76z

Notation that the water formed could exist in the vapor or liquid stage, depending on the temperature and pressure of the combustion products.

Every bit an example consider the stoichiometric combustion of methane (CH_iv) in atmospheric air. Equating the molar coefficients of the reactants and the products we obtain:

Theoretical Air and Air-Fuel Ratio -The minimum amount of air which will let the complete combustion of the fuel is called the Theoretical Air (also referred to as Stoichiometric Air ). In this instance the products do not incorporate any oxygen. If we supply less than theoretical air and then the products could include carbon monoxide (CO), thus it is normal practice to supply more than theoretical air to prevent this occurrence. This Excess Air volition effect in oxygen appearing in the products.

The standard measure of the amount of air used in a combustion process is the Air-Fuel Ratio (AF), defined every bit follows:

Thus considering only the reactants of the methyl hydride combustion with theoretical air presented to a higher place, we obtain:

Solved Trouble 11.one - In this trouble we wish to develop the combustion equation and determine the air-fuel ratio for the complete combustion of due north-Butane (C ₄ H _ten ) with a) theoretical air, and b) 50% excess air.

Analysis of the Products of Combustion - Combustion always occurs at elevated temperatures and nosotros presume that all the products of combustion (including the water vapor) deport equally ideal gases. Since they take different gas constants, it is convenient to use the ideal gas equation of state in terms of the universal gas constant every bit follows:

In the analysis of the products of combustion in that location are a number of items of interest:

• 1) What is the percentage volume of specific products, in item carbon dioxide (CO ₂ ) and carbon monoxide (CO)?

• 2) What is the dew point of the water vapor in the combustion products? This requires evaluation of the partial force per unit area of the water vapor component of the products.

• 3) There are experimental methods of volumetric assay of the products of combustion, commonly done on a Dry Footing , yielding the volume percentage of all the components except the water vapor. This allows a simple method of determining the actual air-fuel ratio and excess air used in a combustion procedure.

For ideal gases we detect that the mole fraction y_i of the i'th component in a mixture of gases at a specific pressure P and temperature T is equal to the volume fraction of that component.
Since from the tooth ideal gas relation: P.V = Due north.R_u.T, we have:

Furthermore, since the sum of the component volumes V_i must equal the total volume V, we have:

Using a like approach we determine the partial pressure of a component using Dalton's Police of Partial Pressures:

Solved Trouble 11.two - In this problem Propane (C ₃ H ₈ ) is burned with 61% excess air, which enters a combustion chamber at 25°C. Assuming consummate combustion and a total pressure of 1 atm (101.32 kPa), determine a) the air-fuel ratio [kg-air/kg-fuel], b) the percentage of carbon dioxide by volume in the products, and c) the dew point temperature of the products.

Solved Problem 11.three - In this trouble Ethane (C ₂ H _half-dozen ) is burned with atmospheric air, and the volumetric analysis of the dry products of combustion yields the following: ten% CO ₂ , 1% CO, 3% O ₂ , and 86% N ₂ . Develop the combustion equation, and determine a) the per centum of excess air, b) the air-fuel ratio, and c) the dew point of the combustion products.

The Start Law Analysis of Combustion - The main purpose of combustion is to produce estrus through a change of enthalpy from the reactants to the products. From the Kickoff Law equation in a command volume, ignoring kinetic and potential energy changes and assuming no work is done, we have:

where the summations are taken over all the products (p) and the reactants (r). N refers to the number of moles of each component and h [kJ/kmol] refers to the molar enthalpy of each component.

Since there are a number of different substances involved we need to plant a common reference state to evaluate the enthalpy, the mutual pick being 25°C and one atm which is commonly denoted with a superscript o. Prof. S. Bhattacharjee of the San Diego State Academy has developed a web based expert arrangement at < www.thermofluids.net > chosen TEST ( T he Due east xpert Due south ystem for T hermodynamics) in which he has included a gear up of ideal gas belongings tables all based on the enthalpy h ^o = 0 at this mutual reference. We take adapted some of these tables specifically for this section, and these tin can exist found in the following link:

Combustion Molar Enthalpy Tables

Every bit an case, consider once again the complete combustion of Marsh gas (CH_iv) with theoretical air:

Notice that in the reactants and the products of the above instance we have bones elements O ₂ and N ₂ equally well every bit compounds CH _four , CO ₂ , and H _two O. When the compound is formed then the enthalpy change is called the Enthalpy of Formation , denoted h _f ^o , and for our case:

Substance	Formula	hfo [kJ/kmol]
Carbon dioxide	COtwo(g)	-393,520
Water Vapor	H2O(g)	-241,820
H2o	H2O(l)	-285,820
Marsh gas	CH4(1000)	-74,850

where (g) refers to gas and (l) refers to liquid.

The negative sign means that the process is Exothermic , i.e. estrus is given off when the compound is formed. Annotation that the enthalpy of germination of basic elements O ₂ and N ₂ is nil.

Consider first the case in which there is sufficient heat transfer such that both the reactants and the products are at 25°C and one atm pressure, and that the h2o production is liquid. Since at that place is no sensible enthalpy change the energy equation becomes:

This heat (Qcv) is called the Enthalpy of Combustion or the Heating Value of the fuel. If the products contain liquid h2o then it is the College Heating Value (as in our example), however if the product contains water vapor then information technology is the Lower Heating Value of the fuel. The enthalpy of combustion is the largest corporeality of oestrus that can be released by a given fuel.

Adiabatic Flame Temperature - The opposite extreme of the above example in which we evaluated the enthalpy of combustion is the case of an adiabatic process in which no heat is released. This results in a meaning temperature increment in the products of combustion (denoted the Adiabatic Flame Temperature ) which tin can only be reduced past an increment in the air-fuel ratio.

Solved Problem 11.4 - Determine the adiabatic flame temperature for the complete combustion of Methane ( CH ₄ ) with 250% theoretical air in an adiabatic command volume.

This equation tin can only be solved by an iterative trial and error procedure using the tables of Sensible Enthalpy vs Temperature for all four components of the products - CO ₂ , H ₂ O, O ₂ , and N _two . A quick approximation to the adiabatic flame temperature can be obtained past bold that the products consist entirely of air. This arroyo was introduced to us past Potter and Somerton in their Schaum's Outline of Thermodynamics for Engineers , in which they assumed all the products to be N ₂ . We detect it more convenient to employ air assuming a representative value of the Specific Heat Capacity of Air : C _p,1000K = 1.142 [kJ/kg.K].

Thus summing all the moles of the products we have:

Using the tables of Sensible Enthalpy vs Temperature we evaluated the enthalpy of all four products at a temperature of 1280K. This resulted in a full enthalpy of 802,410 [kJ/kmol fuel], which is extremely close to the required value, thus justifying this approach.

Problem 11.v - - Determine the adiabatic flame temperature for the complete combustion of Propane ( C ₃ H _viii ) with 250% theoretical air in an adiabatic command volume [ T = 1300K ].

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