Chapter 11: Combustion
 (Thanks to David Bayless for his assistance in writing this section)

Intro - Up to this point the heat Q in all problems and examples was either a given value or was obtained from the First Jurisprudence intercourse. All the same in different heat engines, gas turbines, and steam power plants the rut is obtained from combustion processes, using either solid fire (e.g. coal or wood). liquefied fire (e.g. gasolene, lamp oil, operating theatre diesel fuel), operating theater gaseous fuel (e.g. natural gas Beaver State propane).

In this chapter we introduce the chemistry and thermodynamics of combustion of generic hydrocarbon fuels - (CxHy), in which the oxydizer is the oxygen contained in atmospherical zephyr. Promissory note that we volition non cover the combustion of solid fuels or the complex blends and mixtures of the hydrocarbons which hold up gasolene, kerosene, or Rudolf Christian Karl Diesel fuels.

Atmospheric Air contains approximately 21% oxygen (O 2 ) by volume. The other 79% of "other gases" is mostly N (N 2 ), so we will assume tune to personify tranquil of 21% oxygen and 79% nitrogen past volume. Olibanum each mole of oxygen needed to oxidize the hydrocarbon is accompanied past 79/21 = 3.76 moles of nitrogen. Exploitation this combining the molecular aggregative of air becomes 29 [kg/kmol]. Note that it is assumed that the nitrogen wish not normally undergo some chemical reaction.

The Combustion Summons - The basic combustion process can be described by the fuel (the hydrocarbon) plus oxydizer (air or atomic number 8) titled the Reactants , which undergo a chemical substance swear out while cathartic heat to conformation the Products of combustion so much that mass is conserved. In the simplest combustion process, famous as Stoichiometric Combustion , whol the C in the fuel forms carbon dioxide (CO 2 ) and all the hydrogen forms water (H 2 O) in the products, thus we can write the chemical reaction as follows:


where z is titled the stoichiometric coefficient for the oxidizer (air)

Mark that this reaction yields Little Phoeb unknowns: z, a, b, c, d, thus we want quintuplet equations to solve. Stoichiometric combustion assumes that no excess oxygen exists in the products, therefore d = 0. We obtain the some other quatern equations from balancing the bi of atoms of each component 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 equivalence

Carbon copy (C)

x

a

a = x

Hydrogen (H)

y

2b

b = y/2

Oxygen (O)

2z

2a+b

z = a + b/2

N (N)

2(3.76)z

2c

c = 3.76z

Note that the water formed could be in the vapor or liquid phase, depending on the temperature and coerce of the combustion products.

Every bit an example consider the stoichiometric combustion of methane (CH4) in region air travel. Equating the molar coefficients of the reactants and the products we incur:

Speculative Air and Atmosphere-Fuel Ratio -The minimum amount of air which testament allow the complete burning of the fuel is called the Theoretical Air (also referred to as Ratio Air ). In this encase the products do not moderate any oxygen. If we supply less than theoretical air then the products could include carbon monoxide (Carbon monoxide), thus it is normal practice to add more than a priori flying to prevent this occurrence. This Excess Air will result in oxygen coming into court in the products.

The standard measure of the amount of free-flying victimised in a combustion process is the Air-Fuel Ratio (AF), outlined arsenic follows:

Thence considering but the reactants of the methane combustion with theoretical air presented preceding, we obtain:

Solved Problem 11.1 - In this job we wish to break out the combustion equation and determine the air-fuel ratio for the complete combustion of n-Butane (C 4 H 10 ) with a) theoretic beam, and b) 50% surplusage air.


Analysis of the Products of Combustion - Combustion forever occurs at elevated railroad temperatures and we assume that all the products of combustion (including the water vapour) behave as ideal gases. Since they have different gas constants, it is convenient to use the perfect gas equation of state in terms of the gas constant as follows:

In the analysis of the products of burning there are a number of items of interest:

  • 1) What is the percentage intensity of specific products, in particular carbon copy dioxide (CO 2 ) and carbon paper monoxide (CO)?

  • 2) What is the dew point of the water vapor in the combustion products? This requires evaluation of the partial pressure of the water vapour portion of the products.

  • 3) There are experimental methods of volumetrical analysis of the products of combustion, unremarkably done happening a Dry Groundwork , conciliatory the volume part of all the components except the water vapor. This allows a simple method of determining the actual flying-fuel ratio and excess transmit used in a burning process.

For ideal gases we find that the mole fraction yi of the i'th component in a mixture of gases at a specific pressure P and temperature T is tight to the volume fraction of that component.
Since from the molar perfect gas relation: P.V = N.Ru.T, we have:

Furthermore, since the sum of the element volumes Vi must equalise the total volume V, we have:

Using a similar approach we determine the partial pressure of a factor victimization Law of partial pressures of Partial Pressures:

Solved Trouble 11.2 - In this trouble Propane (C 3 H 8 ) is burned with 61% excess air, which enters a combustion chamber at 25°C. Assuming complete combustion and a total pressure of 1 atm (101.32 kPa), determine a) the air-fire ratio [kilogram-send/kg-fuel], b) the percentage of carbon dioxide by loudness in the products, and c) the dew manoeuvre temperature of the products.

Solved Job 11.3 - In this problem Ethane (C 2 H 6 ) is burned with part air, and the volumetric analysis of the dry products of combustion yields the following: 10% Atomic number 27 2 , 1% CO, 3% O 2 , and 86% N 2 . Develop the burning equality, and determine a) the part of excess air, b) the air-fuel ratio, and c) the dew point of the combustion products.


The First Law Analytic thinking of Combustion - The chief purpose of combustion is to produce heat through a change of enthalpy from the reactants to the products. From the First Police equation in a control volume, ignoring kinetic and P.E. changes and assuming zero 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 for each one component and h [kJ/kmol] refers to the molar enthalpy of each component.

Since there are a count of different substances involved we need to prove a common reference nation to evaluate the heat content, the common selection being 25°C and 1 atm which is commonly denoted with a superscript o. Professor. S. Bhattacharjee of the San Diego State University has developed a web based expert system at < World Wide Web.thermofluids.net > called TEST ( T he E xpert S ystem for T hermodynamics) in which he has included a set of perfect gas property tables all supported the heat content h o = 0 at this common reference. We have adapted some of these tables specifically for this section, and these can be found in the following link:

Combustion Tooth Enthalpy Tables

As an example, view once more the stark burning of Methane (CH4) with theoretical broadcast:

Notice that in the reactants and the products of the in a higher place object lesson we have basic elements O 2 and N 2 as well as compounds CH 4 , Carbon monoxide gas 2 , and H 2 O. When the compound is formed and so the enthalpy alter is called the H of Formation , denoted h f o , and for our example:

Substance

Formula

hfo [kJ/kmol]

CO2

CO2(g)

-393,520

Water Vapor

H2O(g)

-241,820

Water system

H2O(l)

-285,820

Methane

CH4(g)

-74,850

where (g) refers to vaunt and (l) refers to watery.

The negative sign means that the process is Exothermic , i.e. wake is given murder when the compound is formed. Note that the enthalpy of organisation of canonical elements O 2 and N 2 is zero point.

Consider low gear the case in which on that point is sufficient heat transfer much that some the reactants and the products are at 25°C and 1 atm pres, and that the H2O product is clear. Since in that location is nary sensible enthalpy deepen the Department of Energy equation becomes:

This heat (Qcv) is titled the Heat content of Burning or the Heating Rate of the fuel. If the products bear molten water and then information technology is the Higher Heating Value (as in our case), however if the product contains water vapour then it is the Lower Heating Value of the fire. The total heat of combustion is the largest amount of heating system that can exist released by a given fuel.

Adiabatic Flaming Temperature - The opposite extreme of the to a higher place model in which we evaluated the enthalpy of combustion is the showcase of an adiabatic process in which nary heating is released. This results in a probative temperature gain in the products of burning (denoted the Adiabatic Flame Temperature ) which tin can only be reduced by an increase in the air-fire ratio.

Solved Problem 11.4 - Determine the adiabatic flaming temperature for the concluded burning of Methane ( CH 4 ) with 250% hypothetic air in an adiabatic see volume.

This equivalence can only be solved aside an iterative run and error procedure victimization the tables of Sensible Enthalpy vs Temperature for each four components of the products - Carbon monoxide 2 , H 2 O, O 2 , and N 2 . A quick approximation to the adiabatic flaming temperature can Be obtained by assuming that the products consist entirely of ventilate. This approach was introduced to us by Muck about and Somerton in their Schaum's Outline of Thermodynamics for Engineers , in which they assumed all the products to follow N 2 . We find it more convenient to use tune assuming a representative value of the Specific Heat Capacity of Melodic phrase : C p,1000K = 1.142 [kJ/kg.K].

Hence summing each the moles of the products we own:

Using the tables of Sensible H vs Temperature we evaluated the enthalpy of all quaternity products at a temperature of 1280K. This resulted in a total H of 802,410 [kJ/kmol fire], which is extremely close to the obligatory value, thusly justifying this approach.

Problem 11.5 - - Determine the adiabatic flare temperature for the complete burning of Propane ( C 3 H 8 ) with 250% theoretical air in an adiabatic control volume [ T = 1300K ].

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Engineering Thermodynamics by Israel Urieli is licensed under a Inventive Commons Attribution-Noncommercial-Share Alike 3.0 Coalescing States License

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