Combustion Supplemental Notes PDF

Combustion Supplemental Notes Introducing Combustion (Chapter 13.1) ===================================== In *combustion reactions*, rapid oxidation of combustible elements of the fuel results in energy release as combustion products are formed. (Review 13.1, page 806 for chemical reaction termino...

Combustion Supplemental Notes Introducing Combustion (Chapter 13.1) ===================================== In *combustion reactions*, rapid oxidation of combustible elements of the fuel results in energy release as combustion products are formed. (Review 13.1, page 806 for chemical reaction terminology) - Three major combustible elements in common fuels are - Carbon, C such as in methane, ethane, octane - Hydrogen, H - Sulfur, S - Combustion is *complete* when - All carbon present in the fuel is burned to carbon dioxide CO2 - All hydrogen present is burned to water H2O - All sulfur present is burned to sulfur dioxide S2O - All other combustible elements are *fully oxidized* - When these conditions are not fulfilled, combustion is *incomplete*. - We can get other products such as CO, NOx, and other nasty stuff Modeling Combustion Air (13.1.2) -------------------------------- Oxygen is required in every combustion reaction. In most combustion applications, air provides the needed oxygen. The following model of ***dry air*** is used for simplicity: 1. All components of dry air other than oxygen are lumped together with nitrogen. With this idealization, air is considered to be **21% O~2~** and **79% N~2~** on a molar basis. 2. **Accordingly**, when air supplies the oxygen in a combustion reaction, every mole of **O~2~** is accompanied by **0.79/0.21 = 3.76** moles of **N~2~**. 3. The nitrogen present in the air is assumed inert. 4. The molecular weight of dry air is 28.97. When ***moist air*** is used in combustion, the water vapor present in the air should be considered in writing the combustion equation. Air-Fuel Ratio -------------- - The air-fuel ratio is the ratio of the amount of air in a combustion reaction to the amount of fuel. - The air-fuel ratio can be written on a molar basis: or on a mass basis - Conversion between these values is accomplished using the molecular weights of air, ***M*~air~**, and fuel, ***M*~fuel~**, - The fuel-air ratio is the reciprocal of the air-fuel ratio. Step 1 Balancing Reactions ========================== **Example**: Determine the balanced reaction equation for complete combustion of methane (**CH~4~**) with oxygen (**O~2~**). - For complete combustion, the products contain only carbon dioxide and water: where ***a***, ***b***, ***c*** denote the moles of **O~2~**, **CO~2~**, and **H~2~O**, respectively, each per mole of **CH~4~**. - Applying conservation of mass to carbon, hydrogen, and oxygen: - Solving these equations, the ***balanced*** reaction equation is Theoretical Amount of Air (page 808) ------------------------------------ The theoretical amount of air is the minimum amount of air that supplies sufficient oxygen for the complete combustion of all the carbon, hydrogen, and sulfur present in the fuel. - For complete combustion with the theoretical amount of air, the products consist of **CO~2~**, **H~2~O**, and **SO~2~** plus nitrogen present in the reactants. No free oxygen, **O~2~**, appears in the products. - Normally the amount of air supplied is either greater than or less than the theoretical amount. The amount of air actually supplied is commonly expressed as - A **percent of theoretical air** -- e.g., **150%** of theoretical air equals **1.5** times the theoretical amount. - A **percent excess** (**or percent deficiency**) **of air** -- e.g., **50%** excess air equals **150%** of theoretical air. EXAMPLE ------- The balanced chemical equation for complete combustion of methane with the theoretical amount of air is the air-fuel ratio on a molar basis: The balanced chemical equation for complete combustion of methane with 150% theoretical air 1 extra mole of O~2~ and 1.5x extra moles of N~2~ **[\ ]** Dry Product Analysis (top page 811) ----------------------------------- - In practical applications, combustion is generally ***incomplete***. - The products of combustion of actual combustion processes and their relative amounts can be determined only by measurement. - Certain devices for measuring the composition of combustion products report the analysis on a ***dry product analysis*** basis where the mole fractions are given for all products except water. **Example**: Ethanol (**C~2~H~5~OH**) is burned with air to give products with the dry molar analysis **3.16% CO~2~**, **16.6% CO**, **80.24% N~2~**. Determine the balanced chemical reaction. - Basing the solution for convenience on 100 moles of dry products, the reaction equation reads: Per Mole Fuel: divide by 9.88 mole fuel **[\ ]** Step 2 First Law Energy Balance of Reacting Systems =================================================== Example ------- Now we need to find the enthalpies of the reactants and products. Its not so straightforward as you will read. Step 3 Evaluating Properties (13.2.1) ===================================== Energy and Entropy Balances for Reacting Systems ------------------------------------------------ Combustion air and (frequently) products of combustion are each modeled as ideal gas mixtures. The ideal gas mixture principles summarized in Table 13.1 play a role. For reacting systems, the methods used for evaluating specific enthalpy and specific entropy differ fundamentally from the practices used thus far in nonreacting applications. Two new concepts: - Enthalpy of formation - Absolute entropy WHY: example: hydrogen and oxygen enter CV, each as ideal gases, and react to form liquid water according to - To apply an energy balance to the control volume, we might think of using enthalpy data from the ***steam tables*** for liquid water and from Table A-23 for the gases. - However, since those tables use ***arbitrary datums*** to assign enthalpy values, they must be used ***only*** to determine ***differences*** in enthalpy between two states, for then and only then do the arbitrary datums cancel -- see Sec. 3.6.3. - For the case under consideration, **H~2~** and **O~2~** enter the control volume but do not exit, and liquid water exits but does not enter. Accordingly, enthalpy differences from inlet to exit do not arise for each of these substances when applying an energy balance to the control volume. For each it is necessary to assign enthalpy values in a way that the ***common*** datum cancels. This is achieved using the ***enthalpy of formation***. - Like considerations apply when evaluating entropy values for substances in reacting systems; for them ***absolute entropy*** values are required. - An enthalpy datum for the study of reacting systems is established by assigning a value of zero to the enthalpy of **C**, **H~2~**, **N~2~**, **O~2~**, and other stable elements at the ***standard reference state*** defined by ***T*~ref~ = 298.15 K** **(25^o^C)** and ***p*~ref~ = 1atm.** - The enthalpy of a ***compound*** at the standard state equals its ***enthalpy of formation***, denoted. - The enthalpy of formation is the energy released or absorbed when the compound is formed from its elements, the compound and elements all being at ***T*~ref~** and ***p*~ref~**. Evaluating Enthalpy for Reactive Systems ---------------------------------------- - Tables A-25 and A-25E give values of the enthalpy of formation in kJ/kmol and Btu/lbmol, respectively. - As indicated by the table heading these data apply only to the standard state, **298 K**, **1 atm**. - A zero value is assigned to the enthalpy of formation for each of the first four substances (elemental formation). - Nonzero values apply to all other substances, where a minus sign correspond to an ***exothermic*** reaction when the compound is formed from its elements. A positive value of the enthalpy of formation corresponds to an ***endothermic*** reaction when the compound is formed from its elements. - The specific enthalpy of a compound at a state where temperature is ***T*** and pressure is ***p*** is determined from That is, the enthalpy of a compound is composed of - associated with the formation of the compound from its elements. This is obtained from Tables A-25 and A-25E. - associated with the change in state from the standard state to the state where temperature is ***T*** and the pressure is ***p***. Since this term is a difference at fixed composition, it can be evaluated from the ideal gas tables, steam tables, or other tables, as appropriate. - The enthalpy of formation concept enables the control volume energy rate balance to be implemented for engineering applications involving combustion. **[\ ]** Example ------- +---------+---------+---------+---------+---------+---------+---------+ | Compone | T | h\_f | h(T) | H(T\_re | Dh | h | | nt | | | kJ/kmol | f) | | | | | | kJ/kmol | | | kJ/kmol | kJ/kmol | | | | | | kJ/kmol | | | +=========+=========+=========+=========+=========+=========+=========+ | C | 298K | 0 | \- | \- | 0 | 0 | +---------+---------+---------+---------+---------+---------+---------+ | O~2~ | 400K | 0 | 11711 | 8682 | 3029 | 3029 | +---------+---------+---------+---------+---------+---------+---------+ | CO~2~ | 500K | -393520 | 17678 | 9364 | 8314 | -385206 | +---------+---------+---------+---------+---------+---------+---------+ Q\_out = 0 + 3029 -- (-385206 kJ/kmol) Q\_out = 388235 kJ/kmol**[\ ]** Solving Combustion Problems in EES ================================== In EES to solve combustion problems: - Change unit system to molar analysis and Kelvin -- options -- units molar and Kelvin - Use only Molecular formulas for the "fluid" in all property look ups -- either under ideal gas or NASA lists of fluids, e.g. O2, N2, CH4, CO2, CO - The enthalpy function, which find h\_bar is the h\_bar needed for all problems. You do not need to look up h, h\_ref, and the enthalpy of formation, e.g., - h\_bar = enthalpy(CO2, T= 400K) = h\_formation(CO2) + {h(400)-h(298)} - For the fuels, need to check if enthalpy at the reference temperature is correct or if you need to use **enthalpy\_formation(fluid)** function. Depending the phase of the fuel, you will get different answers. Check against Table A-25 - Look up all molecules separately, e.g., not air but O2 and N2 - Set up the energy equation as derived to find Q\_out Example from above as EES (and EES file posted on Canvas) The solution is less than 1% different and that's okay for the work we are doing. A screenshot of a computer Description automatically generated Heating Values of Hydrocarbon Fuels (13.2.3) ============================================ - The ***heating values*** of hydrocarbon fuels have important applications. - The ***heating value*** of a fuel is the difference between the enthalpy of the reactants and the enthalpy of the products when the fuel burns completely with air, reactants and products being at the same temperature ***T*** and pressure ***p***. - That is, the heating value per mole of fuel HV= **where** - **R** denotes the reactants and **P** denotes the products. - The ***n*'s** correspond to the coefficients of the reaction equation, each per mole of fuel. - Two heating values are recognized by name: - The **higher heating value** (HHV) is obtained when all the water formed by combustion is a liquid. - The **lower heating value** (LHV) is obtained when all the water formed by combustion is a vapor. - The higher heating value exceeds the lower heating value by the energy that would be released were all water in the products condensed to liquid. - Heating value data at 298 K, 1 atm are provided in Tables A-25 and A-25E with units of kJ/kg and Btu/lb, respectively. Adiabatic Flame Temperature (13.3) ================================== Consider the reactor at steady state shown in the figure In the absence of work and appreciable kinetic and potential energy effects, the energy liberated on combustion is transferred from the reactor in two ways only - Energy accompanying the exiting combustion products - Heat transfer to the surroundings - The **smaller** the heat transfer to the surroundings, the **greater** the energy carried out with the combustion products and thus the **greater** the temperature ***T*~P~** of the combustion products. - The ***adiabatic flame temperature*** is the temperature that would be achieved by the products in the limit of ***adiabatic*** operation. - The **maximum** adiabatic flame temperature corresponds to complete combustion with the theoretical amount of air. In class lecture will include examples of finding the adiabatic flame temperature and how to evaluate properties using EES. Summary and Process =================== 1. Balance reaction equation. Balance for theoretical amount of air first 2. Set up 1^st^ Law equation to solve to heat out or adiabatic flame temperature 3. Look up properties---tables 4. Solve Solving adiabatic flame temperature in EES ------------------------------------------ Solving for adiabatic flame temperature is much easier in EES because it is a simultaneous and implicit solver and the T\_p is imbedded in the product's enthalpy look-ups. **Example**: Methane gas at **25^o^C**, **1 atm** enters an insulated reactor operating at steady state and burns completely with 100% of theoretical air entering at **25^o^C**, **1 atm**. 1. Balanced combustion equation  2. Derive the 1^st^ Law equation that will solve for the adiabatic flame temperature. Q = 0 A white board with red writing Description automatically generated  3. On the reactants side we only have the fuel enthalpy of formation since the air is at T\_ref. 4. See variable window below solution. You will need to change the guess value for the product temperature from the default (1) to something more appropriate, i.e., 2000-3000K. Mathematically there is more than 1 solution to the adiabatic solution, but only one solution is correct. (The other solution is 1K so clearly wrong for a combustion temperature.) A screenshot of a computer Description automatically generated  As the amount of theoretical air decreases the adiabatic flame temperature Absolute Entropy and the Third Law of Thermodynamics ==================================================== ### Section 13.5 Third Law of Thermodynamics --------------------------- The **third law** deals with the entropy of substances at the absolute zero of temperature. Based on empirical evidence, this law states that the entropy of a pure crystalline substance is zero at the absolute zero of temperature, 0 K or 0°R. Substances not having a pure crystalline structure at absolute zero have a nonzero value of entropy at absolute zero. The experimental evidence on which the third law is based is obtained primarily from studies of chemical reactions at low temperatures and specific heat measurements at temperatures approaching absolute zero. ### ABSOLUTE ENTROPY The importance of the third law is that it provides a datum relative to which the entropy of each substance participating in a reaction can be evaluated so that no ambiguities or conflicts arise. The entropy relative to this datum is called the **absolute entropy.** When we use the tables for ideal gases (Tables A22 and A23), the s-not term (circled) is considered the absolute entropy at temperature T and P\_ref = 1 atm. (entropy in tables A-2 though A-18 are not absolute values) A mathematical equation with blue text Description automatically generated with medium confidence For ideal gas mixtures (as we have in combustion), the pressure needs to be the partial pressure of the component i.  We need to evaluate the entropy of each component of the mixture because the total mixture entropy is the sum of the components as given by A math formula with a plus and a positive symbol Description automatically generated The reason we want to evaluate the entropy so we can determine the entropy production and/or the exergy to do analysis on the exergetic efficiency of burning fuels.  is tabulated in Table A23 and A25. Reminder the partial pressure can be found by the mole fraction *yi* of the component and the total pressure. where the mole fraction is and n is the total number of moles of the mixture. Entropy Balance and determining entropy production -------------------------------------------------- On a *per mole of fuel basis* at Steady State*:* A black and white logo Description automatically generated Where F is for the fuel, R is for the reactants and P is for the products. Example and using EES for entropy --------------------------------- We will use Example 13.9 from the text to show how to do entropy analysis using EES. **Problem Statement:** Liquid Octane and air burn completely with 100% theoretical air. The reactants are at 25C and 1 ATM (reference T and P) and the products are at 2395 K and 1 atm. We need to find the entropy production per mole of fuel. There is no heat transfer to the environment. **Balanced Reaction equation:**  **Entropy Balance** for no heat transfer and entropy of products and reactants. A black text on a white background Description automatically generated And rearranged  **EES Code** Note: For entropy lookups in EES, you need the temperature and partial pressure for each component. The solution and values of the entropy values are differ by less than 1% from the text book solution. A screenshot of a computer Description automatically generated  Note on UNITS: Because of equating yi to Pi in the EES code, there are issues with some units. I will accept these types of unit issues as shown below in the screen shot, BUT you need to also turn in a screen shot of the unit analysis to show what the unit errors are and relate to mole fraction and such. A screenshot of a computer Description automatically generated Gibbs Function (13.5.3, page 843) --------------------------------- The thermodynamic property known as the Gibbs function is  For combusting mixtures to evaluate the Gibbs function, we also define the Gibbs function of formation, which is tabulated for in A25. The Gibbs function of formation is the change in the Gibbs function for the reaction in which the compound is formed from its elements. The Gibbs function at a state other than the standard state is found by adding to the Gibbs function of formation the change in the specific Gibbs function g\_not\_f between the standard state and the state of interest: Where  The Gibbs function of component *i* in an ideal gas mixture is evaluated at the *partial pressure* of component *i* and the mixture temperature. Chemical Exergy =============== #### Section 13.6, 13.7, 13.8 Read page 845 about conceptualizing chemical exergy. It is summarized here. - Chemical exergy is the measure of how far the composition of a system is from that of the exergy reference environment. - We distinguish chemical exergy by e^ch^ - Total exergy is thermomechanical (from before) plus chemical - Chemical formula of a of a substance is CaHbOc, where a, b and c are subscripts corresponding number of molecules of C, H or O - We want to determine the maximum theoretical work per mole of substance CaHbOc entering a control volume and reacting to form CO2, and H2O - Chemical exergy is given by A black text on a white background Description automatically generated Where  and h\_bar and s\_bar are the specific properties. Note that for chemical exergy for a fuel, the reactants include the amount of O2 to have complete combustion to H2O and CO2. And in general the equation based on CaHbOc (page 845 and see page 846 for derivation) A mathematical equations on a white background Description automatically generated Since the properties are the specific properties, we can evaluate them in EES but if you don't have that ability, we can split the entropy into two terms -- the absolute entropy, which is tabulated and the part that is a function of the partial pressure of the component. The y^e^ term is the mole fraction of the component in the environment.  A math equations on a white background Description automatically generated And simplified using the Gibbs function:  Where the Gibbs function is evaluated for each component at P0 and T0 and using the Gibbs function of formation: Note that the y^e^ terms are the mole fractions in of the components in the environment (reference) Standard Chemical Exergy ------------------------ For Hydrocarbons and some other substances, the standard chemical exergy at T0 and P0 are tabulated in TABLE A26 for two different models. We will use Model II for this class. You can read about the derivation of standard chemical exergy for hydrocarbons on pages 849-851 ### Standard Chemical Exergy of Other Substances (13.7.2) For other substances, we consider a reaction of the substance with other substances for which thestandard chemical exergies *are known,* resulting in:  where DELTAG is the change in Gibbs function for the reaction, regarding each substance as separate at temperature *T0* and pressure *p0.* The underlined term is evaluated using the *known* standard chemical exergies, together with the n\'s giving the moles of these reactants and products per mole of the substance whose chemical exergy is being evaluated. Total Exergy (13.8) ------------------- The exergy associated with a specified state of a substance is the sum of two contributions: the thermomechanical contribution and the chemical contribution introduced in this chapter. On a unit mass basis, the **total exergy** is A math equation with black text Description automatically generated with medium confidence And the **total flow exergy** is  Where we can calculate or use TABLE A26 Model II for the chemical exergy, converting it to mass basis using the MW.