ENERGY CONVERSION ES 832a Eric Savory eng.uwo/people/esavory/es832.htm

1. ENERGY CONVERSION ES 832a Eric Savory www.eng.uwo.ca/people/esavory/es832.htm Lecture 6 – Basics of combustion Department of Mechanical and Material Engineering University of Western Ontario

2. Fundamental Concepts of Combustion Theory • The first step in any conversion system is to determine the amount of energy which will be available for conversion. For combustion processes, this information is important for determining the necessary mass rates and the temperatures of the products entering a turbine. • Recall that a turbine is normally designed for a specific point of operation. By far, combustion is still the most common form of energy generation.

3. Combustion Theory • Determine the products of combustion: e.g 2H2 + O2 → 2H2O • Determine the heat released by combustion and the temperature of the products. • Energy balance: internal energy and enthalpy Q = UP0 – UR0 = ∆U0 = ∆Ho if K.E. negligible. [∆U0 is the internal energy of combustion and is always –ve since heat is transferred to the surroundings] • Determine the energy available for conversion.

4. Combustion processes (1) • EXOTHERMIC (produces heat) reactions where mass is conserved. • Most combustion processes use oxygen as oxidant (to burn with). • The substances used before the reaction are known as REACTANTS and those arising from the process are referred to PRODUCTS. • The substance undergoing oxidation is known as a FUEL. • STOICHIOMETRIC mixtures are where the air mass available is exactly that required for combustion. • EXCESS air is the amount of air present above stoichiometric ratios.

5. Combustion processes (2) PRINCIPAL CONSERVATION LAWS: 1. The number of atoms is conserved. 2. Mass is conserved.

6. Relative atomic mass of some common elements Defined so that the element’s mass is scaled such that Carbon-12 is exactly 12.

7. Relative molecular mass of common substances = mass of one molecule of that substance, relative to 1/12th the mass of one atom of Carbon-12 Molar mass M ≈ relative molecular mass, numerically, but has units of kg / kmol

8. 1. Conservation of atoms • Substances consist of atoms in combinations called molecules. The atoms recombine during chemical reactions, but their number remains unchanged. • Example: Combustion of hydrogen: element Hydrogen (H): molecule H2 + element Oxygen (O): molecule O2 Water : molecule H2O Reactants Products nH2 + mO2 → H2O ; n, m = # of molecules

9. To balance the reaction the number of atoms remain the same (mathematically): Hydrogen 2 n = 2 → n = 1 Oxygen 2 m = 1 → m = ½ Hence, H2 + ½ O2 → H2O This could also be written 2H2 + O2 → 2H2O (or any multiple !!)

10. 2. Conservation of Mass • From (1), mass must also be conserved. • Each atom has a characteristic mass which is a property of the isotope. This is usually expressed in terms of the relative molecular mass (either g / mol or kg / kmol : where a mol is NA= 6.022 x 1023 units) For an ideal gas, 1 mol at STP occupies a volume of 23.6 litres. (STP = 15oC, 1 atm) e.g. Hydrogen: H MH ~ 1 1 kg / kmol Oxygen : O MO ~ 16 16 kg / kmol → H2 2 MH = MH2 ~ 2 kg / kmol → O2 2 MO = MO2 ~ 32 kg / kmol → H2O 2 MH2 + ½ MO2 = 18 kg / kmol Hence, 2H2 + O2 → 2 H2O It can be stated that 2 * (MH2) + MO2 → 2 MH2O 4 kg / kmol + 32 kg / kmol → 36 kg / kmol

11. Example • Calculate the mass of oxygen required to burn 1 m3 of fuel propane at 1 bar, 273 K. How much air must be supplied (at these conditions)? • Solution: (a) Need mO2required to burn 1 m3 C3H8 (propane) • To accomplish this task we must first determine the relative amount of reactants and products to burn the propane. This requires setting up the chemical reaction. • The next step is to determine the relative mass of each based on the ratios determined from the chemical reaction. • Then express everything in terms of 1m3 C3H8 (propane) at 1 bar, 273 K. This will require the use of the ideal gas law (or tabled density values) for these conditions. • Finally, since air is used, calculate the other components based on the composition of air. (b) Set-up the primary and secondary relationships.

12. Calculate the relative amounts first in terms of mol: We know the products will be CO2, H2O Note all products are gaseous. n C3H8 + m O2 → p CO2 + q H2O for C: 3n = p; H: 8n = 2q; O: 2m = 2p + q Solving in terms of q: q = 4n, p = 3n = ¾ q m = p + q/2 = 5/4 q n = q/4, m = 5/4 q, p = ¾ q Thus, for n = 1: C3H8 + 5O2 → 3CO2 + 4H2O

13. Calculate the relative mass ratios using the molar mass: Obtain molar mass for each species involved in the chemical reaction MO2 = 32 kg / kmol; MH2O = 18 kg / kmol; Mc = 12 kg / kmol; MH2 = 2 kg / kmol; MC3H8 = 3 MC + 4 MH2 = 44 kg / kmol Hence, for every mol of C3H8 we need 5 mol of O2 or mO2/mC3H8 = 5MO2/1MC3H8 = (5 x 32) kg/kmol / 44 kg/kmol = 3.636 kgO2 / kgC3H8 (from tables: rC3H8 = 2.02 kg/m3 , at 1 bar, 273 K)

14. [Note: from gas constant (R): RC3H8 = R / M = (8.314 J/mol-K) / (44 kg/kmol) = 189 J/kg-K (recall: R is universal gas constant)  = P / RT = (101,300 Pa) / (189 J/kg-K) * (273 K) = 1.97 kg/m3 which is very close to 2.02 kg/m3] mC3H8 =  Vol = 2.02 kg * 1 m3 = 2.02 kg mO2 = 3.636 kgO2/kgC3H8 * 2.02 kgC3H8 = 7.34 kg of oxygen

15. Calculate the total amount of air: The molar (i.e. volumetric) ratios of air composition are approximately: N2 = 79%; Argon (Ar) = 1% and O2 = 20%; For every 1 m3 of propane, we have (from the chemical equation of part (a)): 5m3 O2 + 5(0.79/0.20)m3 N2 + 5(0.01/0.20)m3 Ar = 25 m3 of air. What if the ideal gas equation had been used? We would then predict 7.16 kg of oxygen but the volume of air would remain the same.

16. SUMMARY: Combustion processes are exothermic reactions in which mass and energy are conserved. Molecular ratios are used to determine volumetric and mass ratios. Stoichiometric mixtures are those for which the amount of air is exactly that need for 100% combustion. Excess air is the amount of air above Stoichiometric conditions. Excess air is usually supplied in practice to ensure complete combustion. A deficiency of air when combusting hydrocarbons  Hydrogen is complete but C gives CO2+ CO.

17. Internal energy and enthalpy of combustion A heat conversion system is designed to produce a desired power output. Typically, the engineer is provided with the desired output levels, the turbine thermodynamic and mechanical efficiencies and the operating point (inlet temperature and pressure). The task of the engineer then becomes to determine the amount of energy generation and the mass flow rate of products needed to achieve these conditions. To do so requires that the engineer know • the amount of fuel required; • the amount air required to combust these fuels and the mass flow rate and temperature of the products of combustion.

18. Objective 1. Calculate the rate of energy which is liberated during a combustion process 2. Determine the minimum mass flow rate of air required for combustion 3. Determine the temperature and flow rate of the products.

19. The heat released, Q, during a chemical reaction is a property of the reactants and the process. For a closed system (no-flow) one obtains: Q = (UP2 - UP0) + (UR0 - UR1) + ∆U0 And for an open system (steady-flow): Q = (HP2 - HP0) + (HR0 - HR1) + ∆H0 + ∆K.E. [ Subscripts: P = product; R = reactant; 0, 1, 2 = temperatures T0, T1 and T2 ] (note sign convention: Q > 0 if supplied by surroundings, thus ∆U0 < 0; ∆H0 < 0 ) ∆U0 and ∆H0 are the internal energy and enthalpy of combustion, respectively. ∆K.E. is kinetic energy

20. The enthalpy of combustion is usually given in terms of kJ/kmol, ∆h0, such that: ∆H0 = n*∆h0 where n = kmol of fuel ∆h0: • Is a property of the reaction. It is given at a standard reference (25°C and 1atm) • It is assumed that reactants and products are at the same temperature and pressure • If not tabulated, it can be calculated from the energy of formation, hf, of the individual chemical species at 25°C and 1 atm. ∆h0 = ∑n*hf - ∑n*hf ; n = kmol of species/kmol fuel Products Reactants

21. NOTES: (1) It is important to know the phase (liquid or gas) for reactants, fuel and products since the enthalpy of vaporization, hfg, must be included if present. (2) By definition: ∆u0 = ∆h0 - (∑Pν - ∑Pν) Products Reactants v = specific molar volume, P = pressure For most practical situations: ∆u0 ≈ ∆h0 and so ∆U0 ≈ ∆H0

22. Typically, the reactants are supplied at a temperature T1, the combustion conditions are given at a temperature T0 and products are found at T2. Reactants U or h Products UR1 hR1 1 KE (open System) UR0 hR0 0 UP2 hP2 2 UP0 hP0 2 0 Temp. T T1 T2 T0

23. Rigorously, the enthalpy (or internal energy) at the states 0, 1 and 2 should be obtained from the tabulated thermodynamic values. However, for most engineering design purposes a good approximation is obtained from: or where: Cv or Cp are evaluated at (T0+T1)/2 for reactants and (T0+T2)/2 for products.

24. Example A stoichiometric mixture of air and gaseous methane at 54oC and 2 bar is buried in a 0.1 m3 rigid vessel. The temperature of the products is measured to be 1,529oC. Given that the internal energy of combustion ∆Uo = - 802,310 kJ/kmolat 25oC, calculate the amount of heat rejected to the environment. Hence, information given: To = 25oC = 298 K ∆Uo = -802,310 kJ/kmol T1 = 54oC = 327 K P1 = 2 bar V = 0.1 m3 T2 = 1,529oC = 1,802 K Air is provided: Use 79% N2 , 21% O2

25. Solution We want to find Q = Uo +  (U0 – U1) +  (U2 – U0) To do this we need the mass m of each component and Cv for each at the average temperature (from tables). For m we need to find the total number of moles, N: Recall: n = mol/m3 = P1 / R T1 = 2 bar / [(8.314 J/mol-k) * (327 K)] = 2 x 105 / (8.314 * 327) = 73.57 mol/m3 Hence, N = 0.1 * 73.57 = 7.357 mol R P

26. This consists of CH4 (methane), O2, N2 and so we now need to know how much O2 we have (for stoichiometric conditions): CH4 + 2 O2 → CO2 + 2 H2O For 1 mol of CH4 we have 2 mol of O2 and also 0.79/0.21 * 2 of N2 (because air is provided) NCH4 * (1 + 2 + 0.79/0.21 * 2) = 7.357 mol Hence, NCH4 = 0.699 mol So, NO2 = 2 NCH4 = 1.398 mol NN2 = 5.529 mol NH20 = 1.398 mol NCO2 = 0.699 mol

27. The heat given off by the reaction at 25oC is Uo = - 802,310 kJ/kmol * 0.699 mol * 10-3 kmol/mol = - 560.8 kJ Next we need to determine the internal energy change required to bring the reactants to the “tabulated” reaction temperature and then the products back up to the measured exit temperature. From tables we find

28. Hence, for the reactants (R) and products (P):  (U0 – U1) = NCH4 * CvCH4 * (T0 – T1) + N02 * Cv02 * (T0 – T1) + NN2 * CvN2 * (T0 – T1) = [ NCH4CvCH4 + N02Cv02 + NN2CvN2 ] * (T0 – T1) = [0.699 * 27.8 + 1.398 * 21.1 + 5.529 * 20.8] * (25 – 54) * 10-3 = - 4.75 kJ  (U2 – U0) = NC02 * CvC02 * (T2 – T0) + NH20 * CvH20 * (T2 – T0) + NN2 * CvN2 * (T2 – T0) = [ NC02CvC02 + NH20CvH20 + NN2CvN2 ] * (T0 – T0) = [0.699 * 46.6 + 1.398 * 33.5 + 5.529 * 24.6] * (1529 – 25) * 10-3 = 324.3 kJ R P

29. So the overall balance is Q = Uo +  (U0 – U1) +  (U2 – U0) = -560.8 kJ – 4.75 kJ + 324.3 kJ  - 241 kJ R P

30. Summary • The total energy liberated during a chemical reaction is based on a 1st Law of Thermodynamics balance (for either a closed or an open system). • The heat released during the reaction, ∆H0, is a property of the reaction. • ∆H0 is generally tabulated for a specific condition. Equations of state and the 1st Law of thermodynamics can be used to calculate the heat release to the actual state. It can be calculated from the energy of formation of the chemical species.