ENERGY CONVERSION ES 832a Eric Savory Lecture 7 – Energy of formation and application of the first law Department of Mechanical and Material Engineering University of Western Ontario

Energy of formation and application of the first law Objectives: 1. Energy of formation for calculation of the enthalpy of combustion 2. Practical application of the 1 st law of thermodynamics on a system

(1) Calculation of enthalpy of combustion ∆h 0 from energy (heat) of formation ∆h 0 = ∑n i *h fi - ∑n i *h fi [KJ/kmol f ] Products Reactants where i – denotes the species n i – # mol of species per mol of fuel h fi – energy of formation of species at 25°C,1 atm. (for stable species at 25°C,1atm:h f ≡ 0) Note: always identify the phase of the reactants/products (enthalpy of vaporization, h fg = 0 for naturally occurring gases). Since ∆h 0 is a state relation, it can be calculated using any path of chemical reactions. This allows an easy method for calculating ∆h 0 for imperfect (i.e. non-ideal) reactions.

Example: Combustion of propane (gaseous) C 3 H O 2  3 CO H 2 0 (derived in earlier lecture) For gaseous reactants (with the calculation carried out per kmol f ): For gaseous products (at 25 o C, 1 atm):

For liquid products (at 25 o C, 1 atm):

(2) Application of 1 st law of thermodynamics The first law is used to calculate (estimate) the theoretical yield (energy available for work) of a combustion process. Ideal vs. Real Combustion (i) Combustion efficiency is defined as η co = total # of mol of C converted to CO 2 # of mol of C available in fuel (ii) Efficiency depends on P, T and concentration of species

(iii) For open systems, efficiencies of 90% are typical for stoichiometric conditions. It is, thus, typical to use excess air. Depending on the fuel, excess air of 15% to 30% will yield 98% to 99.5% combustion. (iv) Inefficiencies mainly occur since only part of C is converted to CO 2 (C + ½ O 2 → CO for the rest) (v) Secondary reactions N 2 + XO 2 → NO X (e.g. nitric oxide, NO; nitrogen dioxide, NO 2 ) occur when the temperature in the combustion chamber is too high. Other reactions occur due to contaminants (e.g. S). These are generally a problem with liquid and solid fuels.

Example for you to try before next class ! A combustion process with propane fuel is said to be 98% efficient. Determine the composition of the product gases and exit temperature if the combustor loses 3% of the heat of the combustion to the environment. The following technical data are available: m f = 0.1 kg/s (at 25°C, 1 atm, gaseous) Excess air: 20% Inlet velocity = 30 m/s Outlet velocity = 300 m/s Inlet condition of the reactants 25°C, 1 atm Outlet pressure: 1 atm Assume all products are gaseous

Solution strategy outline: 1. Set up the energy balance (enthalpy) equation 2. Assume air is 79% N 2 and 21% O 2 3. Write out and compute ideal chemical reaction equation 4. Modify this for the fact that the process is 98% efficient 5. Use ideal reaction value for the enthalpy of formation (call it  h’ 0 for the fuel), from earlier in these notes, and  h’’ 0 for CO (from tables), to find the overall value  h 0 (hence, H 0 ) 6. Calculate all the masses (for the reactants and products) 7. Check that mass in = mass out ! [should be 1.97 kg/s] 8. Calculate  H 0, Q and  K.E. for the energy equation 9. Using known inlet conditions T 1 = T 0 = 25 o C use the energy equation to iterate for T 2. Since most mass flow is nitrogen use the Cp of nitrogen at 25 o C as first guess for T Use this T 2 to calculate Cp at (T 2 +T 0 )/2 for all products and calculate the  m Cp and keep iterating to a solution. Find the resources you need to work on this. Don’t worry if you can’t solve it as we will go through it in class – but give it a try. Two iterations gives T 2 = 1757 O C