Energy of Formation & First Law Applications

1. ENERGY CONVERSION ES 832a Eric Savory www.eng.uwo.ca/people/esavory/es832.htm Lecture 7 – Energy of formation and application of the first law Department of Mechanical and Material Engineering University of Western Ontario

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

3. (1) Calculation of enthalpy of combustion ∆h0 from energy (heat) of formation ∆h0 = ∑ni*hfi - ∑ni*hfi [KJ/kmolf] Products Reactants where i – denotes the species ni – # mol of species per mol of fuel hfi – energy of formation of species at 25°C,1 atm. (for stable species at 25°C,1atm: hf ≡ 0) Note: always identify the phase of the reactants/products (enthalpy of vaporization, hfg = 0 for naturally occurring gases). Since ∆h0 is a state relation, it can be calculated using any path of chemical reactions. This allows an easy method for calculating ∆h0 for imperfect (i.e. non-ideal) reactions.

4. Example: Combustion of propane (gaseous) C3H8 + 5 O2 3 CO2 + 4 H20 (derived in earlier lecture) For gaseousreactants (with the calculation carried out per kmolf): For gaseous products (at 25oC, 1 atm):

5. For liquid products (at 25oC, 1 atm):

6. (2) Application of 1st 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 ηco= total # of mol of C converted to CO2 # of mol of C available in fuel (ii) Efficiency depends on P, T and concentration of species

7. (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 CO2 (C + ½ O2 → CO for the rest) (v) Secondary reactions N2 + XO2 → NOX (e.g. nitric oxide, NO; nitrogen dioxide, NO2) 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.

8. 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: mf = 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

9. Solution strategy outline: 1. Set up the energy balance (enthalpy) equation 2. Assume air is 79% N2 and 21% O2 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 h0 (hence, H0) 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 H0, Q and K.E. for the energy equation 9. Using known inlet conditions T1 = T0 = 25oC use the energy equation to iterate for T2. Since most mass flow is nitrogen use the Cp of nitrogen at 25oC as first guess for T2 10. Use this T2 to calculate Cp at (T2+T0)/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 T2 = 1757OC