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) +ve means requires energy to form, -ve means gives off energy when forms 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 8 = 5 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 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 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.

Start with the energy balance equation: Q =  H 0 +  Cp (T 2 – T 0 ) -  Cp (T 1 – T 0 ) +  K.E. where Q = 0.03  H 0 (3% heat loss)  H 0 = ?  h’ 0 = MJ/kmol f (enth. of comb of fuel) Finding  H 0 (assuming 79% N 2 and 21% O 2 ): We have excess air of 20%. Ideal reaction requires 5O 2 and so 20% excess air means and extra 1O 2 Ideal reaction is: C 3 H 8 + 5O 2  3CO 2 + 4H excess air of 1 O * (0.79/0.21) N 2 = O N 2 P R

Actual reaction gives 98% CO 2 yield so: 3 * 0.98 = 2.94 CO 2 with the rest given off as CO. Hence, actual reaction is C 3 H O N2  2.94 CO CO + 4 H O N 2 Computing overall enthalpy of combustion: For C 3 H 8 + 5O 2  3CO 2 + 4H 2 0  h’ 0 = MJ/kmol f For CO 2  CO + ½ O 2  h’’ 0 = 283 MJ/kmol CO (from tables) So  h 0 =  h’ 0 – 0.02 * ( 3  h’’ 0 ) Since 1 -  CO = 0.02 and there are 3 CO 2 / mol f   h 0 = – 0.06 * 283 = MJ/kmol f

We could also write this as:  h 0 = 0.98  h’ (  h’ *  h’’ 0 ) (again because process is 98% efficient and there are 3 CO 2  3 CO + 3/2 O 2 )  h 0 = 0.98 * * ( * 283 ) = MJ / kmol f

Calculate all the masses and the heat rejection given that C 3 H O N2  2.94 CO CO + 4 H O N 2 Since and M f = 44 kg/kmol (3 M C + 8 M H )

Check:  H 0 =  h 0 = 2.27x10 -3 kmol/s * MJ/kmol f = MW So heat loss to surroundings is Q = 0.03  H 0 = MW Computing: where subscripts i = in and o = out, noting that the fuel (0.1 kg/s) is injected into the combustion process so that the mass flow rate of air is 1.87 kg/s.  K.E. = MW – MW = MW We know the inlet conditions are T 1 = 25 o C and T 0 = 25 o C and so  Cp (T 1 – T 0 ) = 0

So, we need to solve or, rearranging in terms of T 2 We don’t know the outlet temperature T 2 and so we need an iteration algorithm to find it: (1) Since most of the mass flow is nitrogen we can use the value of Cp of nitrogen (at T = 25 o C which changes little with T) to get a first guess for T 2.

(2) Calculate T 2 from 1 st Law relationship (3) Use this T 2 to calculate Cp at (T 2 +T 0 )/2 for all products. Call this T 2, (T 2 ) old. (4) Calculate  Cp (5) Find new T 2 = (T 2 ) new (6) If (T 2 ) new - (T 2 ) old < tolerance, the iteration is complete. Otherwise use (T 2 ) new as the new guess and start again from step 2. Tolerance depends on the application but 50 o C is a reasonable rule-of-thumb. So first iteration is: SpeciesMass flow (kg/s)Cp (kJ/kgK) N O CO CO H 2 O

 Cp = kJ/kgK Giving T 2 = 25 + (4.374 MW / kJ/kgK) = 2084 o C Now we need to compute Cp values at (T 2 +T 0 )/2 = 1054 o C  Cp = kJ/kgK Giving T 2 = 25 + (4.374 MW / kJ/kgK) = 1757 o C etc! SpeciesMass flow (kg/s)Cp (kJ/kgK) N O CO CO H 2 O

Example Butane (C 4 H 10 ) is burned with dry “theoretical” air (21%O 2, 79%N 2 ) at an air-fuel ratio of 20. Calculate the percentage of excess air and the volume percentage of CO 2 in the products. Take the molecular weight of air as 29 kg/kmol. What is the reaction equation for theoretical air ? C 4 H (O N 2 )  4 CO H N 2 What is the air-fuel ratio AF th for theoretical air ? AF th = m air = 6.5 * 4.76 * 29 = kg air m fuel 1 * 58 kg fuel

This represents 100% theoretical air. The actual air-fuel ratio is 20 and so what is the percentage of excess air ? % excess air = ( AF act – AF th ) x 100% AF th = (20 – 15.47) x 100 = 29.28% So, actual reaction with % theoretical air is: C 4 H * * (O N 2 )  4 CO H O N 2 Hence, volume percentage of CO 2 using total moles in products of combustion is: %CO 2 = ( 4 / 42.5 ) * 100 = 9.41%

Example Butane (C 4 H 10 ) is burned with 90% dry “theoretical” air (21%O 2, 79%N 2 ). Calculate the volume percentage of CO in the products and the air-fuel ratio. Take the molecular weight of air as 29 kg/kmol. Assume no hydrocarbons in the products. What is the reaction equation for this incomplete combustion ? C 4 H * 6.5 (O N 2 )  a CO H N 2 + b CO 4 = a + b and 11.7 = 2a + 5+ b  a = 2.7, b = 1.3 C 4 H * 6.5 (O N 2 )  2.7 CO H N CO

What is the volume percentage of CO ? %CO = ( 1.3 / 31 ) * 100 = 4.19% So, the air fuel ratio is: AF = m air = 0.9 * 6.5 * 4.76 * 29 = kg air m fuel 1 * 58 kg fuel