Example from Lecture 6 A stoichiometric mixture of air and gaseous methane at 54 o C and 2 bar is buried in a 0.1 m 3 rigid vessel. The temperature of the products is measured to be 1,529 o C. Given that the internal energy of combustion ∆U o = - 802,310 kJ/kmol at 25 o C, calculate the amount of heat rejected to the environment. Hence, information given: T o = 25 o C = 298 K ∆U o = -802,310 kJ/kmol T 1 = 54 o C = 327 K P 1 = 2 bar V = 0.1 m 3 T 2 = 1,529 o C = 1,802 K Air is provided: Use 79% N 2, 21% O 2

Solution We want to find Q =  U o +  (U 0 – U 1 ) +  (U 2 – U 0 ) To do this we need the mass m of each component and C v for each at the average temperature (from tables). For m we need to find the total number of moles, N: Recall: n = mol/m 3 = P 1 / R T 1 = 2 bar / [(8.314 J/mol-k) * (327 K)] = 2 x 10 5 / (8.314 * 327) = mol/m 3 Hence, N = 0.1 * = mol RP

This consists of CH 4 (methane), O 2, N 2 and so we now need to know how much O 2 we have (for stoichiometric conditions). The reaction is: ? For 1 mol of CH 4 we have 2 mol of O 2 and also 0.79/0.21 * 2 of N 2 (because air is provided). So N CH4 * ( ? ) = ? mol Hence, N CH4 = mol So, N O2 = 2 N CH4 = mol N N2 = mol N H20 = mol N CO2 = mol

The heat given off by the reaction at 25 o C is  U o = - 802,310 kJ/kmol * mol * kmol/mol = 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 Units: kJ/kmol-KCH 4 O2O2 N2N2 Reactants Cv [at (25+54)/2] CO 2 H2OH2ON2N2 Products Cv [at ( )/2]

Hence, for the reactants (R) and products (P):  (U 0 – U 1 ) = N CH4 * Cv CH4 * (T 0 – T 1 ) + N 02 * Cv 02 * (T 0 – T 1 ) + N N2 * Cv N2 * (T 0 – T 1 ) = [ N CH4 Cv CH4 + N 02 Cv 02 + N N2 Cv N2 ] * (T 0 – T 1 ) = [0.699 * * * 20.8] * (25 – 54) * = kJ  (U 2 – U 0 ) = N C02 * Cv C02 * (T 2 – T 0 ) + N H20 * Cv H20 * (T 2 – T 0 ) + N N2 * Cv N2 * (T 2 – T 0 ) = [ N C02 Cv C02 + N H20 Cv H20 + N N2 Cv N2 ] * (T 0 – T 0 ) = [0.699 * * * 24.6] * (1529 – 25) * = kJ R P

So the overall balance is Q =  U o +  (U 0 – U 1 ) +  (U 2 – U 0 ) = kJ – 4.75 kJ kJ  kJ RP

Example from Lecture 7 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 ? 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 the reaction equation is: ? 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 (T ave = 25 o C) 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 from Lecture 7 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 ? ? What is the air-fuel ratio AF th for theoretical air ? AF th = m air = ? = kg air m fuel 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: ? Hence, volume percentage of CO 2 using total moles in products of combustion is: %CO 2 = ( 4 / 42.5 ) * 100 = 9.41%