CBE 450 Chemical Reactor Fundamentals Fall, 2009 Homework Assignment #1

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CBE 45 Chemical Reactor Fundamentals Fall, 9 omework Assignment #1 1. Review of Mass Balances of Reacting Systems atural gas (assume pure methane) is being burned in air (assume.79 mol% and.

he flow rate of natural gas is 1 mol/s. he furnace operates at steady state. (a) Draw a diagram with all input and output streams. Indicate known and unknown flowrates.

1 CBE 45 Chemical Reactor Fundamentals Fall, 9 omework Assignment #1 1. Review of Mass Balances of Reacting Systems atural gas (assume pure methane) is being burned in air (assume.79 mol% and.1 mol% O ). Assume complete combustion, which means that all of the natural gas is burned and is converted into carbon dioxide and water. Assume that the oxygen is provided in 1% excess. he flow rate of natural gas is 1 mol/s. he furnace operates at steady state. (a) Draw a diagram with all input and output streams. Indicate known and unknown flowrates. Indicate known and unknown compositions. (b) Determine the flowrate of air. (c) Determine the flowrate and composition of the output gas using atomic balances. (d) Determine the flowrate and composition of the output gas using molecular balances. Solution (a) Draw a diagram with all input and output streams. Indicate known and unknown flowrates. Indicate known and unknown compositions. (b) Determine the flowrate of air. he balanced reaction is C 4 + O CO + O

2 he stoichiometrically required amount of O is two moles of O for every mole of C 4. Since the oxygen is provided in 1% excess, you actually need twice as much, or four moles of O for every mole of C 4. Since you have 1 mol/s of C 4, you need 4 mol/s of O. o determine the total air flowrate,.1 4mol s Ax O A / 4 A mol / s 195mol / s.1 (c) Determine the flowrate and composition of the output gas using atomic balances. We have five unknowns, the flowrate and four compositions of the stack gas. heore, we need five equations. he first four equations are atom balances of the form: accumulation = in out + generation he accumulation term is zero because the furnace is operating at steady state. he generation term is zero because atoms are neither created or destroyed in this furnace. heore, the atom balance is in = out : Ax Sz C: FyC 4 SzCO : 4FyC 4 Sz O O: AxO SzO SzCO Sz O he fifth equation is the sum of the mole fractions is unity ow solve. sum of the mole fractions is unity: 1 zo zco z O z in = out : Sz Ax 155 C: Sz Fy 1 CO C 4

3 : Sz Fy O C O: Sz O AxO SzCO Sz O 4 1 We can multiply the mole fractions sum by S and write S SzO SzCO Sz O Sz mol / s he compositions are thus : z. 755 S C: z CO. 499 S 5 : z O. 998 S 5 O: z O. 998 S 5 (d) Determine the flowrate and composition of the output gas using molecular balances. We again have five unknowns, the flowrate and four compositions of the stack gas. heore, we need five equations. he first four equations are molar balances of the form: accumulation = in out + generation he accumulation term is zero because the furnace is operating at steady state. he generation term is not zero because molecules are created or destroyed in this furnace. heore, the molar balances are = in out + generation he generation term is the number of moles created or destroyed. Since the stoichiometric coefficient of methane in the reaction is one, we can use that as our basis. C 4 generation 1mol / s he generation is negative since C 4 is consumed in the reaction. he other generation terms are reached my taking the magnitude of the C 4 generation term and multiplying it by the stoichiometric coefficient for each component.

4 : Ax Sz CO : Sz (1) 1 CO O: Sz () 1 O : Ax Sz ( ) 1 O O O C 4 : Fy 4 ( 1) 1 C We see that the methane balance was actually used to determine the generation term. he fifth equation is the sum of the mole fractions is unity ow solve. 1 z sum of the mole fractions is unity: O zco z O z : Sz Ax 155 CO : Sz CO 1 O: Sz O O : Sz Ax O O We can multiply the mole fractions sum by S and write S SzO SzCO Sz O Sz mol / s he compositions are thus : C: S 5 z 1 1 S 5 z CO

5 : O: z S 5 O S 5 z O his is the same result as was obtained from the atomic balances, as it must be.. Molecular Description of Reaction Equilibrium Consider the distribution of para-xylene, meta-xylene and ortho-xylene. (a) For an ideal gas at a given temperature, what factors at the molecular level determine the equilibrium distribution of components? (b) At a given temperature, will the mole faction of the component with the lowest enthalpy be (c) At a given temperature, will the mole faction of the component with the lowest entropy be (d) At a given temperature, will the mole faction of the component with the lowest free energy be (e) As the temperature is increased, will the mole fraction of the component with the lowest enthalpy increase or decrease (all other things being equal)? (f) As the temperature is increased, will the mole fraction of the component with the lowest entropy increase or decrease (all other things being equal)? Solution: (a) For an ideal gas at a given temperature, what factors at the molecular level determine the equilibrium distribution of components? At the molecular level, the enthalpy and the entropy can be broken into contributions from translation of the center of the mass of the molecule rotation about the center of mass of the molecule vibration of bonds within the molecule internal rotation of internal rotors like the methyl groups in xylene he total enthalpy and entropy (and thus the free energy) are related to these contribution. hus the equilibrium distribution is based on these contributions. (b) At a given temperature, will the mole faction of the component with the lowest enthalpy be 1 G1 1 S1 S1 1 Keq,1 exp exp exp exp R R R R

6 he enthalpy difference between state 1 and is 1 f,1 f, If state one has the lowest enthalpy, then 1 If all other things are equal (meaning S ), then 1 R 1 and exp 1 1so 1 R A low enthalpy state is favored at equilibrium. (c) At a given temperature, will the mole faction of the component with the lowest entropy be he entropy difference between state 1 and is S1 S f,1 S f, If state one has the lowest entropy, then S 1 If all other things are equal (meaning ), then 1 S1 R S and exp 1 1so 1 R A high entropy state is favored at equilibrium. (d) At a given temperature, will the mole faction of the component with the lowest free energy be he free energy difference between state 1 and is G1 G f,1 G f, If state one has the lowest free energy, then G then 1 G R 1 G and exp 1 1so 1 R

7 A low free energy state is favored at equilibrium. (e) As the temperature is increased, will the mole fraction of the component with the lowest enthalpy increase or decrease (all other things being equal)? If state one has the lowest enthalpy, then 1 and S 1, then the derivative of this distribution with respect to temperature is S1 1 1 exp exp R R R 1 where we neglected the temperature dependence of the enthalpies of formation. So as the temperature increases the relative amount of 1 decreases. In other words, the low energy state is more favored at low temperature. (f) As the temperature is increased, will the mole fraction of the component with the lowest entropy increase or decrease (all other things being equal)? If state one has the lowest entropy, then S 1 and 1, then the derivative of this distribution with respect to temperature is S1 1 1 exp exp R R R 1 where we neglected the temperature dependence of the enthalpies of formation. So as the temperature increases the relative amount of 1 remains unchanged. In other words, the entropic contribution to the distribution of components does not have a temperature dependence. 3. Review Continuum Description of Reaction Equilibrium Consider a batch reactor initially containing nitrogen and hydrogen gases. he volume of the batch reactor is 1 m 3. he initial pressure is 1 atm. he temperature is kept constant at 5 K. he initial mole fractions are.5 and.75. he relevant reaction is 3 3 he erence enthalpies of formation are at a erence temperature of K f, kcal / mol f, kcal / mol f, kcal / mol he erence Gibbs free energies of formation are at a erence temperature of K

8 G f, kcal / mol G f, kcal / mol G f, kcal / mol In this proect, you may assume that the heat capacities are constant, given by p, 7.3 cal / mol / K 6.9 cal / mol K C C p, / C p, cal / mol / K (a) What are the heats of formation at 5 K for each component? C p, i For constant heat capacities d C p, i kcal kcal 1 5K kcal mol f, / 5K mol f, / 5K mol f, / (b) What are the free energies of formation at 5 K for each component. (You may assume that the heat of formation is constant in this part (b) calculation only.) G G d For constant heats of formation, G 1 G 1

9 () 5(1.4).95 kcal mol ( 3.93) 5( 9.8).394 kcal () 5(1.4).96 kcal mol G f, / G f, / G f, 3 / (c) Compute the heat of reaction and free energy of reaction at 5 K on a per mole basis of 3 produced. 1 3 r f, 3 f, f, kcal / mol 1 3 Gr G f, 3 G f, G f, 1.51 kcal / mol (d) Compute the equilibrium coefficient of the reaction at 5 K. mol K eq Gr exp R exp (5) (e) Compute the equilibrium mole fractions of each component at 5 K. For an ideal gas, we can write the equilibrium coefficient in terms of the mole fractions. p Keq x p We can write the mole fractions in terms of the initial number of moles, the moles reacted,, and the total number of moles,, o, x he initial number of moles is pv (1135)(1) o, xi xi xi 4. 4moles R (8.314)(5) x 4.4moles 6.9 o, i o, xi 4.4moles 18.3 o, 3 xi 4.4moles he total number of moles is given by o,

10 o i 4. 4 i he number of each species is o, o o 3 o, 3 3, , he erence pressure is 1135 Pa. he final pressure is (using the ideal gas law) p R V o i R i V Substituting these into equations the equilibrium coefficient 1 p x 3 x 3 Keq 3/ 1/ 3/ 1/ p x x x x K eq p p / / p p / / Solving this single, non-linear algebraic equation, I find 1.37 and mole fractions x x x o,

11 x

Module 5: Combustion Technology. Lecture 32: Fundamentals of thermochemistry

1 P age Module 5: Combustion Technology Lecture 32: Fundamentals of thermochemistry 2 P age Keywords : Heat of formation, enthalpy change, stoichiometric coefficients, exothermic reaction. Thermochemistry