Lecture 6 Defining the equilibrium by minimizing the Gibbs energy Acid dew point temperature of flue gases

Calculating the chemical equilibrium using the condition G = min In lecture 5, example 2 showed how to solve the chemical equilibrium using the equilibrium constant K. Let’s consider the same reaction ( T =1800K and p = 10bar). H2(g) +1/2O2(g) = H2O(g) In the beginning, nH2 = 1mol and nO2 = 0.5mol What is the equilibrium concentration for this reaction when the condition G = min is used? In other words, the Gibbs energy of the system is minimized. We can write at the equilibrium H2(g) +1/2O2(g) = H2O(g) nH2 nO2 nH2O According to Euler’s homogenous theory the Gibbs energy for the system is

Calculating the chemical equilibrium using the condition G = min To minimize the function we must define the objective function as a function of nH2 , nO2 and nH2O and constraints. This is a traditional optimization problem Objective function = kohdefunktio Constraints = reunaehdot

Calculating the chemical equilibrium using the condition G = min, chemical potential i of a species Chemical potential i for a component i at the temperature of T and pressure of p where xi is the mole fraction of a component i xi = ni/ntot and oi(T) can be claculated using the tabulated values for enthalpies and entropy (see Janaf)

Calculating the chemical equilibrium using the condition G = min, the objective function Now, the Gibbs energy (the objective function) for the reaction may be written as follows

Calculating the chemical equilibrium using the condition G = min, constraints In chemical reactions, the number of atoms remains the same. Constraints are defined by creating an atom matrix as follows: Species  H O H2 2 O2 H2O 1 => the number of hydrogen atoms is 21mol = 2mol => the number of oxygen atoms is = 20.5mol = 1mol On the basis of atom matrix and nH2 (in) and nO2 (in) , the constraints become The number of hydrogen atoms The number of oxygen atoms

Calculating the chemical equilibrium using the condition G = min, problem definition and results The final problem may be formulated as follows Constraints Results Species  n, mol x H2O 0,999167 0,99875 H2 0,000833 0,00083 O2 0,000416 0,00042 Total 1,00042 1,00000 The number of hydrogen atoms The number of oxygen atoms There are different mathematical methods to minimize the G function (Gibbs energy), such as the Lagrange Method.

Example: Combustion Calculate the equilibrium concentration for the combustion of CH4 (input flow 1mol/s). Input flows of O2 and N2 are 2.4 and 9.020 mol/s, respectively. Combustion Chamber t = 1350oC p = 1bar  = 1.2 Is an excess air factor nN2 = 3.77nO2 (0.79/0.21*nO2)

Example: Combustion Atom matrix Amounts of atoms Species C H O N CH4 1 O2 2 N2 CO2 H2O CO OH H2

Example: Combustion Formulation of the minimization problem For solid and liquid species the chemical potential would be written as follows:

Example: Combustion Results Chamber t = 1350oC p = 1bar  = 1.2 The following combustion reaction approximates the reality extremely well CH4 + 2.4 O2 + 9.02 N2  CO2 + 2H2O + 0.4O2 + 9.02N2

Calculating the acid dew point temperature of flue gases Some fuels, like coal and oil, usually contain small amounts of Sulphur (S). In a combustion, Sulphur converts to S(s) + O2(g)  SO2(g) and further SO2(g) + 1/2O2(g)  SO3(g) Flue gases also contain H2O(g), and below a certain flue gas temperature T, H2O(g) + SO3(g)  H2SO4 (l) Temperature T is called the acid dew point temperature of flue gases. Below the acid dew point temperature, SO3(g) and H2O(g) form H2SO4 (l) which begins to condense in the boiler. Condensing is not desirable due to highly corrosive properties of H2SO4 (l) => Tout, flue gas > Tacid dew point

Calculation of acid dew point temperature of flue gases At the acid dew point, the equilibrium is valid for the reaction H2O(g) + SO3(g)  H2SO4 (l) ) => H2O(T,p,xH2O) + SO3(T,p,xSO3) = H2SO4(T,p), In several combustion calculatuions, the ideal gas equation of state approximates flues gases accurately enough => the equilibrium can be written Knowing mole fractions xi of each component in flue gases and the total pressure of p the temperature of the acid dew point can be calculated from the equilibrium equation

Example: Acid dew point temperature Flue gas, ptot 1bar Coal xH2O 11.2 % (mole fraction) Boiler Air xSO3 210-3 % CO2, CH4, CO, O2, N2, NOx What is the acid dew point temperature of flue gases?

Example: Acid dew point temperature We can write at the acid dew point temperature The goal is to find a temperature, where this condition is valid. Possible ways to solve the problem 1) The temperature is defined using tabulated values and a graphical solution 2) The temperature is defined by inserting chemical potentials as a function of T in Excel and using a solver tool.

Example: Acid dew point temperature For H2O(g) and SO3(g), the chemical potential is where pH2O = 0.1121bar = 0.112 bar and pSO3 = 210-5 1bar = 210-5 bar For H2SO4(l) the chemical potential is calculated as folllows The pressure term can be ignored, because ptot = po = 1bar

Example: Acid dew point temperature Let’s use tabulated values for each component (subscripts A = H2O, B = SO3, H2SO4 = C) H2O(g)

Example: Acid dew point temperature SO3(g)

Example: Acid dew point temperature H2SO4(l) All values for H2O, SO3 and H2SO4 have been taken from tables

Example: Acid dew point temperature The acid dew point temperature is found graphically by drawing A + B = C as a function of temperature A = H2O B = SO3 C = H2SO4 T = 447K = 174oC If T > 174oC no condensing takes place If T < 174oC condensing begins Note: Go(T) = C – (A + B) < 0 when T 447K => reaction takes place Go(T) = C – (A + B) > 0 when T  447K => no reaction takes place The condition for a chemical reaction is G(B) - G(A) < 0

Acid dew point as a function of SO3 and H2O

Phase rule Three main phases Solid (s) Liquid (l) Gas (g) In several systems, three main phases may be divided into sub-phases such as Solids Crystal 1 Crystal 2 Liquid mixture Water Oil Reverse osmosis where the membrane separates liquids Pure water Salt water

Result of the phase rule Phase rule is proposed by Gibbs and is defined as: M = n – f +2 M = freedom of degrees n = number of components f = number of phases in chemical equilibrium Freedom of degrees refers to the number of independent variables such as temperature and pressure. Let’s assume that n = 1 and M = 1, for example the pressure is fixed but the temperature is an independent variable. The number of phases in chemical equilibrium f = n - M + 2 = 1-1+2 = 2 For example, liquid+vapor or solid+liquid

Phase rule at the triple point of water At the triple point of water (T = 273.16K and p = 611.2 Pa), all three phases are at equilibrium. Is this possible according to the phase rule? Phase rule M = n – f +2 , where n = 1 (water) M = 0, because both the temperature (273.16K) and the pressure (611.2Pa) are fixed. => f = n - M +2 = 1-0+2 = 3 => the phase rule is true.