MATHCAD experts help please this is the temperature that result when a combustib
ID: 2858451 • Letter: M
Question
MATHCAD experts help please
this is the temperature that result when a combustible material is reacted usually with oxygen or air under the following conditions
find the adiabatic temperature of methane using bracket and bisecting method USING MATHCAD 15 OR PRIME and by doing energy balances.
The reaction is carried out without heat exchange with the environment-adiabatic conditions. . All of the combustible material and the oxygen are consumed. As a result, the adiabatic flame temperature To represents the upper temperature limit of a combustion process. Using the following data, determine the To of methane (CH4) burned in air where the heat capacity units are in J/gmol °C. The chemical reaction for this process is given by CH4+2O2 CO2+H2O, which has a heat of reaction of reaction at standard conditions of Hrxn-890.4 kJ/mol. The initial conditions of the methane and oxygen are 25°Cand 1 atm pressure. The heat capacity equation is given by with the values of the constants given in the Table below. co2 | 36.11 | 4.233 x 10-2 1-2887 x 10-5 | 7.464 x 10-9 H201 33.46 | 0.688 x 10-2 | 0.7604 x 10-5 1-3.593x 10-9 N2 | 29.00 | 0.2 199 x 10-2 | 0.5723 x 10-5 |-2.871 x 10-9 Table 1: Heat capacity constants.Explanation / Answer
Solution: I hope it will be helpful to you
The combustion of methane proceeds according to the balanced reaction
CH4+3O2?CO2+2H2O
with a heat of combustion of -890.4 kJ/gmol. For this calculation we choose a basis of 1 gmol ofCH4 because no amount of reactant or product was specified. Next, we apply material balances. Because all of the CH4 is reacted and no excess O2 is used, the mole balances are relatively simple. From the reaction equation, one mole of CO2 and two moles of H2O are formed from the complete combustion of 1 gmol of CH4 and the N2 in the air used leaves in the product gas. Because 3 moles of O2 are required to consume one mole of CH4 and air is approximately 79 mol% N2, 11.2857 moles of N2 (i.e., (3/0.21) 0.79) leave in the product gas.
In [1]:
Now consider the energy balance for this system. There is no external work being done on the system, and because it is adiabatic there is no external heat being produced or added to the system. The reactants enter at 25 C and 1 atm, which we use as the reference state for computing changes in specific energy. The system is operated at constant pressure, so the energy balance can be written as an enthalpy balance
0=?HRxn+?HProd(T)
where the heat liberated by reaction goes into raising the temperature of the product gases. The adiabatic flame temperature is found by solving the enthalpy balance for T.
The basis is 1 gmol of CH4 and all of it reacts, the heat of reaction is simply -890,400 J.
In [2]:
The enthalpy change of the product gases is given by
?HProd(T)=?T25(nCO2Cp,CO2(g)(T)+nH2OCp,H2O(g)(T)+nN2Cp,N2(g)(T))dT
For computational purposes, the temperature dependent heat capacities are represented by anonymous functions. ?HProd(T) is then a function which integrates the heat capacities for a given temperature.
In [3]:
The adiabatic flame temperature is found by finding the temperature for which the right-hand side of the enthalpy balance has a value of zero.
In [4]:
In [5]:
The adiabatic flame temperature is found by using a root-finding algorithm to find a root to the enthalpy balance equation.
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