Equations in flametemp()

date: 2025-02-24

flametemp

Equations

terms = comp(:, 1) + comp(:, 2)/4 - comp(:, 3)/2;

Pv = psat*phi;
nH2Or = (Pv/Pt) * n_air/(1-(Pv/Pt));

cengel pg 727-728, eq 14-9, 14-10

nO2r = n_air;
nN2r = 3.76 * n_air;

%    nCO2       nCO         nH2O        nH2     nO2
        A = [1,         1,          0,          0,      0; % C
             0,         0,          2,          2,      0; % H
             2,         1,          1,          0,      2; % O
             1-xCO2,-xCO2,          0,          0,      0; % CO2 conversion
             0,         0,     1-xH2O,      -xH2O,      0]; % H2O conversion
        B = [nC;        nH;        nO;          0;      0];

h_total_air = hf_air + h_air - hamb_air;

h_N2ideal = (n_r.*h_r - n_p.*h_pBase)./n_p;

t_flame = LagrangeIntp(h_tofind, T_tofind, h_flame);

let xCO2 = nCO2/(nCO2 + nCO) xH2O = nH2O/(nH2O + nH2) the values are given: xCO2 = 0.85; xH2O = 0.93; with the above, balance this equation: 1 CH3OH(l) + 4.5 (O2 + 3.76N2) + 0.086081 H2O => CO2 + CO + H2O + H2 + O2 + N2

HUMID AIR nH2Or = molesH2OinAir(n_air, phi, T, Pt) How many moles of H2O (n_H2O) are in the air under the following conditions? : n_air = 4.5 % Dry air phi = 0.6; % Relative humidity air_temp = 25; % °C pressure = 101.325; % kPa

Pv = psat*phi; nH2Or = (Pv/Pt) * n_air/(1-(Pv/Pt));