P1-12 - 25.407255 25.407255 25.407255 42.455612 gam =-4.3...

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Problem 1.12 Adiabatic Flame Temperature for Natural Gas Combustion Equations Initial values T 1000 1000 2197.99490137 1779.484 y = 0.75 0.75 0.75 0.75 0.7 x = 0.5 0.5 0.5 0.5 1.7 z = 1-y-0.02 0.23 0.23 0.23 0.28 CH4 = if(x<1)then(y*(1-x))else(0) 0.375 0.375 0.375 0 C2H6 = if(x<1)then(z*(1-x))else(0) 0.115 0.115 0.115 0 CO2 = if(x<1)then((y+2*z)*x)else(y+2*z) 0.605 0.605 0.605 1.26 H2O = if(x<1)then((2*y+3*z)*x)else(2*y+3*z) 1.095 1.095 1.095 2.24 N2 = 0.02+3.76*(2*y+7*z/2)*x 4.3534 4.3534 4.3534 15.23296 O2 = if (x < 1) then (0) else ((2 * y + 7 * z / 2) * (x - 1)) 0 0 0 1.666 alp = 3.381 * CH4 + 2.247 * C2H6 + 6.214 * CO2 + 7.256 * H2O + 6.524 * N2 + 6.148 * O2 41.6326516 41.6326516 41.6326516 133.7055 bet = 18.044 * CH4 + 38.201 * C2H6 + 10.396 * CO2 + 2.298 * H2O + 1.25 * N2 + 3.102 * O2
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Unformatted text preview: 25.407255 25.407255 25.407255 42.455612 gam = -4.3 * CH4 - 11.049 * C2H6 - 3.545 * CO2 + 0.283 * H2O - 0.001 * N2 - 0.923 * O2-4.7223284-4.7223284-4.7223284-5.385731 H0 = alp * 298 + bet * 0.001 * 298 * 298 / 2 + gam * 1e-6 * 298 ^ 3 / 3 13493.006523 13493.006523 13493.0065226 41681.84 Hf = alp * T + bet * 0.001 * T ^ 2 / 2 + gam * 1e-6 * T ^ 3 / 3 52762.169633 52762.169633 136166.556523 295030 xx = if (x <= 1) then (x) else (1) 0.5 0.5 0.5 1 f(T) = 212798*y*xx+372820*z*xx+H0-Hf = 0 83404.386889 83404.386889-6.51344E-007 of the cell b20 at zero while changing cell b4 Solution 1 1 Solution is obtained by Goal Seek(see under Tools dropdown menu) by setting the value...
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This note was uploaded on 01/02/2012 for the course CHE 218 taught by Professor Engr.agapay during the Spring '11 term at Mapúa Institute of Technology.

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