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homework principle of combustion
Typology: Essays (university)
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ϕ = ( F A )
F
( 1
) ( 1 5 ) =0. % of excess air = 100 % × 1 − ϕ ϕ = 100 % × 1 −0.
=7.098 %
prob case=1______________5832 uv
rho,m**3/kg= 500
phi= 0.
reac fuel C3H8(L) mole=100.0000 t,k= 298.000 rho,g/cc= 500.000 h,kj/mole=-
oxid Air mole=100.0000 t,k= 298.
output short
output siunits
end REACTANT C3H8(L) HAS BEEN DEFINED FOR THE TEMPERATURE 231.08K ONLY. YOUR TEMPERATURE ASSIGNMENT 298.00 IS MORE THAN 10 K FROM THIS VALUE. (REACT) ERROR IN REACTANTS DATASET (INPUT)