Answer to Question #181525 in Mechanical Engineering for Mohamed sadique

Question

Dry products of combustion of a coal have the following volumetric composition 11 % CO2, 3.5 % O2 and 83.5 % N2, at atmospheric pressure 1 bar and temperature 27°C. Calculate : (i) The gravimetric analysis, (ii) The gas constant for the mixture, (iii) The equivalent molecular weight, (iv) Of the mixture specific heat.

Answer 1

Mr(CO₂) = 44 Kg/Kmol; Mr(O₂) = 32 Kg/Kmol; Mr(N₂) = 28 Kg/Kmol

m(CO₂) = N(CO₂) x Mr(CO₂) = 11 Kmol x 44 Kg/Kmol = 484 Kg

m(O₂) = N(O₂) x Mr(O₂) = 3.5 Kmol x 32 Kg/Kmol = 112 Kg

m(N₂) = N(N₂) x Mr(N₂) = 83.5 Kmol x 28 Kg/Kmol = 2338 Kg

m_total = 484 + 112 + 2338 = 2934 Kg

The mass fraction

mf_CO2 = m(CO₂) / m_total = 484 Kg / 2934 Kg = 0.165

mf_o2 = m(O₂) / m_total = 112 Kg / 2934 Kg = 0.038

mf_N2 = m(N₂) / m_total = 2338 Kg / 2934 Kg = 0.797

Molar mass

Mr_m = m_total / N_total = 2934 Kg / 100 Kmol = 29.34 Kg/Kmol

The gas constant of the mixture

R_m = R_u / Mr_m = 8.314 Kj/(KmolxK) / 29.34 Kg/Kmol = 0.2834 Kj/(KgxK)