Question

Coal is burned at an electric power generation plant at a rate of 150 kg/minute with 8% excess air (dry). The coal has a composition (mass based) of x_C (kg C/coal), x_H (kg H/coal), x_S (kg Sulfur/coal) and x_ash (kg ash/coal). The coal produces 20 kg/minute of ash. The outlet gases, on a dry basis (i.e. we exclude any water from the calculation), have the composition (volume basis): y_CO2=0.1418, y_CO=0.024, y_SO2=0.003, y_O2=0.027, y_N2=0.8042. Calculate the coal composition, air demand and outlet composition on a wet basis.

Answer 1

Solution.

Let's assume that the coal was being burned during 1 minute.

The mass of coal is m(C)+m(H)+m(S)+m(ash) = 150;

m(ash) = 20;

The combustion reactions are

C+O₂ = CO₂

2C+O₂ = 2CO

S+O₂ = SO₂

2H₂+O₂ = 2H₂O

Assuming that the total number of kmoles of gaseous products is n,

m(C) = 12*0.1418n+12*0.024n = 2n

m(S) =32*0.003*n = 0.096n

The amount of oxygen used is (0.027/0.08)n = 0.337n kmoles;

The amount of oxygen used for C and S combustion is (0.1418+0.024/2+0.003)n = 0.157n

The amount of oxygen used for H combustion is 0.337n-0.157n = 0.18n;

m(H) = 2*2*0.18= 0.72n

m(C)+m(H)+m(S) = 150-20

2n+0.096n+0.72n=130;

n = 46.2;

m(C) = 92.4 kg;

m(S) = 4.44 kg;

m(H) = 33.3 kg;

The composition of the coal is

x(C) = 92.4/150 = 0.616 or 61.6 %;

x(S) = 4.44/150 = 0.03, or 3 %;

x(H) = 33.3/150 = 0.222, or 22.2 %;

x(ash) = 20/150 = 0.133, or 13.3 %.

The oxygen demand is 0.337n*1.08 = 0.337*46.2*1.08= 16.82 kmoles, or 16.82*32=538.3 kg. As dry air contains 21% of oxygen by mass, the air demand is 538.3/0.21= 2563 kg/min.