Question

A lignite coal has the following ultimate analysis on the dry basis:

          C = 59.88 N = 1.04 A = 13.51 H = 4.21 S = 1.90 O = 19.46

a) Calculate the theoretical air and actual air per pound of lignite. The combustion is assumed to be complete and the excess air = 18%.

b) Calculate the weight of the flue gas.

Answer 1

ANSWER: (a)

1) Calculate the amount of oxygen needed for te combustion of each element this can be done as:

Write balanced chemical equation (e.g C )

C + O₂ -----------> CO₂

Atomic mass of Carbon = 12g and molecular mass of O2 = 32g

It implies 12 unit of mass of carbon need 32 units of oxygen.

One unit will require = 32/12 = 2.67

Similarly it can be shown that oxgen needed for H, N, and S are 8,1, 4.57 respectivlly. (A has not been defined)

Theoretical air required is

2.67XC% + 8XH% + 1XS% + 4.57XN% - O% = 0.18Kg per Kg Of Coal (NOTE THIS IS WITHOUT A.. PLEASE CHECK THE DEFINITION OF A AND GET THE CORRECT VALUE BY THIS METHOD)

Theoretical air = 0.18/O% = 0.924Kg of oxygen per Kg of coal.

theoretical air for one pound = 0.924 X 446/1000 = 0.412Kg Or 412g

(b)

C +O2 ----------> CO2

N2 + 2O2 --------------> 2NO2

S + O2 --------------------> SO2

H2 +1/2 O2 -------------> H2O

Carbon requires one mole of oxygen or 4.76moles of air (BECAUSE 1MOLE OF OXYGEN = 3.76MOLES OF NITROGEN. HINT: IN AIR O2 = 21% AND N2 = 79%. TAKING OTHER GASES AS NEGLEGIBLE)

Therefore

C + O2 +3.56N2 -------------> CO2 + 3.56N2

total no of moles of air required for one mole of carbon = one mole oxygen (32g) + 3.56moles of nitrogen(99.68g) = 131g of air.

from the same method it can be shown N2 , S, and H2 require 264, 131 and 65.5g of air = 593.5g = wiight of flue gas.