Combustion problems Subject: Coal Fired Power Generation Dr. Shafqat Nawaz

7

Prepared by: M. Nauman Butt M.Phil. Coal Technology (2014 - 16)

Problem – 2

The coal as given in problem 1 is burned with air. Calculate the percentage of excess air used if

the dry flue gas contains:

i. 12% CO2

ii. 8% O2

Solution:

(i) Percentage of excess air ( when 12% CO2 in dry flue gas):

Let say, excess air = e

As, total amount of dry flue gas with theoretical air (from problem 1) = 33.876 Kg – mol

In case of excess air, total amount of dry flue gas = (33.876 + e) Kg – mol

Since CO2 in flue gas is 12% , ������������ �����

���� ��������� ����� × 100 = 12

As, amount of CO2 in flue gas (from problem 1) = 6.325 Kg – mol

So, �.���

��.����� × 100 = 12 , 6.325 = 0.12 (33.876 + e)

6.325 = 4.065 + 0.12 e , �.�����.���

�.��= e

Excess air = e = 18.833 Kg - mol

% excess air = ������ !"

!""�#$!"�% × 100

Air required (from problem 1) = 34.848 Kg – mol

So, % excess air = ��.���

��.��� × 100 = 54.04 %

Hence,

% excess air = 54.04 %

Combustion problems Subject: Coal Fired Power Generation Dr. Shafqat Nawaz

8

Prepared by: M. Nauman Butt M.Phil. Coal Technology (2014 - 16)

(ii) Percentage of excess air ( when 8% O2 in dry flue gas):

Let say, excess air = e

As, total amount of dry flue gas with theoretical air (from problem 1) = 33.876 Kg – mol

In case of excess air, total amount of dry flue gas = (33.876 + e) Kg – mol

Since O2 in flue gas is 8% , ��������&'������ �����

���� ��������� ����� × 100 = 8

As, mol fraction of O2 = ��������&'���

���� ���������&(�����)

i.e. Amount of O2 = (mol fraction of O2 × total amount of excess air)

Amount of O2 = (0.21 × e) Kg – mol

Substituting values:

�.���

(��.�����) × 100 = 8

0.21 e = 0.08 (33.876 + e)

0.21 e = 2.71 + 0.08e

0.21 e – 0.08 e = 2.71

0.13 e = 2.71 , e = �.��

�.��

Excess air = e = 20.846 Kg – mol

% excess air = ������ !"

!""�#$!"�% × 100

Air required (from problem 1) = 34.848 Kg – mol

So, % excess air = ��.���

��.��� × 100 = 59.82 %

Hence,

% excess air = 59.82 %