Carbon combustion efficiency calculation

The carbon combustion efficiency of a system has been expressed as:

hce=C-x100% (1)

whereB and C are the mass fractions of burnt and total carbon in the fuel, respectively. Knowing the flue gas composition, the flue gas composition, fractional excess air and the fuel ultimate analyses of the fuel, B can be determined [20].

This method is particularly appropriate for solid fuels and is described as follows:

Let C, H, O, N, and S be the mass fractions of carbon, hydrogen, oxygen, nitrogen and sulphur, respectively, in the feed.

Подпись: Further define Carbon combustion efficiency calculation Подпись: (2) (3) (4) (5) (6)

Further, let A and B be the mass fractions of unburned and burnt carbon, respectively, in the fuel. Then,

Mass of CO in the flue gas = (28 / 12)PB

Подпись: (7)Подпись:02 consumed to produced C02 + CO = (32 – 16P)B /12

Assuming that H, N, and S present in the fuel are completely converted to H 2 O, NO and SO Respectively,

02 consumed = (16/2)H + (16/14)N + (32/32)S = X1 (9)

S02 produced = (64/ 32)S (10)

NO produced = (30/14)N (11)

Therefore, total О 2 required for stoichiometric combustion of fuel

(32/12)C + (16/2)H + (16/14)N + (32/32)S – О = X 2 (12)

Let Z be the fractional excess air supplied, which is defined as the excess air divided by the stoichiometric air. Therefore,

О 2 supplied = X 2(1 + Z) (13)

Mass of N 2 in the flue gas = (79/12)(28/32) X 2(1 + Z) (14)

О2 consumed during combustion = (32-16P)B /12 + X1 (15)

Mass of О 2 in the flue gas = ( X 2(1 + Z) – ^32 + X1 (16)

Let F be the mass of dry flue gas can also be estimated from the flue gas composition. The flue gas flow rate and composition are not appreciably influenced by neglecting the pres­ence of SO 2and NO in the flue gas. Hence the flue gas may be taken as consisting of CO,

СО2, N 2 and О 2.

Let Y be the mass of dry flue gas per unit mass of C burnt in the fuel. Then,

Y = {44[ СО2] + 32[ О2] + 28[ N2]}/12{[ СО] + [СО2]} (17)

The square brackets represent the volume fraction of the particular chemical species in the flue gas and Y can be simplified to

By substituting

[C^| + [NJ + [co2] + [0J = 1 (19)

mass of dry flue gas per unit mass of the fuel is

F = YB (20)

Substituting F in (A-6.18) into (A-6.17), then the fraction of C burnt, B, can be written as fol­lows:

Подпись: [4.29(1 + z )[(32 16 16 32

Подпись: is - O - 8H + N + SПодпись:Подпись: В =12 )C + WlM + Ы N + І32

цСЕ = (B(21) + Unbwned carbon in ash) / C * 100% (22)