Question

A furnace is fired with fuel oil with partial analysis of 7.6%S and 2.8%N. Orsat analysis of the stack gas shows 9.44% CO2, 1.19%, 0.4% SO2, 0.47% H2, 6.8% O2 and 81.7% N2. Air is supplied at 23℃, 755 mmHg and 85% RH. Calculate a) % excess air; b) Analysis of the fuel oil; c) m3air/kg fuel.

Answer 1

Solution:

Basis: 100 kgmol stack gas SG.

S balance,

(0.076) * (F) * 32 = 0.004(100)

F = 168.42 kg

NF = (0.028) * (168.42) = 4.7158/14 = 0.33684 kgmol
N2 supplied= 81.7 - 0.33684 * (1/2) = 81.53 kmol
O2 supplied = 81.53 * (21/79) = 21.67 kmol

Assume excess be x

O2 x’s = O2theo - CO / 2 − H2 / 2

O2 x’s = 6.8 - 1.19 / 2 − 0.47 / 2 = 5.97 kmoles

% x’s air

% x’s air= 5.97 / (21.67−5.97) x100 = 38.03 %

C balance,

(% C) * (168.42) / 12 = 9.44 +1.19
%C = 0.7574* x *100 = 75.74%

%H + %O = 1 - (0.7574+0.076+0.028)
%H + %O = 0.1386

Assume, x = %O

%H = 0.1386 - x

Let y= O2 in H2O

Actual combustion

C    +   O2 = CO2

9.44    9.44 = 9.44

C    +   1/2O2 = CO

1.19 0.595     1.19

H2    +   1/2O2 = H2O

2y           y    2y

H balance:

(168.42) * (0.1386 - x) = 0.47 * (2) + 4y equation 1

O balance:

((?) * (168.42) / 13) + 21.67 * (2) = 9.44 * (2) + 1.19 + 0.4 * (2) + 6.8 * (2) + 2 * y     equation 2

Solving for x using equation 1 and 2 we get

X = 0.0245 * x * 100 = 2.46%
%H = 0.1386 - 0.0246 = 0.114 * x * 100 = 11.4 %H

Fuel analysis

7.6 %S
2.8 %N
75.74 %C
11.4 %H
2.46 %O

m3air/kg fuel is calculated as

Ps= ?73.649−7258.2296– 7.3037??296 +4.653?10−6(296)2 X 760101325
Ps =20.90mmHg
nH2Oair =(20.90)(0.85)(103.2) / 755−(20.90∗0.85) = 2.487 m3
nair =103.2 + 2.487 = 105.687m3
V = (105.687) * (0.08766)* (296) / (755 / 760) = 2584.11 m3

?3air / kg fuel = 2584.111 / 68.42 = 15.34 m3