1 X 660 MW TPP for Visa Power Limited at Raigarh
(Supercritical Unit)
Date: 02.05.2011
CALCULATION OF FLUE GAS FLOW :
Coal Consumption Rate
As per Visa 600MW unit's Steam Generator Data,
Worst Coal Requirement =
536
TPH
(at 100% BMCR condition)
In absence of 660 MW unit's coal consumption data,
we use coal consumption data of 660MW unit as a certain margin on the above data for coal
Margin considered =
5%
(as advised by Pradoshda on 2.5.11)
Coal Consumption rate =
562.8
TPH
563
TPH (say)
Assumptions
1) Atmospheric moisture is not considered
2) No carbon mono-oxide is present in flue gas as combustion is complete
3) NOx formation is not considered
4) Excess air for combustion is taken as 20%
ULTIMATE ANALYSIS OF COAL (WORST COAL)
Carbon
Hydrogen
Sulphur
Oxygen
Nitrogen
Ash
Moisture
TOTAL:
29.03%
2.15%
0.60%
7.62%
0.60%
44.00%
16.00%
100.00%
Substance/ Weight per
Constituent kg of fuel
12kg Carbon and 32kg Oxygen produce 44kg Carbon di-oxide
2kg Hydrogen and 16kg Oxygen produce 18kg Water
32kg Sulpher and 32kg Oxygen produce 64kg Sulpher di-oxide
Weight of
oxygen per
Weight of
oxygen
Weight of products of combustion
Carbon
Hydrogen
Sulphur
Oxygen
Nitrogen
Ash
Moisture
Total
in kg
kg of
constituent
in kg
required
in kg
CO2
in kg
SO2
in kg
N2
in kg
H2 O
in kg
A
0.2903
0.0215
0.006
0.0762
0.006
0.44
0.16
1
B
2.67
8
1
=(A*B)
0.7741
0.172
0.006
=(44/12)*A
1.0644
=(64/32)*A
=A
=(18/2)*A
0.1935
0.0120
0.006
0.9521
1.0644
0.0120
0.006
0.1600
0.3535
Oxygen required for complete combustion of 1 kg fuel is
Wo2 = 2.67C + 8H + S - O
, where O is the oxygen present in the fuel
Total Oxygen required from atmosphere =
=
So minimum amount of air required
0.9521 - 0.0762 kg
0.8759
kg for per kg of fuel
( 0.8759 x 100 ) / 23.15
= 3.7837
kg
kg
Since, atmospheric air contains 23.15%
( 3.7837 x 76.85 ) / 100
kg
O2 by weight and rest is N2.
=
2.9078
kg
Since, complete combustion of fuel cannot be achieved if only the theoritical air is supplied. Excess air is always
needed for complete combustion.
Considering 20 % excess air for combustion
Total oxygen in excess air =
0.20 x 0.8759
kg
=
0.1752
kg
Nitrogen in the above air =
Nitrogen in excess air =
(0.1752 x 76.85/23.15)
= 0.5816
kg
Total nitrogen in flue gas =(2.9078 + 0.5816 + 0.006)
= 3.4954
kg/kg of fuel
kg
Weight of Flue Gas =
5.1005
kg/kg of fuel burnt
So, for 563TPH coal firing rate as mentioned above,
Flue gas flow rate shall be =
Substance Weight of
constituent
per kg
of fuel
X
CO2
SO2
O2
N2
H2 O
total
1.0644
0.0120
0.1752
3.4954
0.3535
5.1005
molecular
weight
Weight
Mol. Weight
(Mole Fraction)
Z
= X/Y
0.0242
0.0002
0.0055
0.1248
0.0196
0.1743
44
64
32
28
18
2871567.52 kg/hr
Percentage
volume
Z x100
SZ
13.88
0.11
3.14
71.61
11.27
100.00
Now, molecular wt of flue gas (wet basis) :
0.44x13.88+0.64x0.11+0.32x3.14+0.28x71.61+0.18x11.27
= 29.26
and molecular wt of flue gas (dry basis)
(29.26-0.18x11.27)/(1-0.1127)
= 30.7
At N.T.P (00C, 760 mm Hg) 30.7 gm flue gas occupies volume of 22.4 litre
So, 1 tonne flue gas occupies (22.4/30.7)x1000 m 3 = 729.6 m3 at 00C,760 mm Hg
We have to find flue gas flow in Nm3/s where Nm3 corresponds to 250C,760 mm Hg (STP)
So, 1 tonne flue gas occupies 729.6 m 3 = 729.6x298/273 Nm3 = 796.4 Nm3 at 250C,760 mm Hg
Now, Weight of flue gas / Kg of coal burnt =
Taking coal consumption rate 563 t/h, flue gas flow rate will be
5.1005 kg
5.1005x563x1000/3600 kg/s
797.66
Kg/s
Flue gas flow rate in Nm3/s will be
5.1005x563x796.4/3600 Nm3/s
Nm3/s
= 635.3
