Stoichiometric Calc Visa

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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

29.03%Hydrogen 2.15%

Sulphur 0.60% 12kg Carbon and 32kg Oxygen produce 44kg Carbon di-oxide

Oxygen 7.62% 2kg Hydrogen and 16kg Oxygen produce 18kg Water

Nitrogen 0.60% 32kg Sulpher and 32kg Oxygen produce 64kg Sulpher di-oxide

Ash 44.00%

Moisture 16.00%

TOTAL: 100.00%

Substance/ Weight per Weight of Weight of

Constituent kg of fuel oxygen per oxygen Weight of products of combustion


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in kg kg of required CO2 SO2 N2 H2O

constituent in kg in kg in kg in kg in kg

in kg

A B =(A*B) =(44/12)*A =(64/32)*A =A =(18/2)*ACarbon 0.2903 2.67 0.7741 1.0644

Hydrogen 0.0215 8 0.172 0.1935

Sulphur 0.006 1 0.006 0.0120

Oxygen 0.0762

Nitrogen 0.006 0.006

Ash 0.44

Moisture 0.16 0.1600

Total 1 0.9521 1.0644 0.0120 0.006 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 = 0.9521 - 0.0762 kg

= 0.8759 kg for per kg of fuel

So minimum amount of air required = ( 0.8759 x 100 ) / 23.15 kg

= 3.7837 kg

Since, atmospheric air contains 23.15%

Nitrogen in the above air = ( 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 alwaysneeded 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 excess air = (0.1752 x 76.85/23.15)

= 0.5816 kg

Total nitrogen in flue gas (2.9078 + 0.5816 + 0.006) kg

= 3.4954 kg/kg of fuel


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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 = 2871567.52 kg/hr

Substance Weight of molecular Weight Percentage

constituent weight Mol. Weight volume

per kg (Mole Fraction)

of fuel

X Y Z Z x100

= X/Y Z

CO2 1.0644 44 0.0242 13.88

SO2 0.0120 64 0.0002 0.11

O2 0.1752 32 0.0055 3.14

N2 3.4954 28 0.1248 71.61

H2O 0.3535 18 0.0196 11.27

total 5.1005 0.1743 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 m3= 729.6 m

3at 0

0C,760 mm Hg

We have to find flue gas flow in Nm3/s where Nm

3corresponds to 25

0C,760 mm Hg (STP)

So, 1 tonne flue gas occupies 729.6 m3= 729.6x298/273 Nm

3= 796.4 Nm

3at 25

0C,760 mm Hg

Now, Weight of flue gas / Kg of coal burnt = 5.1005 kg

Taking coal consumption rate 563 t/h, flue gas flow rate will be


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5.1005x563x1000/3600 kg/s

797.66 Kg/s

Flue gas flow rate in Nm3/s will be

5.1005x563x796.4/3600 Nm3/s

= 635.3 Nm3/s

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