Compute the mass of wet air required (theoretically) to burn 1 lb. of coal havin

Question

Compute the mass of wet air required (theoretically) to burn 1 lb. of coal having the following composition:

                                Carbon = 66.10%              Hydrogen = 6.03%           Oxygen = 11.91%

                                Nitrogen = 1.04%             Sulfur = 2.40% Ash = 12.52%

                What would be the analysis of the dry flue gases produced?

                If 20% excess air were supplied for combustion of this coal, and the coal was burned at a rate of 3 tons/hr., compute the required capacity of the fan (actual ft3) to supply the combustion air. Assume the air to be at a pressure of 14.7 psi and temperature of 86°F. Assume that the air at 86 F contains 0.013 lb of moisture per lb of dry air.

Explanation / Answer

Basis :1 lb of Coal

Carbon =0.661

The Reaction is C+O2à CO2

12 lb coal requires 32 lb of oxygen

0.661lb coal 0.661*32/12 =1.763 lb oxygen   (1)

CO2 produced = 0.66*44/12= 2.42   (1A)

Hydrogen =6.03/100 =0.0603

The Reaction is

H2+1/2O2-à H2O

2 lbs of H2 requires 16 lbs of Oxygen

0.0603 lb requires 0.0603*16/2 =0.4824 lb oxygen (2)

Suflur =2.4/100 =0.024lb

The reaction is S+O2-àSO2

32 lb suflur requires 32 lb of oxygen

0.024 requires 0.024 lb oxygen (3)

Sulfur dioxide produced = 0.024*64/32 =0.048 (1B)

Total oxygen required = 1.763(from 1) + 0.4824(from 2) + 0.024 (from 3)

                                     =2.27

Oxygen present in coal = 0.1191lbs

Oxygen actually required =2.27-0.1191=2.1509 lbs

Dry Air to be supplied (since air contains 23.3% Oxygen and 77.7% Nitrogen by mass )

= 2.1509/0.233 =9.23 lb Air

Nitrogen in air =9.23*0.767= 7.08 lb

Nitrogen present in the fuel = 1.04/100=0.0104

Total nitrogen = 7.08+0.0104=7.0904 (4)

Air contains 0.013lb moisture/ lb dry air

Moisture in 9.23lb air = 9.23*0.013lb =0.1199 lb air

Wet air supplied = 9.23+0.1199 =9.3499 lb

Dry analysis of flue gases

Gas

Lbs produced

Reference from Eq.

% (weight)

CO2

2.42

1A

100*2.42/9.5584= 25.32

SO2

0.048

1B

100*0.048/9.5584=0.5022

N2

7.0904

4

7.0904/9.5584=74.18

Total

9.5584

Basis : 3 tons/hr

S.No

Chemical

Amount of chemical (tons/hr)

Oxygen required (tons/hr)

Reference

1

Carbon

3*66.1/100 =1.983

Reaction

C+O2-à CO2

1.763*3=5.289

Eq.1

2

Hydrogen

3*6.03/100= 0.1809

H2+1/2O2-à H2O

0.4824*3= 1.4472

Eq.2

3

Sulphur

3*2.4/100= 0.072

=3*2.4/100= 0.072

Eq.3

4

Oxygen

11.91*3/100=0.3573

-0.3573

Oxygen required

=5.289+1.4472+0.072-0.3573=6.4509

Air to be supplied

=6.4509/0.233= 27.69tons/hr

Air actually supplied

27.69*1.2=33.12 tons/hr

Wet air supplied

33.12+33.12*.013=33.55tons/hr

Products

S.No

Name of the gas

Reaction

Amount of gas (tons/hr

Moles of gas (lbmoles/hr)

1.

Nitrogen

No reaction

Nitrogen in air + nitrogen in the fuel

=33.12*0.767+ 3*1.04/100=

25.43424

25.43424*2000/28

=1816.73

2

Oxygen

Combustion

Supplied-reacted= 33.12*0.233-6.4509=1.26606

1.2606*2000/32

=78.7875

3

CO2

C+O2-àCO2

(44/12)*3*66.1/100=7.271 tons/hr

7.271*2000/44=

330.5

4

SO2

S+O2àSO2

(64/32)*3*2.4/100= 0.144 tons/hr

0.144*100/64=

0.225

5

H2O

H2+1/2O2àH2O

+ water present in wet air

(18/2)*3*6.03/100+33.12*0.013

=2.05866 tons/hr

2.05866*2000/18

=228.74

Total

2454.984 lbmoles/hr

From PV= nRT

V=?

P= 14.7 PSI, R= 10.731 59(2)   ft3 psi °R1 lb-mol1, n= 2454.984 T= 86+460= 546

V= 2454.984*10.731*546/14.7= 978507.5 ft3/hr

Gas

Lbs produced

Reference from Eq.

% (weight)

CO2

2.42

1A

100*2.42/9.5584= 25.32

SO2

0.048

1B

100*0.048/9.5584=0.5022

N2

7.0904

4

7.0904/9.5584=74.18

Total

9.5584

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