Secondary fuels

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Secondary fuels

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Secondary Fuels ______________fuels, derived from other fuels, e.g. coal

Advantages over primary fuels: 1) 2) 3) 4) Minimal storage and handling ___________________ Reduced ____________ production Wider _________________ (i.e. market) _______________ heating value obtained

Producer Gas Manufacture Steam air

Primary fuel Gas producer

producer gas

Tar, soot residue

Primary fuel-normally coal Gas producer-equipment Process: 1) Gasification of volatile compounds 2) Primary combustion of C C + O2→CO2 C + ½ O2 → CO 3) Water Gas Formation C + H2O → CO + H2 CO + H2O→ CO2 + H2 In all cases, steam (H2O) is decomposed to H2 and O2

High volatile coals- producer gas production is accompanied by formation of tar and soot _________-condensed volatile combustibles _________- deposited carbon Illustration 1: The average analysis of a certain producer gas is 9.84% CO2, 0.04% O2, 0.18% C2H4, 18.28% CO, 12.90% H2, 3.12% CH4, and 55.64% N2. Determine the following: a. Ratio of net hydrogen to carbon in the primary fuel b. Pounds of steam decomposed per pound of carbon gasified c. Cubic feet of producer gas at dry, standard conditions, per pound of carbon gasified d. Cubic feet of air at dry, standard conditions, per pound of carbon gasified Given:

Steam

air

Primary fuel Gas producer

producer gas

residue Solution: Basis: 100 __________ dry producer gas mol CO2 O2 C 2H4 CO H2 CH4 N2 Total mol C mol H2 mol O2

Assume no N2 in fuel:

N2 balance: N2 from air= mol O2 from air:

mol O2 accounted for (or in exit stream): Total mol O2 in dry producer gas = mol O2 from _____ and from steam decomposition mol O2 from decomposition of steam=

H2O → H2 + ½ O2 Assume all H2O decomposed comes from steam mol steam decomposed =

mol H2 from steam =

mol H2 accounted for (or in exit stream): Assume no unburned combustible in refuse: Total mol H2 in dry producer gas =mol net H2 from fuel and mol H2 from ___________________ mol net H2 from fuel=

Mass net H2 from fuel=

C Balance: mol C from fuel= Mass C in fuel=

Lb net H2 Lb fuel

Mass steam decomposed=

Steam decomposed = Lb C gasified

Assume ideal gas behaviour: ft3 producer gas (dsc) = Lb C gasified

Mol dry air= ft3 air (dsc) Lb C gasified

Illustration 2: A gas producer is being fired with coal the ultimate analysis of which on a dry basis is 75.8% C, 5.4% H, 1.4% N2, 0.8% S and 3.5% ash, and the corresponding volatile combustible matter 36.2%. The heating value is 13,880 BTU/lb. As fired, it contains 4.5% moisture. During the 68.8hr of the test, 174,500 lb of coal was fired and 6,770 lb of ash and clinker withdrawn containing 114.2% combustible matter (practically all carbon). The gas analysis carefully conducted over mercury averaged 13.25% CO2, 16.10% CO, 22.65% H2, 3.5% CH4 44.5% N2 and inerts. Barometer averaged 766 mmHg air temperature 23oC; and partial pressure of water vapour in it, 12 mmHg. The steam supplied to the blast was at 20 psig and 184oC. It was metered and totalled 256,000 lb during the test. The blast pressure was 5.35 in water. The gas left the producer at 820oC, at practically atmospheric pressure, with a dew point of 65.5oC, corresponding to a vapour pressure of 192 mmH2O. Practically, all the sulphur in the coal showed up in the gas as H2S. Calculate the cubic feet (dsc) of gas produced per pound of coal as fired. Schematic Diagram:

Solution: Basis: 100 lb dry coal mass C H2 N2 S Ash O2 total MW mol

Ash balance:

Mass clinker= Mass C in clinker= Or, Mass wet coal=

Mass clinker=

Mass C in clinker=

C balance:

Basis: 100 lbmol dry producer gas mol C + mol S=

Ratio of S to C in the producer gas is the same as the ratio of S to C in the fuel less the C in the refuse

mol S in the producer gas=

mol CO2

mol CO2 H2S CO H2 CH4 N2 Total

Mol C

Mol H2

Mol O2

Assume N2 in coal is negligible (<2%) N2 balance: mol N2 from air= mol O2 from air=

mol dry air=

Air: mol H2O = PH2O mol dry air P dry air mol H2O= Producer gas Oxygen in the producer gas as water vapour is obtainable from the measured dew point

Basis of 100 lb of dry coal Atom O in dry coal

mol H20 as moisture in coal

mol steam

Let x= mol of dry gas produced. Oxygen balance:

Blue water gas- a mixture containing predominantly _______________ and _______________, characterized by blue flame on combustion, the heating value of this gas is too low for economical distribution over wide areas, but it can be increased by means of carburetion. Carburetion-allows gaseous fuel to come into contact with cracked oil vapors in carburettor Illustration 6: Oil used in a carburettor has the following ultimate analysis: 85.7%C, 12.7% H and 0.5% S. The rest, maybe assumed O. The tar and emulsion removed from the carburetted gas contained 20.5% of all the C in the oil used, and the analysis of the dry tars averaged 91.2%C, 6.2% H, and 1.2% S, leaving 1.4% O by difference. The blue gas contained 6.1% CO2, 0.1% O2, 38.3% CO, 50.9% H2, o.5% CH4, and 4.1% N2. The carburetted gas analyzed 5.2% CO2, 5.8% illuminants, 0.1% O2, 34.9% CO 38.1% H2, 10.3% CH4 and 5.6% N2. From these data, compute the pounds of carbon per 1000 ft3 of the final product which came from the oil and the volume of blue gas used at standard conditions. Given:

Solution: Basis: 100lbmol dry blue gas mol 6.1 0.1 mol C mol H2 mol O2

CO2 O2 CO H2 CH4 N2 Total

4.1 100

Assume no N2 in coal Consider gas producer: N2 balance: mol N2 from air= 4.1 lbmol N2

mol O2 from air=

mol O2 from steam decomposition=

mol steam decomposed=

mol H2 from steam decomposition=

mol net H2 from coal=

Assume no combustible left in the residue. coal mol C Net H2

Basis: 100 lb oil Mass C H S O total

Mass C in tar = Mass dry tar=

Tar: mass

Total mass oil gasified

= mass oil supplied-mass tar

The components of the oil gasified can be found by difference: Mass C gasified= Mass H gasified= Mass S gasified= Mass O gasified= Oil gasified: mass C H2 S O2 Total mol net H2 in oil gasified= MW mol

Basis: 100 lbmol dry carburetted gas Mol CO2 C3H6 O2 CO H2 CH4 N2 Total mol C mol H2 mol O2

Consider overall system: N2 balance: mol N2 from air= mol O2 from air= mol O2 from steam decomposition= mol steam decomposed= mol H2 from steam decomposition= mol net H2 from coal=

Basis: 100 lbmol dry carbureted gas Let x=mol dry blue gas Let y= mol C from oil Consider carburettor, C balance:

Net H2 balance:

Assume ideal gas behavior Volume dry carburetted gas (s.c)= 100 lbmol . __________ Basis: 1000 ft3 dry carburetted gas

Illustration 7 and 8 are left for students to study Additional info for coal: Definitions:
As Received (ar): includes Total Moisture (TM) Air Dried (ad): includes Inherent Moisture (IM) only Dry Basis (db): excludes all Moisture Dry Ash Free (daf): excludes all Moisture & Ash

Definitions from: http://www.worldcoal.org/resources/coal-statistics/coal-conversion-statistics/

Other website you may want to look at: http://majarimagazine.com/2008/06/understanding-coal-sample-analysis/

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