A coal with the following composition was burnt in a power plant with 23.2 % exc

Question

A coal with the following composition was burnt in a power plant with 23.2 % excess air. Partial pressure of water vapor in the flue gases determines the dew point of the flue gases. The stack gases must be emitted above dew point. The plant engineer would like to know the partial pressure of water vapor in the flue gases. Calculate the partial pressure of water vapor in mm of Hg. The air drawn into the furnace at that location has 0.021 moles of water vapor per mole of dry air. Coal analysis: Carbon 61.9%; Hydrogen 5.0 %; Nitrogen 1.8%; Sulfur 1.3%. The lab also reported that the coal contained 4.3 % moisture and 8.4% ash. Given 1 atm = 760 mm Hg 147.7810

Explanation / Answer

Basis : 100 gm of coal

Fuel contains : 61.9 gm Coal, 5 gm of Hydrogen, 1.8 gm of nitrogen, sulphur = 1.3gm, moisture = 4.3 gm

Atomic weights : C=12, H2= 2 , N2= 28, H2O= 18 and S= 32

Moles : mass/atomic weight

Moles : C= 61.9/12=5.16, H2= 5/2= 2.5, N2= 1.8/28=0.064, H2O= 4.3/18= 0.24, S =1.3/32=0.041

Oxygen is required for combustion of C, H2 and O2

The reactions are

C+O2-àCO2, moles of O2 required= 5.16, moles of CO2 produced =5.16 ( since all the molar ratios of the reactants ar 1:1:1)

H2+0.5O2-àH2O, mole of O2 required = 2.5/2= 1.25, mole of H2O formed = 1.25

S+O2-àSO2, moles of O2 required = 0.041 moles, moles of SO2 formed =0.041

Total moles of oxygen required = 5.16(CO2)+1.25(H2O) +0.041 =6.451 moles

Dry air contains 79% N2 and 21% O2, moles of dry air= 6.451/0.21=30.7

Air supplied is 23.2% excess, mole of air supplied = 30.7*1.232=37.8 moles

Air contains 0.021moles of moisture per mole of dry air

So moles of moisture = 37.8*0.021 =0.79 moles

Products ( moles) : CO2= 5.16, SO2= 0.041, O2= O2 supplied- O2 consumed = 37.8*0.21- 6.451=1.5 moles

N2= N2 in the air+ N2 in the fuel = 37.8*0.79+0.064=29.93

H2O= H2O formed during combustion+ H2O carried by air+ H2O inthe fuel=

1.25+0.79+0.24=2.28 moles

Mole fraction of water = moles of water vapor/ total moles = 2.28/(2.28+29.93+1.5+0.041+5.16)=0.059

Partial pressure of water vapor = mole fraction* total pressure = 0.059*760=44.84 mm Hg

Dew point is the temperature at which partial pressure= vapor pressure= 44.84 mm Hg

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