100 m3 of a fuel oil are burned per hour with 20 percent excess air in a steam g

Question

100 m3 of a fuel oil are burned per hour with 20 percent excess air in a steam generator. The fuel oil has a density of 926 kg/m3 and the following mass composition: 87% C, 12% H2, 1% S. Air enters the boiler at 100 kPa and 15 °C. Determine the following:

(a). The volumetric flow rate of the air (m3 /min) fed to the boiler;

(b). The flue gas molar composition on a dry basis, reporting all constituents to the nearest 0.001% and assuming complete combustion of the fuel oil; and

(c). The dew point temperature (°C) of the combustion flue gas, if the pressure exiting the boiler stack is 200 kPa.

(d). The mass flow rate of sulfur dioxide (SO2) emitted from the boiler, in metric tons per day, assuming complete combustion of the fuel oil.

Explanation / Answer

The volume of fuel oil burned = 100m^3

The mass of fuel oil burned = Volume X density = 100 X 926 = 92600 Kg

Mass composition = 87% Carbon , 12% hydrogen, and 1% sulphur

so the mass of carbon = 87% X 92600 = 80562 Kg

Mass of hydrogen = 12% X 92600 = 11112 Kg

Mass of sulphur = 1% X 92600 = 926 Kg

a) Volume of air fed = 100m^3 / hour = 100m^3 / 60 minutes

so volumetric flow rate of th air = 1.667 m^3 / minute

b) The burning will be like this

C + O2 --> CO2 (g)

12 grams of carbon will react with 32grams of O2 to give 44 grams / 1 mole of CO2

So 80562 Kg of carbon will react with 214832 Kg of O2 to give = 295394 Kg / 6713.5 Kmoles

H2 + 1/2O2 --> H2O(l)

2 grams will react with 16 grams of O2 to give 18 grams / 1mole of water

so 11112 Kg will react 88896Kg of O2 to give 100008 Kg/ 5556 Kmoles of water

S + O2 --> SO2(g)

32 grams will react with 32 grams of O2 to give 64 grams / 1mole SO2

so 926 Kg will react with 926Kg of O2 to give 1852Kg / 57.875 Kmoles of SO2

So flue gas will consist of carbon dioxide, water (l) and sulphur dioxide

total moles = 6713.5 moles + 5556 + 57.875 moles = 12327.375

Moles on dry basis = 12327.375 - 5556 = 6771.375

Mole % of CO2 = 6713.5 X 100 / 6771.375 = 99.145 %

Mole % SO2 = 57.875 X 100 / 6771.375 = 0.855 %

c) The air entered is 20% excess

so let us calculate the air used for combustion

Oxygen used for complete combustio fuel per hour = 214832 +   88896 + 926 = 304654 Kg

we know that the 23Kg of oxygen is present in 100Kg of air

so air required for 304654 Kg oxygen = 1324582.609 Kg of air

20% excess air = 264916.521 Kg

total grams of air = 1589499.13 Kg

Moles of air = Mass / mol wt = 1589499.13 Kg / 28.97 = 54867.073 K moles

Moles of N2 = 78.084 X 54867.073 / 100 Kmoles = 42842.40 K moles

Moles of oxygen = 1911.23 K moles

Total moles = 42842.40 K moles + 1911.23 K moles + 12327.375 = 57081.005 moles

Mole % will be

Mole % of N2 = 75.045%

Mole % of oxygen = 3.348

Mole % of CO2= 11.761

Mole % of H2O = 9.733

Mole % SO2 = 0.103%

So dew point = 113.964 F

d) Mass flow rate / day

Mass flow rate per hour = 1852Kg / hour

so in 24 hours = 44448 Kg / day

1 Kg = 0.0011tons

so 44448 Kg /day = 48.892 tons /day

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