A gas produced as a by-product from the carbonization of coal has the following composition, mole %: carbon dioxide 4, carbon monoxide 15, hydrogen 50, methane 12, ethane 2, ethylene 4, benzene 2, balance nitrogen. Using the data given in Appendix C (available online at booksite .Elsevier/Towler), calculate the gross and net calorific values of the gas. Give your answer in MJ/m3, at standard temperature and pressure.

6059.63 kcal/kg.

Explanation:

CH4 consist of Hydrogen and carbon. Therefore, Hydrogen in CH4 = 12 × 4 = 48 kg,  carbon in CH4 = 12 × 12 = 144kg.

For ethane, the amount of hydrogen present = 2 × 6 = 12kg and that of carbon in ethane = 2 × 24 = 48kg.

The weight of carbon in CO = 15 × 12 = 18kg and the weight of Hydrogen in CO = 15 × 16 = 240kg.

For hydrogen, its weight in H2 = 50 × 2 = 100kg.

For CO2, carbon has = 4 × 12 = 48 kg and oxygen has = 4 × 32 = 128kg.

For C6H6, carbon has 2 × 72 = 144kg and hydrogen has 2 × 6 = 12kg.

For N2, the amount of nitrogen= 11 × 28 = 308 kg.

For CH2= CH2, carbon has 4 × 24 = 96kg and hydrogen = 4 × 4 = 16kg.

The gross calofiric value = 1/100 [ 8080 C + 34500 + ( H - O/8) + 22405].

Where the total weight = 128 + 180+ 48 + 240 + 100 + 144 + 48 + 12 + 48 + 16 + 96 + 308 + 12 + 144 = 1524 kg.

The percentage by weight of carbon = total weight of carbon/total weight × 100.

The total Weight of carbon= 48 + 180 + 144 + 48 +144 + 96 = 660kg.

The percentage weight of carbon = 660/1524 × 100 = 43.3 %.

The percentage weight of hydrogen = total weight of hydrogen/total weight × 100.

The total weight of hydrogen = 100 + 12 + 48 + 16 +12 = 188.

The percentage weight of Hydrogen = 188/ 1524 × 100 = 12.33%.

Percentage weight of oxygen = total weight of oxygen/total weight × 100.

Percentage weight of oxygen = ( 128 + 240) / 1524 × 100 = 24.15%.

The gross calorific value = 1/100 [ 8080 × 43.3 ,+ 34500 ( 13.33 - 24.15/8) ] = 6711.02 kcal/kg.

Net calorific value = 6711.02 - 0.09 × 12.3 × 587 = 6059.63kcal/kg.

answer: mercury

explanation: