Answer to Question #332536 in Chemical Engineering for Habib

Question

A coal contains 78% carbon, 4% hydrogen, 2% oxygen, 1.8%

sulphur and rest as non-combustibles.

Calculate:

(i) Gross calorific value of the coal in GJ/ton using Dulong’s Formula

337 C + 1442 [H – (O/8)] + 93 S kJ/kg

where C, H, O and S are percentages of carbon, Hydrogen, oxygen and

sulphur respectively.

(ii) Amount of coal needed per day to burn in a 10 MW power plant

working with 32% thermal efficiency (Given 1 kWh = 3.6 MJ).

Answer 1

(i) HCV = 337 C + 1442 [H – (O/8)] + 93 S = 337 × 0.78 + 1442 [0.04 – (0.02/8)] + 93 × 0.018 = 318.609 kJ/kg or 0.319 GJ/ton

(ii) Q` = W/$\eta$ = 10 MW / 0.32 = 31.25 MW

Q = Q`∆t = (31.25 MJ/s)(24×3600 s) = 2.7×10⁶ MJ

m_coal = Q/q = 2.7×10⁶ MJ / 0.319 MJ/kg = 8.46×10⁶ kg

m'_coal = m_coal / ∆t = (8.46×10⁶ kg) / (24×3600 s) = 97.96 kg/s