A sample of coal has the following composition C = 84 %, O = 8.4 %, H = 5.5 %, S = 1.5%, N = 0.6 %, Calculate the higher and lower calorific values of coal.

Given: C = 84%, H = 5.5%, O = 8.4%, S = 1.5%, N = 0.6% (N does not contribute to the heat).

Dulong's formula (C, H, O, S in %):
\( \text{HCV} = \frac{1}{100}\left[8080\,C + 34500\left(H - \frac{O}{8}\right) + 2240\,S\right] \) kcal/kg
\( \text{LCV} = \text{HCV} - 0.09\,H \times 587 \) kcal/kg (latent heat of steam = 587 kcal/kg)

\( H - \frac{O}{8} = 5.5 - \frac{8.4}{8} = 5.5 - 1.05 = 4.45 \)

\( \text{HCV} = \frac{1}{100}\left[8080(84) + 34500(4.45) + 2240(1.5)\right] = \frac{1}{100}\left[678720 + 153525 + 3360\right] = \frac{835605}{100} \)

HCV = 8356.05 kcal/kg

\( \text{LCV} = 8356.05 - 0.09 \times 5.5 \times 587 = 8356.05 - 290.565 \)

LCV = 8065.49 kcal/kg