Calculate the higher and lower calorific values of coal sample containing 84% carbon, 1,5% sulphur, 0.6 nitrogen, 8.4% hydrogen and 5.5% oxygen

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written 6.1 years ago by

vibhavarikulkarni • 520

modified 6.1 years ago by

yashbeer ★ 11k

Solution is as follows:

$\begin{aligned} HCV &= \frac{1}{100} \bigg[8080 C+34500(H-O/8) +2240 S \bigg] \\ &= \frac{1}{100} \bigg[8080 \times 84+34500(8.4-5.5/8) + 2240 \times 1.5 \bigg] \\ &= \frac{1}{100} \bigg[678720 + 34500 (7.71) \times 2240 \times 1.5 \bigg] \\ &= \frac{1}{100} \bigg[678720 + 265995+ 3360 \bigg] \\ &= \frac{1}{100} [948075] \\ &= 9480.75.21 \space kcal/kg \\ \end{aligned}$

$\begin{aligned} LCV &= \bigg[ HCV- 9 \times H/100 H \times 587 \bigg] \\ &= \bigg[9480.75 - 9 \times 5.5/100 \times 587 \bigg] \\ &= 9190.185 \space kcal/kg \end{aligned}$

Answer:

$HCV/GCV = 9480.75.21 \space kcal/kg$

$LCV/NCV = 9190.185 \space kcal/kg$

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