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A coal sample contains C = 70%, O= 5%, H = 23%, N = 0.4%, ash = 0.1%. calculate GCV and NCV of the fuel.
written 5.9 years ago by gravatar for vibhavarikulkarni vibhavarikulkarni • 510 modified 4.7 years ago by gravatar for sanketshingote sanketshingote • 100
advanced engineering chemistry
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written 5.8 years ago by gravatar for vibhavarikulkarni vibhavarikulkarni • 510

Solutions:

$\begin{aligned} HCV &= 1/100[8080 C+34500(H-O/8) +2240 S] \\ &= 1/100[8080 \times 70+34500(23-5/8) +2240 \times 0] \\ &= 1/100 [565600 + 34500 (2.37) \times 2240] \\ &= 1/100 [565600 + 81.765+ 2240] \\ &= 1/100 [567921.765] \\ &= 5679.21 \space kcal/kg \\ \end{aligned}$

$\begin{aligned} LCV &= [HCV- 9 \times H/100 H \times 587] \\ &= [5679.21 - 9 \times 23/100 \times 587] \\ &= 4464.12 kcal/kg \end{aligned}$

Answer:

HCV/GCV = 5679.21 kcal/kg

LCV/NCV = 4464.12 kcal/kg
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