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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.
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Solutions:

HCV=1/100[8080C+34500(HO/8)+2240S]=1/100[8080×70+34500(235/8)+2240×0]=1/100[565600+34500(2.37)×2240]=1/100[565600+81.765+2240]=1/100[567921.765]=5679.21 kcal/kg

LCV=[HCV9×H/100H×587]=[5679.219×23/100×587]=4464.12kcal/kg

Answer:

HCV/GCV = 5679.21 kcal/kg

LCV/NCV = 4464.12 kcal/kg

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