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The assay of particular coal from Illinois is as follows by mass: 67.40 percent...
Jun 15, 2024
Solution by Steps
step 1
To find the higher heating value (HHV) of the coal, we use the formula: HHV=33.94×C+144.45×(HO8)+9.16×S \text{HHV} = 33.94 \times \text{C} + 144.45 \times (\text{H} - \frac{\text{O}}{8}) + 9.16 \times \text{S} where C, H, O, and S are the mass percentages of carbon, hydrogen, oxygen, and sulfur, respectively
step 2
Substitute the given values into the formula: C=67.40%,H=5.31%,O=15.11%,S=2.36% \text{C} = 67.40\%, \text{H} = 5.31\%, \text{O} = 15.11\%, \text{S} = 2.36\% HHV=33.94×67.40+144.45×(5.3115.118)+9.16×2.36 \text{HHV} = 33.94 \times 67.40 + 144.45 \times (5.31 - \frac{15.11}{8}) + 9.16 \times 2.36
step 3
Calculate the intermediate values: 33.94×67.40=2287.196 33.94 \times 67.40 = 2287.196 144.45×(5.311.88875)=144.45×3.42125=494.073 144.45 \times (5.31 - 1.88875) = 144.45 \times 3.42125 = 494.073 9.16×2.36=21.6176 9.16 \times 2.36 = 21.6176
step 4
Sum the intermediate values to find the HHV: HHV=2287.196+494.073+21.6176=2802.8866kJ/kg \text{HHV} = 2287.196 + 494.073 + 21.6176 = 2802.8866 \, \text{kJ/kg}
step 5
To find the lower heating value (LHV), we use the formula: LHV=HHV2.447×H \text{LHV} = \text{HHV} - 2.447 \times \text{H} Substitute the values: LHV=2802.88662.447×5.31 \text{LHV} = 2802.8866 - 2.447 \times 5.31 LHV=2802.886612.99357=2789.89303kJ/kg \text{LHV} = 2802.8866 - 12.99357 = 2789.89303 \, \text{kJ/kg}
Answer
The higher heating value (HHV) of the coal is approximately 2802.89 kJ/kg, and the lower heating value (LHV) is approximately 2789.89 kJ/kg.
Key Concept
Higher and Lower Heating Values
Explanation
The higher heating value (HHV) represents the total energy released when the coal is burned, including the latent heat of vaporization of water. The lower heating value (LHV) excludes this latent heat, providing a measure of the usable energy.
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