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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×(H−O8)+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×(5.31−15.118)+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×(5.31−1.88875)=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=2802.8866 kJ/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=HHV−2.447×H \text{LHV} = \text{HHV} - 2.447 \times \text{H} Substitute the values: LHV=2802.8866−2.447×5.31 \text{LHV} = 2802.8866 - 2.447 \times 5.31 LHV=2802.8866−12.99357=2789.89303 kJ/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.
Solution by Steps
step 1
Calculate the annual electricity consumption difference between the old and new refrigerators: 1500 kWh−375 kWh=1125 kWh 1500 \, \text{kWh} - 375 \, \text{kWh} = 1125 \, \text{kWh}
step 2
Calculate the annual cost savings by multiplying the difference in electricity consumption by the cost per kWh: 1125 kWh×0.12 RM/kWh=135 RM 1125 \, \text{kWh} \times 0.12 \, \text{RM/kWh} = 135 \, \text{RM}
step 3
Calculate the annual CO₂ emissions avoided by multiplying the difference in electricity consumption by the CO₂ emission factor: 1125 kWh×0.7 kg/kWh=787.5 kg 1125 \, \text{kWh} \times 0.7 \, \text{kg/kWh} = 787.5 \, \text{kg}
Answer
The household will save 135 RM per year and avoid 787.5 kg of CO₂ emissions per year.
Key Concept
Energy savings and CO₂ emissions reduction
Explanation
By replacing an old refrigerator with a more energy-efficient model, the household reduces its electricity consumption, leading to cost savings and a reduction in CO₂ emissions.
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