On burning 0.83 g of a solid fuel in a bomb calorimeter, the temperature of 3,500 g of water increased from 26.5 oC to 29.2 oC. In another experiment in the same calorimeter 0.5 g of naphthalene (HCV = 9,688 kcal/kg) was burnt and temperature was increased form 25.5 oC to 29.2 oC. Latent heat of evaporation of water is 587.0 cal/g and the fuel contains 0.7 % hydrogen, calculate its gross and net calorific value.
Answers
Answer:
The air is passed into the cylinder and it is compressed by the piston till it attains ignition temperature of the fuel .As soon as the fuel is injected in to the cylinder, it absorbs heat from the heated air and burns on attaining ignition temperature spontaneously. Ignition delay is the time lag between the start of the fuel injected and its ignition. Straight chain hydrocarbon has shorter ignition delay whereas branched and aromatic hydrocarbon has longer ignition delay.
Answer:
12,601 cal/g
Explanation:
Higher Calorific Value (HCV) is described as the full quantity of warmth produced whilst a unit quantity (mass/volume) of gasoline is burnt completely, and the goods of combustion are cooled to room temperature.
Here, Weight of gasoline (x) = 0.eighty three g; weight of water (W) = 3500 g; water equal of calorimeter
(w) = 385 g; (t2 – t1 ) = (29.2°C – 26.5°C) = 2.7°C;
percent of hydrogen (H) = 0.7%;
Latent warmth of steam = 587 cal/g
HCV of gasoline (H) = (W+w )( t₂-t₁)/x cal/g
= (3500 +358 ) x2.7 /0.eighty three
=12,638 cal/g
Net calorific value = (HCV – 0.09 H × 587) = (12638 – 0.09 × 0.7 × 587)cal/g.
= (12,638–37) cal/g = 12,601 cal/g
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