Q4(c). Gross and net calorific values of coal
Using the following data, find the difference between gross calorific value and net calorific value of a coal sample whose composition (by weight) is: carbon = 85.0%, hydrogen = 10.29% and ash = 4.71%. Mass of the coal sample fired in a bomb calorimeter = 1.5 g; mass of water taken in the calorimeter = 2500 g; mass equivalent of the calorimeter = 2750 g; rise in temperature = 3.0°C; specific heat of water = 1 cal/(g°C); latent heat of condensation of steam = 540 cal/g. Assume all correction factors to be negligible.
Answer in brief
GCV =10 500 cal g⁻¹; the difference GCV - NCV =0.09H×540=500 cal g⁻¹, so NCV =10 000 cal g⁻¹.
Key points
- Heat absorbed: Q=(W+w) s Δ T=(2500+2750)(1)(3.0)=15 750 cal.
- GCV (per g): (15 750)/(1.5)=10 500 cal g⁻¹
- 2H + 1/2 O₂ → H₂O: 1 g H gives 9 g water, so per g coal =9×(10.29)/(100)=0.926 g.
- 0.926×540=500 cal: the heat given back when the steam condenses is counted in the GCV (bomb cools the gases) but is lost in actual use.
The full model answer
Worked out step by step, as it should go in the answer booklet. The solution sheet follows it.
- Heat absorbed: Q=(W+w) s Δ T=(2500+2750)(1)(3.0)=15 750 cal.
- GCV (per g): (15 750)/(1.5)=10 500 cal g⁻¹
- Water formed from H: 2H + 1/2 O₂ → H₂O: 1 g H gives 9 g water, so per g coal =9×(10.29)/(100)=0.926 g.
- Latent heat: 0.926×540=500 cal: the heat given back when the steam condenses is counted in the GCV (bomb cools the gases) but is lost in actual use.
- NCV=GCV-0.09 H×540=10 500-500=10 000 cal g⁻¹. Difference =500 cal g⁻¹ (4.8%).
Coverscalorific valueGross and net calorific values of coal
All parts of Question 4
- 4(a) Ion-exchange capacity of an anion-exchange resin10 marks
- 4(b) Composition of a Ca(OH)₂ sample from its TGA10 marks
- 4(c) Gross and net calorific values of coal10 marks