Engineering Chemistry S Datta

FUELS AND COMBUSTION 405
Water equivalent of bomb calorimeter (W) = 2,200 g.
Rise in temperature (t 2
– t 1
) = 2.42°C; Fuse wire correction = 10 cal.
L = 580 cal/g –1 .
Sol. HCV = ( W + w)( t 2 − t 1 ) − [ acid + fuse ] correction
m
(2,200 + 530) × 2.42 − [50 + 10] –1
= cal/g
0.92
= 7168.5 cal/g
Net Calorific Value (NCV)
= (HCV – 0.09 H × L) cal/g –1
= (7168.5 – 0.09 × 6 × 580) cal/g –1
= 6,855.3 cal/g.
Example 2. A sample of coal was analysed as follows:
Exactly 2.5 g was weighed in a silica crucible, after heating for 1 hr. at 110°C the residue
was weighed to be 2.415 g. Next the crucible was covered with a rented lid and strongly heated
for exactly 7 mins. at 1000°C. The residue was weighed to be 1.528 g. Then the crucible was
heated without cover until a constant weight to 0.245 g was obtained. From the above data
calculate the proximate analysis of coal.
Sol. Moisture in the sample = (2.5 – 2.915)g = 0.085 g
So, % moisture = 0 . 085 × 100 = 3.4
25 .
So, amount of volatile carbonaceous matter (VCM) in the sample = (2.415 – 1.528) g
= 0.887 g.
0.
887 × 100
So, % VCM = = 35.
48%
25 .
Weight of ash
% Ash =
Weight of coal × 100
= 0 . 245 × 100 = 35.
48%
25 .
Fixed carbon = (1.528 – 0.245)g = 1.283 g
1.283 × 100
% Fixed carbon = = 51.32%.
25 .
Example 3. On burning 0.83 g of a solid fuel in bomb calorimeter, the temperature of
3,500 g of water increased from 26.5°C to 29.2°C. Water equivalent of calorimeter and latent
heat of steam are 385 g and 587 cal/g, respectively. If the fuel contains 0.77% H, calculate HCV
and NCV.
Sol. HCV = ( W + w)( t 2 − t 1 )
m
= ( 385 + 3500 )( 29 . 2 −26 . 5 )
−
= 12638 cal g 1
083 .
NCV = (HCV – 0.09 H × L) cal g –1
= (12638 – 0.09 × 0.7 × 587) cal g –1 = 12601 cal g –1 .