Homework SourseDetermined high heating value and low heating value from giv
Determined high heating value and low heating value from given information. (please show your work)
| Name | Mass(g) | Qvad (BTU/lb) | Moisture Content | Ash yield |
| Benzoic Acid | 1.0073 | 11230.5 | 0 | 0 |
| Coal 4W | 0.1022 | 10737.9 | 0.54 | 3.89 |
| Torrified Biomass | 0.1426 | 11321.4 | 0.20 | 1.50 |
Solution
Solution:-
A = Ash content in percentage
M = Moisture content in percentage
Hvap = Latent heat of vaporisationof water
High Heating Value = (100 - A - M)% of Mass of Fuel x Qvad
Lower Heating Value = Higher Heating Value - M % of Mass of Fuel x Hvap
Hvap = 970.335 Btu/lb (From standard Tables)
(a) For Benzoic Acid,
A = 0 and M = 0 Hence, its Higher heating value will be same as Qvad that is 11230.5 Btu/lb
Also, due to absence of moisture content its Lower heating value will be same as higher heating value that is 11230.5 Btu/lb
(b) For, Coal 4W M = 0.54 , A = 3.89, Mass of Fuel = 1.022g = 0.000225312432 lb
High Heating Value = (100 - A - M)% of Mass of Fuel x Qvad
= (100 - 3.89 - 0.54) x 0.000225312432/100 x 10737.9
= 2.3122 Btu (this is for 0.000225312432 lb of fuel)
HHV = 2.3122 Btu / 0.000225312432 lb =10262.211 Btu/lb
Lower Heating Value = Higher Heating Value - M % of Mass of Fuel x Hvap
= 2.3122 - 0.54 x 0.000225312432/100 x 970.335
= 2.3110 Btu (this is for 0.000225312432 lb of fuel)
LHV = 2.3110 Btu / 0.000225312432 lb =10256.2954 Btu/lb
(b) For, Torrified Biomass M = 0.20 , A = 1.50, Mass of Fuel = 0.1426g = 0.000314379186 lb
High Heating Value = (100 - A - M)% of Mass of Fuel x Qvad
= (100 - 1.50 - 0.20) x 0.000314379186/100 x 11321.4
= 3.498 Btu (this is for 0.000314379186 lb of fuel)
HHV = 3.498 Btu / 0.000314379186 lb =11128.9362 Btu/lb
Lower Heating Value = Higher Heating Value - M % of Mass of Fuel x Hvap
= 3.498 - 0.20 x 0.000314379186/100 x 970.335
= 3.4369 Btu (this is for 0.000314379186 lb of fuel)
LHV = 3.4369 Btu / 0.000314379186 lb = 10932.624 Btu/lb
