Calorific Value Computation Based on Ultimate Analysis data
DETERMINING HEATING VALUES OF FUELS:-
The heating values of a fuel can be determined either from chemical analysis or by burning a sample in a calorimeter.
In the former method the calculation is based on an ultimate analysis. It reduces the fuel to its elementary constituents of carbon, hydrogen, oxygen etc.
The proximate analysis determines the percentages of moisture and carbon.
The ultimate analysis when resolves the fuel into its elementary constituents doesn’t reveal how they combine in the fuel. It is given on both moist and dry fuel basis.
When an analysis is given on a moist fuel it is converted to dry basis by dividing the percentage of various constituents by one minus the percentage of moisture.
CALCULATION FROM AN ULTIMATE ANALYSIS
The most commonly used formula is the DULONG’S FORMULA.
Heat units in B.t.u per pound of dry fuel=
Where C is carbon,
H is hydrogen,
O is oxygen and
S is sulphur
Heating value per pound of dry coal=
= 14765 B.t.u
This method gives a satisfactory results. It may be made on a moist or dry basis. The method of converting from a moist to a dry basis is same as that of ultimate analysis.
To compute the efficiency of a boiler the heating value is determined with the help of a fuel calorimeter. In this apparatus the fuel is burned and the heat generated is absorbed by water.
The calorimeter which gives the best results is M.PIERRE.MAHLER.
If the result is not correct we use the PFAUNDLER’S METHOD.
It can be expressed as
Where C= correction in degree centigrade
N= number of intervals over which the correction is made
R= initial radiation
R’= final radiation
T= average temperature for the initial radiation
T’=average temperature for the final radiation
T”=average temperature over period of combustion
EXAMPLE 1. ASSUME A BLAST FURNACE GAS, THE ANALYSIS OF WHICH IN PERCENTAGE BY WEIGHT IS OXYGEN=2.7, CARBON MONOXIDE=19.5, CARBON DIOXIDE=18.7, NITROGEN= 59.1.
Sol:- Here the only combustible gas is the carbon monoxide and the heat value will be
0.195 x 4450 = 867.75 B.t.u per pound
The total volume of air required to burn one pound of this gas will be
0.195 x 30.6 = 5.967 cubic feet
EXAMPLE 2. ASSUME A NATURAL GAS THE ANALYSIS OF WHICH IN PERCENTAGE BY VOLUME IS OXYGEN= 0.40, CARBON MONOXIDE= 0.95, CARBON DIOXIDE= 0.34, OLEFIANT GAS(C2H4)= 0.66, ETHANE(C2H6)=3.55, MARSH GAS(CH4)= 72.15 AND HYDROGEN= 21.95.
Sol:- All except oxygen and carbon dioxide are combustible and the heat per cubic foot will be
From CO= 0.0095 x 347= 3.30
C2H4= 0.0066 x 1675= 11.05
C2H6= 0.0355 x 1862= 66.10
CH4= 0.7215 x 1050= 757.58
H= 0.2195 x 349= 76.61
B.t.u per cubic foot= 414.64
The total air required for combustion of one cubic foot of the gas will be
CO= 0.0095 x 2.39= 0.02
C2H4= 0.0066 x 14.33= 0.09
C2H6= 0.0355 x 16.74= 0.59
CH4= 0.7215 x 9.57= 6.90
H= 0.2195 x 2.41= 0.53