Part 20 (1/2)
The gas is then driven over into the pipette C and a similar operation is carried out The difference between the resulting reading and the first reading gives the percentage of oxygen in the flue gases
The next operation is to drive the gas into the pipette D, the gas being given a final wash in E, and then passed into the pipette C to neutralize any hydrochloric acid fuiven off by the cuprous chloride solution, which, especially if it be old, ases andon the burette less than the true amount
The process allol solution will also absorb carbon dioxide, while the cuprous chloride solution will also absorb oxygen
As the pressure of the gases in the flue is less than the ath the pipe connecting the flue to the apparatus The gas may be drawn into the pipe in the way already described for filling the apparatus, but this is a tedious method For rapid work a rubber bulb aspirator connected to the air outlet of the cock G will enable a new supply of gas to be drawn into the pipe, the apparatus then being filled as already described Another foras fro a fresh supply for each sample
The analysis made by the Orsat apparatus is voluht is required, it can be found from the volues by voluas, and divide the products by the sues by weight Forthe even values of the hts of the gases appearing in an analysis by an Orsat are:
Carbon Dioxide 44 Carbon Monoxide 28 Oxygen 32 Nitrogen 28
Table 33 indicates the as analysis into an analysis by weight
TABLE 33
CONVERSION OF A FLUE GAS analYSIS BY VOLUME TO ONE BY WEIGHT
Colus:
A: analysis by Voluht D: analysis by Weight Per Cent _____________________________________________________________________ | | | | | | | Gas | A | B | C | D | |________________________|_______|___________|________|_______________| | | | | | | | | | | | | | | | | | 5368 | | Carbon Dioxide CO_{2} | 122 | 12+(216) | 5368 | ------ = 177 | | | | | | 30228 | | | | | | | | | | | | 112 | | Carbon Monoxide CO | 4 | 12+16 | 112 | ------ = 4 | | | | | | 30228 | | | | | | | | | | | | 2208 | | Oxygen O | 69 | 216 | 2208 | ------ = 73 | | | | | | 30228 | | | | | | | | | | | | 22540 | | Nitrogen N | 805 | 214 | 22540 | ------ = 746 | | | | | | 30228 | |________________________|_______|___________|________|_______________| | | | | | | | Total | 1000 | | 30228 | 1000 | |________________________|_______|___________|________|_______________|
Application of Formulae and Rules--Pocahontas coal is burned in the furnace, a partial ultien 425 Oxygen 26 Sulphur 16 Ash 60 B t u, per pound dry 14500
The flue gas analysis shows:
_Per Cent_
CO_{2} 107 O 90 CO 00 N (by difference) 803
Deterht (see Table 33), the aht of air per pound of fuel, the weight of flue gas per pound of coal, the heat lost in the chirees Fahrenheit, and the ratio of the air supplied to that theoretically required
Solution: The theoretical weight of air required for perfect combustion, per pound of fuel, from formula (11) will be,
(821 026 016) W = 3456 (---- + (0425 - ----) + ----) = 1088 pounds
( 3 8 8 )
If the aas is 80 per cent, which would allow for 21 per cent of unburned carbon in teras per pound of carbon burned will be from formula (16):
11 107 + 8 90 + 7(0 + 803) W = --------------------------------- = 2342 pounds 3(107 + 0)