# Full text of "The Flow Of Gases In Furnaces"

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```DESIGN OF HOT-BLAST STOVES

309

These values do not allow for heat loss; that is, this
volume of brickwork, when heating, would have a temperature
change from 20° to 50° greater than when on blast, as the
100° change is the change covering the heat transferred to the
blast.

The temperature of the blast increases 830° in the checker.
The average rate of increase assumed will determine the time
allowed for heating and, with the blast volume, the space required
in the checkerwork. Assuming a heating rate of 100° per second,
the time necessary will be 8.3 seconds. Slower or faster rates of
heating will correspondingly affect the time as well as the apuoo
required to contain the blast while heating. The volume of froo
air blown per second is 16 m3 78, its average temperature in tho
checker is 485°, and it is under a pressure of one atmosphere, tho
volume under these conditions being 23 m3 33, which, multiplied
by 8.3, the heating time, fixes the space required as 193 m;j 70.
The total checker volume will be, therefore, 193.70+368.80 =
562 m3 50.

The brick coefficient = 368.80 -*• 562.50 = 0.6557.
The pass coefficient = 193.70-4- 562.80 = 0.3443.

The side of the square for the pass unit = V0.3443 = 0.5868.
The portion of the unit square occupied by the brickwork will
be 1.0000-0.5868 = 0.4132.
As the cooling time of the stove is one hour, it is not desirable
to make the wall thickness between the passes greater than 75 mm;
therefore the side of the unit square will be 75 — 0.4132= 181.5 mm
or, say, 180 mm. The diameter of the square pass will be
180-75 = 105 mm = 4.125 ins.
The area occupied by a checker unit = 0. ISO2 = 0 rn2 0324
The area occupied by the pass          = 0.1052 = 0 rn2 0110
The area occupied by brick              =               Om2 0214
The lineal amount of checker required, based upon the pass
= 193.70-0.0110=17,609 m, or, if based on tho brick = 368.80-f-
0.0214 = 17,230 m. Using the largest of these values arid assuming
a checker height of 25 m, the number of passes = 17,609-r-25.00
= 704.```