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

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```DESIGN OF OPEN-HEARTH FURNACES               259
in which H = the middle ordinate of the parabola, in this case
1 m 200 or 4.00 ft;
*; = the resultant velocity of the two jets uniting to form
the flame;
sin 5 = the sine of the resultant angle of the two jets;
2g = gravitational  constant = 2X9.81 = 19.62  in metric
units or 2X32.2 = 64.4 in English units;
ti = the temperature of the gases within the chamber;
Zm = the temperature of the gases in the jet of flame.
When a furnace is heating up, the stream of flame tends to
follow the roof, until the interior of the chamber becomes raised
to a temperature sufficient to permit it to drop, and the drop of
the flame is an index of the progress made in heating the furnace.
It is likewise desirable to be able to sinter the bottom, when the
furnace, for any reason, is cooler than usual, or the ports are
eroded. The temperature of the jet of flame, tm, may be assumed
as 1800° C. (3272° F.) and that of the gases in the chamber as
800° C. (1472° F.)=fc; velocities of 0=15 m 00, 20 m 00 and
25 m 00 (49.2, 65.6 and 82.0 ft) per second will be tried out in the
formula, solving for sin 5. The resultant angle for these velocities
will be as follows:
For v = 15 m 00 (49.2 ft) per second, 5 = 18° 12'
2; = 20 m 00 (65.6 ft) per second, 5 = 13° 33'
y = 25 m 00 (82.0 ft) per second, 5 = 10° 46'
In solving to obtain these angles
fc = 800° C. (1472° F.)    273+*, = 273+800 = 1073
2m-Z<= 1800-800 = 1000
#=lm200 (4.00ft)    20 = 2X9.81 = 19.62
The formula may now be written:
«? sin* SX1078
n-*-•*>     19.62X1000 '
which becomes                 _____
.    s       /2T794   4.684
sin 5 = A/—~— =-------.
\    v2          v
For v=15 m 00 (49.2 ft) per second, sin 5 = 0.3123
v = 20 m 00 (65.6 ft) per second, sin 5 = 0.2342
v = 25 m 00 (82.0 ft) per second, sin 5 = 0.1873```