THE THEORY OF JETS
69
gas and of the air multiplied by coefficients proportional to their
weight. Now the ratio of the weights (1) in this case is equal to
1.81 X1.069^1.9349 = 37.26 per cent
2.525X1.29 3.2575 62.74 per cent
Therefore for Furnace No. 20, for example, the result is
For the air: 21.8X0.6274 = 13.6773
.For the gas: 29.6X0.3726=11.0290
The diagonal of the parallelogram for these two values is equal
to 24 m 50 per second at an inclination of 27°. The results, which
will be obtained for the other furnaces by similar computations,
are given in the following table:
Number
Average
Average
of
Angle
Velocity
Furnace
ex..
V.
Degrees
Meters per Second
20
27
24.00
21
29
21.20
23
33
17.30
24
25
24.50
25
29
21.00
Inserting the above values in the formula
?;2 sin2 a 273+*,
// --- ~~ pj ~ /N ™~ *
with the values of v and a. given in the table and assuming that
kn=1850° and that H equals the difference in level between the
sill of the gas port and the tapping hole, the temperature U at
which the stream of flame drops to the level of the hearth at the
tapping hole may be computed as follows, in order to determine
the temperature to which the gases filling the furnace must be
raised before it becomes possible to start making the bottom:
(1) It should be remembered that at 0° the volume per second of the gas is
1.81, and its specific weight is 1.069. The volume of the air is 2.525 and its
specific weight 1.29.
