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Full text of "The Flow Of Gases In Furnaces"

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Allowing 2 kg 50 for impressing velocity upon the air, the
formula d = (vt2 -r-20)At can be transposed, as follows:

For the case in hand, <5 = 2 kg 50
and                             v = 14 m 325 per second (47.00 ft per second)
The air and gas velocities have to be corrected by a coefficient,
according to their relative masses and volumes. These coefficients
are derived as follows:
5.32X1.125 = 5.9855 = 0.3523 for gas
8.51X1.293~11.003~0.6477 for air
Therefore, the value of the air component will be
14.325X0.6477 = 9.28.
The area of the air-port will be
45.99-^-14.325 = 3 m2 21 = 34.55 sq ft.
Considerable latitude may be allowed in proportioning the gas-
and air-port velocities and angles, to obtain the necessary resultant
angle and velocity. Theoretically, the entire velocity may be
supplied by either the gas or the air, or it may be divided between
them. The ports may both be given the resultant angle, or the
port angles may differ, provided the angular velocity components
of the air and the gas give the angle and velocity necessary for the
resultant. In practice there are certain structural limitations in
regard to the location of the ports. These limits must be con-
sidered. In this case it will be assumed that the port angles are
the same, being the resultant angle. The resultant velocity is
20 m 00 (65.62 ft) per second. Under these conditions the
velocity component for the gas will be
20.00-9.28 = 10.72;
from which the gas-port velocity will be
10.72 *- 0.3523 = 30 m 43 per second (99.84 ft per second).