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.