mile, Table 30 gives, opposite the radius of 1.1 (which is the nearest to the one we require), the mean velocity of 73·9 feet per minute; hence the discharge is 73.9 × 27·74 = 2050 cubic fee per minute. With a very short channel, allowance should b made for velocity at entry, as explained in (56). Table 30 may also be applied to the calculation of the dis charge, &c., of common pipes running full, or to those of & square or other section, for an illustration of which see (36), also to culverts, &c., partially filled, see (62). (60.) "Openings of Bridges, &c.”—The head lost by a stream in passing through a bridge is principally that due to velocity alone, the length of the channel being in most cases so short as to have little influence on the discharge. The head for velocity may be calculated by Table 15: say we take the case (58) of the stream (Fig. 40) discharging 5575 cubic feet per minute, and passing through an opening at a bridge, say 8 feet wide and 3 feet deep. The area being 8 x 3 = 24 square feet, the velocity 5575 will be = 3.87 feet per second, which, with pointed 24 × 60 piers (Fig. 32) will require by Col. B of Table 15, 3 inches head (A, B in Fig. 37). But, the stream approaches the bridge 89.2 with a mean velocity of 89.2 feet, or a maximum (50) of ⚫84 = 106 feet per minute, or 1·77 foot per second, the head due to which by the same Table is ğ inch. The head at the bridge is, therefore, reduced to 323 inches; with square piers (Fig. 33), the head by Col. C is 33 inches, or at the bridge. 3231 inches. (61.) "Submerged Openings."-The velocity of discharge through a submerged opening A (Fig. 43) is governed by the difference of the level of water at the two sides of it, or by H, and is not affected by the depth below the surface at which it is placed. Table 15 will give the velocity with small heads: thus an aperture 2 feet × 1·5 foot = 3 square feet area, and with H = 5 inches, would, by Col. D of Table 15, discharge 3.2893 x 3 = 9·87 cubic feet per second. (62.) "Discharge by Egg-shaped Culverts."-The discharge of culverts of the common oval or other forms may be calculated by the preceding rules, or by Table 30. The proportions of culverts are arbitrary. Fig. 44 shows a good form, and Table 16 gives the general sizes, areas, &c., when filled to two different depths, so as to adapt the Table to the varying requirements of practice. Say we take the case of a 5-feet culvert ths full of water or 4 feet 2 inches deep, with a fall of 10 inches per mile, then, by Table 16, the hydraulic radius is 1.105, and the area of waterway 10.82 feet; by Table 30 we find that with 1·1 hydraulic radius, and a fall of 10 inches per mile, the mean velocity is 73·9 feet, and the discharge 73.9 × 10·82 = 800 cubic feet per minute. (63.) With very short culverts, allowance must be made for the velocity at entry by Table 15, &c.; thus, in the case just given, if the culvert had been only 45 yards long, the fall due to friction alone would have been, by Table 30, equal to ⚫00568 × 45 = 255 or inch; the mean velocity is 73.9 60 = 1.23 and the maximum 1.23 = 1.46 foot per second, the head due to which by Col. C of Table 15 is about = inch. The total head is therefore, of an inch. To calculate with precision the discharge of short culverts, with a given fall, the method explained in (56) should be followed. (64.) "Head for very Low Velocities."-In ordinary cases Table 30 gives results sufficiently correct for practical purposes with great facility, but with very small velocities experiment has shown that the head is considerably greater than that Table would give. In such cases the more laborious and refined formula of Prony, Saint Venant, and Eytelwein give more correct results. A comparison of these three rules with 96 experiments on the discharge of rivers shows that Eytelwein's rule agrees best with 38 experiments, Saint Venant's with 32, and Prony's with 26. The following is a modification of Eytelwein's rule: C = In which L A Р 896400 × F x A = LX P length of the channel in yards. = cross-sectional area of the stream in square feet. the perimeter, or border of the channel in feet. F = the fall, or difference of level at the two ends of = the channel in inches. C = cubic feet discharged per minute. (65.) Thus, say that we require the discharge by the channel, Fig. 40, 1 mile long, with a fall of 1 inch only, then L = 1760, A = 62.5, P = 31.2, as in (58), and F = 1, and the discharge will be 896400 × 1 × 62.5 +42·8)3 — 6·534) × 62·5 = 1629·8 1760 × 31.2 cubic feet per minute. We may compare this result with that given by the rule in (55), by which the discharge comes out 874520 × 1 × 62.5 1972 1629 1760 x 31-2)+ × 62.5 = 1972 cubic feet minute = per = 1.21, or 21 per cent. difference. But with an increased = head, the difference becomes less, and is reduced practically to nothing with large heads, as shown by Table 17. TABLE 17.—Of the DISCHARGE of an OPEN CHANNEL, Fig. 40, calculated by DIFFERENT RULES. This shows that in all cases where extreme accuracy is desired, the rule in (64) should be used; but that where the fall exceeds 8 or 10 inches per mile, Table 30 gives results sufficiently correct for most practical purposes. (66.) When the discharge is given, to determine the fall, the rule becomes Thus the fall for friction with the same channel, Fig. 40, 2000 yards long to deliver 3000 cubic feet per minute would be +6·584) — 42·8) × 2000 × 81·2 3000 62.5 896400 × 62.5 = 3.26, or 31 inches. Adding the head due to velocity at entry (51), the mean velocity is 48 3000 57 feet per minute, or 62.5 .95 foot per second, the head for which by Col. C of Table 15 is about inch; the total head is therefore 31+1=3 inches. (67.) Table 18 has been calculated by the following modification of Eytelwein's rule: the mean velocity over the whole area in feet per second. area in square feet R the hydraulic radius in feet, or border in feet By this Table approximately correct results may be obtained with less labour than by the rules. 1st. To find the Velocity.-Multiply the area of the channel in square feet by the fall in inches, and divide the product by the border in feet multiplied by the length of the channel in inches: find the nearest number thereto in Col. B of Table 18, and oppo |