Synopsis
This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1914 Excerpt: ...runs at a uniform depth of 1 foot, what will be the velocity V It is first necessary to find E; and as the sectional area is 4x1 = 4 square feet, while the wetted perimeter is 4 + 1 + 1 = 6 feet, we have--4-0 B-J?-0 667. Inserting this value in the old formula, and using again K = 8874, we should find--V = v/8874 x i x 0 0049 = 2899 = 5'38 feet per second, and the discharge would be--Q = 5 38 x 4 0 = 2152 cubic feet per second. To make the same calculations by the logarithmic formula, we may take, for Class 7 in Table 3, the values n = 18, m = 12, and log. n = 5 990; so that--=, and-= 5. Then the logarithm of V and of Q will be found as follows:--s = 0 0049, and log. s = 3 690196 Subtracting log. fi = 5 990000 we have (log. s--log. fi) = T700196 Multiply by § 5 9)8-500980 0-944553 0-944553 E = 0-6667, and log. E =1-823909 whose negative value is--0-176091 multiply by f 2 3)-0352182-0-117394 Add this negative quantity, or subtract 0117394 which gives log. V = 0 827159 So that V = 6-717 feet per second a = 4 00 square feet, and adding log. a = 0-602060 we get log. Q = 1-429219 So that Q = 26-87 cubic feet per second. This result agrees with Bazin's experiments, the conditions being nearly the same as in his Experiment No. 127, where s = 0-0049, R = 0-662, and V was found to be 6 71. In calculating the discharge of any circular pipe or culvert which is running full, we may remember that its sectional area will be a =.;and, as the hydraulic radius E is one-fourth of the diameter d, we may write a = 4jrEa = 12-566Ba. The logarithm of 12-566 is T099209; hence we can express the discharge Q (in cubic feet per second) by--Q = Va = 12-566VB2 or, log. Q = log. V + 2 log. E + 1-099209 We may also note that a supply of 1 cubic foot per second is equivalent to 538,...
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