Sinossi
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. 1902 Excerpt: ...calculations by the logarithmic formula, we may take, for Class 7 in Table 3, the values n = 18, m = 1-2, m 2 15 and log. u = 5990; so that--=, and-=-r. n 3 n 9 Then the logarithm of V and of Q will be found as follows:--s = 0-0049, and log. s = 3690196 Subtracting log. ft = 5-990000 we have (log. s--log. /t) = 1-700190 Multiply by; 5 9)8-500980 0-944553 0-944553 R = 0-6667, and log. R = 1-823909 whose negative value is--0176091 multiply by § 2 3)-0-352182-0-117394 Add this negative quantity, or subtract 0117394 which gives log. V = 0827159 So that V = 6717 feet per second a = 4-00 square feet, and adding log. a = 0-602060 we get log. Q = 1-429219 So that Q = 2687 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 = j. d?; and, as the hydraulic radius R is one-fourth of the diameter d, we may write a = 4irR2 = 12566R2. The logarithm of 12-566 is 1-099209; hence we can express the discharge Q (in cubic feet per second) by--Q = Va = 12566VE2 or, log. Q = log. V + 2 log. R + 1-099209 We may also note that a supply of 1 cubic foot per second is equivalent to 538,440 gallons per day of 24 hours; and the logarithm of this number is 5-731137. The daily supply of a gravitation main will, therefore, be given in gallons by--Q, = 538440 Q = 6766220 VR2 log Q, = log. V + 2 log. R + 1-099209 + 5731137 = log. V + 2 log. R + 6-830346 In the case of an open semicircular section, running full up to the diametral line, the sectional area will be exactly half as great, or a = 2ttr2 = 6-283 Ea; therefo...
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