And again, from (6) we have for pipes running full, In many works and reports the discharge is spoken of per minute, instead of per second: and for this unit of time we have 60 x 39.27 2356.2 as the factor outside; hence (8a) . = Q'=2356 √ × d3 cb. ft. per minute, being the formula used by Beardmore in calculating Table 5, in his work. If, as is not unusual, the diameter of the pipe be given in inches, which call d1, the above equation becomes— for this change in the units of the diameter is equivalent to multiplying the right-hand side of the equation by √12°, or √248832 = 498.83. In order, therefore, that Q' remain unaltered, we must divide the factor 2356 by this, and consequently, 2356498.83 = 4.72, as above. When the length of the pipes is given in yards, which call, as is sometimes done in practice, we have the right-hand side of the last equation multiplied by √3= = 1.732, by which, in order that Q' may remain the same, we must divide the numerical coefficient 4.72, which therefore becomes 2.725, and— = And if in gallons per minute, which call Q′′- Q′ × 6.25, then both sides must be multiplied by 6.25, and we have (8e) Q" = 17.03 Id, h gallons per minute. Again, if we find (though it cannot be said to be very common in practice) that the discharge is expressed in gallons per hour, we have, making G = Q′′ × 60, and multiplying by 60, we find, 60 x 17.03 = 1021.8 In some works we find the above stated as (15d3) which gives a less result. An approximate practical rule of very easy application can be derived from equation (8ƒ), by multiplying by 1000 and adding 2 per cent. All these expressions from (8) have reference to pipes flowing full under pressure. Other formulæ for the mean velocity, generally expressed in words, are in use amongst engineers, which are derived from (5) and those above given, namely, that the mean velocity of water in any pipe or channel that has attained a uniform velocity is nine-tenths of the square root of the product of twice the fall per mile into the hydraulic mean depth; or sometimes thus expressed, 0.92 into a mean proportional between twice the fall per mile and the hydraulic mean depth. These, which would not be given in words but to obviate any disadvantage arising from the student meeting with them so expressed, are consequences of equation (5); for the numerator and denominator of the h fraction ī may be replaced by any numbers having the same ratio. If, then, we make l = 5280, i. e. the number of feet in a mile: the numerator, which we may call f, expresses the fall per mile thus, we have, therefore, h ī = f 5280 ; and from (5) and as 1.38 = 0.92 √2, we have, by substituting this value in (9), And if the velocity be expressed in feet per minute, we have, since 0.92 × 60 = 55.2 v = 55√2ƒH, ft. per minute, the decimal being neglected. Again in Dr. Young's "Natural Philosophy" we find this rule :-"The square of the velocity in any measures per sec. is equal to the product of the fall in 2800 yards into the hydraulic mean depth, all in the same units. For if fi be the fall in 2800 yards, and ƒ that in one mile, as above, then, since 2800 1760 = 1.59, we have f1 = 1.59 x ƒ also 1.59 1.26. If, now, we take the coefficient in equation (10) as being 0.9 instead of 0.92, we may, since 0.9 √2 = 1.27 express it thus: v = 1.27 √ƒH1. In this changing ƒ the fall per mile into f, the fall in 2800 fi yards, we have v = 1.27 X Hy, or v = 1.59 1.27 1.26√ Filly. The quotient of these numbers is so nearly unity that we may assume the equation gives the above rule, and shows that this eminent author used the same coefficient of resistance as has been deduced in equation (5), p. 204. 113. From the formula (8) to (8) for the discharge of pipes running full under pressure, we can, being given any two of the three quantities Q, the inclination, h or d, determine the other. Let it be required to find the diameter of the pipe, which, with a given inclination, shall convey a given quantity of water. Dividing equation (8) by 39.27, and squaring both sides, we have The requisite inclination is found from (11) by dividing both sides by d', and if we multiply both sides by l, we obtain h: so that if the length the water has to be conveyed be also amongst the data, we obtain the head or pressure necessary to force the given quantity along a pipe of known length and diameter (4) 3 h · (b) x 4 - 4 ft. = We cannot, however, fully determine the figure of a rectangular or trapezoidal channel from (7); solving it In this we require, in addition, to be given either S or C, and also the ratio of the slopes of the sides if it be a trapezium; moreover, S and C are so related that, with given slopes, there is a maximum value of S to every given value of C; if S exceed this maximum, the solution is impossible. 114. It is found in practice that certain soils, in every excavation for whatever purpose, require a rate of slope in the sides adapted to the degree of cohesion of the ground, to obviate the danger of slips, which occur when they are too steep: this slope of the banks is, therefore, always found amongst the requisite data in the designing of channels being trapeziums in transverse section. In order that the side slopes of channels, intended to be permanent, may stand without any masonry or dry stone pitching, they should have a slope between the rates of |