If it were required to find the time in which the two surfaces would be at the same level, we should have from (c)—

and this value of x being substituted in the above expression for t, will give

From whence it is evident that for the same value of (H – h) t is the same whether A be the horizontal section of the basin that lowers, and B that of the other, whose surface rises, or vice versa, B that which falls, and A that which rises.

EXAMPLES AND PRACTICAL APPLICATIONS.

CHAPTERS I. AND II.

97. Questions solved by means of the formula m S✓ 2gH=Q, the charge on the centre being represented by H.-In order to obtain a comparative view of the effects resulting from the use of the different coefficients for the discharge through orifices, given in §§ 18 to 43, to which we first confine our attention, let us take a circular orifice of 0.25 ft. in diameter, the area S being therefore 0.252 x .7854.04909 sq. ft., and determine: First, the discharge through it in some given time, as 40 minutes, with a constant charge of, suppose, 9 ft. above the centre of the orifice; and, Secondly, with the same orifice and charge, seek the different intervals of time required to discharge a given volume of water, as 1000 cubic feet. As the charge is so great compared with the diameter in the above data, we may use the formula (§ 14)—

Q = mST √2g√ H,

in which H is the charge on the centre. mentioned above we calculate the value of

ST √29 √ H,

In the first case

which becomes .04909 × 40 × 60 × 8.024 × 3 = 2836.067 cb. ft., and multiply it by the several values of m, as is done below. For the second case we have

the value of the first factor of the left-hand side is

which must be divided also by the successive values of m to obtain T, the time required to discharge the given quantity.

(11.) Required the discharge in six minutes, through a rectangular sluice 3 ft. by 1 ft., the side 3 ft. long being horizontal, the depth to the sill from the surface being 7 ft., and m being equal to 0.62.

Here 0.62 × 3 sq. ft. x 8.024 √6.5 = Q,

and √6.5 = 2.5495 may be taken equal to 2.55,

hence and

Q=38.06 cb. ft. per sec.,

Q=6× 60 × 38.06-13701.6 cubic feet.

X

(III.) A reservoir having at full water a depth of 40 feet over the centre of the discharging sluice, whose area is 2 ft. horizontal by 1.5 ft. vertical when fully opened :-Required the discharge at that depth, and also when the water has sunk to the heads, 30 ft., 20 ft., and 10 ft., the value of m being taken at 0.62 in each case,- -we have

S = 1.5 × 23 sq. ft., and √40, 30, √20, and √10, being respectively 6.324, 5.477, 4.472, 3.162. We must multiply these numbers successively by—

0.62 x 8.024 × 3 = 14.92464,

which is the same in each. Hence, for 40 ft. head the discharge is 94.384 per sec.; for 30 ft., 81.742 cb. ft.; for 20 ft., 66.743 cb. ft.; and for 10 ft., 47.192, or the half of that for 40 ft.; 3.162 being necessarily half 6.324, as they are the roots of numbers in the ratio of 1 to 4. This question points out the fact that leakages of sluices in lock-gates, &c., increase far less rapidly than the head, being, in fact, as the square roots of the charges. (Vide Smeaton's Reports, vol. i., pp. 196-9.)

(IV.) What is the discharge through a circular pipe 4 ft. diameter in the embankment of a reservoir, the head upon the centre being 90 ft., m being taken equal to 0.60? In this

case

hence

S=(4) × 0.7854 = 12.5664 and

90=9.487,

0.6 × 12.5664 × 8.024 × 9.487 = 573.9 cb. ft. per sec.

(v.) A rectangular sluice, sides 4 ft. horizontal and 3 ft. vertical, having a charge of 20 ft. on the centre, is raised 1.5 ft.: required the discharge per sec., and also when fully opened. We have the value of S in the first instance one-half that in the second, but the heads are 20.75 ft. and 20 ft. respectively; and taking m = 0.62, we have first

0.62 × 8.024 × 6 × √ 20.75 (= 4.5552) = 135.97 cb. ft.;

=

and secondly,

L

0.62 × 8.024 × 12 × √ 20 (= 4.472) = 266.97 cb. ft. The double of the former would be 271.94 cb. ft.

(VI.) In cities in which water is supplied at high pressure, and constant service, it is sometimes usual to give the water to manufactories and works through a very small orifice, perforated in a disc, which is closed up secure from any possibility of unfair interference. Calculate the discharge through an orifice .0089 in diameter for 24 hours, the head being 129 ft. and m equal to 0.62; we have—

log. m + log. S+ log. (2 g) + log. H+ log. 86400" = log. Q, the log. of S being 2 log. .0089 + log. .7854,—we have thusQ'=303.655 cb.ft.

(VII.) Suppose the pressure on the mains to be 150 ft. of water, and the diameter of the orifice .02 ft.: required the quantity delivered in 24 hours, the coefficient of discharge being 0.62. The ✔ 150 being equal to 12.247, and—

we have

S=(.02)2x.7854 = .00031416,

Q=24" x 3600" x 0.62 x .00031416 × 8.024 × 12.247

(VIII.) What must be the diameter of the orifice to give 600 cb. ft. per diem, the head on the main being 100 feet? Here

As the exact adjustment of this diameter would be nearly impossible, the process is somewhat tentative.

(IX.) In the sluices constructed in tidal harbours for scouring away at low water the silt that generally accumulates in them, we obtain examples on a very large scale of the discharge of water through orifices.

This simple remedy for a defect that had rendered nearly useless some of the most important tidal harbours on the coast of England, not having the advantage of any sufficient natural streams to keep them open, was introduced by the great Smeaton from his personal observation of the practice in the Low Countries (vide Reports, vol. ii. p. 202-209). A bank thrown across

some part covered at high tide impounded the water allowed to enter during the rise of the tide, and which at low water is discharged very rapidly through sluices constructed in this embankment, the sills of which are at low water of springs, or as low as possible.

The practice subsequently fell into disrepute, as it was found that the area of the back-water was itself soon silted up; but the same engineer adopted the simple and efficient remedy of dividing the back water into two separate areas by a second bank at or about perpendicular to the first mentioned, and by occasionally using one of these to cleanse the other, they were both, as well as the harbour itself, kept clear. Ramsgate and Dover are well-known examples (vide Smeaton's Reports and Sir J. Rennie on Harbours); from which last-mentioned work we take an example from the description of Hartlepool Harbour, on the coast of Durham.

Each sluice was 3 feet wide and 6.33 feet high, having a charge estimated at 10 ft. on the average. From the detailed plans of these works given by Sir J. Rennie, we may consider the coefficient 0.600 applicable; hence

0.600 × 3 × 6.33 × 8.024 x 10 = 289.14 cb. ft. per sec.

is the discharge for each sluice; and as it is also stated, that the total area of the scouring sluices was 366 sq. ft., of which 24 sq. ft. was given by four sluices, each 3 ft. by 2, in the lock-gates, which communicated with the back-water or slake, we have 342 sq. ft. left for those through the embankment; and each of these being 3 × 6.33 19 sq. ft., we have their number 18, i. e. 34219; and the discharge for one being 289.14 cb. ft., the total discharge is 18 x 289.14 = 5204.52 cb.ft. per sec., or 312271 cb. ft. per minute. Now the back-water containing 15420000 cb. ft., it could be discharged in about 50 minutes (15420000 312271). It is essential that the back-water should be discharged rapidly before the rising tide diminishes the force of the artificial scouring action.

(x.) The widely different statements as to the efficiency of hydraulic engines as prime movers, some being asserted to give as high as 80, and others 60 per cent. of the power used, may in some cases, perhaps, be traced to a false estimate of the quantity of water discharged upon the prime mover used; for unless this be actually gauged, it must be calculated, and some coefficient used. In the case of undershot wheels with sloping sluices, as in Poncelet wheels, the bottom and sides being in continuation of the channel of supply, the coefficient is