basin, and the lower, of a given area, receive the discharge; required the time in which it reaches the level of the upper basin or a given height. This problem is the inverse of that in § 89, in which the surface of the water descended with a uniformly retarded motion. In the present case, the surface of the lower basin rises with a uniformly retarded motion.

Let H represent the charge AC (Fig. 32) at the commencement, and h the

A

F

These latter formulæ are of some importance: they serve to determine the time in which canal locks, &c., may be filled, and to assign the area of sluice-way required to fill a certain volume in a given time.

96. The Level of the Water being variable in each Vessel.-We now come to the third case that can arise, namely, when two reservoirs of different level communicate with each other, each being limited in area and receiving no supply, and thus one surface descends as the other rises. Such is the case of the two basins K and L (Fig. 33), communicating by a wide pipe EF, provided with a sluice-door or cock at G. Before the opening of this sluice-door the level of the water is at AB in the first reservoir, and CO in the second. At the end of a certain

АП

time after the opening of the communication it has descended to MN in the first, and has ascended to PQ in the second. It is required to determine the relation between these M two heights at a given time, or, vice versa, from

B

N

Р

K

the given difference in the respective heights,

G

C

Fig. 33.

to determine the time

corresponding to a given discharge.

Let f equal the given time, BE = H, CF = h, NE = x, PF = y, A = horizontal section of the first vessel, and B that of the second, s = section of the pipe of communication: in the coefficient m we must include the resistance of the water passing through this pipe. Whilst the water has risen in the second basin by the quantity dy, during the instant dt, it will have lowered in the other by dx; and remembering that x diminishes while y and increase, we have Adx Bdy and (§ 14),

=

The first equation being integrated, remembering that H, y=h: we have—

when x =

and substituting this value of y in the preceding equation

I

(b), integrating, and observing that H = x when t = o, we have

t=

2A √B

mS√ 2g (A + B) {√B (H-h)- √ (A + B) x − AH-Bh

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

x = y =

AH+Bh
A+B

;

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) the time 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

ON CHAPTERS I. AND II.

97. THE following Rules, approximately true, may be found useful in every-day practice. It is important to know how they are derived, and thus be able to reproduce them, as no book may be at hand for reference, and the memory may fail. They all depend upon the volume and weight of water in relation to the weights and measures of the United Kingdom.

The statical pressure, i. e. of still water, in any pipe, or on the bottom of a tank, is qp, equal to 3lbs. per square inch for every 7 feet head. Thus a main laid across a valley is, let us suppose, at the lowest part, 130 feet below the surface from which it is supplied. From the rule 1307 18.57 and 3 x 18.57 55.7 lbs. per square inch; this result is about theдth part too small, it should be 56.26 lbs.

=

=

=

If, on the other hand, we had a known pressure of water, suppose of 38.5 lbs. per square inch, to determine the vertical head in feet; by the Rule 38.5 3 12.833 and 7 × 12.833 = 89.83 ft. The exact result is 88.956 ft, so that when the pressure is given, the result is about theth part too large.

Since a cubic inch of water weighs 252.458 grains, a column one foot high and one square inch in base weighs 3029.496 grains, which, divided by 7000 to reduce it to

or 3.03 lbs. for 7 feet, which is one per

In the same manner the longitudinal bursting pressure of water in a pipe per inch of length is found by multiplying the diameter in inches into the pressure per square inch-that is, × H ft. x D ins. Thus, if the diameter of a pipe be 26 inches, and H, as above, we have (using the exact result) 26 × 56.26 = 1462.8 lbs.

When computing the resistance against the plunger of a forcing pump in motion, it is usual to take half the height in feet for the pressure per square inch-that is, 3.5ths of a lb. av. per ft. of height.

7

Thus, to force water

to the height of 47 feet we have 23 lbs. per square inch resistance; this gives a fair allowance for friction, passing through valves, &c.

In pumping engines for mines it is useful to be able readily to compute the total weight of water in the vertical pipe at any lift, from that per yard or per fathom (= 6 feet). For this purpose;

Square the diameter in inches and the result is nearly equal to the lbs. per yard vertical, and for the fathom multiply this by 2; or per foot use 3 as a divisor.

=

Thus, in a pipe 13 inches in diameter and rising 40 fathoms we have 169 (= 132) x 2 = 338 lbs. and 40 × 338 × 13520 lbs. The exact multiplier is 2.0454, giving a result a little more than 2 per cent. greater than the approximate rule. In all these the number of gallons is found by cutting off, from the number expressing the lbs. weight, one figure for decimals; thus in the length of 40 fathoms of the above pipe we have 1352 gallons, to which, adding 2 per cent., or 27, we have 1380 gal