To find the quantity of Water that will be discharged through an orifice, or pipe, in the side or bottom of a Vessel.

Area of orifice, sq. in. X

No. corresponding to height of surface above orifice, as per table

= Cubic feet discharged per minute.

 Height ofSurface aboveOrifice. Multiplier. Ft. 1 2.25 2 3.2 4 4.5 6 544 8 64 10 7 1 12 7.8 14 84 16 9.
 Height ofSurface aboveOrifice. Multiplier. Ft. 18 9.5 20 10. 22 10.5 24 11. 26 11.5 28 12. 30 12.3 32 12 7 35 13.3
 Height ofSurface aboveOrifice. Multiplier. Ft. 40 14.2 45 15.1 50 16. 60 17.4 70 18.8 80 20.1 90 21.3 100 22.5

To find the size of hole necessary to discharge a given quantity of Water under a given head.

Cubic ft. water discharged / No. corresponding to height, as per table = Area of orifice, sq. in.

To find the height necessary to discharge a given quantity through a given orifice.

Cubic ft. water discharged / Area orifice, sq. inches. = No. corresp. to height, as per table.

The velocity of Water issuing from an orifice in the side or bottom of a vessel being ascertained to be as follows;

√Height ft. surface above orifice X 5.4 =

Velocity of water, ft. per second.

√Height ft. X Area orifice, ft. X 324 =

Cubic ft. discharged per minute.

√Height ft. X Area orifice, ins. X 2.2 = Do. Do.

It may be observed, that the above rules represent the actual quantities that will be delivered through a hole cut in the plate; if a short pipe be attached, the quantity will be increased, the greatest delivery with a straight pipe being attained with a length equal to 4 diameters, and being 1-3 more than the delivery through the plain hole; the quantity gradually decreasing as the length of pipe is increased, till, with a length equal to 60 diameters the discharge again equals the discharge through the plain orifice. If a taper pipe be attached the delivery will be still greater, being 1½ times the delivery through the plain orifice; and it is probable that if a pipe with curved decreasing taper were to be tried, the delivery through it would be equal to the theoretical discharge, which is about 1.65 the actual discharge through a plain hole.

To find the quantity of Water that will run through any orifice, the top of which is level with the surface of water as over a sluice or dam.

√Height, ft. from water surface to bot-tom of orifice or top of dam

X

Area of water passage, sq. ft.

= Cub. ft. discharged per minute. Or,

Two-thirds Area of water passage, sq. ins X No. corresponding to height as per table, = Cub. ft. discharged per minute.

To find the time in which a Vessel will empty itself through a given orifice.

√Height ft. surface above orifice x Area water surface, sq, ins. /Area orifice, sq. in. X 3.7

= Time required, seconds.

The above rules are founded on Bank's experiments.