SPOUTING FLUIDS.] MECHANICAL PHILOSOPHY HYDRODYNAMICS. 



765 



tained in the time of falling from D to A, and to multiply 

 2 B D by the resulting number, in order to get the 

 horizontal range A P. Now the times of falling through 

 any spaces are as the square roots of those spaces 

 (DYNAMICS, page 736); therefore the required multiply - 



ing number is 



DA 



. : 



consequently, 



/DA 

 Range AP = 2BD Xy/ B g = 2 V BD'DA. 



If, therefore, with centre C a semicircle be described 

 upon A B, and the perpendicular D F be drawn, the 

 range of A P will be twice the length of this perpendicu 

 lar. (Euc., XIII., Book VI.) 



In like manner, the range of the fluid spouting from E 

 will be twice the length of the perpendicular E K ; and 

 as these perpendiculars are equal, being equidistant from 

 C, the ranges from D and E are equal. 



As the perpendicular C L from the centre C is the 

 longest that can be drawn from A B to the arc of the 

 semicircle, the range A R = 2 C L, of the fluid spouting 

 from C, is the greatest of all ranges. 



These theoretical results are fully confirmed by ex- 

 periment, when the orifices in the vessel are small, so 

 that the vena contracta may be sufficiently near the 

 yessel to be regarded, without sensible error, as at the 

 orifice itself. 



As the perpendicular D F is the sine of the arc B F, 

 we see that the horizontal range from any orifice, is twice 

 the sine of the arc of a circle whose diameter is the 

 depth of the fluid, and whose versed sine is the depth 

 of the orifice. By the depth of the fluid may be under- 

 stood the depth of the horizontal plane on which the 

 range is measured below the upper surface of the fluid ; 

 for whether the bottom of the vessel be in this plane, 

 or be raised above it, makes no difference. The range 

 from any orifice, measured on any horizontal plane, is 

 the same, provided the distance between that orifice and 

 the surface be the same, whether the fluid extend down 

 to that plane or not. 



PROBLEM. A fluid issues from the side of a vessel 

 through an oblique jet : to determine the horizontal 

 xaiige. 



. Let a be the angle of direction of the jet above the 

 horizontal line, and h the height of the surface of the 

 fluid above the orifice : then (DYNAMICS, page 739) the 

 range r on the horizontal plane through the orifice will 

 be r = 2h sin. 2. This is the greatest when sin. 2u is 

 greatest ; that is, when a = 45 , the range then being 

 R = 2fc. 



The velocity with which a fluid spouts from a small 

 orifice, as the surface of the fluid descends, varies as 

 the square root of the height of the surface above the 

 hole. 



For the velocity, at any instant, is always equal to 

 that acquired by falling from the surface to the orifice ; 

 and this velocity varies as the square root of the space 

 fallen through : hence the velocity varies as the square 

 root of the height of the surface above the hole. The 

 velocity, therefore, is retarded as the surface descends, 

 but not uniformly retarded, unless the horizontal sec- 

 tions of the vessel are all equal in area ; for the more 

 extensive the section or the surface of the fluid at 

 any instant of the descent, the more slowly will it 

 increase its distance from the top of the vessel, and the 

 less, therefore, will the velocity of the issuing fluid be 

 retarded. 



But if the horizontal sections of the vessel be all equal, 

 then the velocity of the issuing fluid, as likewise the 

 velocity of the descending surface, will be uniformly 

 retarded. 



For let V be the velocity of the descending surface at 

 any instant, and D that at the orifice : let also the areas 

 of the surface and orifice be A and o : then (page 763) 



V:r ::a:A.\V=|-t. 



A 



Now, as shown above, v varies as the square root of 



the space s between the descending surface and the 

 orifice : we may put, therefore, c / for v, and write 



, T ae . / aV 



V= A - V* = \/ -JJB = V2.A 



by putting 2/ for (^ ) 2 - As this agrees with the gene- 



ral expressions for the velocity in uniformly accelerating 

 or retarding forces (DYNAMICS, page 736), we infer that 

 the velocity V is uniformly retarded ; and since v is V 

 multiplied by an invariable number, we conclude that v, 

 the velocity of the spouting fluid, is also uniformly re- 

 tarded. Consequently, the volumes or quantities of 

 fluid discharged in equal times, decrease as the numbers 

 1, 3, 5, 7, <fec., taken in reverse order. (DYNAMICS, page 

 736). 



If a vessel of uniform horizontal section be kept con- 

 stantly full, then twice the quantity which the vessel 

 holds, will run out from a hole in the bottom, in the 

 same time that the vessel would empty itself. 



For the surface of the descending fluid is uniformly 

 retarded as the fluid runs out ; and when it arrives at the 

 bottom, the velocity is destroyed : the space which the 

 surface would describe in the same time, with the first 

 velocity of it uniformly continued, would be twice the 

 actual space described ; and the quantity discharged in 

 any time when the vessel is kept constantly full, is the 

 same as what would be discharged if the surface de- 

 scended uniformly with the first velocity : hence the 

 quantity discharged in the one case is double that dis- 

 charged in the other. 



PROBLEM. To find the time in which a vessel of uni- 

 form horizontal section will empty itself through a small 

 orifice in the bottom. 



Let A be the area of any horizontal section of the 

 fluid, h the height of the vessel, and a the area of the 

 orifice. Then the quantity of fluid that would be dis- 

 charged in the time t of emptying, if the first velocity 

 were to continue uniform, would be twice the contents 

 of the vessel ; that is, the quantity would be 2AA ; so 

 that if c be the velocity with which the fluid would then 

 uniformly issue, we should have 



atv 



Oil. 



- 

 av 



which therefore expresses the number of seconds occupied 

 in emptying the vessel through the orifice. 



It follows from this expression for t, that the times of 

 emptying cylinders or prisms through equal orifices in 

 their bases, vary as A */h. If the bases of the vessels 

 be equal in area, then the times vary as Jh. If the 

 altitudes are equal, the times vary as the bases. The 

 time of emptying any altitude h h' of the vessel, is 

 found by substracting from the time of emptying the 

 whole height h, the time of emptying the height h' : 

 thus, since 



and t'= J 

 " 9 



the number of seconds elapsed during the descent of the 

 surface of the fluid from the height h to the height h'. 



The time in which the surface would descend from the 

 height h to the height h', or the time in which the fluid 

 occupying this interval would run out, if the vessel were 

 kept constantly full, by equation (1), page 764, is 



Q 





Hence the time of discharging the proposed quantity 

 Q, when there is no supply from without, is to the time 

 of discharging the same quantity when the vessel is kept 

 constantly full, as 



