566 HYDRAULICS AND ITS APPLICATIONS 



While in the second case, when v = 6'64, we have 

 ^ = 60 -685 X 2 



= 58*63 feet. 



.-. Fall in level at stand pipe = '77 feet. 



The mean height in stand pipe = 59'0 feet, so that if A is its area, 

 the energy leaving during the transition 



= -81 X 59 X 62-4 X A foot Ibs. 

 = 2,830 A foot Ibs. 



The energy entering the casing from the penstock in the same interval 

 of time, ti seconds, is given by 

 h 



62-4 av \h-- f -\ dt 

 ( 2 g m j 



and, assuming uniform acceleration so that - - = a, and writing 



f i 



2-24 



v = i\ -f- a t = 4*4 -\- . t, this becomes 



62-4 a h 



o ( 



= 784 ti [ (60 X 4-4) + (30 X 2'24) 



1 ( / 3 \ 2'24^ ) ~~\ 



- 6^4 { (4'4 8 ) + ( 2 X 4'4 2 X 2'24 J + (4'4 X 2'24 2 ) + -j- j J 



= 784 ^ \ 264 + 67-2 - 2'7 } 

 = 257,500 ti foot Ibs. 



/. For speed to be maintained with the increased load, we mus 

 have : 



2,830 A + 257,500 ti = - ^-^ X h (I 



= 309,000 



or ti = '055 A. 



Thus the acceleration of the supply column, corresponding to any 

 increase in load, varies inversely as the area of the stand pipe. 



2'24 2'24 



With a 4 foot stand pipe a = = i2'87 X '055 



= 3-16 f.s.s. 



