2fiO HYDRAULICS AND ITS APPLICATIONS 



= i ' 5 + 422 ' 4 + 2 ' 2 + 4 " 2 + 27 ' 033 ' 6 



= 27,479 -^ feet. 



^ r/ 



Since the pressures at the entrance B and exit J.) are those 

 corresponding to their depths below the free surfaces, viz., 10 feet and 

 2 feet, we have, neglecting the velocity before and after leaving the 

 pipe- 



Head before entering pipe = Z B -f 10 = Z A . 



Head after leaving pipe = Z D -}- 2 = Z K . 



.'. Loss of head in pipe = Z A Z E = 220 feet. 



.*. 27,479 - = 220 

 & 9 



= ' 717 feefc per second ' 



.*. v a = 2*868 feet per second. 

 .*. Quantity flowing per second = - ^ cubic feet, 



= "141 cubic feet per second. 



In gallons per minute this gives { '141 x 6*24 X 60}. 

 = 52-8 gallons per minute. 



If it be required to find the diameter of pipe necessary to give a 

 certain discharge between two reservoirs, the difference of level in the two 

 reservoirs being fixed, on expressing the fact that the total difference of 

 head is equal to the sum of the pipe losses, we have, if 



I = length ; d = diameter of pipe ; v = velocity of flow ; 



H difference of head in reservoirs ; Q = quantity required per 

 second in cubic feet. 



rj K' v* , . 4/Z?; 2 , K" v* . 



H = -~ (at entrance) + -^ -7- (in friction) H - (at exit). 



a Cf A (f (ii & 



If we have any bends or obstructions in the pipe, the losses caused by 



these may all be expressed as K 1 " |-, so that if K = (K' + K" + K'") 



. 2 9 



we have 



Also Q = v X area of pipe, .'. v = 4 Q -=- TT $ 



, ,/4'-{A- + 4 f' 

 2 // T, - d* ( d 



an equation from which the value of d may be obtained by graphical 

 solution, by trial, or bv successive approximations. 



