696 HYDKAULICS AND ITS APPLICATIONS 



Determine Qi, Q s and Q d , and also the pressure at the throat of the 

 mixing cone and in the plane of the orifices. 



From equation (4) we have : 



+ -6 X 10} 



(f) 



- {'2 X 40 + -8 X 15 + -6 X 10} ~ = v d * - 



Again, from (5), 



(6 v d - -S^) 2 = -16 W - 64-4 X 55} 

 .-. Vl * + 2 t?i t? d - 8 v d z = 4,718. (ii) 

 Substituting for v d * in (ii) from (i) : 



v? + 2 vi V^i 2 - 2,790 - 3 fa 2 - 2,790) = 4,718, 



.-. n 2 - 1,826 = n *'i 2 - 2,790. 

 Squaring both sides, we get, on reduction : 

 862 v? = 3,333,000 

 or ?! = 3,866 



.. i'x = 62*2 feet per second. 

 Substituting this value in (i) : 



v d = V 3,866 - 2,790 = 32'8 feet per second ; 

 while from equation (3) : 



v s = J 3,866 3,542 = 18'0 feet per second. 

 .*. Qi =. *2 vi = 12*4 cubic feet per second. 

 Q d = -6 v d = 19 '7 cubic feet per second. 

 Q s = -4 v s = 7*2 cubic feet per second. 



Again, since ^ + |^ = /M, : 



-^ = 40 - 60-1 = - 20-1 feet of water. 

 W 



n 



while since h d = , -j- ^- : 



.'. & = 10 - 16-7 = - 6-7 feet of water. 

 The actual height through which the water may be forced by the pump 



/ 2 \ 



is less than the value h d given by j -|p + |p- 1- feet, because of the 



loss of energy by eddy formation in the diverging discharge pipe, and 

 may amount to between *6 and *7 h d . 



This loss of energy is proportional to v d 2 , and will therefore increase 

 since the necessary value for v d increases as h d increases. Consequently, 



