FLOW IN OPEN CHANNELS 325 



/. -82 = '001 (- k) - 1-027 { 1-996 - T4'29 } 

 .-. -262 = -001 1 2 



k = 262 feet. 



If, at the entrance, the depth is less than 2*68 feet, as might occur 

 if this flume were fed from a sluice, we have Case 1 (a) repeated, and 

 the depth would finally increase up to 3*08 feet. With an open 

 channel leading directly out of a reservoir, the required discharge 

 could only be obtained by having the depth at entrance equal to or 

 greater than H. 



ART. 90. CHANNEL WITH HORIZONTAL BED. 



Here i is zero, so that equation (6) of the last article ceases to apply. 

 Making i = o in equation (3), p. 310, we get 



/ v 2 P 



dh 2g- A 



<n-r~ v 2 



ah 



O 2 

 Writing v z = -J-g we nave > if the channel is rectangular and broad, so 



P 1 

 that T = f (sensibly), 



Integrating between the limits li and 1% we get 



k - 1* = j fa* - /< 2 4 ) - 2 (h - h,) (3) 



from which the difference in level (hi h 2 ) at any two points distant 

 li /2 from each other, may be calculated when a given quantity Q cubic 

 feet per second is flowing along the channel. 

 Expressing (2) as 



^ - 1 (4) 



dl 2 j g 6" fe 3 _ ' 



/ l y a 



we have, for ^-y to be infinite, the condition 



g 62 ft* __ Q2 _ ^2 b 2 



.-. v 2 = g h, 



