FLOW IN OPEN CHANNELS 299 



A 3 

 Again, for Q to be a maximum, ' is to be a maximum, so that 



/. 3 P A 2 d A A*d'P = 0. (2) 



Where A is fixed, d A = and both conditions are satisfied if d P = 0. 



EXAMPLE. 

 (1) Rectangular Channel. Breadth 2 b depth d. 



Here A = 2 b d : P = 2 b + 2 d = ^ -f 2 d 



d 



Putting A = 2 b d and equating j-Tj} ^ zero ' we ^ ave ^ ^- 



73 



Putting the full breadth = B, we have d = , i.e., for maximum flow 



the depth must equal one-half the breadth. 

 We then have Q = C /y ^L . i 



8 

 = C 



(1) 

 4 " 



(2) Trapezoidal Channel. Fig. 129. 



Let b = half bottom breadth ; d = depth ; s = cotangent of angle of 

 slope of sides. 



Then A = 2 b d + s d*. 



P = 2 (b + d x/1 + s 2 > 



For Q to be a maximum with a given area of channel or for the cost 



d P 



of construction to be a minimum, it is necessary that -J-TJ. = 0. 



tl (ci) 



t A <, 3 - \ 



But P 



dP 



