| Fig. 117. |
Let fig. 117 represent the cross section of the channel. From the symmetry of the section, only half the channel need be considered. Let obac be any section suitable for the minimum flow, and let it be required to find the curve beg for the upper part of the channel so that the mean velocity shall be constant. Take o as origin of coordinates, and let de, fg be two levels of the water above ob.
Let ob = b/2; de = y, fg = y + dy, od = x, of = x + dx; eg = ds.
The condition to be satisfied is that
v = c √ (mi)
should be constant, whether the water-level is at ob, de, or fg. Consequently
m = constant = k
for all three sections, and can be found from the section obac. Hence also
| Increment of section | = | y dx | = k. |
| Increment of perimeter | ds |
y2 dx2 = k2 ds2 = k2 (dx2 + dy2) and dx = k dy / √ (y2 − k2).