The wetted perimeter = 2πr, that is, the circumference of the circle formed by the pipe.
Therefore hydraulic mean depth = h = πr2 ∕ 2πr = ¼;
Similarly when the pipe runs half full—
- πr2 ∕ 2
- h = ——— = ¼
- 2πr ∕ 2
The solution of problems where a smaller arc of a circle is occupied by fluid requires trigonometrical methods, and is not usually needed in practice.
The quantity of fluid discharged in a given time is represented by the product of the sectional area of the stream into its velocity. The greater the hydraulic mean depth the greater is the velocity, if the inclination remains the same.
The velocity of flow is determined by Eytelwein’s formula, which states that the mean velocity per second of a stream of water similar in form to those now under consideration is nine-tenths of a mean proportional between the hydraulic mean depth and the fall in two English miles, if the channel were prolonged so far.
- Thus if f = the fall (in feet) in two miles,
- h = hydraulic mean depth in feet,
- V = mean velocity per second,
- Then V = ·9√(hf),
- or if v = velocity per minute, then
- v = 55√(hf).
It is more convenient to let f = fall in one mile.
Then the formula becomes v = 55√(h × 2f).