Travel Time of Flow Along a Segment (Length over Velocity)

How long does water take to get from one point to another along a flow path? The time is the distance divided by the speed, $T_t = \ell / V$, where $\ell$ is the distance between the two points along the flow path and $V$ is the average velocity of the flow between them. Make the path longer in proportion and the time rises in proportion; a faster flow shortens it in inverse proportion. The relation holds in any consistent units, which is why the page lets you pick them freely.

The Natural Resources Conservation Service's National Engineering Handbook, Part 630, chapter 15 (Time of Concentration), prints it for feet, feet per second and hours as equation 15-1, which divides the length by 3,600 times the velocity. The 3,600 is only the number of seconds in an hour, and the unit picker supplies it, so the bare ratio loses nothing.

Where this matters is the time of concentration. The flow path is cut into stretches of sheet flow, shallow concentrated flow and channel flow, each with its own velocity, and each stretch is a separate $\ell / V$. The travel times add up to the time for runoff to travel from the hydraulically most distant point to the outlet. So the care in the method goes into $V$, which must be the average velocity of that stretch, and into where you cut the path, not into the division.

The handbook's velocity-method example shows the numbers. Reach 2 is 6,000 ft at 5.2 ft/s, which gives 0.32 h (exactly 0.3205). In Table 15-6, 2,400 ft at 3.6 ft/s gives 0.19 h, 2,800 ft at 3.8 ft/s gives 0.20 h and 900 ft at 6.1 ft/s gives 0.04 h; the handbook prints these to two places. The page's own example is metric: 1,000 m at 2 m/s is 500 s, or 0.13889 h in the unit the handbook uses. Run the other way, 0.5 h at 4 ft/s is a path of 7,200 ft.