Sewage Force-Main Scour Velocity
Whether a sewage force main runs fast enough to keep solids from settling out.
Example
You enter
- Pump flow (gpm) 50
- Force-main inside diameter (in) 2
You get
- Velocity 5.106
- Largest ID holding 2 ft/s 3.20 in
Details, formula, and sources
V = 0.4085 Q / d^2 (ft/s, gpm, in), vs the ~2 ft/s minimum scour to keep solids suspended. 50 GPM in a 2 in main -> 5.11 ft/s (scours); the largest ID still holding 2 ft/s is 3.20 in, so a 4 in main (1.28 ft/s) lets solids settle. Ten States Standards; the state criteria and pump curve govern.
V = 0.4085 Q / d^2 (ft/s, Q gpm, d in); d_max = sqrt(0.4085 Q / 2); scours when V >= 2 ft/s.
The force-main scour-velocity criterion (about 2 ft/s minimum) from the Ten States Standards (Recommended Standards for Wastewater Facilities), by name.
The V = 0.4085 Q / d^2 velocity relation is public arithmetic; the 2 ft/s minimum scour velocity is the published Ten States Standards criterion.
Estimate. AHJ and licensed professional govern.
Field names used by the API: gpm, id_in, velocity_fps, d_max_scour_in
- Velocity V = 0.4085 Q / d^2 (ft/s, gpm, in)continuity
- Scour minimum about 2 ft/s to keep solids suspendedTen States Standards
- Pump curve governs the actual velocity varies with the pump operating pointscope of this tile