A sewage ejector will discharge through a 2-inch pipe. Take these values as given: the discharge piping must maintain a velocity of at least 2 feet per second, the inside cross-sectional area of the 2-inch pipe is 0.0233 square foot, and 1 cubic foot per second equals 448.8 gallons per minute. What is the smallest pump capacity that satisfies the velocity requirement?
- A10.5 gallons per minute
- 20.91408 gallons per minute using the given conversion factors
- C31.4 gallons per minute
- D41.8 gallons per minute
Where to look it up
Explicit exercise assumptions; not a universal code requirement.
Why B is correct
Using the supplied values, Q = area × velocity = 0.0233 × 2 = 0.0466 cubic foot per second. Convert to GPM: 0.0466 × 448.8 = 20.91408. This is the minimum flow under the stated assumptions; do not round it downward when selecting equipment. Check delivered pump flow at the actual system head. Higher flow alone does not establish short cycling or water hammer, and this calculation does not verify other pump or code requirements.