Storm Lateral Lacks Required Slope and Subsoil Drain Is Misconnected
A commercial warehouse has a flat roof area of 8,500 sq ft. The design rainfall is 3.2 in./hr for a 60-minute storm, and the roof runoff coefficient is 0.95. Four roof drains connect to a single 6-in. PVC storm drain that runs 160 ft to the city storm main. The building storm drain invert at the building is 101.25 ft; the city main invert at the connection is 100.00 ft. The contractor proposes to use the existing 1.25 ft of fall and says the 6-in. pipe is large enough because it is bigger than the roof drain outlets. The site also has a subsoil drain around the foundation that currently connects to the sanitary sewer because the storm connection is too shallow. The plumbing inspector has asked for storm drainage calculations and a corrected connection before approving the underground. The code requires a minimum slope of 1/8 in./ft for horizontal storm drainage and prohibits subsoil drainage from connecting to the sanitary sewer.
1. Calculate the peak design storm flow in gallons per minute.
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Use the rational-method form for small roof areas: Q = 0.0104 x C x i x A, where Q is in gpm, C is the runoff coefficient, i is rainfall intensity in in./hr, and A is area in sq ft. The 0.0104 factor converts inches per hour over square feet to gallons per minute.\n\nQ = 0.0104 x 0.95 x 3.2 x 8,500. First, 0.95 x 3.2 = 3.04; then 3.04 x 8,500 = 25,840; then 25,840 x 0.0104 = 268.7 gpm. The design flow is therefore about 269 gpm. This is the flow the storm drain and roof drains must handle for the stated design storm. If the roof drains are individually rated, each drain and leader must also be checked; a large horizontal pipe does not guarantee that the roof drains themselves can accept the flow.
2. Evaluate the proposed slope and the capacity of the 6-in. storm drain. What correction is needed?
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The required fall over 160 ft at 1/8 in./ft is 160 x 0.125 = 20 in., or 1.67 ft. The available fall is 101.25 - 100.00 = 1.25 ft, or 15 in. The actual slope is 15 in. / 160 ft = 0.09375 in./ft, which is less than the required 0.125 in./ft. The installation is therefore short by about 5 in. of fall.\n\nCapacity also fails. Using Manning's equation for 6-in. PVC with n = 0.012, full-flow area A = 0.196 sq ft, hydraulic radius R = 0.125 ft, and S = 0.09375/12 = 0.0078125 ft/ft, Q is about 0.54 cfs, or about 241 gpm. That is below the 269-gpm design flow. At the required 1/8 in./ft slope, the same pipe can carry about 278 gpm, which would be adequate.\n\nThe correction is to raise the building storm drain invert by at least about 5 in., relocate or shorten the run, or obtain an approved engineered alternative. Increasing the pipe size alone does not cure a slope violation, and the contractor's statement that 6-in. is larger than the roof drain outlets is not a capacity calculation.
3. What must be done with the foundation subsoil drain connection?
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The subsoil drain cannot remain connected to the sanitary sewer. Foundation and subsoil drainage is clear water and must be handled by storm drainage, an approved sump and pump, or another approved disposal method. Connecting it to the sanitary sewer adds clear water to the sanitary system and violates the separation required by the plumbing code.\n\nIf the storm main is too shallow for gravity flow, use a subsoil sump with a pump discharging to storm drainage where approved, or another approved outlet. The discharge must not create a cross-connection or discharge onto a neighboring property. Provide the required backwater valve or check valve if the storm connection can surcharge, and verify local amendments for subsoil drainage. The inspector should require a corrected underground drawing before the storm and subsoil systems are covered.