An Undersized Storm Line and a Utility Crossing Conflict
A 400-ft storm sewer must carry 6.5 cfs from a new parking lot to an existing 30-in RCP outfall whose invert is fixed at elevation 96.0. The inlet at the lot needs an invert of 98.0, so only 2.0 ft of fall is available over the 400 ft. The engineer has drawn 15-in RCP at 0.005 ft/ft with n = 0.013. A 12-in sanitary sewer crosses the alignment at station 3+20 with an invert of elevation 98.9; the local standard requires 18 in of vertical separation between the outside surfaces of the pipes. Cover over the storm pipe is 2.7 ft at the upstream end, adequate for H-20 loading. The contractor has already ordered the 15-in pipe, and the parking lot is scheduled to open in eight weeks.
1. Check the capacity of the 15-in pipe and state the minimum acceptable size at the available slope.
Read the worked answer
Manning: Q = (1.486/n) A R^(2/3) S^(1/2).
For 15-in RCP: D = 1.25 ft, A = 1.227 sf, R = D/4 = 0.3125 ft, R^(2/3) = 0.461, and the square root of S = 0.0707. Q = (1.486/0.013)(1.227)(0.461)(0.0707) = 4.6 cfs at full flow. Required is 6.5 cfs, so the 15-in is about 30 percent undersized even when flowing full; the pipe will surcharge and back water into the lot inlet.
For 18-in RCP: A = 1.767 sf, R = 0.375 ft, R^(2/3) = 0.520. Q = 114.3 x 1.767 x 0.520 x 0.0707 = 7.4 cfs, so 6.5 cfs is 88 percent of capacity. Full-flow velocity is 7.4/1.767 = 4.2 fps, above the 2 fps scour threshold and well below the 10 to 15 fps abrasion range for concrete.
So 18-in at 0.005 is the minimum acceptable pipe, but it consumes most of the available fall with nothing left for inlet control. Reject the 15-in order; it cannot be fixed after installation by cleaning.
2. Evaluate the sanitary sewer crossing at station 3+20 and recommend a resolution.
Read the worked answer
Check the separation. Storm invert at station 3+20 = 98.0 - 0.005(320) = 96.40; with an 18-in pipe the crown is 97.90. The sanitary invert is 98.90, so clearance between the outside surfaces is 1.00 ft, 6 in less than the 18-in standard.
You cannot lower the storm locally. The outfall invert is fixed at 96.0, and at station 3+20 the storm invert is 96.40; dropping another 0.5 ft puts it at 95.90, below the outfall, so the pipe would flow uphill. Moving the crossing downstream gains only 0.005 ft per foot; even at station 4+00 clearance is 1.4 ft, still short. The conflict therefore cannot be resolved within the storm line's own geometry.
The practical resolutions are to relocate the sanitary sewer reach, usually the cheapest if it is shallow and the utility owner will accept a revised profile, or to request a reduced separation with concrete encasement of both pipes, which many standards allow down to 12 in. Take the relocation to the utility owner immediately; eight weeks is not enough time for a variance.
3. The outfall is an existing 30-in main. Describe the outlet conditions and the protection you would specify.
Read the worked answer
At 6.5 cfs in the 18-in pipe the flow depth is slightly less than full and velocity is about 4 fps. The Froude number, V divided by the square root of gD, is 4.2 / the square root of (32.2 x 1.5) = 0.60, so the flow is subcritical but near the transition. The 30-in outfall at 0.005 has far more capacity than the 18-in delivers, so there is no tailwater to drown the outlet jet and the discharge will leave the pipe at an angle into the main.
Provide a riprap apron sized to the outlet velocity and the pipe diameter: geotextile beneath a 1.5-ft blanket of stone with d50 in the 6- to 9-in range, extending at least 3 pipe diameters beyond the end, about 5 ft, with a cutoff trench at the downstream toe. A headwall and flared end section protect the pipe and reduce entrance loss, and a trash rack at the upstream inlet keeps the 18-in from plugging.
Confirm that the connection is crown-to-crown or better and not a submerged discharge, and check the existing main's capacity with the added 6.5 cfs.