Peak Flow Reduced by Upstream Storage (Estimate)
Also known as Qr = Qp - s / (80 tp) · culvert storage routing estimate · reduced peak flow from storage behind a culvert · is storage routing worth doing · routed peak flow estimate · peak attenuation by ponding · HDS 5 storage routing step 3 · estimate reduced outflow discharge · how much does ponding cut the design flood · reduced design discharge for a culvert
Worked example: same culvert in SI: 6.23 m3/s, 7,400 m3, 75 min → 4.9967 m3/s. Press Try an example to run it live, then adjust anything.
Enter your known values, leave one input blank, and solves for the missing one. Tap a variable’s symbol to see what it means, with a typical value. Change any unit and the answer follows.
Where Peak Flow Reduced by Upstream Storage (Estimate) fits
- QᵣRational Method Peak Runoff · SCS Triangular Unit Hydrograph Peak
- QₚRational Method Peak Runoff · SCS Triangular Unit Hydrograph Peak
- sDetention Storage from a Routing Step · Muskingum Channel Storage
- tₚSCS Triangular Unit Hydrograph Peak
Peak Flow Reduced by Upstream Storage (Estimate) explained
Before spending the effort on a full storage routing, it helps to know whether the storage upstream of a culvert will do much. This formula gives that quick answer. It takes the unrouted peak flow , the volume that can be stored upstream of the crossing up to the allowable headwater, and the time to peak of the inflow hydrograph, and it returns the reduced peak flow that the culvert would have to pass. The subtracted term, a volume over a time, is a flow.
The Federal Highway Administration's culvert manual (HDS 5) prints the rule as for cubic feet per second, cubic feet and minutes; the 80 belongs to those units and carries the 60 seconds in a minute. With the time in seconds, as this page takes it, the constant is 80/60 = 4/3. The manual does not say where its 80 comes from, but one reading is ours: written as with , the 80 matches the volume of a triangular hydrograph whose base is 2.67 times the time to peak (60 × 2.67 / 2 = 80.1), so the rule takes off the peak the share of the runoff that the storage can hold back.
The manual's own example is the check. With 220 ft³/s of peak flow, 261,360 ft³ of storage and a time to peak of 75 min, the reduced peak is 176 ft³/s (176.44 before rounding), which the manual calls a 20 percent reduction (19.8 by the numbers). The same culvert in SI, with 6.23 m³/s, 7,400 m³ and 75 min, gives 4.9967 m³/s. Run the other way, the storage that takes 220 ft³/s down to 176 ft³/s over 75 min is 80 × 75 × 44 = 264,000 ft³.
This is an estimate for deciding whether to route, not the routing. In the same example the full storage routing of the hydrograph gives a peak outflow of 150 ft³/s, lower than the 176 ft³/s the shortcut returns. The manual notes that the storage needed for a significant cut in the peak is typically quite large, and that leaving storage out only makes the barrel somewhat larger, which is a margin of safety. If the estimate shows a worthwhile cut, the full level-pool routing is the next step.
Peak Flow Reduced by Upstream Storage (Estimate) formula
- = Routed (reduced) peak flow (m³/s)
- = Unrouted peak flow (m³/s)
- = Storage available at the allowable headwater (m³)
- = Time to peak of the inflow hydrograph (s)
- US Department of Transportation, Federal Highway Administration, Hydraulic Design of Highway Culverts, Third Edition, Hydraulic Design Series No. 5 (HDS 5), publication FHWA-HIF-12-026, April 2012, Design Guideline 4 Storage Routing, section DG 4.3.1 step 3 (page DG4.4): Qr = Qp - s/(80 tp) = 220 - (261,360 ft3)/[(80)(75)] = 176 ft3/s, a 20 percent reduction; symbols Qr, QP, s and tp in the List of Symbols; section 2.1.5 Basics of Storage Routing (page 2.5). Read it
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formula.expert. (2026, October 30). Peak Flow Reduced by Upstream Storage (Estimate) Calculator. https://formula.expert/solvers/peak-flow-reduction-upstream-storage-estimate