Many people have asked how the HEC-RAS culvert routines compare to the USGS culvert procedures outlined in the publication "Measurement of Peak Discharge at Culverts by Indirect Methods" (Bodhaine, 1978), and if the HEC-RAS software would give the same or similar answers to their culvert analysis program (CAP). To prove that HEC-RAS could handle all 6 flow types outlined in the USGS publication, we put models together to replicate 8 of the example data sets in the back of the USGS publication mentioned above. HEC-RAS was able to compute similar upstream water surface elevations, as reported in the USGS publication, for all of the example data sets except number 8. However, we believe the reported result for example problem number 8 is questionable. This will be explained in more detail below. The 8 examples were put together in HEC-RAS by entering the culvert geometry, and all other properties provided in the publication.
Here is a table of the HEC-RAS and USGS answers for the 8 example problems:
Example Number | Flow Rate Q
(cfs) | USGS Water Surface
(ft) | HEC-RAS Water Surface (ft) | HEC-RAS
Flow
Classification | USGS Flow
Classification |
|---|
1 | 729 | 12.00 | 11.89 | Inlet | 1 |
2 | 530 | 10.00 | 10.68 | Inlet | 1 |
3 | 268 | 6.00 | 6.00 | Outlet | 2 |
4 | 523 | 8.19 | 8.88 | Outlet | 2 |
5 | 251 | 6.00 | 5.94 | Outlet | 3 |
6 | 125 | 7.00 | 7.17 | Outlet | 4 |
7 | 120 | 8.00 | 8.14 | Inlet | 5 |
8* | 209 | 8.00 | 11.00 | Inlet | 6 |
*Note: We think the answer shown in the USGS publication, for example number 8, is questionable. Here is why:
Example 8 is for a circular concrete culvert that is 4.0 feet in diameter. The Manning's n is 0.012, and the culvert has a beveled entrance. The resulting flow rate computed in the example is 209 cfs. The culvert invert is set at an elevation of 1.0 ft at the upstream end, and the top of the culvert is at 5.0 feet inside elevation at the upstream end.
The Culvert Area is A = 12.5664 sq. ft.
Therefore V = Q/A = 209/12.5664 = 16.63 ft/s inside the culvert at the upstream end.
The velocity head is V2/2g = (16.63)2/ (2x32.3) = 4.3 feet of velocity head.
Therefore, the energy at the upstream inside end of the barrel must be at least 5.0 + 4.3 = 9.3 feet of energy head. The energy upstream, outside of the barrel, will be this energy plus an entrance losses to get the flow into the barrel, plus friction losses. Therefore the upstream energy will be greater than 9.3 feet. HEC-RAS computed an upstream energy of 11.0 feet, which we believe is more correct than the 8.00 feet reported in example 8 in the USGS report.
The USGS results of 8.0 feet is based on the assumption of this culvert acting as a Syphon for this particular flow rate. The HEC-RAS computations did not fill the barrel, so we do not think the culvert will act as a Syphon. If the culvert were to act like a Syphon, then the answer would be closer to the USGS culvert routines answer.
Additionally HEC-RAS was able to reproduce all six of the USGS flow classification types. For Type 6, the example 8 problem did not flow as a full barrel for HEC-RAS. However, we took the same data set and lowered the upstream invert to an elevation of 0.0 feet, which put the culvert on a horizontal slope. HEC-RAS did compute that the flow was following the USGS Type 6 classification for this culvert.
We have therefore concluded that HEC-RAS can handle all 6 of the USGS Culvert flow classifications, and can reproduce the results of the CAP program within a reasonable tolerance (Except for example 8, which is in not been resolved at this time). It is also believed that most of the differences in the results are due to the fact they the two programs use different: empirical coefficients entrance losses; friction slope computations (therefore different friction losses); and exit losses.