Hydraulics

Hydraulic Review & Basic Pipe Sizing Design Aid

This information has been brought together in this format so that an easy review of a drainage system can be quick and almost instantaneous to tell if it is operating under a free flow condition as well as this could be used as a very rudimentary design tool for pipe sizing for a simple drainage run like a driveway culvert. Examples are shown below. Also, any pipe larger than 60 inches is classified as a bridge and thus the Bridge Design section should be consulted.

Table 1 below is based on separate sections of the Road Design Manual brought together for a quick and easy reference. It is also based on a pipe Manning’s n = 0.012. As per Chapter 6 of the Road Design Manual, this would be the Manning’s n value used for reinforced concrete pipe (RCP) as well as high density polyethylene pipe (HDPE) [Figure 6-27, Road Design Manual, pg. 6-53]. The third column shows the minimum slope needed in order for a pipe flowing full (open channel flow, i.e. the HGL is below the top of the pipe) to maintain a minimum cleanout velocity of 3 ft/s [Figure 6-3, Road Design Manual, pg. 6-5]. The fourth and fifth columns show a conveyance factor (K) which can be used for the pipes in order to quickly calculate a rate of flow. Again, this would be based on a pipe flowing full under open channel flow conditions. One very important facet to consider is that this does not take into account any tailwater conditions. Discussion of the K factor can be referenced from the Road Design Manual on page 6-26, formula 6.18. K factor values can be referenced from the Road Design Manual on page 6-53, Figure 6-25.

Table 1.

Pipe Size (in)	Elliptical Equivalent (in)	Minimum Slope* (ft/ft)	K Factor (circular)	K Factor (elliptical)	Full Flow Capacity ** Qmax (cfs)
12	-	0.0038	38.7	-	2.4
15	-	0.0028	70.2	-	3.7
18	14x23	0.0022	114	116	5.3
21	-	0.0018	172	-	7.2
24	19x30	0.0015	246	252	9.4
27	22x34	0.0013	337	339	11.9
30	24x38	0.0011	446	456	14.7
33	27x42	0.0010	575	607	18.1
36	29x45	0.0009	725	746	21.6
42	34x53	0.0007	1093	1156	28.9
48	38x60	0.0006	1560	1565	37.7
54	43x68	0.0005	2136	2196	47.7
60	-	0.0004	2829	-	58.9

* - minimum slope required for pipe flowing full to maintain 3 ft/s cleanout velocity

** - circular pipe full flow capacity based on minimum slope

Pipe Full Flow Capacity (based on open channel flow) = Q_max (cfs) = K Factor x [Slope (ft/ft)]^1/2

Example Problem 1:
The following is an example of a drainage system review. For the following drainage system, determine if it is operating under an open channel design condition. This basic analysis can be used on an existing or proposed closed drainage system. If not operating as an open channel system (i.e., possible pressure flow condition / the HGL is above the top of the pipe), then a discussion can be had with the designer about trying to redesign as a free flow system as well as a more in-depth analysis would need to be conducted to determine that the HGL is below appropriate inlets; hence, avoiding surcharges. For an existing system, this could be a quick check to see if it is potentially undersized. Basically, there will be a comparison between the Q value calculated here versus the designed Q value for the pipe run being studied. It is not necessary to calculate this for every pipe in the system. In order to help expedite, check a few pipes at the most downstream portion as that is where any problems would potentially occur first.

Check pipes 601, 602, and 603. (Note: information shown here is different from the actual plans for example purposes).
Information obtained from the plans:
P601 – 54” RCP, S = 0.008; calculated Q from drainage report = 199.21 cfs
P602 – 54” RCP, S = 0.010; calculated Q from drainage report = 198.41 cfs
P603 – 60” RCP, S = 0.009; calculated Q from drainage report = 197.81 cfs

The first thing that stands out for questioning is, how does the flow decrease going downstream as there are no other outlets before the very end of the pipe run? There very well could be a logical explanation, but for review purposes, it is worth asking about.

Calculated Q_max (for an open channel design) = K Factor x (Pipe Slope)^1/2
P601 = 2136 x (0.008)1/2 = 191.04 cfs ~ 191 cfs
P602 = 2136 x (0.010)1/2 = 213.6 cfs ~ 214 cfs
P603 = 2829 x (0.009)1/2 = 268.38 cfs ~ 268 cfs

From the quick analysis completed, the calculated Q_max values for pipes 602 and 603 show a larger capacity than what the Q values from the drainage report are showing; hence, they should be okay. Pipe 601 is showing a larger Q from the drainage report than what the calculated Q_max is for open channel design. It is very negligible, but worth double checking to ensure no potential surcharging from upstream inlets.

Example Problem 2:
Size a driveway culvert based on the following calculated runoff and field data. Tailwater should not be a problem at this location.

1. Runoff value of the 10 year storm event going to the proposed culvert = Q = 17 cfs
2. Best slope achievable as per field measurement = S = 0.005 ft/ft

In order to keep this proposed pipe as an open channel flow design, Q = K Factor x S^1/2
Solve for the K Factor based on the known Q value:
17 cfs = K Factor x (0.005)^1/2
K Factor = 17 cfs / (0.005)^1/2
K factor = 17 cfs / (.0707)
K Factor = 240.5

In referencing Table 1, the K Factor calculated is greater than a 21 inch circular pipe, but less than a 24 inch circular pipe; therefore, a 24 inch circular (or 19 x 30 elliptical) should be sufficient for installation. Of course, cover over the pipe, any utility conflicts, and possible right of way concerns still need to be considered before installation. Elliptical pipes are very useful if encountering cover challenges.

Pipe Size (in)	Elliptical Equivalent (in)	Minimum Slope* (ft/ft)	K Factor (circular)	K Factor (elliptical)
21	-	0.0018	172	-
24	19x30	0.0015	246	252

Circular Qmax	Elliptical Qmax
Qmax=246x(0.005)1/2	Qmax=252x(0.005)1/2
Qmax=246x0.0707	Qmax=252x0.0707
Qmax=17.39 cfs	Qmax=17.82 cfs

References

Delaware DOT Road Design Manual, Chapter 6