Designing Swales Norman W. Garrick CE 4410 Spring 2017 Lecture 15

Swale Design Design Constraints The swale should be able to carry the design flow without overtopping or eroding. Erosion occurs if the velocity of flow is too high. Conversely, if the velocity is too low siltation will occur. Silt deposited in this way will change the characteristics of the swale and will need to be removed during maintenance.

Carlton Landing, Oklahoma Grass versus Rip rap Grass is preferable but, if the velocity is too high then a turf reinforced matting or rip rap is used to prevent erosion Carlton Landing, Oklahoma Source: Tom Low

Swale Design Velocity in Swale The flow velocity is affected by the following characteristics of the swale: Slope of the swale Area or volume of water Frictional resistance of the surface If any of these factors change, then the velocity could change, leading to erosion or siltation. Note that the surface of the vegetative swale is a living thing! In other words, the frictional resistance changes with the height of the grass

Swale Design Peak Run-off One important parameter for design is the peak run-off, which can be determined using the Rational Method Typically for a swale, a ten year return storm is used.

Swale Design Shape of the Channel The swale maybe parabolic, trapezoidal or triangular. The side-slope is often kept gentle to allow easy mowing. The recommended side slope is no more that 3:1 The cross-section of a swale is usually characterized by D – depth of the swale W – width of the swale

Swale Design Two Equations are needed for swale design Manning’s Equation (page 234-235 in Text) Continuity Equation (page 253 in Text)

Swale Design Manning’s Equation The equation gives the relationship between the flow velocity, and i) gravity and ii) the resistance to flow offered by the roughness of the channel. V = (1.485/n) R2/3 S1/2 V – flow velocity in ft/s n – Manning’s roughness coefficient for open channels (Table 13.2 R – Hydraulic Radius in ft {R = X-section A of Flow (ft2)/Wetted perimeter (ft)} S – Channel Slope (ft/ft) <<< not percentage

Swale Design Dimensions of a Parabolic Swale Cross-section Area: A = 2/3 WD Top width: W2 = W1(D2/D1)0.5

Swale Design Hydraulic Radius of a Parabolic Swale Hydraulic Radius = Cross-section Area/Wetted Perimeter Hydraulic Radius: R = W2D/(1.5W2+4D2)

Swale Design Continuity Equation The equation gives the relationship between the flow volume, and i) cross-section area and ii) flow velocity. q = AV q – flow in cu. ft./s A – cross-sectional area of flow in ft2 V – flow velocity in ft/s

Swale Design Example Problem A grassed swale with slope of 4% is must carry a peak runoff of 50 cu. ft. The soil is easily eroded. Design a vegetated swale with a parabolic cross section. The vegetative cover will be o good stand of bluegrass sod, mowed to be kept at 2 in. height.

Swale Design Example Problem Design parabolic swale? We need i) A, ii) W, iii) D To get A we use the Continuity formula q = AV What V? Permissible V from Table 13.1

Swale Design Example Problem Permissible V from Table 13.1 To get Permissible V Need >>> S = 0.04 Need the cover >>> bluegrass sod, easily erodible T 13.1 >>> 5 ft/s (note that the table gives ft/s in bracket!) q = AV >>>> A = 50 cu. ft. per second / 5 ft per second Area of swale = 10 ft2

Swale Design Example Problem Now we need W and D To get W and D we need R Manning’s Equation V = (1.485/n) R2/3 S1/2 V and S are known…. We need n

Swale Design Example Problem V and S are known…. We need n Table 12.2 gives n, Manning Roughness Coefficient Vegetated Channel Flow >>> 0.03 to 0.08 ??? Grass is kept low (2 inches) so we are in the low end of the scale Select n = 0.04

Swale Design Example Problem V = (1.485/n) R2/3 S1/2 Re-arrange to give R on the left of equation R2/3 = V(n/1.485)(1/S1/2) = 5(0.04/1.485)(1/0.04^0.5) = 0.673 R = 0.673^3/2 R = 0.55

Swale Design Example Problem We have A and R A = 2/3 WD R = W2D/(1.5W2+4D2) We can now get W and D BY TRIAL and ERROR First try, D = 1.5 R Try D = 1.5 * 0.55 = 0.825 ft

Swale Design Example Problem Try D = 1.5 * 0.55 = 0.825 ft Therefore, since A = 2/3 WD 10 = 2/3 * 0.825 * W W = 18.18 ft Check if the trial D and trial W give a trial R that is close to calculated R of 0.55

Swale Design Example Problem Check if the trial D and trial W give a trial R that is close to calculated R of 0.55 Trial R = W2D/(1.5W2+4D2) = 0.547 ft This is very close to calculated R So OK We can use A = 10 ft, W = 18.18 ft, D = 0.545 ft