Culvert Hydraulics

Culverts are conduits used to convey water through highway embankments. Culvert hydraulics is governed by FHWA Hydraulic Design Series No. 5 (HDS-5) guidelines. Designing or analyzing a culvert requires determining whether the flow is governed by Inlet Control or Outlet Control.

Control Types

1. Inlet Control

Inlet control occurs when the flow capacity of the culvert is limited by the entrance geometry (cross-sectional area, shape, and inlet edge configuration) and the headwater depth ($HW$).

• Key Factor: The barrel slope, length, and roughness do not affect culvert capacity.
• Flow Condition: Flow passes through critical depth ($y_c$) at or near the inlet, and flow inside the barrel is supercritical.
• Equations (from NCEES handbook):
  • Unsubmerged ($HW/D < 1.2$): $$\frac{HW_i}{D} = \frac{H_c}{D} + K \left[\frac{Q}{A D^{0.5}}\right]^M - 0.5 S_o$$
  • Submerged ($HW/D \ge 1.2$): $$\frac{HW_i}{D} = c \left[\frac{Q}{A D^{0.5}}\right]^2 + Y - 0.5 S_o$$ (where $H_c$ is specific head at critical depth, $S_o$ is culvert barrel slope, and $K, M, c, Y$ are inlet constants based on shape and edge configuration).

2. Outlet Control

Outlet control occurs when the flow capacity is limited by downstream conditions, barrel friction, entrance losses, and tailwater depth ($TW$).

• Key Factors: Barrel length ($L$), roughness ($n$), slope ($S_o$), tailwater depth ($TW$), and entrance loss coefficient ($K_e$) all influence capacity.
• Flow Condition: Flow in the barrel is subcritical, and the culvert is often flowing full or nearly full.
• Energy Equation for Outlet Control: $$HW = H + h_o - L \cdot S_o$$

Where:

• $HW$ = Headwater depth above inlet invert (feet)
• $H$ = Total head loss (feet) through the culvert: $$H = h_e + h_f + h_v = \left[ 1 + K_e + \frac{29.16 \cdot n^2 \cdot L}{R^{1.33}} \right] \frac{V^2}{2g}$$ (where $K_e$ is the entrance loss coefficient, $n$ is Manning’s roughness, $L$ is length in feet, $R$ is hydraulic radius in feet, and $V$ is velocity in ft/s).
• $h_o$ = Downstream tailwater height above outlet invert, calculated as: $$h_o = \max\left(TW, \frac{y_c + D}{2}\right)$$ (where $y_c$ is critical depth, and $D$ is pipe diameter).
• $L \cdot S_o$ = Elevation drop along the culvert length (feet).

Entrance Loss Coefficients ($K_e$)

The entrance loss coefficient ($K_e$) is a major factor in outlet control design:

Worked Example 1: Inlet Control Headwater Depth

A $36\text{-inch}$ ($3.0\text{-foot}$) concrete circular culvert drains $Q = 35\text{ cfs}$ under inlet control conditions. The culvert is unsubmerged. Using the NCEES handbook inlet control parameters for a concrete pipe with a headwall and square edge:

• $K = 0.0098$
• $M = 2.0$
• Critical depth for this flow is $y_c = 1.95\text{ ft}$
• Specific head at critical depth is $H_c = y_c + \frac{V_c^2}{2g} = 2.65\text{ ft}$
• The culvert slope is $S_o = 1.0\% = 0.01\text{ ft/ft}$

Calculate the required headwater depth ($HW$) in feet.

Solution:

1. Calculate the Area ($A$) and Diameter ($D$):

   • $D = 3.0\text{ feet}$
   • $A = \frac{\pi D^2}{4} = \frac{\pi (3.0)^2}{4} \approx 7.07\text{ ft}^2$

2. Apply the Unsubmerged Inlet Control Equation:

   $$\frac{HW_i}{D} = \frac{H_c}{D} + K \left[\frac{Q}{A D^{0.5}}\right]^M - 0.5 S_o$$

3. Evaluate the terms:

   • $\frac{H_c}{D} = \frac{2.65}{3.0} = 0.883$
   • $\frac{Q}{A D^{0.5}} = \frac{35}{7.07 \cdot 3.0^{0.5}} = \frac{35}{7.07 \cdot 1.732} = \frac{35}{12.245} \approx 2.858$
   • $K \left[\frac{Q}{A D^{0.5}}\right]^M = 0.0098 \cdot (2.858)^{2.0} = 0.0098 \cdot 8.168 \approx 0.080$
   • $- 0.5 S_o = -0.5 \cdot 0.01 = -0.005$

4. Solve for $HW_i$:

   $$\frac{HW_i}{D} = 0.883 + 0.080 - 0.005 = 0.958$$
   $$HW_i = 0.958 \cdot D = 0.958 \cdot 3.0\text{ ft} \approx 2.87\text{ feet}$$

Worked Example 2: Outlet Control Headwater Depth

A $24\text{-inch}$ ($2.0\text{-foot}$) concrete culvert ($n = 0.012$) of length $L = 120\text{ feet}$ is laid on a slope $S_o = 0.005\text{ ft/ft}$.

• Design flow is $Q = 12.0\text{ cfs}$
• The inlet has a square-edge headwall ($K_e = 0.5$)
• Tailwater depth above the outlet invert is $TW = 2.2\text{ feet}$ (which is greater than the pipe diameter $D = 2.0\text{ feet}$, forcing outlet control full flow)
• Acceleration of gravity $g = 32.2\text{ ft/s}^2$

Determine the headwater depth ($HW$) required above the inlet invert.

Solution:

1. Verify full flow properties:

   • $D = 2.0\text{ feet}$
   • $A = \frac{\pi (2.0)^2}{4} = 3.142\text{ ft}^2$
   • $P_w = \pi D = 6.283\text{ feet}$
   • $R = \frac{D}{4} = 0.50\text{ feet}$
   • Velocity $V = \frac{Q}{A} = \frac{12.0}{3.142} \approx 3.82\text{ ft/s}$

2. Calculate head loss ($H$):

   $$H = \left[ 1 + K_e + \frac{29.16 \cdot n^2 \cdot L}{R^{1.33}} \right] \frac{V^2}{2g}$$
   • $1 + K_e = 1 + 0.5 = 1.5$
   • $\frac{29.16 \cdot (0.012)^2 \cdot 120}{(0.5)^{1.33}} = \frac{29.16 \cdot 0.000144 \cdot 120}{0.3976} = \frac{0.5039}{0.3976} \approx 1.267$
   • Velocity Head $\frac{V^2}{2g} = \frac{(3.82)^2}{2 \cdot 32.2} = \frac{14.592}{64.4} \approx 0.227\text{ feet}$

   Total loss:

   $$H = [1.5 + 1.267] \cdot 0.227 = 2.767 \cdot 0.227 \approx 0.63\text{ feet}$$

3. Determine $h_o$: Since $TW = 2.2\text{ feet}$ (pipe is submerged downstream):

   $$h_o = TW = 2.2\text{ feet}$$

4. Calculate Headwater ($HW$):

   $$HW = H + h_o - L \cdot S_o$$
   $$HW = 0.63\text{ ft} + 2.2\text{ ft} - (120\text{ ft} \cdot 0.005\text{ ft/ft})$$
   $$HW = 2.83 - 0.60 = 2.23\text{ feet}$$

Technical Pitfalls

• Inlet vs. Outlet Control Choice: In practice, you must calculate $HW$ for both inlet and outlet control. The larger of the two headwater depths governs the design. On the exam, read carefully to see if the control condition is pre-defined or if you must check both.
• Elevation Drop ($L \cdot S_o$): Do not forget to subtract the elevation drop ($L \cdot S_o$) in the outlet control energy equation. Leaving it out makes the required headwater appear higher than it is.
• Entrance Losses ($K_e$): Pay close attention to the pipe material and inlet edge condition to select the correct $K_e$ from the reference handbook.