Energy Losses and Dissipation

Hydraulic structures must be designed to manage energy losses and prevent erosion. High-velocity discharge from culverts or channels can cause scour at outfalls, which is prevented using energy dissipation structures (like riprap aprons or stilling basins) and by managing transitions.

Minor Head Losses

Minor head losses occur due to changes in velocity, flow direction, or cross-sectional area at inlets, bends, expansions, contractions, and manholes.

Governing Equation:

$$h_L = K \cdot \frac{V^2}{2g}$$

Where:

  • $h_L$ = Head loss (feet or meters)
  • $K$ = Minor loss coefficient (dimensionless)
  • $V$ = Velocity of flow (ft/s or m/s)
  • $g$ = Acceleration of gravity ($32.2\text{ ft/s}^2$ or $9.81\text{ m/s}^2$)

Hydraulic Jumps in Rectangular Channels

A Hydraulic Jump occurs when supercritical flow ($Fr_1 > 1.0$) transitions abruptly to subcritical flow ($Fr_2 < 1.0$). This transition creates high turbulence and dissipates a significant amount of kinetic energy.

Sequent Depth (Conjugate Depth) Equation:

The relationship between the depth before the jump ($y_1$) and the depth after the jump ($y_2$) in a horizontal rectangular channel is:

$$\frac{y_2}{y_1} = \frac{1}{2} \left( \sqrt{1 + 8 Fr_1^2} - 1 \right)$$

Where:

  • $y_1$ = Upstream (supercritical) depth (feet or meters)
  • $y_2$ = Downstream (subcritical) depth (feet or meters)
  • $Fr_1$ = Upstream Froude number: $$Fr_1 = \frac{V_1}{\sqrt{g \cdot y_1}}$$

Head Loss in a Jump ($\Delta E$):

The energy dissipated during a hydraulic jump is:

$$H_L = \Delta E = E_1 - E_2 = \frac{(y_2 - y_1)^3}{4 \cdot y_1 \cdot y_2}$$

Energy Dissipators

Energy dissipators are structured outlets designed to reduce velocities before runoff enters natural watercourses.

  • Riprap Aprons: A placed bed of rock riprap at the culvert outlet. The median stone size ($d_{50}$) is determined based on the outlet velocity and culvert diameter/flow rate (typically sized using FHWA HEC-14 charts or equations).
  • Stilling Basins: Reinforced concrete structures containing chute blocks, baffle blocks, and end sills to force a hydraulic jump within the basin and stabilize the flow.

Worked Example 1: Hydraulic Jump and Energy Loss

A rectangular drainage channel with a bottom width of $6.0\text{ ft}$ carries a flow of $Q = 72\text{ cfs}$. At a transition, the flow depth is measured to be $y_1 = 0.50\text{ ft}$. A hydraulic jump occurs just downstream of the transition.

1. Calculate the conjugate depth ($y_2$) downstream of the jump. 2. Determine the energy head loss ($H_L$) across the jump.

Solution:

Step 1: Calculate upstream velocity ($V_1$) and Froude number ($Fr_1$)

  • Width $b = 6.0\text{ ft}$, Depth $y_1 = 0.50\text{ ft}$
  • Area $A_1 = b \cdot y_1 = 6.0 \cdot 0.50 = 3.0\text{ ft}^2$
  • Velocity $V_1 = \frac{Q}{A_1} = \frac{72}{3.0} = 24.0\text{ ft/s}$

Calculate $Fr_1$:

$$Fr_1 = \frac{V_1}{\sqrt{g \cdot y_1}} = \frac{24.0}{\sqrt{32.2 \cdot 0.50}} = \frac{24.0}{\sqrt{16.1}} = \frac{24.0}{4.0125} \approx 5.98$$

Note: Since $Fr_1 \approx 5.98 > 1.0$, the flow is supercritical, validating the occurrence of a hydraulic jump.

Step 2: Calculate conjugate depth ($y_2$)

$$\frac{y_2}{y_1} = \frac{1}{2} \left( \sqrt{1 + 8 Fr_1^2} - 1 \right)$$

$$\frac{y_2}{0.50} = 0.5 \cdot \left( \sqrt{1 + 8 \cdot (5.98)^2} - 1 \right)$$
  • $(5.98)^2 \approx 35.76$
  • $8 \cdot 35.76 = 286.08$
  • $\sqrt{1 + 286.08} = \sqrt{287.08} \approx 16.94$
  • $\sqrt{287.08} - 1 = 15.94$

Solve for $y_2$:

$$\frac{y_2}{0.50} = 0.5 \cdot 15.94 = 7.97$$

$$y_2 = 7.97 \cdot 0.50 \approx 3.99\text{ feet}$$

Step 3: Calculate energy loss ($H_L$)

$$H_L = \frac{(y_2 - y_1)^3}{4 \cdot y_1 \cdot y_2}$$

$$H_L = \frac{(3.99 - 0.50)^3}{4 \cdot 0.50 \cdot 3.99} = \frac{(3.49)^3}{7.98}$$
  • $(3.49)^3 \approx 42.51$ $$H_L = \frac{42.51}{7.98} \approx 5.33\text{ feet of head loss}$$

Worked Example 2: Minor Losses in Pipe Junction

A $12\text{-inch}$ storm sewer pipe carries flow at $V = 6.0\text{ ft/s}$. The pipe passes through a manhole junction that causes a minor loss of $K = 0.35$ for a straight-through configuration.

Calculate the head loss ($h_L$) at the manhole.

Solution:

$$h_L = K \cdot \frac{V^2}{2g}$$

$$h_L = 0.35 \cdot \frac{6.0^2}{2 \cdot 32.2} = 0.35 \cdot \frac{36.0}{64.4}$$

$$h_L = 0.35 \cdot 0.559 \approx 0.20\text{ feet}$$

Technical Pitfalls

  • Sequent Depth vs Alternate Depth: Do not confuse sequent (or conjugate) depths with alternate depths. Conjugate depths represent the depths before and after a hydraulic jump (where momentum is conserved, but energy is lost). Alternate depths are depths that share the exact same specific energy ($E_1 = E_2$), which occurs under frictionless transitions without a jump.
  • Unit of g: Ensure you use $g = 32.2\text{ ft/s}^2$ for USCS or $g = 9.81\text{ m/s}^2$ for SI units.
  • Froude Number calculation: Double check that $Fr_1$ is calculated using the upstream depth $y_1$ and upstream velocity $V_1$.