Open Drainage Channel Hydraulic Cross-Section Optimization (Environmental & Civil Engineering) Solution

A water resources and civil environmental engineer at the UNLV Howard R. Hughes College of Engineering is designing an open, rectangular reinforced-concrete flood control drainage channel for the Clark County Regional Flood Control District in the Las Vegas valley.

Hydrologic peak runoff calculations dictate that the channel must maintain a fixed cross-sectional water flow area of:

where $x$ is the bottom channel width (in meters, $\text{m}$) and $y$ is the vertical water depth (in meters, $\text{m}$).

To minimize excavation costs, frictional head loss along the fluid boundary, and the expensive shotcrete concrete liner material across the bottom bed and both vertical sidewalls, the engineer must minimize the wetted perimeter:

1. Formulate the objective function $P(x, y)$ and the constraint equation $g(x, y) = k$.
2. Compute the gradient vectors $\nabla P(x, y)$ and $\nabla g(x, y)$.
3. Formulate the Lagrange multiplier vector equation $\nabla P = \lambda \nabla g$ and equate the individual scalar components.
4. Eliminate $\lambda$ to establish the optimal geometric aspect ratio between channel bottom width $x$ and fluid depth $y$.
5. Substitute into the flow area constraint to calculate the optimal channel dimensions $(x_{\text{opt}}, y_{\text{opt}})$ and the minimum wetted perimeter $P_{\min}$.
6. Verify why this optimal hydraulic ratio ($x = 2y$, where width equals twice the depth) represents the classic "most efficient rectangular open channel" in civil fluid mechanics.

Step-by-Step Solution:

1. Formulate Functions:

   • Objective function to minimize: Wetted perimeter $P(x, y) = x + 2y$ (base plus two vertical wet sidewalls).
   • Constraint equation: Fixed wet flow area $g(x, y) = xy = 18$, where $x > 0$ and $y > 0$.

2. Compute Gradient Vectors: State the Two-Variable Gradient Formula (defined in The Gradient Vector):

   $$\nabla P(x, y) = \left\langle \frac{\partial P}{\partial x}, \frac{\partial P}{\partial y} \right\rangle = \langle 1, 2 \rangle = \hat{i} + 2\hat{j}$$
   $$\nabla g(x, y) = \left\langle \frac{\partial g}{\partial x}, \frac{\partial g}{\partial y} \right\rangle = \langle y, x \rangle = y\hat{i} + x\hat{j}$$

3. Formulate Lagrange Multiplier System: State the Lagrange Multiplier Vector System (defined in Lagrange Multipliers):

   $$\nabla P = \lambda \nabla g \quad \text{and} \quad xy = 18$$
   Equate scalar components:
   $$\hat{i}: 1 = \lambda y$$
   $$\hat{j}: 2 = \lambda x$$
   $$\text{Constraint}: xy = 18$$

4. Eliminate $\lambda$ and Determine Geometric Aspect Ratio: Because physical dimensions require depth $y > 0$ and width $x > 0$, we can safely solve for $\lambda$ without dividing by zero: From $\hat{i}$:

   $$\lambda = \frac{1}{y}$$
   Substitute $\lambda = \frac{1}{y}$ into $\hat{j}$:
   $$2 = \left(\frac{1}{y}\right)x \implies 2 = \frac{x}{y} \implies x = 2y$$
   The optimal bottom width must be exactly twice the water depth ($x = 2y$).

5. Solve for Optimal Channel Dimensions: Substitute $x = 2y$ into the flow area constraint $xy = 18$:

   $$(2y)y = 18 \implies 2y^2 = 18 \implies y^2 = 9 \implies y = 3\text{ m}$$
   (Discarding negative root $y = -3$ since physical depth must be positive). Now compute the channel bottom width:
   $$x = 2(3) = 6\text{ m}$$
   The minimum wetted perimeter is:
   $$P_{\min} = x + 2y = 6 + 2(3) = 6 + 6 = 12\text{ m}$$

6. Hydraulic Engineering Interpretation: In civil fluid mechanics, Manning's equation states that fluid discharge velocity increases with hydraulic radius $R_h = A / P$. For a fixed cross-sectional area $A$, minimizing the wetted perimeter $P$ directly maximizes the hydraulic radius $R_h$, maximizing volumetric conveyance capacity while minimizing frictional shear resistance against the rough concrete liner walls. The aspect ratio $x = 2y$ (a half-square profile) is the universal theoretically optimal cross-section for open rectangular channels.