Open-Channel Cross-Sections For Maximum Flow Area

Why Optimize an Open Channel’s Cross-Section?

An open channel carries water with a free surface exposed to the air. Its cross-section determines how much water can occupy the channel at a given depth. For a rectangular channel, the cross-section is described by its bottom width and water depth.

Suppose the available lining material fixes the length of the channel boundary that is in contact with water. A design question is then: how should that fixed wetted perimeter be divided between the bottom and the two sides to enclose the largest cross-sectional area?

This is an area optimization problem, not a complete model of flow rate. The area through which water flows is important, but discharge also depends on other hydraulic conditions. Here, the objective is specifically to maximize cross-sectional area for a fixed wetted perimeter.

Model and Definitions

Let \(b\) be the channel’s bottom width and \(h\) its water depth. Assume the sides are vertical, the channel is open at the top, and \(b>0\) and \(h>0\). The free surface is not part of the wetted perimeter: water touches the bottom and the two vertical sides only.

Definition: For an open rectangular channel with bottom width \(b\) and depth \(h\), the wetted perimeter is \(P=b+2h\), and the cross-sectional flow area is \(A=bh\). In this problem, \(P\) is fixed and positive.

Because \(P\) is fixed, the dimensions must satisfy \(b+2h=P\). The feasible positive dimensions have \(0<h<P/2\), with \(b=P-2h\). The strict inequalities ensure that both dimensions are positive.

The optimization problem is therefore to maximize \(A=bh\) over all positive \(b,h\) satisfying \(b+2h=P\). The constraint makes one dimension dependent on the other, so the problem can be reduced to a single-variable function.

Deriving the Optimal Dimensions

Substitute \(b=P-2h\) into the area formula. This gives a quadratic function of depth. Completing the square shows both the maximum value and the unique depth where it occurs.

$$ A(h)=bh=(P-2h)h=Ph-2h^2 $$

Rewrite this expression by completing the square:

A(h) =-2\left(h-\frac{P}{4}\right)^2+\frac{P^2}{8}.

A square is always nonnegative, so \(-2(h-P/4)^2\leq 0\). Thus \(A(h)\leq P^2/8\), with equality exactly when \(h=P/4\). This depth is feasible because \(P>0\), so \(0<P/4<P/2\). The corresponding width is \(b=P-2(P/4)=P/2\).

Theorem: Among positive rectangular open-channel dimensions with fixed wetted perimeter \(P>0\), the unique maximum cross-sectional area occurs at \(b=P/2\) and \(h=P/4\). The maximum area is \(P^2/8\), and the optimal width-to-depth ratio is \(b:h=2:1\).

This argument proves a global maximum, not merely a local one: every feasible depth satisfies \(A(h)\leq P^2/8\), and the equality condition identifies the single maximizing design. At the limiting endpoints \(h=0\) or \(h=P/2\), one dimension vanishes and the area is zero; these degenerate cases are excluded by the requirement that both dimensions be positive.

Why the Ratio Is Two to One

The ratio can also be understood directly from the perimeter constraint. The two side lengths together use \(2h\) units of wetted perimeter. At the optimum, \(b=P/2\) and \(2h=P/2\), so the bottom length equals the combined length of the two sides. Since there are two equal sides, this means the bottom width is twice the depth.

Equivalently, set \(x=b\) and \(y=2h\). Then \(x+y=P\), while \(A=bh=xy/2\). For a fixed sum, the product \(xy\) is greatest when \(x=y\), as follows from \((x-y)^2\geq 0\). Equality gives \(b=2h\), and the perimeter constraint then gives \(b=P/2\) and \(h=P/4\). This is the same optimum expressed in terms of how the perimeter is divided between the bottom and the two sides.

Worked Examples

Worked Example: Optimize a Channel with 20 Metres of Wetted Perimeter

A rectangular open channel has fixed wetted perimeter \(P=20\text{ m}\). Find its area-maximizing width and depth, then verify the perimeter and area.

From the optimal-dimension formulas, the width is

b=\frac{P}{2}=\frac{20\text{ m}}{2}=10\text{ m}.

The depth is

h=\frac{P}{4}=\frac{20\text{ m}}{4}=5\text{ m}.

The area is therefore

A=bh=(10\text{ m})(5\text{ m})=50\text{ m}^2.

Check the wetted perimeter: \(b+2h=10\text{ m}+2(5\text{ m})=20\text{ m}\), as required. The formula for the maximum area gives \(P^2/8=(20\text{ m})^2/8=400\text{ m}^2/8=50\text{ m}^2\), matching the calculated area. The dimensions also satisfy \(b=2h\).

Worked Example: Compare a Proposed Design with the Maximum

A channel has \(P=30\text{ ft}\) of wetted perimeter. A proposed design has bottom width \(b=18\text{ ft}\) and depth \(h=6\text{ ft}\). Determine whether it maximizes area and calculate the area difference from the optimum.

First check the constraint:

b+2h =18\text{ ft}+2(6\text{ ft}) =18\text{ ft}+12\text{ ft} =30\text{ ft}.

So the proposed dimensions are feasible. Their area is

A_{\mathrm{proposed}} =(18\text{ ft})(6\text{ ft}) =108\text{ ft}^2.

For \(P=30\text{ ft}\), the maximizing dimensions are

b_{\mathrm{opt}}=\frac{30\text{ ft}}{2}=15\text{ ft}, \qquad h_{\mathrm{opt}}=\frac{30\text{ ft}}{4}=7.5\text{ ft}.

Their area is \((15\text{ ft})(7.5\text{ ft})=112.5\text{ ft}^2\). The maximum-area formula independently gives \((30\text{ ft})^2/8=900\text{ ft}^2/8=112.5\text{ ft}^2\). Thus the proposed design falls short by \(112.5\text{ ft}^2-108\text{ ft}^2=4.5\text{ ft}^2\). It is feasible, but not optimal: its width-to-depth ratio is \(18:6=3:1\), rather than \(2:1\).

Worked Example: Find the Perimeter Needed for a Maximum Area of 50 Square Metres

Suppose a rectangular open channel is to have maximum cross-sectional area \(50\text{ m}^2\). Find the wetted perimeter and the dimensions that achieve this maximum.

For an optimized channel, \(A_{\max}=P^2/8\). Set this equal to the target area and solve for the positive perimeter:

50\text{ m}^2=\frac{P^2}{8}, \qquad P^2=400\text{ m}^2, \qquad P=20\text{ m}.

Only the positive square root is relevant because a perimeter is positive. The corresponding optimal dimensions are

b=\frac{P}{2}=\frac{20\text{ m}}{2}=10\text{ m}, \qquad h=\frac{P}{4}=\frac{20\text{ m}}{4}=5\text{ m}.

Check that the dimensions use the calculated perimeter: \(10\text{ m}+2(5\text{ m})=20\text{ m}\). Their area is \((10\text{ m})(5\text{ m})=50\text{ m}^2\), which matches the target. The ratio is \(10:5=2:1\), as required for the maximum.

A Common Modeling Mistake

A frequent error is to count the open top as part of the wetted perimeter. For example, one might incorrectly write \(P=2b+2h\) and then conclude that the maximum-area design has \(b=h\). That perimeter formula describes a closed rectangle, with all four sides included. In an open channel, the top is a free surface rather than a boundary in contact with the channel lining, so the correct constraint is \(P=b+2h\). Using that constraint gives \(b=2h\), not \(b=h\).

It is also important not to interpret the result as saying that this ratio always maximizes water discharge. The result establishes the greatest cross-sectional area for a fixed wetted perimeter under the rectangular, vertical-sided model. A conclusion about discharge would require a model for how flow depends on the channel and its operating conditions.

When to Use This Result

Use this optimization when the design objective is maximum rectangular cross-sectional area and the available wetted perimeter is fixed. The model requires a positive fixed perimeter, a horizontal bottom, vertical sides, and an open top. Under those assumptions, the calculation is direct: halve the perimeter to get the width, take one quarter of the perimeter to get the depth, and divide the perimeter squared by eight to get the maximum area.

The ratio \(2:1\) is independent of the particular perimeter. Increasing \(P\) scales both optimal dimensions in direct proportion, while the maximum area scales with \(P^2\). That scaling follows from the formulas and reflects that area has units of length squared, whereas width, depth, and perimeter have units of length.

Check Your Understanding

Use the constraint and maximum-area result to answer these questions.

Why does the wetted perimeter of an open rectangular channel equal \(b+2h\), rather than \(2b+2h\)?
For a fixed wetted perimeter, what width and depth maximize the cross-sectional area?
A channel has wetted perimeter \(P=16\text{ m}\). Find its optimal width, depth, and maximum area.
A proposed design has \(b=12\text{ ft}\) and \(h=4\text{ ft}\). Does its width-to-depth ratio match the maximum-area ratio?
Does the maximum-area result alone prove that a channel has maximum discharge? Explain briefly.