# How To Calculate The Wetted Perimeter

Many factors affect the flow of water through a river or pipe, and one of the most important of these is the hydraulic radius. This depends on the total cross-sectional area of the enclosure and what’s called the wetted perimeter, which essentially tells you how much of the walls of the enclosure are in contact with water.

Calculating the wetted perimeter isn’t always easy, because it depends strongly on the shape of the reservoir and the water level. If you can’t directly measure the wetted perimeter, you’ll need to estimate it using a shape that roughly matches the shape of the reservoir.

## What Is a Wetted Perimeter?

The wetted perimeter of a river or other container of water is part of the perimeter of the cross-sectional area of the container. To be more precise, it’s the part of the cross-sectional area in direct contact with the water, so it extends all along the bed of the water, and up the sides to the point that corresponds to the surface of the water.

Finding this out is a little different from calculating the cross-sectional area of the container, then, even though there are some similarities in terms of the information you need.

## Calculating the Wetted Perimeter – General

To calculate the wetted perimeter, you need to either estimate it or measure the length of each of the sides of the river or container in contact with the water. The general formula for a wetted perimeter P is:

P= \sum_i l_i

Where li is the length of the side i, and the sum runs over all of the sides in contact with the water. This formula is quite straightforward to use in principle, but in practice actually finding the information you need isn’t easy. If you’re actually in the location of the water and the surfaces in contact with it, the simplest way to find the wetted perimeter is to physically measure all of the relevant sides and add them together.

In some cases, though – such as for a river – this presents its own problems, and estimating the perimeter might be a more practical way to address the problem.

## Approximating as a Trapezoid

In many cases, the cross-sectional area for water in a river can be approximated as forming a trapezoid, with the shorter side as the base along the river bed. The formula for finding the wetted perimeter in this case is:

P = b + 2 \Bigg(\bigg(\frac{(T - b)}{2}\bigg)^2 + h^2\Bigg)^{1/2}

Where b is the length of the base, T is the length of the top (from bank to bank) and h is the height of the water. Again, finding the values for these might not be easy, but you can estimate if it’s difficult to obtain the information otherwise.

## Approximating as a Rectangle

A rectangle is simpler to calculate the wetted perimeter for, but most natural water flows have angled banks and so would be better approximated as a trapezoid. However, if you do have a reservoir that can be approximated as a rectangle, the math is much easier:

P = b + 2h

Where b is the base and h is the height of the water.

## Approximating as a Circle

If you’re considering the flow of water through a pipe, or through another shape you think can be accurately approximated as the cross-section of part of a circle, you can calculate the wetted perimeter using the formula for the length of an arc of a circle.

If you’re calculating for a pipe, you’ll likely know the diameter (and therefore the radius) of the pipe from its specifications, which makes the process much easier. The formula for the length of the arc (with the angle measured in radians) is:

P = θr

Where θ is the angle at the center of the circle covered by the arc containing water and r is the radius. For example, if the water fills up half of the circular cross-section, this is π radians, giving a wetted perimeter of π_r_ = π_d_ / 2, where d is the diameter of the pipe.

In other words, as you would expect, the wetted perimeter in this case is half the circumference of the circle. Given that there are 2π radians in a circle, a full pipe would have a wetted perimeter of 2 π_r_ – the circumference of the circle.

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