# How to Derive an Expression for Average Cross-Sectional Velocity

## What is the given velocity distribution in the problem P.1.3?

Let the point velocity in a wide rectangular channel be expressed as v = 2.5vx In 30z ks. How can we derive an expression for the average cross-sectional velocity, V, based on this velocity distribution?

## Step-by-Step Guide to Derive the Average Cross-Sectional Velocity

To derive an expression for the average cross-sectional velocity, V, based on the given velocity distribution, follow these steps:

- Start with the given velocity distribution equation: v = 2.5vx ln(30z ks), where v is the point velocity.
- Integrate the velocity distribution over the cross-section of the channel and divide by the flow depth, y.
- Apply the hint provided: v = 0 at z = k^3/30.
- Simplify the integral using the hint provided in the problem.
- Substitute the variables and integrate the equation.
- Evaluate the integral at the upper and lower limits.
- Divide the result by the flow depth, y, to find the average cross-sectional velocity, V.

## Detailed Explanation

The expression for the average cross-sectional velocity, V, based on the given velocity distribution v = 2.5vx ln(30z ks) can be derived by following the steps outlined above.

Integrating the velocity distribution equation over the cross-section of the channel and applying the provided hint allows us to simplify the integral and find the average cross-sectional velocity, V.

By evaluating the integral at the upper and lower limits and dividing the result by the flow depth, y, we can obtain the expression for the average cross-sectional velocity, V. This helps in understanding the flow dynamics in the wide rectangular channel based on the given velocity distribution.