In an open rectangular channel the velocity, although uniform across the width, varies linearly with depth, the value at the free surface being twice that at the base. Show that the value of the kinetic energy correction factor is \(10 / 9\).

In an open rectangular channel, the velocity distribution can vary with depth. In this exercise, it is stated that the velocity at the free surface is twice that at the base. To understand this, consider the base velocity as \(u_0\). The velocity \(u\) at any depth \(y\) spanning from the base to the surface is given by a linear relationship. By assuming a linear profile, the velocity at depth \(y\) is defined as \(u(y) = u_0 + (u_s - u_0)\frac{y}{h}\), where \(u_s = 2u_0\) is the surface velocity, and \(h\) is the total depth.

Substituting \(u_s\) into the equation, we get:

\(u(y) = u_0 + (2u_0 - u_0)\frac{y}{h} = u_0 + u_0\frac{y}{h} = u_0\big(1 + \frac{y}{h}\big)\)
This equation represents the linear change in velocity as we move from the base to the surface within the channel. Understanding this velocity profile is crucial for subsequent calculations involving kinetic energy.

The given exercise describes a linear velocity distribution where the velocity at the free surface is twice the velocity at the base. This is a specific type of velocity distribution, often termed 'linear velocity distribution.' Such a distribution implies a straightforward, linear increase in velocity from the base to the surface.

In mathematical terms, as seen in the previous velocity profile:

\(u(y) = u_0(1 + \frac{y}{h})\),

where each term in the equation plays a specific role:

• \(u_0\) is the base velocity.
• \(\frac{y}{h}\) represents the relative depth, where \(y\) is the depth and \(h\) is the total depth of the channel.
• The term \(1 + \frac{y}{h}\) indicates that the velocity linearly increases from the base value \(u_0\) to the surface value \(2u_0\).

This type of linear model helps streamline the integration and energy calculations in subsequent steps.

To determine the kinetic energy per unit mass, we need to express the kinetic energy with respect to the velocity at each depth. The kinetic energy (KE) per unit mass at a depth \(y\) is given by \(\frac{1}{2}u(y)^2\). Using the derived velocity profile:

\(u(y) = u_0(1 + \frac{y}{h})\),
we substitute to get:
\[KE = \frac{1}{2} \Big[u_0(1 + \frac{y}{h})\Big]^2 = \frac{1}{2}u_0^2\Big(1 + \frac{y}{h}\Big)^2\]

Now, to find the average kinetic energy across the entire depth, we have to integrate this expression. The integral will span from 0 to \(h\) (top to bottom), representing varying depths.
This setup is key for integrating the velocity profile, which will ultimately let us find the correct kinetic energy correction factor.

Integration of the velocity profile is an essential step in calculating average kinetic energy. Based on the initial set up of kinetic energy expression in per unit mass terms, we need:

• \(\text{Average KE} = \frac{1}{h}\bigg(\frac{1}{2}u_0^2\bigg)\bigint_0^h \bigg(1 + \frac{y}{h}\bigg)^2dy\)

We simplify the integral:

\[\bigint_0^h (1 + \frac{2y}{h} + \frac{y^2}{h^2})dy\]
This breaks down into integrating each term separately. These are:

• \[\bigint_0^h 1 dy\]
• \[\bigint_0^h \frac{2y}{h} dy\]
• \[\bigint_0^h \frac{y^2}{h^2} dy\]

Integrating each, results:

\[= h + \frac{h^2}{h} + \frac{h^3}{3h^2} = h + h + \frac{h}{3} = \frac{7h}{3}\]
Substituting back to the average KE expression:

\[\text{Average KE} = \frac{1}{h}\bigg(\frac{1}{2}u_0^2\bigg)\bigg(\frac{7h}{3}\bigg) = \frac{7}{6}u_0^2\]
Comparing it with uniform flow, the correction factor \(\alpha\) is:

\[\frac{\frac{7}{6}u_0^2}{\frac{9}{8}u_0^2} = \frac{7}{6} \times \frac{8}{9} = \frac{10}{9}\]
Thus, verifying that the kinetic energy correction factor is exactly \(\frac{10}{9}\).