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Chapter 4: Problem 34

Water enters a wide, flat channel of height \(2 h\) with a uniform velocity of \(2.5 \mathrm{m} / \mathrm{s}\). At the channel outlet the velocity distribution is given by \\[ \frac{u}{u_{\max }}=1-\left(\frac{y}{h}\right)^{2} \\] where \(y\) is measured from the centerline of the channel. Determine the exit centerline velocity, \(u_{\max }\)

Short Answer

Expert verified
The exit centerline velocity, or maximum exit velocity, \(u_{max}\) is \(3.75 \, m/s\).

Step by step solution

01

Identify Given Information

The provided information includes the initial velocity of the water entering the channel, noted as \(2.5 \, m/s\), and the height of the channel, denoted as \(2h\). We are also given the velocity distribution at the channel's outlet, represented by the equation \(\frac{u}{u_{\max }}=1-\left(\frac{y}{h}\right)^{2}\), where \(y\) measures from the centerline of the channel.
02

Apply the Principle of Mass Conservation

In fluid dynamics, the mass flow rate must remain constant along the channel, assuming no losses. Therefore, this principle states that the product of the area and velocity at the inlet should be equal to the product of the area and the average velocity at the outlet. The area of the wide channel can be considered constant, so we just match the velocities.
03

Compute for the Average Exit Velocity

Having established the conservation of mass flow rate, the next step requires finding the average exit velocity (\(u_{avg}\)) from the given velocity distribution. For a flat channel, we integrate the velocity profile over the cross-sectional area and then divide by the total area. Here is the formula: \[ u_{avg} = \frac{1}{2h} \int_{-h}^{h} u dy \]. Substitute the velocity distribution equation to \[u = u_{max}(1-(\frac{y}{h})^2)\]. The integral simplifies to: \[ u_{avg} = \frac{u_{max}}{2h} \int_{-h}^{h} (1-\frac{y^2}{h^2}) dy \]
04

Evaluate the Integral

Now, evaluate the integral, which gives \(u_{avg} = \frac{2u_{max}}{3}\). This is the equation that relates the average exit velocity to the maximum exit velocity.
05

Determine the Maximum Exit Velocity

From mass conservation, as noted in Step 2, we know that the inlet velocity is equal to the average outlet velocity. Therefore, we set \(2.5 \, m/s = \frac{2u_{max}}{3}\). Solving this equation for \(u_{max}\), we get \(u_{max} = 3.75 \, m/s\)

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