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Chapter 10: Problem 21

Water flows in a rectangular channel with a flowrate per unit width of \(q=1.5 \mathrm{m}^{2} / \mathrm{s}\) and a depth of \(0.5 \mathrm{m}\) at section \((1) .\) The head loss between sections (1) and (2) is \(0.03 \mathrm{m}\). Plot the specific energy diagram for this flow and locate states (1) and (2) on this diagram. Is it possible to have a head loss of \(0.06 \mathrm{m} ?\) Explain.

Short Answer

Expert verified
The specific energy at sections (1) and (2) is 1.436 m and 1.406 m respectively. As for the possibility of sustaining a head loss of 0.06 m, it is not possible with the given flow conditions since it would result in a specific energy less than the minimum specific energy for this flow condition.

Step by step solution

01

Calculation of Velocity at Section (1)

The first step is to calculate the velocity at section (1). Velocity, \( V \), can be calculated using the formula \( V = q/y \), where \( q \) is the flowrate per unit width and \( y \) is the depth. Substituting the given values, we have \( V = 1.5/0.5 = 3 \mathrm{m/s} \).
02

Calculation of Specific Energy at Section (1)

Next, we find the specific energy at section (1). The specific energy, \( E \), can be calculated using the formula \( E = y + V^2/2g \), where \( g \) is the gravity constant (9.81 \( \mathrm{m/s^2} \)). Substituting the known values, we get \( E = 0.5 + 3^2/2*9.81 = 1.436 \mathrm{m} \).
03

Calculation of Specific Energy at Section (2)

Now, we find the specific energy at section (2). As we know, the head loss between sections (1) and (2) is 0.03 m. Thus, the specific energy at section (2) should be the specific energy at section (1) minus the head loss. So, \( E = 1.436 - 0.03 = 1.406 \mathrm{m} \).
04

Plotting the Specific Energy Diagram and Points

The specific energy diagram plots Depth vs Specific Energy. Plot the curve typically starting from (0,0). Marking points for section (1) and (2) we have the coordinates (1.436, 0.5) and (1.406, 0.5) respectively.
05

Analysis of Increased Head Loss

Now, if the head loss were 0.06 m, the specific energy at section (2) would become \( E = 1.436 - 0.06 = 1.376 \mathrm{m} \). This is less than the minimum specific energy possible for this flow condition. Hence, a head loss of 0.06 m is not sustainable for given conditions.

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