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Chapter 9: Problem 3

Show that for circular pipes of diameter, \(D,\) the hydraulic radius, \(R,\) is related to the pipe diameter by \(R=D / 4\).

Short Answer

Expert verified
The hydraulic radius, \( R = \frac{D}{4} \), is confirmed for a full circular pipe.

Step by step solution

01

Understand the Hydraulic Radius

The hydraulic radius, \( R \), is defined as the cross-sectional area of flow \( A \) divided by the wetted perimeter \( P \). For a full circular pipe, the flow covers the entire circular cross-section.
02

Calculate the Cross-Sectional Area

For a circular pipe with diameter \( D \), the radius \( r \) is \( \frac{D}{2} \). The cross-sectional area \( A \) is calculated as:\[ A = \pi r^2 = \pi \left( \frac{D}{2} \right)^2 = \frac{\pi D^2}{4} \]
03

Calculate the Wetted Perimeter

The wetted perimeter \( P \) for a full circular pipe is the circumference of the circle, given by:\[ P = \pi D \]
04

Substitute and Simplify

Substitute \( A \) and \( P \) into the formula for hydraulic radius:\[ R = \frac{A}{P} = \frac{\frac{\pi D^2}{4}}{\pi D} \]Simplifying the expression, we cancel \( \pi \) and \( D \):\[ R = \frac{D^2}{4D} = \frac{D}{4} \]
05

Conclusion

We have shown through substitution and simplification that the hydraulic radius \( R \) is \( \frac{D}{4} \) for a full circular pipe with diameter \( D \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Circular Pipes
Circular pipes are commonly used in various engineering applications, especially for transporting fluids like water, oil, and gas. The primary advantage of circular pipes is their ability to withstand internal pressure equally across the pipe's surface. This is because the circular shape evenly distributes stress, which prevents localized weaknesses. Circular pipes are also efficient in terms of fluid flow because they have a smaller perimeter for a given cross-sectional area, which reduces frictional resistance.
  • The shape helps in minimizing structural stress due to internal pressure.
  • Efficient in transporting fluids due to optimized flow characteristics.
  • Widely used because manufacturing and installation are relatively straightforward and cost-effective.
Understanding circular pipes in terms of their geometry is essential for analyzing their flow properties and hydraulic characteristics.
Cross-Sectional Area
The cross-sectional area of a circular pipe is crucial in determining its carrying capacity. It represents the area through which the fluid flows, and for a circular pipe with diameter, the area is defined using the formula for the area of a circle. For a pipe with diameter \( D \), the radius \( r \) is \( \frac{D}{2} \).
The formula for the cross-sectional area \( A \) is:\[ A = \pi r^2 = \pi \left( \frac{D}{2} \right)^2 = \frac{\pi D^2}{4} \]
  • This formula helps to assess how much fluid can pass through the pipe at any given time.
  • A larger cross-sectional area allows for a greater volume of fluid flow, assuming all other conditions are constant.
Understanding the cross-sectional area is essential when calculating the hydraulic radius and for further fluid mechanics analysis.
Wetted Perimeter
The wetted perimeter in the context of circular pipes describes the portion of the pipe's circumference that is in contact with the fluid. For a pipe completely filled with fluid, the wetted perimeter is simply the entire circumference of the circle.
The formula for the wetted perimeter \( P \) is:\[ P = \pi D \]
  • For full circular pipes, the entire perimeter is wetted.
  • The perimeter influences the frictional forces acting on the flow on the pipe surface.
  • It is an important factor in calculating the hydraulic radius, which influences how efficiently fluid can flow through the pipe.
Knowing the wetted perimeter is essential for analyzing fluid dynamics in pipes and determining flow resistance.
Diameter
The diameter of a pipe is a fundamental dimension that significantly impacts its flow characteristics and structural behavior. In a circular pipe, the diameter is the distance across the circle through its center.
  • The diameter directly affects the cross-sectional area, as it is squared in the area equation.
  • A larger diameter leads to a higher cross-sectional area and, consequently, a greater flow capacity.
  • It is also directly used in calculating the hydraulic radius, as seen in understanding the relationship \( R = \frac{D}{4} \).
Recognizing the role of pipe diameter helps in the selection of appropriate pipe sizes for specific applications, ensuring efficient fluid flow and structural integrity.

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Most popular questions from this chapter

Water flows in a horizontal trapezoidal channel at \(21 \mathrm{~m}^{3} / \mathrm{s}\), where the bottom width of the channel is \(2 \mathrm{~m}\), side slopes are \(1: 1,\) and the depth of flow is \(1 \mathrm{~m} .\) Calculate the downstream depth required for a hydraulic jump to form at this location. What would be the energy loss in the jump?

Given that hydraulically rough flow conditions occur in open channels when \(u_{*} k_{s} / \nu \geq\) 70 , show that this condition can be expressed for water in terms of Manning parameters as $$ n^{6} \sqrt{R S_{0}} \geq 7.9 \times 10^{-14} $$ If a concrete-lined rectangular channel with a bottom width of \(5 \mathrm{~m}\) is constructed on a slope of \(0.05 \%\) and Manning's \(n\) is estimated to be \(0.013,\) determine the minimum flow depth for hydraulically rough flow conditions to exist.

Water flows at \(6.2 \mathrm{~m}^{3} / \mathrm{s}\) in a rectangular channel of width \(5 \mathrm{~m}\) and depth of flow of \(1.5 \mathrm{~m}\). If the channel width is decreased by \(0.50 \mathrm{~m}\) and the bottom of the channel is raised by \(0.15 \mathrm{~m}\), what is the depth of flow in the constriction?

At a particular river cross section, the elevation of the bottom of the river is \(103.75 \mathrm{~m}\) and the elevation of the river bank is \(106.43 \mathrm{~m}\). The river has an approximately trapezoidal cross section with side slopes of \(3: 1(\mathrm{H}: \mathrm{V})\) and a bottom width of \(18 \mathrm{~m} .\) The longitudinal slope of the river and the adjacent floodplain is \(2 \%,\) and the Manning roughness coefficient of the river is estimated as \(0.07 .\) A bridge is to be placed 120 m downstream of the river section, and over this distance, the river will be made to transition to a bottom width of \(9 \mathrm{~m}\) while maintaining the same side slope. Under design conditions, the flow in the river is \(12 \mathrm{~m}^{3} / \mathrm{s}\) and the depth of flow at the upstream section is \(1.60 \mathrm{~m}\). (a) What will be the depth of flow at the bridge section? (b) Will the floodplain be flooded at the bridge location? Explain.

It has been shown that in fully turbulent flow, Manning's \(n\) can be related to the height, \(d\), of the roughness projections by the relation $$ n=0.039 d^{\frac{1}{6}} $$ where \(d\) is in meters. If the estimated roughness height in a channel is \(30 \mathrm{~mm}\), determine the percentage error in \(n\) resulting from a \(70 \%\) error in estimating \(d\).
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