/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 2 In this problem, you use momentu... [FREE SOLUTION] | 91影视

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In this problem, you use momentum flux (Problem \(7.25\) ) to estimate the drag force on a sphere of radius \(r\) in a flow at low Reynolds number \((\operatorname{Re} \ll 1)\). If \(\operatorname{Re} \ll 1\), the boundary layer (Section \(7.3 .4)\) - the region over which the fluid velocity changes from zero to the free-stream velocity \(v\) -is comparable in thickness to \(r\). Using that information, estimate the viscous drag force on the sphere.

Short Answer

Expert verified
In a low Reynolds number flow where \(\operatorname{Re} \ll 1\), the viscous drag force (F_D) acting on a sphere of radius r can be estimated using the momentum flux and boundary layer thickness concepts. The viscous drag force can be calculated using the following formula: \( F_D = 4 \pi \rho v^2 r^3 \) Where \(\rho\) is the fluid density, and \(v\) is the free-stream velocity.

Step by step solution

01

Define the variables

In this problem, we have the following variables: - r: the radius of the sphere - Re: Reynolds number - v: free-stream velocity - F_D: viscous drag force
02

Estimate the momentum flux across the boundary layer

Momentum flux (蠁) is defined as the momentum transfer per unit area per unit time. In this case, the momentum flux across the boundary layer can be estimated as: \( \phi = \rho v^2 d \) where: - 蟻: fluid density - d: boundary layer thickness We know that the boundary layer thickness (d) is comparable to the radius (r) in this case, so we have: \( \phi = \rho v^2 r \)
03

Calculate the drag force on the sphere

To estimate the drag force on the sphere, we need to consider the momentum flux acting over the entire surface of the sphere. The surface area of the sphere is given by the following equation: \( A = 4 \pi r^2 \) The drag force (F_D) can be obtained by multiplying the momentum flux (蠁) by the surface area (A) of the sphere: \( F_D = \phi A \) Substitute the value of 蠁 and A from the previous steps: \( F_D = (\rho v^2 r)(4 \pi r^2) \)
04

Simplify the expression for drag force

Now, simplify the expression for the drag force: \( F_D = 4 \pi \rho v^2 r^3 \) This is the estimated viscous drag force acting on a sphere of radius r in a flow with low Reynolds number.

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Key Concepts

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

Viscous Drag Force
Viscous drag force is a fundamental concept in fluid dynamics that refers to the resistance experienced by an object moving through a fluid. It arises from the interaction between the object and the layers of fluid sliding over it.
This drag force is significant in conditions like low Reynolds numbers, where the flow is highly influenced by viscous forces rather than inertial forces. The formula for viscous drag force involves parameters such as the fluid velocity, fluid viscosity, and surface area over which the fluid interacts with the object. In the case of the sphere discussed in the exercise, the viscous drag force equation derived is:
  • The basic form is proportional to surface area and the square of velocity:
    • \[ F_D = 4 \pi \rho v^2 r^3 \]
  • It indicates that for a fluid flowing around a sphere, the drag force not only depends on the speed of the flow and radius of the sphere, but also on how these interact across the surface of the sphere.
Understanding the viscous drag force provides insight into how objects of various shapes and sizes are affected by fluid flows in practical scenarios like engineering and environmental sciences.
Low Reynolds Number
The Reynolds number is crucial for determining the flow regime in which an object interacts with a fluid. It is a dimensionless quantity that compares inertial forces to viscous forces. When dealing with low Reynolds numbers, defined by \( \text{Re} \ll 1 \), the flow characteristics differ sharply from high Reynolds number scenarios.
In low Reynolds number conditions:
  • The flow is generally smooth and laminar.
  • Viscous forces dominate over inertial forces, meaning the fluid flows steadily without chaotic turbulence.
  • The movement of the fluid is highly predictable, which makes the calculation of viscous drag force more straightforward.
The concept is best visualized in instances like microscopic organisms swimming in water, where the size and speed are extremely low, emphasizing the viscous effects. For the exercise, the fluid's interaction with the sphere at low Reynolds number implies the boundary layer also plays a critical role in flow behavior.
Boundary Layer Thickness
The boundary layer is the thin region adjacent to a body's surface in a flowing fluid where velocity changes from zero (at the surface due to the no-slip condition) to the free-stream velocity. Boundary layer thickness is a critical parameter in determining the region's influence on overall flow and drag force.
If we have a scenario such as described in the exercise, where the thickness of the boundary layer is comparable to the radius of the sphere, several key points arise:
  • The boundary layer contributes significantly to the viscous drag because it occupies a large portion around the surface of the object.
  • In conditions like low Reynolds number, the boundary layer doesn鈥檛 separate as easily, making the flow within it more stable and the analysis simpler.
  • The momentum transfer, or momentum flux, across this boundary layer is crucial for accurately estimating the drag force (viscous).
Comprehending the boundary layer's behavior and thickness allows for the prediction of flow patterns, surface wear, and design efficiencies in applications ranging from aerodynamics to chemical processing.

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Most popular questions from this chapter

You are trying to estimate the area of a rectangular field. Your plausible ranges for its width and length are \(1 \ldots 10\) meters 10...100 meters, respectively. a. What are the midpoints of the two plausible ranges? b. What is the midpoint of the plausible range for the area? c. What is the too-pessimistic range for the area, obtained by multiplying the corresponding endpoints? d. What is the actual plausible range for the area, based on combining log- normal distributions? This range should be narrower than the pessimistic range in part (c)! e. How do the results change if the ranges are instead \(2 \ldots 20\) meters for the width and \(20 \ldots 200\) meters for the length?

Based on the boundary-layer thickness \(\delta \sim \sqrt{\nu t}\), estimate the Reynolds number in a boundary layer on an object of size \(L\) (a length) moving in a fluid at speed \(v\).

If a quantity a has the plausible range \(1 \ldots 4\), and the quantity \(b\) has the plausible range \(10 \ldots 40\), what are the plausible ranges for \(a b\) and for \(a / b\) ?

For an object moving through a fluid, the Reynolds number is defined as \(v L / \nu\), where \(v\) is the object's speed, \(L\) is its size (a length), and \(\nu\) is the fluid's kinematic viscosity. Show that the Reynolds number has the physical interpretation $$ \frac{\text { momentum-diffusion time over a distance comparable to the size } L}{\text { fluid-transport time over a distance comparable to the size } L} $$

Why does mercury (Hg) have such a low thermal conductivity for a metal?

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