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How fast can a raindrop \((r=1.5 \mathrm{~mm})\) fall through air if the flow around it is to be close to turbulent - that is, for \(N_{R}\) close to 10 ? For air, \(\eta=1.8 \times 10^{-5} \mathrm{~Pa} \cdot \mathrm{s}\) and \(\rho=1.29 \mathrm{~kg} / \mathrm{m}^{3}\).

Short Answer

Expert verified
The raindrop can fall at approximately 0.0465 m/s for nearly turbulent flow.

Step by step solution

01

Understanding the Problem

To find the speed of the raindrop when the Reynolds number is near 10, we need to use the relationship between the Reynolds number, flow speed, the radius of the raindrop, and properties of air.
02

Reynolds Number Formula

The formula for the Reynolds number for a sphere is given by: \[ N_{R} = \frac{2 r v \rho}{\eta} \] where \( r \) is the radius of the raindrop, \( v \) is the velocity, \( \rho \) is the density of the fluid (air), and \( \eta \) is the dynamic viscosity.
03

Substitute Known Values

Substitute the known values into the Reynolds number equation \((N_{R}=10, r=1.5\,\mathrm{mm}=1.5\times 10^{-3}\,\mathrm{m}, \rho=1.29\,\mathrm{kg/m^3}, \eta=1.8\times 10^{-5}\,\mathrm{Pa\cdot s})\): \[ 10 = \frac{2 \times 1.5 \times 10^{-3} \times v \times 1.29}{1.8 \times 10^{-5}} \]
04

Solve for Velocity \(v\)

Rearrange the equation to solve for \(v\): \[ v = \frac{10 \times 1.8 \times 10^{-5}}{2 \times 1.5 \times 10^{-3} \times 1.29} \] Calculate: \[ v \approx 0.0465 \text{ m/s} \]

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Key Concepts

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

Fluid Dynamics
Fluid dynamics is a part of fluid mechanics, which deals with the study of fluids (liquids and gases) in motion. The field combines several principles of physics and mathematics to understand how fluids flow and the forces that act upon them. It’s crucial in many natural phenomena and technical applications.

Within fluid dynamics, there are various flow types categorized by their behavior, such as laminar and turbulent flows. Understanding these can help solve problems like predicting weather patterns or determining the speed of a raindrop. The Navier-Stokes equations are the core equations in fluid dynamics, allowing us to describe fluid flow under different conditions.
  • Applications: From designing aircraft and predicting weather patterns to understanding ocean currents.
  • Importance: Helps in reducing energy loss in fluid transport systems.
  • Related Concepts: It ties closely with pressure, viscosity, and turbulence.
Studying fluid dynamics also includes examining the interactions between fluid elements, boundaries, and surfaces they contact, which can lead to understanding how forces like lift or drag are generated.
Turbulent Flow
Turbulent flow is when the fluid flow exhibits chaotic changes in pressure and velocity. Unlike a smooth and orderly laminar flow, turbulence is characterized by irregular, swirling motion. This chaotic behavior increases the energy required for movement and affects how substances mix within the fluid.

The Reynolds number, a dimensionless parameter, is critical in determining flow type. In the raindrop problem:
  • If the Reynolds number is low (typically < 2300 for pipe flow), the flow tends to be laminar.
  • If it's higher, the flow shifts towards being turbulent.
A turbulent flow, while seemingly disorderly, is extremely efficient in mixing and heat transfer. Therefore, engineers often need to predict whether flow will be turbulent or not to optimize processes that involve heat exchangers, and mixing tanks or control pollution in the atmosphere.
Dynamic Viscosity
Dynamic viscosity is a property of fluids that indicates how resistant the fluid is to deform under shear stress. It fundamentally measures a fluid's internal friction. When you stir honey, it resists being moved due to its high viscosity, unlike water which flows easily.

In mathematical terms, dynamic viscosity \(\eta\) is a factor in the Reynolds number equation, directly affecting whether flow becomes turbulent or stays laminar. For the air where a raindrop falls:
  • The dynamic viscosity of air is 1.8 × 10^{-5} Pa·s, indicating a relatively low resistance to shear.
  • Higher viscosities indicate thicker materials (like oil), implying a need for more force to move them.
Understanding viscosity is vital for designing systems where control of flow rates or force predictions are needed, like in pipes, through pumps, and around blades in aerodynamics.
Velocity Calculation
Velocity calculation becomes essential when determining the speed of objects moving in fluids, like a raindrop falling through the air. To find this speed when considering turbulence, one needs to use the Reynolds number formula, incorporating the raindrop's radius and the air's density and dynamic viscosity.

The formula gives us the relationship between these factors:
\[ N_{R} = \frac{2 r v \rho}{\eta} \]

Solving for the velocity, \(v\), involves substituting in known values for the raindrop's radius, the fluid's density, and dynamic viscosity. By rearranging and solving the equation:
  • The velocity is calculated based on the force balance, aiming for a specific type of flow, close to turbulence (Re here being around 10).
  • Velocity influences not just how fast the raindrop falls, but also the impact effects when it hits a surface.
This simple formula allows us to predict and understand numerous real-world phenomena related to movement within different fluids.

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Most popular questions from this chapter

Under the same pressure differential, compare the flow of water through a pipe to the flow of SAE No. 10 oil. \(\eta\) for water is \(0.801\) \(\mathrm{cP} ; \eta\) for the oil is \(200 \mathrm{cP}\). From Poiseuille's Law, \(J \propto 1 / \eta .\) Therefore, since everything else cancels, $$ \frac{J_{\text {water }}}{J_{\text {oil }}}=\frac{200 \mathrm{cP}}{0.801 \mathrm{cP}}=250 $$ The flow of water is 250 times as large as that of the oil under the same pressure differential.

A venturi meter equipped with a differential mercury manometer is shown in. At the inlet, point-1, the diameter is \(12 \mathrm{~cm}\), while at the throat, point-2, the diameter is \(6.0 \mathrm{~cm}\). What is the flow \(J\) of water through the meter if the mercury manometer reading is 22 \(\mathrm{cm}\) ? The density of mercury is \(13.6 \mathrm{~g} / \mathrm{cm}^{3}\). From the manometer reading (remembering that \(1 \mathrm{~g} / \mathrm{cm}^{3}=1000\) \(\mathrm{kg} / \mathrm{m}^{3}\) ): \(P_{1}-P_{2}=\rho g h=\left(13600 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(0.22 \mathrm{~m})=2.93 \times 10^{4} \mathrm{~N} / \mathrm{m}^{2}\) Since \(J=v_{1} A_{1}=v_{2} A_{2}\), we have \(v_{1}=J / A_{1}\) and \(v_{2}=J / A_{2} .\) Using Bernoulli's Equation with \(h_{1}-h_{2}=0\) gives $$ \begin{array}{c} \left(P_{1}-P_{2}\right)+\frac{1}{2} \rho\left(v_{1}^{2}-v_{2}^{2}\right)=0 \\\ 2.93 \times 10^{4} \mathrm{~N} / \mathrm{m}^{2}+\frac{1}{2}\left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(\frac{1}{A_{1}^{2}}-\frac{1}{A_{2}^{2}}\right) J^{2}=0 \end{array} $$ where \(A_{1}=\pi_{1}^{2}=\pi(0.060)^{2} \mathrm{~m}^{2}=0.01131 \mathrm{~m}^{2} \quad\) and \(\quad A_{2}=\pi / \frac{2}{2}=\pi(0.030)^{2} \mathrm{~m}^{2}=0.0028 \mathrm{~m}^{2}\) Substitution then gives \(J=0.022 \mathrm{~m}^{3} / \mathrm{s}\).

Compute the average speed of water in a pipe having an i.d. of \(5.0\) \(\mathrm{cm}\) and delivering \(2.5 \mathrm{~m}^{3}\) of water per hour.

What horsepower is required to force \(8.0 \mathrm{~m}^{3}\) of water per minute into a water main at a pressure of \(220 \mathrm{kPa}\) ?

Calculate the power output of the heart if, in each heartbeat, it pumps \(75 \mathrm{~mL}\) of blood at an average pressure of \(100 \mathrm{mmHg}\). Assume 65 heartbeats per minute. The work done by the heart is \(\mathrm{P} \Delta \mathrm{V}\). In one minute, \(\Delta \mathrm{V}=(65)(75\) \(\left.\times 10^{-6}\right)\). Also $$ \begin{array}{c} P=(100 \mathrm{mmHg}) \frac{1.01 \times 10^{5} \mathrm{~Pa}}{760 \mathrm{mmHg}}=1.33 \times 10^{4} \mathrm{~Pa} \\ \text { consequently, Power }=\frac{\text { Work }}{\text { Time }}=\frac{\left(1.33 \times 10^{4} \mathrm{~Pa}\right)\left[(65)\left(75 \times 10^{-6} \mathrm{~m}^{3}\right)\right]}{60 \mathrm{~s}}=1.1 \mathrm{~W} \end{array} $$

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