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Chapter 3: Problem 101

Measurements in the outer regions of a hurricane indicate a pressure of \(101 \mathrm{kPa}\) and velocities of \(6 \mathrm{~m} / \mathrm{s}\) and \(3 \mathrm{~m} / \mathrm{s}\) in the radial and tangential directions, respectively. If the pressure at the center of the hurricane is estimated as \(98 \mathrm{kPa}\), what wind speed do you expect at the center?

Short Answer

Expert verified
The expect wind speed at the center of the hurricane is approximately 70.28 m/s.

Step by step solution

01

Understand the Problem

We need to find the wind speed at the center of a hurricane, given measurements of pressure and velocity in the outer regions. Specifically, the outer pressure is \(101 \mathrm{kPa}\), and the velocity components in the radial and tangential directions are 6 m/s and 3 m/s, respectively. The pressure at the center is \(98 \mathrm{kPa}\).
02

Apply Bernoulli's Principle (Simplified)

For an incompressible and steady flow, Bernoulli's equation can help relate the pressure and velocity at two points. Usually, Bernoulli's Principle states: \[ P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2 \] where \( P \) is the pressure, \( \rho \) is the fluid density (approximately constant), and \( v \) is the velocity magnitude. We'll use this to compare the center and the outer region.
03

Calculate Velocity Magnitude at the Outer Region

The velocity magnitude \( v_1 \) at the outer region is given by: \[ v_1 = \sqrt{v_{radial}^2 + v_{tangential}^2} = \sqrt{6^2 + 3^2} = \sqrt{36 + 9} = \sqrt{45} \approx 6.71 \mathrm{~m/s} \].
04

Calculate Velocity at the Center Using Bernoulli's Principle

We rearrange the Bernoulli equation to solve for the wind speed at the center, \( v_2 \): \[ v_2 = \sqrt{2 \left( \frac{P_1 - P_2}{\rho} \right) + v_1^2} \]. Assume air density, \( \rho \approx 1.225 \mathrm{~kg/m^3} \). Substitute \(P_1 = 101,000 \mathrm{~Pa} \), \(P_2 = 98,000 \mathrm{~Pa} \), and \(v_1 = 6.71\, \mathrm{m/s} \).
05

Perform the Calculation

Substitute these values into the equation: \[ v_2 = \sqrt{2 \left( \frac{101,000 - 98,000}{1.225} \right) + 6.71^2} \approx \sqrt{2 \left( \frac{3,000}{1.225} \right) + 45} \]. Calculate the inner expression: \(\frac{3,000}{1.225} \approx 2448.98\) so \[ v_2 \approx \sqrt{2 \cdot 2448.98 + 45} \approx \sqrt{4942.96} \approx 70.28 \mathrm{~m/s} \].

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Key Concepts

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

Hurricane Wind Speed Calculation
Calculating the wind speed at the center of a hurricane involves understanding how pressure differences can create faster winds in the core than in the outer regions. In the exercise, we are given the outer pressure of a hurricane at 101 kPa and the estimated center pressure at 98 kPa. These pressure differences are manipulated to determine wind speed changes due to the influence of fluid dynamics principles, specifically Bernoulli's Principle.
  • The wind velocity is first calculated for the outer regions using radial and tangential velocity components.
  • Then, Bernoulli鈥檚 equation is applied to find the velocity at the center using these pressure differences between the outer and central parts.
  • The differences in pressure lead to a significant increase in wind speed moving towards the center.
This approach shows the immense power and speed changes encountered in hurricanes caused by alignment of specific atmospheric conditions.
Fluid Dynamics
Fluid dynamics is the study of fluids in motion. In hurricanes, air flows behave like fluid, driven by pressure and velocity changes. To find the wind speed at different parts of a hurricane, understanding fluid motions and their governing laws is essential. In particular, using Bernoulli鈥檚 equation simplifies the understanding of how pressure and velocity interact within fluid systems.
  • Fluid dynamics allows us to predict how air behaves under pressure differences in a hurricane.
  • Bernoulli's Principle is crucial as it expresses the conservation of energy for a flowing fluid.
  • Assuming the air behaves as an incompressible fluid (where the density remains constant), Bernoulli's equation directly links changes in pressure to changes in the velocity of the fluid.
Understanding these principles is key to predicting and explaining the behavior of wind speeds in the context of complex storm systems like hurricanes.
Pressure and Velocity Relationship
The interplay between pressure and velocity is crucial in understanding the behavior of fluids, such as air in a hurricane. According to Bernoulli's Principle, an increase in fluid velocity occurs simultaneously with a decrease in pressure. This principle can be used to predict the wind speed at the center of a hurricane.
  • As pressure decreases towards the center of a hurricane, the velocity of the wind increases.
  • Bernoulli鈥檚 equation is an expression of energy conservation in fluid flows, relating kinetic and potential energies in terms of velocity and pressure.
  • By rearranging Bernoulli's equation, it is possible to solve for unknown velocities if the pressures at two points are known.
The calculated result reflects how the energy conservation principles underline the natural processes occurring within a hurricane, leading to high-speed winds at its core due to the associated drop in pressure.

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Most popular questions from this chapter

A fluid particle follows a circular streamline that has a radius of \(2 \mathrm{~m}\). The magnitude of the velocity, \(V\), of a fluid particle on the streamline varies with time according to the relation \(V=(0.8+2.5 t) \mathrm{m} / \mathrm{s}\), where \(t\) is the time in seconds. What are the normal and tangential components of the acceleration of a fluid particle located on the streamline at \(t=3 \mathrm{~s}\) ?

An atmospheric disturbance can be approximated by a circular rotation of air around a central location. If the pressure and velocity measured \(110 \mathrm{~m}\) from the center are \(95.0 \mathrm{kPa}\) and \(110 \mathrm{~m} / \mathrm{s}\), respectively, estimate the distance from the center to where the pressure is equal to the atmospheric pressure of \(101 \mathrm{kPa}\). Assume air at \(20^{\circ} \mathrm{C}\).

In a uniform velocity field, the velocity does not change spatially. Consider the uniform unsteady velocity field, \(\mathbf{v}\), given by $$ \mathbf{v}=a \cos \omega t \mathbf{i}+b \sin \omega t \mathbf{j} \mathbf{m} / \mathrm{s} $$ where \(a\) and \(b\) are the amplitudes of the velocity fluctuations and \(\omega\) is the frequency of the fluctuations. In a particular case, \(a=0.15 \mathrm{~m} / \mathrm{s}, b=0.2 \mathrm{~m} / \mathrm{s}, \omega=2 \pi / T,\) and \(T=30 \mathrm{~h} .\) Determine the acceleration as a function of time at all points within the velocity field. At what times, if any, are the acceleration equal to zero?

It is known that when a liquid with an initially horizontal surface is rotated in a cylindrical container, the rise of the liquid surface on the wall of the container is equal to the fall of the liquid surface at the center of the container. Consider an \(850-\mathrm{mm}-\) diameter cylindrical container with an initial depth of water equal to \(1.2 \mathrm{~m}\). At what rotational speed does the water surface at the center of the container intersect the bottom of the container?

The velocity field in a two-dimensional flow is given by $$ \mathbf{v}=(2+1.5 x+2.1 y) \mathbf{i}+(1.8-3 x+4 y) \mathbf{j} $$ (a) Calculate the acceleration field and (b) identify any stagnation points in the flow field.
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