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Magazine Chapter 3: Problem 100
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 \(20^{\circ} \mathrm{C}\).
Short Answer
Expert verified
The distance where pressure is 101 kPa is approximately 115 meters.
Step by step solution
01
Understanding the Problem
We are given that an atmospheric disturbance is creating a circular motion of air, and we need to find out where the pressure equals the atmospheric pressure of 101 kPa. We have measurements at a known distance (110 m) and need to find a different distance (r) that results in a pressure of 101 kPa.
02
Bernoulli's Equation for Circular Motion
Use Bernoulli's equation to relate pressures and velocities at two radial distances from the center of rotation. Bernoulli's equation for two points in a steady flow is given by: \[ P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2 \]Where \( P_1 = 95.0 \text{ kPa} \), \( v_1 = 110 \text{ m/s} \), and \( P_2 = 101 \text{ kPa} \). We need to find \( v_2 \), the velocity at the desired distance r.
03
Calculating Air Density
The density of air (\( \rho \)) at 20掳C can be approximated using the ideal gas law: \[ \rho = \frac{P}{RT} \] where \( R = 287 \text{ J/kg K} \) for air, and \( T = 293 \text{ K} \) (given \( 20^{\circ} \text{C} = 293 \text{ K} \)). Using standard atmospheric pressure \( P = 101325 \text{ Pa} \), calculate \( \rho \):\[ \rho = \frac{101325}{287 \times 293} \approx 1.2041 \text{ kg/m}^3 \].
04
Solve for the Velocity at the New Distance
Rearrange Bernoulli's equation to solve for \( v_2 \):\[ v_2 = \sqrt{v_1^2 + \frac{2(P_1 - P_2)}{\rho}} \]Substitute \( v_1 = 110 \text{ m/s} \), \( P_1 = 95000 \text{ Pa} \), \( P_2 = 101000 \text{ Pa} \), and \( \rho = 1.2041 \text{ kg/m}^3 \):\[ v_2 = \sqrt{110^2 + \frac{2(95000 - 101000)}{1.2041}} \approx \sqrt{110^2 - 8313.28} \approx 105.27 \text{ m/s} \].
05
Finding the New Distance
Using the conservation of angular momentum for circular motion \( r_1 v_1 = r_2 v_2 \), solve for the radial distance \( r_2 \) where the velocity is \( v_2 \):\[ r_2 = \frac{r_1 v_1}{v_2} \]Substitute \( r_1 = 110 \text{ m} \), \( v_1 = 110 \text{ m/s} \), and \( v_2 \approx 105.27 \text{ m/s} \):\[ r_2 = \frac{110 \times 110}{105.27} \approx 115.0 \text{ m} \].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Atmospheric Pressure
Atmospheric pressure is the force exerted by the weight of the atmosphere above a surface. At sea level, it is typically measured at approximately 101 kPa. This pressure can vary based on weather conditions, altitude, and temperature. Understanding atmospheric pressure is essential in meteorology and various fields, as it affects weather patterns and atmospheric phenomena.
In the context of Bernoulli's equation, changes in atmospheric pressure play a critical role. As air moves in a circular motion around a disturbance, different pressures are observed at different distances. Using these pressure differences allows us to learn more about the rotation's characteristics. For example, by examining where pressure reaches atmospheric levels, we can estimate changes in velocity and position due to the circular motion of air.
In exercises like the given one, atmospheric pressure is used as a reference point. By calculating where pressure equals the standard atmospheric level (101 kPa), one can find significant insights about the motion of air in such disturbances.
In the context of Bernoulli's equation, changes in atmospheric pressure play a critical role. As air moves in a circular motion around a disturbance, different pressures are observed at different distances. Using these pressure differences allows us to learn more about the rotation's characteristics. For example, by examining where pressure reaches atmospheric levels, we can estimate changes in velocity and position due to the circular motion of air.
In exercises like the given one, atmospheric pressure is used as a reference point. By calculating where pressure equals the standard atmospheric level (101 kPa), one can find significant insights about the motion of air in such disturbances.
Circular Motion
Circular motion occurs when an object moves in a path along a circle. In terms of atmospheric phenomena, air often flows in circular paths due to disturbances like cyclones or tornadoes. Understanding circular motion helps in predicting weather patterns like wind speeds and directions.
In the problem, the air moves in a circular path around a center point. The speed of this motion helps define key properties such as the centrifugal force affecting the air particles. Bernoulli's equation can be applied to establish relationships between pressure and velocity at various points along the circle. For instance, if velocity diminishes as one moves further from the center, the pressure usually rises to maintain steady-state conditions.
In the problem, the air moves in a circular path around a center point. The speed of this motion helps define key properties such as the centrifugal force affecting the air particles. Bernoulli's equation can be applied to establish relationships between pressure and velocity at various points along the circle. For instance, if velocity diminishes as one moves further from the center, the pressure usually rises to maintain steady-state conditions.
- Velocity: The speed at which an object moves along its circular path. High velocity near the center provokes lower pressure, and vice versa. The exercise utilizes this to determine distances based on changes in air velocity.
- Radius: A critical measure since it influences both velocity and pressure in circular motion. Larger radii correspond to slower velocities if angular momentum is conserved.
Ideal Gas Law
The ideal gas law is a fundamental principle in thermodynamics that relates pressure, volume, and temperature of an ideal gas. Represented as \( PV = nRT \), it allows for the calculation of one property if the others are known.
Here, the ideal gas law enables the calculation of air density, a necessary variable in Bernoulli's equation. By knowing the temperature (20掳C) and atmospheric pressure, the density can be found, which then aids in understanding how pressure and velocity relate in the circular motion of air.
In practical exercises like the one given, understanding the density of air is critical. It helps determine how variables like pressure and velocity interact. Moreover, assuming air behaves as an ideal gas simplifies calculations, leading to more straightforward and applicable solutions.
Here, the ideal gas law enables the calculation of air density, a necessary variable in Bernoulli's equation. By knowing the temperature (20掳C) and atmospheric pressure, the density can be found, which then aids in understanding how pressure and velocity relate in the circular motion of air.
In practical exercises like the one given, understanding the density of air is critical. It helps determine how variables like pressure and velocity interact. Moreover, assuming air behaves as an ideal gas simplifies calculations, leading to more straightforward and applicable solutions.
- R: The specific gas constant for dry air. In this case, it is 287 J/kg K.
- Temperature: Must be understood in Kelvin for accurate calculations within the ideal gas law.
Angular Momentum
Angular momentum is a measure of the quantity of rotation an object has, taking into account its mass, shape, and speed. In circular motion, it is a conserved quantity, meaning it remains constant unless acted upon by an external force.
In the problem, angular momentum helps determine how changes in distance, speed, and pressure are interrelated. If a portion of air in the cyclone moves outward, it tends to slow down if no additional forces act upon it. This is due to conservation of angular momentum, which can be formulated as \( r_1v_1 = r_2v_2 \).
Understanding this concept is crucial because it underpins the dynamics of circular movement in physics. In atmospheric disturbances, retaining the angular momentum means that any change in radius or velocity must reciprocally adjust to maintain the system's equilibrium. For instance, moving outward in the problem effectively reduces speed, leading to an increase in pressure until reaching atmospheric levels.
In the problem, angular momentum helps determine how changes in distance, speed, and pressure are interrelated. If a portion of air in the cyclone moves outward, it tends to slow down if no additional forces act upon it. This is due to conservation of angular momentum, which can be formulated as \( r_1v_1 = r_2v_2 \).
Understanding this concept is crucial because it underpins the dynamics of circular movement in physics. In atmospheric disturbances, retaining the angular momentum means that any change in radius or velocity must reciprocally adjust to maintain the system's equilibrium. For instance, moving outward in the problem effectively reduces speed, leading to an increase in pressure until reaching atmospheric levels.
