/*! 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 55 A fluid of density \(\rho\) circ... [FREE SOLUTION] | 91Ó°ÊÓ

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A fluid of density \(\rho\) circulates around a central point such that the velocity, \(V\), at any distance \(r\) from the central point is given by $$ V=\frac{C}{r} $$ where \(C\) is a constant. The pressure is equal to \(p_{0}\) at a distance \(r_{0}\) from the central point. Determine an expression for the pressure distribution in any horizontal plane in terms of \(r,\) where \(C, p_{0}, r_{0}\) are parameters of the pressure distribution.

Short Answer

Expert verified
Pressure is given by \( p = p_0 + \frac{1}{2} \rho \left(\frac{C}{r_0}\right)^2 - \frac{1}{2} \rho \left(\frac{C}{r}\right)^2 \).

Step by step solution

01

Set Up Bernoulli's Equation

We start by invoking Bernoulli's equation for an incompressible and steady flow in a horizontal plane. Bernoulli's equation is written as: \[ p + \frac{1}{2} \rho V^2 = \text{constant} \] In terms of the given variables, the velocity \( V \) is \( \frac{C}{r} \), making the equation: \[ p + \frac{1}{2} \rho \left(\frac{C}{r}\right)^2 = \text{constant} \] We'll take the condition at \(r = r_0\) where the pressure is \(p_0\) to find the constant.
02

Find the Specific Constant

At \(r = r_0\), the pressure is \(p_0\), and the velocity is \( \frac{C}{r_0} \). Substituting these values into the Bernoulli equation gives: \[ p_0 + \frac{1}{2} \rho \left(\frac{C}{r_0}\right)^2 = \text{constant} \] This equation will help us determine the pressure \(p\) at any distance \(r\).
03

Express Pressure in Terms of Distance

Substitute \( \text{constant} \) back into Bernoulli's equation to find pressure \( p \) at any \( r \):\[ p + \frac{1}{2} \rho \left(\frac{C}{r}\right)^2 = p_0 + \frac{1}{2} \rho \left(\frac{C}{r_0}\right)^2 \]Rearrange this equation to express \(p\):\[ p = p_0 + \frac{1}{2} \rho \left(\frac{C}{r_0}\right)^2 - \frac{1}{2} \rho \left(\frac{C}{r}\right)^2 \]This expression represents the pressure distribution as a function of \(r\).

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Key Concepts

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

Pressure Distribution
Pressure distribution refers to how pressure varies across different points in a fluid. When dealing with fluid dynamics, understanding pressure distribution is key to predicting and analyzing fluid behavior in various systems. In the case of a fluid rotating around a central point, such as the exercise provided, the pressure changes depend on the radial distance from that point. This is because the velocity of the fluid, denoted by \( V = \frac{C}{r} \), affects how pressure is distributed.

In simple terms, in a rotating fluid, closer to the central point, velocity is higher which can cause pressure to be lower due to kinetic energy considerations in Bernoulli's equation. Conversely, further from the center, velocity decreases causing pressure to potentially increase if other conditions remain constant.

### Key Takeaways on Pressure Distribution:
  • Pressure distribution is critical for analyzing fluid behavior.
  • It depends on radial distance in a rotating system.
  • Higher velocity zones can result in lower pressures.
Fluid Dynamics
Fluid dynamics is the study of how liquids and gases move and the forces acting on them, using principles like Bernoulli's equation. In the exercise, we see an example of fluid dynamics in action with a fluid system where the velocity is inversely proportional to the radial distance from a central point.

Bernoulli's equation, \( p + \frac{1}{2} \rho V^2 = \text{constant} \), helps us understand and predict how pressure and velocity interplay within a fluid. For a rotating fluid, the kinetic energy associated with velocity affects the pressure seen at any point within the flow. This relationship is why we can use Bernoulli's equation to solve for pressure distribution around the circular flow. The equation tells us that as velocity increases, pressure decreases, given a constant energy in the system.

### Key Concepts in Fluid Dynamics:
  • Interactions between pressure and velocity are foundational.
  • Bernoulli's equation allows prediction of flow behavior.
  • Rotational and linear flow exhibit different dynamic characteristics.
Velocity Field
A velocity field describes how the velocity of a fluid varies in space. In our exercise, the velocity field is defined by the equation \( V = \frac{C}{r} \), indicating that velocity depends on the radial distance \( r \). In essence, as you move away from the center, velocity decreases due to the inverse relationship.

Velocity fields are crucial for understanding fluid movement and dynamics. They allow us to visualize how fluid particles travel across different regions. By studying the velocity field, engineers can design systems that better control and utilize fluid behaviors, such as minimizing drag in aerodynamic designs or optimizing pipe flow systems.

### Important Aspects of Velocity Fields:
  • Velocity varies with spatial coordinates, impacting flow dynamics.
  • They help visualize fluid motion and design efficient systems.
  • Understanding velocity fields can improve predictions in engineering applications.

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Most popular questions from this chapter

The velocity distribution in a pipe with a circular cross section under turbulent flow conditions can be estimated by the relation $$ v(r)=V_{0}\left(1-\frac{r}{R}\right)^{\frac{1}{7}} $$ where \(v(r)\) is the velocity at a distance \(r\) from the centerline of the pipe, \(V_{0}\) is the centerline velocity, and \(R\) is the radius of the pipe. (a) Calculate the average velocity and the volume flow rate in the pipe in terms of \(V_{0} .\) Express your answers in rational form. (b) Based on the result in part (a), assess the extent to which the velocity can be assumed to be constant across the cross section.

Consider the case in which an ideal fluid flows through a horizontal conduit. (a) Determine the acceleration of the fluid as a function of the pressure gradient and the density of the fluid. (b) If the fluid flowing in the conduit is water at \(25^{\circ} \mathrm{C}\) and the pressure decreases at a rate of \(1.5 \mathrm{kPa} / \mathrm{m}\) in the flow direction, at what rate is the fluid accelerating? (c) What pressure gradient is required to accelerate water at a rate of \(6 \mathrm{~m} / \mathrm{s}^{2} ?\)

A 2 -m-long tank with a trapezoidal cross section is drained from the bottom by a rectangular (slot) opening in the bottom center of the tank. The drain slot extends over the entire \(2-\mathrm{m}\) length of the tank and is \(1 \mathrm{~cm}\) wide. The slot opening has been smoothed such that it has a discharge coefficient approximately equal to unity. The trapezoidal cross section of the tank has a bottom width of \(1 \mathrm{~m}\) and side slopes that are at \(60^{\circ}\) to the horizontal. If the initial depth of liquid in the tank is \(1 \mathrm{~m}\), how long does it take to drain the tank completely? Comment on the influence of the density of the liquid on the drain time.

A hose discharges at ground level and is inclined at \(60^{\circ} .\) If water exits the hose at \(15 \mathrm{~m} / \mathrm{s}\), what is the maximum height the jet attains and what is the velocity at that height?

A waterjet cutter that is capable of slicing metals and other materials uses a small nozzle to generate a 0.08 -mm-diameter jet at a speed of \(650 \mathrm{~m} / \mathrm{s}\). Assuming that the velocity head in the source conduit is negligible compared with the velocity head in the cutting jet, what pressure and water supply rate must be generated in the source conduit to operate the waterjet cutter?

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