/*! 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 66 A torpedo \(8 \mathrm{m}\) below... [FREE SOLUTION] | 91Ó°ÊÓ

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A torpedo \(8 \mathrm{m}\) below the surface in \(20^{\circ} \mathrm{C}\) seawater cavitates at a speed of \(21 \mathrm{m} / \mathrm{s}\) when atmospheric pressure is 101 kPa. If Reynolds number and Froude number effects are negligible, at what speed will it cavitate when running at a depth of \(20 \mathrm{m} ?\) At what depth should it be to avoid cavitation at \(30 \mathrm{m} / \mathrm{s} ?\)

Short Answer

Expert verified
At 20 m depth, cavitation speed is higher than 21 m/s. To avoid cavitation at 30 m/s, depth should be greater than 20 m.

Step by step solution

01

Understand Cavitation Conditions

Cavitation occurs when local water pressure drops to the vapor pressure of water. The pressure at the depth of the torpedo is the sum of atmospheric pressure and pressure due to water column above it.
02

Calculate Initial Depth Pressure

At 8 m below the surface, calculate the pressure using the formula: \( P = P_{atm} + \rho g h \). Here, \( \rho \) is the density of seawater (approximately 1025 kg/m³), \( g \) is gravity (9.81 m/s²), and \( h = 8 \) m.
03

Calculate Pressure at 21 m/s

Convert the atmospheric pressure to water depth equivalent using \( P_{atm} = 101 \text{ kPa} = 101,000 \text{ Pa} \). Then calculate the local pressure 8 m deep, which is \( 101,000 + 1025 \times 9.81 \times 8 \).
04

Calculate Velocity at New Depth

When the torpedo is at a new depth of 20 m, the pressure is \( 101,000 + 1025 \times 9.81 \times 20 \). Determine the speed at which cavitation will occur by solving the equation for velocity in terms of new pressure.
05

Calculate Required Depth for Prevention

For a speed of 30 m/s, calculate the required depth to avoid cavitation: \( P_{new} = P_{atm} + \rho g h_{new} \). Solve for \( h_{new} \), ensuring pressure exceeds that needed to balance the vapor and velocity pressures at 30 m/s.

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Key Concepts

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

Cavitation
Cavitation is a fascinating yet challenging concept in fluid mechanics. It occurs when the local pressure of a fluid drops below its vapor pressure, causing the formation of vapor bubbles within the fluid. These bubbles can travel with the flow and eventually collapse, potentially causing damage to surfaces or mechanical components. In situations like the one described with the torpedo, cavitation is a concern because the rapid movement through seawater can create low-pressure areas. If the local pressure becomes equal to or less than the vapor pressure of seawater, cavitation will occur.
  • Occurs when local pressure < vapor pressure
  • Can cause physical damage through bubble collapse
  • Important in marine engineering and fluid transport systems
Understanding cavitation is crucial for designing systems that move through water efficiently and without damaging themselves.
Pressure Calculations
Pressure calculations are vital in determining the conditions under which cavitation occurs. In the context of the given exercise, pressure at a given depth is influenced by both atmospheric pressure and the pressure due to the water column above.The total pressure at a certain depth can be calculated using the formula:\[ P = P_{atm} + \rho g h \]where:
  • \(P\) is the pressure at depth
  • \(P_{atm}\) is the atmospheric pressure
  • \(\rho\) is the fluid density
  • \(g\) is the gravitational acceleration
  • \(h\) is the depth below the surface
This equation helps predict at what speed and depth the torpedo will cavitate by assessing the needed pressure to prevent it.
Seawater Density
Seawater density plays a crucial role in pressure and buoyancy calculations. In fluid mechanics, it's important to recognize that density affects how pressure increases with depth.Seawater has a typical density of approximately \(1025 \text{ kg/m}^3\), and this value is required in pressure calculations to determine forces acting on submerged objects like torpedoes.
  • Density influences pressure calculations through \(\rho g h\)
  • Key factor in buoyancy and object's ability to remain submerged
  • Varies slightly with salinity and temperature
By understanding how seawater density influences these calculations, engineers can design underwater vehicles and structures that operate safely and effectively.
Vapor Pressure
Vapor pressure is the pressure at which a liquid's molecules are in equilibrium with its gas phase at a given temperature. For water, this means the pressure where water can vaporize into steam in the presence of liquid water.In marine applications, knowing the vapor pressure of water at different temperatures is critical, as it helps determine when cavitation might occur.
  • At \(20^\circ C\), water's vapor pressure is lower than boiling point pressure
  • Vapor pressure varies with temperature, impacting cavitation risk
  • Ensures system designs account for bubble formation and phase change
In calculations similar to those in the exercise, engineers ensure the operating pressure is above vapor pressure to avoid undesired effects.
Atmospheric Pressure
Atmospheric pressure is a factor contributing to the overall pressure exerted on submerged objects. It is the pressure exerted by the weight of air above the Earth's surface, approximately 101 kPa at sea level.This pressure combines with the hydrostatic pressure from a water column to form the total pressure acting on an object underwater.
  • Standard atmospheric pressure at sea level is \(101 \text{ kPa}\)
  • Influences calculations of submerged pressure with \(P_{atm} + \rho g h\)
  • Needed to determine cavitation limits and operational depths
When calculating pressures at different depths, it is crucial to include atmospheric pressure to ensure complete and accurate pressure assessments.

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Most popular questions from this chapter

In turbulent flow near a flat wall, the local velocity \(u\) varies only with distance \(y\) from the wall, wall shear stress \(\tau_{w},\) and fluid properties \(\rho\) and \(\mu .\) The following data were taken in the University of Rhode Island wind tunnel for airflow, \(\rho=0.0023\) slug \(/ \mathrm{ft}^{3}, \mu=3.81 \mathrm{E}-7 \mathrm{slug} /(\mathrm{ft} \cdot \mathrm{s})\) and \(\tau_{w}=0.029 \mathrm{lbf} / \mathrm{ft}^{2}\) $$\begin{array}{l|l|l|l|l|l|l} y, \text { in } & 0.021 & 0.035 & 0.055 & 0.080 & 0.12 & 0.16 \\ \hline u, \mathrm{ft} / \mathrm{s} & 50.6 & 54.2 & 57.6 & 59.7 & 63.5 & 65.9 \end{array}$$ (a) Plot these data in the form of dimensionless \(u\) versus dimensionless \(y,\) and suggest a suitable power-law curve fit. (b) Suppose that the tunnel speed is increased until \(u=90\) ft/s at \(y=0.11\) in. Estimate the new wall shear stress, in \(1 \mathrm{bf} / \mathrm{ft}^{2}\)

A dam spillway is to be tested by using Froude scaling with a one-thirtieth- scale model. The model flow has an average velocity of \(0.6 \mathrm{m} / \mathrm{s}\) and a volume flow of \(0.05 \mathrm{m}^{3} / \mathrm{s} .\) What will the velocity and flow of the prototype be? If the measured force on a certain part of the model is \(1.5 \mathrm{N}\), what will the corresponding force on the prototype be?

The wall shear stress \(\tau_{w}\) in a boundary layer is assumed to be a function of stream velocity \(U,\) boundary layer thickness \(\delta,\) local turbulence velocity \(u^{\prime},\) density \(\rho,\) and local pressure gradient \(d p / d x .\) Using \((\rho, U, \delta)\) as repeating variables, rewrite this relationship as a dimensionless function.

When tested in water at \(20^{\circ} \mathrm{C}\) flwing at \(2 \mathrm{m} / \mathrm{s}\), an \(8-\mathrm{cm}\) diameter sphere has a measured drag of 5 N. What will be the velocity and drag force on a 1.5 -m-diameter weather balloon moored in sea-level standard air under dynamically similar conditions?

The pressure drop per unit length \(\Delta p / L\) in smooth pipe flow is known to be a function only of the average velocity \(V,\) diameter \(D,\) and fluid properties \(\rho\) and \(\mu .\) The following data were obtained for \(\mathrm{Aw}\) of water at \(20^{\circ} \mathrm{C}\) in an 8 -cm-diameter pipe \(50 \mathrm{m}\) long: $$\begin{array}{l|l|l|l|l} Q, \mathrm{m}^{3} / \mathrm{s} & 0.005 & 0.01 & 0.015 & 0.020 \\ \hline \Delta p, \mathrm{Pa} & 5800 & 20,300 & 42,100 & 70,800 \end{array}$$ Verify that these data are slightly outside the range of Fig. \(5.10 .\) What is a suitable power-law curve fit for the present data? Use these data to estimate the pressure drop for Aw of kerosene at \(20^{\circ} \mathrm{C}\) in a smooth pipe of diameter \(5 \mathrm{cm}\) and length \(200 \mathrm{m}\) if the flow rate is \(50 \mathrm{m}^{3} / \mathrm{h}\)

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