/*! 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 47 At a certain point in a pipe, ai... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 47

At a certain point in a pipe, air flows steadily with a velocity of \(150 \mathrm{m} / \mathrm{s}\) and has a static pressure of \(70 \mathrm{kPa}\) and a static temperature of \(4^{\circ} \mathrm{C}\). The flow is adiabatic and frictionless.

Short Answer

Expert verified
Based on the given exercise, it can be concluded that if the flow of air in the pipe is steady, adiabatic, and frictionless, there are no changes in static pressure and temperature. Thus, their values remain at 70 kPa and 4 degrees Celsius respectively.

Step by step solution

01

Convert Given Units

First, convert the given units into more workable ones. The static pressure is given in kiloPascals (kPa) but it is easier to work with Pascals (Pa), so convert it by multiplying by 1000. Likewise, convert the temperature from degrees Celsius to Kelvin by adding 273.15 to it. Thus, the static pressure becomes \(70,000 \, Pa\) and the static temperature \(277.15 \, K\).
02

Apply Bernoulli's Equation

Next, apply Bernoulli's equation, which states that the total energy in a steadily flowing fluid system is constant along the flow path. The formula is: \[\frac{1}{2}蟻v^{2} + 蟻gh + p = constant\] where 蟻 is the fluid density, \(v\) is the fluid velocity, \(g\) is acceleration due to gravity, \(h\) is height, and \(p\) is pressure. Given that there are no height changes in the pipe and the flow has a constant velocity, the equation simplifies to: \[\frac{1}{2}蟻v_{1}^{2} + p_{1} = \frac{1}{2}蟻v_{2}^{2} + p_{2}\] where the subscript '1' refers to initial conditions and '2' refers to final conditions. As velocities are constant, the final pressure \(p_{2}\) becomes: \(p_{2}=p_{1}-\frac{1}{2}蟻(v_{1}^{2}-v_{2}^{2})\). Here, it's given that the flow is adiabatic and frictionless, meaning there are no changes in velocity, so \(v_{1} = v_{2}\). Therefore, the pressure remains constant.
03

Determine the Change in Temperature

The change in temperature can be determined using the ideal gas law which states that the pressure of a gas is directly proportional to its temperature if the volume and quantity of gas are held constant, written as \(p=蟻RT\), where \(R\) is the specific gas constant. Isolated and substituting in the adiabatic, inviscid flow conditions where pressure \(p\) is constant this equation rewrites as \(T_{2}=T_{1}\), which indicates, that the temperature remains constant as well.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Bernoulli's equation
Bernoulli's equation is a fundamental principle in fluid mechanics that describes the behavior of a fluid as it flows through a system. It describes how energy is conserved through different forms in a steady, incompressible flow. This equation can be very useful when examining fluid behavior in pipes.
  • The equation is expressed as: \[\frac{1}{2} \rho v^2 + \rho gh + p = \text{constant}\]
  • Here, \(\rho\) represents fluid density, \(v\) is the velocity of the fluid, \(g\) is the acceleration due to gravity, \(h\) is the height of the fluid above a reference point, and \(p\) is the pressure.
In the given exercise, since the flow's height isn't changing and the flow is frictionless, the potential energy term, \(\rho gh\), can be ignored. The equation, therefore, becomes a relationship between kinetic energy and pressure. This simplification happens because frictionless and adiabatic flow means no converter energy into heat or potential energy.
adiabatic flow
Adiabatic flow refers to a process where no heat is exchanged with the surroundings. In fluid mechanics, it is a key concept when dealing with gas flow conditions like the one given in the exercise. Characteristics of adiabatic flow:
  • No heat is transferred to or from the fluid, which makes energy conservation particularly simple.
  • For flows that are adiabatic and frictionless, like in the exercise, there are no dissipation effects to account for.
  • The energy in the flow remains constant, which works well with Bernoulli's equation to predict changes in other properties like pressure and temperature.
This makes the analysis using Bernoulli's equation straightforward since the only variables are those directly involved in flow properties鈥攙elocity and pressure鈥攁nd not external heat exchange.
ideal gas law
The ideal gas law is an equation of state for a hypothetical gas considered ideal under specific conditions. It's particularly useful in relating pressure, volume, and temperature of a gas. The equation is given by:\[ p = \rho R T \]In this formula, \(p\) stands for pressure, \(\rho\) is the gas density, \(R\) is the specific gas constant, and \(T\) is the temperature.
  • This law helps to connect the conditions outlined in Bernoulli鈥檚 equation with temperature changes in practical scenarios.
  • Adiabatic conditions imply no external heat transport, facilitating direct application of the ideal gas law.
  • In the exercise, since no changes in volume occur, the pressure and temperature remain constant. Hence, it aligns with the nature of an ideal gas under adiabatic flow conditions.
inviscid flow
Inviscid flow assumes there are no viscous forces acting within the flow of a fluid. This is an idealization, but it is quite useful in simplifying complex fluid dynamics problems. Core ideas of inviscid flow include:
  • The assumption of no viscosity means that energy is not lost to frictional forces, allowing for simplifications in energy conservation equations like Bernoulli鈥檚.
  • It makes the mathematical analysis of fluid flow much more straightforward, focusing only on density, pressure, and velocity.
  • This concept is fundamental when evaluating high-energy flows or in situations where friction forces are negligible compared to other forces acting within the system.
In this exercise, assuming inviscid flow complements the use of Bernoulli鈥檚 equation and allows the simplification of the problem, ensuring that the flow properties can be directly analyzed without accounting for dissipative forces.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Determine the Mach number of a car moving in standard air at a speed of (a) \(25 \mathrm{mph}\) (b) \(55 \mathrm{mph},\) and (c) \(100 \mathrm{mph}\)

Air flows adiabatically between two sections in a constant area pipe. At upstream section \((1), p_{0,1}=100\) psia, \(T_{0,1}=600^{\circ} \mathrm{R}\) and \(\mathrm{Ma}_{1}=0.5 .\) At downstream section \((2),\) the flow is choked. Estimate the magnitude of the force per anit cross-sectional area exerted by the inside wall of the pipe on the fluid between sections (1) and (2).

The flow blockage associated with the use of an intrusive probe can be important. Determine the percentage increase in section velocity corresponding to a \(0.5 \%\) reduction in flow area due to probe blockage for airflow if the section area is \(1.0 \mathrm{m}^{2}, T_{0}=\) \(20^{\circ} \mathrm{C},\) and the unblocked flow Mach numbers are (a) \(\mathrm{Ma}=0.2\) (b) \(\mathrm{Ma}=0.8\) (c) \(\mathrm{Ma}=1.5,(\mathrm{d}) \mathrm{Ma}=30\)

The stagnation pressure in a Mach 2 wind tunnel operating with air is 900 kPa. A 1.0 -cm-diameter sphere positioned in the wind tunnel has a drag coefficient of \(0.95 .\) Calculate the drag force on the sphare.

An air heater in a large coal-fired steam generator heats fresh air entering the steam generator by cooling flue gas leaving the steam generator. One million lbm/hr of air at \(100^{\circ} \mathrm{F}\) and 1.1 million lbm/hr of flue gas at \(720^{\circ} \mathrm{F}\) enters the air heater. The flue gas leaves at \(310^{\circ} \mathrm{F}\). Flue gas has \(c_{p}=0.26 \mathrm{Btu} / \mathrm{lbm}^{\circ} \mathrm{F}\) and \(k=1.39 .\) Pressure changes are small and may be neglected. Calculate the temperature of the air leaving the air heater and the total entropy change for the process.
See all solutions

Recommended explanations on Physics Textbooks

Electricity

Read Explanation

Famous Physicists

Read Explanation

Oscillations

Read Explanation

Fluids

Read Explanation

Physics of Motion

Read Explanation

Work Energy and Power

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.