Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 11: Problem 57

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).

Short Answer

Expert verified
The force per unit cross-sectional area exerted by the inside wall of the pipe on the fluid is the difference in pressures at the upstream and downstream, calculated by using adiabatic and isentropic flow relationships.

Step by step solution

01

Assume the conditions for choked flow and Adiabatic flow

When considering an adiabatic process with negligible shaft work, choked flow occurs when the Mach number \(Ma = 1\). Therefore, it can be assumed that at downstream section (2), \(\mathrm{Ma}_{2}=1\). Also, the given ambient conditions at the upstream are, \(p_{0,1}=100\) psia and \(T_{0,1}=600^{\circ} R\)
02

Calculate the ideal gas constant

The ideal gas constant (R) can be obtained if we assume that the working fluid is air, which is a diatomic molecule. Thus, air can be seen as calorically perfect and the ideal gas constant for air is \(R = 53.3 \, lb_f \cdot ft / (lb_m \cdot ^{\circ}R)\).
03

Apply the isentropic flow relationships

Isentropic flow relations can be used to calculate the pressure and temperature at the downstream (2) from the given conditions at the upstream flow. For a calorically perfect gas undergoing an isentropic process, the following relationships give the static to stagnation conditions: \(T_1/T_{0,1} = 1 + ((\gamma - 1)/2) * Ma_1^2\) and \(T_2/T_{0,2} = 1 + ((\gamma - 1)/2) * Ma_2^2\), where \(\gamma=1.4\) is the specific heat ratio for a calorically perfect gas. Solving these for \(T_1\) and \(T_2\), we get the temperatures at upstream and downstream.
04

Evaluate the temperature ratio

With these temperatures \(T_1\) and \(T_2\) obtained, the ratio \(T_{0,2} / T_{0,1}\) can be evaluated.
05

Calculate the static pressure at downstream

Applying the isentropic flow relations again, this time to obtain pressure relations at the upstream and downstream - the static pressures can be obtained. They can be calculated as \(p_1/p_{0,1}= \left[1 + ((\gamma - 1)/2) * Ma_1^2\right]^{\frac{-\gamma}{\gamma -1}}\) and \(p_2/p_{0,2}= \left[1 + ((\gamma - 1)/2) * Ma_2^2\right]^{\frac{-\gamma}{\gamma -1}}\). From these, we can compute the pressures \(p_1\) and \(p_2\).
06

Estimate the magnitude of the force

The force exerted by the inside wall of the pipe on the fluid is related to the change in momentum of the fluid due to the change in pressure between sections 1 and 2. It can be written as \(F/A = p_1 - p_2\), where A is the constant cross-sectional area of the pipe. Knowing \(p_1\) and \(p_2\), the force can be estimated.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

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.

Air flows steadily and isentropically from standard atmospheric conditions to a receiver pipe through a converging duct. The cross-sectional area of the throat of the converging duct is \(0.05 \mathrm{ft}^{2}\). Determine the mass flowrate through the duct if the receiver pressure is (a) 10 psia, (b) 5 psia. Sketch temperature-entropy diagrams for situations (a) and (b). Verify results obtained with values from the appropriate graph in Appendix D with calculations involving ideal gas equations. Is condensation of water vapor a concern? Explain.

Prove that, in Rayleigh flow, the Mach number at the point of maximum temperature is \(1 / \sqrt{k}\).

Air enters a frictionless, constant area duct with \(\mathrm{Ma}=2.5\) \(T_{0}=20^{\circ} \mathrm{C},\) and \(p_{0}=101 \mathrm{kPa}(\mathrm{abs}) .\) The gas is decelerated by heating until a normal shock occurs where the local Mach number is \(1.3 .\) Downstream of the shock, the subsonic fow is accelerated with heating until it exits with a Mach number of \(0.9 .\) Determine the static temperature and pressure, the stagnation temperature and pressure, and the fluid velocity at the duct entrance, just upstream and downstream of the normal shock, and at the duct exit. Sketch the temperature- entropy diagram for this flow.

Waste gas \(\left(\mathrm{CO}_{2}\right)\) is vented to outer space from a spacecraft through a circular pipe \(0.2 \mathrm{m}\) long. The pressure and temperature in the spacecraft are \(35 \mathrm{kPa}\) and \(25^{\circ} \mathrm{C}\). The gas must be vented at the rate of \(0.01 \mathrm{kg} / \mathrm{s}\). The friction factor for the flow in the pipe is given by \\[\begin{array}{c}f=64 / \mathrm{Re}, \quad \mathrm{Re}<5000 \\\f=0.013, \quad \mathrm{Re}>5000, \quad \mathrm{Re}=\frac{ \dot{m}}{\pi \mu D}\end{array}\\] The viscosity \((\mu)\) is \(4 \times 10^{-4} \mathrm{N} \cdot \mathrm{s} / \mathrm{m}^{2}\). Determine the required pipe diameter.
See all solutions

Recommended explanations on Physics Textbooks

Thermodynamics

Read Explanation

Linear Momentum

Read Explanation

Solid State Physics

Read Explanation

Fluids

Read Explanation

Circular Motion and Gravitation

Read Explanation

Astrophysics

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.

Sign-up for free