/*! 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 70 A flow nozzle is a device for me... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 70

A flow nozzle is a device for measuring the flow rate in a pipe. This particular nozzle is to be used to measure low-speed air flow for which compressibility may be neglected. During operation, the pressures \(p_{1}\) and \(p_{2}\) are recorded, as well as upstream temperature, \(T_{1}\). Find the mass flow rate in terms of \(\Delta p=p_{2}-p_{1}\) and \(T_{1},\) the gas constant for air, and device diameters \(D_{1}\) and \(D_{2}\). Assume the flow is frictionless. Will the actual flow be more or less than this predicted flow? Why?

Short Answer

Expert verified
The mass flow rate of air can be calculated using the derived equation: \(m= \sqrt{\Delta p . ((\frac{\pi^{2}}{2}) . (\frac{D_{1}^{4}D_{2}^{4}}{(D_{1}^{4}- D_{2}^{4})}) . R . T_{1}})\). However, it's important to note that the actual flow rate might be less than the calculated rate due to factors that have been neglected in the calculation, such as friction, heat transfer, or minor losses.

Step by step solution

01

Use Bernoulli鈥檚 equation

Using Bernoulli鈥檚 equation for point 1 and 2, we get:\n \(\frac{p_{1}}{蟻} + \frac{1}{2}u_{1}^{2}= \frac{p_{2}}{蟻} + \frac{1}{2}u_{2}^{2}\)\nGiven that the flow is frictionless we neglect the loss of energy due to friction and that the nozzle is horizontal, which makes the energy head (the potential energy) remain constant
02

Express velocity

Express velocity \(u_{2}\) and \(u_{1}\) in terms of mass flow rate (\(m=\rho Au\)):\nSubstitute \(u=\frac{m}{\rho A}\) in the Bernouilli's equation, we obtain:\n\((p_{1}-p_{2})= \frac{1}{2蟻}( \frac{m^{2}}{\pi^{2}(D_{2}/2)^{4}}- \frac{m^{2}}{\pi^{2}(D_{1}/2)^{4}})\) or simplify it to\n\((p_{1} - p_{2}) = \frac{1}{2蟻} \left[\left( \frac{m}{\pi D_{2}^{2}/4} \right)^{2} - \left( \frac{m}{\pi D_{1}^{2}/4} \right)^{2} \right] = \Delta p\)
03

Use Ideal Gas Law

To find the density values, use the Ideal Gas Law: \( 蟻 = \frac{p}{RT}\), so the density at point 1 is: \( 蟻_{1} = \frac{p_{1}}{RT_{1}}\)
04

Substitute into equation

Substitute it into the simplified equation, solve for m to get the mass flow rate, which is airlarge textour unknown. We get:\n\(m= \sqrt{\Delta p . ((\frac{\pi^{2}}{2}) . (\frac{D_{1}^{4}D_{2}^{4}}{(D_{1}^{4}- D_{2}^{4})}) . R . T_{1}})\)

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!

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

Consider the flow field with velocity given by \(\vec{V}=\) \(A x \sin (2 \pi \omega t) \hat{i}-A y \sin (2 \pi \omega t) \hat{j},\) where \(A=2 \mathrm{s}^{-1}\) and \(\omega=1 \mathrm{s}^{-1}\) The fluid density is \(2 \mathrm{kg} / \mathrm{m}^{3}\). Find expressions for the local acceleration, the convective acceleration, and the total acceleration. Evaluate these at point (1,1) at \(t=0,0.5,\) and 1 seconds. Evaluate \(\nabla p\) at the same point and times.

Consider the flow field represented by the velocity potential \(\quad \phi=A x+B x^{2}-B y^{2}, \quad\) where \(\quad A=1 \quad \mathrm{m} \cdot \mathrm{s}^{-1}\) \(B=1 \mathrm{m}^{-1} \cdot \mathrm{s}^{-1},\) and the coordinates are measured in meters. Obtain expressions for the velocity field and the stream function. Calculate the pressure difference between the ori\(\operatorname{gin}\) and point \((x, y)=(1,2)\)

The inlet contraction and test section of a laboratory wind tunnel are shown. The air speed in the test section is \(U=50 \mathrm{m} / \mathrm{s} .\) A total- head tube pointed upstream indicates that the stagnation pressure on the test section centerline is \(10 \mathrm{mm}\) of water below atmospheric. The laboratory is maintained at atmospheric pressure and a temperature of \(-5^{\circ} \mathrm{C}\) Evaluate the dynamic pressure on the centerline of the wind tunnel test section. Compute the static pressure at the same point. Qualitatively compare the static pressure at the tunnel wall with that at the centerline. Explain why the two may not be identical.

A mercury barometer is carried in a car on a day when there is no wind. The temperature is \(20^{\circ} \mathrm{C}\) and the corrected barometer height is \(761 \mathrm{mm}\) of mercury. One window is open slightly as the car travels at \(105 \mathrm{km} / \mathrm{hr}\). The barometer reading in the moving car is \(5 \mathrm{mm}\) lower than when the car is stationary. Explain what is happening. Calculate the local speed of the air flowing past the window, relative to the automobile.

A rectangular computer chip floats on a thin layer of air, \(h=0.5 \mathrm{mm}\) thick, above a porous surface. The chip width is \(b=40 \mathrm{mm},\) as shown. Its length, \(L,\) is very long in the direction perpendicular to the diagram. There is no flow in the \(z\) direction. Assume flow in the \(x\) direction in the gap under the chip is uniform. Flow is incompressible, and frictional effects may be neglected. Use a suitably chosen control volume to show that \(U(x)=q x / h\) in the gap. Find a general expression for the \((2 \mathrm{D})\) acceleration of a fluid particle in the gap in terms of \(q, h, x,\) and \(y\) Obtain an expression for the pressure gradient \(\partial p / \partial x\). Assuming atmospheric pressure on the chip upper surface, find an expression for the net pressure force on the chip; is it directed upward or downward? Explain. Find the required flow rate \(q\) \(\left(\mathrm{m}^{3} / \mathrm{s} / \mathrm{m}^{2}\right)\) and the maximum velocity, if the mass per unit length of the chip is \(0.005 \mathrm{kg} / \mathrm{m} .\) Plot the pressure distribution as part of your explanation of the direction of the net force.
See all solutions

Recommended explanations on Physics Textbooks

Measurements

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Modern Physics

Read Explanation

Oscillations

Read Explanation

Scientific Method Physics

Read Explanation

Further Mechanics and Thermal Physics

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.