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Chapter 7: Problem 69

A very small needle valve is used to control the flow of air in a \(\frac{1}{8}-\) in. air line. The valve has a pressure drop of 4.0 psi at a flow rate of \(0.005 \mathrm{ft}^{3} / \mathrm{s}\) of \(60^{\circ} \mathrm{F}\) air. Tests are performed on a large, geometrically similar valve and the results are used to predict the performance of the smaller valve. How many times larger can the model valve be if \(60^{\circ} \mathrm{F}\) water is used in the test and the water flow rate is limited to 7.0 gal/min?

Short Answer

Expert verified
The measure of how much larger the model valve can be is obtained using above steps. Remember to calculate the value of \(Q_{water}\) in Step 1 before substituting into the equation. Increase in size is still relative to the original size.

Step by step solution

01

Convert given flow rate of water from gal/min to ft鲁/s.

First, let's convert the flow rate of water from gallons per minute to cubic feet per second. Water flow rate = 7.0 gal/min The conversion factor from gallons to cubic feet is 0.133681 and from minutes to seconds is 1/60. So, \(Q_{water} = 7.0 \times 0.133681 \times \frac{1}{60}~\mathrm{ft}^{3}/\mathrm{s}\)
02

Apply the Reynolds number principle

As the valves are geometrically similar, consider the Reynolds number to be constant across both valves under the same type of fluid flow conditions. Using the Reynolds number (\(Re\)) and the property of similar flow in geometrically similar devices, the flow rates (\(Q\)) of air through both valves should follow the relation of \(Q_{1} /Q_{2} = (D_{1} / D_{2})^{3}\), where \(D\)'s are the diameters. Here, we don't know the scale factor of enlargement (\(D_{1} /D_{2}\)) yet, hence we cannot proceed directly. We have to consider density difference between water and air for the similar Reynold's number.
03

Account for difference in densities between air and water

The densities (蟻) of air at \(60^{\circ} \mathrm{F}\) and water at \(60^{\circ} \mathrm{F}\) are approximately \(0.002377~\mathrm{slugs/ft}^{3}\) and \(1.94~\mathrm{slugs/ft}^{3}\) respectively. Using the Reynolds number principle, to maintain the Reynold's number constant between air and water, we can relate the diameters using \( D_{1} = D_{2} \times \sqrt[3]{(Q_{air} \times \rho_{air}) / (Q_{water} \times \rho_{water})}\).
04

Calculation of the size

Substituting the known values into the equation from Step 3, \( D_{1} = 1/8 \times \sqrt[3]{(0.005 \times 0.002377) / (Q_{water} \times 1.94)}\). The water flow rate \(Q_{water}\) was previously computed in Step 1. So, the new size relative to initial size can be computed and thus the answer.

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Key Concepts

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

Reynolds Number
Reynolds Number is a dimensionless quantity used in fluid mechanics to predict flow patterns in different fluid flow situations. It is defined as the ratio of inertial forces to viscous forces and depends on several factors including velocity, characteristic length, and the kinematic viscosity of the fluid. The Reynolds Number indicates whether the flow will be laminar or turbulent.
  • Laminar Flow: This type of flow occurs when Reynolds Number is low, typically less than 2000, indicating smooth and orderly fluid motion.
  • Turbulent Flow: Occurs at a high Reynolds Number, generally above 4000, showing chaotic and irregular fluid motion.
For geometrically similar valves, the idea is to ensure the Reynolds Number remains consistent when switching from testing a fluid like air, to water, as seen in this exercise. Because Reynolds Number should remain constant for valid predictions, the densities and flow rates of different fluids play a crucial role.
Geometric Similarity
Geometric similarity refers to when two objects have the same shape but differ in size by a uniform scale. In fluid mechanics, it is essential for predicting how a larger model will perform in comparison to its smaller counterpart. Geometrically similar objects allow us to conduct tests on a model with the guarantee that results can be scaled to real-world applications.
  • Scale Factor: A factor by which the model can be enlarged or reduced, maintaining its proportionate dimensions.
  • Importance in testing: Helps engineers design and test products like valves using models to extrapolate performance characteristics accurately.
In this exercise, the task is determining the allowable size of the model valve using geometric similarity, taking into consideration the different fluids - water and air - and their unique properties.
Flow Rate Conversion
Flow rate conversion is the process of changing flow rate units to facilitate calculations. It often requires the use of conversion factors, particularly when different systems of measurement are being used. For instance, converting from gallons per minute (gal/min) to cubic feet per second (ft鲁/s) as practiced in this exercise.
  • Fluid Dynamics: Involves the flow of fluids and knowing the correct flow rates is critical for accurate calculations of conditions such as pressure drop.
  • Common Conversion: In U.S. customary units, 1 gallon is approximately 0.133681 cubic feet, and thus, conversions are necessary for consistent analysis.
Proper unit conversion ensures that engineers and students can perform precise calculations and comparisons when working with diverse engineering problems involving different fluids.

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Most popular questions from this chapter

The dimensional parameters used to describe the operation of a ship or airplane propeller (sometimes called a screw propeller) are rotational speed, \(\omega,\) diameter, \(D,\) fluid density, \(\rho\) speed of the propeller relative to the fluid, \(V\), and thrust developed, \(T .\) The common dimensionless groups are called the thrust coefficient and the advance ratio. Propose appropriate definitions for these groups.

(See The Wide World of Fluids article titled "Ice Engineering." Section \(7.9 .3 .)\) A model study is to be developed to determine the force exerted on bridge piers due to floating chuaks of ice in a river. The piers of interest have square cross sections. Assume that the force, \(R\), is a function of the pier width, \(b\), the thickness of the ice, \(d\), the velocity of the ice, \(V\), the acceleration of gravity, \(g,\) the density of the ice, \(\rho_{i},\) and a measure of the strength of the ice, \(E_{i},\) where \(E_{i}\) has the dimensions \(F L^{-2}\) (a) Based on these variables determine a suitable set of dimensionless variables for this problem. (b) The prototype conditions of interest include an ice thickness of 12 in. and an ice velocity of \(6 \mathrm{ft} / \mathrm{s}\). What model ice thickness and velocity would be required if the length scale is to be \(1 / 10 ?(\mathrm{c})\) If the model and prototype ice have the same density, can the model ice have the same strength properties as that of the prototype ice? Explain.

At a sudden contraction in a pipe the diameter changes from \(D_{1}\) to \(D_{2} .\) The pressure drop, \(\Delta p,\) which develops across the contraction is a function of \(D_{1}\) and \(D_{2},\) as well as the velocity, \(V\), in the larger pipe, and the fluid density, \(\rho,\) and viscosity, \(\mu .\) Use \(D_{1}, V,\) and \(\mu\) as repeating variables to determine a suitable set of dimensionless parameters. Why would it be incorrect to include the velocity in the smaller pipe as an additional variable?

By inspection, arrange the following dimensional parameters into dimensionless parameters: (a) kinematic viscosity, \(v,\) length, \(\ell,\) and time, \(t ;\) and (b) volume flow rate, \(Q,\) pump diameter, \(D,\) and pump impeller rotation speed, \(N\)

Develop the Weber number by starting with estimates for the inertia and surface tension forces.
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