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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 ft³/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 (ρ) 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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