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Chapter 15: Problem 33

Water flows through a 2.5 -cm-diameter pipe at \(1.8 \mathrm{m} / \mathrm{s}\). If the pipe narrows to 2.0 -cm diameter, what's the flow speed in the constriction?

Short Answer

Expert verified
The flow speed in the narrow section of the pipe is approximately \(2.8 m/s\).

Step by step solution

01

Calculate the cross-sectional area of the wider section

We can use the formula for the area of a circle \(A = \pi r^2\) to calculate the cross-sectional area \(A_1\) of the wider section of the pipe. The radius is half of the diameter. The diameter of the wider section is 2.5 cm, so the radius is 1.25 cm, or 0.0125 m. Therefore, the cross-sectional area \(A_1\) is \(\pi (0.0125 m)^2\).
02

Calculate the cross-sectional area of the narrower section

Similarly, we use the formula for the area of a circle to calculate the cross-sectional area \(A_2\) of the narrower section. The diameter of the narrower section is 2.0 cm, so the radius is 1.0 cm, or 0.01 m. Therefore, the cross-sectional area \(A_2\) is \(\pi (0.01 m)^2\).
03

Use the equation of continuity to solve for \(V_2\)

We know that \(A_1V_1 = A_2V_2\). Plug in the known values: \(\pi (0.0125 m)^2 * 1.8 m/s = \pi (0.01 m)^2 * V_2\). Therefore, \(V_2 = \frac{(0.0125 m)^2 * 1.8 m/s}{(0.01 m)^2}\)

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Most popular questions from this chapter

A 1.0 -cm-diameter venturi flowmeter is inserted in a \(2.0-\mathrm{cm}-\) diameter pipe carrying water (density \(1000 \mathrm{kg} / \mathrm{m}^{3}\) ). Find (a) the flow speed in the pipe and (b) the volume flow rate if the pressure difference between venturi and unconstricted pipe is \(17 \mathrm{kPa}\)

A solid sphere of radius \(R\) and mass \(M\) has density \(\rho\) that varies with distance \(r\) from the center: \(\rho=\rho_{0} e^{-r / R} .\) Find an expression for the central density \(\rho_{0}\) in terms of \(M\) and \(R\)

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