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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 constriction is about 2.79 m/s.

Step by step solution

01

Determine the initial and final cross-sectional areas

The cross-sectional area \(A\) of a pipe can be calculated using the formula for the area of a circle, \(A = \pi r^2\), where \(r\) is the radius of the circle. Here, the radius r is half the diameter of the circle. So, we have \(A_1 = \pi (2.5 cm / 2)^2 = 1.225 cm^2\) and \(A_2 = \pi (2.0 cm / 2)^2 = 0.785 cm^2\). Now, we just need to convert cm^2 to m^2 by multiplying by \(10^{-4}\). So, we have \(A_1 = 1.225 × 10^{-4} m^2\) and \(A_2 = 0.785 × 10^{-4} m^2\).
02

Substitute values into the continuity equation

We can use the equation of continuity \(A_1v_1 = A_2v_2\) to find the final speed \(v_2\). Substituting the given values and the ones we found, we get \(1.225 × 10^{-4} m^2 × 1.8 m/s = 0.785 × 10^{-4} m^2 × v_2\)
03

Solve for the final speed

Solving for \(v_2\), we have \(v_2 = A_1v_1 / A_2 = (1.225 × 10^{-4} m^2 × 1.8 m/s) / (0.785 × 10^{-4} m^2) = 2.79 m/s\). Therefore, the flow speed in the constriction is approximately 2.79 m/s.

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