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Chapter 9: Problem 44

Water entering a house flows with a speed of \(0.20 \mathrm{m} / \mathrm{s}\) through a pipe of \(1.0 \mathrm{cm}\) inside radius. What is the speed of the water at a point where the pipe tapers to a radius of \(2.5 \mathrm{mm} ?\)

Short Answer

Expert verified
Answer: The speed of water at the point where the pipe tapers to a radius of 2.5 mm is 3.2 m/s.

Step by step solution

01

Write down the given quantities and the continuity equation

The given quantities are: Speed of water at point 1: \(v_1 = 0.20 \ \mathrm{m/s}\) Radius of pipe at point 1: \(r_1 = 1.0 \ \mathrm{cm}\) Radius of pipe at point 2: \(r_2 = 2.5 \ \mathrm{mm}\) We need to find the speed of water at point 2, \(v_2\). The continuity equation states that the product of the area of the pipe and the speed of the water is constant at any two points in the pipe, so: $$A_1v_1 = A_2v_2$$
02

Convert units and calculate the pipe areas

Convert the given radii into meters: \(r_1 = 1.0 \ \mathrm{cm} \times \frac{1 \ \mathrm{m}}{100 \ \mathrm{cm}} = 0.01 \ \mathrm{m}\) \(r_2 = 2.5 \ \mathrm{mm} \times \frac{1 \ \mathrm{m}}{1000 \ \mathrm{mm}} = 0.0025 \ \mathrm{m}\) Calculate the areas of the pipe at points 1 and 2: $$A_1 = \pi r_1^2 = \pi (0.01 \ \mathrm{m})^2 = 0.0001 \pi \ \mathrm{m^2}$$ $$A_2 = \pi r_2^2 = \pi (0.0025 \ \mathrm{m})^2 = 0.00000625 \pi \ \mathrm{m^2}$$
03

Use the continuity equation to find the speed of water at point 2

Plug the values of \(A_1\), \(A_2\), and \(v_1\) into the continuity equation: $$0.0001 \pi \ v_1 = 0.00000625 \pi \ v_2$$ Solve for \(v_2\): $$v_2 = \frac{0.0001 \pi \ v_1}{0.00000625 \pi} = \frac{0.0001}{0.00000625} \times v_1 = 16 \times 0.20 \ \mathrm{m/s} = 3.2 \ \mathrm{m/s}$$ So, the speed of water at the point where the pipe tapers to a radius of 2.5 mm is \(3.2 \ \mathrm{m/s}\).

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