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Chapter 16: Problem 10

Water is moving with a speed of \(5.18 \mathrm{~m} / \mathrm{s}\) through a pipe with a cross-sectional area of \(4.20 \mathrm{~cm}^{2} .\) The water gradually descends \(9.66 \mathrm{~m}\) as the pipe increases in area to \(7.60 \mathrm{~cm}^{2} .(a)\) What is the speed of flow at the lower level? ( \(b\) ) The pressure at the upper level is \(152 \mathrm{kPa}\); find the pressure at the lower level

Short Answer

Expert verified
The speed of flow at the lower level is 2.87 m/s and the pressure at the lower level is 158.27 kPa

Step by step solution

01

Conversion of Units

Convert the cross-sectional areas to square meters (from square centimeter). \( A_1 = 4.20 \times 10^{-4} \)m² and \( A_2 = 7.60 \times 10^{-4} \) m².
02

Find Speed at Lower Level using Continuity Equation

The Continuity equation states that the mass flow rate must remain constant in a fluid pipe. It can be formulated as \(A_1 v_1 = A_2 v_2\). Solve this equation for \(v_2\) (Speed at lower level), \(v_2\) = \( \frac{A_1 \times v_1}{A_2} \)
03

Calculate the Speed

Insert the values into the above equation to find the speed at the lower point, \(v_2\) = \( \frac{4.20 \times 10^{-4} \times 5.18}{7.60 \times 10^{-4}} \) = 2.87 m/s
04

Find Pressure at Lower Level using Bernoulli's Equation

Reshape Bernoulli's equation and solve for \(P_2\), \(P_2 = P_1 + \frac{1}{2} \rho v_1^2 + \rho g h_1 - \frac{1}{2} \rho v_2^2 - \rho g h_2\)
05

Calculate Pressure at Lower Point

Insert the values into the above equation to find the pressure at the lower point, remembering to convert pressures to Pascals (1kPa = 1000 Pa), \(P_2 = 152000 + \frac{1}{2} \times 1000 \times {5.18}^2 + 1000 \times 9.8 \times 9.66 - \frac{1}{2} \times 1000 \times {2.87}^2 - 1000 \times 9.8 \times 9.66 \)= 158265 Pa = 158.27 kPa

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Most popular questions from this chapter

Water is pumped steadily out of a flooded basement at a speed of \(5.30 \mathrm{~m} / \mathrm{s}\) through a uniform hose of radius \(9.70 \mathrm{~mm}\). The hose passes out through a window \(2.90 \mathrm{~m}\) above the water line. How much power is supplied by the pump?

A Venturi tube has a pipe diameter of \(25.4 \mathrm{~cm}\) and a throat diameter of \(11.3 \mathrm{~cm}\). The water pressure in the pipe is \(57.1 \mathrm{kPa}\) and in the throat is \(32.6 \mathrm{kPa}\). Calculate the volume flux of water through the tube.

Air flows over the top of an airplane wing, area \(A\), with speed \(v_{1}\) and past the underside of the wing with speed \(v_{\mathrm{u}} .\) Show that Bemoulli's equation predicts that the upward lift force \(L\) on the wing will be $$ L=\frac{1}{2} \rho A\left(v_{\mathrm{t}}^{2}-v_{\mathrm{u}}^{2}\right) $$ where \(\rho\) is the density of the air. (Hint: Apply Bernoulli's equation to a streamline passing just over the upper wing surface and to a streamline passing just beneath the lower wing surface. Can you justify setting the constants for the two streamlines equal?)

A liquid flows through a horizontal pipe whose inner radius is \(2.52 \mathrm{~cm}\). The pipe bends upward through a height of \(11.5 \mathrm{~m}\) where it widens and joins another horizontal pipe of inner radius \(6.14 \mathrm{~cm}\). What must the volume flux be if the pressure in the two horizontal pipes is the same?

A Pitot tube is mounted on an airplane wing to determine the speed of the plane relative to the air, which has a density of \(1.03 \mathrm{~kg} / \mathrm{m}^{3} .\) The tube contains alcohol and indicates a level difference of \(26.2 \mathrm{~cm} .\) What is the plane's speed relative to the air? The density of alcohol is \(810 \mathrm{~kg} / \mathrm{m}^{3}\).
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