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Chapter 16: Problem 13

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?

Short Answer

Expert verified
The volume flux (Q) can be calculated by substituting the found cross-sectional areas and the given height into the formula derived from Bernoulli's equation.

Step by step solution

01

Calculation of the cross-sectional area of the two pipes

We calculate the cross-sectional area using the formula: \( A = \pi r^2 \). Letting \( A_1 \) be the area of the first pipe, we have \( A_1 = \pi (0.0252 m)^2 \) We do the same for the second pipe, denoting the area as \( A_2 \): \( A_2 = \pi (0.0614 m)^2 \)
02

Set up Bernoulli's equation

We apply Bernoulli's equation, setting the height \( h_1 \) to 0 and the height \( h_2 \) at 11.5 m, and given the pressures at 1 and 2 are equal. Hence the Bernoulli’s equation becomes, \( \frac{1}{2} \rho v_1^2 = \frac{1}{2} \rho v_2^2 + \rho g h_2 \)
03

Substitute for velocities in terms of volume flux

We write down the velocity of the fluid in the two pipes in terms of the volume flux \( Q \) and their corresponding cross-sectional areas. Hence, \( v_1 = Q/A_1 \) and \( v_2 = Q/A_2 \). Substituting these into the equation from Step 2, we get \( \frac{1}{2} \rho (Q / A_1)^2 = \frac{1}{2} \rho (Q / A_2)^2 + \rho g h_2 \). This equation can be solved for Q.
04

Solve for Q

Surface area is a quadratic, we get \( (Q / A_1)^2 - (Q / A_2)^2 = 2 g h_2 \) By rearranging, we find: \( Q = \sqrt{\frac{A_1^2 A_2^2}{A_1^2 - A_2^2} 2 g h_2} \)
05

Substitute the calculated areas and given height into the formula

Substitute \( A_1 \), \( A_2 \) and \( h_2 \ = 11.5 m \) into the equation and calculate the values to find the volume flux \( Q \)

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Most popular questions from this chapter

In a hurricane, the air (density \(1.2 \mathrm{~kg} / \mathrm{m}^{3}\) ) is blowing over the roof of a house at a speed of \(110 \mathrm{~km} / \mathrm{h}\). (a) What is the pressure difference between inside and outside that tends to lift the roof? (b) What would be the lifting force on a roof of area \(93 \mathrm{~m}^{2} ?\)

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?)

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