/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 49 Water flows in a circular duct. ... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 49

Water flows in a circular duct. At one section the diameter is \(0.3 \mathrm{m}\), the static pressure is \(260 \mathrm{kPa}\) (gage), the velocity is \(3 \mathrm{m} / \mathrm{s},\) and the elevation is \(10 \mathrm{m}\) above ground level. At a section downstream at ground level, the duct diameter is \(0.15 \mathrm{m}\) Find the gage pressure at the downstream section if frictional effects may be neglected.

Short Answer

Expert verified
To find the gauge pressure at the downstream section, apply Bernoulli's equation and continuity equation to find the velocity of the fluid at the downstream and then use these values to solve for the pressure at that position. Use the given parameters and calculate, remembering to convert units where necessary. The solution is the calculated pressure at the second point.

Step by step solution

01

Bernoulli's equation

Bernoulli's equation is defined as the sum of pressure energy, kinetic energy, and potential energy constancy between two points in a streamline flow of the fluid. If frictional effects are neglected and the fluid is incompressible, we can apply Bernoulli's principle which is written as: \[P_1 + \frac{1}{2}\rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho gh_2\] where \(P_1\) and \(P_2\) are the pressures at the two points, \(v_1\) and \(v_2\) are speeds at the two points, \(h_1\) and \(h_2\) are heights at the two points regarding to a certain reference level, \(\rho\) is the fluid density, \(g\) is the gravitation constant.
02

Application of the continuity equation

The mass flow rate must remain constant in the duct. We express it using a continuity equation as follows: \(A_1v_1 = A_2v_2\) where \(A_1 = \frac{\pi}{4}D_1^2\) and \(A_2 = \frac{\pi}{4}D_2^2\) are the areas of the cross section of the duct at the two points and \(D_1\) and \(D_2\) are diameters at the first and second points. By substituting areas into the equation we can solve for \(v_2\) as follows: \[v_2 = v_1\left(\frac{D_1}{D_2}\right)^2\] Given \(v_1 = 3 m/s\), \(D_1 = 0.3 m\), and \(D_2 = 0.15 m\), we calculate the velocity \(v_2\) at the second point.
03

Calculate the pressure at the second point

We rearrange the Bernouilli's equation for \(P_2\) and substitute known values (Also, note that because the second location is at ground level, \(h_2 = 0 m\)). The fluid here is water, with a density (\(\rho\)) of \(1000 kg/m^3\), and acceleration due to gravity (\(g\)) is \(9.81 m^2/s\). We now know the values for \(P_1\), \(v_1\), \(h_1\), \(v_2\), and \(h_2\), we can now solve for \(P_2\): \[P_2 = P_1 + \frac{1}{2}\rho v_1^2 + \rho gh_1 - \frac{1}{2}\rho v_2^2\] Calculate the result and remember to convert result to kPa by diving by 1000.

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