/*! 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 41 A porous round tube with \(D=60 ... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 41

A porous round tube with \(D=60 \mathrm{mm}\) carries water. The inlet velocity is uniform with \(V_{1}=7.0 \mathrm{m} / \mathrm{s}\). Water flows radially and axisymmetrically outward through the porous walls with velocity distribution \\[ v=V_{0}\left[1-\left(\frac{x}{L}\right)^{2}\right] \\] where \(V_{0}=0.03 \mathrm{m} / \mathrm{s}\) and \(L=0.950 \mathrm{m} .\) Calculate the mass flow rate inside the tube at \(x=L\)

Short Answer

Expert verified
The mass flow rate inside the tube at \(x = L\) is \(0 \, \mathrm{kg}/\mathrm{s}\).

Step by step solution

01

Identify the necessary parameters for calculation

We know the diameter of the tube \(D\), the initial velocity \(V_1\), the velocity distribution of water \(v\), the radially outwards velocity \(V_0\), and the length of the tube \(L\). We need to find the mass flow rate, which is given by the formula \(\dot{m}= \rho \cdot A \cdot v\) where \(\dot{m}\) is the mass flow rate, \(A\) is the cross-sectional area of the tube, \(\rho\) is the density of the fluid (water in this case, which is typically \(1000 \, \mathrm{kg}/\mathrm{m}^3\)), and \(v\) is the fluid velocity.
02

Calculate the cross-sectional area

The cross-sectional area \(A\) of a tube is given by \(\pi \cdot (D/2)^2\). Given that \(D = 60 \, \mathrm{mm} = 0.06 \, \mathrm{m}\), we can calculate \(A = \pi \cdot (0.06/2)^2 = 0.00283 \, \mathrm{m}^2\).
03

Substitute \(x = L\) into the velocity distribution formula

The equation for the velocity distribution in the tube is given as \(v = V_0 [1 - (x/L)^2 ]\). Substituting \(x = L\) into the equation simplifies it to \(v = V_0 [1 - (1)^2] = 0 \, \mathrm{m}/\mathrm{s}\)
04

Calculate the mass flow rate

Using our formula for mass flow rate \(\dot{m}= \rho \cdot A \cdot v\), and substituting in our known values: \(\rho = 1000 \, \mathrm{kg}/\mathrm{m}^3\), \(A = 0.00283 \, \mathrm{m}^2\), and \(v = 0\), we find that \(\dot{m} = 1000 \cdot 0.00283 \cdot 0 = 0\).

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Most popular questions from this chapter

A classroom demonstration of linear momentum is planned, using a water-jet propulsion system for a cart traveling on a horizontal linear air track. The track is \(5 \mathrm{m}\) long, and the cart mass is 155 g. The objective of the design is to obtain the best performance for the cart, using 1 L of water contained in an open cylindrical tank made from plastic sheet with density of \(0.0819 \mathrm{g} / \mathrm{cm}^{2}\). For stability, the maximum height of the water tank cannot exceed \(0.5 \mathrm{m}\). The diameter of the smoothly rounded water jet may not exceed 10 percent of the tank diameter. Determine the best dimensions for the tank and the water jet by modeling the system performance. Using a numerical method such as the Euler method (see Section 5.5 ), plot acceleration, velocity, and distance as functions of time. Find the optimum dimensions of the water tank and jet opening from the tank. Discuss the limitations on your anaIysis. Discuss how the assumptions affect the predicted performance of the cart. Would the actual performance of the cart be better or worse than predicted? Why? What factors account for the difference(s)?

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