/*! 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 9 A flow is described by the veloc... [FREE SOLUTION] | 91Ó°ÊÓ

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A flow is described by the velocity field \(\vec{V}=(A x+B)\) is \((-A y) j,\) where \(A=10\) ft/s/ft and \(B=20\) ft/s. Plot a few streamlines in the \(x y\) plane, including the one that passes through the point \((x, y)=(1,2)\).

Short Answer

Expert verified
The streamlines of the flow described by the velocity field \(\vec{V}=(Ax+B)\) is \((-Ay) j,\) where \(A=10\) ft/s/ft and \(B=20\) ft/s are represented by the equation \[0.1 \ln |10x + 20| = -0.1 \ln |y| + \ln{3} + \ln\sqrt{2}\]. This equation is obtained by solving a system of differential equations and can be plotted to visualize the streamlines, including the one passing through the point \((1,2)\).

Step by step solution

01

Establishing The Equations

The fluid particle's pathlines (streamlines) are represented mathematically by the integrals: \[ dx / V_x = dy / V_y \]This can be inputted with the given values to get:\[ dx / (10x + 20) = dy / (-10y)\].
02

Solving the Differential Equation

Upon separating the variables and integrating, we get:\[\int{dx / (10x + 20)} = - \int{dy / 10y}\]which simplifies and solves to:\[ 0.1 \ln |10x + 20| = -0.1 \ln |y| + C \]Here, \(C\) is the constant of integration which can be calculated using the given point \((1,2)\).
03

Calculating The Constant Of Integration

In order to find the value of \(C\), we substitute the given point \((1,2)\) into the equation.This gives us,\[0.1 \ln |10*1 + 20| = -0.1 \ln |2| + C \]solving will give us\[C = 0.1 \ln{30} - (-0.1 \ln{2} ) \]Therefore, we find \(C = \ln{3} + \ln{\sqrt{2}}\)
04

Final Streamline Equation

Substituting the calculated value of \(C\) back into our equation from step 2, gives the final streamline equation. Therefore, the streamlines of the flow are given by:\[0.1 \ln |10x + 20| = -0.1 \ln |y| + \ln{3} + \ln\sqrt{2}\]Now we can plot these streamlines to visualize the flow.
05

Plotting Streamlines

Using the derived streamline equation, with several different values for \(C\) (which represents different streamlines) and a fixed value for \(x\), one can plot the corresponding \(y\) values over a defined range of \(x\). The streamline that passes through the given point \((1,2)\) corresponds to the value of \(C\) derived in step 3.

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Most popular questions from this chapter

A velocity field is given by \(\vec{V}=a x t \hat{i}-b y \hat{j},\) where \(a=0.1 \mathrm{s}^{-2}\) and \(b=1 \mathrm{s}^{-1}\). For the particle that passes through the point \((x, y)=(1,1)\) at instant \(t=0\) s, plot the pathline during the interval from \(t=0\) to \(t=3\) s. Compare with the streamlines plotted through the same point at the instants \(t=0,1,\) and \(2 \mathrm{s}\).

For the velocity fields given below, determine: a. whether the flow field is one- , two-, or three-dimensional, and why. b. whether the flow is steady or unsteady, and why. (The quantities \(a\) and \(b\) are constants.) (1) \(\vec{V}=\left[a y^{2} e^{-b t}\right]\) (2) \(\vec{V}=a x^{2} i+b x j+c k\) (3) \(\vec{V}=\)axyi\(-\)byt (4) \(\vec{V}=a x \hat{i}-b y j+c t \hat{k}\) (5) \(\bar{V}=\left[a e^{-\ln x}\right] i+b t^{2}\) (6) \(\vec{V}=a\left(x^{2}+y^{2}\right)^{1 / 2}\left(1 / z^{3}\right) \hat{k}\) (7) \(\vec{V}=(a x+t) i-b y^{2}\) (8) \(\vec{V}=a x^{2} \hat{i}+b x z j+c y k\)

Consider the flow ficld \(\vec{V}=a x t \hat{i}+b \tilde{j},\) where \(a=1 / 4 s^{-2}\) and \(b=1 / 3 \mathrm{m} / \mathrm{s}\). Coordinates are measured in meters. For the particle that passes through the point \((x, y)=(1,2)\) at the instant \(t=0,\) plot the pathline during the time interval from \(t=0\) to 3 s. Compare this pathline with the streakline through the same point at the instant \(t=3\) s.

Some experimental data for the viscosity of helium at 1 atm are $$\begin{array}{cccccc}\boldsymbol{T},^{\circ} \mathbf{C} & 0 & 100 & 200 & 300 & 400 \\ \boldsymbol{\mu}, \mathbf{N} \cdot \mathbf{s} / \mathbf{m}^{2}\left(\times \mathbf{1 0}^{5}\right) & 1.86 & 2.31 & 2.72 & 3.11 & 3.46 \end{array}$$ Using the approach described in Appendix A.3, correlate these data to the empirical Sutherland equation \\[\mu=\frac{b T^{1 / 2}}{1+S / T}\\] (where \(T\) is in kelvin) and obtain values for constants \(b\) and \(S\).

Magnet wire is to be coated with varnish for insulation by drawing it through a circular die of \(1.0 \mathrm{mm}\) diameter. The wire diameter is \(0.9 \mathrm{mm}\) and it is centered in the die. The varnish \((\mu=20\) centipoise) completely fills the space between the wire and the die for a length of \(50 \mathrm{mm}\) The wire is drawn through the die at a speed of \(50 \mathrm{m} / \mathrm{s}\) Determine the force required to pull the wire.

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