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Chapter 4: Problem 23

The velocity components of \(u\) and \(v\) of a two-dimensional flow are given by \\[ u=a x+\frac{b x}{x^{2} y^{2}} \quad \text { and } \quad v=a y+\frac{b y}{x^{2} y^{2}} \\] where \(a\) and \(b\) are constants. Calculate the acceleration.

Short Answer

Expert verified
The solution to this exercise involves four main steps: Differentiating the velocity components with respect to time and space, calculating the local and convective accelerations, and finally summing everything up to obtain the acceleration. The exact values will depend on the constants \(a\) and \(b\) and variables \(x\) and \(y\), which were not given in the exercise.

Step by step solution

01

Differentiate \(u\) and \(v\) with respect to \(t\)

Since the velocity components \(u\) and \(v\) don't explicitly depend on time \(t\), their partial derivatives with respect to \(t\) (i.e., \( \frac{du}{dt} \) and \( \frac{dv}{dt}\)) will both be zero.
02

Differentiate \(u\) and \(v\) with respect to \(x\) and \(y\)

Next, calculate the derivatives of the x-component of velocity, \(u\), with respect to \(x\) and \(y\). Similarly, calculate the derivatives of the y-component of velocity, \(v\) with respect to \(x\) and \(y\). This will result in four different terms.
03

Calculate local and convective acceleration

After having calculated the derivatives in the previous step, apply them into the formula for acceleration. Compute the local accelerations \(\frac{du}{dt}\) and \(\frac{dv}{dt}\) by multiplying the obtained time derivatives of velocity with zero (as shown in step 1). Compute the convective accelerations \(u \frac{du}{dx} + v \frac{du}{dy}\) and \(u \frac{dv}{dx} + v \frac{dv}{dy}\) using the derivative values obtained in step 2.
04

Sum up the contributions

Now, add both contributions - the local and convective accelerations - together to receive the total acceleration vector. Separate the final answer for acceleration into its x and y components.

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Most popular questions from this chapter

From calculus, one obtains the following formula (Leibnitz rule) for the time derivative of an integral that contains time in both the integrand and the limits of the integration: \\[ \frac{d}{d t} \int_{x_{1}(t)}^{x_{2}(t)} f(x, t) d x=\int_{x_{1}}^{x_{2}} \frac{\partial f}{\partial t} d x+f\left(x_{2}, t\right) \frac{d x_{2}}{d t}-f\left(x_{1}, t\right) \frac{d x_{1}}{d t} \\] Discuss how this formula is related to the time derivative of the total amount of a property in a system and to the Reynolds transport theorem.

A fluid particle flowing along a stagnation streamline, as shown in Video \(V 4.9\) and Fig. \(P 4.38,\) slows down as it approaches the stagnation point. Measurements of the dye flow in the video indicate that the location of a particle starting on the stagnation streamline a distance \(s=0.6 \mathrm{ft}\) upstream of the stagnation point at \(t=0\) is given approximately by \(s=0.6 e^{-0.5 y},\) where \(t\) is in seconds and \(s\) is in feet. (a) Determine the speed of a fluid particle es a function of time, \(V_{\text {particle }}(t),\) as it flows along the streamline. (b) Determine the speed of the fluid as a function of position along the streamline, \(V=V(s)\) (c) Determine the fluid acceleration along the streamline as a function of position, \(a_{s}=a_{s}(s)\)

A constant-density fluid flows in the converging, twodimensional channel shown in Fig. P4.20. The width perpendicular to the paper is quite large compared to the channel height. The velocity in the \(z\) direction is zero. The channel half-height, \(Y\), and the fluid \(x\) velocity, \(u,\) are given by \\[ Y=\frac{Y_{0}}{1+x / \ell} \quad \text { and } \quad u=u_{0}\left(1+\frac{x}{\ell}\right)\left[1-\left(\frac{y}{Y}\right)^{2}\right] \\] Where \(x, y, Y,\) and \(\ell\) are in meters, \(u\) is in \(\mathrm{m} / \mathrm{s}, u_{0}=1.0 \mathrm{m} / \mathrm{s},\) and \(Y_{0}=1.0 \mathrm{m} .\) (a) Is this flow steady or unsteady? Is it one-dimensional, two-dimensional, or three- dimensional? (b) Plot the velocity distribution \(u(y)\) at \(x / \ell=0,0.5,\) and \(1.0 .\) Use \(y / Y\) values of 0 \\[ \pm 0.2,\pm 0.4,\pm 0.6,\pm 0.8, \text { and }\pm 1.0 \\]

A test car is traveling along a level road at \(88 \mathrm{km} / \mathrm{hr}\). In order to study the acceleration characteristics of a newly installed engine, the car accelerates at its maximum possible rate. The test crew records the following velocities at various locations along the level road: $$\begin{array}{lll} x=0 & V=88.5 \mathrm{km} / \mathrm{hr} \\ x=0.1 \mathrm{km} & V=93.1 \mathrm{km} / \mathrm{hr} \\ x=0.2 \mathrm{km} & V=98.3 \mathrm{km} / \mathrm{hr} \\ x=0.3 \mathrm{km} & V=104.0 \mathrm{km} / \mathrm{hr} \\ x=0.4 \mathrm{km} & V=110.3 \mathrm{km} / \mathrm{hr} \\ x=0.5 \mathrm{km} & V=117.2 \mathrm{km} / \mathrm{hr} \\ x=1.0 \mathrm{km} & V=164.5 \mathrm{km} / \mathrm{hr} \end{array}$$ A preliminary study shows that these data follow an equation of the form \(V=A\left(1+e^{B x}\right),\) where \(A\) and \(B\) are positive constants. Find \(A\) and \(B\) and a Lagrangian expression \(V=V\left(V_{0}, t\right),\) where \(V_{0}\) is the car velocity at time \(t=0\) when \(x=0\)

The components of a velocity ficld are given by \(u=x+y\) \(v=x y^{3}+16,\) and \(w=0 .\) Determine the location of any stagnation points \((\mathbf{V}=0)\) in the flow field.
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