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Chapter 6: Problem 7

Consider the flow field with velocity given by \(\vec{V}=\) \(A x \sin (2 \pi \omega t) \hat{i}-A y \sin (2 \pi \omega t) \hat{j},\) where \(A=2 \mathrm{s}^{-1}\) and \(\omega=1 \mathrm{s}^{-1}\) The fluid density is \(2 \mathrm{kg} / \mathrm{m}^{3}\). Find expressions for the local acceleration, the convective acceleration, and the total acceleration. Evaluate these at point (1,1) at \(t=0,0.5,\) and 1 seconds. Evaluate \(\nabla p\) at the same point and times.

Short Answer

Expert verified
The local acceleration is given by \( A\omega\cos(2\pi\omega t)(x\hat{i} - y\hat{j}) \), the convective acceleration by \( -A^2 x\sin(2\pi \omega t)\hat{i} - A^2 y\sin(2\pi \omega t)\hat{j} \), and the total acceleration by the sum of these two. At each time point and position, this sum yields a specific value of total acceleration, and multiplication by fluid density and negation gives the gradient of pressure at that point.

Step by step solution

01

Calculation of Local Acceleration

Local acceleration can be calculated as the derivative of the velocity field with respect to time:\[ \frac{d\vec{V}}{dt} = A\omega\cos(2\pi\omega t)(x\hat{i} - y\hat{j}) \]
02

Calculation of Convective Acceleration

Convective acceleration is calculated as the product of velocity and its gradient with respect to space:\[ \vec{V} \cdot \nabla\vec{V} = -A^2 x\sin(2\pi \omega t)\hat{i} - A^2 y\sin(2\pi \omega t)\hat{j} \]
03

Calculation of Total Acceleration

The total acceleration is the sum of local acceleration and convective acceleration:\[ \vec{a} = \frac{d\vec{V}}{dt} + \vec{V} \cdot \nabla\vec{V} \] Substitute the expressions from Steps 1 and 2 into this expression to get total acceleration.
04

Evaluating Acceleration at Different Times

Plug in the coordinates (1,1) and time values (0, 0.5, and 1 second) into the total acceleration formula from step 3 in order to obtain the specific values of acceleration at these points.
05

Calculate the Gradient of Pressure

The pressure gradient can be found using the equation of fluid motion that relates pressure gradient to fluid density and total acceleration:\[ \nabla p = -\rho \vec{a} \] Substitute the given fluid density and the total acceleration from Step 4 into this equation to find \( \nabla p \) at each time point.

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Most popular questions from this chapter

Consider the flow field with velocity given by \(\vec{V}=\left[A\left(x^{2}-y^{2}\right)-3 B x\right] \hat{i}-[2 A x y-3 B y] \hat{j},\) where \(A=1 \mathrm{ft}^{-1}\) \(\mathrm{s}^{-1}, B=1 \mathrm{s}^{-1},\) and the coordinates are measured in feet. The density is 2 slug/ft \(^{3}\) and gravity acts in the negative \(y\) direction. Determine the acceleration of a fluid particle and the pressure gradient at point \((x, y)=(1,1)\)

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