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Chapter 6: Problem 5

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 \((x, y)=(1,1)\)

Short Answer

Expert verified
The acceleration of a fluid particle at point (1,1) is calculated to be (-4 ft/s虏 , -1 ft/s虏) and the pressure gradient at same point is computed to be (4 slug/ft.s虏 , -3 slug/ft.s虏).

Step by step solution

01

Compute Velocity gradient

Firstly, take spatial derivatives of the velocity vector \(V(x, y) = (A(x^2 - y^2) - 3Bx, -2Axy-3By)\) to get the velocity gradient 鈭嘨.
02

Calculate Material Derivative for Acceleration

Secondly, use the velocity gradient 鈭嘨 calculated in Step 1 and compute the acceleration of a fluid particle by taking the material derivative \(D\vec{V}/Dt = 鈭俓vec{V}/鈭倀 + (鈭嘰vec{V} . \vec{V})\). In this steady flow problem, 鈭俓vec{V}/鈭倀 = 0 because with time-independent vector field, there's no temporal evolution of \(\vec{V}\). The acceleration will then take the form of \(D\vec{V}/Dt = (\vec{V} . 鈭�)\vec{V}\).
03

Apply Euler鈥檚 Equation for Pressure gradient

Next, utilize Euler鈥檚 equation of fluid dynamics \(D\vec{V}/Dt = -\nabla p / 蟻 + \vec{g}\) to find the pressure gradient \(-\nabla p\). Substitute the value for acceleration obtained in Step 2 to Euler鈥檚 equation. Here, \(\vec{g} = -g \hat{y}\) indicating gravity acts in negative y-direction. This will provide the pressure gradient which indicates how pressure p changes spatially in x and y directions at point (1,1).

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Most popular questions from this chapter

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