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Problem 3 (30 pts). A 2-D flow field between two infinitely large horizontal plates is initially static (V = 0). The fluid co
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Problem 3 (30 pts). A 2-D flow field between two infinitely large horizontal plates is initially static V = 0). The fluid con
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Answer #1

Solution:-

Navier - stokes theorem in two dimension

let say x

\rho g_x-\frac{\partial p}{\partial x}+\mu(\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}+\frac{\partial^2 u}{\partial z^2})=\rho\vec{a_x}

a_x=\frac{\partial u}{\partial t}+u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}+w\frac{\partial u}{\partial z}

Assuming

(1) no body forces in x - direction

(2) steady state

(3) \frac{\partial^2 u}{\partial x^2}=\frac{\partial^2 u}{\partial z^2}\approx 0

so, navier stokes theorem will reduced to

\frac{dp}{dx}=\mu\frac{\partial^2 u}{\partial y^2}

on integrating it we will get

u=\frac{1}{2\mu}\frac{dp}{dx}y^2+c_1y+c_2

Boundary conditions

u = 0 at y = 0

so, 0 = 0 + 0 + c_2

c_2 = 0

so,  u=\frac{1}{2\mu}\frac{dp}{dx}y^2+c_1y

and y = h, u = U

so,  U=\frac{1}{2\mu}\frac{dp}{dx}h^2+c_1h

C_1=\frac{U}{h}-\frac{1}{2\mu}\frac{dp}{dx}\cdot h

hence

u=\frac{1}{2\mu}\frac{dp}{dx}\cdot y^2+(\frac{U}{h}-\frac{1}{2\mu}\frac{dp}{dx}\cdot h)y

on re-arranging terms

\frac{u}{U}=\frac{y}{h}-\frac{h^2}{2\mu U}\frac{df}{dx}\frac{y}{h}\cdot (1-\frac{y}{h})

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