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QUESTION-3: For a fully developed laminar flow between two parallel flat plates with no gravity effects (see figure); yo W<sub>x</sub>(y) 0 -yo 1- Show that the momentum equation for steady-state flow with constant density and viscosity takes the form: $\frac{dp}{dx} = \mu \frac{d^2W_x}{dy^2}$ where $p$ is the pressure, $\mu$ is the dynamic viscosity, and $W_x$ is the velocity component in the x-direction parallel to the plates. 2- If the distance between the plates is 2$y_0$ show that the velocity profile is given by: $W_x(y) = \frac{3}{2}W_m \left[1 - \left(\frac{y}{y_0}\right)^2\right]$ where $W_m$ is the averaged velocity. 

          QUESTION-3:
For a fully developed laminar flow between two parallel flat plates with no gravity effects (see
figure);
yo
W<sub>x</sub>(y)
0
-yo
1- Show that the momentum equation for steady-state flow with constant density and viscosity
takes the form:
$\frac{dp}{dx} = \mu \frac{d^2W_x}{dy^2}$
where $p$ is the pressure, $\mu$ is the dynamic viscosity, and $W_x$ is the velocity component in the
x-direction parallel to the plates.
2- If the distance between the plates is 2$y_0$ show that the velocity profile is given by:
$W_x(y) = \frac{3}{2}W_m \left[1 - \left(\frac{y}{y_0}\right)^2\right]$
where $W_m$ is the averaged velocity.
Show more…
QUESTION-3: For a fully developed laminar flow between two parallel flat plates with no gravity effects (see figure); yo W<sub>x</sub>(y) 0 -yo 1- Show that the momentum equation for steady-state flow with constant density and viscosity takes the form: (dp)/(dx) = μ(d^2Wx)/(dy^2) where p is the pressure, μ is the dynamic viscosity, and Wx is the velocity component in the x-direction parallel to the plates. 2- If the distance between the plates is 2y0 show that the velocity profile is given by: Wx(y) = (3)/(2)Wm [1 - ((y)/(y0))^2] where Wm is the averaged velocity.

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Chemistry: Structure and Properties
Chemistry: Structure and Properties
Nivaldo Tro 2nd Edition
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QUESTION-3 For a fully developed laminar flow between two parallel flat plates with no gravity effects (see figure); yo W() 0 -yo 1- Show that the momentum equation for steady-state flow with constant density and viscosity takes the form dp = aW dx where p is the pressure, μ is the dynamic viscosity, and W is the velocity component in the x-direction parallel to the plates. 2- If the distance between the plates is 2yo, show that the velocity profile is given by w = wm[1- (y/yo)^2] where Wm is the averaged velocity.
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Transcript

-
00:01 Let us have a look on the question.
00:02 We have given here the viscous incompressible fluid inside this tube and this surface of downward is moving with the v1 velocity in this direction and this surface, upward surface is moving in this direction with the v2 velocity.
00:15 So, first we have to find the navier -stokes equation for the laminar flow of this equation.
00:19 So, what is the navier -stokes equation? so, we know this is equal to d square v by dy square equal to simply 1 by mu d rho by dx.
00:31 This is the navier -stokes equation.
00:34 Now, according to question already we have given that this is d rho by dx equal to 0.
00:40 First of all understand what are the list terms also.
00:43 This is mu is the dynamic viscosity, y is equal to vertical distance of particular point of interest, v is the velocity at y and dp by dx is equal to pressure gradient in the x direction.
00:53 And we have given that already here the d rho by dx equal to 0.
00:58 This is dp, sorry not d rho, this is dp, dp by dx.
01:03 This is equal to 0.
01:05 So, we can say here directly d square y by dy square is equal to 0.
01:11 Now, think about this thing that if i am talking that i have one value dv by dy and i am just differentiating this term, i am just differentiating this term and i am getting the differentiation of this value is 0.
01:25 Means this value was initially a constant, means we know the differentiation of the constant is 0, means dv by dy is the constant term.
01:37 I can say it is the c1 constant.
01:39 Now, i am just integrating both side with the dy.
01:42 Now, i am integrating this term dv by dy, both side with dy.
01:48 This is also c1 dy.
01:50 Now, dy is just cut y, this dy.
01:53 Here only present is dv only and here only c1 and dy...
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