Chapter 10, Problem 29P

Consider again the slipper the slipper-pad bearing of Prob. 10-26. (a) List appropriate boundary conditions on u. (b) Solve the creeping flow approximation of the x-momentum equation to obtain an
expression for u as a function of y (and indirectly as a function of x through h and dP/dx, which are functions of x). Yen mav assume that P is not a function of y. Your final expression should be written as a $u(x,y)=f\left(y,h,dP/dx,V,\text{and}\mu \right)$ . Name the two distinct components of the velocity profile in vour result. (c) Nondimensionalize vour expression for u using these appropriate scales: $x*=x/L,y*=y/h_0,h*=h/h_0,u*=u/V,\text{and}P\text{*=}\left(P-P_0\right)h_0{}^2/\mu VL$ ??

To determine

(a)

The list of boundary conditions that can be applied at u(0, V)

Answer to Problem 29P

  $\begin{array}{l}\text{For}\text{all}\text{value}\text{of x}\\ \text{1}\text{.}\\ \text{At y}\text{= 0,}\\ \text{u}\text{=}\text{V}\\ 2.\\ \text{At y}\text{= h,}\\ \text{u}\text{=}0\end{array}$

The boundary condition that can be applied on u(0, V) are

First boundary condition:

  $\begin{array}{l}\text{For}\text{all}\text{value}\text{of x:}\\ \text{At y}\text{= 0,}\\ \text{u}\text{=}\text{V}\end{array}$

Second boundary condition:

  $\begin{array}{l}\text{For}\text{all}\text{value}\text{of x:}\\ \text{At y}\text{= h,}\\ \text{u}\text{=}0\end{array}$

We observe that h is a function of x.

Explanation of Solution

The boundary condition that can be applied on u(0, V) are

First boundary condition:

  $\begin{array}{l}\text{For}\text{all}\text{value}\text{of x:}\\ \text{At y}\text{= 0,}\\ \text{u}\text{=}\text{V}\end{array}$

Second boundary condition:

  $\begin{array}{l}\text{For}\text{all}\text{value}\text{of x:}\\ \text{At y}\text{= h,}\\ \text{u}\text{=}0\end{array}$

We observe that h is a function of x.

To determine

(b)

An expression for u as a function of y.

Answer to Problem 29P

An expression for u as a function of y is $\text{u}\left(\text{x,y}\right)=\text{V}\left(\text{1-}\frac{\text{y}}{\text{h}}\right)+\frac{{\text{h}}^{\text{2}}}{\text{2μ}}\frac{\partial \text{P}}{\partial \text{x}}\frac{\text{y}}{\text{h}}\left(\frac{\text{y}}{\text{h}}\text{-}\text{1}\right)$

We need to use momentum of x-component.

  $\begin{array}{l}\frac{\partial P}{\partial x}\approx \mu {\nabla }^{2}u\\ \frac{\partial P}{\partial x}\approx \mu \left(\frac{{\partial }^{2}u}{\partial x^2}+\frac{{\partial }^{2}u}{\partial y^2}+\frac{{\partial }^{2}u}{\partial z^2}\right)\\ \text{taking}\text{the}\text{value}\text{of}\text{y-component}\\ \frac{{\partial }^{2}u}{\partial y^2}=\frac{1}{\mu }\frac{\partial P}{\partial x}\end{array}$

On integration of x-momentum equation,

  $\frac{\partial u}{\partial y}=\frac{1}{\mu }\frac{\partial P}{\partial x}y+\text{f}\left(\text{x}\right)$

Again, on second integration of x-momentum equation,

We get,

  $\text{u}=\frac{1}{2\mu }\frac{\partial \text{P}}{\partial \text{x}}{\text{y}}^{\text{2}}+\text{yf}\left(\text{x}\right)\text{+}\text{<msup><mn>f</mn><mo>′</mo></msup>}\left(\text{x}\right)$

Applying boundary condition to find the values of two unknown function present,

  $\begin{array}{l}\text{For}\text{all}\text{value}\text{of x}\\ \text{1}\text{.}\\ \text{At y}\text{= 0,}\\ \text{u}\text{=}\text{V}\\ \\ \text{<msup><mn>f</mn><mo>′</mo></msup>}\left(\text{x}\right)=\text{V}\\ 2.\\ \text{At y}\text{= h,}\\ \text{u}\text{=}0\\ \text{f}\left(\text{x}\right)=\frac{-\text{V}-\frac{1}{2\mu }\frac{\partial \text{P}}{\partial \text{x}}{\text{h}}^{\text{2}}}{\text{h}}\end{array}$

Now, we can obtain the final expression for u as a function of y,

  $\text{u}\left(\text{x,y}\right)=\text{V}\left(\text{1-}\frac{\text{y}}{\text{h}}\right)+\frac{{\text{h}}^{\text{2}}}{\text{2μ}}\frac{\partial \text{P}}{\partial \text{x}}\frac{\text{y}}{\text{h}}\left(\frac{\text{y}}{\text{h}}\text{-}\text{1}\right)$

The components of velocities present in the final expression of u as a function of y are:

1. Couette flow
2. Poiseuille flow

Couette flow is developed by the axial movement of the wall present at the bottom. Poiseuille flow is developed by the presence of pressure gradient.

Explanation of Solution

We need to use momentum of x-component.

  $\begin{array}{l}\frac{\partial P}{\partial x}\approx \mu {\nabla }^{2}u\\ \frac{\partial P}{\partial x}\approx \mu \left(\frac{{\partial }^{2}u}{\partial x^2}+\frac{{\partial }^{2}u}{\partial y^2}+\frac{{\partial }^{2}u}{\partial z^2}\right)\\ \text{taking}\text{the}\text{value}\text{of}\text{y-component}\\ \frac{{\partial }^{2}u}{\partial y^2}=\frac{1}{\mu }\frac{\partial P}{\partial x}\end{array}$

On integration of x-momentum equation,

  $\frac{\partial u}{\partial y}=\frac{1}{\mu }\frac{\partial P}{\partial x}y+\text{f}\left(\text{x}\right)$

Again, on second integration of x-momentum equation,

We get,

  $\text{u}=\frac{1}{2\mu }\frac{\partial \text{P}}{\partial \text{x}}{\text{y}}^{\text{2}}+\text{yf}\left(\text{x}\right)\text{+}\text{<msup><mn>f</mn><mo>′</mo></msup>}\left(\text{x}\right)$

Applying boundary condition to find the values of two unknown function present,

  $\begin{array}{l}\text{For}\text{all}\text{value}\text{of x}\\ \text{1}\text{.}\\ \text{At y}\text{= 0,}\\ \text{u}\text{=}\text{V}\\ \\ \text{<msup><mn>f</mn><mo>′</mo></msup>}\left(\text{x}\right)=\text{V}\\ 2.\\ \text{At y}\text{= h,}\\ \text{u}\text{=}0\\ \text{f}\left(\text{x}\right)=\frac{-\text{V}-\frac{1}{2\mu }\frac{\partial \text{P}}{\partial \text{x}}{\text{h}}^{\text{2}}}{\text{h}}\end{array}$

Now, we can obtain the final expression for u as a function of y,

  $\text{u}\left(\text{x,y}\right)=\text{V}\left(\text{1-}\frac{\text{y}}{\text{h}}\right)+\frac{{\text{h}}^{\text{2}}}{\text{2μ}}\frac{\partial \text{P}}{\partial \text{x}}\frac{\text{y}}{\text{h}}\left(\frac{\text{y}}{\text{h}}\text{-}\text{1}\right)$

The components of velocities present in the final expression of u as a function of y are:

1. Couette flow
2. Poiseuille flow

Couette flow is developed by the axial movement of the wall present at the bottom. Poiseuille flow is developed by the presence of pressure gradient.

To determine

(c)

The nondimensionalized form of expression obtained at u(0, V)

Answer to Problem 29P

  ${\text{u}}^{\ast }=\left({\text{1-y}}^{\text{*}}\right)+\frac{{\text{h}}^{\text{*}}{}^{\text{2}}}{\text{2}}\frac{{\text{dP}}^{\ast }}{{\text{dx}}^{\text{*}}}{\text{y}}^{\text{*}}\left({\text{y}}^{\text{*}}\text{-}\text{1}\right)$

In the question, following scales are to be used to obtain the nondimensionalized form of the u as a function of y.

  $\begin{array}{l}{\text{x}}^{\text{*}}\text{=}\frac{\text{x}}{\text{L}}\text{,}\\ {\text{y}}^{\text{*}}\text{=}\frac{\text{y}}{{\text{h}}_{\text{0}}}\text{,}\\ {\text{h}}^{\text{*}}\text{=}\frac{\text{h}}{{\text{h}}_{\text{0}}}\text{,}\\ {\text{u}}^{\text{*}}\text{=}\frac{\text{u}}{\text{V}}\text{,}\\ {\text{P}}^{\text{*}}\text{=}\frac{\left({\text{P-P}}_{\text{0}}\right){\text{h}}_{\text{0}}{}^{\text{2}}}{\text{μVL}}\end{array}$

On substituting the values of the length, velocity and pressure scale in the derived expression of u(x, y),

  $\text{u}\left(\text{x,y}\right)=\text{V}\left(\text{1-}\frac{\text{y}}{\text{h}}\right)+\frac{{\text{h}}^{\text{2}}}{\text{2μ}}\frac{\partial \text{P}}{\partial \text{x}}\frac{\text{y}}{\text{h}}\left(\frac{\text{y}}{\text{h}}\text{-}\text{1}\right)$

We get,

  ${\text{u}}^{\ast }=\left({\text{1-y}}^{\text{*}}\right)+\frac{{\text{h}}^{\text{*}}{}^{\text{2}}}{\text{2}}\frac{{\text{dP}}^{\ast }}{{\text{dx}}^{\text{*}}}{\text{y}}^{\text{*}}\left({\text{y}}^{\text{*}}\text{-}\text{1}\right)$

Explanation of Solution

In the question, following scales are to be used to obtain the nondimensionalized form of the u as a function of y.

  $\begin{array}{l}{\text{x}}^{\text{*}}\text{=}\frac{\text{x}}{\text{L}}\text{,}\\ {\text{y}}^{\text{*}}\text{=}\frac{\text{y}}{{\text{h}}_{\text{0}}}\text{,}\\ {\text{h}}^{\text{*}}\text{=}\frac{\text{h}}{{\text{h}}_{\text{0}}}\text{,}\\ {\text{u}}^{\text{*}}\text{=}\frac{\text{u}}{\text{V}}\text{,}\\ {\text{P}}^{\text{*}}\text{=}\frac{\left({\text{P-P}}_{\text{0}}\right){\text{h}}_{\text{0}}{}^{\text{2}}}{\text{μVL}}\end{array}$

On substituting the values of the length, velocity and pressure scale in the derived expression of u(x, y),

  $\text{u}\left(\text{x,y}\right)=\text{V}\left(\text{1-}\frac{\text{y}}{\text{h}}\right)+\frac{{\text{h}}^{\text{2}}}{\text{2μ}}\frac{\partial \text{P}}{\partial \text{x}}\frac{\text{y}}{\text{h}}\left(\frac{\text{y}}{\text{h}}\text{-}\text{1}\right)$

We get,

  ${\text{u}}^{\ast }=\left({\text{1-y}}^{\text{*}}\right)+\frac{{\text{h}}^{\text{*}}{}^{\text{2}}}{\text{2}}\frac{{\text{dP}}^{\ast }}{{\text{dx}}^{\text{*}}}{\text{y}}^{\text{*}}\left({\text{y}}^{\text{*}}\text{-}\text{1}\right)$