Chapter 9, Problem 102P

Consider dimensionless velocity distribution in Couette flow (which is also called generalized Couette flow) with an applied pressure gradient which is obtained in the following from in Example 9-16 as
   $\begin{array}{l}u*=y+\frac{1}{2}P*y*\left(y*-1\right)\\ u*=\frac{u}{v}y*=\frac{y}{h}P*=\frac{h^2\partial P}{\mu V\partial x}\end{array}$

where $u,V,\partial P/\partial x$ , and $h$ represent fluid velocity, upper plate velocity, pressure gradient, and distance between parallel plats, respectively. Also, u*,y* and P* repressent dimensionless velocity, dimensionless distance between the plates, and dimensionless pressure gradient, respectivly. (a) Explain why the velocity distribution and Posiseuille flow with a para bolic velocity distribution. (b) Show that if $P*>2$ , backflow begins at the lower wall and it never occures at the upper wall. Plot u* versus y* for this situation. (c) Find the possition and magnitude of maximim dimensionless velocity.

FIGURE P9-102

To determine

(a)

The velocity distribution is a superposition of Couette flow with a linear velocity distribution and Poiseuille flow with a parabolic velocity distribution.

Explanation of Solution

Given information:

The following figure shows that two infinite plates.

  

  Figure-(1)

At the point of wall and fluid, the velocity of the fluid is equal to zero.

Write the expression for the pressure gradient.

  $\frac{\partial P}{\partial x}=\frac{P_2-P_1}{x_2-x_1}$   ....... (I)

Here, the arbitrary locations along the $x$ axis is $x_1$ and $x_2$, the pressure at the exit is $P_2$ and the pressure at the inlet is $P_1$.

Write the expression for $y$ momentum equations.

  $\frac{{\partial }^{2}u}{\partial y^2}=\frac{1}{\mu }\frac{\partial P}{\partial y}$   ....... (II)

Here, the viscosity is $\mu$.

Write the expression for $z$ momentum equation.

  $\frac{\partial P}{\partial z}=-\rho g$   ....... (III)

Here, the density of the fluid is $\rho$ and the acceleration due to gravity is $g$.

Integrate Equation (II) with respect to $y$.

  $\begin{array}{c}{\displaystyle \int \frac{{\partial }^{2}u}{\partial y^2}}={\displaystyle \int \frac{1}{\mu }\frac{\partial P}{\partial y}}\\ \frac{\partial u}{\partial y}=\frac{1}{\mu }\frac{\partial P}{\partial y}y+C_1\end{array}$   ...... (IV)

Here, the constant is $C_1$.

Integrate Equation (IV) with respect to $y$.

  $\begin{array}{c}{\displaystyle \int \frac{\partial u}{\partial y}}={\displaystyle \int \frac{1}{\mu }\frac{\partial P}{\partial y}y+C_1}\\ u=\frac{1}{\mu }\frac{\partial P}{\partial y}\frac{y^{1+1}}{1+1}+C_{1}y+C_2\\ u=\frac{1}{2\mu }\frac{\partial P}{\partial y}{y}^2+C_{1}y+C_2\end{array}$   ....... (V)

Here, the constant is $C_2$.

Integrate Equation (III) with respect to $z$.

  $\begin{array}{c}{\displaystyle \int \frac{\partial P}{\partial z}}={\displaystyle \int -\rho g}\\ z=-\rho g{z}^{0+1}+C_3\end{array}$   ....... (VI)

Here, the constant is $C_3$.

The following figure represents planar Poiseuile flow.

  

  Figure (2)

Write the expression for dimensionless form of velocity field.

  $u*=y*+\frac{1}{2}P*y*\left(y*-1\right)$   ....... (VII)

Here, the non- dimensional velocity along the arbitrary location along the $y$ axis is $y*$ and non dimensional pressure is $P*$.

The Equation (VII) indicates that the pressure gradient is positive. If the both walls are stationary and pressure gradient is also there, then the flow should be planar Poiseuile flow.

Write the expression for non- dimensional velocity.

  $u*=\frac{u}{V}$   ...... (VIII)

Write the expression for non- dimensional pressure.

  $P*=\frac{h^2}{\mu V}\left(\frac{\partial P}{\partial x}\right)$   ....... (IX)

Write the expression for non -dimensional distance $y*$.

  $y*=\frac{y}{h}$   ...... (X)

Calculation:

Substitute $0$ for $x$ and $V$ for $u$ in Equation (V).

  $\begin{array}{l}0=\frac{1}{2\mu }\frac{\partial P}{\partial y}{\left(0\right)}^2+C_1\left(0\right)+C_2\\ 0=0+0+C_2\\ C_2=0\end{array}$

Substitute $h$ for $y$, $0$ for $C_2$ and $V$ for $u$ in Equation (V).

  $\begin{array}{l}V=\frac{1}{2\mu }\frac{\partial P}{\partial y}{\left(h\right)}^2+C_1\left(h\right)+0\\ V-\frac{1}{2\mu }\frac{\partial P}{\partial y}{\left(h\right)}^2=C_1\left(h\right)\\ C_1=\frac{V-\frac{1}{2\mu }\frac{\partial P}{\partial y}{\left(h\right)}^2}{h}\\ C_1=\frac{V}{h}-\frac{1}{2\mu }\frac{\partial P}{\partial y}h\end{array}$

Substitute $\frac{V}{h}-\frac{1}{2\mu }\frac{\partial P}{\partial y}h$ for $C_1$ and $0$ for $C_2$ in Equation (V).

  $\begin{array}{c}u=\frac{1}{2\mu }\frac{\partial P}{\partial y}{y}^2+\left(\frac{V}{h}-\frac{1}{2\mu }\frac{\partial P}{\partial y}h\right)y+0\\ =\frac{1}{2\mu }\frac{\partial P}{\partial y}{y}^2+\frac{Vy}{h}-\frac{1}{2\mu }\frac{\partial P}{\partial y}hy\\ =\frac{Vy}{h}-\frac{1}{2\mu }\frac{\partial P}{\partial y}\left(y^2-hy\right)\end{array}$

The Equations (VII) indicates super position of the linear velocity profile $u=0$ at the bottom plate to $u=V$ at the top plate and a distribution that depends on the magnitude of the applied pressure gradient. If the pressure gradient is zero, the profile of the parabolic portion is linear.

Substitute $\frac{y}{h}$ for $y*$, $\frac{u}{V}$ for $u*$ and $\frac{h^2}{\mu V}\left(\frac{\partial P}{\partial x}\right)$ for $P*$ in Equation (VII).

  $\begin{array}{l}\frac{u}{V}=\frac{y}{h}+\frac{1}{2}\left(\frac{h^2}{\mu V}\left(\frac{\partial P}{\partial x}\right)\right)\left(\frac{y}{h}\right)\left(\frac{y}{h}-1\right)\\ \frac{u}{V}=\frac{y}{h}+\frac{1}{2}\left(\frac{h^2}{\mu V}\left(\frac{\partial P}{\partial x}\right)\right)\left(\frac{y^2-y^2}{h^2}\right)\\ \frac{u}{V}=\frac{y}{h}+\left(\frac{1}{2\mu V}\left(\frac{\partial P}{\partial x}\right)\right)\\ u=\frac{Vy}{h}+\frac{1}{2\mu }\left(\frac{\partial P}{\partial x}\right)\end{array}$

To determine

(b)

The plot the $y*$ and $u*$ for the given situation.

Answer to Problem 102P

The following Figure (4) represents the $y*$ and $u*$.

  

Explanation of Solution

At the point of wall and fluid, the velocity of the fluid is equal to zero.

Write the expression for the pressure gradient.

  $\frac{\partial P}{\partial x}=\frac{P_2-P_1}{x_2-x_1}$   ....... (XI)

Here, the arbitrary locations along the $x$ axis is $x_1$ and $x_2$, the pressure at the exit is $P_2$ and the pressure at the inlet is $P_1$.

The following figure represents planar Poiseuile flow.

  

  Figure (2)

Write the expression for dimensionless form of velocity field.

  $u*=y*+\frac{1}{2}P*y*\left(y*-1\right)$   ....... (XII)

Here, the non dimensional velocities is $u*$, the arbitrary locations along the $y$ axis is $y*$ and non dimensional pressure is $P*$.

The Equation (VII) indicates that the pressure gradient is positive. If the both walls are stationary and pressure gradient is also there, then the flow should be planar Poiseuile flow.

Write the expression for non -dimensional velocity.

  $u*=\frac{u}{V}$   ...... (XIII)

Write the expression for non -dimensional pressure.

  $P*=\frac{h^2}{\mu V}\left(\frac{\partial P}{\partial x}\right)$   ....... (XIV)

The following figure represents the Couette flow between two parallel plates.

  

  Figure-(3)

The pressure gradient is less than two, then pressure is decreasing in $x$ direction, causing flow is pushed from left to right, the lower plate stationary and upper plate is moving. If the pressure gradient is greater than two, then pressure is increasing in $x$ direction, causing flow is pushed from right to left, then top plate stationary.

Write the expression for non dimensional distance $y*$.

  $y*=\frac{y}{h}$   ...... (XV)

Substitute $2$ for $y*$ and $0$ for $u*$ in Equation (II).

  $\begin{array}{c}0=2+\frac{1}{2}P*\left(2\right)\left(2-1\right)\\ P*=-2\end{array}$

The following table represents the pressure gradient and x or y non- dimensional distance.

$y*$	$u*$	$P*$
$2$	$0$	$-2$
$-2$	$0$	$0.66667$
$3$	$0$	$-0.6667$

The following figure represents between $y*$ and $u*$.

  

  Figure (4)

The Figure (4) indicates that the pressure gradient is positive. If the both walls are stationary and pressure gradient is also there, then the flow should be planar Poiseuile flow.

Conclusion:

The following Figure (4) represents the $y*$ and $u*$.

  

To determine

(c)

The position and magnitude of maximum dimensionless velocity.

Answer to Problem 102P

The expression for the pressure of fluid 1 is $-{\rho }_{1}gz+P_o$.

The expression for the pressure of fluid $2$ is $P_o+g{\rho }_2{h}_1-gz\left({\rho }_1+{\rho }_2\right)$.

Explanation of Solution

Assume, at the point of wall and fluid, the velocity of the fluid is equal to zero.

Write the expression for velocity of the fluid 1,

  $u_1=0$   ....... (XVI)

Here, the velocity of fluid 1 is $u_1$.

Assume, the velocity of the fluid 2 at the free surface of the wall is equal to the velocity of the moving plates.

Write the expressions for the velocity of fluid 2.

  $\begin{array}{c}{u}_2=V\\ =h_1+h_2\end{array}$   ....... (XVII)

Here, the velocity of fluid 2 is $u_2$, the velocity of the flow is $V$, the height of the fluid 1 from the wall is $h_1$ and the height of the fluid 2 from the interface is $h_2$.

Write the expression for velocity at interface.

  $u_2=u_1$   ....... (XVIII)

Write the expression for rate of shear stress.

  $\tau =\mu \frac{du}{dz}$   ....... (XIX)

Here, the kinematic coefficient of fluid is $\mu$ and the rate of shear stress is $\tau$.

Write the expression for the shear stress acting on fluid $1$.

  ${\tau }_1={\mu }_1\frac{d{u}_1}{dz}$   ...... (XX)

Here, the kinematic coefficient of fluid 1 is ${\mu }_1$ and the shear stress acting on the fluid 1 is ${\tau }_1$.

Write the expression for the shear stress acting on fluid 2.

  ${\tau }_2={\mu }_2\frac{d{u}_2}{dz}$   ....... (XXI)

Here, the kinematic coefficient of fluid 2 is ${\mu }_2$ and the shear stress acting on the fluid 2 is ${\tau }_2$.

Write the expression for the rate of shear stress at interface.

  ${\mu }_1\frac{d{u}_1}{dz}={\mu }_2\frac{d{u}_2}{dz}$   ....... (XXII)

Write the expression for pressure at the bottom of the flow,

  $P=P_0$   ...... (XXIII)

Here, the pressure is $P$ and the pressure at the bottom of the flow is $P_0$.

Write the expression for the pressure at the interface of fluid 1.

  $P=P_1$   ...... (XXIV)

Here, the pressure at the fluid 1 is $P_1$.

Write the expression for the pressure at the interface of fluid 2.

  $P=P_2$   ....... (XXV)

Here, the pressure at the fluid 1 is $P_2$.

At the interface of the fluid the pressure cannot have discontinuity and the surface is ignored.

Write the expression for the pressure at interface of fluid.

  $P_1=P_2$   ....... (XXVI)

Here, the velocity of flow for fluid $2$ is $u_2$.

Write the expression for $z$ momentum equation of flow of fluid 1.

  $\frac{d{P}_1}{dz}=-{\rho }_{1}g$   ...... (XXVII)

Here, the density of the fluid 1 is ${\rho }_1$ and the acceleration due to gravity is $g$.

Write the expression for $z$ momentum equation of flow of fluid 2.

  $\frac{d{P}_2}{dz}=-{\rho }_{2}g$   ...... (XXVIII)

Here, the density of the fluid 2 is ${\rho }_2$.

Calculation:

Integrate Equation (XXVII) with respect to $z$.

  $\begin{array}{l}{\displaystyle \int \frac{d{P}_1}{dz}}={\displaystyle \int -{\rho }_{1}g}\\ P_1=-{\rho }_{1}gz+C_5\end{array}$   ...... (XXIX)

Here, the constant is $C_5$.

Substitute $P_o$ for $P_1$ and $0$ for $z$ in Equation (XIV).

  $\begin{array}{l}{P}_o=-{\rho }_{1}g\left(0\right)+C_5\\ C_5=P_o\end{array}$   ...... (XXX)

Substitute $P_o$ for $C_5$ in Equation (XXIX).

  $P_1=-{\rho }_{1}gz+P_o$   ....... (XXXI)

Integrate Equation (XXVIII) with respect to $z$.

  $\begin{array}{l}{\displaystyle \int \frac{d{P}_2}{dz}}={\displaystyle \int -{\rho }_{2}g}\\ P_2=-{\rho }_{2}gz+C_6\end{array}$   ....... (XXXII)

Here, the constant is $C_6$.

Substitute $P_1$ for $P_2$ and $h_1$ for $z$ in Equation (XXXII).

  $P_1=-{\rho }_{2}g{h}_1+C_6$   ....... (XXXIII)

Substitute $-{\rho }_{1}gz+P_o$ for $P_1$ in Equation (XXXIII).

  $\begin{array}{l}-{\rho }_{1}gz+P_o=-{\rho }_{2}g{h}_1+C_6\\ C_6=-{\rho }_{1}gz+P_o+{\rho }_{2}g{h}_1\\ C_6=P_o+g\left({\rho }_2{h}_1-{\rho }_{1}z\right)\end{array}$   ....... (XXXIV)

Substitute $P_o+g\left({\rho }_2{h}_1-{\rho }_{1}z\right)$ for $C_6$ in Equation (XXXII).

  $\begin{array}{c}{P}_2=-{\rho }_{2}gz+P_o+g\left({\rho }_2{h}_1-{\rho }_{1}z\right)\\ =P_o+g{\rho }_2{h}_1-g{\rho }_{1}z-{\rho }_{2}gz\\ =P_o+g{\rho }_2{h}_1-gz\left({\rho }_1+{\rho }_2\right)\end{array}$

Conclusion:

The expression for the pressure of fluid 1 is $-{\rho }_{1}gz+P_o$.

The expression for the pressure of fluid 2 is $P_o+g{\rho }_2{h}_1-gz\left({\rho }_1+{\rho }_2\right)$.

To determine

(d)

The position and magnitude of the maximum dimensionless velocity.

Answer to Problem 102P

The position of the maximum dimensionless velocity is $B\frac{\Delta p}{\mu }\left(r_0{}^2-r^2\right)$.

The magnitude of the maximum dimensionless velocity is $\sqrt{{\left(B\frac{\Delta p}{\mu }\left(r_0{}^2-r^2\right)\right)}^2+{\left(u_{\max }\left(1-{\left(\frac{r}{r_0}\right)}^2\right)\right)}^2}$.

Explanation of Solution

Given information:

Write the expression for the position of the velocity.

  $u=B\frac{\Delta p}{\mu }\left(r_0{}^2-r^2\right)$   ....... (XXXV)

Here, the viscosity of the fluid is $\mu$, the change in pressure is $\Delta p$, the dimension constant is $B$, the circular outer radius is and the radius of the positions is $r$.

The following figure represents positions of the velocities.

  

  Figure-(5)

Write the expression for the radial velocity

  $u\left(r\right)=u_{\max }\left(1-{\left(\frac{r}{r_0}\right)}^2\right)$   ....... (XXXVI)

Here, the maximum velocity is $u_{\max }$ and the radial distance is $u\left(r\right)$.

The following figure represents the magnitude of the velocity.

  

  Figure-(6)

Write the expression for magnitude.

  $M=\sqrt{u{\left(r\right)}^2+u^2}$   ...... (XXXVII)

Here the magnitude is $M$.

Substitute $B\frac{\Delta p}{\mu }\left(r_0{}^2-r^2\right)$ for $u$ and $u_{\max }\left(1-{\left(\frac{r}{r_0}\right)}^2\right)$ for $u\left(r\right)$ in Equation (XXXVII).

  $M=\sqrt{{\left(B\frac{\Delta p}{\mu }\left(r_0{}^2-r^2\right)\right)}^2+{\left(u_{\max }\left(1-{\left(\frac{r}{r_0}\right)}^2\right)\right)}^2}$

Conclusion:

The position of the maximum dimensionless velocity is $B\frac{\Delta p}{\mu }\left(r_0{}^2-r^2\right)$.

The magnitude of the maximum dimensionless velocity is $\sqrt{{\left(B\frac{\Delta p}{\mu }\left(r_0{}^2-r^2\right)\right)}^2+{\left(u_{\max }\left(1-{\left(\frac{r}{r_0}\right)}^2\right)\right)}^2}$.