Laminar flow between two inclined plates. Combined Couette and Poiseuille flow of a newtonian fluid

 Here, two different cases shall be considered from a simple situation. The flow between two inclined plates with one fixed wall aand one movable wall shall be considered for the calculation of the velocity distribution and the flow rate. This arrangement of the plates may be useed to study the Poiseuille and the Couette flow, separately or as a whole.

The nature of the parallel, Poiseuille and Couette, flow

For instance, we make reference to the sketch shown in Fig. 1. The fluid is newtonian and is travelling down.

Figure 1. Sketch for the flow of a newtonian fluid between two inclined plates which may be useful for the study of the Poiseuille or Couette flows, or even both.

In order to give some kind of order, a list of comments and/or observations and/or assumptions are to be made and listed below:

• the fluid moves slowly so that the flow regime can be assumed to be laminar. This assumption is at the core of study of this situation;
• since the system is formed by plane rectangular walls and no curvature seems to be part of it the cartesian coordinates $(x,y,z)$ will be used for this small study;
• the walls are equally spaced, on $z$, along coordinate $x$;
• notice that the axis-$z$ is perpendicular to the orientation of the walls;
• both walls have identical inclination $\chi$;
• the fluid moves only along the axial direction from $x=0$ to $x=L$. This is, the fluid moves parallel to the walls;
• the fluid moves downwards between the two plates as indicated by the arrow in Fig. 1;
• the fluid does not move vertically, perpendicular to the plates, nor transversally, on the $y$-axis. From a transversal view, the fluid does not move outside the screen nor towards the screen. Also, the fluid does not move in other direction that along the $x$-axis;
• considering that the vectorial function representing the fluid velocity is $\textbf{v}=(v_x,v_y, v_z)$ and in agreement with the above statements $v_x \ne 0$, $v_y =0$ and $v_z = 0$ follow;
• the component $v_x$ does not vanish and may depend on $(x,y,z,t)$. In this case, it only depends on $z$. One arrives to this conclusion as follows. The continuity equation assures that the fluid velocity at $x=0$ and $x=L$ is the same, so that it does not depends on $x$. Besides, it is easy to realize that the fluid velocity remains unchanged under $y$-variations;

Figure 2. Small sketch for the analysis of the components of the velocity

• stationary conditions are considered. This means that there is no acceleration and that the temporal derivatives of $\textbf{v}$ are zero or that there is no temporal dependency;
• the pressure difference between $x=0$ and $x=L$ may set the fluid in motion. In other words, pressure can be exerted by a pump and by the action of gravity;
• the pressure $p$ is in general a function of $(x, y, z,t)$; but in this case it only depends on $x$. Actually, pressure only acts as a projection over the $x$-axis;
• the pressure gradient generated externally by a pump is not shown in Fig. 1. However, since the fluid moves downwards the acceleration due to gravity has a contribution effect;
• the effect of temperature on the fluid properties is very small and will be discarded. Then, $\rho$ and $\mu$ are constants.

The equations to calculate the velocity distribution

In order to determine the velocity distribution two equations should be considered: one defining the type of fluid and another one that models the fluid motion under any circumstance. These equations are,

The fuid constitutive equation, 

Which corresponds to a newtonian fluid. This equation can be expressed, in cartesian coordinates as follows,

$\mathbf{\tau}=-\mu \left[\nabla \mathbf{v} + \left(\nabla \mathbf{v}\right)^T\right]$        Eq. (01)

where $\mu$ is the kinematic viscosity. Notice that Eq. (01) is in fact a matricial equation because the term,

$\nabla \mathbf{v}$

can not be written, in the familiar manner, as a vector but as a matrix of size $3 \times 3$. The matrix in the right hand side of Eq. (01) contains much more information than a simple vector. Equation (01) can be written in extended way as follows,

$\begin{bmatrix}
\tau_{xx} & \tau_{xy} & \tau_{xz}\\
\tau_{yx} & \tau_{yy} & \tau_{yz}\\
\tau_{zx} & \tau_{zy} & \tau_{zz}
\end{bmatrix}=-\mu
\begin{bmatrix}
2\dfrac{\partial v_x}{\partial x} & \dfrac{\partial v_y}{\partial x}+\dfrac{\partial v_x}{\partial y} & \dfrac{\partial v_x}{\partial z}+\dfrac{\partial v_z}{\partial x} \\
\dfrac{\partial v_y}{\partial x}+\dfrac{\partial v_x}{\partial y} & 2\dfrac{\partial v_y}{\partial y} & \dfrac{\partial v_z}{\partial y}+\dfrac{\partial v_y}{\partial z} \\
\dfrac{\partial v_x}{\partial z}+\dfrac{\partial v_z}{\partial x} & \dfrac{\partial v_z}{\partial y}+\dfrac{\partial v_y}{\partial z} & 2\dfrac{\partial v_z}{\partial z}
\end{bmatrix}$        Eq. (02)

The momentum balance equation

Which in cartesian coordinates $(x,y,z)$ is written as:

$\rho\left(\frac{\partial v_x}{\partial t}+v_x\frac{\partial v_x}{\partial x}+v_y\frac{\partial v_x}{\partial y}+v_z\frac{\partial v_x}{\partial z}\right)=-\frac{\partial p}{\partial x}-\left[\frac{\partial \tau_{xx}}{\partial x}+\frac{\partial \tau_{yx}}{\partial y}+\frac{\partial \tau_{zx}}{\partial z}\right]+\rho g_x$        Eq. (03)

$\rho\left(\frac{\partial v_y}{\partial t}+v_x\frac{\partial v_y}{\partial x}+v_y\frac{\partial v_y}{\partial y}+v_z\frac{\partial v_y}{\partial z}\right)=-\frac{\partial p}{\partial y}-\left[\frac{\partial \tau_{xy}}{\partial x}+\frac{\partial \tau_{yy}}{\partial y}+\frac{\partial \tau_{zy}}{\partial z}\right]+\rho g_y$        Eq. (04)

$\rho\left(\frac{\partial v_z}{\partial t}+v_x\frac{\partial v_z}{\partial x}+v_y\frac{\partial v_z}{\partial y}+v_z\frac{\partial v_z}{\partial z}\right)=-\frac{\partial p}{\partial z}-\left[\frac{\partial \tau_{xz}}{\partial x}+\frac{\partial \tau_{yz}}{\partial y}+\frac{\partial \tau_{zz}}{\partial z}\right]+\rho g_z $        Eq. (05)

The constitutive and the momentum balance equations form a system of differential equations for the stress $\mathbf{\tau}$ and the fluid velocity $\mathbf{v}$. Of course, $p$ appears in Eqs. (03-05) but this is to be handled later.

Equations (02-05) can be simplified by using the above listed observations to the physics of the problem. First, Eq. (02) is simplified to get,

$\begin{bmatrix}
\tau_{xx} & \tau_{xy} & \tau_{xz}\\
\tau_{yx} & \tau_{yy} & \tau_{yz}\\
\tau_{zx} & \tau_{zy} & \tau_{zz}
\end{bmatrix}
=-\mu
\begin{bmatrix}
0 & 0 & \dfrac{\partial v_x}{\partial z} \\
0 & 0 & 0 \\
\dfrac{\partial v_x}{\partial z} & 0 & 0
\end{bmatrix}$        Eq. (06)

From Eq. (06) it can be seen that all elements in the stress matrix are zero but $\tau_{xz}$ and $\tau_{zx}$. This is comes out after element-to-element comparison between the left hand side and the right hand side matrices. The resulting equation for the shear stress is,

$\tau_{xz}=-\mu \dfrac{dv_x}{dz}$        Eq. (07)

Since $v_x$ only depends on a single variable it can not have partial derivatives.

The same process of simplification followed to obtain Eq. (06) from Eq. (02) is used again to simplify the components of the momentum balance equation given in Eqs. (03)-(05). From the $x$ and $z$ components given in Eqs. (04)-(05) all terms vanish. From the $x$ component given in Eq. (03) only the following terms remain,

$0=-\dfrac{dp(x)}{dx}-\dfrac{d\tau_{zx}}{dz}+\rho g_x$        Eq. (08)

In Eq. (08) the coefficient $g_x$ represents the $x$ component of the gravity vector wich can be calculated by using the following arrangement obtained from the main schematics shown at the beginning of this post. It follows that $g_x = \mathbf{g} \cos \chi$.

Fig. 3 Schematics to represent the $g_x$ component of the acceleration due to gravity $\mathbf{g}$

On the other hand, Eqs. (7) and (8) may be combined. The result is,

$0=-\dfrac{dp(x)}{dx} + \mu\dfrac{d^2 v_{x}}{dz^2}-\rho g \cos \chi$       Eq. (9)

Notice that $\mathbf{g}=-g$. Equation (9) is in fact a constant coefficients ordinary differential equation with a polynomial inhomogeneity that can be easily solved. Its solution is,

$v_x=\dfrac{1}{\mu}\left[ \dfrac{dp(x)}{dx}+\rho g \cos \chi \right]\dfrac{z^2}{2} + C_1z + C2$       Eq. (10)

The integration constants in solution Eq. (10) are calculated from a set of boundary conditions. One of these demands that the fluid velocity at the bottom wall should be zero due to the no-slip condition to a motionless wall.

The other condition is located at the top wall. In this case, the top wall is moving so that the the fluid velocity should be equal to that of the tp wall. In other words,

$v_x=0$ at $z=-\dfrac{H}{2}$

$v_x=U$ at $z=\dfrac{H}{2}$        Eqs. (11)

where $U$ is the velocity  of the top  wall. Evaluation of the boundary conditions Eq. (11) leads to a set of algebraic equations,

$0=\dfrac{1}{\mu}\left[ \dfrac{dp(x)}{dx}+\rho g \cos \chi \right]\dfrac{H^2}{8} - C_1\dfrac{H}{2} + C2$


$U=\dfrac{1}{\mu}\left[ \dfrac{dp(x)}{dx}+\rho g \cos \chi \right]\dfrac{H^2}{8} + C_1\dfrac{H}{2} + C2$       Eqs. (12)

in which $C_1$ and $C_2$ are the unknown. These are easily found to be,

$C_1=\dfrac{U}{H}$

$C_2=\dfrac{1}{2}\biggl\{ U - \dfrac{1}{\mu}\left[ \dfrac{dp(x)}{dx}+\rho g \cos \chi \right]\dfrac{H^2}{4} \biggr\}$    Eqs. (13)

Further substitution of $C_1$ and $C_2$ given in Eqs. (13) into solution for $v_x$ in Eq. (10) gives,

$v_x=\dfrac{1}{\mu}\left[ \dfrac{dp(x)}{dx}+\rho g \cos \chi \right] \left(\dfrac{z^2}{2}- \dfrac{H^2}{8}\right) + \left(\dfrac{z}{H} + \dfrac{1}{2}\right)U$    Eq. (14)

In Eq. (14) pressure is still unknow and a comment is worth to make: the whole term in the expression in square brakets correspond to all pressure contributions: hydrostatic and due to gravity.

A discussion of Couette and Poiseuille flows

You may have already noticed that the velocity distribution presented in Eq. (14) combines two contributions for the fluid motion: one due to pressure gradients and another one due to the top wall motion.

If the fluid motions were only due to presure gradients, this is with $U=0$, the resulting flow would be called plane Poiseuille flow and $v_x$ would be,

$v_x=\dfrac{1}{\mu}\left[ \dfrac{dp(x)}{dx}+\rho g \cos \chi \right] \left(\dfrac{z^2}{2}- \dfrac{H^2}{8}\right) $    Eq. (14a)

On the other hand, if fluid motions were only due to the top wall motions the resulting flow would be called Coutte flow and $v_x$ would be,

$v_x= \left(\dfrac{z}{H} + \dfrac{1}{2}\right)U$   Eq. (14b)

About the pressure gradient calculation

In fact, the velocity Eq. (14) is a first order ordinary differential equation for $p(x)$ and can be solved by separation of variables. Its solution is,

$p(x)=\Biggl\{\dfrac{\mu\left[ v_x-\left( \dfrac{z}{H}+\dfrac{1}{2} \right)U \right]}{\dfrac{z^2}{2}-\dfrac{H^2}{8}}-\rho\cos\chi \Biggr\}x + C_3$    Eq. (16)

In order to determine contant of integration $C_3$ in Eq. (16) a boundary condition is required. Two boundary conditions arise: one at $x=0$ where pressure is $p_0$ and another one at $x=L$ where pressure is $p_L$. The difficulty with this is that there are two different values for $C_3$.

$p=p_0$ at $x=0$

$p=p_L$ at $x=L$    Eqs. (17)

If boundary conditions Eqs. (17) are evaluated on the pressure solution Eq. (16) two expressions for $C_3$ arise. However, these can be combined to obtain the pressure gradient defined as,

$\dfrac{\Delta P}{L}=\dfrac{p_0-p_L}{L}=\dfrac{\mu\left[ v_x-\left( \dfrac{z}{H}+\dfrac{1}{2} \right)U \right]}{\dfrac{z^2}{2}-\dfrac{H^2}{8}}-\rho\cos\chi$        Eq. (18)

which is a more tractable form of the pressure gradient. Notice that from the ordinary differential equation for pressure (not shown) it can be deduced that,

$\dfrac{dp}{dx}=\dfrac{\Delta P}{L}$       Eq. (19)

so that the velocity can be now written in terms of parameters of a more simple nature. This is,

$v_x=\dfrac{1}{\mu}\left[ -\dfrac{\Delta P}{L}+\rho g \cos \chi \right] \left(\dfrac{z^2}{2}- \dfrac{H^2}{8}\right) + \left(\dfrac{z}{H} + \dfrac{1}{2}\right)U$    Eq. (20)

Notice that $p(x)$ and $v_x(z)$ are closely related. Changes in one modifies the other one. Remember that $p(x)$ is only the hydrostatic pressure contribution.

The flow rate between two plates

Once the fluid velocity or velocity distribution across the fluid layer, the flow rate can be calculated. For this, we should consider the fluid flowing through a channel rather than through a geometry as shown in Fig. 1.

The channel will have cross section as shown below,

Figure 4. Cross section of the channel for flow rate calculations

Next, the flow rate can now be calculated as a double integral for the velocity $v_x$,

$Q=\int^{-W/2}_{W_2} \int^{H/2}_{-H/2} v_x\,dz\,dy$

$Q=W \int^{H/2}_{-H/2} v_x\,dz$

$v_x$ from Eq. (14) is now substituted

$Q=W \int^{H/2}_{-H/2} \bigg\{\dfrac{1}{\mu}\left[ \dfrac{dp(x)}{dx}+\rho g \cos \chi \right] \left(\dfrac{z^2}{2}- \dfrac{H^2}{8}\right) + \left(\dfrac{z}{H} + \dfrac{1}{2}\right)U\bigg\}\,dz$

$Q=W \dfrac{1}{\mu}\left[ \dfrac{dp(x)}{dx}+\rho g \cos \chi \right] \int^{H/2}_{-H/2}\left(\dfrac{z^2}{2}- \dfrac{H^2}{8}\right)\,dz + WU \int^{H/2}_{-H/2} \left(\dfrac{z}{H} + \dfrac{1}{2}\right) \,dz$

$Q=-\dfrac{WH^3}{12}\dfrac{1}{\mu}\left[ \dfrac{dp(x)}{dx}+\rho g \cos \chi \right] + \dfrac{WUH}{2}$    Eq. (21)

This is the end of the solution.

Any question? Write in the comments and I shall try to help.

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• The flow of a plastic Bingham fluid through an inclined pipe
• The Poiseuille flow of a power law fluid through an inclined pipe
• Flow of a newtonian fluid in the annular region between two vertical cylinders
• Understanding the mechanics of the Bernoulli equation
• Practical situation. A polymeric fluid flowing through a pipe
• Practical situation. A polyisoprene solution flowing through a pipe

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Ildebrando.