Question 11.5: Consider the flow of an incompressible Newtonian fluid between...

We are asked to find the stresses in a given flow and to use the momentum equation to gain insight into the nature of the flow. Figure 11.4 serves as the sketch for this fluid system. To find the stresses we will use the Newtonian constitutive relationships, Eqs. 11.6, with the known velocity components. The only nonzero velocity component is u, and the density is constant. Since u is a function of y alone, all x and z spatial derivatives of u are zero. We can begin by confirming that the velocity field satisfies the continuity equation for an incompressible fluid. From Eq. 11.2a we find

\left(\frac{\partial \rho }{\partial t}+u\frac{\partial \rho }{\partial x}+v\frac{\partial \rho }{\partial y}+w\frac{\partial \rho }{\partial z} \right)+\rho \left(\frac{\partial u}{\partial x} +\frac{\partial v}{\partial y} +\frac{\partial w}{\partial z} \right)=0                              (11.2a)

\frac{\partial u}{\partial x} +\frac{\partial v}{\partial y} +\frac{\partial w}{\partial z}=\frac{\partial}{\partial x} (u(y))+\frac{\partial}{\partial y}(0)+\frac{\partial}{\partial z}(0)=0

Inserting the velocity components into the Newtonian constitutive relationships, Eqs. 11.6, we find

σ_{xx} =−p−\frac{2}{3} µ(∇ • u)+2µ\frac{\partial u}{\partial x} =−p−\frac{2}{3} µ(0)+2µ(0) =−p

σ_{yy} =−p−\frac{2}{3} µ(∇ • u)+2µ\frac{\partial v}{\partial y} =−p−\frac{2}{3} µ(0)+2µ(0) =−p

σ_{zz} =−p−\frac{2}{3} µ(∇ • u)+2µ\frac{\partial w}{\partial z} =−p−\frac{2}{3} µ(0)+2µ(0) =−p

σ_{xy}=σ_{yx}=\mu \left(\frac{\partial u}{\partial y} +\frac{\partial v}{\partial x}\right)=\mu \left[\frac{\partial}{\partial y}\left(U\frac{y}{h}\right)+\frac{\partial}{\partial x} (0) \right]=\mu \frac{U}{h}

σ_{zy}=σ_{yz}=\mu \left(\frac{\partial w}{\partial y} +\frac{\partial v}{\partial z}\right)=\mu \left[\frac{\partial}{\partial y}\left(0\right)+\frac{\partial}{\partial z} (0) \right]=0

σ_{zx}=σ_{xz}=\mu \left(\frac{\partial w}{\partial x} +\frac{\partial u}{\partial z}\right)=\mu \left[\frac{\partial}{\partial x}\left(0\right)+\frac{\partial}{\partial z}\left(\frac{Uy}{h}\right) \right]=0

We see that the only nonzero shear stress is σ_xy = σ_yx = µ(U/h), a constant. Note, however, that we do not know the pressure distribution because this was not given to us. We can use these results to write the stress tensor in matrix form as

σ=\left(\begin{matrix} -p&\frac{\mu U}{h}&0\\\frac{\mu U}{h}&-p&0 \\0&0&-p \end{matrix} \right)

realizing that the pressure is an unknown function of the coordinates.

To find out what the momentum equation can tell us about this flow, we will substitute the velocity field into the momentum equations (Eqs. 11.5). The flow is steady, so time derivatives of the velocity components are zero. The only body force is gravity; thus the body force per unit mass is f = (0, −g, 0). The momentum equations become

\rho \left(\frac{\partial u }{\partial t}+u\frac{\partial u }{\partial x}+v\frac{\partial u}{\partial y}+w\frac{\partial u }{\partial z} \right)=\rho f_x+\left(\frac{\partial σ_{xx}}{\partial x} +\frac{\partial σ_{yx}}{\partial y} +\frac{\partial σ_{zx}}{\partial z} \right)                        (11.5a)

\rho \left(\frac{\partial v }{\partial t}+u\frac{\partial v }{\partial x}+v\frac{\partial v}{\partial y}+w\frac{\partial v }{\partial z} \right)=\rho f_y+\left(\frac{\partial σ_{xy}}{\partial x} +\frac{\partial σ_{yy}}{\partial y} +\frac{\partial σ_{zy}}{\partial z} \right)                        (11.5b)

\rho \left(\frac{\partial w }{\partial t}+u\frac{\partial w}{\partial x}+v\frac{\partial w}{\partial y}+w\frac{\partial w}{\partial z} \right)=\rho f_z+\left(\frac{\partial σ_{xz}}{\partial x} +\frac{\partial σ_{yz}}{\partial y} +\frac{\partial σ_{zz}}{\partial z} \right)                          (11.5c)

\rho \left[0+u(0)+(0)\frac{∂u}{ ∂y} +(0)(0)\right] =\rho (0)+\left(\frac{\partial σ_{xx}}{\partial x} +\frac{\partial σ_{yx}}{\partial y} +\frac{\partial σ_{zx}}{\partial z} \right)

ρ[0 + u(0) + (0)(0) + (0)(0)] = ρ(−g) + \left(\frac{\partial σ_{xy}}{\partial x} +\frac{\partial σ_{yy}}{\partial y} +\frac{\partial σ_{zy}}{\partial z} \right)

ρ[0 + u(0) + (0)(0) + (0)(0)] = ρ(0) + \left(\frac{\partial σ_{xz}}{\partial x} +\frac{\partial σ_{yz}}{\partial y} +\frac{\partial σ_{zz}}{\partial z} \right)

We conclude that the derivatives of the various stresses obey the following equations:

\left(\frac{\partial σ_{xx}}{\partial x} +\frac{\partial σ_{yx}}{\partial y} +\frac{\partial σ_{zx}}{\partial z} \right)=0

ρ(−g)+ \left(\frac{\partial σ_{xy}}{\partial x} +\frac{\partial σ_{yy}}{\partial y} +\frac{\partial σ_{zy}}{\partial z} \right)=0

\left(\frac{\partial σ_{xz}}{\partial x} +\frac{\partial σ_{yz}}{\partial y} +\frac{\partial σ_{zz}}{\partial z} \right)=0

Inserting the stresses found earlier into these reduced momentum equations, we find

\left(\frac{\partial σ_{xx}}{\partial x} +\frac{\partial σ_{yx}}{\partial y} +\frac{\partial σ_{zx}}{\partial z} \right)=\left[\frac{\partial}{\partial x} (-p)+\frac{\partial}{\partial y}\left(\frac{\mu U}{h} \right)+\frac{\partial}{\partial z}(0) \right]=-\frac{\partial p}{\partial x}=0

ρ(−g)+ \left(\frac{\partial σ_{xy}}{\partial x} +\frac{\partial σ_{yy}}{\partial y} +\frac{\partial σ_{zy}}{\partial z} \right)=ρ(−g)+\left[\frac{\partial}{\partial x}\left(\frac{\mu U}{h} \right)+\frac{\partial}{\partial y}(-p)+\frac{\partial}{\partial z}(0) \right]
=ρg-\frac{\partial p}{\partial y}=0

\left(\frac{\partial σ_{xz}}{\partial x} +\frac{\partial σ_{yz}}{\partial y} +\frac{\partial σ_{zz}}{\partial z} \right) \left[\frac{\partial}{\partial x} (0)+\frac{\partial}{\partial y}\left(0\right)+\frac{\partial}{\partial z}(-p) \right]=-\frac{\partial p}{\partial z}=0

From the first and last of these equations we conclude that the pressure does not vary in the x or z directions. Integrating the remaining equation ∂p/∂y = − ρg, noting that the density is constant, and evaluating the constant of integration on the top plate at y = h, we find p(y) = p_h − ρg(y − h). Thus, the momentum equations have shown that the pressure distribution in this shear flow is unchanged from the hydrostatic pressure distribution that would exist in the absence of flow. Notice that both the momentum equations and the constitutive relations are needed to solve this (typical) flow problem.