Consider the ﬂow of an incompressible Newtonian ﬂuid between parallel plates with the top plate moving as shown in Figure 11.4. The velocity ﬁeld is u = U(y/h)i, where U is the speed of the moving plate and h is the gap between the plates. Find the stresses in this ﬂow. What can the momentum

Question

Consider the ﬂow of an incompressible Newtonian ﬂuid between parallel plates with the top plate moving as shown in Figure 11.4. The velocity ﬁeld is u = U(y/h)i, where U is the speed of the moving plate and h is the gap between the plates. Find the stresses in this ﬂow. What can the momentum equation tell us about this ﬂow?

Accepted Answer

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

[latex]\left(\frac{\partial ho }{\partial t}+u\frac{\partial ho }{\partial x}+v\frac{\partial ho }{\partial y}+w\frac{\partial ho }{\partial z} ight)+ ho \left(\frac{\partial u}{\partial x} +\frac{\partial v}{\partial y} +\frac{\partial w}{\partial z} ight)=0[/latex] (11.2a)

[latex]\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[/latex]

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

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

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

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

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

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

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

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

[latex]σ=\left(\begin{matrix} -p&\frac{\mu U}{h}&0\\frac{\mu U}{h}&-p&0 \0&0&-p \end{matrix} ight) [/latex]

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

[latex] ho \left(\frac{\partial u }{\partial t}+u\frac{\partial u }{\partial x}+v\frac{\partial u}{\partial y}+w\frac{\partial u }{\partial z} ight)= ho f_x+\left(\frac{\partial σ_{xx}}{\partial x} +\frac{\partial σ_{yx}}{\partial y} +\frac{\partial σ_{zx}}{\partial z} ight)[/latex] (11.5a)

[latex] ho \left(\frac{\partial v }{\partial t}+u\frac{\partial v }{\partial x}+v\frac{\partial v}{\partial y}+w\frac{\partial v }{\partial z} ight)= ho f_y+\left(\frac{\partial σ_{xy}}{\partial x} +\frac{\partial σ_{yy}}{\partial y} +\frac{\partial σ_{zy}}{\partial z} ight)[/latex] (11.5b)

[latex] ho \left(\frac{\partial w }{\partial t}+u\frac{\partial w}{\partial x}+v\frac{\partial w}{\partial y}+w\frac{\partial w}{\partial z} ight)= ho f_z+\left(\frac{\partial σ_{xz}}{\partial x} +\frac{\partial σ_{yz}}{\partial y} +\frac{\partial σ_{zz}}{\partial z} ight)[/latex] (11.5c)

[latex] ho \left[0+u(0)+(0)\frac{∂u}{ ∂y} +(0)(0) ight] = ho (0)+\left(\frac{\partial σ_{xx}}{\partial x} +\frac{\partial σ_{yx}}{\partial y} +\frac{\partial σ_{zx}}{\partial z} ight)[/latex]

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

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

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

[latex]\left(\frac{\partial σ_{xx}}{\partial x} +\frac{\partial σ_{yx}}{\partial y} +\frac{\partial σ_{zx}}{\partial z} ight)=0[/latex]

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

[latex]\left(\frac{\partial σ_{xz}}{\partial x} +\frac{\partial σ_{yz}}{\partial y} +\frac{\partial σ_{zz}}{\partial z} ight)=0[/latex]

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

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

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

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

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.

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

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