### 物理代写|流体力学代写Fluid Mechanics代考|CHE2161

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## 物理代写|流体力学代写Fluid Mechanics代考|Sign of the Shear Stress

One can see that the shear stress is negative above the $x$-axis and is positive below it. This needs some explanation. First of all, as the pressure drop tries to move the fluid in the positive $x$-direction, the shear stresses on the two walls oppose it. See Fig. 1.1. While a more detailed description of Cauchy’s stress principle is provided in Chap. 3, at present it is sufficient to assume that the stress tensor $\mathbf{T}$ in the fluid is symmetric and two-dimensional, given in matrix form through:
$$\mathbf{T}=\left[\begin{array}{ll} T_{11} & T_{12} \ T_{21} & T_{22} \end{array}\right], \quad T_{12}=T_{21}$$
On the plane $y=H$, the external unit normal $\mathbf{n}=\mathbf{j}$ is oriented towards the positive $y$-direction. Cauchy’s stress principle says that the external traction vector $\mathbf{t}$ on this plane is given by $\mathbf{t}=\mathbf{T n}$. So,$$\mathbf{t}=\left[\begin{array}{ll} T_{11} & T_{12} \ T_{21} & T_{22} \end{array}\right]\left[\begin{array}{l} 0 \ 1 \end{array}\right]=\left[\begin{array}{l} T_{12} \ T_{22} \end{array}\right]=\left[\begin{array}{c} -\sigma_{w} \ T_{22} \end{array}\right]$$
where $\sigma_{w}$ is the magnitude of the shear stress at the wall. Since this external stress points in the negative $x$-direction, the shear stress $T_{12}<0$ in the fluid. This negative value persists till it changes from a negative to a positive value, as one moves from the plane $y=H$ to the plane $y=-H$. Now, why is the shear stress on the plane $y=-H$ positive? This is because on this plane, the external unit normal is given by $\mathbf{n}=-\mathbf{j}$. So, the external traction vector is given by $$\mathbf{t}=\left[\begin{array}{ll} T_{11} & T_{12} \ T_{21} & T_{22} \end{array}\right]\left[\begin{array}{c} 0 \ -1 \end{array}\right]=\left[\begin{array}{c} -T_{12} \ T_{22} \end{array}\right]=\left[\begin{array}{c} -\sigma_{w} \ T_{22} \end{array}\right] .$$ Obviously, the shear stress $T_{12}>0$ here.
Once again, note that the sign of the shear stress is independent of the constitutive equation and applies to all continuous media.

## 物理代写|流体力学代写Fluid Mechanics代考|Critical Pressure Drop and the Constitutive Relation

Now, let the pressure drop $G$ be increased slowly. The shear stress will grow in magnitude till the magnitude of the wall shear stress, $\sigma_{w}$, equals the yield stress, $\tau_{y}$, of the fluid. That is $\sigma_{w}=\tau_{y}$. Consider the axial force acting on the fluid over a cube of height $2 H$ in the $y$-direction, unit width in the z-direction and unit length in the $x$-direction. This force is given by $2 G H$. Opposing it are the forces on the boundaries of the channel at the top and bottom. Per unit length in the $x$-direction and unit width in the $z$-direction, these forces are given by $2 \tau_{y}$. Thus, the flow is incipient when the critical pressure drop per unit length is given by
$$G_{c}=\frac{\tau_{y}}{H} .$$
Note that the fluid does not flow till this critical value has been exceeded. If the pressure drop per unit length $G$ is increased beyond $G_{c}$, the fluid will flow with the yielding occurring at the wall at first. Assuming that the transient effects have died away and that the flow is steady, there will be a boundary layer of the Bingham fluid moving as a liquid, while away from the wall, the Bingham material will flow as a solid plug; these phenomena require some explanation.

The yield stress and the adherence condition at the wall together prevent the Bingham fluid from undergoing a deformation, i.e., shearing, till the magnitude of the shear stress at the wall, due to the applied pressure gradient, exceeds the yield stress. Elsewhere in the flow domain, the yield stress prevents the fluid from undergoing a deformation, i.e., shearing, where the magnitude of the shear stress is less than or equal to the yield stress. From Figs. 1.1 and 1.2, one sees that this situation arises in a symmetrical region around the centre of the channel. Since there is no fixed boundary at the centre, the only way the fluid can undergo zero deformation is to move as a solid plug.

## 物理代写|流体力学代写Fluid Mechanics代考|Sign of the Shear Stress

$$\mathbf{T}=\left[\begin{array}{lll} T_{11} & T_{12} T_{21} & T_{22} \end{array}\right], \quad T_{12}=T_{21}$$

## 物理代写|流体力学代写Fluid Mechanics代考|Critical Pressure Drop and the Constitutive Relation

$$G_{c}=\frac{\tau_{y}}{H} .$$

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