### 物理代写|流体力学代写Fluid Mechanics代考|AMME2261

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• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

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

Since the velocity field satisfies a linear differential equation, it would appear that linearity would prevail and the principle of superposition should apply. That is, if $u_{1}=u_{1}\left(y, G_{1}\right), u_{2}=u_{2}\left(y, G_{2}\right)$ are two velocity fields under the constant pressure drops per unit length $G_{1}, G_{2}$ respectively, superposition would mean that $u\left(y, G_{1}+\right.$ $\left.G_{2}\right)=u_{1}\left(y, G_{1}\right)+u_{2}\left(y, G_{2}\right)$. However, this is false because the location of the yield surface is not a linear function of the pressure drop, and the vanishing of the shear rate at the yield surface is crucial in determining the velocity field. To be precise, let the locations of the yield surfaces under the pressure drops $G_{1}$ and $G_{2}$ be $h_{1}$ and $h_{2}$ respectively. Thus,
$$h_{1}=\frac{\tau_{y}}{G_{1}}, \quad h_{2}=\frac{\tau_{y}}{G_{2}} .$$
However, the yield surface due to the pressure drop $\left(G_{1}+G_{2}\right)$ is located at $h$, given by
$$h=\frac{\tau_{y}}{G_{1}+G_{2}} \neq h_{1}+h_{2} .$$
A different way of understanding the nonlinearity is to look at Eq. (1.4.5). Without loss of generality, let $G_{1} \geq G_{2}$, and consider $y$ such that this point lies within the yielded zone whether the pressure drop per unit length is $G_{2}, G_{1}$, or $G_{1}+G_{2}$. That is
$$\frac{\tau_{y}}{G_{1}+G_{2}}<\frac{\tau_{y}}{G_{1}} \leq \frac{\tau_{y}}{G_{2}}<y<H .$$
Given this,
\begin{aligned} u_{1}\left(y, G_{1}\right)+u_{2}\left(y, G_{2}\right) &=\frac{G_{1}+G_{2}}{2 \eta}\left(H^{2}-y^{2}\right)-2 \frac{\tau_{y}}{\eta}(H-y) \ u\left(y, G_{1}+G_{2}\right) &=\frac{G_{1}+G_{2}}{2 \eta}\left(H^{2}-y^{2}\right)-\frac{\tau_{y}}{\eta}(H-y) \ & \neq u_{1}\left(y, G_{1}\right)+u_{2}\left(y, G_{2}\right) \end{aligned}
In fact, $u(y, 2 G) \neq 2 u(y, G)$. This loss of linearity rules out the application of Laplace transform methods to solve initial-boundary value problems in the flows of Bingham fluids; for additional reasons, see Sect. 6.3.

## 物理代写|流体力学代写Fluid Mechanics代考|Non-dimensionalisation

There are two distinct combinations of length, time and velocity scales associated with a Bingham fluid. One is the intrinsic set arising from the material properties, viz. the density $\rho$, the viscosity $\eta$ of the fluid, and the yield stress $\tau_{y}$. The second is induced by a given flow and we shall return to this later.

It is simple to note that an intrinsic mass $M$, length $L$ and time $T$ scales are given by
$$M \sim \eta^{3} / \sqrt{\rho \tau_{y}^{3}}, \quad L \sim \sqrt{\eta^{2} / \rho \tau_{y}}, \quad T \sim \eta / \tau_{y} .$$
The characteristic velocity $U$ derived from the above length and time scales is:
$$U \sim \sqrt{\tau_{y} / \rho} .$$
This can lead to a physically meaningless situation whereby a fluid at rest can have a non-zero characteristic velocity. Since $\tau_{y} / \eta \sim T^{-1}$, one can multiply it by a measure of length, such as the radius of a pipe or that of the edge of a square, and obtain a characteristic velocity [1]. Similarly, one can assume that a force acts on a fluid in the absence of any motion. Hence, only the flow induced non-dimensional entities are to be preferred.

That is, when solving initial-boundary value problems, it is preferable to replace the length, velocity, the pressure and stresses, and the time scale through nondimensional quantities, which are induced by the flow under consideration. As an example, consider the flow in a channel. Using the width $H$ of the channel and a characteristic velocity $U$ related to the flow rate, say, one can render the $(x, y)$ coordinates, the velocity $u$, and time $t$ into non-dimensional forms as follows:
$$\tilde{x}=x / H, \quad \tilde{y}=y / H, \quad \tilde{u}=u / U, \quad \tilde{t}=U t / H .$$
As far as the pressure $p$, and the wall shear stress $\sigma_{w}$ are concerned, we need a characteristic stress. This is provided by $\eta U / H$, for $U / H$ has the dimension of shear rate. Thus, one obtains
$$\tilde{p}=p H / \eta U, \quad \tilde{\sigma}{w}=\sigma{w} H / \eta U$$

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

$u\left(y, G_{1}+G_{2}\right)=u_{1}\left(y, G_{1}\right)+u_{2}\left(y, G_{2}\right)$. 然而，这是错误的，因为屈服面的位置不是压降的线性函数，并 且屈服面处剪切速率的消失对于确定速度场至关重要。准确地说，让屈服面在压力下降下的位置 $G_{1}$ 和 $G_{2}$ 是 $h_{1}$ 和 $h_{2}$ 分别。因此，
$$h_{1}=\frac{\tau_{y}}{G_{1}}, \quad h_{2}=\frac{\tau_{y}}{G_{2}} .$$

$$h=\frac{\tau_{y}}{G_{1}+G_{2}} \neq h_{1}+h_{2} .$$

$$\frac{\tau_{y}}{G_{1}+G_{2}}<\frac{\tau_{y}}{G_{1}} \leq \frac{\tau_{y}}{G_{2}}<y<H .$$

$$u_{1}\left(y, G_{1}\right)+u_{2}\left(y, G_{2}\right)=\frac{G_{1}+G_{2}}{2 \eta}\left(H^{2}-y^{2}\right)-2 \frac{\tau_{y}}{\eta}(H-y) u\left(y, G_{1}+G_{2}\right)=\frac{G_{1}+G_{2}}{2 \eta}\left(H^{2}\right.$$

## 物理代写|流体力学代写Fluid Mechanics代考|Non-dimensionalisation

$$M \sim \eta^{3} / \sqrt{\rho \tau_{y}^{3}}, \quad L \sim \sqrt{\eta^{2} / \rho \tau_{y}}, \quad T \sim \eta / \tau_{y} .$$

$$U \sim \sqrt{\tau_{y} / \rho} .$$

$$\tilde{x}=x / H, \quad \tilde{y}=y / H, \quad \tilde{u}=u / U, \quad \tilde{t}=U t / H .$$

$$\tilde{p}=p H / \eta U, \quad \tilde{\sigma} w=\sigma w H / \eta U$$

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