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Unsteady Stagnation-Point Flow of Oldroyd-B Fluid along a Stretching Sheet with Magnetic Field
DOI: 10.12677/MP.2021.111001, PDF, HTML, XML, 下载: 552  浏览: 1,341  国家自然科学基金支持

Abstract: This paper examines the influence of magnetic field on unsteady stagnation-point flow of Oldroyd-B fluid towards a stretching sheet. Suitable similarity transformations are put into use to yield the ordinary differential equation, which are dealt with double-parameter transformation expansion method with base function method (DPTEM-BF). Impacts of various physical parameters on the velocity field are explored via graphs. It is noteworthy that as relaxation time parameter enlarges, the viscous force of fluid increases, which causes larger resistance to fluid flow. The increase of retardation time parameter increases both the velocity and momentum boundary layer thickness. Larger magnetic parameter corresponds to the larger Lorentz force, which impedes fluid flow and thus slows it down.

1. 引言

2. 数学模型

Figure 1. The physical flow diagram

Oldroyd-B流体的非稳态驻点流动控制方程为：

$\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=0$ (1)

$\begin{array}{c}\frac{\partial u}{\partial t}+u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}=-{\lambda }_{1}\left(\frac{{\partial }^{2}u}{\partial {t}^{2}}+2u\frac{{\partial }^{2}u}{\partial x\partial t}+2v\frac{{\partial }^{2}u}{\partial y\partial t}+{u}^{2}\frac{{\partial }^{2}u}{\partial {x}^{2}}+2uv\frac{{\partial }^{2}u}{\partial x\partial y}+{v}^{2}\frac{{\partial }^{2}u}{\partial {y}^{2}}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\nu \frac{{\partial }^{2}u}{\partial {y}^{2}}+{\lambda }_{2}\nu \left(\frac{{\partial }^{3}u}{\partial {y}^{2}\partial t}+v\frac{{\partial }^{3}u}{\partial {y}^{3}}+u\frac{{\partial }^{3}u}{\partial x\partial {y}^{2}}-\frac{\partial u}{\partial x}\frac{{\partial }^{2}u}{\partial {y}^{2}}-\frac{\partial u}{\partial y}\frac{{\partial }^{2}v}{\partial {y}^{2}}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{+}\frac{\partial {u}_{e}}{\partial t}+{u}_{e}\frac{\partial {u}_{e}}{\partial x}+{\lambda }_{1}{u}_{e}^{2}\frac{{\partial }^{2}{u}_{e}}{\partial {x}^{2}}-\frac{\sigma {B}_{0}^{2}}{\rho }\left[u-{u}_{e}+{\lambda }_{1}\left(\frac{\partial u}{\partial t}+v\frac{\partial u}{\partial y}-\frac{\partial {u}_{e}}{\partial t}\right)\right]\end{array}$ (2)

$y=0$ 时， $u={u}_{w}\left(x,t\right)=\frac{bx}{1-at}$$v=0$ (3)

$y\to \infty$ 时， $u={u}_{e}\left(x,t\right)=\frac{cx}{1-at}$ (4)

$u=\frac{bx}{\left(1-at\right)}{f}^{\prime }\left(\eta \right),v=-\sqrt{\frac{\nu b}{\left(1-at\right)}}f\left(\eta \right),\eta =\sqrt{\frac{b}{\nu \left(1-at\right)}}y,\psi =\sqrt{\frac{\nu b}{\left(1-at\right)}}xf\left(\eta \right)$ (5)

$\begin{array}{l}S\left({f}^{\prime }+\frac{1}{2}{f}^{″}\eta \right)+{\left({f}^{\prime }\right)}^{2}-f{f}^{″}+{\beta }_{1}\left\{{S}^{2}\left(2{f}^{\prime }+\frac{7}{4}{f}^{″}\eta +\frac{1}{4}{f}^{‴}{\eta }^{2}\right)+2S\left[{\left({f}^{\prime }\right)}^{2}+\frac{1}{2}{f}^{\prime }{f}^{″}\eta \right]\\ \text{ }-S\left(3f{f}^{″}+f{f}^{‴}\eta \right)-2f{f}^{\prime }{f}^{″}+{f}^{2}{f}^{‴}\right\}-{f}^{‴}-{\beta }_{2}\left[S\left(2{f}^{‴}+\frac{1}{2}{f}^{iv}\eta \right)-f{f}^{iv}+{\left({f}^{″}\right)}^{2}\right]\\ \text{ }-AS-{A}^{2}+M\left\{\left({f}^{\prime }-A\right)+{\beta }_{1}\left[S\left({f}^{\prime }+\frac{1}{2}{f}^{″}\eta \right)-f{f}^{″}-SA\right]\right\}\end{array}$ (6)

$\eta =0$ 时， $f\left(0\right)=0$${f}^{\prime }\left(0\right)=1$ (7)

$\eta \to \infty$ 时， ${f}^{\prime }\left(\infty \right)=A$ (8)

3. DPTEM-BF方法求解

$f\left(\eta \right)={\epsilon }^{4}F\left(\xi \right)+f\left(0\right)+{f}^{\prime }\left(0\right)\eta +\frac{{f}^{″}\left(0\right)}{2!}{\eta }^{2}+\frac{{f}^{‴}\left(0\right)}{3!}{\eta }^{3}$$\xi ={\epsilon }^{-1}\eta$ (9)

${f}^{″}\left(0\right)={a}_{1}$${f}^{‴}\left(0\right)={a}_{2}$ ，其中a1，a2均为未知常数，F(ξ)为变量ξ的函数。

$F\left(0\right)=0$${F}^{\prime }\left(0\right)=0$${F}^{″}\left(0\right)=0$${F}^{‴}\left(0\right)=0$ (10)

$F\left(\xi \right)={F}_{0}\left(\xi \right)+\underset{i=1}{\overset{\infty }{\sum }}{F}_{i}\left(\xi \right){\epsilon }^{i}$ (11)

${F}_{i}\left(0\right)=0$${{F}^{\prime }}_{i}\left(0\right)=0$${F}_{i}{}^{\prime \text{​}\prime }\left(0\right)=0$${F}_{i}{}^{\prime \text{​}\prime \text{​}\prime }\left(0\right)=0$$i=0,\text{1},\text{2},\text{3,}\cdots$ (12)

$f\left(\eta \right)\approx {f}_{{N}_{1},{N}_{2}}\left(\eta \right)={f}_{0,0}\left(\eta \right)+\underset{j=3}{\overset{{N}_{1}}{\sum }}\underset{i=1}{\overset{{N}_{2}}{\sum }}{a}_{ij}{f}_{i,j}\left(\eta \right)={f}_{0,0}\left(\eta \right)+\underset{j=3}{\overset{{N}_{1}}{\sum }}\underset{i=1}{\overset{{N}_{2}}{\sum }}{a}_{ij}{\eta }^{j}{\text{e}}^{i{a}_{0}\eta }$ (13)

${f}_{0,0}\left(\eta \right)={m}_{1}+s\eta -{m}_{1}{\text{e}}^{{a}_{0}\eta }+\left(1-s+{a}_{0}{m}_{1}\right)\eta {\text{e}}^{{a}_{0}\eta }+{m}_{2}{\eta }^{2}{\text{e}}^{{a}_{0}\eta }$ (14)

${N}_{\text{1}}={N}_{\text{2}}=\text{3}$ ，其中m1，m2，a0，a13，a23，a33为假设的未知数，最后使用牛顿迭代法便可以计算出每个未知参数的值，将其带入函数f(η)的表达式，就可以得到原微分方程的近似解析解。

4. 结果与讨论

Table 1. Comparison of − f ″ ( 0 ) with different values of S while β 1 = β 2 = A = M = 0

Figure 2. Influence of β1 on the velocity field ${f}^{\prime }\left( \eta \right)$

Figure 3. Influence of β2 on the velocity field ${f}^{\prime }\left( \eta \right)$

Figure 4. Influence of S on the velocity field ${f}^{\prime }\left( \eta \right)$

Figure 5. Influence of A on the velocity field ${f}^{\prime }\left( \eta \right)$

Figure 6. Influence of M on the velocity field ${f}^{\prime }\left( \eta \right)$

5. 总结