*2.4.2 Unsteady flow of a viscous fluid between parallel walls*

As a practical example, consider an unsteady flow of a viscous fluid between parallel walls. Let a viscous fluid fill the entire space between horizontal planes located at a certain distance from each other. Let the lower plane be stationary all the time, and the upper one starts to move to the right at a constant speed. We neglect the action of gravity and assume that the pressure is constant everywhere. The flow is assumed to be directed parallel to the x-axis. Then the equation of motion of a viscous fluid in dimensionless variables has the form.

$$\frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial \mathbf{y}^2} \tag{39}$$

The exact solution of the equation under the conditions:

$$
\mu(\mathbf{0}, \boldsymbol{\uprho}) = \mathbf{0}, \;\mu(t, \mathbf{0}) = \mathbf{0}, \;\mu(t, \mathbf{1}) = \mathbf{1}.
$$

looks like:

$$u = \mathcal{y} + \frac{2}{\pi} \sum\_{k=1}^{\infty} \frac{(-1)^k}{k} \sin(k\pi\mathcal{y}) \exp\left(-k^2 \pi^2 t\right) \tag{40}$$

Let us replace (39) with the difference relation:

$$\frac{u\_1(t,y) - u\_1(0,y)}{t - 0} = \frac{2}{1 - 0} \left[ \frac{u\_1(t, 1) - u\_1(t, y)}{1 - y} - \frac{u\_1(t, y) - u\_1(t, 0)}{y - 0} \right]$$

From here, considering the boundary conditions, we obtain an approximate solution in the form [15]:

$$u\_1(t, \mathbf{y}) = \frac{2t\mathbf{y}}{\mathbf{y}(\mathbf{1} - \mathbf{y}) + \mathbf{2}t} \tag{41}$$

Note that, lim*t*!<sup>∞</sup> *u t*ð Þ¼ , *<sup>y</sup>* lim*t*!<sup>∞</sup> *<sup>u</sup>*1ð Þ¼ *<sup>t</sup>*, *<sup>y</sup> <sup>y</sup>:*

There is another approach to obtaining an approximate solution to Eq. (39). We replace (39) with the following equation:

$$\frac{du\_2}{dt} = \frac{2}{1-0} \left[ \frac{u\_2(t, 1) - u\_2(t, y)}{1 - y} - \frac{u\_2(t, y) - u\_2(t, 0)}{y - 0} \right] \tag{42}$$

Considering Eq. (42) *y* as a parameter, and solving it, we get

$$u\_2(t, \mathbf{y}) = \mathbf{y} \left( \mathbf{1} - \exp\left(\frac{2t}{\mathbf{y}(\mathbf{1} - \mathbf{y})}\right) \right) \tag{43}$$

This shows that partial approximation gives the best result (**Figures 23** and **24**).

### *2.4.3 Non-stationary convection-diffusion differential equation*

Consider the equation

$$\frac{\partial \Phi}{\partial t} + \frac{\partial \Phi}{\partial \mathbf{x}} = \frac{1}{Pe} \frac{\partial^2 \Phi}{\partial \mathbf{x}^2} + f(\mathbf{x}, t), \tag{44}$$

Under appropriate boundary and initial conditions. We approximate Eq. (44) as follows

$$\begin{aligned} \frac{U(\mathbf{x},t) - U(\mathbf{x},0)}{t} + \frac{U(\mathbf{x},t) - U(W,t)}{\mathbf{x} - W} &= \\ \frac{1}{Pe} \frac{2}{(E - W)} \left( \frac{U(E,t) - U(\mathbf{x},t)}{E - \mathbf{x}} - \frac{U(\mathbf{x},t) - U(W,t)}{\mathbf{x} - W} \right) &+ f(\mathbf{x},t), \end{aligned} \tag{45}$$

(45) is an implicit difference scheme with a moving node. In this case, the convective term was approximated by the scheme against the flow, and the diffusion term, as usual, with the second order of accuracy.

The comparison obtained with the help of (45) of the approximate solution with the exact solution (44) under the conditions *<sup>U</sup>*<sup>0</sup>ð Þ¼ *<sup>x</sup> <sup>x</sup>*<sup>2</sup> <sup>þ</sup> *<sup>x</sup>*, *UW*ðÞ¼ *<sup>t</sup>* 0, *UE*ðÞ¼ *<sup>t</sup>* <sup>1</sup> <sup>þ</sup> *<sup>e</sup>*�*Pe t*, *f x*ð Þ¼ , *<sup>t</sup>* ð Þ <sup>1</sup> <sup>þ</sup> *Pe x <sup>e</sup>*�*Pe t* <sup>þ</sup> <sup>2</sup>*<sup>x</sup>* � <sup>2</sup>*=Pe* is shown in **Figure 25**. Exact solution <sup>ð</sup>*<sup>W</sup>* <sup>¼</sup> 0, *<sup>E</sup>* <sup>¼</sup> <sup>1</sup><sup>Þ</sup> *<sup>Ф</sup>*ð Þ¼ *<sup>x</sup>*, *<sup>t</sup> <sup>x</sup>*<sup>2</sup> <sup>þ</sup> *<sup>e</sup>*�*Pe tx*. In **Figure 25**, the solid curves are the exact solution, the dotted curves are the approximate solution, and the graphs correspond to the sections *x* ¼ 0, 1; *x* ¼ 0, 5; *x* ¼ 0, 8. **Figure 26** same results corresponding to *t* ¼ 1; *t* ¼ 5; *t* ¼ 10*:*

**Figures 25** and **26** show the acceptability of the approximate solution for the MNM.

It should be noted that with increasing *Pe* the discrepancy between the exact and approximate solutions increases. On **Figures 27** and **28** compare the same problem with *Pe* ¼ 2.

*Moving Node Method for Differential Equations DOI: http://dx.doi.org/10.5772/intechopen.107340*

**Figure 24.**

*Comparison of (39), (41), and (43) on the section y = 0.8. Blue line on (41) black on (43), red fine.*

**Figure 25.** *Comparison solution Pe* ¼ 0, 5*.*

Thus, the MMN makes it possible to obtain an approximate analytical solution.

#### **2.5 MNM for two-dimensional boundary value problems**

Now let us consider the application of MMN to two-dimensional boundary value problems to obtain rough approximate solutions of DE.

**Figure 26.** *Comparison solution Pe* ¼ 0, 5*.*

**Figure 27.** *Comparison solution. Pe* ¼ 2*.*

Consider a convex closed two-dimensional region (**Figure 29**). P point inside area. If P changes position inside the region, the boundary points *E*, *N*,*W*, *S* change their positions while being on the border of the region.

*Moving Node Method for Differential Equations DOI: http://dx.doi.org/10.5772/intechopen.107340*

When studying stationary processes of various physical natures (oscillations, heat conduction, diffusion, hydrodynamics, etc.), one usually leads to equations of the elliptic type. The most common equation of this type is the Poisson equation.

There are various approximate-analytical and numerical methods for the equation of mathematical physics.

Consider the two-dimensional Poisson equation in a rectangle ð Þ *x*, *y* ∈½ �� *W*, *E* ½ � *S*, *N*

$$
\Delta U(\mathbf{x}, \mathbf{y}) = f(\mathbf{x}, \mathbf{y}), \tag{46}
$$

with boundary conditions.

$$U(W, \mathbf{y}) = U\_W(\mathbf{y}), U(E, \mathbf{y}) = U\_E(\mathbf{y}), U(\mathbf{x}, \mathbb{S}) = U\_{\mathbb{S}}(\mathbf{x}), U(\mathbf{x}, N) = U\_N(\mathbf{x}).\tag{47}$$

Take an arbitrary point in the rectangle approximation of second-order partial derivatives:

$$\frac{\partial^2 U}{\partial \mathbf{x}^2} \approx \frac{2}{E - W} \left[ \frac{U\_E(\mathbf{y}) - u(\mathbf{x}, \mathbf{y})}{E - \mathbf{x}} + \frac{u(\mathbf{x}, \mathbf{y}) - U\_W(\mathbf{y})}{\mathbf{x} - W} \right],\tag{48}$$

$$\frac{\partial^2 U}{\partial \mathbf{y}^2} \approx \frac{2}{N - \mathbb{S}} \left[ \frac{U\_N(\mathbf{x}) - u(\mathbf{x}, \mathbf{y})}{N - \mathbf{y}} + \frac{u(\mathbf{x}, \mathbf{y}) - U\_S(\mathbf{x})}{\mathbf{y} - \mathbf{S}} \right]. \tag{49}$$

Substituting (46) and (49) into (46), and solving, the resulting equation with respect to *u x*ð Þ , *y* , we have

$$u(\mathbf{x},\mathbf{y}) = \frac{1}{(E-\mathbf{x})(\mathbf{x}-\mathbf{W}) + (N-\mathbf{y})(\mathbf{y}-\mathbf{S})} \cdot \left[\frac{(N-\mathbf{y})(\mathbf{y}-\mathbf{S})}{E-\mathbf{W}}((\mathbf{x}-\mathbf{W})U\_E + (E-\mathbf{x})U\_W) + \mathbf{V}\right] \tag{5}$$

$$\left[1 + \frac{(E-\mathbf{x})(\mathbf{x}-\mathbf{W})}{N-\mathbf{S}}((\mathbf{y}-\mathbf{S})U\_N + (N-\mathbf{y})U\_S)\right] + \frac{(E-\mathbf{x})(\mathbf{x}-\mathbf{W})(N-\mathbf{y})(\mathbf{y}-\mathbf{S})}{2((E-\mathbf{x})(\mathbf{x}-\mathbf{W}) + (N-\mathbf{y})(\mathbf{y}-\mathbf{S}))}f(\mathbf{x},\mathbf{y}) \tag{6}$$

This is the approximate analytical solution of the Poisson equation in a rectangle. (49) satisfies the boundary conditions. Due to the fact that (48) and (49) is an approximate relation for the approximation of the second derivatives (50) is an approximate solution. Nevertheless, (50) gives an acceptable solution to many practical problems.

Consider examples.

### *2.5.1 Test problems*


$$
\mu(\mathbf{x}, \boldsymbol{y}) = \frac{\mathbf{x}\mathbf{y}(\mathbf{y} - \mathbf{x}) + \mathbf{y}^4(\mathbf{1} - \mathbf{y}) + \mathbf{x}^4(\mathbf{x} - \mathbf{1})}{\mathbf{y}(\mathbf{y} - \mathbf{1}) + \mathbf{x}(\mathbf{x} - \mathbf{1})}.
$$

The maximum absolute difference between the exact and approximate solutions calculated by points *xi* ¼ 1 þ *ih*, *yj* ¼ *jh*, *i*, *j* ¼ 1, 2, … *n*, *h* ¼ 0, 1 is 0.048.

If we approximate the right side based on the control volume [35], the approximate solution has the form:

$$u(\mathbf{x}, \mathbf{y}) = \frac{\mathbf{x}\mathbf{y}(\mathbf{x} - \mathbf{y})(\mathbf{1} - \mathbf{3}(\mathbf{x} + \mathbf{y}) + \mathbf{3}\mathbf{x}\mathbf{y}) + 2\mathbf{y}^4(\mathbf{1} - \mathbf{y}) + 2\mathbf{x}^4(\mathbf{x} - \mathbf{1})}{2[\mathbf{y}(\mathbf{y} - \mathbf{1}) + \mathbf{x}(\mathbf{x} - \mathbf{1})]}$$

and the maximum absolute difference between the exact and approximate solutions calculated by points *xi* ¼ 1 þ *ih*, *yj* ¼ *jh*, *i*, *j* ¼ 1, 2, … *n*, *h* ¼ 0, 1 is 0.024.

#### *2.5.2 Flow in an ellipsoidal pipe*

The equation describing the one-dimensional flow in an ellipsoidal tube of a viscous fluid has the form:

$$\frac{\partial^2 U}{\partial y^2} + \frac{\partial^2 U}{\partial z^2} = -\frac{\Delta p}{\mu l} \tag{51}$$

Here u is the flow rate, μ is the flow viscosity, Δp/l (Δp/l = const) is the pressure drop. Eq. (51) is considered in the area *<sup>y</sup>*<sup>2</sup> *<sup>a</sup>*<sup>2</sup> <sup>þ</sup> *<sup>z</sup>*<sup>2</sup> *<sup>b</sup>*<sup>2</sup> ≤1 (section of an ellipsoidal pipe, **Figure 30**), and the boundary condition is the no-slip condition (U = 0).

Eq. (51) is replaced by the difference

$$\frac{2}{y\_E - y\_W} \left( \frac{U\_E - u}{y\_E - y} - \frac{u - U\_W}{y - y\_W} \right) + \frac{2}{z\_N - z\_S} \left( \frac{U\_N - u}{z\_N - z} - \frac{u - U\_S}{z - z\_S} \right) = -\frac{\Delta p}{\mu l}.$$

Hence, given that

$$\mathbf{z}\_N = b\sqrt{1 - y^2/a^2}, \ \mathbf{z}\_S = -b\sqrt{1 - y^2/a^2}, \ \ y\_E = a\sqrt{1 - z^2/b^2}, \ \ y\_W = -a\sqrt{1 - z^2/b^2}.$$

*Moving Node Method for Differential Equations DOI: http://dx.doi.org/10.5772/intechopen.107340*

**Figure 28.** *Comparison solution. Pe* ¼ 2*.*

**Figure 29.** *The convex closed two-dimensional region.*

**Figure 30.** *Ellipsoidal pipe section.*

we get

$$u = \frac{a^2b^2}{2\left(a^2 + b^2\right)} \left(1 - \frac{y^2}{a^2} - \frac{z^2}{b^2}\right) \frac{\Delta p}{\mu l}$$

coinciding with the exact solution.

#### *2.5.3 Two-dimensional temperature field in a solid*

This problem is reduced to solving an equation Δ*T* ¼ 0, with boundary conditions

$$T(\mathbf{0}, \mathbf{y}) = \mathbf{0}, T(\mathbf{1}, \mathbf{y}) = \mathbf{0}, T(\mathbf{x}, \mathbf{0}) = T\_0, T(\mathbf{x}, \mathbf{1}) = \mathbf{0}.$$

Exact solution to the problem

$$T(\mathbf{x}, \boldsymbol{\chi}) = \sum\_{n=1}^{\infty} A\_n \sin(n\pi \mathbf{x}) sh \left[ n\pi (\boldsymbol{\chi} - \mathbf{1}) \right]$$

where *<sup>А</sup><sup>n</sup>* <sup>¼</sup> <sup>2</sup>*T*<sup>0</sup> *nπ* ð Þ �<sup>1</sup> *<sup>n</sup>* ½ � �<sup>1</sup> *sh n*ð Þ *<sup>π</sup>* . Approximate solution

$$T(\mathbf{x}, \boldsymbol{\chi}) = \frac{(\mathbf{1} - \boldsymbol{\chi})\mathbf{x}(\mathbf{1} - \boldsymbol{\chi})T\_0}{\mathbf{x}(\mathbf{1} - \boldsymbol{\chi}) + \boldsymbol{\chi}(\mathbf{1} - \boldsymbol{\chi})}.$$

The maximum absolute difference between the exact and approximate solutions calculated by points *xi* ¼ *ih*, *yj* ¼ *jh*, *i*, *j* ¼ 1, 2, … *n*, *h* ¼ 0, 01 is 0.015.

Thus, the method presented here allows for obtaining solutions to Dirichlet problems. To improve the solution, the mesh refinement technique can be used.

To increase the accuracy, increase the number of moved nodes. When the number of nodes to be moved is four, we get.

*<sup>u</sup>*4ð Þ¼ *<sup>x</sup>*, *<sup>y</sup>* <sup>1</sup> � <sup>4</sup> *A* þ *B B*2 8 *x* 1�*x* 2 � �<sup>2</sup> <sup>þ</sup> *<sup>B</sup>* þ 1 � *x x* 2 � �<sup>2</sup> <sup>þ</sup> *<sup>B</sup>* ! þ *A*2 8 *y* 1�*y* 2 � �<sup>2</sup> þ *A* þ 1 � *y y* 2 � �<sup>2</sup> <sup>þ</sup> *<sup>A</sup>* 0 B@ 1 CA 2 6 4 3 7 5 8 >< >: 9 >= >; �1 � <sup>4</sup> *A* þ *B B* 2 1 2 *x* <sup>2</sup> *Bub*ð Þþ *<sup>y</sup>* 1 � *x* 2 � �<sup>2</sup> *yud* 1 þ *x* 2 � � þ ð Þ 1 � *y uc* 1 þ *x* 2 ! � � � � 1�*x* 2 � �<sup>2</sup> <sup>þ</sup> *<sup>B</sup>* þ 2 6 6 6 6 4 8 >>>>< >>>>: þ 1 4 ð Þ <sup>1</sup> � *<sup>x</sup> Bua*ð Þþ *<sup>y</sup> <sup>x</sup>* 2 � �<sup>2</sup> *yud x* 2 � � <sup>þ</sup> ð Þ <sup>1</sup> � *<sup>y</sup> uc x* 2 � � � � � � *x* 2 � �<sup>2</sup> <sup>þ</sup> *<sup>B</sup>* 3 5þ þ *A* 2 1 2 *y A* <sup>2</sup> *ud*ð Þþ *<sup>x</sup>* <sup>1</sup> � *<sup>y</sup>* 2 � �<sup>2</sup> *xub* 1 þ *y* 2 � � þ ð Þ 1 � *x ua* 1 þ *y* 2 ! � � � � 1�*y* 2 � �<sup>2</sup> þ *A* þ 2 6 6 6 6 4 þ 1 2 ð Þ <sup>1</sup> � *<sup>y</sup> Auc*ð Þþ *<sup>x</sup> <sup>y</sup>* 2 � �<sup>2</sup> *xub y* 2 � � <sup>þ</sup> ð Þ <sup>1</sup> � *<sup>x</sup> ua y* 2 � � � � � � *y* 2 � �<sup>2</sup> <sup>þ</sup> *<sup>A</sup>* 3 5 9 = ;

The maximum absolute difference between the exact and approximate solutions, calculated by points *xi* ¼ *ih*, *yj* ¼ *jh*, *i*, *j* ¼ 1, 2, … *n*, *h* ¼ 0, 1, is 0.14 according to the formula with one moving node, and when calculating with five moving nodes, it is 0.07.

#### *2.5.4 Flow in a rectangular pipe*

Eq. (51) also describes the flow of an incompressible viscous fluid in a rectangular pipe. Let us denote the height of the rectangle parallel to the axis *Oz* as 2*h*, and the base parallel to the axis *Oy* as – 2*σh*, where *σ* is any positive constant. We draw the axis through the center of the rectangle and direct it downstream.

Let us transform Eq. (51) into a dimensionless form. For the scale of lengths, we take the height, *h*, and for the scale of speeds—the value *h*<sup>2</sup> *=μ* � Δ*p=l*. We introduce the following dimensionless quantities:

$$Y = \mathfrak{y}/h, \quad Z = \mathfrak{z}/h, \quad V = U\mu l/(h^2 \Delta p)$$

Substituting into (51), we obtain

$$\frac{\partial^2 V}{\partial Y^2} + \frac{\partial^2 V}{\partial Z^2} = -\mathbf{1} \tag{52}$$

Boundary conditions for (52)

$$V(Y, -1) = 0, \ V(Y, 1) = 0, \ V(-\sigma, Z) = 0, \ V(\sigma, Z) = 0 \tag{53}$$

Eq. (52) is replaced by a difference equation and taking into account the boundary condition (53) we have

$$\frac{2}{2\sigma} \left( \frac{-V}{\sigma - Y} - \frac{V}{Y + \sigma} \right) + \frac{2}{1 + 1} \left( \frac{-V}{1 - Z} - \frac{V}{Z + 1} \right) = -1.$$

From here we determine the approximate analytical solution:

$$V = \frac{1}{2} \frac{\left(\sigma^2 - Y^2\right)\left(1 - Z^2\right)}{1 - Z^2 + \sigma^2 - y^2} \tag{54}$$

The exact solution of the problem has the form:

$$\mu = \frac{16\sigma^2}{\pi^3} \sum\_{n=0}^{\infty} \frac{\left(-1\right)^n}{\left(2n+1\right)^3} \left[1 - \frac{ch\left(\frac{2n+1}{2}\frac{\pi}{\sigma}Y\right)}{ch\left(\frac{2n+1}{2}\frac{\pi}{\sigma}\right)}\right] \cos\left(\frac{2n+1}{2}\frac{\pi}{\sigma}Z\right),$$

**Figure 31** shows a comparison of the exact and approximate solutions on the crosssection *x* ¼ 0 for *σ* ¼ 1. The maximum absolute difference between the exact and approximate solutions is 0.045.

To increase the accuracy of the approximate solution in Eq. (52), we approximate only one of the terms. For example, we approximate Eq. (52) as follows:

$$\frac{2}{2\sigma} \left( \frac{-V}{\sigma - Y} - \frac{V}{Y + \sigma} \right) + \frac{\partial^2 V}{\partial Z^2} = -1 \tag{55}$$

*Numerical Simulation – Advanced Techniques for Science and Engineering*

**Figure 31.** *Comparison of the solution on the section according to (54).*

We got an ordinary differential equation, we consider the variable *Y* in Eq. (55) as a parameter. We solve Eq. (55) with constant coefficients, considering the boundary conditions, we find an approximate solution

$$V = C\_1 \exp\left(\sqrt{k}Z\right) + C\_2 \exp\left(-\sqrt{k}Z\right) + \frac{1}{k} \,. \tag{56}$$

Here *<sup>k</sup>* <sup>¼</sup> <sup>2</sup>*=*ð Þ ð Þ *<sup>σ</sup>* � *<sup>Y</sup>* ð Þ *<sup>Y</sup>* <sup>þ</sup> *<sup>σ</sup>* ,*С*<sup>2</sup> ¼ � <sup>1</sup> *k* exp ffiffi *k* p ð Þ� exp � ffiffi *k* <sup>p</sup> ð Þ exp 2 ffiffi *k* <sup>p</sup> ð Þ� exp �<sup>2</sup> ffiffi *k* <sup>p</sup> ð Þ, *<sup>С</sup>*<sup>1</sup> ¼ �*С*<sup>2</sup> � exp 2 ffiffiffi *k* � � <sup>p</sup> � <sup>1</sup> *<sup>k</sup>* exp � ffiffiffi *k* � � <sup>p</sup> *:*

**Figure 31** shows a comparison of the exact approximate solution obtained based on (56) on the cross-section *x* ¼ 0 at *σ* ¼ 1. A comparison of **Figures 31** and **32** shows that the calculation by formula (56) gives a more accurate result. The maximum absolute difference between the exact and approximate solutions is equal to that obtained by (56) and equals 0.024. In **Figures 31** and **32** solid curves are the exact solution.

**Figure 32.** *Comparison of the solution on the section according to (56).*

*Moving Node Method for Differential Equations DOI: http://dx.doi.org/10.5772/intechopen.107340*

**Figure 33.** *Coordinate systems and the region of solution.*

*2.5.5 Flow at the inlet section of the pipe*

With appropriate simplifications, the flow of a viscous incompressible fluid in a dimensionless form is described by the following differential equation:

$$u\frac{\partial u}{\partial x} = -\frac{1}{\text{Re}}N + \frac{1}{\text{Re}}\left(\frac{\partial^2 u}{\partial \mathbf{y}^2} + \frac{\partial^2 u}{\partial \mathbf{x}^2}\right),\tag{57}$$

Here, N = �12 is the pressure drop, Re is the Reynolds number. The equations are considered in the area *D* : �0, 5<*x*<0*:*5, 0<*y* <*L:* (**Figure 33**). Boundary conditions for (57):

$$\begin{aligned} u(0, y) &= 1; \ u(L, y) = 1, 5(1 - 4y^2); \\ u(\varkappa, -0, 5) &= 0; \ u(\varkappa, 0.5) = 0. \end{aligned}$$

The convective term is linearizable

$$u\frac{\partial u}{\partial x} \approx \frac{\partial u}{\partial x}.$$

Approximating in (57) by the liquid volume ð Þ *y* þ 0, 5 *=*2 <*y*<ð Þ *y* � 0, 5 *=*2, we obtain an ordinary equation, solving which we obtain an approximate solution:

$$u = C\_1 \exp(k\_1 x) + C\_2 \exp(k\_2 x) - \frac{1 - 4y^2}{8} N,\tag{58}$$

where *<sup>k</sup>*1,2 <sup>¼</sup> Re <sup>2</sup> 1 � ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi <sup>1</sup> <sup>þ</sup> <sup>32</sup> Re <sup>2</sup> <sup>1</sup>�4*y*<sup>2</sup> ð Þ � � <sup>q</sup> *:*

For comparison, solutions (57) were also made with the numerical method.

**Figure 34** shows the velocity profiles obtained on the basis of an approximate solution. The solid curve to the section *x* ¼ 0, 1, and the pointed curve to *x* ¼ 0, 5, the dotted one corresponds to the section *x* ¼ 3*:* **Figure 35** shows a comparison of the approximate and numerical solution of Eq. (57). The solid lines correspond to the solution (58), and the dotted lines correspond to the numerical solution (velocity profiles are given for the cross-section *x* ¼ 0, 1 and *x* ¼ 0, 5).

#### **2.6 Solution of the flow problem in the combined region**

Exact solution. Let a liquid flow in a flat pipe partially filled with a porous medium. The lower part of the horizontal pipe is filled with a porous medium of height h

**Figure 34.** *Approximate solution based on (58). Velocity profiles corresponding to sections x* ¼ 0, 1; 0, 5; 3*: Re = 1, L = 5.*

(pipe height H). Considering the flow to be one-dimensional and stationary, we obtain from the Rakhmatulin equation [16, 17], we obtain

$$
\mu \frac{du}{dy} \left( f \frac{du}{dy} \right) - Ku = f \frac{dp}{d\mathfrak{x}}.\tag{59}
$$

In (59) for the parameter K, we use the Kozeny-Karman relation as adopted in porous media:

$$K = \frac{\mu \cdot f^2}{k}.\tag{60}$$

where the *<sup>k</sup>* <sup>¼</sup> *<sup>d</sup>*<sup>2</sup> *f* 3 150 1ð Þ �*<sup>f</sup>* <sup>2</sup> , permeability, d is the characteristic size of the porous medium.

Let us pass to dimensionless variables assuming *u* ¼ *uU*, *y* ¼ *yH*, *x* ¼ *xH*, *p* ¼ *ρU*<sup>2</sup> Re *p:* Then Eq. (59) in dimensionless form for *f* ¼ *const*, has the form:

$$\frac{d^2\overline{u}}{d\overline{y}^2} - A\overline{u} = \frac{d\overline{p}}{d\overline{x}}.\tag{61}$$

Here *<sup>A</sup>* <sup>¼</sup> <sup>180</sup>ð Þ *<sup>H</sup>=<sup>d</sup>* <sup>2</sup> ð Þ <sup>1</sup> � *<sup>f</sup>* <sup>2</sup> *=f* 2 . In the free zone, the one-dimensional flow satisfies the equation

$$\frac{d^2\overline{u}}{d\overline{y}^2} = \frac{d\overline{p}}{d\overline{\varpi}}.\tag{62}$$

In the future, in Eqs. (61) and (62), we release the dash above the variables.

Eq. (61) is considered when 0 <*y*<*h*0, and Eq. (62) *h*<sup>0</sup> <*y*< 1*:* Equations are solved under the following boundary conditions.

No-slip conditions for Eq. (61) to the lower walls, and for Eq. (62) to the upper walls:

$$
\mu(\mathbf{0}) = \mathbf{0}, \ \mu(\mathbf{1}) = \mathbf{0}.\tag{63}
$$

In the inner boundary region, we set the conditions for the continuity of the flow and the equality of the shear stress:

$$u(h\_0 - \mathbf{0}) = u(h\_0 + \mathbf{0}), \ \frac{du(h\_0 - \mathbf{0})}{dy} = \frac{du(h\_0 + \mathbf{0})}{dy}.\tag{64}$$

It is easy to obtain an analytical solution of (61) and (62) under the given boundary conditions. **Figure 36** shows an analytical solution. The dimensionless pressure difference is adopted *dp dx* ¼ �12, so that it corresponds to the flow without a porous layer. The dotted line corresponds to the solution obtained with a porosity of 0.3, and the dotted-dotted line is 0.5.

Numerical solution. Consider eq. (61) for the entire region and set

$$f = \begin{cases} \varepsilon \, npu \, 0 < y < h\_0 \\ 1 \, npu \, h\_0 \le y < 1 \end{cases} \tag{65}$$

In this case, Eq. (61) in the pure region takes the form (62). Thus, Eq. (61) can be used in the entire area, with porosity (65), while the interboundary conditions are satisfied automatically (in the porous layer, the porosity is taken equal to ε). For this purpose, a finite-difference approximation of Eq. (61) was compiled and calculated using the sweep method in the combined region. **Figure 37** presents the results of numerical calculations (solid curves are the analytical solution, and point data are the numerical results). This shows that it is possible to perform a thorough calculation without highlighting the interboundary condition.

Approximate analytical solution using a moving node. Using the moving node method, one can find an approximate analytical solution to the problem.

**Figure 35.** *Comparison of approximate and numerical solution. Re = 1, L = 5.*

**Figure 36.** *Exact solution: velocity distributions for different porosity values.*

**Figure 37.** *Comparison of exact and numerical results for different porosity values.*

Eqs. (61) and (62) is approximated by difference relations:

$$\frac{2}{h\_0} \left( \frac{u\_G - u}{h\_0 - \chi} - \frac{u}{\chi} \right) - Au = \frac{dp}{d\chi}.\tag{66}$$

*Moving Node Method for Differential Equations DOI: http://dx.doi.org/10.5772/intechopen.107340*

$$\frac{2}{1-h\_0} \left( \frac{-u}{1-y} - \frac{u-u\_G}{y-h\_0} \right) = \frac{dp}{dx}.\tag{67}$$

In the difference Eqs. (66) and (67) no-slip boundary conditions are used. In Eqs. (66) and (67) *uG*— the value of the unknown function on the inner boundary. To find *uG*, we use the second interboundary condition (64). We put in (66) *y* ! *h*<sup>0</sup> � 0, then we have

$$\frac{2}{h\_0} \left( \frac{du}{dy}|\_{h\_0 - 0} - \frac{u}{h\_0} \right) - Au\_G = \frac{dp}{d\mathbf{x}}.\tag{68}$$

If in (67) *y* ! *h*<sup>0</sup> þ 0 then we have:

$$\frac{2}{1-h\_0} \left( \frac{-u\_G}{1-h\_0} - \frac{du}{dy}|\_{h\_0+0} \right) = \frac{dp}{d\infty}.\tag{69}$$

Using (64), we obtain

$$u\_G = -\frac{h\_0(\mathbf{1} - h\_0)}{2 + h\_0^2(\mathbf{1} - h\_0)} \frac{dp}{d\mathbf{x}}.\tag{70}$$

Using (70) from (66) and (67) we determine the distribution of velocities in the porous

$$u = -\frac{\mathcal{Y}}{2 + A\mathbf{y}(h\_0 - \mathbf{y})} \left( h\_0 - \mathcal{Y} + \frac{2(\mathbf{1} - h\_0)}{2 + h\_0^2(\mathbf{1} - h\_0)} \right) \frac{dp}{d\mathbf{x}}.\tag{71}$$

**Figure 38.** *Solution comparison.*

and free zone

$$u = -\frac{\mathbf{1} - \mathbf{y}}{\mathbf{1} - h\_0} \left( \frac{(\mathbf{1} - h\_0)(\mathbf{y} - h\_0)}{2} + \frac{h\_0(\mathbf{1} - h\_0)}{2 + Ah\_0^2(\mathbf{1} - h\_0)} \right) \frac{dp}{d\mathbf{x}}.\tag{72}$$

**Figure 38** compares the exact and approximate solutions (the solid line is the exact solution, and the dotted line is the approximate one obtained using (71) and (72) at *A* ¼ 40000, *h*<sup>0</sup> ¼ 0, 2).
