Gravity Driven Flow Past the Bottom with Small Waviness

Abstract

We propose an introductory study of gravity driven Stokesian flow past the wavy bottom, based on Adler’s et al. papers. In examples the waviness is described by a sinus function and its amplitude is small, up to O(ε ²). A correction to Hagen-Poiseuille’s type free-flow solution is found. A contribution of capillary surface tension is discussed.

Appendices

Appendix 1: Natural Units for the Problem

Let the module of velocity of the main stream can be characterized by the quantity u, and the geometrical properties of the flow can be denoted by l. Then any flow is specified by three parameters u, l and the viscosity η. The only one dimensionless quantity which can be formed from the above three is called Reynolds’ number

$$\displaystyle \begin{aligned} R \equiv \frac{\rho}{\eta} \, u l{} \end{aligned} $$

(A.1)

Navier-Stokes’ equation is considerably simplified in the case of flow with small Reynolds number R. For steady flow of an incompressible fluid with constant viscosity η this equation reads

$$\displaystyle \begin{aligned} (\boldsymbol{v} \cdot \nabla) \boldsymbol{v} = \, - \, \frac{1}{\rho} \, \nabla p \, + \, \boldsymbol{b} \, + \, \frac{\eta}{\rho} \, \Delta \boldsymbol{v} \end{aligned} $$

(A.2)

The term (v ⋅∇)v is of the order of magnitude of (u ²/l), and the quantity (η/ρ) Δv is of the order of ηu/(ρl ²). The ratio of two is just Reynolds’ number R. Hence the term (v ⋅∇)v may be neglected for small R. Then the equation of motion reduces to a linear equation

$$\displaystyle \begin{aligned} \eta \, \Delta \boldsymbol{v} \, - \, \nabla p + \, \boldsymbol{b} \, = 0 \end{aligned} $$

(A.3)

known as Stokes’ equation, which together with the equation of continuity (1) and appropriate boundary conditions completely determines the motion, [7].

We notice yet that the smallness of R is not necessary to linearize Stokes’ equation in the case of laminar flow with the velocity v = (v, 0, 0), if v does not depend on x ₁. In this case the nonlinear term at the left hand side of Eq. (A.2) identically vanishes.

In analogy with Reynolds’ number R, we take as a natural for this problem the depth h of the fluid as the unit of length x̆. As the unit of velocity v̆ we accept three times repeated the mean velocity 3 ⋅ v ^mean. The natural unit of pressure p̆ is the x component of the hydrostatic pressure at the bottom, what assures the unit value of the x component of the body force. Thus, cf. Fig. 1,

$$\displaystyle \begin{aligned} \begin{array}{l} \displaystyle \breve{x} = h \\ \displaystyle \breve{v} = 3 \cdot v^{\mathrm{mean}} \, = \,\frac{\rho}{\eta} \, h^2 \,g \,\sin \alpha \\ \displaystyle \breve{p} = \,\frac{\eta}{h} \,3 \, v^{\mathrm{mean}} = \rho h g \sin \alpha \end{array} \end{aligned} $$

(A.4)

Consequently, the unit of time is

$$\displaystyle \begin{aligned} \breve{t} = \frac{\breve{x}}{\breve{v}} \, = \, \frac{h}{3 \cdot v^{\mathrm{mean}}} \, = \, \frac{\eta}{\rho h g \sin \alpha} \end{aligned} $$

(A.5)

Then new variables $\widetilde x, \widetilde v$ and $\widetilde p$ are introduced:

$$\displaystyle \begin{aligned} \widetilde x \, \equiv \, \frac{x}{\breve{x}}, \quad \widetilde v \, \equiv \, \frac{v}{\breve{v}} \quad \mathrm{and} \quad \widetilde p \, \equiv \, \frac{p}{\breve{p}} \end{aligned} $$

(A.6)

In the new variables the zeroth approximation of Stokes’ equation has the form

$$\displaystyle \begin{aligned} \frac{\partial^2 \widetilde v}{\partial \widetilde y^2} - \frac{\partial \widetilde p}{\partial \widetilde x} + 1 = 0 \qquad \mathrm{and} \qquad \frac{\partial \widetilde p}{\partial \widetilde y} + \mathrm{ctg} \alpha = 0 \end{aligned}$$

cf. Eqs. (1.5).

We take into consideration a typical laboratory experiment with h = 10 cm. If we approximate the gravity acceleration g as 1000 cm/s², then for the water with the density ρ = 1 g/cm³ the relation between the natural units and the SI units is

$$\displaystyle \begin{aligned} \begin{array}{l} \displaystyle \breve{x} = 10\,\mathrm{cm} \\ \displaystyle \breve{v} = \frac{1 \,\frac{\mathrm{g}}{\mathrm{cm}^3}}{0.01 \, \frac{\mathrm{g}}{\text{cm s}}} \cdot \frac{10^2\,\mathrm{cm}^2}{3} \cdot 1000 \frac{\mathrm{cm}}{\mathrm{s}^2} \cdot \sin \alpha \, = \, \frac{10}{3} \cdot 10^6 \cdot \frac{\mathrm{cm}}{\mathrm{s}} \cdot \sin \alpha \\ \displaystyle \breve{p} = 1 \frac{\mathrm{g}}{\mathrm{cm}^3} \cdot 10\,\mathrm{cm} \cdot 1000 \frac{\mathrm{cm}}{\mathrm{s}^2} \cdot \sin \alpha = 10^4 \frac{\text{g cm}}{\mathrm{s}^2} \cdot \frac{1}{\mathrm{cm}^2} \cdot \sin \alpha \\ \displaystyle \breve{t} = \, \frac{ 0.01 \cdot \frac{\mathrm{g}}{\text{cm s}}} { 1 \frac{\mathrm{g}}{\mathrm{cm}^3} \cdot 10\,\mathrm{cm} \cdot 1000 \frac{\text{g cm}}{\mathrm{s}^2} \cdot \sin \alpha } = \, \frac{ 10^{- 6} \mathrm{s}}{\sin \alpha} \end{array} \end{aligned} $$

(A.7)

In peculiar, we have also

$$\displaystyle \begin{aligned} 1 \cdot \frac{\text{g cm}}{\mathrm{s}^2} \cdot \frac{1}{\mathrm{cm}^2} \, = \, \breve{p} \cdot \frac{10^{- 4}}{\sin \alpha} \end{aligned}$$

and

$$\displaystyle \begin{aligned} 1 s \, = \, \breve{t} \cdot 10^6 \cdot \sin \alpha \end{aligned}$$

Then, the viscosity of the water in these new units is equal to one

$$\displaystyle \begin{aligned} \eta^{\mathrm{H}_2\mathrm{O}} \, = \, 0.01 \cdot \frac{\mathrm{g}}{\text{cm s}} \, = \, 0.01 \cdot \frac{1}{\mathrm{cm}^{2}} \,\frac{\text{g cm}}{\mathrm{s}^2} \cdot s \, = \, 1 \cdot \breve{p} \cdot \breve{t} {} \end{aligned} $$

(A.8)

In our example we accept

$$\displaystyle \begin{aligned} \alpha = \frac{100\,\mathrm{m}}{1000\,\mathrm{km}} = 10^{- 4} \end{aligned} $$

(A.9)

what is an order of the mean slope of the great world rivers profile, cf. the Appendix 3. Then

$$\displaystyle \begin{aligned} \begin{array}{l} \displaystyle \breve{x} = 10\,\mathrm{cm} \\ \displaystyle \breve{v} = \,10 \cdot 10^2 \cdot \frac{\mathrm{cm}}{\mathrm{s}} \/ = \, 10 \cdot \frac{\mathrm{m}}{\mathrm{s}} \\ \displaystyle \breve{p} = \, 1 \cdot \frac{\text{g cm}}{\mathrm{s}^2} \cdot \frac{1}{\mathrm{cm}^2} \\ \displaystyle \breve{t} = \, 0.01 \, \mathrm{s} \end{array} \end{aligned} $$

(A.10)

In these units, we find

$$\displaystyle \begin{aligned} \breve{x} \cdot \frac{\breve{p}}{\breve{v}} = \frac{1}{100} \cdot \frac{\mathrm{g}}{\text{cm s}} \end{aligned}$$

or

$$\displaystyle \begin{aligned} 1\,\mathrm{P} \, \equiv \, 1 \cdot \frac{\mathrm{g}}{\text{cm s}} \, = \, 100 \cdot \breve{x} \cdot \frac{\breve{p}}{\breve{v}}\end{aligned} $$

It is conspicuous for this angle of slope the smallness of the pressure unit. Note that p̆ = 1 dyn/cm² = 0.1 Pa ≈ 10⁻⁶ atm.

As it was mentioned above, the viscosity of water at 20^∘C is almost exactly 1 cP, and equals 1 in our natural units, cf. Eq. (A.8).

Appendix 2: Elaboration of the Example

1.1 General Solution

By equation (2.7) for m = 1 we have

$$\displaystyle \begin{aligned} \begin{array}{l} \displaystyle u_1(x, y)= \alpha_1(y) \, (A \, \cos x \, - \, B \, \sin x) \\ \displaystyle v_1(x, y)= \beta_1(y) \, (A \, \sin x \, + \, B \, \cos x) \\ p_1(x, y)= \gamma_1(y) \, (A \, \sin x \, + \, B \, \cos x) \end{array}{} \end{aligned} $$

(A.11)

where α ₁, β ₁, and γ ₁ are given by Eqs. (3.8)–(3.10) and the superscripts 1 were omitted. After Eqs. (3.7) we find

$$\displaystyle \begin{aligned} \begin{array}{l} \displaystyle u_{11} \, = \, -\left\{ [C_1 + (C_1 y + C_2) ] \, \mathrm{e}^{y} \, + \, [C_3 - (C_3 y + C_4 ] \,\mathrm{e}^{- y} \right\}B \\ \displaystyle u_{12} \, = \,\left\{ [C_1 + (C_1 y + C_2) ] \, \mathrm{e}^{y} \, + \, [C_3 - (C_3 y + C_4 ] \,\mathrm{e}^{- y} \right\}A \\ \displaystyle v_{11}=\left\{ (C_1 y + C_2)\, \mathrm{e}^{y} \, + \, (C_3 y + C_4)\, \mathrm{e}^{- y} \right\}A \\ v_{12}=\left\{ (C_1 y + C_2)\, \mathrm{e}^{y} \, + \, (C_3 y + C_4)\, \mathrm{e}^{- y} \right\}B \\ p_{11}= (2 C_1 \, \mathrm{e}^{y} \, + \, 2 C_3 \,\mathrm{e}^{- y})A,\qquad p_{12}= (2 C_1 \, \mathrm{e}^{y} \, + \, 2 C_3 \,\mathrm{e}^{- y})B \end{array}{} \end{aligned} $$

(A.12)

1.2 Boundary Conditions for the Example

If the expansions (2.4) are limited up to O(ε ²), only the terms with m = 0, 1 must be left, and the boundary conditions (2.5) reduce to the following ones

$$\displaystyle \begin{aligned} \begin{array}{l} \displaystyle u(x, \varepsilon a \cos x) \doteq u_0(x, 0) + \varepsilon \left( u_1(x, 0) + a \cos x \cdot \left. \frac{\partial u_0(x,y)}{\partial y}\right|{}_{y=0} \right) \\ \displaystyle v(x, \varepsilon a \cos x) \doteq v_0(x, 0) + \varepsilon \left(v_1(x, 0) + a \cos x \cdot \left. \frac{\partial v_0(x,y)}{\partial y}\right|{}_{y=0} \right) \\ \displaystyle p(x, \varepsilon a \cos x) \doteq p_0(x, 0) + \varepsilon \left(p_1(x, 0) + a \cos x \cdot \left. \frac{\partial p_0(x,y)}{\partial y}\right|{}_{y=0} \right) \end{array}{} \end{aligned} $$

(A.13)

where ≐ means the asymptotic equality with the accuracy O(ε ²). This symbol was applied in [2].

Next, we submit Fourier’s series (2.7) into Maclaurin’s expansions (2.5), taken at the bottom boundary up to O(ε ²) approximation, and receive

$$\displaystyle \begin{aligned} \begin{array}{l} \displaystyle u(x, \varepsilon \,a \, \cos x) \, \doteq \, u_0(x,0) \, + \, \\ \displaystyle \varepsilon \, \left( \left. \alpha_1(y)\right|{}_{y=0} \, \frac{\partial}{\partial x} \, (A \, \sin x \, + \, B \, \cos x) \right) \\ \displaystyle v(x, \varepsilon \,a \, \cos x) \, \doteq \, v_0(x,0) \, + \, \\ \displaystyle \varepsilon \, \left( \left. \beta_1^{(1)}(y)\right|{}_{y=0} \, (A \, \sin x \, + \, B \, \cos x) \right) \\ \displaystyle p(x, \varepsilon \,a \, \cos x) \, \doteq \, p_0(x,0) \, + \\ \displaystyle \varepsilon \, \left( \left. \gamma_1^{(1)}(y)\right|{}_{y=0} \, (A \, \sin x \, + \, B \, \cos x) \right) \end{array}{} \end{aligned} $$

(A.14)

The subscripts and coefficients at x are put s = 1. Above, u ₀(x, 0) and v ₀(x, 0) are known and equal to zero, while p ₀(x, 0) must be found from a boundary problem at the free surface of the flowing stream.

From the solutions (2.12)–(2.13) we get

$$\displaystyle \begin{aligned} \begin{array}{ll} \displaystyle \left.\alpha_1^{(1)}(y)\right|{}_{y=0} & = \, C_1 + \, C_2 \, + \, C_3 - \, C_4 \\ \displaystyle \left.\beta_1^{(1)}(y)\right|{}_{y=0} & = \, C_2 \, + \, C_4 \\ \displaystyle \left. \frac{\partial}{\partial y} \, \beta_1^{(1)}(y)\right|{}_{y=0} & = \, C_1 + C_2 \, + \, C_3 - C_4 \\ \displaystyle \left.\gamma_1^{(1)}(y)\right|{}_{y=0} & = \, 2 \,( C_1 \, + \, C_3 ) \\ \displaystyle \left. \frac{\partial}{\partial y} \, \gamma_1^{(1)}(y)\right|{}_{y=0} & = \,2 \, C_1 \, - \, 2 \, C_3 \end{array}{} \end{aligned} $$

(A.15)

Here also the superscripts at the coefficients C ₁, C ₂, C ₃ and C ₄ were omitted. Substituting (A.13) into (A.12) gives

$$\displaystyle \begin{aligned} \begin{array}{l} \displaystyle u(x, \varepsilon \,a \, \cos x) \, \doteq \, \\ \displaystyle \varepsilon \, \left\{ (C_1 + \, C_2 \, + \, C_3 - \, C_4 ) (A \, \cos x - B \, \sin x) \, + \, \cos x \right\} \\ \displaystyle v(x, \varepsilon \,a \, \cos x) \, \doteq \displaystyle \varepsilon \, (C_2 + C_4) \, (A \, \sin x \, + \, B \, \cos x) \\ \displaystyle p(x, \varepsilon \,a \, \cos x) \, \doteq \, p_0(x,0) \, + \varepsilon \, 2 \,( C_1 \, + \, C_3 ) \, (A \, \sin x \, + \, B \, \cos x) \end{array}{} \end{aligned} $$

(A.16)

1.3 Finding Constants

At the bottom

$$\displaystyle \begin{aligned} u(x, \varepsilon \,a \, \cos x) \, = \, 0 \end{aligned} $$

(A.17)

and

$$\displaystyle \begin{aligned} v(x, \varepsilon \,a \, \cos x) \,= \, 0 \end{aligned} $$

(A.18)

where both $u(x, \varepsilon \,a \, \cos x)$ and $v(x, \varepsilon \,a \, \cos x)$ are given by (A.16). Also, in good approximation (if surface undulations were neglected)

$$\displaystyle \begin{aligned} p(x, \varepsilon \,a \, \cos x) \,= \, p_{\mathrm{A}} \, + \, \mathrm{ctg} \, \alpha \end{aligned} $$

(A.19)

These conditions should be satisfied separately at each power of ε. We apply the integral relations

$$\displaystyle \begin{aligned} \int_{- \pi}^\pi \cos k x \sin l x \mathrm{d} x \, = \,0 \quad \mathrm{and} \quad \int_{- \pi}^\pi \cos k x \cos l x \mathrm{d} x \, = \, \pi \, \delta_{k l} \end{aligned}$$

and successively obtain.

From Eqs. (A.14) and (A.15)

$$\displaystyle \begin{aligned} (C_1 + \, C_2 \, + \, C_3 - \, C_4 ) \, A \, = \, - \, 1 \end{aligned} $$

(A.20)

and

$$\displaystyle \begin{aligned} B = 0 \end{aligned} $$

(A.21)

From Eqs. (A.14) and (A.16)

$$\displaystyle \begin{aligned} C_2 + C_4 = 0{} \end{aligned} $$

(A.22)

After the set (3.2) and Eq. (1.9) we have

$$\displaystyle \begin{aligned} p=p_{\mathrm{A}} +(1-y)\,\mathrm{ctg}\alpha+2\varepsilon(C_1\mathrm{e}^y+C_3\mathrm{e}^{-y})A\sin x{} \end{aligned} $$

(A.23)

To assure the value p = p _A at y = 1 we should have

$$\displaystyle \begin{aligned} C_1\mathrm{e}+C_3\mathrm{e}^{-1}=0{} \end{aligned} $$

(A.24)

If we assume that transversal velocity component v vanishes at the stream surface (y = 1)

$$\displaystyle \begin{aligned} (C_1+C_2)\mathrm{e}+(C_3+C_4)\mathrm{e}^{-1}=0{} \end{aligned} $$

(A.25)

Comparing (A.22), (A.24) and (A.25) we observed that

$$\displaystyle \begin{aligned} C_2=0=C_4 \end{aligned} $$

(A.26)

Moreover,

$$\displaystyle \begin{aligned} AC_1=\frac{\mbox{e}^2}{\mbox{e}^2-1} \qquad {\mathrm{{and}}}\quad AC_3=\frac{1}{\mbox{e}^2-1}\end{aligned} $$

(A.27)

Appendix 3: Slopes of the River Beds: Typical Examples

What concerns the lower Nile river, the city Aswan is elevated 194 m above the sea level, and from this town to the Mediterranean sea the river flows about 1200 km yet.

What concerns the lower Amazon river, the city Manaus is on 92 m elevation and 1500 km away from the Atlantic ocean. In approximation the mean slope of the Amazon profile is

$$\displaystyle \begin{aligned} \alpha \approx \frac{60\,\mathrm{m}}{ 1500\,\mathrm{km}} \, = \,0.4 \times 10^{- 4}\end{aligned} $$

The similar numbers are obtained for other rivers such as Volga, Oka, Mississippi and Wisła (Vistula) below Warsaw. The mean slope of the Wisłok river below Rzeszów until the mouth in Dȩbno is about 14 m / 60 km ≈ 2 × 10⁻⁴.

For comparison, the slope of the bridge Pont du Gard, which descends 2.5 cm in 456 m has the approximate value of the gradient 0.5 × 10⁻⁴. The bridge is part of the Nîmes aqueduct. The aqueduct was built probably around the reign of the emperor Claudius (41–54 AD). It required constant maintenance by circitores, workers responsible for the aqueduct’s upkeep, who crawled along the conduit scrubbing the walls clean and removing any vegetation, cf. [11].

Appendix 4: Estimation of the Upper Waviness

After the formula for differentiation of a definite integral whose limits are functions of the differential variable, Eq. (4.1) can be written in the form

$$\displaystyle \begin{aligned} I_1 + I_2 + I_3 = 0\end{aligned} $$

(A.28)

Here

$$\displaystyle \begin{aligned} \begin{array}{l} \displaystyle I_1 = \, - \, \int_{\varepsilon \cos x}^{1 + \zeta(x)} \{ - \mathrm{e}^2 \mathrm{e}^{- y} \, + \, \mathrm{e}^y \, + \, \mathrm{e}^2 \, \mathrm{e}^{- y} y \, + \, \mathrm{e}^ y y\} \, \sin x \,\mathrm{d} y \\ \displaystyle I_2 = \frac{\partial \zeta}{\partial x} \, \{ - \mathrm{e}^2 \mathrm{e}^{- (1 + \zeta)} \, + \, \mathrm{e}^{1 + \zeta} \, + \, \mathrm{e}^2 \, \mathrm{e}^{- (1 + \zeta)} (1 + \zeta) \, + \, \mathrm{e}^{1 + \zeta} (1 + \zeta) \} \, \cos x \\ \displaystyle I_3 = \, - \, \varepsilon \sin x \, \{ - \mathrm{e}^2 \mathrm{e}^{- \varepsilon \cos x} \, + \, \mathrm{e}^{\varepsilon \cos x} \, + \, \mathrm{e}^2 \, \mathrm{e}^{- (\varepsilon \cos x)} (\varepsilon \cos x) \\ +\, \displaystyle \mathrm{e}^{\varepsilon \cos x}(\varepsilon \cos x) \} \, \cos x \end{array} \end{aligned}$$

After integration we get

$$\displaystyle \begin{aligned} I_1 = \, - \,\left. \left\{ \mathrm{e}^2 \mathrm{e}^{- y} \, + \, \mathrm{e}^y \, -\, \mathrm{e}^2 \, \mathrm{e}^{- y} (y +1) \, + \, \mathrm{e}^ y(y -1) \right\}\right|{}_{\varepsilon \cos x}^{1 + \zeta(x)} \, \cdot \, \sin x\end{aligned} $$

or

$$\displaystyle \begin{aligned} I_1 = \, - \, \left\{ \mathrm{e} \cdot 2 \, \zeta \, ( 1 + \zeta) \, + \, [ \mathrm{e}^2 - 1 - ( \mathrm{e}^2 + 1) \, \varepsilon \cos x ] \, \varepsilon \cos x \right\} \cdot \,\sin x\end{aligned} $$

Keeping the terms of O(ε) and O(ζ) only we have

$$\displaystyle \begin{aligned} I_1 = \, - \, \left\{ 2 \mathrm{e} \zeta \, \, + \, ( \mathrm{e}^2 - 1 ) \, \varepsilon \cos x \right\} \cdot \,\sin x \end{aligned}$$

Within this accuracy we get

$$\displaystyle \begin{aligned} I_2 = \frac{\mathrm{d} \zeta}{\mathrm{d} x} \cdot \mathrm{e} \cdot (3 + 2 \zeta) \, \cos x \end{aligned}$$

and

$$\displaystyle \begin{aligned} I_3 = \, - \, \varepsilon \sin x \cdot (1 - \mathrm{e}^2 ) \, \cos x\end{aligned} $$

If we leave out the term I ₁, we have equality I ₂ + I ₃ = 0, which, after integration gives $3 \zeta + \zeta ^2 = \varepsilon \cdot \left ( \mathrm {e} \, - \, {1}/{\mathrm {e}} \right ) \cos x,$ and after linearization

$$\displaystyle \begin{aligned} \zeta = \frac{ \varepsilon}{3} \cdot \left( \mathrm{e} \, - \, \frac{1}{\mathrm{e}} \right) \cos x \end{aligned}$$

where e ≈ 2.71828. In this case O(ζ)≃ O(ε).

Appendix 5: Young-Laplace’s Equation

Young-Laplace’s equation is a statement of normal stress balance for static fluids meeting at an interface

$$\displaystyle \begin{aligned}p_1 - p_2 = \sigma \left({\frac {1}{R_{1}}}+{\frac {1}{R_{2}}}\right) \end{aligned}$$

where p ₁ − p ₂ is the pressure difference across the fluid interface, σ is the surface tension, and R ₁ and R ₂ are the principal radii of the interface surface curvature. In our case p ₁ = p is the pressure inside of the liquid, and p ₂ = p _A is the atmospheric pressure. Young-Laplace’s equation can be written as

$$\displaystyle \begin{aligned} \frac{p_1}{\breve p } - \frac{ p_2}{\breve p} = \frac{\sigma}{ {\breve p} h} \, \left({\frac {h}{R_{1}}}+{\frac {h}{R_{2}}}\right) \end{aligned}$$

The value of surface tension of water against air σ = 72.5 g/s², and in our natural units, cf. the Appendix 1,

$$\displaystyle \begin{aligned} \widetilde \sigma = \frac{\sigma}{ {\breve p} h} = \frac{72.5}{10^5 \cdot \sin \alpha} \approx 7.25 \end{aligned}$$

For the droplet of radius 1 mm, p − p _A = 0.0014 atm, and for the droplet of radius 0.1 mm, p − p _A = 0.0144 atm, with p meaning the pressure inside of the liquid, and p _A is the atmospheric pressure.