Modelo de datos físicos

Inviscid Incompressible Flows

3.1 Integrals of Motion

The motion of inviscid incompressible fluid flows is governed by the Euler equations

∂V

∂t + V· ∇

V= f−1

ρ∇p (momentum equation), (3.1.1a) divV = 0 (continuity equation), (3.1.1b) which are obtained from the Navier–Stokes equations (1.7.5) by setting the viscosity coefficient ν to zero.¹

An alternative form of writing the momentum equation arises from using the Lamb formula (see Problem 2 in Exercises 4)

V· ∇

V= ω× V + ∇

V² 2



. (3.1.2)

Here ω = curl V and V is the modulus of the velocity vector V. Substitution of (3.1.2) into (3.1.1) leads to the so-called Gromeko–Lamb form of the momentum equation:

∂V

∂t + ω× V + ∇

V² 2



= f−1

ρ∇p. (3.1.3)

In further analysis, we will often use simplifications arising from the assumption that the body force f is potential.

Definition 3.1 A body force field f is termed potential (or conservative) if there exists a scalar function U such that

f =−∇U. (3.1.4)

The gravitational force clearly belongs to this class. If we introduce a Cartesian co-ordinate system with x and y lying in the horizontal plane and z directed vertically upwards, then equation (3.1.4) may be written in coordinate-decomposition form as

0 =−∂U

∂x, 0 =−∂U

∂y, −g = −∂U

∂z,

1Of course, the Euler equations were deduced significantly earlier than the Navier–Stokes equa-tions. In fact, this was done by Euler as early as in 1755.

leading to

U = gz + C, (3.1.5)

where C is an arbitrary constant.

We shall start our analysis of inviscid incompressible fluid motion by introducing a number of important integrals of the momentum equation. The first of these is the Bernoulli integral, which allows one to find the pressure distribution over a flow field with known velocity distribution.

3.1.1 Bernoulli integral

Theorem 3.1 Let the fluid flow be steady and the body force potential. Then along any streamline, the Bernoulli integral

V² 2 +p

ρ+ U = H, (3.1.6)

holds, where H remains constant along each streamline but might be different for dif-ferent streamlines.

Proof Since the flow is steady (∂V/∂t = 0) and the body force is potential (f =

−∇U), we can write the momentum equation (3.1.3) as

ω× V + ∇

V² 2 +p

ρ+ U



= 0. (3.1.7)

Let line L be a streamline. The unit vector τ tangent to L is parallel to the velocity vector V, and therefore is perpendicular to ω× V, which ensures that the scalar product of τ and ω× V is zero. Hence, multiplying equation (3.1.7) by τ , we have

τ· ∇

V² 2 +p

ρ+ U



= 0.

This proves that the expression in parentheses really stays constant alongL. ✷ 3.1.2 Kelvin’s Circulation Theorem

Kelvin’s (1869) Circulation Theorem concerns the circulation of the velocity vector along a fluid contour. To introduce the notion of a fluid contour, let us consider a closed contour C0 that is fixed with respect to the coordinate system Oxyz as shown in Figure 3.1. Let us then ‘mark’ all the fluid particles that happen to be on the contour C0 at an initial instant t0. We shall follow these particles as they move in space with increasing time t. Considered together, they constitute a fluid contour C that, owing to the continuity of the flow field, will remain closed for t > t0.

The circulation along this contour is given by the integral (1.4.18):

Γ(t) = I

C

V· dr. (3.1.8)

3.1. Integrals of Motion 131

Fig. 3.1: Fluid contour.

Let us differentiate this integral with respect to time t. To perform the differentiation, it is convenient to introduce the Lagrangian position vector

r= r(t, r0). (3.1.9)

Taking into account that, at t = t0, all the fluid particles in which we are interested lie on C0, we can ‘mark’ them using the distance s measured along C0from, say, point M (see Figure 3.1). Correspondingly, we will write (3.1.9) as

r= r(t, s).

At each instant t, the variation of the position vector along C is given by dr = ∂r

∂sds. (3.1.10)

Using (3.1.10) in (3.1.8) yields

Γ(t) = ZL

0

 V·∂r

∂s



ds, (3.1.11)

where V = V

t, r(t, s)

and L is the entire length of the initial contour C0. Differen-tiation of (3.1.11) results in

dΓ dt =

ZL

0

DV Dt ·∂r

∂s

 ds +

ZL

0



V· ∂²r

∂s∂t



ds. (3.1.12)

Taking into account that

V= ∂r

∂t, M

s C

C0

x y

z

O

we can calculate the second integral in (3.1.12) as

and, since s = L and s = 0 represent the same point on contour C, we can conclude that this integral is zero.

Turning to the first integral in (3.1.12), we shall assume that the body force f is potential. In this case, using (3.1.4) in (3.1.1a) yields

DV

Observing that the point of integration, after making a full circle along C, returns to its original position, we can conclude that the integral (3.1.14) is also zero.

This proves the following statement, known as Kelvin’s Circulation Theorem.

Theorem 3.2 In an inviscid incompressible fluid flow, the circulation Γ along any closed fluid contour does not change with time; i.e.

dΓ dt = 0, provided that the body force is potential.²

This theorem plays an important role in fluid dynamics; it lays a foundation for potential flow theory. As we shall see, many inviscid flows may be treated as potential, in which case the Euler equations (3.1.1) reduce to the simpler Laplace equation. In fact, it is primarily because of this simplification that significant progress in inviscid flow theory has been achieved.

Before turning to a description of potential flow theory, we shall give here two examples that demonstrate how the Kelvin theorem may be used. The first concerns fluid flows that start from rest. If at an initial instant t0the flow velocity V = 0, then the circulation along any closed contour C0 is zero. According to Kelvin’s Circulation Theorem, it will remain zero after the fluid is put into motion, and at any time t > t0,

I

C

V· dr = 0, (3.1.15)

where C is the fluid contour originating from C0.

2In particular, this theorem may be used when there is no body force at all.

3.1. Integrals of Motion 133

σ C

n

x y

z

O

Fig. 3.2: Surface σ to which Stokes’s theorem is applied.

Let us now consider a surface σ based on C as shown in Figure 3.2. We denote the unit vector normal to σ by n and the area of a small element of σ by ds. Then, using Stokes’s theorem, we have

I

C

V· dr

= ZZ

σ

curl V· n
ds =

ZZ

σ

ω· n
ds,

and it follows that ZZ

σ

ω· n
ds = 0.

Taking into account the arbitrariness of C, we can conclude that any flow of this kind is irrotational, i.e. free of vorticity:

ω= curl V = 0. (3.1.16)

In the second example, we consider a rigid body placed into a uniform stream with free-stream velocity V∞, as sketched in Figure 3.3. Assuming that there are no

C0

C

V^∞

body

Fig. 3.3: Fluid contour motion in flow past a rigid body.

recirculating regions (these may form if the flow separates from the body surface), one can claim that any closed contour C in the flow field ‘originates’ from the corresponding contour C0 in the oncoming flow. Since on C0the fluid velocity is constant, V = V∞, the circulation along C0 is zero. According to Kelvin’s Circulation Theorem, it will remain zero as C0 deforms into C. Hence, equation (3.1.15) holds again, and using Stokes’s theorem, we can prove that the flow is irrotational.

3.1.3 Cauchy–Lagrange integral

It will be shown in Section 3.2 that, for any irrotational flow satisfying equation (3.1.16), there exists a scalar function ϕ(t, r), called the velocity potential, such that

V=∇ϕ. (3.1.17)

Hence, irrotational flows are also termed potential flows. In order to relate the pressure p to the velocity V =|V| in a potential flow, steady or unsteady, the Cauchy–Lagrange theorem may be used.

Theorem 3.3 Suppose that the flow of an inviscid incompressible fluid is potential, V =∇ϕ. Suppose further that the body force has potential U, such that f = −∇U.

Then the Cauchy–Lagrange integral holds:

∂ϕ

∂t +V² 2 +p

ρ+ U =F(t) (3.1.18)

with the functionF(t) being independent of position in the flow field.

Proof Under the conditions of the theorem, we can write the momentum equation (3.1.3) as

∇

∂ϕ

∂t



+ ω× V + ∇

V² 2



=−∇U − ∇

p ρ



. (3.1.19)

Substitution of (3.1.17) into (1.4.15) shows that any potential flow is irrotational, i.e.

ω= 0. Thus, equation (3.1.19) may be written as

∇

∂ϕ

∂t +V²

2 + U +p ρ



= 0,

which immediately proves that (3.1.18) is valid. ✷

If the conditions of both Theorem 3.1 and Theorem 3.3 hold simultaneously, namely,

• the flow considered is steady,

• the body force is potential,

• and the flow is irrotational,

then the function H in (3.1.6) becomes independent of the streamline considered and the functionF(t) is independent of time. Hence, we can write

V² 2 +p

ρ+ U = C, (3.1.20)

where C is a true constant.

3.1. Integrals of Motion 135 Exercises 7

1. Consider a pipe with cross-sectional area A placed vertically into the Earth’s gravitational field g and connected to a horizontal pipe whose cross-sectional area is A^′ = kA, as shown in Figure 3.4. The vertical pipe is filled with water to height h0; shutters F F^′ and GG^′ are used to prevent the water flowing into the horizontal pipe. Find the time that is required to discharge the vertical pipe after the shutters are opened. Both vertical and horizontal pipes are open to the atmosphere.

x y

O g

F

F^′

G

G^′

−l(t) l(t)

h(t)

Fig. 3.4: Geometrical layout of the problem.

To solve this problem, denote the height of the water column in the vertical pipe at time t by h(t), and use the Euler equations (3.1.1) to determine the pressure distribution in this pipe between y = 0 and y = h(t).

Repeat the analysis for the horizontal pipe, and find the pressure distribution between x = 0 and x = l(t).

Taking into account that at point O the pressure should be the same for the horizontal and vertical pipes, show that the function h(t) satisfies the differential equation

 1− 1

4k²



hh^′′+ h0

4k²h^′′+ gh = 0.

Solve this equation assuming that k = A^′/A = 0.5.

Hint: You may assume that the water motion in both pipes is one-dimensional.

2. Prove that the Bernoulli integral (3.1.6) holds not only along streamlines but also along vortex lines.

3. An open barrel has a small orifice in the wall just above the base. It is filled with water to height h and placed into the Earth’s gravitational field as shown in Figure 3.5. Using Bernoulli’s equation, show that the velocity V in the jet that

g

h

V

Fig. 3.5: Water escaping through a hole in a barrel.

forms as the water flows through the orifice may be calculated using the Torricelli formula

V =p 2gh.

4. A thin spherical shell of radius R0 is filled with water vapour and placed in deep water. At initial instant t = 0, the shell is destroyed and the vapour comes in contact with the water (see Figure 3.6).

Show that the time tc that it takes for the bubble to collapse to a point may be calculated as

tc=

R0

Z

0

s dR 2∆p

3ρ

R³₀ R³ − 1

 , (3.1.21)

Here ∆p = p∞− p⁰, with p∞ denoting the pressure far from the centre of the sphere and p0the vapour pressure.

p∞

p0

R(t)

Fig. 3.6: Vapour bubble in deep water.

3.1. Integrals of Motion 137 You may assume that the flow may be treated as inviscid. You may also assume that the flow is symmetrical with respect to the bubble centre and that gravitational effects are too small to affect the collapse process. You may finally assume that the pressure p_∞ far from the bubble and the vapour pressure p0

inside the bubble remain constant during the collapse process. The governing Euler equations may be obtained by setting ν = 0 in the Navier–Stokes equations (1.8.48).

(a) Start your analysis with the continuity equation, and deduce that r²Vr is a function of time only, say f (t):

r²Vr= f (t). (3.1.22)

Then, using the impermeability condition on the bubble surface, r = R(t), show that

R²dR

dt = f (t). (3.1.23)

(b) Now, solve (3.1.22) for Vr, and using it in the radial momentum equation, deduce that

−f^′(t) r +1

2

f (t)2

r⁴ =−p

ρ+ F (t). (3.1.24)

Taking into account that the pressure far from the bubble centre is p∞ and the pressure on the bubble surface is p0, deduce that

−f^′(t) R(t) +1

2

f (t)2

R(t)4 =p∞− p⁰

ρ . (3.1.25)

(c) Equations (3.1.23) and (3.1.25) involve two unknown function, f (t) and R(t).

In order to solve these equations, represent the function f (t) in the form f (t) = Φ

R(t) , and deduce that

Φ =± s

R

C−2∆p 3ρ R³

,

where C is a constant. Then return to equation (3.1.23) and show that the time that it takes for the bubble to collapse to a point is given by (3.1.21).

5. Consider inviscid incompressible fluid flow through a symmetrical diffuser as shown in Figure 3.7. The diffuser expands from cross-section AA^′, whose area is S1, to cross-section BB^′ with area S2. The fluid velocity at BB^′ is V .

Owing to the symmetry of the diffuser geometry, the total pressure force acting upon the inner surface of the diffuser (called the thrust) is directed along the diffuser axis. Using the integral momentum equation (1.7.30), find the value of the thrust

T =− ZZ

Sc

pnxds,

where nxis the projection of n on the x-axis drawn along the axis of the diffuser.

A A^′

B B^′

Fig. 3.7: Flow through a diffuser.

Suggestion: In the inviscid flow, the surface stress pn is represented by the pressure only, and is given by (1.2.14):

pn =−pn, (3.1.26)

where n is the unit vector normal to the control surface Sc. With (3.1.26), the integral momentum equation (1.7.30) assumes the form

ZZ

Sc

hρ V· n

V+ pni

ds = 0. (3.1.27)

Choose the control surface Sc to be composed of the two walls of the diffuser, AB and A^′B^′, and the cross-sections AA^′ and BB^′ shown by the dashed lines in Figure 3.7, and consider the projection of the momentum equation (3.1.27) on the x-axis.

6. A jet of width d impinges upon a flat wall at an angle α as shown in Figure 3.8.

V

d α

d2

d1

Fig. 3.8: Interaction of a jet with a flat wall.

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