European Journal of Mechanics B/Fluids 47 (2014) 202–210

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European Journal of Mechanics B/Fluids

journal homepage: www.elsevier.com/locate/ejmflu

Incompressible impulsive wall impact of liquid bodiesPeder A. Tyvand a, Knut Magnus Solbakken a, Karina Bakkeløkken Hjelmervik b,∗

a Department of Mathematical Sciences and Technology, Norwegian University of Life Sciences, 1432 Ås, Norwayb Faculty of Technology and Maritime Science, Buskerud and Vestfold University College, 3103 Tønsberg, Norway

a r t i c l e i n f o

Article history:Available online 13 April 2014

Keywords:ImpactIncompressibleLiquidPotential flow

a b s t r a c t

Analytical leading-order solutions are given for various liquid bodies in translational motion that hit aplane wall in impulsive impact at constant velocity. The initial velocity field and the associated virtualmasses are calculated for selected liquid bodies of uniform density. We consider 2D and 3D wedges, acone, the semi-circle and the hemisphere. Sideways force impulses are obtained for 2D bodies. The lossof mechanical energy during impact is calculated for a 2D wedge.

© 2014 Elsevier Masson SAS. All rights reserved.

1. Introduction

The present paper is concerned with incompressible impulsiveimpact of liquid bodies on plane walls. The theory of finite fluidmasses impacting on solid walls was initiated and later followedup by Cooker and Peregrine [1–3]. Incompressible liquid impacthas technological importance for breakingwaves that suddenly hitcoastal and marine structures.

In analogywith impulsive sloshing in open containers [4], thereare four basic time scales for liquid impact on rigid walls: (i) Theacoustic time scale. (ii) The incompressible impulsive time scale.(iii) The gravitational time scale. (iv) The viscous time scale.

The acoustic time scale for liquid impact can usually be disre-garded because it is of the order of milliseconds, and the resultingdisplacements in the liquid will be negligible. The incompressibleimpulsive flow lacks an explicit time scale, so it sets in immediatelyafter the acoustic stage is finished. Gravitational and viscous effectsafter impact need time to develop, and these will not influence theinitial impulsive flow.

The topic of incompressible impulsive impact of fluid bodieson plane walls was already introduced by Milne-Thomson [5]. Theexact solution for a cone was given in this classic textbook, but itwas presented in a modest way as an exercise for the reader. Theeven simpler solution for a 2D wedge was not mentioned in thebook, but it has been explored by Cooker [6].

We take an academic approach to the topic of incompress-ible impulsive impact. The idealization of a flat impact means thatthere is a finite area of instantaneous collision between the fluid

∗ Corresponding author.E-mail address: karina.hjelmervik@hbv.no (K.B. Hjelmervik).

http://dx.doi.org/10.1016/j.euromechflu.2014.03.0180997-7546/© 2014 Elsevier Masson SAS. All rights reserved.

body and the solid wall. Our idealized approach complements theexisting research of liquid impact, which has had an applied tech-nological scope since the first paper by Cooker and Peregrine [1].The focus has been on the understanding of extreme loads on har-bors and marine constructions due to impacting seawater. Theseliquid masses have their origin in breaking ocean waves. Incom-pressible liquid impact on rigid walls has not yet become estab-lished as an academic topic in its own right. The present work isa modest attempt to improve this situation. In establishing a topicacademically, one should explore systematically the simplest non-trivial cases, and develop analytical theories from first principles.Moreover, one should provide identity to this particular scientifictopic by clarifying its relations to the neighboring branches of sci-ence.

Incompressible impact of finite fluid bodies on plane rigid wallsis linked to three fields of engineering science:

(1) The broad field of slamming, which is concerned with solidbodies collidingwith fluidmasses. This is awell-established field ofresearch, see the review articles by Korobkin and Pukhnachov [7]and Faltinsen, Landrini and Greco [8]. The latter authors identifiedliquid impact as a subfield of slamming, and followed it up in aseparate paper [9]. Our work is restricted to the idealized case offluid bodies that are at rest before being hit by a plane wall inuniform motion.

(2) Impulsive free-surface flows, driven by moving walls or ob-jects that are located inside the fluid. The topic most relevant hereis impulsive sloshing [4], where a container with fluid in hydro-static equilibrium is suddenly put into motion. A generalizationcould merge impulsive sloshing with the present topic of impact-ing liquid bodies. This is achieved if we abandon the restrictions ofhaving either a flat wall of impact or an initially flat liquid surface.Insteadwe assume two types of curved boundaries that confine the

P.A. Tyvand et al. / European Journal of Mechanics B/Fluids 47 (2014) 202–210 203

total fluid mass: (i) Parts of the boundary is a curved surface thatis free to move. (ii) The remaining boundary is a curved solid wallbeing forced impulsively into motion.

(3) Dam breaking, or more generally gravitational collapse,where liquid bodies resting on a horizontal plane are instanta-neously released to flow freely under gravity. This topic was stud-ied by Penney and Thornhill [10], with accompanying experimentsreported by Martin and Moyce [11]. These problems relate to fluidimpact problems where the relative motion between the liquidmasses and the planes of impact take place with constant accel-eration instead of constant velocity. In the present work we willdemonstrate a mathematical analogy between dam breaking andliquid impact on plane walls, but its validity is limited to the initialflow only.

The potential flow theory of incompressible dam breaking istroublesome because of the mathematical singularity that appearsat the waterline where a vertical liquid face meets a horizontalwall. The traditional shallow-water theory of dam breaking [12]tries to avoid the singularity by taking the inaccurate assumption ofhydrostatic pressure. Stansby et al. [13] simulated the full potentialtheory of wave breaking, but had to invent numerical artifacts inorder to avoid the waterline singularity. The only consistent wayof dealing with the contact line singularity is to take the singularflow field as the outer field in a matched-asymptotics sense, anddevelop inner solutions to resolve the singularity [14].

The present paper is restricted to the development of leading-order outer solutions for a variety of liquid body geometries.We donot solve any gravitational collapse problem, but a closely relatedmathematical problem: incompressible impulsive impact of liquidbodies on flatwalls. In the presentworkweonly consider the initialimpact flow, studying liquid bodies for which an exact analyticalsolution can be found. 3D cylinders are excluded from the presentwork, and will be analyzed in a second paper.

The incompressible impulsive impact on flat walls is a specialcase of the more general pressure–impulse theory of liquidimpact [1–3]. The type of impact that we consider lets a finite flatportion of the liquid surface hit the flat wall head on.

As pointed out by Greco, Landrini and Faltinsen [9], the the-ory of impacting liquid bodies may apply to green-water phenom-ena of seawater on ship decks. Nielsen and Mayer [15] consideredmild green-water flows with a gradual start, where the fullNavier–Stokes equations are appropriate. The presentwork appliesonly to instantaneous flows due to sudden impact, where gravityand viscosity will be negligible.

2. Model assumptions and formulation

A finite liquid body in 2D or 3D is considered, being initially atrest. At time t = 0 this inviscid and incompressible fluid body is hitby a plane wall z = 0 that is in uniformmotion with velocityW inthe z direction. Lord Kelvin’s circulation theorem implies that theinduced inviscid flow is irrotational. The constant density of theliquid body is ρ. The assumption of incompressible flow implies∇ · v = 0 leading to Laplace’s equation

∇2φ = 0 (1)

for the velocity potential φ, where the velocity field is v = ∇φ.The horizontal wall z = Z(t) moves vertically and hits the flat

front of the stagnant fluid body at t = 0, inducing immediateincompressible flow in the fluid body. The position of the movingwall is

Z(t) = H(t)(Z1t + Z2t2 + Z3t3 + · · ·), (2)

where H(t) is the Heaviside unit step function. The contact be-tween the fluid body and themovingwall starts instantaneously att = 0+, with a finite initial contact area. The flowdue to themoving

wall with position z = Z(t) may be expressed by the small-timeexpansions of the potential and the pressure

φ(x, y, z, t) = H(t)(φ0(x, y, z)+ tφ1(x, y, z)

+ t2φ2(x, y, z)+ · · ·), (3)p(x, y, z, t) = p−1(x, y, z)δ(t)+ H(t)(p0(x, y, z)

+ tp1(x, y, z)+ t2p2(x, y, z)+ · · ·). (4)

δ(t) denotes Dirac’s delta function. Possible waterline singularitiesmay limit the validity of these Taylor series to one term only.

The Bernoulli equation is given by

∂φ

∂t+

pρ

+|v|2

2+ gz = 0. (5)

Here we have included a gravitational acceleration g in the −zdirection, even though the leading-order impulsive flow iswithoutgravity. The fluid body is surrounded by vanishing density. Thekinematic condition at the bottom plane in forced motion is

∂φ

∂z=

dZdt, z = Z(t). (6)

The flow is impulsively started during 0 < t < 0+. IntegratingBernoulli’s equation (5) over this infinitesimal time interval gives

φ = 0, t = 0+, along the free contour of the fluid body. (7)

We will solve the leading-order flow problem at t = 0+. Then weintroduce W = Z1 and put all Zn = 0 for n > 1. We will solve amixed boundary value problem with Laplace’s equation for φ0

∇2φ0 = 0, (8)

applying an inhomogeneous Neumann condition at the bottom

∂φ0

∂z= Z1 = W , z = 0, (9)

and the homogeneous Dirichlet condition φ0 = 0 along the opencontour. Alternatively we may write φ0 = Wz + φ′

0 and solve amixed problem for φ′

0.The potential φ′ represents the flow immediately after impact

in a (primed) coordinate system (x′, y′, z ′) defined by

(x′, y′, z ′) = (x, y, z − Wt), (10)

implying φ′= φ − Wz. In the primed reference system, the liquid

body impacts at the static wall z ′= 0. Before the impact at t = 0,

the liquid body moves with a uniform translational velocity W inthe −z ′ direction.

In the original (unprimed) coordinate system the liquid body atrest is hit by the wall that moves in the z direction with velocityW . The impulsive pressure is given by p−1 = −ρφ0, which givesthe force impulse

F−1z = −ρ

Azφ0(x, y, 0)dxdy = mzW . (11)

Az is the set of points on the wall that is initially wetted by the im-pact.We define the virtualmassmz for the impulsivemotion of thefluid body forced by the impacting wall. The total fluid mass isM .

For a general 3D body, there are four lateral flow components, inthe +x, −x, +y and −y directions, all of these perpendicular to theplane of impact z = 0. We denote the virtual masses for the lateralflow in these directions by mx, m−x, my, m−y, respectively. All ofthese virtual masses are defined by an appropriate effective fluidmomentum divided by W . There is no lateral momentum beforethe wall impact. Therefore it follows from Newton’s third law andthe conservation of momentum thatmx = m−x andmy = m−y.

204 P.A. Tyvand et al. / European Journal of Mechanics B/Fluids 47 (2014) 202–210

For a body that is symmetric with respect to a plane x = x,there will be a repulsive force impulse equal to mxW between thetwo half bodies for x < 0 and x > 0, see [6]. The virtual masses are

mx = m−x = −ρ

W

Axφ0(x, y, z)dydz, (12)

where Ax is the set of points of the initial fluid body in the planex = x. A similar formula applies to a 3D fluid body that is symmetricabout a plane y = y. For a 2D fluid body described in the xz plane,there is a transverse (sideways) force impulse on the sidewall inthe y direction, given by

F−1y = −ρ

Ayφ0(x, 0, z)dxdz. (13)

If the 2D flow takes place in relatively narrow gap 0 < y < Y , thisimpulsive sideways force may exceed the force F−1z on the planeof impact z = 0.

We will investigate the local flow at the waterline where anopen boundarymakes a slope angle αwith the impactingwall. Thedynamic condition of zero tangential flow implies the followingvelocity at the waterline

∇φ0|waterline = W (k + n tanα), 0 < α ≤ π/2. (14)

Here k denotes the unit vector in the z direction. The unit normalvector n lies in the impacting plane z = 0 and points out fromthe waterline contour. Eq. (14) is obtained by decomposing thenormal velocity at the free boundary in the horizontal and verticaldirections. It gives a flow singularity (infinite horizontal velocity)at the waterline in the case α = π/2, where the open boundary ofthe fluid body meets the wall at a right angle.

3. Kinetic energy integrals

In a coordinate system where the liquid is at rest before it is hitby a moving wall, the kinetic energy K0 after impact is

K0 =ρ

2

V

∇φ0 · ∇φ0 dV , (15)

where V denotes the liquid volume. The Gauss theorem combinedwith an equipotential free boundary and the bottom condition (9)imply

K0 = −ρW2

Azφ0(x, y, 0)dxdy =

WF−1z

2=

12mzW 2. (16)

Here we have inserted from Eq. (11). In the primed referencesystem the liquid body has uniform velocity −W in the z directionbefore it hits the solid wall z = 0 at time t = 0. The kinetic energybefore impact is then

K =12MW 2. (17)

The kinetic energy K ′ (in the primed system) of the liquid massimmediately after the impact is given by

K ′=ρ

2

V

∇φ′

0 · ∇φ′

0 dV = K − K0. (18)

Here we have inserted φ′

0 = φ0 − Wz and elaborated the integral(18). The loss of kinetic energy is ∆K = K − K ′

= K0. ByEqs. (16)–(17) we find

∆KK

=K0

K=

mz

M, (19)

noting thatmz has already been defined in Eq. (11).

4. Loss of mechanical energy during impact

The Bernoulli equation without gravity is

∂φ

∂t+

pρ

+|v|2

2= 0, (20)

where the scalar field p/ρ represents a stored energy per unit ofmass.Wewill study energy losses in the primed coordinate systemwhere thewall z ′

= 0 is at rest. The totalmechanical energy beforeimpact is K = MW 2/2, while its value E ′

0 immediately after impactis

E ′

0 =ρ

2

V

∇φ′

0 · ∇φ′

0 dV +

Vp′

0dV = −ρ

Vφ′

1dV , (21)

provided a first-order potential exists. The pressure impulse p−1does not influence the loss of mechanical energy: ∆E = K − E ′

0.For this energy loss we need to know φ′

1. In the system where thewall is at rest we have

∂φ′

1

∂z= 0, z ′

= 0. (22)

The free-surface dynamic condition p = 0 is expressed by

φ′

1 = −|∇φ′

0|2

2= −

|∇φ0 − Wk|2

2,

at the initial free liquid boundary. (23)

We need φ′

1 to calculate the energy loss during impact, but it canonly be found if the zeroth-order flow field is regular in the wholefluid domain. In the present paper, this is true only for the 2Dwedge and the cone.

5. On gravitational collapse or dam breaking

The remaining paper is devoted to a constant velocity ofimpulsive impact

Z1 = W , with Zj = 0 for j > 1. (24)

In this section we will turn our attention to gravitational collapseof liquid bodies [10]. We will link our impact problem withgravitational collapse where an acceleration is turned on at t = 0

Z2 = constant, with Zj = 0 for j = 2. (25)

This gravitational collapse problem applies a reference system atrest with the undisturbed fluid. The kinematic bottom boundarycondition is

∂φ1

∂z= 2Z2, z = 0, (26)

while the other boundaries are open equipotential boundaries. Thesteady pressure is given by p0 = −ρφ1, producing the steady force

F0 = −ρ

Aφ1(x, y, 0)dA (27)

with A denoting the bottom area in initial contact with the movingwall.

The following transformations link the gravitational collapseproblem (Z2 = 0) to the impulsive impact problem (Z1 = 0)

(φ1, p0)2Z2

=(φ0, p−1)

Z1, (28)

and from this pressure transformation we find

F02Z2

=F−1

Z1= mz . (29)

P.A. Tyvand et al. / European Journal of Mechanics B/Fluids 47 (2014) 202–210 205

The wrapped fluid body resting on a plane z = 0 is released togravitational influence at t = 0+ (dam breaking). Immediately theweight on the bottom z = 0 drops from Mg to mzg . The boundaryconditions for φ′

1 are

φ′

1 = −gz, along the free boundary of fluid body, (30)

combinedwith an impermeable condition at the horizontal bottom

∂φ′

1

∂z= 0, z = 0. (31)

Gravitational collapse is physically equivalent to the impact prob-lem with the bottom plane accelerating plane instead of movingwith constant velocity.

∂φ1

∂z= 2Z2, z = 0, (32)

combined with the constant-pressure condition along the freeboundary

φ1 = 0, along the free boundary of fluid body. (33)

The potentials φ′

1 and φ1 express the same flow in two differentcoordinate systems. The transformations linking these two equiv-alent flow problems are

φ′

1 = φ1 − gz, 2Z2 = g, (34)

which can be combined with Eq. (28) to give

φ1

g=φ0

W,

φ′

1

g=φ1

g− z =

φ0

W− z =

φ′

0

W,

p0g

=p−1

W.

(35)

These transformations give us freedom to choose among the fourpotentials φ0, φ′

0, φ1, φ′

1 for solving the boundary-value problemfor the initial flow.

Eq. (14) applies to gravitational collapse, and it tells that theoutward horizontal acceleration of thewaterline is n ·φ1 = g tanαat time t = 0+.

6. A wedge body in 2D

The simplest of all solutions is that of a 2D wedge body, shapedas a right-angle isosceles triangle of fluid with sides given by z =

h±x. Thewidth of the triangle is 2h, and its height is h. The velocitypotential is

φ0 =W2h(x2 − (z − h)2), (36)

known from [6]. Fig. 1 shows this triangle with the initial isobars,flow field and streamline pattern in the coordinate system wherethe moving wall hits the liquid body. The ratio between the virtualmassmz and the total liquid massM is

mz

M=

23

=∆KK, (37)

being equal to the relative loss of kinetic energy in the impactingfluid body (Eq. (19)). During the impact between the liquid massM and the wall, the wall exerts a force impulse mzW = (2/3)MWin the z direction on the wall. Momentum conservation impliesthat the liquid mass will have a net momentum −MW/3 in the z ′

direction immediately after the impact, in the (primed) coordinatesystem where the wall is at rest. This is verified by integrating thenet momentum in the primed coordinate system

ρ

V

∂φ0

∂z− W

dV = −

13MW . (38)

Fig. 1. The 2D wedge body (bold lines), its velocity potential isobars (black),streamlines (red dotted), and current vectors (arrows) of impulsive wall impact.

In the related problem of gravitational collapse (Section 3above), the net weight on the bottom plane suddenly drops fromthe total liquid weightMg to (2/3)Mg . The wedge apex will expe-rience zero pressure gradient, so it will start falling with the grav-itational acceleration.

The virtual masses in the ±x directions are mx = m−x = M/6,see [6]. Let us assume that this 2D wedge body has a finite con-stant width b in y direction, constrained by two walls y = 0 andy = b. The sideways impulsive force on each wall during impact isindependent of b and given by

F−1y =ρ

6Wh3. (39)

We can calculate the total mechanical energy just after impact.First we have to determine the first-order potential φ′

1 in theprimed reference system where the bottom is at rest. We mayreplace the bottom condition at z ′

= 0with a symmetry condition,so that the triangular domain is replaced by a square. This gives thefollowing Dirichlet boundary conditions

φ′

1 = −W 2

2h2((x)′2 + (z)′2), z ′

= ±x′± h, (40)

along the four sides of the square under consideration. In orderto solve this problem, we introduce a (X, Z) coordinate systemrotated an angle π/4 so that the square boundaries are given byX = ±h/

√2, Z = ±h/

√2. The coordinate transformations are

√2x′

= X−Z,√2z ′

= X+Z . TheDirichlet conditions (40) are now

φ′

1 = −W 2

2h2(X2

+ Z2), X = ±h/√2, Z = ±h/

√2. (41)

The Fourier series solution is found to be

φ′

1 = −W 2

4

∞n=0

An(cos(knX) cosh(knZ)+ cosh(knX) cos(knZ)),

(42)

with the wave numbers kn =√2(n + 1/2)π/h and the Fourier

coefficients

An =8(−1)n

cosh((n + 1/2)π)

1

(2n + 1)π−

4(2n + 1)3π3

. (43)

The integration of the total mechanical energy after the impact canbe performed in the (X, Z) system. Due to symmetry we have fromEq. (21)

2E ′

0 = −ρb h/

√2

−h/√2

h/√2

−h/√2φ′

1(X, Z)dXdZ . (44)

206 P.A. Tyvand et al. / European Journal of Mechanics B/Fluids 47 (2014) 202–210

Fig. 2. The geometry of the 3D wedge body.

The result of this integration is

E ′

0 = 16ρbh2W 2∞n=0

1

(2n + 1)3π3−

4(2n + 1)5π5

× tanh((n + 1/2)π). (45)

After summing the series, we compare the result with the kineticenergy before the impact and find E ′

0 = 0.61449K , which leads tothe final result for the loss of mechanical energy during impact

∆E = 0.38551K = 0.57827∆K . (46)

This calculation shows that the total energy loss (∆E) is muchsmaller than the loss of kinetic energy (∆K ) alone. In this partic-ular case, more than 40% of the lost kinetic energy is immediatelystored as a steady pressure.

7. A finite wedge body in 3D

We can develop a Fourier series solution for a 3D wedge bodywith an isosceles triangular cross section, constricted by two ver-tical endwalls y = 0 and y = b. The cross section in the x, z planeis the triangle with sides given by z = h ± x, which is described inthe previous section. See the sketch in Fig. 2.

The forced bottom boundary condition is given by

∂φ0

∂z= W , z = 0,−h < x < h, 0 < y < b (47)

and the constant-pressure condition along the sidewalls gives

φ0 = 0, z = −h ± x, 0 < y < b, (48)φ0 = 0, y = 0 and y = b. (49)

It is convenient to superpose two 2D solutions φ01 and φ02

φ0(x, y, z) = φ01(y, z)+ φ02(x, z). (50)

The boundary value problem for φ01 is determined by a fictitiouslid at z = h

∂φ01

∂z= 0, z = h, (51)

combined with a periodic extension of the bottom velocity

∂φ01

∂z=

4Wπ

∞m=1

sin(kmy)2m − 1

, z = 0 (52)

where km = (2m − 1)π/b. The solution of this boundary valueproblem is

φ01(x, z) = −4Wπ

∞m=1

sin(kmy) cosh(km(z − h))(2m − 1)km sinh(kmh)

. (53)

The boundary value problem for φ02 is determined by the antisym-metry conditions along the slopes z = h ± x, which requires a

uniform normal velocity out of the fictitious vertical boundariesx = ±h. This compensates for the inward flux at z = 0 and leadsto the condition similar to Eq. (52)

∂φ02

∂x=

4Wπ

∞m=1

sin(kmy)2m − 1

, x = h. (54)

This potential is symmetric around the y axis.

∂φ02

∂x= 0, x = 0. (55)

The solution of this boundary value problem is analogous to (53)

φ02(x, y) =4Wπ

∞m=1

sin(kmy) cosh(kmx)(2m − 1)km sinh(kmh)

, (56)

completing the velocity potential for the finite wedge body

φ0(x, y, z) =4Wπ

∞m=1

sin(kmy)(cosh(kmx)− cosh(km(z − h)))(2m − 1)km sinh(kmh)

.

(57)

The outward flow at the waterlines confirms Eq. (14). Fig. 3 showsthe flow field and isobars for selected cross sections of the 3Dwedge body.

The mass of the 3D wedge body is given by M = ρbh2. Thevirtual masses in x, y, and z direction are given by

mx =8ρb3

π4

∞m=1

sinh(kmh)− kmh(2m − 1)4 sinh(kmh)

, (58)

my =8ρπ

∞n=1

(−1)n+1

nk3n

2

sinh(knh)− 2 coth(knh)+ knh

, (59)

mz =16ρb3

π4

∞m=1

h coth(kmh)− 1(2m − 1)4

, (60)

where km = (2m − 1)π/b. Fig. 4 shows the virtual masses for the3D wedge body.

8. A half-wedge in 2D

Another 2D body for which an analytical solution is found,is the isosceles triangle that comprises half the 2D wedge bodytreated above. For the compactness of the mathematical solution,we choose to place the origin in the tip of the fluid body. The im-pacting boundary is given by z = −h, with 0 < x < h. Thetwo other sides of the triangle are free boundaries. One of theseis aligned along the z axis, where −h < z < 0. The other freeboundary is the hypotenuse of the triangle, given by z = −xwhere0 < x < h. The 2D half-wedge is the only asymmetric body thatwe consider. Its potential is

φ0 =

∞n=1

An(sin(knx) cosh(kn(z − h))+ sin(knz) cosh(kn(x + h))

− sin(knx) cosh(kn(z + h))− sin(knz) cosh(kn(x − h))),(61)

where kn = (n − 1/2)π/h. The Fourier coefficients are given by

An = −8Wh

π2(2n − 1)2 sinh(2knh). (62)

Fig. 5 shows the flow field and isobars for half-wedge 2D body.The virtual mass in the z direction ismz = 0.2120M , calculated

with 100 terms in the Fourier series solution. Since the body isnot symmetric, it is difficult to integrate separately the virtual

P.A. Tyvand et al. / European Journal of Mechanics B/Fluids 47 (2014) 202–210 207

Fig. 3. The 3D wedge body (bold lines) in the yz-plane at selected x-values (left) and in the xz-plane at selected y-values (right) together with its velocity potential isobars(solid lines) and current vectors (arrows) of impulsive wall impact.

masses in the ±x directions. An alternative method is to evaluatethe following integral numerically

A

∂φ0

∂x

dxdz = 2mx

ρb, (63)

where A is the impact area in the x, z plane, and b is the thicknessof the fluid body in the y direction. The factor 2 follows from oursimultaneous integration of the contributions in the −x and +xdirections. The result of this integration is

mx = m−x = 0.1353M, (64)

calculated with 100 terms in the Fourier series solution. Thesideways force impulse on the channel sidewalls is found to be

F−1y = 0.035669ρWh3. (65)

It is computed by inserting Eq. (61) into Eq. (13) and truncating thesum after 60 terms. This sideways force impulse on an individualhalf-wedge is only one fifth of that for a compact 2D wedge, givenby Eq. (39).

9. A cone with half opening angle arctan(√2)

The simplest 3D solution is that of a liquid cone with half open-ing angle arctan(

√2), given by Milne-Thomson [5]. The cone has

height h, with a circular base of radius h√2. The velocity potential

is

φ0 =W4h(r2 − 2(z − h)2), (66)

Fig. 4. The virtualmasses of impulsivewall impact (solid lines) on a 3Dwedge bodyand the corresponding asymptotes (dotted lines).

where we have introduced the radial coordinate r =x2 + y2.

The potential (66) confirms the waterline condition (14). Stokes’streamfunction is

Ψ0 = −W2h

r2(z − h). (67)

Fig. 6 shows a section through the axis of this cone, with theinitial isobars and flow field in the coordinate system where thewall hits the liquid body. The relative virtual mass is equal to the

208 P.A. Tyvand et al. / European Journal of Mechanics B/Fluids 47 (2014) 202–210

Fig. 5. The 2D half-wedge body (bold lines), its velocity potential isobars (black),streamlines (red dotted), and current vectors (arrows) of impulsive wall impact.

Fig. 6. The cone body (bold lines), its velocity potential isobars (solid lines) andcurrent vectors (arrows) of impulsive wall impact.

fractional loss of kinetic energy

mz

M=

34

=∆KK. (68)

This energy loss is relatively greater for the cone than for the 2Dwedge. The virtual masses in the transverse directions of the wallare

mx = my =M

23/2π= 0.11254M. (69)

Gravitational collapse of such a liquid cone makes the net weighton the bottom plane drop instantaneously from Mg to (3/4)Mg .The collapsing cone shrinks (inward flow) for h/2 < z < h andexpands (outward flow) for 0 < z < h/2.

A potential that generates a class of elliptical cones of height his

φ0 =W4h((1 + K)x2 + (1 − K)y2 − 2(z − h)2). (70)

K may take a value between K = 0 (axisymmetric cone) and K = 1(2D wedge).

Fig. 7. The semicircle (bold lines), its velocity potential isobars (black), streamlines(red dotted), and current vectors (arrows) of impulsive wall impact.

10. The 2D semicircle

Wewill also solve the problem of the 2D semicircle with radiusR. The corresponding problem of gravitational collapse is knownfrom Penney and Thornhill [10], who applied polar coordinates(r, θ) = (

√x2 + z2, arctan(z/x)).

The boundary-value problem for the potential φ0(r, θ) is

φ0 = 0, r = R, (71)∂φ0

∂z= W , z = 0. (72)

It is better to solve the problem in terms of the modified potentialφ′

0 = φ0 − Wz, obeying the boundary conditions

φ′

0 = −WR sin θ, r = R, (73)

∂φ′

0

∂z= 0, z = 0. (74)

The symmetry condition (74) is incorporated into the followingexpansion

φ′

0 = −2πWR

1 − 2

∞n=1

cos(2nθ)4n2 − 1

, r = R, (75)

giving the solution

φ0/W = r sin θ −2πR

1 − 2

∞n=1

rR

2n cos(2nθ)4n2 − 1

. (76)

The logarithmic singularity in the velocity field at z = 0 is knownfrom the corresponding dam-break solution by Penney and Thorn-hill [10]. The streamfunction is

ψ0/W = r cos θ −4πR

∞n=1

rR

2n sin(2nθ)4n2 − 1

. (77)

Fig. 7 shows the initial isobars and flow field in the coordinatesystemwhere the liquid body is hit by themovingwall. The relativevirtual mass is

mz

M=

8π2

−16π2

∞n=1

1(2n + 1)2(2n − 1)

= 1 −4π2

= 0.59472 =∆KK, (78)

valid for the z direction. The other virtual massmx is given by

mx

M=

4π2

−1π

−8π2

∞n=1

(−1)n

(2n + 1)2(2n − 1)= 0.16858. (79)

P.A. Tyvand et al. / European Journal of Mechanics B/Fluids 47 (2014) 202–210 209

The sideways impulsive force on the sidewalls is

F−1y =ρ

3WR3, (80)

pointing in the y direction.The semicircle is our first example that has a rounded peak

with a nonzero velocity. Its initial upward velocity is (∂φ0/∂r)(R, π/2) = 0.36338W , found by truncating the Fourier series after2 million terms.

11. The hemisphere

Our final case is a hemisphere with radius R. The correspondingdam-break problem is known from Penney and Thornhill [10].We apply spherical coordinates (r, θ) = (

x2 + y2 + z2, arctan

(x2 + y2/z)). The boundary conditions for φ0(r, θ) are

φ0 = 0, r = R, (81)∂φ0

∂z= W , z = 0. (82)

We solve the modified potential φ′

0 = φ0 − Wz, with boundaryconditions

φ′

0 = −WR cos θ, r = R, (83)

∂φ′

0

∂z= 0, z = 0. (84)

The symmetry condition (84) is incorporated by writing

φ′

0 = −WR| cos θ |, r = R, (85)

thereby extending the validity of Eq. (83). The solution has the form

φ0/W = r cos θ − R∞n=0

Bn

rR

2nP2n(cos θ), (86)

introducing the Legendre polynomial P2n(cos θ). The coefficientsBn are

∞n=0

BnP2n(ζ ) = |ζ |, −1 < ζ < 1. (87)

By the orthogonality of the Legendre polynomials we find

Bn = (4n + 1) 1

0ζP2n(ζ )dζ =

(4n + 1)√π

4Γ (3/2 − n)Γ (2 + n),

n = 0, 1, 2, . . . (88)

The corresponding Stokes streamfunction Ψ0(r, θ) is given by

Ψ0/W =r2

2sin2 θ + 2rR

∞n=0

nBn

4n + 1

rR

2n× (P2n+1(cos θ)− P2n−1(cos θ)). (89)

Fig. 8 shows the initial isobars and flow field in the coordinatesystemwhere the liquid body is hit by the wall. The relative virtualmass is

mz

M=

32

∞n=0

BnP2n(0)n + 1

=3√π

8

∞n=0

4n + 1(n + 1)Γ (3/2 − n)Γ (2 + n)

P2n(0). (90)

The virtual masses in the x and y directions are equal and given by

mx = my = ρR3

∞n=0

Bn

n + 1

π/2

0P2n(cos θ)dθ −

23

. (91)

Fig. 8. The hemisphere (bold lines), its velocity potential isobars (black) andstreamlines (red dotted) of impulsive wall impact.

Truncation after 23 terms in the series gives the approximatevirtual massesmx

M,my

M,mz

M

= (0.1101, 0.1101, 0.4539), (92)

which are smaller values than those for the semicircular cylinder.The hemisphere is our second example that has a rounded

peak with a nonzero velocity. Its initial upward velocity is(∂φ0/∂r)(R, 0) = 0.17168W , found by truncating the Fourier se-ries after 30 million terms. This upward velocity is only about onehalf of that for the semicircle, which is plausible because a hemi-sphere has a relatively greater surface area available for the up-ward motion, compared with the semicircle.

12. Conclusions

The present work is a contribution to the literature of incom-pressible impact where a finite liquid mass of uniform density col-lides with a rigid wall. We are concerned with the initial impulsiveflow of liquid bodies that have a flat impacting area in head-on im-pulsive encounter with a flat wall. For the fluid bodies under con-sideration, we have calculated the virtual masses for the flow inthe normal and lateral directions to the impacting wall. The lateralvirtual masses are found by integration of the impulsive pressuredistribution, as long as the fluid body is symmetric. We have in-cluded one asymmetric body where an integration of the lateralmomentum was needed in order to compute the correspondingvirtual mass.

It has been shown by mathematical transformations that aninitial flow field in a fluid body impacting impulsively on a flatwall is equivalent to initial gravitational collapse (dam breaking)of the same liquid body that is released to flow freely undergravity.We solve only leading-order outer problems in amatched-asymptotics sense. When the flow evolves beyond the early stagesinvestigated here, impulsive impact for a given liquid shape willevolve quite differently from the corresponding dam-break flow.The transformations between these two types of flow is limitedonly the leading-order outer flow problems.

A simple relationship is established between the slope angleat which the fluid body meets the wall at the waterline and thetangential velocity, as long as this angle is smaller than π/2. Ifthis slope angle is π/2, the liquid surface meets the wall at a rightangle, and the flowwill be singular according to the present theory.Korobkin and Yilmaz [14] demonstrated how this flow singularitymay be resolved in terms of matched asymptotic expansions.

The incompressible inviscid impact flow in the liquid bodiesdoes not have any entropy production. Still it is not meaningfulto postulate energy conservation for fluid bodies impacting on awall. This is because it is impossible to isolate a physical systemfor which energy conservation could be claimed. On the otherhand, the principle of momentum conservation is fully valid andapplicable.

210 P.A. Tyvand et al. / European Journal of Mechanics B/Fluids 47 (2014) 202–210

As a tacit assumption,we have considered only inelastic impact.This means that the force impulse from the impacting liquid bodyis completely absorbed by thewall. The force impulse of restitutionis taken to be zero, which means that there is no bounce back(rebound). The wall absorbs a force impulse received from theliquid momentum, but does not give any momentum back to theliquid. Rebounding walls with partly elastic restitution are leftfor later work. Momentum conservation is the key principle foranalyzing such problems.

Anunderrated fact concerning free-surface flows is that an earlyimpulsively started flow has an explicit time arrow. An equipo-tential free-boundary condition applies to the initial impulsiveflow, because all forces during the abrupt start are pressure forcesperpendicular to the free boundary. The equipotential dynamicfree-boundary condition lacks a time derivative, implying that thesystem is only of first order in time through the kinematic free-boundary condition. This first-order system has a time arrow, be-cause only a system that is of second order in time can be fullyreversible. The initial time arrow of an impulsive free-boundaryflow implies that it is not possible to stop such a flow in the sameimpulsive manner as it was started.

Acknowledgments

This work is supported by a grant from the Oslo Fjord Alliance.Dr. M.J. Cooker is thanked for valuable suggestions.

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