A Water Fountain

Problem B: A Water FountainTeam 743

(Dated: 16 November 2014)

The characteristics of the flow of a vertically directed conic jet is investigated using a derivation of theWeber number and the Reynolds number, along with the equations of motion found using the Navier-Stokesequation. We predict a distribution of droplet sizes based on the competition between external forces andsurface tension, and we perform an experiment to justify the reasoning in selecting the Rosin-Rammler particlesize distribution. Discrete computational methods are used to solve for the motion of the particles in thisdistribution, by which we find the median radius of the distribution, the mean square deviation from the bulkjet, and the overall radial distribution for a fountain with controlled jet angle, flow rate, and nozzle width.

PACS numbers: 05.40.-a, 47.11.+j, 47.20.MaKeywords: Liquid jet, droplet formation, discrete element methods

A Water Fountain 2

CONTENTS

I. Introduction 2A. Background 2B. Approach 3

II. Theory 3A. Fluid Dynamics 3

1. Drag force 32. High Reynolds number drag force 43. Droplet radius 54. Terminal velocity 6

B. Lagrangian Formalism 71. Modified Euler-Lagrange with

dissipation 72. Single Particle Trajectory 93. Maximum Height 94. Energy loss due to dissipative forces 10

III. Computational Model 10A. Initializing and evolving deterministic

trajectories 10B. Statistical analysis of trajectories 11C. Model of N particles moving with varying

initial parameters 11

IV. Experimental Methods 11A. Drop radius experiment 12

V. Analysis of Results 12

VI. Conclusion 13

References 13

A. Navier-Stokes 13

B. C++ Numerical Integration Program 15

I. INTRODUCTION

The behaviors of liquid jets are rich with dynamicalphenomena, ranging in scale from intermolecular inter-actions to macroscopic jet deformation and break-up.Because of the coupling of large scale effects to micro-scopic fluctuations, the theoretical treatment of a ver-tically propagating water jet requires careful statisticalanalysis in conjunction with the classical methods ofstudying fluids. Experimentally, emergent effects suchas flow instability and atomization in jets have been thesubject of extensive research, as there is significant in-dustrial interest in the properties of fluid jets. Such ap-plications range from fuel jets and printing jets, in whichmaximum uniform atomization is preferable, to the cre-ation of thin fibers by cooling jets of melted material, inwhich jet stability is desirable.

A. Background

Fluids have intrigued great minds in physics for sev-eral centuries. The work of Claude Navier and GabrielStokes essentially did for fluid mechanics what Newton’sequations did for classical mechanics, in that the evo-lution of the state of a well understood body of fluidcould be solved for using equations of motion. Perhapsdue to their dynamical complexity, fluids provide a verybroad and difficult set of problems for physicists, math-ematicians, and engineers. Among others, a particularlyinteresting case is the concept of a jet, in which one fluidmoves rapidly along a well defined, contiguous trajectorythrough another fluid. Jets have intrinsic physical quirksthat arise due to the boundary of two fluids moving withrespect to each other.

The breakdown of a jet is one such trait that is at-tributed to the propagation and amplification of surfaceinstabilities in the jet1. These instabilities may resultfrom the internal fluctuations in the flow itself or fromimpinging environmental factors. Eventually, the ampli-tude of the surface wave exceeds the width of the jet,causing the jet to break. In addition, these instabilitiesmay arise as a result of the interactions of the fluid withthe nozzle.

The study of jet breakdown has been expanded tothe case of liquid sheets2. These breakdowns are dis-tinct from the analogous phenomena in columnar jetsbecause the surface waves of a circular jet form a singlering, whereas surface deformations in a sheet may not beclosed at all.

The case of a vertically directed conic jet forms an in-teresting dynamical system that is well described as acurviplanar flow under the influence of gravity and in-terfacial forces from the surrounding fluid. When thesystem is framed as a flow of discrete particles, one canfind interesting and surprising relations between observ-able variables such as the root mean square radius, whichmeasures the spreading of the sheet jet, to the distribu-tion of particle sizes created by the nozzle geometry. Weuse the derivation of the Weber number and the Reynolds

A Water Fountain 3

number to demonstrate the connection of several usefulquantitative observables and qualitative characteristicsof the fountain flow to the design of the nozzle and thechoice of flow rate and angle.

The specific parameters that can be controlled in sucha fountain are the axial angle of the cone, φ, the volumet-ric flow rate of the fountain, dVdt , and the nozzle size, δr,which has been shown to correspond directly to the sizeof the droplets that the jet forms3. Utilizing control ofonly these three variables, a wide variety of fountain dis-tributions and time of flight properties can be obtained.

B. Approach

The trajectories of the water flow can be divided intotwo regimes: a deterministic one dominated by iner-tial forces, and a stochastic regime for which the iner-tial terms are negligible. During the breakup of a wa-ter jet, the stream breaks up to form droplets of varioussizes and energies4. These droplets are bound by inter-molecular forces, and they are broken by kinetic energytransfer from the bulk jet and from the surrounding gas.Other phenomena that might affect droplet size includedroplet agglomeration and evaporation. By solving forthese forces, we are able to find a mean particle size aboutwhich we can estimate a distribution of probable particlesizes. Integrating over this domain of sizes using deter-ministic and stochastic methods, we can find a reasonableapproximation to the radial distribution function for theoutput of a conic jet.

Outside of a certain spatial domain that is very closeto the nozzle, the droplet interaction cross-section of thedroplets becomes negligible as the area swept out by thefountain grows. This allows us to simulate the particlesusing discrete element methods and the equations of mo-tion for particles in a fluid. In this region, droplets aretreated as projectiles, and the spray of the fountain ismodelled as a diffusing monodisperse aerosol. The iner-tial regime for particles moving through a fluid is derivedfrom the Navier-Stokes equations, and total particle ac-cumulation along the radial axis is found for both massivedroplets dominated by deterministic forces as well as forsmall droplets dominated by viscous terms.

This method partitioning the dynamics into two dis-tinct modes is inspired by the symmetry breaking thatoccurs in nature when the bulk properties of fluidsmust compete with the particle nature of the constituentmolecules. In the conic fountain, the force of surfacetension that binds a droplet or a sheet together mustovercome drag forces and unstable configurations. Asthe Weber number of the bulk fluid in this fountain sys-tem crosses the critical boundary created by these twocompeting influences, two qualitatively distinct behav-iors emerge. Each of these situations demands a differ-ent mathematical approach. We expand on these twoapproaches in the following sections, and use a computa-tional model to investigate the physical implications and

accuracy of our mathematics.

II. THEORY

We begin our mathematical treatment with the goalof finding some quantitative representation of the twoqualitative regimes that we wish to simulate. We thenfind the drag force on a particle in the inertial regimeby applying dimensional analysis to the Navier-Stokesequation in terms of the Reynolds number. The Webernumber is then introduced by relating the surface tensionpressure and the Bernoulli pressure. Using our results,we predict the mean droplet size for our flow and itsterminal velocity.

A. Fluid Dynamics

1. Drag force

In order to understand the final distribution of thefountain flow, we must first understand the trajectories ofindividual water droplets. To solve for these trajectories,we must make reasonable assumptions about the forcesacting on the fluid, which are functions of the sizes andvelocities of the droplets. By this relation, an individualdroplet’s size directly affects its final radial displacement.The weighted sum of the distribution of droplet sizes willgive the final radial distribution of the water jet.

We therefore consider our droplets to be particlesmoving through a fluid while being acted upon by aviscous force, F . We first assume that this force will bea function of viscosity µ, fluid density ρ, the particlediameter `, and the velocity of the particle v, or

F = f1(v, `, ρ, µ). (1)

To assist in the solution of the problem, we will utilizethe Buckingham π theorem, which will allow us to solvefor the equations of force on the particle using dimen-sional analysis and nondimensionalization techniques.Let the physical system depend on n parameters, whichwe explicitly refer to in equation (1). We will label theseparameters as a1, . . . , an, and we will say that theyhave fundamental dimensions d1, . . . , dk.

[ai] = dαi,1

1 · · · dαi,kn . (2)

Our physical system from (1), f(a1, . . . , an), isinvariant under some change of dimensional scalesdj → sjdj : j ∈ [1, k]. Hence, f(a1, . . . , an) may berecast in the form

g(π1, . . . , πn) = 0, (3)

A Water Fountain 4

where πi are the dimensionless combinations of ai andm = n − rank(aij) ≤ n − k. Our system has 5 param-eters, F, ρ,v, `, µ, with 3 fundamental dimensions ofmass (M), length (L), and time (T). According to thetheorem, we will then have (5 − 3) = 2 dimensionlesscombinations.

To balance the units, we first note that the viscosityhas units of mass per length per time, or

[µ] =M

LT. (4)

To find a functional form for the force, we may use di-mensional analysis. As long as the units of the forceterm equal those of the related parameters, we may di-vide terms in our force function until we remove all un-necessary dimensional dependencies. Initially, we haveunits

F ρ v ` µMLT 2

ML3

LT L M

LT .

If we divide both sides by ρ, cancelling units in the Fand µ terms,

Fρ ρ v ` µ

ρL4

T 2ML3

LT L L2

T .

We see that density is the only term that is dependenton mass. Because we cannot balance the mass term us-ing any other parameter, we conclude that the secondcolumn will not be one of the dimensionless terms in theBuckingham π relation. Then we have

Fρ v ` µ

ρL4

T 2LT L L2

T .

We now divide by our length ` to the fourth power. Thisyields

Fρ`4

v` ` µ

ρ`21T 2

1T L 1

T .

By the same reasoning used to eliminate the ρ column,we eliminate the ` column,

Fρ`4

v`

µρ`2

1T 2

1T

1T .

And finally we multiply by length squared over velocitysquared to give

Fρv2`2

v`

µρv`

1 1T 1.

where a length term on the right hand side is absorbedinto one of the unused parameters. For the unitsto balance, the two dimensionless columns must befunctions of one another, as stated in the Buckingham π

theorem. Thus,

F

ρv2`2= g

(µ

ρv`

), (5)

where g is some function of the right hand side argument.We recognize that this argument is the reciprocal ofthe Reynolds number, which is the ratio of inertial toviscous forces. We shall denote the Reynolds numberquantity Re. Our force is then

F = ρv2`2g (Re) . (6)

Here we have shown that the drag force acting on a par-ticle should be proportional to the square of the velocity,v2, the density of the fluid, ρ, and some area, `2, whichwe will take to be proportional to the cross-sectional areaof the particle.

2. High Reynolds number drag force

Implementing the Navier-Stokes equation (see appendixA), we have

ρ

[∂v

∂t+ v (∇ · v)

]= −∇P + µ∇2v (7)

at a pressure P , viscosity µ, density ρ, and velocity v.We may substitute the Reynolds number

Re =ρv`

µ(8)

and using

dv

dt=∂v

∂t+ v (∇ · v) (9)

to rewrite equation (7) to give

µ

v`Re

dv

dt= − µ

ρv`Re∇P +

ρv`

Re∇2v, (10)

where the coefficient of ∇P has simply been multipliedby (Re)(Re)−1 = 1.

For particle with a high Reynolds number, the equa-tions of motion will be dominated by inertial terms. Aswe have an ordinary differential equation, we can usethe method of dominant balance to approximate localsolutions to the motion of the high Reynolds numberparticles. There must be some relation between theterms proportional to first order terms of Re, since terms

A Water Fountain 5

proportional to Re−1 will be negligible:

µ

v`Re

dv

dt∝ µ

ρv`Re∇P

dv

dt∝ 1

ρ∇P

∂v

∂t+ (∇ · v) · v ∝ 1

ρ∇P.

(11)

Now consider a frame where the droplet is stationaryand the fluid is moving about it. We expect inertialterms to dominate the dynamics for a high Reynoldsnumber, which means that the droplet velocity will bereasonably constant. We assume that the velocity hasno explicit time dependence, ∂v

∂t = 0. Thus,

(∇ · v) · v ∝ 1

ρ∇P. (12)

To nondimensionalize the equation, we introduce di-mensional scaling units Sρ, S`, SP , and Sv for density,length, pressure, and velocity, respectively. We definethese variables such that

xi = Sixi, (13)

meaning that the variable xi may be decomposed intonondimensionalized components, xi, and a dimensionalscaling factor, Si. Using these factors to scale equation(12), we find that

S2v

S`=

1

Sρ

SPS`

SρS2v = SP .

(14)

since the dimensions must be equal. Then

SρS2vS

2` = SPS

2` . (15)

Since we earlier found

F ∝ ρv2`2, (16)

we see that

F ∝ SρS2vS

2` . (17)

By using the earlier expression for SP , we finally see the

proportionality

F ∝ ∆P`2 (18)

for high Reynolds number and an order of magnitudeplace holder ∆P ≡ SP .

3. Droplet radius

The water will feel competing forces due to surface ten-sion and pressures as it travels. The Weber number is ameasure of the balance between inertial terms and sur-face tension terms in the force equations for the droplet,similar to the Reynolds number that relates the inertialterms of a fluid flow to the viscous terms.

FIG. 1. The forces endured by a water droplet.

The Weber number is given by

We =ρv2`

σ, (19)

where ρ is the density, v the velocity, and σ the surfacetension. For flows with a low Weber number, surfacetension dominates the dynamics of the jet and a contin-uous stream can be expected. As the relative velocity ofthe drops increases, either by an increase in the fountainflow rate or by the introduction of wind, the Webernumber increases. For Weber numbers on the order of 10to roughly 300, the forces of the surrounding fluid tendto break the stream into discrete droplets and mist5. Tobetter understand the origin of the Weber number, wecan consider two pressures acting on a the droplet. Onepressure, the Young-Laplace pressure6, will exist due tothe surface tension. It is given by

PY L =σ

r. (20)

where r is the radius of the droplet. The other pressure

A Water Fountain 6

that a droplet will feel is the Bernoulli pressure, which isdue to the kinetic energy of the droplet. The Bernoulliequation is given by

1

2ρv2 + ρgz + p0 = constant (21)

at height z under the influence of gravitational acceler-ation g. Along the stream in a fluid, we may omit thegravitational term since it will not vary greatly from oneparticle to the next. Then we may simplify the Bernoulliequation:

static pressure + dynamic pressure = total pressure,(22)

where we regard the dynamic pressure as 12ρv

2. Let

ρ′ ≡ 12ρ. Then we call the Bernoulli pressure

PB = ρ′v2. (23)

The Weber number is the ratio of equations (23) and(20),

We =PBPY L

. (24)

The Weber number may now be used to quantitativelydetermine the traits of the droplets formed during jetbreakdown.

To determine the parameters required for dropletformation, we want the inertial and surface tension pres-sures to be approximately equal. At this critical point,the Weber number is on the order of unity. Then we state

1 ' ρv2`

σ, (25)

which implies

` ' σ

ρv2. (26)

We know7 the surface tension of water at room tempera-ture to be σ = 7.2 ·10−2N

m and the density ρw = 1000 kgm3 .

To give a preliminary indication of droplet radius at areasonable velocity, we will make order of magnitudeestimates. We wish to find the change in the length scaleover the course of the trajectory as the velocity changes.We assume that the water will form a droplet nearlyimmediately during flight, during a spatial intervalon the order of a centimeter. Then by rough energy

conservation in the presence of dissipative forces,

mgh ' 1

2mv2, (27)

where m is the mass of the droplet. Then

v2 ' 2gh, (28)

which means when the droplet has dropped 1 centimeter,its velocity squared will be on the order of

v2 ' 2 · 10m s−2 · 1

100m =

1

5m2 s−2. (29)

Substituting this estimate into our expression for criticallength scale gives

` ' 7.2 · 10−2N m−1

1000kg m−3 · 15m2 s−2

= 3.6 · 10−3m. (30)

The estimated critical length scale will be on the orderof 3.6 mm at the time of droplet formation.

4. Terminal velocity

As the droplet falls, its velocity will be limited by airdrag and buoyancy. When these forces balance with thegravitational acceleration, the droplet achieves terminalvelocity. Knowledge of the terminal velocity givesimportant information about the radial distribution,since the terminal velocity will be dependent on the sizeof the droplets. Terminal velocity itself only becomesrelevant in the high Reynolds number regime. Weshowed in section II A 1 that the force must be of theform in equation (6). We restate this force and multiplyby π to relate the `2 term to the cross-sectional area ofour droplet. Defining our coordinates such that +z isthe positive vertical direction,

FD = −1

2cDv

2ρairπ`2, (31)

where cD is the drag coefficient. At Reynolds numberson the order of 103, this coefficient8 is close to cD ' 0.5.Then our drag force is near

FD ' −π

4ρairv

2`2. (32)

We also have the force due to gravity,

A Water Fountain 7

Fg = mg

=4

3π (`)

3ρwg

=4π

3ρwg`

3,

(33)

where m is the mass of the droplet and ρ is the densityof water. Finally, we have the contribution due tobuoyancy. This term will be given by Archimedes’principle,

FB =4

3π`3ρairg. (34)

The addition of buoyant and gravitational forces give

Fg + FB =4

3π`3g (ρw − ρair) , (35)

but seeing that the density of air is much less thanthe density of water, we will disregard buoyancy. ByNewton’s second law, the net force of the particle canbe found by summing the external forces acting on theparticle,

mz = Fg + FD

=4π

3ρwg`

3 − π

4ρairv

2`2,(36)

We may substitute m = 43π`

3ρw and v = z for aspherical, vertically falling droplet to yield

4

3π`3ρwv =

4π

3ρwg`

3 − π

4ρairv

2`2

v = g − 3

4π`3ρw

π

4ρairv

2`2

v = g − 3

16`

ρairρw

v2.

(37)

Again, it is useful to introduce scaling factors v = Sv vand t = Stt, where v and t are nondimensional. Thenwe have the relation

SvSt

˙v

g= 1− 3

16g`

ρairρw

S2v v

2. (38)

We may define our scale factors such that the coefficientsin front of the dimensionless variables are equal to unity.Taking the coefficient of v2,

3

16g`

ρairρw

S2v = 1 (39)

which means

Sv = 4

√g`

3

ρairρw

. (40)

Next, taking the coefficient of ¯v,

SvStg

= 1 (41)

which upon substitution of Sv gives

St =1

4g

√g`

3

ρwρair

. (42)

These scaling factors indicate the the order of magnitudefor the terminal velocity and an estimate of the timerequired to reach it. We then have

Sv ' 4

√10m s−2 · (3.6 · 10−3m)

3· 1000kg m−3

1.2kg m−3

' 4√

10 m s−1

' 12.6m s−1.

(43)

for a rough estimate of the terminal velocity. The timeis takes to reach this velocity is on the order of

St '12.6m s−1

10m s−2

' 1.26s.

(44)

B. Lagrangian Formalism

1. Modified Euler-Lagrange with dissipation

As the radius of the jet cone increases, we assume thatthe interaction of water droplets becomes negligible. Theonly external forces acting on the particle are then airdrag, the gravitational force, and buoyancy. As earlier,we will ignore the buoyancy term as it is negligiblecompared to the gravitational force.

Consider a system of N droplet particles. The forceFi may be expressed as the negative gradient of somescalar potential,

F iP = − ∂

∂xiV (x1, ...,xN ). (45)

A Water Fountain 8

By the chain rule, we may express our gradient in termsof generalized coordinates qα to give

F iP = −

N∑α=1

∂V

∂qα∂qα

∂xi. (46)

Assume there exists an invertible Jacobian that trans-forms our system from the xi coordinate system to theqα. We may then decompose the external forces on oursystem and sum them, giving the partial of the potentialsuch that

N∑i=1

F iP ·

∂xi∂qα

= − ∂V∂qα

. (47)

Now we introduce stochastic conditions by consideringan external force F i

D that cannot be derived from adeterministic potential, V (x1,x2, ...,xN ). Define a termfor the external dissipative force

Dα ≡N∑i=1

F iD ·

∂xi∂qα

. (48)

The total forces acting on the system are assumed toobey superposition, since nonlinear effects are weak hereand will be absorbed into the noise terms later.

F i = F iP + F i

D (49)

Using (49), we rewrite (47) to be

N∑i=1

F i · ∂xi∂qα

= − ∂V∂qα

+Dα. (50)

By imposing Hamilton’s principle, we find an Euler-Lagrange equation with a Rayleigh dissipation force,9

d

dt

∂L

∂qα− ∂L

∂qα= Dα (51)

with Lagrangian L that describes the dynamics of thedroplets. We now require the dissipative forces, FD,to be functions of the generalized coordinate velocities.This is a reasonable assumption, as particle collisionsare assumed to be the mechanism for the dissipation ofenergy as an object moves through a fluid. Followingthe convention of Jose and Saletan in their ClassicalDynamics text9, assume the ith force will be composedexplicitly of the velocity and implicitly in another

function gi(vi):

FD = −gi(vi)vi. (52)

This assumption is useful, since different order regimesof the Reynolds number will depend on different powersof the velocity. Substituting (52) into the dissipativeforces gives us

Dα = −N∑i=1

gi(vi)vi ·∂vi∂qα

. (53)

To solve for the dot product term on the right hand side,we note that for some variable β,

∂

∂β(vi · vi) = 2vi ·

∂vi∂β

(54)

and

∂

∂β(vi · vi) =

∂

∂β

(v2i

)= 2vi

∂vi∂β

. (55)

Thus, we can show that

vi ·∂vi

∂β= vi

∂vi∂β

. (56)

With (56) we may write our dissipative term in the form

Dα = −N∑i=1

gi(vi)vi∂vi∂qα

. (57)

Now we define the Rayleigh dissipative function F , suchthat

Dα ≡ −∂F∂qα

. (58)

Then we may state this function explicitly, using anintegration variable z for the velocity:

F =

N∑i=1

∫ vi

0

gi(z)zdz. (59)

The modified Euler-Lagrange equation (51) now reads

0 =d

dt

∂L

∂qα− ∂L

∂qα+∂F∂qα

=d

dt

∂L

∂qα− ∂L

∂qα+

∂

∂qα

N∑i=1

∫ vi

0

gi(z)zdz.

(60)

A Water Fountain 9

For high Reynolds number, we found the drag force to be

F ' 1

2cDρairπ`

2v2. (61)

The drag coefficient8 cD is approximately cD ' 12 , the

density of air10 is about 1.2 kg m−3, and the criticallength ` ' 3.6 · 10−3 m, as found in section II A 3. Now,our force is

F ' 1.2π

4

(3.6 · 10−3

)2v2

' (1.22 · 10−3v2)N.(62)

Using the general form derived above and definingα0 ≡ 1.22 · 10−3, we define our g function to be

g(vi) ≡ α0vi, (63)

keeping in mind that only one velocity term is absorbedinto the g function. The modified Euler-Lagrangeequation is now

0 =d

dt

∂L

∂qα− ∂L

∂qα+

∂

∂qα

N∑i=1

∫ vi

0

gi(v′i)v′idv′i

' d

dt

∂L

∂qα− ∂L

∂qα+ α0

∂

∂qα

∫ v

0

v′2dv′

' d

dt

∂L

∂qα− ∂L

∂qα+ α0

∂

∂qα

(v3

3

)' d

dt

∂L

∂qα− ∂L

∂qα+ α0

(v2 ∂v

∂qα

).

(64)

The Lagrangian for this system will be of the form

L =1

2m · (qα)

2 − V (qα). (65)

2. Single Particle Trajectory

We now consider droplets in the high Reynolds numberregime that have left the spatial domain where dropletinteractions are more frequent. Additionally, we assumethat the total height of the jet is not great enough forthere to be any significant difference in the gravitationalpotential difference between the top and the bottom ofthe jet. Under these assumptions, we can project thesystem into a azimuthally symmetric distribution. Let x

be the horizontal radial component and let z be the ver-tical component. The Lagrangian for this configuration is

L =1

2m(x2 + z2

)−mgz. (66)

We will have two modified Euler-Lagrange equationsfor each coordinate. Let α ≡ α0

m . The Euler-Lagrangeequations are

x = −α(x2 + z2

) ∂

∂x

(x2 + z2

)1/2(67)

and

z = −g − α(x2 + z2

) ∂∂z

(x2 + z2

)1/2. (68)

These equations result in

x = −2αx(x2 + z2

)1/2(69)

and

z = −g − 2αz(x2 + z2

)1/2. (70)

Combining these equations gives the result

z −(z

x

)x = −g. (71)

3. Maximum Height

With the differential equation found above, we canestimate the maximum height of a water droplet fromthe fountain. First we shall introduce scaling factors Sx,Sz, and St which give

SzS2t

¨x− SzSx

SxS2t

(˙z˙x

)¨x = −g

SzS2t

[¨x−

(˙z˙x

)¨x

]= −g.

(72)

For this equation to hold true, the scale on the left mustbe balanced by the gravitational acceleration by themethod of dominant balance. Thus

SzS2t

' g, (73)

A Water Fountain 10

and therefore

Sz ' gS2t

' 9.8S2t .

(74)

The height we expect a particle to reach is approximatelyone order of magnitude larger than the square of the timespent on the trajectory.

4. Energy loss due to dissipative forces

With the modified Euler-Lagrange equations, we canfind the change in energy of the system. The Hamil-tonian is the Legendre transform of the Lagrangian,and it represents the total energy of the system whenconserved. Any change in the Hamiltonian over timemust be due to some gain or loss in the energy of thedroplets. Using Einstein’s index notation to sum over re-peated indices, we write the change in energy over time as

dE

dt=

d

dt

(qα∂L

∂qα−L

)= qα

∂L

∂qα+ qα

d

dt

∂L

∂qα− dL

dt

= qα(d

dt

∂L

∂qα− ∂L

∂qα

)= −qα ∂F

∂qα

= −qα ∂

∂qα

N∑i=1

∫ vi

0

gi(z)zdz.

(75)

Then the energy dissipation for our system is

dE

dt= −α0

3

∂v3

∂qα

= −α0

3

(∂

∂x+

∂

∂z

)(x2 + z2

)3/2= −α0 (x+ z)

(x2 + z2

)1/2.

(76)

III. COMPUTATIONAL MODEL

Using the force equations derived from section II, weconstruct a computational model, written in C++, thatsolves for the motion of discrete particles. To confirm theagreement of our model with the rough predictions madein section II A 4, we simulated a droplet falling from restand determined the terminal velocity. The radius of the

water droplet was set to rw = 2 · 10−3m, whereas ourlength scale in section II A 3 was ` ≈ 3.6 · 10−3m. Thecomputational simulation yielded a value of 13.1m s−1,in close agreement with the theoretically predicted valueof 12.6m s−1 found in II A 4.

In the model, we introduce stochasticity to the deter-ministic trajectories by adding noise terms to the angle ofejection, the particle size, and the velocity. The motiva-tion for this “front-loaded stochasticity” method comesfrom the geometry of the jet. As the radius of the coneincreases, the cross-section particle density, shown in Fig-ure 5, goes like r−2. This is because mass is conserved asthe ring expands, and the cross-sectional area increaseslike r2. In addition, particles deviate slightly from theoriginal trajectory determined by φ0, the initial angle ofthe cone.

FIG. 2. A representation of the distribution along a comovingreference frame with respect to the conic sheet.

We ran large numbers of trajectories using the C++ pro-gram, systematically varying those parameters which wetook to be related to the design of the fountain and ob-serving the way in which the particle trajectories behavedwhen constrained by the physics. Specifically, we variedthe mean droplet radius, which depends on the nozzlegeometry3, the initial velocity magnitude, which dependson the flow rate, and the initial velocity angle with re-spect to the vertical axis, which depends on the angle ofthe nozzle aperture.

A. Initializing and evolving deterministic trajectories

Our model simulates the trajectory of individual par-ticles moving under the influence of drag, buoyancy, andgravity, with adjustable parameters for the initial size,initial velocity, and initial angle. Once again, buoyancywas neglected. The randomness in the initial states ofthe droplets was embedded into the dynamics by usinga Guassian distribution for the angle and initial velocityof the droplets.

A Water Fountain 11

We used and compared two seperate distributionsfor the radius of the particles. The first was a sim-ple Gaussian and the second was the Rosin-Rammlerdistribution11. The latter distribution is an applicationof the Weibull distribution, and it accounts for the factthat droplets whose radii are smaller than the optimalvalue found by the Weber number are more probablethan those whose radii are larger by the same degree.The peak of the drop radius distribution was centeredusing this optimal value, as calculated in section II A 3.The peak of the initial velocity and angle distribution canbe chosen arbitrarily, as these are parameters controlledby the observer.

The standard deviation in the radius Gaussian waschosen in order to compare our models with those of othergroups in the literature.1112 The distribution of particlesizes and velocities result from a number of competingfactors and dynamical effects.

Since the equations of motion are fairly simple, we usea moderate timestep of ∆t = 0.05s for integration in or-der to run larger numbers of trajectories. We employedthe forward Euler method of integration, which is tradi-tionally a very fast, albeit dirty method. This methodcan be justified since the error for forward Euler is pro-portional to ∆t2, and total error accumulation will beinsignificant since we are not integrating for long-timetrajectories. At every timestep, the future position of thedroplet was calculated by updating the velocity using theequations of force, then using this velocity to extrapolatethe trajectory to the next time step. Since drag is theonly force that varies with particle velocity, this was theonly force term updated at every step.

B. Statistical analysis of trajectories

Each design-related parameter was systematically con-trolled throughout a series of simulations. The numberof particles simulated was made large enough to createa realistic distribution for each run. During the simula-tion, the radial positions and times of impact correspondto the position and time data of particles as their heightdrops below zero. The radial probability was found bycreating “bins” which keep a count of the number of par-ticles that fall in a certain interval, or bin size, δr. Thisbin size can be specified to get varying levels of reso-lution. We choose to have a resolution on the order ofcentimeters, since the width of the jet is assumed to beon the order of this length.

In order to show the difference between the two par-ticle size distributions chosen for the model, we ran twosimulations of 1·105 particles. The particle radius was ini-tialized using either the Gaussian or the Rosin-Rammlerdistribution. Initial velocities were distributed over aGuassian centered at v = 20m s−1, with a standard devi-ation of σv = 3m s−1. The initial angles were distributedover a Gaussian with a peak at θ = 35, and a standarddeviation of σθ = 4 respectivly.

For these simulations, the particle distribution func-tion is the only independent variable. The adjustableparameter in the Rosin-Rammler distribution was set tounity. Both distributions were centered at r = 3.6mm,a value chosen because of the results of both the sectionII A 3 theory and the experiment in IV A. The resultsof the simulations can be found in Figures 6 and 7 inAppendix B. The median radius for a fountain with aGaussian distribution was found to be 13.35m, with a3.31m root-mean-square deviation from the original tra-jectory. The Rosin-Rammler distribution under the sameconditions gave a median radius of 12.25m with an RMSdeviation of 2.06m. We choose to use the Rosin-Rammlerdistribution for the radii in the following simulations.

C. Model of N particles moving with varying initialparameters

Our next step was to find the median radius as a func-tion of the initial angle of the jet. We started our simu-lation at θ = 10 and iterated in steps of 2 to θ = 70.At each angle we simulated 1 · 104 particles with the ve-locites distributed about 20m s−1, the radii about 3.6mm,and the angle distribution was centered about the chosenangle.

FIG. 3. The median radius in meters plotted against the ini-tial angle in radians. The error bar for each point is the RMSdeviation for each landing distribution (see Appendix Fig.6).We observe a region of maximum median radius between 33

and 42.

IV. EXPERIMENTAL METHODS

In order support the validity of the predictions madein section II A 3, we assembled an apparatus to createa distribution of water droplets moving downwards in agravitational field. A container of water was fastened ata known height, h, above the ground. Water released ina jet from height h formed droplets at a variable distancefrom the nozzle depending on the flow rate.

A Water Fountain 12

FIG. 4. Plots of the radial distributions as the initial angle of the fountain is varied. The velocity for all trajectories wascentered at 20m s−1

A. Drop radius experiment

By measuring the drop count rate, the volumetric flowrate, and the time spent in a single trial, it is possibleto establish a mean volume for each water droplet ina constant dripping flow. We measured the drop rateby hand while the apparatus was allowed to leak into abeaker.

FIG. 5. The apparatus used to form the droplets measures inthe experiment.

To balance noise in the drop count, we utilized the lawof large numbers and performed trials for one minute ata time. With the drop count rate on the order of s−1,and the total counted number on the order of 100 countsper trial, reasonable averages were achieved over severaltrials.

Since it is likely that the nozzle has a great deal ofinfluence over the size of the droplets formed, we con-ducted trials where the nozzle produced a stream that

broke up into droplets during flight. In order to mea-sure the high drop rate created by this configuration, weused the SlowPro video app in conjunction with iPhonevideo capability. With the app, we took a recording ofthe droplets hitting a hollow cup and slowed it to 25% ofthe original frame rate. For our high rate runs, we wereable to measure up to a drop rate of 7 s−1.

Our lower bound drop rate was 2.73 s−1. At this droprate, 25ml, or 25 ·10−6m−3, of water accumulated in 60s.The volume per drop was found to be 0.152 · 10−6m−3.Approximating these drops as spheres, the volume ofthe drop is equal to

V =4

3πr3, (77)

and using this formula, we find the average droplet ra-dius to be 0.0034m, which is in close agreement to thetheoretical radius of highest probability found in sectionII A 3.

For the upper bound drop rate of 7s−1, 65 · 10−6m−3

of water accumulated in 60s. The volume per dropletwas found to be 0.155 · 10−6m−3, yielding a volume of0.0033m. We see that in this experiment, doubling thedrop rate had little to no effect on the mean droplet ra-dius, and we beleive that this strongly hints at the sig-nificance of the optimal radius of a droplet found by bal-ancing the Weber number.

V. ANALYSIS OF RESULTS

We used a first principles view of fluid mechanicsand Lagrangian formalism to motivate a computationalmodel for a conic fountain. What the computational sim-ulations for this system show is that the radial distribu-tion of the jet flow is complicated by the simultaneoustransport of large, inertial particles and small, indeter-ministic ones.

A Water Fountain 13

One interesting quirk of this fountain stems from thesignificance of the drag force, the relevance of which isrelated to the mass to area ratio of the water droplets.The subtlety of this relation is shown in the change in thespread of the jet, shown by the RMS deviation, as theparticle size is varied. For small particles, drag forcesdominate the trajectory, but they do so in such a waythat the particles don’t travel very far. This means thatthe spread of a jet of small particles is limited to thespeed of diffusion.

As the radius of the droplet is increased, drag forcesbecome less effective. The RMS deviation increases asthe particles travel further, since the spread of the jet isno longer limited to the speed of diffusion. However, thedrag force is also the main mechanism for creating thespread in the distribution, and decreasing the magnitudeof the acceleration caused by drag forces has the effect ofdecreasing the spread.

The competition of these effects creates a specific par-ticle radius value that generates the maximum possiblespread. Amazingly, computational simulations show thatthis optimal radius happens to be the radius predictedin section II A 3, which is also the most probable ra-dius found in the droplet experiment in section IV A.It is clear that droplet size is a critical component in thespread of a fountain, and nature seems to prefer its waterdroplets to be approximately 3 mm in radius.

In addition to this, we show that the deviation of thejet of a fountain is also heavily dependent on the angleat which the jet starts. For fountains with very narrowangles, the drag force acting on the particles only servesto reduce the total height of the fountain. When theangle is wide, however, the drag force slows small dropletsdown much moreso than the larger ones. This causesthe distance between the landing coordinate of the smalldroplets to be much shorter than those of the large ones,creating a slightly inward spread of the distribution.

The key weakness of our approach lies in our neglect ofextremely small droplets and water vapor. For the bulk ofthe calculations, we assume that the trajectories of parti-cles behave inertially, and that the intermolecular forcesmanifest themselves through perturbations in these tra-jectories. This argument breaks down for droplets thatare too small to overcome viscous forces in the fluid.

To create a correction for this oversight, the distribu-tion of the smallest droplets would have to be predictedprobabilistically. A certain amount of mist will be gen-erated at each point along the trajectory of the jet, andthis amount will likely be proportional to the velocity ofthe jet, to the droplet density, and in small part to thevapor pressure of the surrounding fluid. As this mist isgenerated, most of it will diffuse uniformly. The heightof the jet at the point where the mist is created will de-termine how far the mist diffuses, since the mist at thetop of the arc will be in the air for the longest period.

The rate of mist accumulation at a given δr can be cal-culated by integrating over the contributions from eacharc length segment of the jet trajectory. The contribu-

tions of one element of the jet would be proportional tothe velocity at that point and the density, since vaporpressure would be approximately constant throughout.The contribution of this misting distribution functionwould then be added to the final radial distribution.

VI. CONCLUSION

In conclusion, we analyzed the problem of a conic liq-uid jet using stochastic methods derived from the equa-tions of motion for particles moving in an incompress-ible fluid. The symmetry of the conic jet is broken bysurface instabilities introduced by the nozzle and fluctu-ations in the molecular interactions. Taking advantageof the probabilistic behavior of large numbers of weaklyinteracting droplets, we found that noisy fluctuations inthe trajectories of these particles can be related to theenergy of the flow that is lost to dissipative forces.

REFERENCES

1L. Rayleigh, “On the instability of jets,” in Proceedings of theLondon Mathematical Society, Vol. 10 (1878).

2J. York, H. Stubbs, and M. Tek, “The mechanism of disintegra-tion of liquid sheets,” Trans. ASME 75, 1279–1286 (1953).

3e. a. Z Wang, “Experimental and theoretical investigation ofa hollow conic liquid jet atomization,” in 31st AIAA/AS-ME/SAE/ASEE Joint Propulsion Conference and Exhibit(1995).

4S. P. Lin and R. D. Reitz, “Drop and spray formation from aliquid jet,” Annual Review of Fluid Mechanics 30, 85–105 (1998),http://dx.doi.org/10.1146/annurev.fluid.30.1.85.

5A. Frohn, Dynamics of droplets (Springer, Berlin New York,2000).

6P. S. marquis de Laplace, Trait de Mcanique Cleste, Vol. 4 (1805).7N. Lange, Lange’s Handbook of chemistry (McGraw-Hill, NewYork, 1973).

8B. McCormick, Aerodynamics, aeronautics, and flight mechanics(Wiley, New York, 1979).

9J. Jose and E. Saletan, Classical Dynamics: A ContemporaryApproach (Cambridge University Press, 1998) pp. 129–130.

10Manual of the ICAO standard atmosphere extended to 80 kilo-metres (262 500 feet) = Manuel de l’atmosphere type OACI :elargie jusqu’a 80 kilometres (262 500 pieds) = Manual de laatmosfera tipo de la OACI : ampliada hasta 80 kilometros (262500 pies (ICAO, Montreal, 2002).

11P. of Droplet Size and V. D. in Droplet Formation Region ofLiquid Spray, Entropy , 1490–1495 (2010).

12B. Tsinghua University, “Numerical simulation study on the ef-fects of fountain on around thermal environment,” InternationalBuilding Performance Simulation Association , 3–4 (2007).

13M. C. Y. E. F. Goedde, “Experiments on liquid jet instability,”Journal of Fluid Mechanics 40 (1970).

14J. Plateau, Statique exprimentale et thorique des liquides soumisaux seules forces molculaires (Gauthier-Villars, 1873).

Appendix A: Navier-Stokes

Here we derive Navier-Stokes as used in this paperfollowing the conventions and methods of Jose and

A Water Fountain 14

Saletan9. Consider a region Ω(t) of a fluid. Wewill concern ourselves with the result of applyingNewton’s second law to a fixed mass MΩ within thismoving volume Ω(t). Newton’s second law may be stated

FΩ = PΩ, (A1)

where FΩ is the total force on this region Ω(t), andPΩ is the total momentum in Ω(t). Assume there areno external forces on the body, and name a parameterp(x, t) the pressure in some region at some time. Thenforce is then

FΩ = −∮∂Ω

pdΣ, (A2)

where dΣ is the outward normal surface element alongthis volume. By the divergence theorem, we may writethis closed loop integral as

FΩ = −∫

Ω

∇pd3x. (A3)

The momentum PΩ on the other hand may be writtenas the integral of the momentum density in this regionρv stated as

PΩ =

∫Ω

ρvd3x. (A4)

The time derivative of this integral is equal to the force.For some scalar function W (x, t), this derivative wouldbe

d

dt

∫Ω

W (x, t)d3x =

∫Ω

[∂W

∂t+∇ · (Wv)

]d3x. (A5)

Write this W function to be

W ≡ ρf(x, t). (A6)

Then we have the relation

d

dt

∫Ω

ρfd3x

=

∫Ω

ρ∂f

∂t+ f

[∂ρ

∂t+∇ · (ρv)

]+ ρv · ∇f

d3x.

(A7)

By the conservation of mass we know

∂ρ

∂t= −∇ · (ρv) , (A8)

so we’re left with

d

dt

∫Ω

ρfd3x

=

∫Ω

ρ∂f

∂t+ ρv · ∇f

d3x.

(A9)

We may regard the right hand side in terms of thesubstantial derivative

Dt =∂

∂t+ v · ∇, (A10)

which means

d

dt

∫Ω

ρfd3x =

∫Ω

ρDtfd3x, (A11)

and equation known as the transport theorem. Withthis theorem under the constraint of Newton’s secondlaw we may reach Euler’s equation for an ideal fluid:

FΩ = −∫

Ω

∇pd3x =

∫Ω

ρvd3x = PΩ

−3∑i=1

∂p

∂i=

3∑i=1

ρ

(∂vi∂t

+ v · ∇vi)

−∇p = ρ

[∂v

∂t+ (v · ∇)v

].

(A12)

In light of the substantial derivative,

−∇ρ = ρDtv. (A13)

With the continuity equation we may express thisequality in terms of the momentum-stress tensor of thefluid

∂

∂t(ρvi) +

∂

∂kΠik = 0, (A14)

where

Πik = pδik + ρvivk. (A15)

We may interpret this result as a conserved currentdensity. Assume our fluid is incompressible. Then thedensity ρ does not depend on the pressure p. Also we

A Water Fountain 15

have

∂ρ

∂t= 0 (A16)

and

∇ρ = 0. (A17)

With these constraints we get

∇ · v = 0. (A18)

Introduce the viscosity stress tensor such that

Πik = pδik + ρvivk − σ′ik ≡ σik + ρvivk. (A19)

Taylor expansion to first order gives this viscosity stresstensor as

σ′ik = µ

(∂vi∂k

+∂vk∂i− 2

3δik

∂vj∂j

)+ ξδik

∂vj∂j

, (A20)

where µ is the viscosity. Then our conserved currentdensity equation reads

ρ∂vi∂t

= − ∂

∂k

[pδik + ρvivk − νρ

(∂vi∂k

+∂vk∂i

)]= −∂p

∂i− ρvk

∂vi∂k

+ νρ∂2vi∂k2

.

(A21)

Sum over i and repeated indices to give the Navier-Stokes equation

ρ∂v

∂r+ (v · ∇)v = −∇p+ µ∇2v. (A22)

Appendix B: C++ Numerical Integration Program

The software used to simulate these particles is calledROOT and helped us make these pretty graphs.

FIG. 6. The distribution of the landing distance of 100,000particles where the radii randomized under the Rosin-Rammler distribution with the parameters stated in sec-tionIII B. Assuming azimuthal independence the radius ofmedian circle is just the median of this plot which was 12.25mwith an RMS of 2.06. The average time spent by each particlespent in the air is ≈ 1.5sec.

FIG. 7. The distribution of the landing distance of 100,000particles where the radii randomized under the Gaussian dis-tribution with the parameters stated in sectionIII B. Assum-ing azmuthal independence the radius of median circle is justthe median of this plot which was 13.25m with an RMS of3.31 The average time spent by each particle spent in the airis ≈ 1.6sec

#include <iostream>#include <fstream>#include <s t r i ng>

A Water Fountain 16

#include <TROOT. h>#include <TCanvas . h>#include <TCutG. h>#include <TMath . h>#include <TMultiGraph . h>#include <TGraph . h>#include <TGraph2D . h>#include <TF3 . h>#include <TVector3>

// Doub le t time = DT * NUM STEPS;//////PARTICLE CLASS/*Each d r o p l e t w i l l have a p o s i t i o n vector , v e l o c i t y vector , and f o r c e v e c t o r−an i n i t i a l ang le− a rad ius

*/class P a r t i c l e protected :

Double t pos [ 2 ] ;Double t v e l [ 2 ] ;Double t f o r c e [ 2 ] ;Double t theta ;// Doub le t *out [ 2 ] ;

public :// P a r t i c l e ( ) ;void setPos ( Double t x , Double t y ) ;void s e tVe l ( Double t vx , Double t vy ) ;void s e tForce ( Double t fx , Double t fy ) ;void setTheta ( Double t ang le ) ;Double t getPosx ( ) ;Double t getPosy ( ) ;Double t getVelx ( ) ;Double t getVely ( ) ;Double t getForcex ( ) ;Double t getForcey ( ) ;Double t getTheta ( ) ;

;

void P a r t i c l e : : setPos ( Double t x , Double t y ) pos [ 0 ] = x ;pos [ 1 ] = y ;

void P a r t i c l e : : s e tVe l ( Double t vx , Double t vy ) ve l [ 0 ] = vx ;v e l [ 1 ] = vy ;

void P a r t i c l e : : s e tForce ( Double t fx , Double t fy ) f o r c e [ 0 ] = fx ;f o r c e [ 1 ] = fy ;

void P a r t i c l e : : setTheta ( Double t ang le ) theta = angle ;

Double t P a r t i c l e : : getPosx ( )

Double t out = pos [ 0 ] ;return out ;

Double t P a r t i c l e : : getPosy ( ) return pos [ 1 ] ;

Double t P a r t i c l e : : getVelx ( ) Double t out = ve l [ 0 ] ;

A Water Fountain 17

return out ;

Double t P a r t i c l e : : getVely ( ) return ve l [ 1 ] ;

Double t P a r t i c l e : : getForcex ( ) Double t out = f o r c e [ 0 ] ;return out ;

Double t P a r t i c l e : : getForcey ( ) return f o r c e [ 1 ] ;

Double t P a r t i c l e : : getTheta ( ) return theta ;

// /////////////////////END OF PARTICLE CLASS

#include ” P a r t i c l e . h”

using namespace std ;

#define LIFT( r , rho ) ( (4/3) * 3.14159 * ( r * r * r ) * ( rho ) * 9 . 8 1 ) ;#define DRAGH( r , dCoeff , v e l o c i t y , rho ) ( . 5 *( rho ) *( v e l o c i t y * v e l o c i t y ) * dCoef f * 3.14159 * ( r * r ) ) ;//#d e f i n e#define DT . 0 0 1 ;#define NUM STEPS10 ;#define MASS( r , rho ) ( (4/3) * 3.14159 * ( r * r * r ) * ( rho ) ) ;#define V E( eg ,m) ( s q r t (2 * ( eg ) / m) ) ;

//Update s h i t#define NEW VEL(v , F ,M, t ) ( v + ( (F* t ) / M) ) ;#define NEW POS(x , v , t , a ) ( x + ( v* t ) + ( . 5 * a * t * t ) ) ;

#define NUM PART 10000;#define SNR s q r t (NUM PART) ;#define PI 3 . 14159 ;#define RAD . 0 0 3 6 ; // mm#define VEL 10 ; // m/ s#define ANGLE ( PI* 7 / 3 6 ) ;#define SIGMA ( PI / 6 0 ) ;#define SIGMAR . 0 0 0 2 ; // .5mm#define SIGMAV 3 ; // 7 m/ s

//STEP SIZES ! ! ! ! ! !#define DTH( PI / 9 0 ) ;#define DR . 0 0 0 2 ; //10ˆ−4 meter increment#define DV 2 ;

//AMOUNT PARAMETERS WILL BE CHANGED#define NUM ANGLES 30 ;#define NUM RAD 1 ;#define NUM VEL 20 ;

#define FITPARAM 1 ;

int System ( ) Double t rho a i r , rho water , dCoeff , r ad iu s ;Double t F l , F g , F dx , F dy ;Double t Energy , r 0 , v i ; // found using .002 rad iusDouble t x 0 , y 0 , v i ;Double t fdrag , dt , time , dh , t s c a l e ;

Double t x , y , nvy , nvx , v e l ;Double t M w;

A Water Fountain 18

Double t nbins = 200 ;

rho water = 1000 ; // kg*mˆ−3r h o a i r = 1 . 2 ; // kg*mˆ−3dCoef f = . 4 7 ; // drag c o e f fdt = . 0 5 ;dh = 1 ;t s c a l e = 15 ;time = 0 ;t o t a l t = 0 ;

// r 0 = . 0 0 2 ; // l e n g t h s c a l e//TGraph2D * p l o t = new TGraph2D ( ) ;c0 = new TCanvas ( ”c1” , ” mult igraph L3” ,200 , 10 , 700 , 500 ) ;c0−>SetFrameFi l lColor ( 3 0 ) ;TGraph *g1 [NUM PART] ;P a r t i c l e *ap [NUM PART] ;// const I n t t param ;

// ////////////////////////////////////////////////////ALLL THE ERRORBAR PARAMETERS;Double t MEDIAN[NUM VEL] ;// Doub le t Angle [ param ] ;Double t Ve loc i ty [NUM VEL] ;Double t ex [NUM VEL] ;Double t RMS[NUM VEL] ;Double t peak = 0 ;

//TMultiGraph *mg = new TMultiGraph ( ) ;//TH1D *h1= new TH1D(” h1 ” ,” Postion1 ” ,400 ,0 ,30) ;//h1−>GetXaxis()−>S e t T i t l e (” Pos i t ion ” ) ;/// p l o t s landning p o s i t i o n s

TRandom * angleGenerator = new TRandom ( ) ;TRandom * rad iusGenerator = new TRandom ( ) ;TRandom * ve l o c i tyGene ra to r = new TRandom ( ) ;

for ( int th = 0 ; th < NUM VEL; th++)

TH1D *h1= new TH1D( ”h1” , ” Post ion1 ” , nbins , 0 , 1 0 0 ) ; // r e s t a r t histogram each run//VARYING ANGLEDouble t theta = ANGLE; //+ ( th * DTH) ;// cout << ”Theta ”<< th << ” = ” << t h e t a << end l ;

// v i = ve loc i tyGenera tor−>Gaus(VEL,SIGMAV) ; // m/ s//Energy = .5 * (4/3) * PI * ( r 0 * r 0 * r 0 ) * rho water * ( v i * v i ) ; // Energy s c a l e

//VARYING RADIUS at 35 degreesDouble t rads = RAD; //+ ( r * DR) ;TF1 * f 1 = new TF1( ” f1 ” , ”RR( x ) ” , rads , 4 * ( rads * . 2 ) ) ;

Double t ave l = VEL + ( th * DV) ;cout << ” v e l o c i t y o f : ” << ave l ;/* rad ius = radiusGenerator−>Gaus(RAD,SIGMAR

M w = MASS( radius , rho water ) ;// f i n d i n g v e l c o i t y to match Energy s c a l e// v i = V E( Energy ,M w) ;// cout << ” V e l o c i t y at rad ius ” << rad ius << ” i s :” << v i << end l ;F g = LIFT( radius , rho water ) ;F l= LIFT( radius , r h o a i r ) ; */// cout << ”Radius ” << r << ” = ” << rads << end l ;

// /////////////////////////////////////////// D i s t r i b u t i o n o f a n g l e sfor ( int i = 0 ; i < NUM PART; i++) ap [ i ] = new P a r t i c l e ( ) ;ap [ i ]−>setTheta ( angleGenerator−>Gaus ( theta , SIGMA) ) ;v i = ve loc i tyGenerator−>Gaus ( avel ,SIGMAV) ; // random v e l o c i t y m/ s

A Water Fountain 19

ap [ i ]−> s e tVe l (TMath : : Cos ( ap [ i ]−>getTheta ( ) ) * v i ,TMath : : Sin ( ap [ i ]−>getTheta ( ) ) * v i ) ;// cout << ap [ i ]−>getTheta () << end l ;// /////////////////////////////////////////

for ( int j = 0 ; j < NUM PART; j++)g1 [ j ] = new TGraph ( ) ; // make new graphx 0 = 0 ; y 0 = 0 ; // ( j * dh ) ;// make a p a r t i c l eap [ j ]−>setPos ( x 0 , y 0 ) ; // S t a r trad iu s = radiusGenerator−>Gaus ( rads , rads * . 2 ) ; //random rad ius mi f ( rad iu s > rads ) // Re− f i t// cout << ” Before ” << rad ius << end l ;rad iu s = f1−>GetRandom ( ) ;// cout << rad ius << end l ;M w = MASS( radius , rho water ) ;F g = LIFT( radius , rho water ) ;F l= LIFT( radius , r h o a i r ) ;//ap [ j ]−>setTheta (TMath : : ATan( ap [ j ]−>ge tVe ly ()/ ap [ j ]−>ge tVe lx ( ) ) ) ;// cout << ” s t a r t r = ” << rad ius << end l ;

// ///////////////////////////////////////////////////////////////////////////////////////////*Drop le t s are s e p e r a t e and non−i n t e r a c t i n g with i n i t i a l p o s i t i o n s and v e l o c i t i e s*/

for ( int i = ( j * t s c a l e /dt ) ; i < ( ( j +1) * t s c a l e /dt ) ; i++)i f ( ap [ j ]−>getPosy ( ) >= 0 ) // /////////Moving upi f ( ap [ j ]−>getVely ( ) > 0) // cout << ap [ j ]−>ge tVe ly () << end l ;//g1 [ j ]−>SetPoint ( ( I n t t ) i , ap [ j ]−>getPosx ( ) , ap [ j ]−>getPosy ( ) ) ; // p l o t current p o s i t i o n// ////////////////////////////UPDATE FORCESve l = s q r t ( ap [ j ]−>getVelx ( ) * ap [ j ]−>getVelx ( ) + ap [ j ]−>getVely ( )* ap [ j ]−>getVely ( ) ) ; // v e l o c i t y// cout << ” V e l o c i t y o f : ” << v e l << end l ;fd rag = DRAGH( radius , dCoeff , ve l , r h o a i r ) ;F dx = fdrag * TMath : : Cos ( ap [ j ]−>getTheta ( ) ) ;F dy = fdrag * TMath : : Sin ( ap [ j ]−>getTheta ( ) ) ;i f ( F dx >= .00001 | | ap [ j ]−>getVelx ( ) > 0 ) // non−zero x−f o r c efx = −1 * F dx ; else fx = 0 ;fy = −1 * ( F dy + F g ) ;// /////////////////////////// cout << ” f o r c e =:” << f y << end l ;// cout << ” t h e t a =: ” << ap [ j ]−>getTheta ( ) ;// cout << ap [ j ]−>getTheta () << end l ;i f (TMath : : Abs ( fy ) > . 00001 | | ap [ j ]−>getVely ( ) > 0) // non−zero y f o r c eap [ j ]−> s e tForce ( fx , fy ) ;// cout << time << end l ; else // termina l v e l o c i t yap [ j ]−> s e tForce ( fx , 0 ) ;fy = 0 ;// //////////////////////////// UPDATE VELOCITYi f ( fx == 0) nvx = 0 ; else nvx = NEW VEL( ap [ j ]−>getVelx ( ) , fx ,M w, dt ) ;

i f (TMath : : Abs ( fy ) > . 0000000001) nvy = NEW VEL( ap [ j ]−>getVely ( ) , fy ,M w, dt ) ;// cout << nvx << end l ;

A Water Fountain 20

ap [ j ]−> s e tVe l ( nvx , nvy ) ;// ////////////////////////

// /////////////////////////////////////// UPDATE POSITIONx = NEW POS( ap [ j ]−>getPosx ( ) , ap [ j ]−>getVelx ( ) , dt , fx ) ;y = NEW POS( ap [ j ]−>getPosy ( ) , ap [ j ]−>getVely ( ) , dt , fy ) ;ap [ j ]−>setPos (x , y ) ;// //////////////////////////////

// UPDATE ANGLEap [ j ]−>setTheta (TMath : : ATan( (TMath : : Abs ( ap [ j ]−>getVely ( ) ) ) / ap [ j ]−>getVelx ( ) ) ) ;time += dt ;// //////////////////////////////////////////////// /////////////////////////////////////////////// /////////Moving downi f ( ap [ j ]−>getVely ( ) <= 0) // cout << ap [ j ]−>ge tVe ly () << end l ;//g1 [ j ]−>SetPoint ( ( I n t t ) i , ap [ j ]−>getPosx ( ) , ap [ j ]−>getPosy ( ) ) ; // p l o t current p o s i t i o n// ////////////////////////////UPDATE FORCESve l = s q r t ( ap [ j ]−>getVelx ( ) * ap [ j ]−>getVelx ( ) + ap [ j ]−>getVely ( )* ap [ j ]−>getVely ( ) ) ; // v e l o c i t y// cout << ” V e l o c i t y o f : ” << v e l << end l ;fd rag = DRAGH( radius , dCoeff , ve l , r h o a i r ) ;F dx = fdrag * TMath : : Cos ( ap [ j ]−>getTheta ( ) ) ;F dy = fdrag * TMath : : Sin ( ap [ j ]−>getTheta ( ) ) ;i f ( F dx >= .00001 | | ap [ j ]−>getVelx ( ) > 0 ) // non−zero x−f o r c efx = −1 * F dx ; else fx = 0 ;fy = F dy − F g ;// /////////////////////////// cout << ” f o r c e =:” << f y << end l ;// cout << ” t h e t a =: ” << ap [ j ]−>getTheta ( ) ;// cout << ap [ j ]−>getTheta () << end l ;i f (TMath : : Abs ( fy ) > . 00001 | | ap [ j ]−>getVely ( ) > 0) // non−zero y f o r c eap [ j ]−> s e tForce ( fx , fy ) ;// cout << time << end l ; else // termina l v e l o c i t yap [ j ]−> s e tForce ( fx , 0 ) ;fy = 0 ;// //////////////////////////// UPDATE VELOCITYi f ( fx == 0) nvx = 0 ; else nvx = NEW VEL( ap [ j ]−>getVelx ( ) , fx ,M w, dt ) ;

i f (TMath : : Abs ( fy ) > . 0000000001) nvy = NEW VEL( ap [ j ]−>getVely ( ) , fy ,M w, dt ) ;// cout << nvx << end l ;ap [ j ]−> s e tVe l ( nvx , nvy ) ;// ////////////////////////

// /////////////////////////////////////// UPDATE POSITIONx = NEW POS( ap [ j ]−>getPosx ( ) , ap [ j ]−>getVelx ( ) , dt , fx ) ;y = NEW POS( ap [ j ]−>getPosy ( ) , ap [ j ]−>getVely ( ) , dt , fy ) ;ap [ j ]−>setPos (x , y ) ;// //////////////////////////////

// UPDATE ANGLEap [ j ]−>setTheta (TMath : : ATan( (TMath : : Abs ( ap [ j ]−>getVely ( ) ) ) / ap [ j ]−>getVelx ( ) ) ) ;time += dt ;

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//g1 [ j ]−>SetPoint ( ( I n t t ) j , ap [ j ]−>getPosx ( ) , 0 ) ;// cout << ” Pos i t ion at : ” << ap [ j ]−>getPosx () << end l ;// cout << time << end l ;h1−>F i l l ( ap [ j ]−>getPosx ( ) , 1 ) ;t o t a l t = t o t a l t + time ;time = 0 ;// p o s i t i o n s [ j ] = ap [ j ]−>getPosx ( ) ;// index [ j ] = j ;// //////////////////////////////////////////////////////////////////////////////////////////*mg−>Add( g1 [ j ] ) ;//g1 [ j ]−>Draw ( ) ;g1 [ j ]−>SetMarkerColor ( j ) ;g1 [ j ]−>SetMarkerSty le ( 7 ) ;g1 [ j ]−>SetLineColor ( j ) ; *///g1 [ j ]−>S e t T i t l e (”Y vs X” ) ;//g1 [ j ]−>GetXaxis()−>S e t T i t l e (”X Pos i t ion ” ) ;//g1 [ j ]−>GetYaxis()−>S e t T i t l e (”Y Pos i t ion ” ) ;// cout << F g << end l ;t o t a l t = t o t a l t / NUM PART;cout << ” Total time : ” << t o t a l t << endl ;for ( int a = 0 ; a < nbins ; a++) Double t temp = h1−>GetBinContent ( a ) ;h1−>SetBinContent ( a , temp/NUM PART) ;/*h1−>Draw ( ” ” ) ;//h1−>Fi t (” gaus ” ) ;h1−>FitPanel ( ) ;h1−>GetXaxis()−>S e t T i t l e (” Distance o f P a r t i c l e Travel ( meters ) ” ) ;h1−>GetYaxis()−>S e t T i t l e (” P r o b a b i l i t y ( Normalized to 1 ) ” ) ;h1−>S e t T i t l e (” Landing Distance D i s t r i b u t i o n With RR” ) ; */MEDIAN[ th ] = Median ( h1 ) ;// cout << ”MEADIAN IS : ” << Median ( h1 ) << end l ;// cout << ”RMS ” << h1−>GetRMS ( ) ;i f (MEDIAN[ th ] > peak ) peak = MEDIAN[ th ] ;cout << ”New Peak at : ” << peak << endl ;Ve loc i ty [ th ] = ave l ;ex [ th ] = 0 ;RMS[ th ] = h1−>GetRMS ( ) ;// d e l e t e h1 ;

// ////////////////////////////////////////////////////////////// cout << ” Total time i s : ” << time << ” ” << end l ;/*mg−>Draw(”APL” ) ;mg−>S e t T i t l e (”Y vs X” ) ;mg−>GetYaxis()−>S e t T i t l e (”Y Pos i t ion ” ) ;mg−>GetXaxis()−>S e t T i t l e (”X Pos i t ion ” ) ; *//*c0−>(2);TH1D *h1 = new TH1D(” h1 ” ,” Postion1 ” ,100 ,60 ,80) ;h1−>GetXaxis()−>S e t T i t l e (” Pos i t ion ” ) ;f o r ( i n t i = 0; i < NUM PART; i++) h1−>F i l l ( ap [ j ]−>getPosx ( ) ) ;*/TGraphErrors * gr = new TGraphErrors (NUM VEL, Veloc i ty ,MEDIAN, ex ,RMS) ;gr−>GetXaxis()−>S e t T i t l e ( ” Droplet Ve loc i ty m/ s ” ) ;gr−>GetYaxis()−>S e t T i t l e ( ”R median ( meters ) ” ) ;gr−>S e t T i t l e ( ”Median Fountain Radius vs I n i t i a l Ve loc i ty ” ) ;

gr−>SetMarkerColor ( 4 ) ;gr−>SetMarkerStyle ( 2 1 ) ;gr−>Draw( ”ALP” ) ;//c1 −> SaveAs (”HISTO. ps ” ) ;

// cout << ”MEADIAN IS : ” << Median ( h1 ) << end l ;cout << ”SNR : ” << SNR << endl ;return 0 ;

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double Median ( const TH1D * h1 ) // h t t p :// root . cern . ch/phpBB3/ v i e w t o p i c . php? f=3&t =7802

int n = h1−>GetXaxis()−>GetNbins ( ) ;s td : : vector<double> x (n ) ;h1−>GetXaxis()−>GetCenter ( &x [ 0 ] ) ;const double * y = h1−>GetArray ( ) ;// exc lude underf low / o v e r f l o w s from bin content array yreturn TMath : : Median (n , &x [ 0 ] , &y [ 1 ] ) ;

Double t RR( Double t x ) return = exp(−(TMath : : Power ( x/RAD, ( Double t ) FITPARAM) ) ) ;