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A1 horizontal substrate
创建日期：2024-04-20 15:33:24
所属分类：物理Physical
标签： A1
horizontal substrate

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Modelling the flow of a thin film of fluid over a

horizontal substrate
Introduction
The flow of thin films of viscous fluids, such as honey or syrup, can be effectively modelled
by identifying the dominant forces driving the flow and applying the approximations of
lubrication theory, which hold as long as the thickness of the layer of fluid is much smaller
than its horizontal extent. A schematic diagram of a typical thin film of viscous fluid,
of thickness H, spreading under gravity over a horizontal substrate is shown in figure 1.
Such flows are also referred to as viscous gravity currents, and are common in the world
around us, ranging from the spread of honey over morning toast, to lava flows and the
flow of glacial ice sheets [1, 5, 6, 7, for example].
Assuming that vertical shear stress provides the dominant resistance to the flow,
and that inertial effects and the effects of surface tension are negligible, the flow can be
shown to be governed by the following momentum balance equation
0 = −∇p− ρgez + µ
∂2u
∂z2
, (1)
where the velocity field u = (u, 0, 0) is horizontal, and ez = (0, 0, 1) is the unit basis
vector in the z-direction. Here, the pressure p and velocity u may, in general, depend
on all spatial coordinates (x, y, z) as well as on time t, whereas the density ρ and dy-
namic viscosity µ are constants which describe the material properties of the fluid. The
gravitational acceleration is denoted by g. Axes are chosen so that the fluid flows along
the x-direction, z = 0 describes the horizontal substrate, and z = H describes the upper
surface of the thin film of viscous fluid as shown in figure 1. The viscous fluid is supplied
at a specified flux at x = 0, and its leading edge is given by x = xN (t). The setup is
fully two-dimensional, with no variation along the y-direction.
Equation (1) is a version of the so-called Stokes equations, used to describe the flow
of viscous fluids.
xN(t)
H(x,t)

x
z
inflow
Figure 1: Schematic diagram illustrating the side profile of a thin film of viscous fluid
spreading under gravity over a rigid horizontal substrate.
1
Problem 1: hydrostatic pressure
Solve for the pressure by considering the z-component of (1) and applying the boundary
condition that the pressure attains its atmospheric value p0 at the upper surface of the
viscous fluid, that is
p = p0 at z = H. (2)
Problem 2: equation for the horizontal velocity
By substituting your expression for the pressure into the x-component of (1) and as-
suming that the atmospheric pressure p0 is constant, show that the horizontal velocity
u satisfies
∂2u
∂z2
=
ρg
µ
∂H
∂x
. (3)
Problem 3: solving the Stokes equations
Noting that the right-hand side of (3) is independent of z, integrate both side of (3)
twice in z in order to obtain an expression for u in terms of two arbitrary constants, A
and B.
Problem 4: velocity field
Solve for A and B by applying the boundary condition that the upper surface is stress
free, that is,
∂u
∂z
= 0 at z = H, (4)
and that the fluid does not slip at the contact line with the substrate, that is,
u = 0 at z = 0. (5)
The latter is also known as the no-slip condition.
Problem 5: depth-integrated flux
The depth-integrated flux of fluid per unit width is given by
q =
∫ H
0
u dz. (6)
After substituting in your determined values of A and B into your expression for the
velocity, show that the depth integrated flux is given by
q = −CHk
∂H
∂x
. (7)
where C and k are constants that you should determine.
2
Problem 6: Similarity solution
Along with the equation (7) specifying the flux q, the deformation of the upper surface
is governed by the mass conservation equation
∂H
∂t
= −
∂q
∂x
, (8)
along with the source flux boundary condition
q = Q at x = 0, (9)
the vanishing frontal thickness condition
H = 0 at x = xN (t), (10)
and the global mass conservation condition
∫ xN
0
H dx = Qt. (11)
By rescaling the surface height H as
H(x, t) = αtaF (η) (12)
where
η =
x
βtb
and xN = ηNβt
b, (13)
for some constant ηN , find the values of a, b, α and β so that the governing equations
(7) and (8) reduce to the single ordinary differential equation
1
5
F −
4
5
η
dF
dη
=
1
3
d
dη
(
F 3
dF
dη
)
, (14)
and recast the boundary and integral conditions (9)–(11) in terms of only F and its
derivatives. No physical constants, such as µ, ρ, Q, g, and ηN
1 and no time variable t
should appear in these reduced boundary and integral conditions. If any of these appear
in your boundary and integral conditions, that means that your values of a, b, α and
β should be adjusted. A similar, though differently scaled, derivation can be found in
[4] for Newtonian viscous fluids and in [8] for non-Newtonian, power-law fluids. A more
general treatment of self-similar problems, and ways of obtaining similarity solutions,
can be found in [2].
Note: the solution F is known as a similarity solution and the variable η is known
as a similarity variable. Note also that the front x = xN corresponds to η = ηN under
the transformation (13).
1
ηN will actually appear, but only in your transformed version of (10) and (11).
3
Problem 7: Asymptotic solution near the front
In the neighbourhood of the front η = ηN (or, equivalently, x = xN ), the thickness of the
liquid layer is small, while its slope is large. Find an asymptotic solution by following
the steps outlined below.
(a) By changing variables so that
η = ηN − ǫX, (15)
where ǫ is a small parameter, and rescaling F as F = ǫ1/3G, determine an ordinary
differential equation forG in terms ofX only (no η should appear in your equation).
(b) Identify any terms that are small in comparison to the remaining terms, when
ǫ→ 0.
(c) Formulate an approximate equation for G by deleting the small terms identified
above. Your approximate equation should have no ǫ’s and no η’s in it; just ηN and
G as a function of X. If you have any ǫ’s in your equation, that means that you
either did not delete all the small terms, or you did not divide both sides of the
equation by the right power of ǫ.
(d) Show that the solution to this approximate equation is of the form
G ∼ γXr (16)
where γ and r are constants that you should determine.
(e) Recast your approximate solution for G in terms of F and η. This is known as an
asymptotic solution, and it is valid near the front η = ηN .
(f) For what value of ηN does your asymptotic solution for F satisfy the source flux
boundary condition (the version of (9) recast in similarity coordinates for Problem
5)?
(g) Plot the asymptotic solution in Mathematica, or another program, as a function
of η over the domain [0, ηN ] for the value of ηN found in part (f), above.
A more general overview of asymptotics methods, and ways of obtaining asymptotic
solutions to differential equations, can be found in [3].
References

Project A2
A numerical treatment of similarity solutions
in the real world: thin film flows
Introduction
Many physical phenomena exhibit so-called self-similar behaviour, in which the physical
system grows or decays in time while retaining its general shape and structure, albeit
in rescaled (self-similar) coordinates. For such systems, any initial irregularities (as
formulated via initial conditions) eventually evolve towards a self-similar regime, in which
these initial irregularies become ‘forgotten’ with time. This project will lead you through
an example of such a physical system by examining the fluid mechanics of a thin film
of viscous fluid spreading under gravity over a horizontal, rigid substrate. Such flows
are also referred to as viscous gravity currents, and are common in the world around
us, ranging from the spread of honey over morning toast, to lava flows and the flow of
glacial ice sheets [1, 5, 6, 7, for example].
The key feature behind such flows is that they involve very viscous fluids, such as
honey, or syrup, and as such, viscous forces determine the flow, and any inertial effects
are negligible. Under appropriate assumptions, including that the flow is long and thin,
and resisted dominantly by vertical shear stresses, and that inertial effects and the effects
of surface tension are negligible, the depth-integrated flux of viscous fluid per unit width
can be shown to be given by
q = −
ρg
3µ
H3
∂H
∂x
, (1)
where ρ, µ and H = H(x, t) are the density, dynamic viscosity, and thickness of the
viscous fluid, respectively, and g is the acceleration due to gravity. Here, the flow is fully
two-dimensional, with axes chosen so that the fluid flows along the x-direction, z = 0
describes the horizontal substrate, z = H describes the upper surface of the thin film
of viscous fluid, and there is no variation along the y-direction. The viscous fluid is
supplied at a specified flux at x = 0, and its leading edge, or frontal position, is given
by x = xN (t) as shown in figure 1.
xN(t)
H(x,t)

x
z
inflow
Figure 1: Schematic diagram illustrating the side profile of a thin film of viscous fluid
spreading under gravity over a rigid horizontal substrate.
1
Along with the equation (1) specifying the flux q, the deformation of the upper
surface is governed by the mass conservation equation
∂H
∂t
= −
∂q
∂x
, (2)
along with the source flux boundary condition
q = Q at x = 0, (3)
the vanishing frontal thickness condition
H = 0 at x = xN (t), (4)
and the global mass conservation condition
∫ xN
0
H dx = Qt. (5)
Problem 1: Similarity solution
By rescaling the surface height H as
H(x, t) = αtaF (η) (6)
where the similarity variable η is given by
η =
x
βtb
and xN = ηNβt
b, (7)
for some constant ηN , find the values of a, b, α and β so that the governing equations
(1) and (2) reduce to the single ordinary differential equation
1
5
F −
4
5
η
dF
dη
=
1
3
d
dη
(
F 3
dF
dη
)
, (8)
and recast the boundary and integral conditions (3)–(5) in terms of only F and its
derivatives. No physical constants, such as µ, ρ, Q, g, and ηN
1 and no time variable t
should appear in these reduced boundary and integral conditions. If any of these appear
in your boundary and integral conditions, that means that your values of a, b, α and
β should be adjusted. A similar, though differently scaled, derivation can be found in
[4] for Newtonian viscous fluids and in [8] for non-Newtonian, power-law fluids. A more
general treatment of self-similar problems, and ways of obtaining similarity solutions,
can be found in [2].
Note: the solution F is known as a similarity solution and the variable η is known
as a similarity variable. Note also that the front x = xN corresponds to η = ηN under
the transformation (7).
1
ηN will actually appear, but only in your transformed version of (4) and (5).
2
Problem 2: Asymptotic solution near the front
In the neighbourhood of the front η = ηN (or, equivalently, x = xN ), the thickness of the
liquid layer is small, while its slope is large. Find an asymptotic solution by following
the steps outlined below.
(a) By changing variables so that
η = ηN − ǫX, (9)
where ǫ is a small parameter, and rescaling F as F = ǫ1/3G, determine an ordinary
differential equation forG in terms ofX only (no η should appear in your equation).
(b) Identify any terms that are small in comparison to the remaining terms, when
ǫ→ 0.
(c) By deleting the small terms identified above, show that G satisfies the following
approximate equation
4
5
ηNG
′ =
1
3
(G3G′)′. (10)
(d) Show, by substitution, that this approximate equation has a solution of the form
G ∼ γXr (11)
where γ and r are constants that you should determine.
(e) Recast your approximate solution for G in terms of F and η, and call it Fasym.
You should have no ǫ’s in your formula for Fasym. This is known as an asymptotic
solution, and it is valid near the front η = ηN .
A more general overview of asymptotics methods, and ways of obtaining asymptotic
solutions to differential equations, can be found in [3].
Problem 3: Numerical solution – free boundary problem
Follow the steps outlined below to solve the ordinary differential equation (8), subject
to appropriate boundary conditions, in Mathematica.
(a) Near the front of the viscous fluid (that is, near η = ηN ), the slope of the solution
is negative and large, approaching −∞ as η → ηN . All numerical solvers, no
matter how good, run into problems when the desired solution becomes infinite at
any point in the domain. To eliminate such problems, instead of solving over the
entire interval [0, ηN ], you should instead solve over a shorter interval [0, ηN − ǫ],
where ǫ is a small number. You should use the asymptotic solution Fasym found
in Problem 2(e), as the solution for the interval [ηN − ǫ, ηN ] – the remaining part
of the interval. The asymptotic solution will also be useful in order to formulate
appropriate boundary conditions at η = ηN − ǫ, in order for the numerical solver
3
to be able to give you the correct numerical solution. To do so, substitute your
formula for the asymptotic solution Fasym into the boundary conditions formulated
below:
F (ηN − ǫ) = Fasym(ηN − ǫ) (12)
F ′(ηN − ǫ) = F
′
asym(ηN − ǫ) (13)
(b) Use the Mathematica built-in numerical differential equation solver NDSolve to
solve the ordinary differential equation (8) subject to the boundary conditions
(12) and (13) over the interval [0, ηN − ǫ], assuming that ηN = 1 and ǫ = 0.001,
for now. You will want to be able to change the parameters ηN and ǫ later on,
so you should keep these parameters general when typing in (8), (12) and (13)
into Mathematica, and only assign a specific value to them in NDSolve using a
substitution rule – see
(c) Once you obtain the solution, plot the surface height F (η) as a function of η over
the interval [0, ηN − ǫ], save it (right-click the plot in Mathematica, and select Save
Graphic As) and insert your plot into your write-up.
(d) Use your numerical solution obtained in part (c) above to evaluate the flux
qsim = −
1
3
F 3(0)F ′(0) (14)
at η = 0.
(e) According to one of your boundary conditions obtained for Problem 1, this flux,
given in (14), should be 1. As the answer you obtained in part (d), above, is not
equal to 1, this means that the parameter ηN should be adjusted. Recall we have
just arbitrarily chosen ηN = 1 in part (c), above, for the sake of obtaining just
some numerical solution. You should now manually adjust (or write a short script
to programatically adjust) the value of ηN so that the boundary condition
qsim = −
1
3
F 3(0)F ′(0) = 1 (15)
is satisfied at η = 0, to within a tolerance of 10−3. State the value of ηN that
you have found. Plot the surface height F (η) as a function of η over the interval
[0, ηN − ǫ], save it and insert your plot into your write-up.
(f) Overlay on top of your previous plot a plot of the asymptotic solution Fasym over
the whole interval [0, ηN ]. Save this plot and add it to your write-up.
(g) Comment on how well the asymptotic solution approximates the numerical solution
over the entire interval. In which part of the interval is the fit best?
4

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MOEC0021 STA 106 ENGG 1300A ENGG 1300 OLET1139 STAT 3503 STAT 206 FEM11149 ELEN 4720 COMP20003 SOLA1070 CS574 C11CS LINB29H3 MAST90083 ECON7200 STAT 610 LINB29 DSM-5 COMP3161 EEEN40010 MATH 4431 K353 STA 3386 IERG4300 COMS W4167 ENG5298 AMME 2302 CS3910 COMPSCI 320 STATS 369 GR5242 COSC 328 CSCI435 MGTS1301 INFS2200 COMP3900 CIS 375 Module AE03 COMP3121 CS260 HAM503 MGMT5050 INT3075 COSC 285 GEOL0029 GEGE2X01 STAT1201 MATH1061 MMF1941H STAT 5060 CS 520 COMP 4320 MSIN0104 000T FINS3648 ECMT2160 ECON4003 Phys 129L CSE 142 CS 3130 MGEB01 EECS 340 Finance (20443/20444) ECMM149 COMP5310 EEEN60180 CSC 111 AT2 CSCI 4430 STAB57 MA/CSC 427 MATH322 D100 CSCI544 CITS3004 CI6227 CSE 565 APCO 1P01 852L1 ELEC S411F COMP3207 STATS330 ECON7530 FIT5149 CSI4104 CMPT 371 COMP201 SEHS4696 FI4003 MATH6182 CG1 WS MATH401 MGT 205 QTM 110 COMP 0137 MACE12201 FIT5210 ECE 544 BISM7206 HUDM 6055 ECON3203 STATS 210 MSCI 570 MATH6173 COMPSCI4039 Accounting COM3503 AOE2024 ECON213 PSYC10004 Control Data 100/200A PHIL2617 PHYS3116 MIE250 BCPM0008 ECN 620 QBUS6830 ECSE-415 ECON5102 MGMT20005 MAT 653 ECO 480 COMP 5/600 CSSE 5600 ME 571 7CCMMS61 CS 550 COMP2865 MATH3075 STATS 506 QTM 220 ISYS2120 FIT5106 STAT 153 MKTG3112 ECON2102 ELE2018 COMP3031 MSIN0010 SCC461 MATH237 CS246 CSC3065 CSE 101 ICTWEB411 COMP9313 ECOS 2025 FN620 MATH11111 MATH 3018 KE1068 FINA 6112 C++ MISM 6202 MSIN0105 CSC8631 MECO6935 COMP4121 ACCT2011 DPBS1150 EEL 3701C F19PL OPM 661 CSE 310 CS918 BIOL30001 ACCT3563 BANK3600 MET CS622 MATH367 ACCT5907 Continuing COMP5048 COMP61021 CS3334 GSOE9210 CSCI-570 COMP 5322 HW 2 ALY-6020 IEOR 4706 DATA 200 EBUS504 CGE12412 COMP4108 7SSMM800 ECMM171P MATE50002 CS225 RS 6003 BH2274 ELEC3111 MAT6669 ECON-UA 227 PHIL 1012 7SSMM603 TB028A P6 ECS708 CSC 427 CS6476 CSE 250A A1A2 ITS61504 EECS 376 model HW6 UV0010 ECON 6074 MG4F7 CMP-7030Y INFO3008 BFIP013 ECON 5068 MATH6002 CS3483 PSTAT 131 C31FM CSOR W4246 AAEC 5307 DATA ELEC362 COMSM0047 ECN6006 MA20224 BST351 MATH 323 MATH323 AA2021 COMP3811 COMP2013 IB9X60 INVP001 EMATM0051 COMP220 BLAW1002 E20 TCP8001 ECON61001 EART40005 ECONM1022 MATH 365 MATH 367 ELEC6218 COMP 5812M EDUC 5061M ELEC ENG 1101 equation MATH6174 ST334 C4 EC9570 A33648 EBUS603 ECON47815 stata WFMS0001 PEM102 INF6060 ionically ELECTRICAL MATH 375 MMW 12 MKTG860 ECON60401 ALY6010 SOST20131 Introduction COM3524 MME2 GGR305 JNB 538 EECS101 COMM5030 IEOR E4601 2B CSC336 EN2315 LAW 432 CSC263 MATH2001 OLET1143 CSCB63 SIT772 COMM1170 STAT 441 MS930 INFT3050 GLBH0030 Stat 120B MAST10007 MGEC61 Machine GGR 252H1S ECON7030 Financial Marketing MGMT30019 ECON5120 SDSC 8009 CSC165H1S DSCI 551 3081W ACM41000 ECA 5304 MATH 1064 ECSE444 DATA 604 COMP7105 MIE1615 ECON4004 ENV200H1S EC481 MKT 306 EC1B1 MATH08051 ALY6140 CS260C CSE30 MSIN0226 CS4103 CSC3064 AM16 COMP11212 CSE4101 PSTAT 174 Math 380W CNT 5106C STAT 310 FIN2028 CS5014 STAT 425 COMP 4102A ELEC2140 MANF9544 QRM 9 QRM 8 ECON 457 COMPSCI 4091 MATH 523 DNSC-4211 PEAS 6000 MSIN0180 COMP2119 A2A HT 2022 IB93F0 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ACCOUNTING BUSS1040 CHEM 181 MAT136H1 GY 6183 COMP3170 AGRI90089 PHYSICS 1002 ECON10003 IEOR E4004 ACS61012 MATH 40550 MKTG2113 TELE4653 ENSC1004 ECE 232E MK726 MTRX2700 SHEET 3 CSCA08 COMP30027 Math 2XX3 statistic MMAN3200 BISM7213 ECMM109 ECON 2022 FNCE 20005 GR5067 CSCB09 ME 3017 CS 5100 PHYS 2903 COMS 4771 COMP222 COMP 1046 CSC148H1 ISYS2038 FINS2615 OT5206 COMP2048 A6 MECH5770M MA826/20 CMPT 726 STA457 INFS6071 INFS588 integral ECON 216 IRDR 0017 MA323 PA 15213 PSYC2001 ENV222 Statistics ACCT5930 ACCT12 LAND2121 ME 5405 COMM1180 TELE3113 A7 ECE 568 ACCT2019 INFS1200 ENGG1340 CS 1400 ASST3 FINM7409 SWEN30006 INFO2222 FIN20150 SOSS1000 CSSE7231 FINS5513 COMS3200 CECN 702 ECON4060H MN3515 HW2 PSYA02H3 ECON 2112 COMP90051 CS 1652 MA30059 AF1605 MKTG7501 ECO-6003B COM4521 CTM ECON3350 COMP 3804 ELEC9762 BISM 7255 ST22 MATH20014 MNE3122 COMP343 MATH5165 ELE7029 ENGR30002 SEEM3500 ECON7350 RISK5002 FINM7405 COMP90059 MOS 3320B STAT 431 BISM7221 MATH3014 MTH783P EBA 35302 ACS133 BUS2206 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FN1024 MATH0094 MANG 6554 ECOS3003 ELEC6252 C2511C PHY328 EC1060 ECON32202 COMP4002-E1 MATH3063 EFIMM0124 QBUS2310 ECON4411 FUNDAMENTALS ACCT20001 18YX RPM COMP 2711 QBUS6600 BUSS5221 ST3189 FIT1047 MAT344 EGB375 MATH60013 MATH96011 COMP3557 MAST20029 ECM159 EAE1011 ENSCEN 105 SECTION A STA 141C SOSE 2022 PAM006 COMP0048 M40007 ISOM2500 CMPE 142 A06472 MCC150 ACST 8083 GEOG6087 ECON920 FINM7008 ENVENG 4110 COMPSCI 361 ENGRCEE 21 PHYS 1112 ACF2100 ECMM108 COMPS267F MAT00050M MAST30025 CSIT970 MSCI 224 INT202 PUBL0054 MSDM5004 BMAN 21020 BMAN21020 MGMT1002 MATH11150 LAWS7023 ACTL30002 CIVL3340 MATH1002 ECON 8069 COMPSCI 351 IBUS7302 STAT4528 CHEM1101 STAT3925 BUSN 7008 Comp 3120 ECOS2040 SWEN90010 ENGN6528 ECM158 ECON20022 COMP6452 Calculation CITS1401 S1889112 BUS 351 ACTL30008 FIT2091 ECON6002 IBUS6020 PHI 001 ECON30290 MATH7501 ECON10004 MFIN7017 ISYS90043 LAWS7012 ECON30025 COMP30023 COMM1190 ECOM20001 GSBS6009 STAT 337 EMET8002 ECOM90001 ECOM30001 FINC 2012 BUSINESS ENGCV512 ECO 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