h y x β

Fluid Mechanics Sample Exam 2
Please work problems 7) and 8). Then attempt any 3 of the remaining
problems in 1) through 6). Please do not work with other students – this is a
learning exercise. Earnest attempts at solutions will be given much more weight
than simply copying or sharing solutions.
1) A fluid flows steadily down a plane surface inclined at an angle β as shown.
h
y
x
β
a) Assuming that changes in film thickness are small, perform a mass balance
between two points x and x + dx to show that
∂
∂x
Z
h
udy = 0
(1)
0
where u is the x-velocity component and h is the film thickness. Based on this
equation, we infer (using Liebnitz’ rule) that
∂u
=0
∂x
(2)
within the film.
b) Use the result in (2) and the two-dimensional continuity equation
∂u ∂v
+
=0
∂x ∂y
(3)
to show that the vertical velocity component v is zero:
v=0
1
(4)
c) Assuming that the shear stress at film’s upper surface is zero, determine the
velocity distribution u = u(y), within the film. Note that in this coordinate
system, the x-momentum equation assumes the form
∂u
ρ u
∂x
+v
∂2u
∂u ∂2u +
+ ρgsinβ
=µ
∂y
∂x2
∂y 2
(5)
d) The pressure gradient term ∂P
∂x in the x-momentum equation above was not
included. Use the y-momentum equation to show that P varies linearly (i.e.,
hydrostatically) in the y-direction. In addition, differentiate the x-momentum
equation (including the pressure gradient term) with respect to x to show that
P,xx = 0. From these results and based on the assumption that pressure does not
vary in the atmosphere above the film, show that P,x = 0.
2
2) Large air bubbles moving within vertically-oriented, water-filled, circular tubes
are observed to travel at a constant speed U while maintaining an essentially fixed,
bullet-like shape, as shown below.
a) Assuming that viscous effects are negligible and noting that the water’s velocity
at the bubble’s upper tip is equal to the bubble’s velocity, what is the speed of the
water along the air-water interface as a function of the vertical distance x below
the bubble’s tip?
b) Using conservation of mass, derive a relationship giving d as a function of x.
R
x
d
U
3
3) A rocket is launched from the ground with a full fuel tank. The speed of burnt
fuel relative to the nozzle remains constant at Vjet , while the fuel’s density
remains fixed at ρf . The weight of the unfueled rocket is WR .
a) Define a moving control volume that encloses and moves with the rocket. Starting with the integral form of the conservation of linear momentum applied to the
contents of the control volume (i.e., the rocket and the unburnt fuel within the
rocket), show that in the vertical direction, the momentum equation is given by
dV
1
d
(M V ) + MR
= ρf Ajet Vjet (Vjet − V ) − M g − WR − CD ρa V 2 AR
dt
dt
2
where
M = M (t) = mass of unburnt fuel in rocket at time t
V = V (t) = instantaneous vertical velocity of the rocket (relative to the earth)
MR = WR /g = the (fixed) mass of the rocket (without fuel)
ρf = fluid density
Ajet = the cross-sectional area of the rocket’s nozzle
g = gravitational acceleration
WR = weight of the rocket
Cd = the drag coefficient for air flow past the rocket
ρa = density of air
AR is the rocket’s projected area in the direction of motion
It is assumed that the flow of burnt fuel out of the nozzle is uniform across the
cross-sectional area Ajet and that the velocity of the fuel within the rocket is equal
to the rocket’s instantaneous velocity. Note that the total instantaneous drag D
on the rocket is given by D = Cd 21 ρa V 2 AR . In addition, recall that the surface
integral giving the flux of momentum out of/into a moving control volume is given
by
Z
S
ρu(w − u) · n
ˆ dS
where S is the surface enclosing the control volume, u is the fluid velocity across
S (relative to a fixed reference frame), w is the velocity of the surface S (again
relative to a fixed reference frame), and n
ˆ is the outward unit normal to S.
b) Use the same moving control volume and show that the integral form of the
conservation of mass leads to the following equation:
d
(M ) = −ρf Ajet Vjet
dt
4
g
V(t)
M(t)
A
5
jet
4) An incompressible, Newtonian fluid of density ρ and viscosity µ flows in the
channel shown. The upper and lower walls are porous, allowing fluid to be injected
through the lower boundary and removed from the upper boundary at a constant
velocity V. The channel flow is driven by a pressure gradient in the horizontal
direction.
a) Assuming that the channel width in and out of the page is much larger than
the channel height and that the flow is fully-developed in the x -direction, show
that the vertical velocity component is constant across the channel.
b) It can be shown that under these conditions, the pressure gradient in the x direction is constant. Letting ∂P/∂x = G = constant, and neglecting gravity,
determine the x -component of velocity. State your answer in terms of G, ρ, and
µ.
V
y
2h
x
V
6
u
5) Consider fully-developed, pressure-driven flow of a Newtonian fluid in a wide
channel (i.e., width into the page is much greater then height h)..
a) Assuming that the pressure gradient in the flow direction is constant and equal
to G, simplify the Navier-Stokes equations and show that the velocity profile within
the channel is given by
y
Gh2 y 2
( 2− )
u(y) =
2µ h
h
where h is the channel height and µ is the fluid viscosity.
b) If the channel walls are maintained at a temperature To , and frictional heating
effects are important, what is the temperature distribution within the channel and
what is the heat flux into either wall? Assume that temperature variations in the
x-direction are negligible.
To
h
y
T
o
Aside: The energy equation for a fluid having constant thermophysical properties
is given by:
ρCp (T,t +uT,x +vT,y +wT,z ) = k(T,xx +T,yy +T,zz ) + ρφ
where T is the fluid temperature field, Cp is the specific heat, k is the thermal
conductivity, and φ = µ(ui , j + uj , i)(ui , j + uj , i)/2, is the viscous dissipation
function.
7
6) Textbook (from Panton) problems 7.1 and 7.3.
8
7) Consider the modified steady Couette flow problem depicted below. The up˜ , while a steady cross-flow exists,
per porous plate is pulled at constant speed, U
passing from the upper porous plate to the lower porous plate. [In this problem,
dimensional quantities will be denoted with tildes while dimensionless quantities
have no tilde.] The speed of the flow leaving the top and entering the bottom is
v˜s .
a) By first introducing reasonable assumptions concerning the nature of the flow
(i.e., by simplifying the dimensional governing equations) followed by nondimensionalization, show that the continuity and x-momentum equations assume the
respective dimensionless forms:
∂v
=0
∂y
and
v
∂u
1 ∂2u
=
∂y
Re ∂y 2
Here, the length scale in both the x and y directions is chosen to be the gap
˜ the velocity scale in both directions is chosen as U
˜ , and the Reynolds
thickness, d,
˜ ν.
˜ d/˜
number is defined as Re = U
b) Show that the boundary conditions, when stated in dimensionless form, are
given by
v(0) = v(1) = −vs
u(0) = 0
u(1) = 1
˜.
where vs = v˜s /U
c) Based on the above development, show that the vertical dimensionless velocity
component is constant and given by:
v(y) = −vs
d) Given the solution for v(y), re-express the x-momentum equation as
ǫ
∂u ∂ 2 u
+ 2 =0
∂y
∂y
˜ ν . Thus, we are focusing on the case where the cross-stream
where ǫ = vs Re = v˜s d/˜
flow is small relative to the plate speed. Then assume an asymptotic solution of
the form
u(y) = uo (y) + ǫu1 (y) + ǫ2 u2 (y) + O(ǫ3 )
9
e) Show that the leading order solution, i.e., uo (y), is given by
uo (y) = y
Give a quick physical interpretation of this solution; does the solution make intuitive/physical sense?
f) Show that the first order solution, i.e., u1 (y), is given by
u1 (y) = y(1 − y)/2
Using the last two results, we then at arrive at the following asymptotic solution,
valid to order ǫ2 :
ǫ
u(y) = y + y(1 − y) + O(ǫ2 )
2
Do a quick check to confirm that this solution satisfies the given boundary conditions.
g) Solve in exact form the above dimensionless momentum equation, subject to
the given bc’s on u. Show that the exact solution is given by
u(y) =
1 − e−ǫy
1 − e−ǫ
h) Finally, for small ǫ, expand the exact solution and show that the first two
terms in this expansion are identical to the first two terms that we obtained for the
asymptotic solution. With regard to asymptotic solutions, this level of verification
is typically not possible; however, in this case it provides a nice demonstration of
the power of the technique.
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8) Waves on internal fluid-fluid interfaces. In answering the following questions,
refer to the discussion provided on the attached pages.
a) Derive the condition in Eq. 7.104 by first writing Bernoulli’s equation within
each fluid and applying it at the interface z = ζ. Second, equate the pressures (i.e.,
neglect viscous normal stresses and surface tension) across the interface. Third,
linearize and then transfer the equation to z = 0.
b) Expressing the equation describing the interface as F (x, z, t) = z − ζ(x, t) = 0,
use the procedure we used in class to derive the full kinematic condition at the
interface z = ζ(x, t). Note that this condition applies to both velocity potentials
φ1 and φ2 , i.e., you’ll obtain two conditions (each of the same basic form).
c) Using linearization, followed by transfer of boundary conditions to z = 0, show
that the full kinematic conditions derived in part b) simplify to the forms given
in Eq. 7.103 (which represents the linearized kinematic conditions for both fluid
layers).
d) Using the assumed wave-like solutions for φ1 and φ2 given at the top of page
234, show that these assumed forms satisfy far field bc’s (i.e., conditions at large
positive and negative y), as well as the equation (i.e., Laplace’s equation Eq.
7.100), governing both velocity potentials.
e) Using the linearized, transfered kinematic conditions from part c) show that
the constants A and B are as given.
f) Using the condition derived in part a), obtain the dispersion relation between
the wave frequency, ω, and the wave number, k, as given by eqn. 7.105.
g) Briefly discuss some of the physical implications of the dispersion relationship
obtained in f). Finally, in the case where ρ1 << ρ2 , e.g., air over water, √
show that
the above dispersion relation simplifies to the ’deep-water’ limit ω = gk, that
we obtained in class.
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