Download zero. Ans. (b) P4.8 When a valve is opened, fluid flows in the

Chapter 4
307
Differential Relations for a Fluid Particle
P4.7 Consider a sphere of radius R immersed in a uniform stream Uo, as shown in
Fig. P4.7. According to the theory of Chap. 8, the fluid velocity along streamline AB is given by
R3
i
x3
ui Uo 1
V
Fig. P4.7
Find (a) the position of maximum fluid acceleration along AB and (b) the time required
for a fluid particle to travel from A to B. Note that x is negative along line AB.
Solution: (a) Along this streamline, the fluid acceleration is one-dimensional:
du
dt
u
u
x
U o (1 R 3 /x 3 )( 3U o R 3 /x 4 )
3U o R 3 (x
4
R 3 x 7 ) for x
R
The maximum occurs where d(ax)/dx 0, or at x –(7R3/4)1/3 –1.205R Ans. (a)
(b) The time required to move along this path from A to B is computed from
u
or: U o t
R
dx
dt
3
3
U o (1 R /x ), or:
4R
x
R
(x R)2
ln 2
6
x Rx R 2
dx
1 R 3 /x 3
t
U o dt,
0
R
R
3
tan
1
2x R
R 3
4R
It takes an infinite time to actually reach the stagnation point, where the velocity is
zero. Ans. (b)
P4.8 When a valve is opened, fluid flows in the expansion duct of Fig. P4.8 according
to the approximation
V
iU 1
x
Ut
tanh
2L
L
308
Solutions Manual
Fluid Mechanics, Fifth Edition
Find (a) the fluid acceleration at (x, t) (L, L/U) and (b) the time for which the fluid
acceleration at x L is zero. Why does the fluid acceleration become negative after
condition (b)?
Fig. P4.8
Solution: This is a one-dimensional unsteady flow. The acceleration is
ax
u
t
u
u
x
U 1
x U
Ut
sech 2
2L L
L
U2
x
Ut
(1
)[sech 2
L
2L
L
At (x, t)
(L, L/U), ax
U 1
x
2L
U
Ut
tanh
2L
L
1
Ut
tanh 2
]
2
L
(U2/L)(1/2)[sech2(1) – 0.5tanh2 (1)]
0.0650 U2/L Ans. (a)
The acceleration becomes zero when
The acceleration starts off positive, then goes through zero and turns negative as the
negative convective acceleration overtakes the decaying positive local acceleration.
P4.9 An idealized incompressible flow has the proposed three-dimensional velocity
distribution
V
4xy2i
f(y)j – zy2k
Chapter 4
Solution: With no
variation and no v , the equation of continuity (4.9) becomes
vz
z
1
(r vr )
r r
or :
317
Differential Relations for a Fluid Particle
r
(r vr )
0
1
(r vr )
r r
B r ; Integrate : r vr
Finally, vr
B
r
2
z
( Bz ) ,
B 2
r
2
f ( z)
r
f ( z)
Ans.
The “function of integration”, f(z), is arbitrary, at least until boundary conditions are set.
__________________________________________________________________________
P4.23 A tank volume V contains gas at conditions ( o, po, To). At time t 0 it is
punctured by a small hole of area A. According to the theory of Chap. 9, the mass flow
out of such a hole is approximately proportional to A and to the tank pressure. If the tank
temperature is assumed constant and the gas is ideal, find an expression for the variation
of density within the tank.
Solution: This problem is a realistic approximation of the “blowdown” of a highpressure tank, where the exit mass flow is choked and thus proportional to tank pressure.
For a control volume enclosing the tank and cutting through the exit jet, the mass relation is
d
d
(m tank ) mexit 0, or:
( )
mexit
C p A, where C constant
dt
dt
Introduce
p
RTo
p(t)
and separate variables:
The solution is an exponential decay of tank density: p
po
dp
p
CRTo A
t
dt
o
po exp(–CRToAt/V ). Ans.
P4.24
For incompressible laminar flow between parallel plates (see Fig. 4.12b),
the flow is two-dimensional (v 0) if the walls are porous. A special case solution is
u ( A Bx) (h 2 y 2 ) , where A and B are constants. (a) Find a general formula for
velocity v if v = 0 at y = 0. (b) What is the value of the constant B if v = vw at y = +h?
Solution: (a) Use the equation of continuity to find the velocity v:
318
Solutions Manual
v
y
u
x
Integrate : v
( B)(h 2
B (h 2
Fluid Mechanics, Fifth Edition
y2 )
y 2 )dy
If v 0 at y 0, then f ( x)
B(h2 y
v
0.
y3
)
3
f ( x)
y3
)
3
B(h 2 y
Ans.(a )
(b) Just simply introduce this boundary condition into the answer to part (a):
v( y
h)
vw
B ( h3
h3
) ,
3
hence
B
P4.25 An incompressible flow in polar
coordinates is given by
b
vr K cos 1 2
r
b
v
K sin 1 2
r
Does this field satisfy continuity? For
consistency, what should the dimensions of
constants K and b be? Sketch the surface
where vr 0 and interpret.
3 vw
Ans.(b)
2 h3
Fig. P4.25
Solution: Substitute into plane polar coordinate continuity:
1
1 v
(rv r )
r r
r
0
1
K cos
r r
r
b
r
1
r
K sin
1
b
r2
0 Satisfied
The dimensions of K must be velocity, {K} {L/T}, and b must be area, {b} {L2}. The
surfaces where vr = 0 are the y-axis and the circle r = b, as shown above. The pattern
represents inviscid flow of a uniform stream past a circular cylinder (Chap. 8).
Solution for problem 6).
∂()
∂y
a). According to the continuity equation (two-d flow,
= 0)
∂u ∂w
+
=0
∂x
∂z
we have
! "
#$
2z
∂u
∂
z2
√ − 2
=−
U0
∂x
∂x
C x C x
#
"
" #
1
1
U0 z 2 ∂
2U0 z ∂
√
+
= −
C ∂x
C 2 ∂x x
x
U0 z 2 −2
U0 z − 3
x 2−
x
=
C
C2
Solving the above partial differential equation we have
∂w
∂z
= −
3
U0 x− 2 z 2
U0 x−2 z 3
−
+ C1 (x)
2C
3C 2
Due to the no-slip boundary condition, w(x, z = 0) = 0. Subsitute it into the above general solution we have
C1 (x) = 0, hence the expression for vertical velocity is
w(x, z) =
3
U0 x− 2 z 2
U0 x−2 z 3
w(x, z) =
−
2C
3C 2
b). As assumed by this question, we have δ(x = 1) = 0.011 m, thus
0.011 = C × 11/2 =⇒ C = 0.011 m1/2
At x = 1 we can define w1 (z) = w(x = 1, z), such that
w1 (z) =
Thus we have
U0 z 3
U0 z 2
−
2C
3C 2
dw1
U0 z U0 z 2
=
−
dz
C
C2
2
2U0 z
d w1
U0
−
=
dz 2
C
C2
and
U0 %
z&
dw1
= 0 =⇒
z 1−
=0
dz
C
C
Therefore w1 (z) have maximum or minimum values at z = 0 and z = C. Since we have
'
d2 w1 ''
U0
>0
=
'
2
dz z=0
C
'
d2 w1 ''
U0
U0
U0
−2
=−
<0
=
'
2
dz z=C
C
C
C
Thus the maximun value of w1 (z) is w1 (z = C), hence
w(x = 1, z)max
U0 C
U0 C
−
2
3
=
w(x = 1, z = C) =
=
3 × 0.011
U0 C
=
= 0.0055 m/s
6
6
1
326
Solutions Manual
p
Fluid Mechanics, Fifth Edition
p(0,0,0) – gz – ( /2)K2(x2
y2
4z2)
Ans. (b)
Note that the last term is identical to ( /2)(u2 v2 w2), in other words, Bernoulli’s
equation.
(c) For irrotational flow, the curl of the velocity field must be zero:
V
i(0 – 0)
j(0 – 0)
k(0 – 0)
0 Yes, irrotational. Ans. (c)
P4.35 From the Navier-Stokes equations
for incompressible flow in polar coordinates
(App. E for cylindrical coordinates), find
the most general case of purely circulating
motion (r), r
z 0, for flow with no
slip between two fixed concentric cylinders,
as in Fig. P4.35.
υθ (r)
r
r=a
No slip
r=b
Fig. P4.35
Solution: The preliminary work for this
problem is identical to Prob. 4.32 on an earlier page. That is, there are two possible
solutions for purely circulating motion (r), hence
v
C1r
C2
, subject to v (a) 0
r
This requires C1
C2
0, or v
C1a C2 /a and v (b) 0
0 (no steady motion possible between fixed walls) Ans.
y
g
P4.36 A constant-thickness film of
viscous liquid flows in laminar motion
down a plate inclined at angle , as in Fig.
P4.36. The velocity profile is
u
Cy(2h – y)
v
w
C1b C2 /b
0
Find the constant C in terms of the specific
weight and viscosity and the angle . Find
the volume flux Q per unit width in terms
of these parameters.
h
(y)
θ
x
Fig. P4.36
Solution: There is atmospheric pressure all along the surface at y h, hence p/ x
The x-momentum equation can easily be evaluated from the known velocity profile:
u
u
x
u
y
v
p
x
2
gx
u, or: 0
g sin
2
Solve for C
0
g sin + ( 2C)
Ans. (a)
The flow rate per unit width is found by integrating the velocity profile and using C:
h
Q
h
u dy
0
gh 3 sin
3
2 3
Ch
3
Cy(2h y) dy
0
per unit width
Ans. (b)
P4.37 A viscous liquid of constant
density and viscosity falls due to gravity
between two parallel plates a distance 2h
apart, as in the figure. The flow is fully
developed, that is, w w(x) only. There
are no pressure gradients, only gravity. Set
up and solve the Navier-Stokes equation
for the velocity profile w(x).
Solution: Only the z-component of NavierStokes is relevant:
dw
dt
0
Fig. P4.37
d2w
, or: w
dx 2
g
g
, w( h )
w( h) 0 (no-slip)
The solution is very similar to Eqs. (4.142) to (4.143) of the text:
w
P4.38
g
2
(h 2
x 2 ) Ans.
Show that the incompressible flow distribution, in cylindrical coordinates,
vr
0
v
C rn
vz
0
0.
Chapter 4
331
Differential Relations for a Fluid Particle
Before integrating again, note that dT/dy 0 at y h/2 (the symmetry condition), so
C1 –h3/6. Now integrate once more:
16 u2max 2 y 2
h
2
kh 4
T
If T
Tw at y
0 and at y
T
Tw
h, then C2
2h
y3
3
y4
3
C1y
C2
Tw. The final solution is:
8 u 2max y
k
3h
y2
h2
4y 3
3h 3
2y 4
3h 4
Ans.
This is exactly the same solution as Problem P4.40 above, except that, here, the
coordinate y is measured from the boo tom wall rather than the centerline.
P4.42 Suppose that we wish to analyze the rotating, partly-full cylinder of Fig. 2.23 as a
spin-up problem, starting from rest and continuing until solid-body-rotation is achieved.
What are the appropriate boundary and initial conditions for this problem?
Solution: Let V
conditions are
V(r, z, t). The initial condition is: V(r, z, 0)
0. The boundary
Along the side walls: v (R, z, t) R , vr(R, z, t) 0, vz(R, z, t) 0.
At the bottom, z 0: v (r, 0, t) r , vr(r, 0, t) 0, vz(r, 0, t) 0.
: p patm, rz
At the free surface, z
z 0.
P4.43 For the draining liquid film of Fig.
P4.36, what are the appropriate boundary
conditions (a) at the bottom y 0 and (b) at
the surface y h?
Fig. P4.36
Solution: The physically realistic conditions at the upper and lower surfaces are:
(a) at the bottom, y 0, no-slip: u(0) 0 Ans. (a)
u
u
( h) 0 Ans. (b)
(b) At the surface, y h, no shear stress,
0, or
y
y
Solution: With vz
fcn(r) only, the Navier-Stokes z-momentum relation is
dvz
dt
or:
1 d dvz
r
r dr
dr
g
p
z
0
vz ,
gr 2
4
, Integrate twice: vz
vz
(b)
r
gb2
r
ln
2
a
The final solution is v z
b
The flow rate is Q
2
g
The proper B.C. are: u(a) 0 (no-slip) and
ga 4
8
vz 2 r dr
a
b
a
where
g
4
3
u(y, t) and p/ x
u
u
u
t
x
or:
u
t
v
u
0 0
t
C1 ln(r) C2
0 (no free-surface shear stress)
4
r2
1 4
a2
2
Ans.
4
4
ln
,
Ans.
P4.85 A flat plate of essentially infinite
width and breadth oscillates sinusoidally in its
own plane beneath a viscous fluid, as in
Fig. P4.85. The fluid is at rest far above the
plate. Making as many simplifying assumptions as you can, set up the governing
differential equation and boundary conditions
for finding the velocity field u in the fluid. Do
not solve (if you can solve it immediately,
you might be able to get exempted from the
balance of this course with credit).
Solution: Assume u
361
Differential Relations for a Fluid Particle
Chapter 4
Fig. P4.85
0. The x-momentum relation is
u
y
2
p
x
u
x2
gx
2
u
,
y2
2
0 0
u
, or, finally:
y2
0
2
u
y2
subject to: u(0, t)
U o sin
t
and u
,t
0.
Ans.
368
Solutions Manual
1 d dv
(r
)
r dr
dr
As r
, v
v
r
2
,
Fluid Mechanics, Fifth Edition
Solution : v
0 , hence C1
C1 r
C2
r
0
C2
R2
; C2
R 2 ; Finally, v
Ans.
R
r
Rotating a cylinder in a large expanse of fluid sets up (eventually) a potential vortex flow.
At r
R, v
R
________________________________________________________________________
*P4.95
Two immiscible liquids of
V
equal thickness h are being sheared
h
y
2,
2
between a fixed and a moving plate,
1,
h
as in Fig. P4.95. Gravity is neglected,
1
x
Fixed
and there is no variation with x.
Fig. P4.95
Find an expression for (a) the velocity at the
interface; and (b) the shear stress in each fluid. Assume steady laminar flow.
Solution: Treat this as a Ch. 4 problem (not Ch. 1), use continuity and Navier-Stokes:
u
v
v
0
0 ; thus v const 0 for no slip at the walls
x
y
y
This tells us that there is no velocity v, hence we need only consider u(y) in Navier-Stokes:
Continuity :
1, 2 (u
u
x
u
v )
y
p
x
Thus
2
1, 2 (
u
u
2
x2
a
y
by
u
) or : 0
2
0
0
1, 2 (0
d 2u
dy 2
)
369
Differential Relations for a Fluid Particle
Chapter 4
The velocity profiles are linear in y but have a different slope in each layer. Let uI be the
velocity at the interface. (a) The shear stress is the same in each layer:
1
uI
h
V
uI
2
Solve for u I
h
2
1
V
Ans.(a )
2
(b) In terms of the upper plate velocity, V, the shear stress is
1
(
2
1
)
2
V
h
Ans.(b)
________________________________________________________________________
P4.96 Reconsider Prob. P1.44 and calculate (a) the inner shear stress and (b) the power
required, if the exact laminar-flow formula, Eq. (4.140) is used. (c) Determine whether this
flow pattern is stable. [HINT: The shear stress in (r, ) coordinates is not like plane flow.]
Solution: The exact laminar-flow velocity is Eq. (4.140), and the shear stress is Eq. (D.9):
v
i ri [
(
r
(ro / r ) ( r / ro )
]
(ro / ri ) (ri / ro )
dv
dr
v
)
r
[
i ri
(ro / ri ) (ri / ro )
]
2ro
r2
Recall the data from Prob. P1.44: ri = 5 cm, ro = 6 cm, L = 120 cm, = 0.86 kg/m-s (SAE
50W oil), and i = 900 rev/min = 94.25 rad/s. At the inner cylinder,
inner
[
i ri
(ro / ri ) (ri / ro )
]
2ro
ri2
(0.86)[
94.25(0.05)
2(0.06)
]
0.06 / 0.05 0.05 / 0.06 (0.05) 2
531 Pa Ans.(a)
The moment and power required are
M
i2
ri2 L
Power
(531) 2 (0.05)2 (1.20)
i
M
10.0 N
(94.25 rad / s)(10.0 N
m)
m
943 watts
Ans.(b)