Chapter Solutions, 15
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15.1 The first three parts of this problem start with Eq. 15.7 and simply drop the
unnecessary parts
(Vx ) ( Vy )
0=
+
(a)
x
y
(Vx ) (Vy )
0=
+
(b)
Even though the flow is unsteady, is a constant, so
x
y
its time derivative is zero.
( Vy )
=
(c)
t
y
(d)
Here we start with Eq. C-13 and drop the unnecessary parts.
1 ( rVr ) 1 ( V )
0=
+
r r
r
(e) for (a) two parallel plates, steady flow of a compressible fluid along a line
source parallel to one of the axes, flow out at the edges. For (b) the same as (a) but with
an incompressible fluid (e.g. a liquid) with the inlet flow rate not steady, for (c) starting
or stopping flow of a compressible fluid in a pipe, for (d) a thrust bearing with central
fluid addition.
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15.2 Starting with Eq. C.13 we see that
(a) The terms are all zero. This looks odd, but the radial and circumferential
velocities are zero, and the local value of the axial velocity does not change with axial
distance.
(b) Again the terms are all zero. While the axial velocity does change with axial
distance, its product with the density does not.
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dVx V0 dx V0
V2
x
=
= Vx = 0 1 +
dt
L dt
L
L
L
This shows that while the flow is steady, at any point in the flow (in this case along the
centerline, but also true at any point) the fluid is being accelerated.
dx
x V
L L dx
L
(b)
Vx =
= V0 1 + = 0 (L + x );
0 L + x = V0 ln 2 = t
dt
L L
V0
See the next problem!
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15.3 (a)
15.4
(a) Starting with Eq. C.13 for cylindrical coordinates, with the z coordinate
1 (rVr ) (Vx )
0=
+
replaced by an x,
we see that for the flow to be speeding up in
r r
x
the x direction it must have a radial component which is negative, i.e. there must be an
inflow towards the center as the fluid accelerates. This is obviously what happens in any
Solutions, Fluid Mechanics for Chemical Engineers, Third Edition, Chapter 15, page 1
converging nozzle, like a garden hose nozzle. From the stated value of Vx we can work
V
1 ( rVr )
(Vx )
=
= 0.
backward to see that
r r
x
L
(a) For the velocity to increase linearly with length, the area must decrease
linearly with length which means that the square of the diameter must decrease linearly
with length, or the diameter must decrease with the square root of the length; the nozzle
shape is a parabola. The shape is contrary to our intuition, concave outward, with the
curvature increasing in the flow direction. The flow is sketched
below
D=D
0
D=D / sqrt 2
0
x=L
x =0
Centerline
s
One flow vector is shown. It has an x component which increases from left to right, and
a radial inward component. From the above Eq., we can see that
V
1 ( rVr )
V r2
V
= 0 ;
(rVr ) = 0 rr ; rVr = 0 + const.
r r
L
L
L2
at the centerline, where r = 0 and Vr is not zero, we have 0 = 0 + const. so that
Vr
throughout the flow Vr = 0
L2
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15.5
(a) Starting with Eq 15.17, 0 =
(Vx ) (Vz )
+
x
z
(b) Assuming that at any x , Vx is independent of z we can make a material
V0 z0
z0
dV
.,
and
balance, finding that Vx A = const. , Vx = V0
=
dx ( z0 x) 2
z0 x
d V dV
V0 z 0
=
=
dz
dx ( z 0 x )2
dz
V0 z 0 z
dVz = V0z0 ( z0 x )2 ; Vz = (z0 x )2 + const. , when z = 0, Vz = 0, so the
V0 z 0 z
constant is zero and Vz =
(z 0 x )2
Solutions, Fluid Mechanics for Chemical Engineers, Third Edition, Chapter 15, page 2
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V x
x
dVx Vx
=
+ x
+ etc. but
= Vx , so that this is
t y , z , t
dt t x ,y ,z x y ,z,t t y, z,t
,
Vx
dVx Vx
=
+
V + etc. and similarly for the y and z terms.
dt t x ,y ,z x y ,z,t x
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15.6
The only difference is that the boundary condition, Vx = 0 @ z = h is replaced by
dP h
dP h Vwall
Vx = Vwall @ z = h . This changes C1 from C1 =
to C1 =
.
dx 2
dx 2
h
Substituting this in Eq 15.C and rearranging, changes Eq. 15.E to
z (dP / dx )
z
1 dP 2
(z zh ) + Vwall =
Vx =
(zh z2 ) + Vwall
h
2
h
2 dx
This is simply the solution to Ex. 15.1 with an added term for the flow caused by the
moving upper wall.
15.7
To make the plot we divide both sides of the equation by Vwall and factor a h2 out of the
left side to get
Vx
h 2 dP z2 z z
+
=
Vwall 2 Vwall dx h 2 h h
h 2 dP
which is plotted below for the three values of
2 Vwall dx
Solutions, Fluid Mechanics for Chemical Engineers, Third Edition, Chapter 15, page 3
1
0.8
h 2 dP
=0
2 Vwall dx
0.6
z
h
0.4
h 2 dP
=3
2 Vwall dx
h 2 dP
= 3
2 Vwall dx
0.2
0
-0.5
0
0.5
1
1.5
Vx
Vwall
We see that if there is a large enough positive pressure gradient, then the flow in the
center will be in the opposite direction of the motion of the sliding plate. If the pressure
gradient is negative (the more normal situation) the two flow terms add.
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Q ( dP / dx )lh3 / 12 ( dP / dx ) h2
=
=
A
lh
12
The maximum velocity occurs at h/2
2
h 2
( dP / dx ) h
h = ( dP / dx )h
Vmax =
2
2
8
2
Vavg etc. / 12
=
= 0.6667
etc. / 8
Vmax
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15.8*
Vavg =
15.9 We begin with the z component of the Navier-Stokes equation, Eq. C.17
V
2V 2V 2V
V
V
V
P
z + Vx z + Vy z + Vz z = gz
+ 2z + 2z + 2z
x
y
z
z
y
z
t
x
2V
and drop the terms which are zero, finding 0 = g + 2z
x
Here we have taken the direction perpendicular to the wall as the x direction, and noted
that the gravity vector points in the negative z direction. (The flow will turn out to be in
the negative z direction, so Vz will be negative). By two integrations we get
Solutions, Fluid Mechanics for Chemical Engineers, Third Edition, Chapter 15, page 4
g x 2
= Vz + C1 x + C2
2
At the wall, Vz = 0 so that C2 = 0 . To get the other boundary condition we observe that
at the free surface ( x = x ) the shear stress must be 0 (but see the next problem!), so that
dVz
g
= 0 = C1
x substituting this value of C1, and simplifying, we find
dx
g x 2
gx
x
xx =
x
Vz =
2
2
To find the volumetric flow rate we consider a distance l in the y direction, and then
x
x
x g x 2
gl x 3 x 2 x
x x ldx =
Q = 0 VzdA = 0
2 0
2
6
3
3
gl( x ) 1 1
gl ( x )
=
=
6 2
3
The minus sign appears because the flow is in the minus z direction. One may show that
the equation for Vz gives negative values for all x < x .
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15.10 Start with the solution to the previous problem. The only change is that the
dV
g
second boundary condition changes from z = 0 = C1
x to
dx
A
dVz
g
= = C1
x
dx
Here we show because it not obvious which sign is appropriate. We will settle that
later. Solving this equation for C1 and inserting the value, we find
1 x2
Vz =
g xx Ax
2
The first term in brackets is negative for all x < x . and for the stated condition that the
air is flowing in the upward (positive z) direction, the second term must be positive
1 x2
upward, and thus the must be a +, Vz = g xx + Ax
2
To get the volumetric flow rate, we repeat the calculation in the above problem with this
equation for the velocity, finding
x
x
Ax
x
x g x 2
Al x 2
gl x 3 x 2 x
ldx =
+
Q = 0 VzdA = 0 x x +
2
2 0
6
2 0
=
gl ( x )3 Al(x ) 2
+
3
2
Comparing this to the previous problem, we see that for A = 0, the results are the same.
For small value of A the velocity is always negative and Q is negative, indicating flow in
the downward direction. For large values of A the velocity and Q become positive; the
high shear stress caused by the gas flow drags the flow upward. This latter case
approximates the annular flow pattern in Fig. 12.3, with the modification that the fluid
Solutions, Fluid Mechanics for Chemical Engineers, Third Edition, Chapter 15, page 5
viscosity is now high enough that the liquid flow is laminar. This is observed in some
rising-film evaporator. A brief Google search revealed that a evaporator manufacturers
list this as a type of product they supply, but indicate that it normally operates in
turbulent flow. They give few specific examples of applications in which it is
advantageous over other types of evaporators.
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15.11 This is the same as Prob. 15.9 with the second boundary condition being that Vx =
g (x )2
g x
= 0 + C1 x + 0 ; C1 =
0 at x = x. This leads to
2
2
g 2
Vz =
(x x x ) This velocity is negative for all x < x .
2
x
3
x
x g
gl x 3 x 2 x
2
Q = 0 VzdA = 0
(x xx )ldx = 2 3 2 = gl ( x )
2
12
0
This is negative because the flow is downward. Comparing this to the solution to Ex.
15.1 (the horizontal equivalent of the same flow, we see that we have replaced
(dP / dx ) by g . Other than that, it is the same problem.
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15.12 Going back to Prob. 15.9, we see that if we take the z direction parallel to the
flow, then gz = g cos . By inspection, we get the same answers in as 15.9 with that
change, i.e.
x
g cos x 2
gx cos
xx =
Vz =
x , and
2
2
gl ( x )3 cos
Q =
3
If the plate is vertical, the cos term = 1, and we get Prob. 15.9 back. If the plate is
horizontal, cos = 0, and there is no gravity-driven flow.
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g( x )2
so that
15.13* See the solution to Prob. 15.9. At the surface Vz =
2
2
lbm
ft 0.003ft
80 3 32 2
2
g(x )
cPft s
ft
s 12 60 s
=
=
= 857cP
0.1 ft
2Vz
min 6.72 10 4 lbm
2
12 min
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Solutions, Fluid Mechanics for Chemical Engineers, Third Edition, Chapter 15, page 6
15.14 For both parts of the problem we use as our system
a small vertical strip, as sketched at the right. The film
thickness increases with distance down the plate (the
minus z direction). The distance into the paper, (the y
direction ) is l.
x
z
Thus, the material balance has the volume
Vmaterial balance reginon = l z x . Here l z is constant down
the plate, independent of time, but x is a function of
both position and time.
(a) We go back to Ch. 3 and write that
accumulation = flow in flow out
x x
x
l z
= l Vz dx
l Vzdx
0
0
t
at z= z
at z = z + z
x
We now simplify the problem greatly by writing l Vzdx
lVz, average x , which
0
at z = z
x
= Vz, averagex
Vz , average x
makes the above become z
at z = z
at z =z + z
t
Vz, average x
x
Dividing by z and taking the limit as z goes to zero
=
t
z
which is the proper form of the material balance. This is one of the cases in which we
find Eq. 3.6 much more useful than the equations in Sec. 15.2. That is the case for
almost all unsteady state problems.
(b) To apply the momentum balance, we look back to Prob. 15.9, and copy
Q
gl ( x )3
g( x )2
Q =
Vz , average =
=
;
l x
3
3
g( x )2
x
Vz, average x
3
g(x ) 2 x
=
=
z
z
z
2
x g( x ) x
( z)
=
which has solution x =
g t
t
z
This problem is presented as drainage down a non-moving wall. A more common
problem is the drainage down a surface being pulled out of a coating bath, e.g. a
continuous strip of sheet metal, coated by passing it through a bath of paint. That
solution starts with this result, and extends it, see J. J. van Rossum, "Viscous Lifting and
Drainage of Liquids", Appl. Sci. Research, A7, 122-144 (1958).
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[
[
]
[
]
[
]
]
15.15 Starting with Eq. C.17 we observe that the flow is steady, and has no velocity
component in the y direction, (along the slot). The pressure is practically constant (the
pressure of the surrounding air), and the velocity and velocity gradients in the x direction
(perpendicular to the slot) are very small compared to the velocity in the z direction
(which is negative because the flow is downward). Thus
Solutions, Fluid Mechanics for Chemical Engineers, Third Edition, Chapter 15, page 7
0 + Vx
2 Vz2
Vz
V
2V
+ 0 + Vz z = g 0 + 2 + 0 + 2z
x
z
z
x
By material balance for a constant-density fluid (Eq. 15.8) we see that
x
Vx = Vz
z
2 Vz2 2 V
Vz
V
>> z and g >> 2 + 2z so that,
Except near the slot, z >> x so that
z
x
z
x
2
V
V2 V
Vz z g , z z initial = g(z)
( z is negative! )
z
2
2
If the slot is thin, then x is much less than z, except near the outlet from the tank. So near
the tank there is a significant inward velocity, but as the film falls, the inward velocity is
much less than the downward velocity.
To solve the problem we would need to know the dimensions of the slot, and the
initial velocity distribution. As the fluid falls, the effect of viscosity is to change the
initial velocity distribution, whatever it is, to a planar velocity distribution (if the shear
stress due to the adjacent air is negligible.)
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15.16 Starting with Eq. 15.26 we observe that Vy = Vz = 0 at all times and locations,
and that their derivatives are also zero. The x component of the velocity depends only on
y, not on z or x. There is no pressure gradient, and gravity plays no role, so the
2V
V
V
2V
remaining terms are x = + 2x , more commonly written x = 2x .
t
y
t
y
The same differential equation with the same boundary conditions is solved in all heat
transfer books, where it is the solution for the unsteady-state temperatures in a semiinfinite plane, initially at a uniform temperature, with the surface at x = 0 suddenly raised
to some other temperature. The equivalent mass transfer by diffusion problem is solved
in most mass-transfer books. This same equation reappears in Chapters 17, 19 and 20.
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15.17 Here the velocities are all in the z direction, so we drop the subscript on V. Then
repeating Sec. 6.3 for this cylindrical shell, we find
dV
dV
2
2 dV
P [(r + dr) 2 r 2 ]+ 2 ( r + dr)
r
x = x [( r + dr ) r ]
dr
dt
dr
2
2
2
2
2
We observe that [( r + dr ) r ]= r + 2rdr + (dr ) r 2rdr
Making this substitution twice and dividing by x we find
P
dV
dV
dV
2rdr + 2 (r + dr)
r
= 2rdr
x
dr
dr
dt
dV
dV
P
dV
Next we divide by 2rdr
+
(r + dr )
r
=
x rdr
dr
dt
dr
Solutions, Fluid Mechanics for Chemical Engineers, Third Edition, Chapter 15, page 8
d dV
r Making this substitution leads to Eq. 15.29.
r dr dr
One could also find the results in Sec 6.3 by making the steady-state assumption, which
dV
term
drops the
dt
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The middle term is
15.18 Start with Eq. C.17 and observe that Vy = Vx = 0 at all times and locations, and
2V
V
P
that their derivatives are also zero. Then z + 0 + 0 + 0 = 0
+ 2z + 0 + 0
t
z
x
or
V
P
2V
z = + 2z which is quite similar in form to Eq. 15.29.
t
z
x
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15.19 Vmax is proportional to (-dP/dx). Thus, for a given overall pressure difference,
doubling the pipe length will reduce all velocities by a factor of 2. Fig 15.4 assumes that
the pipe is very long relative to the entrance region.
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15.20* From Fig 15.4 we read that t / R 2 0.3 so that the time is (0.3/0.05) = 6 times
the time for 0.2 of the steady state velocity, or 1.78 s.
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m
s = 100 To apply Fig. 15.7 we observe that it shows the
15.21`` (a) R =
2
4 m
10
s
centerline velocities. 95% of the velocity at infinite distance corresponds to
Vz / V0 = 2 0.95 = 1.90
We have a choice of several curves on Fig. 15.7. Two of them show values of
z / D0R 0.03 while the third shows z / D0R 0.037 . Using the first value, we have
z = 0.03 0.01 m 100 = 0.03 m = 3 cm
(b) Using z / D0R 0.03 we enter Fig 15.8 and read the two curves as 0.5 and
0.8. For this flow
kg m 2 N s 2 Pa m 2
z V2
= 4 0.8 998.2 3 1
= 1.6 kPa
P = 4 favg
m s kg m
N
D0 2
(c) is the same as (b) with the 0.8 replaced by an 0.5, leading to
2
2
2
2
favg z V = 4 0.5 998.2 kg 1 m N s Pa m = 1.0 kPa
P = 4
m 3 s kg m
N
D0 2
Both of these calculated pressure drops are small, but the one taking the entrance effect
into account is 160% of that for Poiseuille flow.
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(0.01 m ) 1
Solutions, Fluid Mechanics for Chemical Engineers, Third Edition, Chapter 15, page 9
1
(rV ) dividing out the
r r r
1
(rV ) integrating once gives 0 = 1 (rV ) + C1
, we have 0 =
r r
r r r
2
r
multiplying through by r and integrating again gives 0 = rV + C1
+ C2 or
2
r
C
V = C1 + 2
2
r
R2
r
R 2 C1
R C2
r
At r = R, V = 0 ,
, V = C1 C1
0 = C1 +
=
2
2
R
2r
2
r
2 kR
C1 =
,
At r = kR, V = kR ,
1
R k
k
2
2 kR 1
R
R2
k 2
r Multiplying by minus 1 and
V =
r = 2
1
r k 1
r
R k 2
k
rearranging gives Eq. 1.AH.
(b) For the case with the outer cylinder rotating, again with angular velocity
k 2R 2
kR C2
r
k 2 R 2 C1
r
we have at r = kR, V = 0 ; 0 = C1 +
; V = C1 C1
=
2 kR
2
2r
2
r
2
k 2R 2
r
. Removing the minus
At r = R, V = R , C1 =
; V =
r
(1 k 2 )
(1 k 2 )
k 2R2
signs produces V = 2
r .
(k 1) r
d
d V
= r into 1.AH we have
(c) Substituting = r
dr r
dr
2
2
k R r
2
r k 2 d R 2 r k 2 2 R 2
d k 1 r
2 1 = 2
=r
= 2
dr
r
k 1 r3
k 1 dr r
2
kRk 2 2 R 2
substituting r = kR produces = 2
(kR)3 = k 2 1 Clearing the minus sign
k 1
produces Eq. 1.AJ.
(d) The torque on the inner and outer cylinders must be equal and opposite.
Ignoring the sign change we can say that = ( Ar ) inside = ( A r )outside The areas are
2
outside rinside
2
=
proportional to r so that
= k . Substituting r = R in the above
inside routside
the value at
r k 2 2 R 2 2k 2
3 = 2
. The
equation for produces = 2
= k2
k 1 r k 1
the inner cylinder
15.22 (a) In Eq. C.19 all of the terms are zero except
Solutions, Fluid Mechanics for Chemical Engineers, Third Edition, Chapter 15, page 10
calculated viscosity is the ratio of these quantities. The k2 terms cancel, and the
computed viscosity is the same.
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Solutions, Fluid Mechanics for Chemical Engineers, Third Edition, Chapter 15, page 11