PROBLEM SOLUTIONS
CHAPTER 1. PRELIMINARY CONCEPTS
1-1
A 1.4-cm diameter sphere placed in a freestream at 18 m/s at 20°C and 1 atm. Compute the
diameter Reynolds number for 3 cases:
(a) Air: Table A-2 - at 20°C, = 1.205 kg/m3 , = 1.81 E-5 Pa-s. Then
Re D = VD/ =
(1.205)(18 )( 0.014 ) = 16, 800
(Ans.)
1.81E-5
(b) Water: Table A-1 - at 20°C, = 998 kg/m3 , = 1.002 mPa-s:
ReD = ( 998)(18)( 0.014) / ( 0.001002) = 251,000
(Ans.)
(c) Hydrogen: Table A-3, M = 2.016, then R = 8313/M = 4124 m2 /s 2 -K. Thus estimate
= p/RT = (101350 ) / ( 4124 )( 293) = 0.0838 kg/m3. From Table 1-2 for hydrogen,
o ( T/To ) = (8.411E-6)( 293/273)
n
068
= 8.83 E-6 Pa-s
Then ReD = ( 0.0838)(18)( 0.014) / (8.83 E-6) = 2,400
(Ans.)
At what wind velocity will an 8-mm-diameter wire “sing” at middle C (256 Hz)?
For air at 20°C, assume v 1.5E-5 m 2/s. From Fig. 1-8 guess a vortex-shedding Strouhal
number of 0.2 [check the Reynolds number afterward]. Then
fD/U 0.2 = ( 256 )( 0.008) /U, or U 10.24 m/s. At this speed the Reynolds number is
1-2
ReD = UD/v = (10.24)( 0.008) / 1.5E-5 = 5400. This is nicely in the range where fD/U = 0.2.
Perhaps we could iterate just a little more closely to obtain
fD/U 0.205, Re = UD/v 5300, or U = 10.0 m/s
(Ans.)
1-3
If U = 12 m/s in Prob. 1-2 above, what is the wire drag in N/m?
For air assume = 1.205 kg/m3 and v = 1.5E-5 m2/s. The Reynolds number is
ReD = UD/v = (12)( 0.008) / ( l.5E-5) = 6400
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-1-
From Fig. 1-9 at this Reynolds number, estimate a drag coefficient of 1.1. Then
2
1
Fdrag = CD V2 ( DL ) = 1.1( 0.5)(1.205)(12 ) ( 0.008)(1.0 ) = 0.76 N /m
2
(Ans.)
1-4
Given, without proof, the Poiseuille-paraboloid laminar-pipe-flow formula from
Chap. 3, u = (C/)(R 2 − r 2 ), find the wall shear stress if u max = 30 m/s, D = 1 cm, and
= 0.3 kg/(m-s). [The exact analysis will be given in Sect. 3-3.1.]
Examining the formula, we see that the maximum velocity occurs on the centerline:
(
u mx = u ( r = 0 ) = CR 2 / = 30 m/s = C ( 0.005 ) / ( 0.3) , or: C = 3.6E5 N/ m 2 -s 2
2
)
With C thus known for this data, we may evaluate wall shear stress by differentiation:
wall =
u
r
r =0
2RC
=
= 2RC = 2 ( 0.005 )( 3.6E5 ) = 3600 Pa
(Ans.)
We should check the Reynolds number Re D but we don’t know the density. But “oil” is usually
in the range 900 kg/m3. Then ReD = u max D/ = ( 900 )(30 )( 0.01) / ( 0.3) 900, which is
well within the laminar-flow range.
1-5
Glycerin at 20 C is confined between two large parallel plates. One plate is fixed and the
other moves parallel at 17 mm/s . The distance between the plates is 3 mm . Assuming
no-slip, estimate the shear stress in the glycerin, in Pa.
Solution: Glycerin at 20 C is confined between two large parallel plates. One plate is fixed and
the other moves parallel at V = 17 mm/s . The distance h between the plates is 3 mm.
u=V
Moving plate
h
Glycerin
u=0
Fixed plate
For glycerin at 20 C , the viscosity = 1.5 kg/m s .
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-2-
Assuming no-slip, the shear stress =
1-6
V
h
=
(1.5 kg/m s ) (17 10−3 m/s )
( 3 10
−3
m/s
)
= 8.5 Pa .
(Ans.)
Given a plane unsteady viscous flow in polar coordinates:
v r = 0; v =
r 2
C
1
−
exp
−
r
4vt
Compute the vorticity and sketch some profiles of vorticity and velocity.
From Appendix B, the vorticity is
r2
1
C
z =
( rv ) = exp −
r r
2vt
4vt
The instantaneous velocity and vorticity profiles are plotted at top. At t = 0, the flow is a “line”
vortex, irrotational everywhere except at the origin ( = ) .
1-7
Given the two-dimensional unsteady flow u = x/ (1+t ) , v = y/ (1+2t ) , find the equation
for the streamlines which pass through the point (x 0 , y0 ) at time ( t = 0 ) . From the geometric
requirement for two-dimensional streamlines at any instant,
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dy
v y (1 + t )
= =
dx streamline u x (1 + 2t )
Holding time t constant, we may separate the variables and integrate to obtain y = C x n , where
n = (1 + t ) / (1 + 2t ) . To satisfy the initial condition, we must have C = y0 / ( x0 ) . The final result
for the (unsteady) streamlines is
n
n
y x
1+ t
= , n =
y0 x 0
1 + 2t
(Ans.)
1-8
For the inviscid streamline approaching the forward stagnation point of the cylinder in
Fig. 1-5, evaluate the strain rates and the time to go from ( 2R, ) to ( R, )
(
)
(
)
From Eqs. (1-2), v r = U 1 − R 2 /r 2 cosθ, v = − U 1 + R 2 /r 2 sinθ
Then, from Appendix B, evaluate the normal and shear-strain rates along the line = :
vr 2U R 2 cos
2U R 2
rr =
=
|= = −
r
r3
r3
2U R 2 cos
2U R 2
1 v vr
=
+ =−
|= = +
r
r
r3
r3
r =
1 vr v v 4U R 2 sin
+
−
=
|= = 0
r
r
r
r3
(Ans. a)
The particle moving toward the stagnation point gets shorter in the “r” direction and fatter by the
same amount in the “ ” direction, thus maintaining constant volume for this incompressible flow.
The shear strain rate is zero because we are on a line of symmetry.
For part (b), by definition, the radial velocity along the stagnation line ( = ) is
vr =
(
dr
= − U 1 − R 2 /r 2
dt
)
We may separate the variables and integrate to find the time of travel between ( 2R ) and ( R ) :
R
− U t =
R
R r − R
r 2 − R 2 = r + 2 ln r + R = −
2R
2R
r 2dr
(Ans. b)
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-4-
It takes infinite time to actually reach the stagnation point, where V = 0 .
1-9
Use the approximate equation of state for water,
p/po ( A + 1) ( /o ) − A
n
with A 3000, n = 7
to compute the following quantities for water, po = 1 atm, o = 998 kg/m3 :
(a) the pressure required to double the density of water:
p/po = ( 3000 + 1)( 2.0) − 3000 = 381128, or: p = 381,000 atm
7
(Ans. a)
(b) the bulk modulus K of water at 1 atm. By definition,
K =
dp
nn −1
|T = po ( A + 1) n = npo ( A + 1) = 21007 po at 1 atm.
d
o
Thus the bulk modulus is K 21,007 atm 2.13 E9 Pa
(Ans. b)
(c) the speed of sound at 1 atm:
a 1 atm = ( K /o )
1/2
1/2
2.13 E9 Pa
=
3
998 kg/m
= 1460 m /s
(Ans. c)
These are accurate estimates of the measured compressibility and sound speed of water.
1-10 As shown, a plate slides down an incline on a film of oil of viscosity = 5E-4 slug/ft-s
(a) Estimate the terminal sliding velocity V*:
Acceleration is zero, so
W sinθ = A =
V
A
y
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-5-
where A is the plate area touching the oil film.
Assuming a linear velocity profile, V = V* and y = the film thickness, hence
V*
V
W sinθ = 40sin ( 30 ) =
A = ( 0.0005 )
9
0.005/12 ( )
y
in English units.
Hence solve for V* ( terminal ) = 1.85 ft /s
(Ans. a)
(b) Estimate the time for the plate to accelerate from rest to 99% terminal velocity: If x is down
the incline, then a dynamic force balance gives
V
Fx = W sinθ − y A =
or:
W dV
,
g dt
dV gA
+
V = g sinθ
dt W y
The solution to this first-order linear ordinary differential equation is
gA
4.605W y
V = V* 1 − exp −
t = 0.99V* if t* =
gA
Wy
For our data, then, the time to reach 99% of terminal velocity is
t* =
1-11
4.605 ( 40 )( 0.005/12 )
= 0.53 sec
( 32.2)( 0.0005)( 9.0)
(Ans. b)
Estimate the viscosity of nitrogen at 86 MPa and 49°C. From Appendix A-3, for N 2 , read
Tc = 226R = 126K, pc = 33.5 atm, c = 18.0 E-6 Pa-s. At this high pressure, we cannot use
“low density” formulas but rather must use Fig. 1-17. Compute ratios:
T 49 + 273
p
86E6
=
= 2.55;
=
= 25.3; Read
2.5 0.1
Tc
126
pc 33.5 (101350 )
c
Then our estimate is = 2.5 c = 2.5 (18.0) = 45 ± 2 μPa-s
(Ans.)
The agreement with the measured value (also 45 Pa-s ) is excellent.
1-12 Estimate the thermal conductivity of helium at 420°C and 1 atm. This is truly “lowdensity”, since p pc and T Tc . A power-law estimate would be based on 0°C:
k k o ( T/To )
n
420 + 273
( 0.142W/m-K )
273
072
0.278 W /m-K
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-6-
Alternately, we could use the kinetic theory formula, Eq. (1-41). From Appendix A-5 for helium,
= 2.551 Å, T = 10.22K, and M = 4.003. First use Eq. (1-34) to compute
420 + 273
v 1.147
10.22
−0.145
.20
420 + 273
+
+ 0.5
10.22
= 0.6226
[check with Table 1-1]
Our estimate from (1-41) then is
k=
0.0833 T
2 v
M
=
( 693)
= 0.27 W /m-K
( 2.551)2 ( 0.6226 ) 4.003
0.0833
The agreement with the experimental value of 0.28 W/m-K is good for both estimates.
1-13
According to Table C-5 and Fig. 1-15, at what pressure is the viscosity of CO2 equal to
approximately 30 10−5 Pa s when the temperature is 800°R ?
Solution: T = 800°R = 444.4444 K , Tc = 548°R = 304.4444 K , Tr =
= 30 10−5 Pa s , c = 3.43 10−5 Pa s , r =
c
=
T 444.4444 K
=
= 1.46
Tc 304.4444 K
30 10−5 Pa s
= 8.75
34.3 10−6 Pa s
Thus, pr = 25 (from Fig. 1-15).
Then, pressure p = pc pr = ( 72.9 atm) 25 = 1822.5 atm .
(Ans.)
1-14 Fit the given viscosity-vs-temperature data for ammonia gas to power-law and Sutherlandlaw formulas.
(a) The power-law is an excellent fit to this data. Taking To = 300K, we obtain, by least-squares
to a log ( ) vs. log ( T ) plot,
1.051
T
o To
0.3% for To = 300K
(Ans. a)
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-7-
(b) The Sutherland-law is not an especially good fit to the data, which has rising with T at an
increasing rate. It may be fit to least squares by minimizing the functional
( ) (
* 3/2
T
1 + S*
i
*
i − T* + S*
i =1
i
6
)
2
* T * S
*
where = , T = T ,S = T
o
o
o
for the given six data points. The minimum is found by differentiating the functional with respect
to S* , v with the result S* = 1.91, or:
Sbest fit 573°K
(Ans. b)
The error is 2.4%, or eight rimes more than the power-law fit.
1-15
Experimental data for the viscosity of helium at low pressure are as follows:
T, °C
0
100
200
300
400
500
µ, Pa·s 1.87 × 10–5 2.32 × 10–5 2.73 × 10–5 3.12 × 10–5 3.48 × 10–5 3.48 × 10–5
Fit these values to a suitable formula.
Solution: Experimental data for the viscosity of helium in Kelvin scale at low pressure are as
follows:
T, K
273
373
473
573
673
773
µ, Pa·s 1.87 × 10–5 2.32 × 10–5 2.73 × 10–5 3.12 × 10–5 3.48 × 10–5 3.48 × 10–5
n
T
= .
Using power law curve,
0 T0
For 0 = 1.87 10−5 Pa s and T0 = 273 K
T 373 K 473 K 573 K 673 K 773 K
n 0.691
Therefore, the mean value
n=
0.688
5
n
i =1 i
5
0.69
0.688
0.597
= 0.671 ( 4 % accuracy for 250 K T 1000 K ).
(Ans.)
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-8-
1-16 Analyze newtonian flow between parallel plates (Fig. 1-15) with a finite slip velocity
u = l ( du/dy ) at both walls.
The velocity profile is still linear, but with slip at both walls the slope is less, as shown in
the sketch:
du/dy = ( V − 2u ) /h
Introducing u from the slip relation, we obtain
du
V
V
=
or: w =
at both walls
dy h + 2l
h + 2l
(Ans.)
1-17 Derive Eq. (1-106) from a balance of forces on the differential surface-area alement shown
in the problem.
Since the sliver of area is negligibly thin, it has no weight. The pressures act on a projected
surface area dSx dSy . The surface tension forces are slanted slightly upward, at angles
(dx /2) and (dy /2), respectively. The force balance is
(
)
2T dSysin ( dx /2 ) + 2T dSxsin dy /2 + ( p − pa ) dSx dSy = 0
For differentially small angles, sin ( d) = d. Clean up this equation and rearrange:
d
dy
p = pa − T x +
= pa − T
dSx dSy
1
1
+
Rx Ry
(Ans.)
since, by definition, d/dS = 1/R, where R is the radius of curvature.
1-18 Two bubbles of radii R1 and R 2 coalesce isothermally into a single bubble R 3. Find the
radius of the new (single) bubble.
Because of surface tension, the pressure inside a bubble (which has two surfaces) is higher
than ambient, p = po + 4T /R. Assuming that no interior-bubble air mass escapes during the
coalescence, m1 + m2 = m3 , or, for an ideal isothermal gas of temperature T,
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-9-
po + 4I /R1 4 3 po + 4I /R2 4 3 po + 4I /R3 4 3
R1 +
R2 =
R3
T
3
T
3
T
3
where is the gas constant. Canceling common terms and cleaning up, we have
(
)
(
po R33 + 4I R32 = po R13 + R23 + 4I R12 + R22
)
(Ans.)
This must be solved numerically or algebraically for the new radius R3 .
1-19 In Prob. 1-1, if the temperature, sphere size, and velocity remain the same for air flow, at
what air pressure will the Reynolds number Re D be equal to 10,000?
Solution: From Prob. 1 −1 , T = 293 K, D = 0.014 m, V = 18 m/s, and = 1.81E-5 Pa-s. Use the
specified Reynolds number to compute the required air density:
Re D =
VD (18 m/s )( 0.014 m )
=
= 10, 000 Solve = 0.718 kg /m3
1.81E -5 kg /m-s
Ideal gas: = 0.718 =
p
p
=
, Solve for p = 60400 Pa 60 kPa
RT ( 287 )( 293)
(Ans.)
1-20 A solid cylinder of mass m, radius R, and length L falls concentrically through a vertical
tube of radius R + R, where R R. The tube is filled with gas of viscosity and mean free
path . Neglect fluid forces on the front and back faces of the cylinder and consider only shear
stress in the annular region, assuming a linear velocity profile. Find an analytic expression for the
terminal velocity of fall, V, of the cylinder (a) for no-slip; (b) with slip, Eq. (1-91).
Solution: (a) For no-slip, the shear stress in the thin annular region between cylinders is
=
u
V
V
=
, then W = mg = Fshear = Awall =
( 2 RL )
y
R
R
Solve for Vno-slip =
mg R
2 RL
Ans. (a)
(b) For slip, modify the shear stress (see Prob. 1.16 for another example):
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-10-
u =
du
V -2 u
du
V
V
=
, or:
=
, =
dy
R
dy R + 2
R + 2
As above in part (a), mg = Aw , Vslip =
1-21
mg ( R + 2
2 RL
)
Ans. (b)
Solve P1-20 for the terminal fall velocity for no-slip if the cylinder is aluminum, with
diameter 4 cm and length 10 cm . The tube has a diameter of 4.02 cm and is filled with
argon gas at 20 C .
Solution: For no-slip, the shear stress 𝜏 in the thin annular region is =
u
V
=
.
y
R
V
Then, W = mg = Fshear = Awall =
( 2 RL ) .
R
Al = 2710 kg/m 3 , R = 0.02 m , L = 0.1 m , g = 9.81 m/s2 ; then, mg = Al R 2 L = 3.3408 N .
R = ( R + R ) − R = 0.0201 m − 0.02 m = 0.0001 m
Ar = 2.24 10−5 Pa s
Therefore, Vno − slip =
mg R
2 RL Ar
=
( 3.3408 N )( 0.0001 m )
( 0.01256 m )( 2.24 10
2
−5
Pa s )
= 1187.4431 m/s
(Ans.)
1-22 In Fig. Pl-22 a disk rotates steadily inside a disk-shaped container filled with oil of viscosity
. Assume linear velocity profiles with no-slip and neglect stress on the outer edges of the disk.
Find a formula for the torque M required to drive the disk.
Fig. Pl-22
Solution: The disk tangential velocity varies with radius, V = r, hence the local shear stress is
= r /h on the top and bottom of the disk. The torque on a circular strip dr wide is
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-11-
r
dM = ( dA) r ( 2 sides ) = 2r
2 r dr
h
or: M = 4
h
R
r dr =
3
0
R 4
h
Ans.
1-23 Show, from Eq. (1-86), that the coefficient of thermal expansion of a perfect gas is given
by = 1/T . Use this approximation to estimate of ammonia gas ( NH3 ) at 20°C and 1 atm and
compare with the accepted value from a data reference.
Solution: Introduce the ideal-gas law into the definition of :
1
1 p
1 −p 1 1
=− =−
=
=
=−
T p
T RT p
RT 2 T T
Ans.
It doesn’t matter what gas we are considering, ammonia or carbon dioxide or whatever, the ideal
gas approximation predicts = 1/T = 1/293K = 0.00341 K -1.
Ans.
This estimate is very close to estimates for ammonia in the literature, e.g., White (1988).
1-24 The rotating-cylinder viscometer in Fig. P1-24 shears the fluid in a narrow clearance r,
as shown. Assuming a linear velocity distribution in the gaps, if the driving torque M is measured,
find an expression for μ by (a) neglecting, and (b) including the bottom friction.
Fig. Pl-24
Solution: (a) Analyze the annular region only. The shear stress equals ( du/dy ) ( R /R ) .
The shear force on the cylinder side is normal to the radius, and the driving moment must be
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-12-
M = RdF = R ( dAw ) =
2
0
R L
R
R
RLd = 2
R
R
3
Solve for =
M R
Ans. (a)
2 R3 L
(b) On the bottom, the shear stress varies linearly with radius:
R
M bottom
R
2 3
2 R 4
r
= r dAw = r
r dr =
2 rdr =
R
R 0
4R
0
Thus M total =
2 R3 L 2 R 4
M R
+
, Solve =
R
4R
2 R3 ( L + R /4 )
Ans. (b)
1-25 Consider 1 m3 of a fluid at 20°C and 1 atm. For an isothermal process, calculate the final
density and the energy, in joules, required to compress the fluid until the pressure is 10 atm, for
(a) air; and (b) water. Discuss the difference in results.
Solution: (a) The work done is − pd , where is the volume. From the ideal-gas law,
p = mRT . Thus
2
W1−2 = − pd = −
1
mRT
p
d = −mRT ln 2 = p11 ln 2
1
p1
( )
10
= (101350Pa ) 1m3 ln = 233,000 J
1
Ans. (a)
(b) For water, we could use the compressed-liquid tables, but we can estimate the (very small)
result from the bulk modulus K = ( dp /d )T = 2.23E9 Pa for water, Eq. (1-84). The change in
volume of the water is very small when the change in pressure is only 9 atm:
−
p
K
1m3 ) ( 9 )(101350 Pa )
(
=−
−0.0041 m3
2.23E 9 Pa
A slightly more accurate estimate from Prob. 1-9, or from the compressed-liquid tables, gives
−0.00042 m3. Then the work required to compress water from 1 atm to 10 atm is
(
)
W1−2 = − pd − pavg = − ( 5.5 )(101350 Pa ) −0.00042 m3 230 J
Ans. (b)
This is 1000 times less than Ans.(a) for air above, since water is nearly incompressible.
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-13-
1-26 Equal layers of two immiscible fluids
are being sheared between a moving and a fixed
plate, as in Fig. P1-26. Assuming linear velocity
profiles, find an expression for the interface
velocity U as a function of V , 1, and 2 .
Fig. P1-26
Solution: The shear stress is the same in each layer:
1 = 1
V −U
U
= 2 = 2
,
h /2
h /2
solve for
U=
1
V
1 + 2
Ans.
1-27 Utilize the inviscid-flow solution of flow past a cylinder, Eqs. (1-3), to (a) find the location
and value of the maximum fluid acceleration amax along the cylinder surface. Is your result valid
for gases and liquids? (b) Apply your formula for amax to air flow at 10 m/s past a cylinder of
diameter 1 cm and express your result as a ratio compared to the acceleration of gravity. Discuss
what your result implies about the ability of fluids to withstand acceleration.
Solution: Along the cylinder surface, r = R, and Eqs. (1-3) reduce to vr = 0 and v = −2U sin .
Thus, along the surface, the absolute velocity is V = 2U sin ( s /R ) , where s is the arc length along
the surface, measured from the front stagnation point. There is a convective acceleration given by
a =V
dV
s 2U
s
= 2U sin cos
ds
R R
R
(a) The acceleration is a maximum at = 135, or s/R = /4. Thus amax = 2U 2 /R. Ans. (a)
This result is valid for all fluids, gases or liquids, in the inviscid approximation.
(b) For the given data, R = 0.005 m, U = 10 m/s, compute, independent of fluid properties,
amax = 2 (10 m/s ) / ( 0.005 m) = 40,000 m/s2 4080 g’s
2
Ans. (b)
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-14-
The lesson is that fluids have no fear of huge accelerations that would defeat a human being.
1-28
The coefficient of thermal expansion is defined as
=−
1
T
p
Determine for an ideal gas with p = RT . Show your work in detail.
Solution: Given, the coefficient for thermal expansion is = −
1
T
p
Using p = RT for ideal gas,
=−
1-29
1 p
1
p
1 RT 1
= − −
=
= − −
2
T RT p
RT
RT 2 T
(Ans.)
2
a,
3
and Newton’s expression of the wall shear stress as a function of the velocity gradient,
u
u
w = , express Maxwell’s slip velocity, uw = ,
y w
y w
Starting with Maxwell’s low-density approximation of the viscosity, namely,
(a) as a function of the shear stress, density, and speed of sound a ;
(b) as a function of the Mach number, the mean-flow velocity U , and the skin friction coefficient,
2
C f = w2 .
U
Solution: Given, Maxwell’s low-density approximation of viscosity is
density,
= mean free path, and a = speed of sound.
2
a , where =
3
u
Newton’s wall shear stress 𝜏𝑤 as a function of velocity gradient is w = .
y w
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Hill LLC.
-15-
Slip velocity 𝑢𝑤 is uw = w , and constant =
.
w 3 w
u
w
Therefore, uw =
.
=
=
2
y w
a 2 a
3
(Ans.)
Dividing by mean-flow velocity U and arranging gives
Mach number is Ma =
Therefore, uw =
1-30
uw 3 U 2 w
=
.
U 4 a U 2
U
2
, and skin friction coefficient is C f = w2 .
U
a
3
Ma U C f .
4
(Ans.)
Consider a hydraulic lift with a 50 cm diameter shaft sliding inside a housing with an
inside diameter of 50.02 cm . If the shaft travels at 0.25 m/s , calculate the shaft
resistance to motion per unit length. You may use water as the working fluid.
V
Solution: Shaft resistance F to motion is F = Awall =
( 2 RL ) .
R
R = 0.25 m , L = 1 m , V = 0.25 m/s
R = ( R + R ) − R = 0.251 m − 0.25 m = 0.001 m
water = 1.02 10 −5 Pa s (at 20 C and 1 atm )
Funit -length
1-31
(1.02 10
=
−5
Pa s ) ( 0.25 m/s ) 2 ( 0.25 m )(1 m )
( 0.001 m )
= 4.0055 10−3 N
(Ans.)
Consider a thin air gap of 1 mm that is formed between two parallel surfaces that are
maintained at 20 C and 40 C , respectively. In the case of a quiescent medium (say still
air), calculate the heat transfer rate across the gap per unit area.
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Hill LLC.
-16-
Solution: The rate of heat transfer per unit area (in one-dimensional space) is
T ( x + x ) − T ( x )
q −k
.
x
The parallel plates are at 20 C and 40 C . Then, k needs to be determined for 30 C .
n
Using power law curve, kair
T
303 K
= k0 = ( 0.0241 W/m K )
273 K
T0
0.81
313 K − 293 K
2
Therefore, q − ( 0.0262 W/m K )
= 524 W/m .
0.001 m
1-32
= 0.0262 W/m K .
(Ans.)
In the presence of viscosity, the pressure drop associated with a fully developed laminar
motion in a horizontal tube of length L and diameter D may be evaluated analytically.
One finds:
p = p1 − p2 =
128 LQ
D4
1
where stands for the dynamic viscosity and Q = D 2V denotes the volumetric flow rate.
4
Show that the corresponding head loss may be written as
hL =
p1 − p2
L V2
= f lam
g
D 2g
What value of f lam do you obtain?
Solution: Consider fully developed flow and apply steady-flow energy equation between section 1
and section 2.
p
p
V2
V2
+
+
z
=
+
+ z + hS − hL
2g
2g
g
2 g
1
Use z1 = z2 (horizontal), V1 = V2 (constant cross-section), 1 = 2 (same velocity profile), hS = 0
p − p2 p
=
(no pump); to simplify as hL = 1
… (1)
g
g
1
2
Q
D
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Hill LLC.
-17-
L
Empirical data on viscous losses in straight sections of pipe are correlated by the dimensionless
Darcy friction factor f =
p
D
… (2)
1
2 L
V
2
From equation (2), p = f
1
L
V 2
… (3)
2
D
Combining equations (1) and (3) gives hL = f lam
L V2
(where f = f lam , for laminar flow).
D 2g
(Ans.)
In the presence of viscosity, the pressure drop associated with a fully developed laminar motion in
128 LQ
a horizontal tube of length L and diameter D is p = p1 − p2 =
… (4)
D4
1
Dynamic viscosity is , and volumetric flow rate is Q = D 2V … (5)
4
For fully developed laminar flow, using equations (4) and (5) in equation (2) gives
f lam
1-33
1
128 L D 2V
4
2 D = 64 = 64
4
D
V 2 L VD Re D
(Ans.)
A time-dependent, two-dimensional motion has three velocity components that are
given by
u=
x
1 + at
v=
y
1 + bt
w=0
where a and b are pure constants. The objective of this problem is to compare and contrast the
streamlines in this flow with the pathlines of the fluid particles.
(a) Find the equations governing the streamline that passes through the point (1,1) at time t .
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Hill LLC.
-18-
(b) Calculate the path of a particle that starts at r0 = ( x0 , y0 ) = (1,1) at t = 0 . Determine the
location of a particle at t = 1 , denoted as r1 .
(c) Use the results of part (a) to determine the condition under which the streamlines and
pathlines coincide.
Solution: The geometric requirement for two-dimensional streamlines is as follows:
v dy y (1 + at )
=
=
u dx x (1 + bt )
By separating the variables,
dy 1 + at dx
=
.
y 1 + bt x
1 + at
By integrating, ln ( y ) =
ln ( x ) + ln ( C ) .
1 + bt
1+ at
Solving this gives y = Cx 1+bt (where C is the integration constant).
The condition of the streamline passing through the point (1, 1) at time t is C = 1 must be
satisfied.
Therefore, the governing equation is y = x
1+ at
1+bt
… (1)
(Ans.)
The rate of change of x component of particle velocity with respect to time is
dx
x
=
.
dt 1 + at
1
By separating the variables and integrating, x = C1 (1 + at ) a (where C1 is the integration
constant).
The rate of change of y component of particle velocity with respect to time is
dy
y
=
.
dt 1 + bt
1
By separating the variables and integrating, y = C2 (1 + bt ) b (where C2 is the integration
constant).
At t = 0 , x = x0 = 1 = C1 and y = y0 = 1 = C2 .
1
Therefore, the path of the particle is
(1 + bt ) b
y=
1
(1 + at ) a
x … (2)
(Ans.)
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Hill LLC.
-19-
1
1
Therefore, at t = 1 , x = x1 = (1 + a ) a and y = y1 = (1 + b ) b .
1
1
Then, r1 = ( x1 , y1 ) = (1 + a ) a , (1 + b ) b .
(Ans.)
Comparing equations (1) and (2), we can say that the condition under which the streamlines
coincide with pathlines is a = b = 0 .
(Ans.)
1-34
A tornado may be simulated as a two-part circulating flow in cylindrical coordinates,
with vr = vz = 0 ,
r ( r R )
v = R 2
(r R)
r
(a) Calculate the divergence of the velocity. Is the flow compressible or incompressible?
(b) Determine the vorticity. Is the flow rotational or irrotational?
(c) Determine the strain rates in each segment of the flow. What is the sum of the three normal
strain rates?
Solution:
(1)
vϴ
(2)
R
Divergence
of
velocity
is
v =
r
1
1
( rvr ) +
( v ) + ( vz ) = 0
r r
r
z
(incompressible).
(Ans.)
1 ( rv ) 1 vr
1 vz v
vr vz
−
−
−
Vorticity is = v =
ez .
e +
er +
r
z r
r z
r r
Only nonzero component of vorticity is z =
1 ( rv )
.
r r
For segment (1), z = 2 (nonzero; thus rotational).
(Ans.)
(Ans.)
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Hill LLC.
-20-
For segment (2), z = 0 (irrotational).
(Ans.)
1 v v
Only nonzero component of tangential strain rate is r = − .
2 r
r
(Ans.)
For segment (1), r = 0 .
(Ans.)
For segment (2), r = −
R2
.
r2
(Ans.)
The sum of normal strain rate components is rr + + zz =
vr vr 1 v
+ +
r r r
vz
=0.
+
z
(Ans.)
1-35
In modeling the motion of an 8-meter diameter tornado rotating at an angular speed of
at the point of maximum swirl, it is possible to use the Maicke–Majdalani profile
(Maicke and Majdalani 2009) as a piecewise approximation for which vr = vz = 0 and
the tangential velocity is given by
16 r 1 − ln ( r 2 )
v ( r ) = 16
r
0 r 1 (inner, forced vortex segment)
r 1 (outer, free vortex segment)
(a) State whether the flow is 1D, 2D, or 3D; steady or unsteady; and specify v ( r ) as r → .
(b) Calculate the divergence of the velocity. Is the flow compressible or incompressible?
(c) Determine the vorticity. Is the flow rotational or irrotational?
(d) Determine the strain rates and the shear stresses in the inner and outer flow segments.
(e) What is the limit of v ( r ) as r → 0 ? Hint: In taking the limit, it is helpful to remember that
u'
( ln u ) = and that, in the inner segment, the tangential velocity can be rewritten as
u
'
1 − ln ( r 2 )
v = 16
−1
r
Solution:
vϴ
(1)
0
(2)
1
r
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Hill LLC.
-21-
The velocity varies with respect to the radial distance r from the centerline and is independent of
the axial distance z or of the angular position . This represents a typical onedimensional flow.
(Ans.)
Since
vr v vz
=
=
= 0 , the flow is time invariant (steady).
t
t
t
(Ans.)
As r → , v ( r ) → 0 .
(Ans.)
Divergence of velocity is v =
1
1
( rvr ) +
( v ) + ( vz ) = 0 (incompressible).
r r
r
z
1 vz v
−
Vorticity is = v =
r z
vr vz
−
er +
z r
Only nonzero component of vorticity is z =
1 ( rv ) 1 vr
−
e +
r
r
r
(Ans.)
ez .
1 ( rv )
.
r r
1
For segment (1), z = 32 ln 2 (rotational).
r
(Ans.)
For segment (2), z = 0 (irrotational).
(Ans.)
1 v v
Only nonzero component of tangential strain rate is r = −
2 r
r
.
For segment (1), r = −16 .
For segment (2), r = −
(Ans.)
16
.
r2
(Ans.)
Then, shear stress for segment (1) is r = −32 and segment (2) is r = −
32
.
r2
(Ans.)
As r → 0 , v ( r ) → 0 .
(Ans.)
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Hill LLC.
-22-
1-36
The Taylor profile, which has been used to describe the bulk gaseous motion in planar,
slab rocket chambers (Maicke and Majdalani 2008), corresponds to a self-similar profile
in porous channels that bears symmetry with respect to the chamber’s midsection plane.
Using normalized Cartesian coordinates, the streamfunction may be written as
1
= x sin y ; 0 y 1 , and 0 x l , where l represents the aspect ratio of the
2
chamber (i.e., the length of the porous surface normalized by the chamber half height). In
this problem, the velocity vector, normalized by the wall injection speed, may be
expressed as V( x, y) = ui + vj .
(a) Determine the axial and normal velocity profiles using
d
d
and
v=−
dy
dx
(b) Evaluate the velocity divergence and determine whether the motion is compressible or
incompressible.
u=
(c) Evaluate the vorticity and determine whether the motion is rotational or irrotational.
Solution: The axial and normal velocity profiles of the stream function are given as follows:
u=
d
d x
1
1
= − sin y
=
cos y , and v = −
dx
dy
2
2
2
Divergence of velocity is
V ( x, y ) =
2x
1
( u ) + ( v ) = − sin y
y
x
4
2
(Ans.)
(compressible)
(Ans.)
Only nonzero vorticity component is
v u 2 x 1
z = Vz ( x, y ) = − =
sin y
4
2
x y
(Ans.)
(rotational)
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Hill LLC.
-23-
1-37
In the Taylor flow problem described above, determine the following:
(a) The strain rates.
(b) The Lagrangian time-dependent coordinates x j ( t ) and y j ( t ) for a particle j that enters the
porous chamber at the sidewall where y j = 1 and x j = X j at t = 0 . Recall that
dx j
dt
=u
dy j
and
dt
=v
(c) The pathlines of a particle j entering the chamber at X j = 1, 2, 3, 4, 5 . Display your results
in the ( x, y ) plane.
Solution: Only nonzero components of strain rate are as follows:
xx =
2x 1
u
1
1
sin y
= cos y , yy = − cos y , and xy = −
8
x 2
2
2
2
2
The rate of change of x j ( t ) is u =
1
1
= x j cos y j .
dt 2
2
dx j
By separating variables and integrating, x j = C1e
The rate of change of y j ( t ) is v =
(Ans.)
1
1
t cos y j
2
2
1
is the integration constant).
C1
, ( ln
1
= − sin y j .
dt
2
dy j
−t
1
By separating variables and integrating, y j = tan C2e 2 , ( ln is the integration
C2
constant).
4
−1
Given, at t = 0 , ( x j , y j ) = ( X j ,1) .
Then, C1 = X j , and C2 = 1 .
Therefore, the time-dependent coordinates for the particle are
1
1
t cos y j
2
2
x j = X je
and y j =
−t
tan −1 e 2 .
4
(Ans.)
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Hill LLC.
-24-
The equation of the pathline is x j = X j
(Ans.)
1
tan y j
4
1
cos y j
2
1
.
Pathlines of the particle j entering the chamber at X j = 1, 2, 3, 4, 5 are plotted on the ( x, y )
plane as follows:
Xj = 1
Xj = 2
Xj = 3
Xj = 4
y
Xj = 5
x
1-38
(Ans.)
The Taylor–Culick profile, which describes the bulk gaseous motion in solid rocket
motors (Culick 1966), corresponds to an axisymmetric, self-similar profile in porous
tubes. Using normalized cylindrical coordinates, the streamfunction may be written as
1
= z sin r 2 ; 0 r 1, and 0 z l , where l represents the aspect ratio of the
2
motor (i.e., the length of the porous surface normalized by the chamber radius). In this
problem, the velocity vector, normalized by the wall injection speed, may be expressed
as V(r , z ) = vr er + vz e z .
(a) Determine the axial and radial velocity profiles using Stokes’ definition:
vz =
1
r r
and
vr = −
1
r z
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Hill LLC.
-25-
(b) Evaluate the velocity divergence and determine whether the motion is compressible or
incompressible.
(c) Evaluate the vorticity and determine whether the motion is rotational or irrotational.
Solution: The axial and normal velocity profiles of the stream function are given as follows:
vz =
1
1
1 1
1
= z cos r 2 and vr = −
= − sin r 2
r r
r z
r
2
2
Divergence of velocity is V ( r , z ) =
(Ans.)
1
1
( rvr ) +
( v ) + ( vz ) = 0 (incompressible).
r r
r
z
(Ans.)
1 ( rv ) 1 vr
1 vz v
vr vz
−
−
−
Vorticity is given by = V ( r , z ) =
e +
er +
r
z r
r z
r r
ez .
Only nonzero vorticity component is
=
1-39
vr vz
1
−
= 2 zr sin r 2 (rotational)
z r
2
(Ans.)
In the Taylor–Culick flow problem described above, determine the following:
(a) The strain rates.
(b) The Lagrangian time-dependent coordinates rj ( t ) and z j ( t ) for a particle j that enters the
cylindrical rocket chamber at the sidewall where rj = 1 and z j = Z j at t = 0 . Recall that
drj
dt
= vr
and
dz j
dt
= vz
(c) The pathlines of a particle j entering the chamber at Z j = 1, 2, 3, 4, 5 . Display your results
in the ( r , z ) plane.
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Hill LLC.
-26-
Solution: Only nonzero components of strain rate are as follows:
rr =
vr
1
1
1
= − cos r 2 + 2 sin r 2 ,
r
2
r
2
1 v v
v
1
zz = z = cos r 2 , and zr = r + z
2 z r
z
2
(Ans.)
The rate of change of rj ( t ) is vr =
drj
dt
=−
1
1
sin rj 2 .
rj
2
By separating variables and integrating, rj =
The rate of change of z j ( t ) is vz =
vr 1 v
1
1
+
= − 2 sin r 2 ,
r r
r
2
1 2
1 2
= − zr sin r
2
2
=
1
tan −1 ( C1e− t ) , ( ln is integration constant).
C1
4
1
= z j cos r 2 .
dt
2
dz j
1
2
t cos r 2
By separating variables and integrating, z j = C2e
1
is integration constant).
C2
, ( ln
Given, at t = 0 , ( rj , z j ) = (1, Z j ) . Then, C1 = 1 , and C2 = Z j .
Therefore, the time-dependent coordinates for the particle are
rj =
4
tan
−1
( e ) and z
1
2
t cos r 2
− t
j
= Z je
The equation of the pathline is z j = Z j
(Ans.)
(Ans.)
1
tan r j 2
4
1
1
cos r j 2
2
.
Pathlines of the particle j entering the chamber at Z j = 1, 2, 3, 4, 5 are plotted on the ( r , z )
plane as follows:
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Hill LLC.
-27-
Zj = 1
Zj = 2
Zj = 3
Zj = 4
rj
Zj = 5
zj
1-40
(Ans.)
The Vyas–Majdalani profile, which describes the helical motion of a cyclonic chamber
(Vyas and Majdalani 2006), consists of a three-component velocity profile,
V ( r , z ) = vr e r + v e + vz e z , where
vr = −U
r2
z
a r2
a
sin 2 , v = U , and vz = 2U cos 2 ;
a
r
r
a
a
0 r a
0 z L
Here a denotes the radius of the cyclonic chamber, U stands for the average tangential velocity
at r = a , and represents a dimensionless offset swirl parameter that gauges the
relative importance of axial and tangential speeds.
(a) Is the flow one-dimensional, two-dimensional, or three-dimensional?
(b) Is the flow steady or unsteady?
(c) Calculate the divergence of the velocity. Is the flow compressible or incompressible?
(d) Determine the vorticity. Is the flow rotational or irrotational?
(e) Determine the strain rates. What is the sum of the three normal strain rates?
(f) Assuming a circular opening of radius r = b = a
2 at z = L , calculate the volumetric flow
rate by integrating the axial velocity from r = 0 to r = a
2.
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Hill LLC.
-28-
Solution: The velocity varies with respect to the radial distance r from the centerline and axial
distance z but it is independent of the angular position ; which corresponds to a
typical two-dimensional flow.
(Ans.)
Since
vr v vz
=
=
= 0 , the flow is time invariant (steady).
t
t
t
Divergence of velocity is V ( r , z ) =
(Ans.)
1
1
( rvr ) +
( v ) + ( vz ) .
r r
r
z
r2
r2
2
2
V ( r , z ) = − U cos 2 + 0 + U cos 2 = 0 (incompressible)
a
a
a
a
1 ( rv ) 1 vr
1 vz v
vr vz
−
−
−
Vorticity is = v =
e +
er +
r
z r
r z
r r
Only nonzero component of vorticity is = −
(Ans.)
ez
r2
vz
4
= − 3 2Uzr sin 2 (rotational).
r
a
a
(Ans.)
Only nonzero components of tangential strain rate are
1 1 v
v
v
1
a
a
a
r
r =
+ − = 0− 2 U − 2 U = − 2 U
2 r r
r 2
r
r
r
(Ans.)
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Hill LLC.
-29-
1 v
v
1
z
r2
z
r2
zr = r + z = 0 − 4 2U 3 r sin 2 = −2 2U 3 r sin 2
2 z r 2
a
a
a
a
(Ans.)
The sum of the normal strain rate components is
vr vr 1 v
+ +
r r r
rr + + zz =
vz
=0
+
z
(Ans.)
The volumetric flow rate is given by
a
Q=
2
r =0
1-41
a
vz dA =
2
r =0
2
r2
z
2U cos 2 d ( r 2 ) = 2 aUL cos d = 2 aUL
a
a
=0
(Ans.)
The Maicke–Majdalani profile, which may be used to describe the motion of an
unbounded tornado (Maicke and Majdalani 2009), can be expressed as a simple
piecewise approximation for which vr = vz = 0 and
r
r 2
1
−
ln
2
2 2 ; r XR
X
X R
v ( r ) =
R 2
r XR
r ;
where R denotes the radius at which the wind’s angular speed may be measured and X
represents the fraction of the radius R where the free vortex behavior ceases.
(a) Is the flow one-dimensional, two-dimensional, or three-dimensional?
(b) Is the flow steady or unsteady?
(c) Calculate the divergence of the velocity. Is the flow compressible or incompressible?
(d) Determine the vorticity. Is the flow rotational or irrotational?
(e) Determine the strain rates and the shear stresses in each segment of the flow.
(f) What is the limiting value of v ( r ) as r → 0 ?
Solution: The velocity varies with respect to the radial distance r from the centerline and is
independent of the axial distance z or of the angular position . This represents a
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Hill LLC.
-30-
typical one-dimensional flow.
(Ans.)
Since
vr v vz
=
=
= 0 , the flow is time invariant (steady).
t
t
t
Divergence of velocity is v =
1
1
( rvr ) +
( v ) + ( vz ) = 0 (incompressible).
r r
r
z
1 vz v
−
Vorticity is = v =
r z
vr vz
−
er +
z r
Only nonzero component of vorticity is z =
For segment ( r XR ), z = −
(Ans.)
1 ( rv ) 1 vr
−
e +
r
r r
(Ans.)
ez .
1 ( rv )
.
r r
2 r 2
ln
(rotational).
X 2 X 2 R2
(Ans.)
For segment ( r XR ), z = 0 (irrotational).
(Ans.)
1 v v
Only nonzero component of tangential strain rate is r = − .
2 r
r
For segment ( r XR ), r = −
For segment ( r XR ), r = −
X2
.
R 2
.
r2
(Ans.)
(Ans.)
Then, shear stress is given as follows:
For segment ( r XR ), r = −
2
.
X2
(Ans.)
For segment ( r XR ), r = −
2 R 2
.
r2
(Ans.)
As r → 0 , v ( r ) → 0 .
(Ans.)
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Hill LLC.
-31-
1-42
Assuming to be a scalar and V a vector, evaluate the following quantities:
(a)
(b) ( ) ( )
(c) ( ) ( )
(d) ( )
(e) ( V )
Solution: = 0
(Ans.)
i+
j+
k
z x
x y
y z
( ) ( ) = 2
2
( ) ( ) = + +
x y z
2
(Ans.)
2
(Ans.)
2 2 2
( ) = 2 + 2 + 2
x
y
z
(Ans.)
( V ) = 0
(Ans.)
1-43
Assuming U , V , and W are Cartesian vectors, prove the following identities:
1
(a) ( V ) V = ( V V ) − V ( V )
2
(Lamb’s vector identity)
(b) ( V ) = ( V ) − 2 V
(c) ( U V ) = V ( U ) − U ( V )
(d) ( U V ) W = U ( V W )
(mixed vector scalar product)
(e) U ( V W ) = ( U W ) V − ( U V ) W
(vector triple product)
(f) ( U V ) ( U V ) = ( U U )( V V ) − ( U V )( U V )
(Lagrange’s identity)
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(g) ( U − V ) ( U + V ) = 2U V
(h) ( V ) = V + V
(i) ( V ) = V + ( ) V
(j) ( UV ) = ( U ) V + U (V )
Solution:
1
( V V ) − V ( V ) = V V − V V − ( V ) V = ( V ) V
2
( V ) − 2 V = ( V ) − ( V ) − ( V ) = ( V )
(U V) =
(Ans.)
(Ans.)
(VzU y − VyU z ) + y (VxU z − VzU x ) + z (VyU x − VxU y )
x
U z U y
U y U x Vz Vy
Vy Vx
U x U z
Vx Vz
= Vx
−
−
−
−
−
−
+ Vy
− U x
+U y
+ Vz
+Uz
z
x
y y
z
x
y
z
z
x
x
y
= V ( U ) − U ( V )
(Ans.)
( U V ) W = (VzU y − VyU z )Wx + (VxU z − VzU x )Wy + (VyU x − VxU y )Wz
= (VzWy − VyWz )U x + (VxWz − VzWx )U y + (VyWx − VxWy )U z
= U (V W)
(Ans.)
U (V W)
= (U x i + U y j + U z k ) (VzWy − VyWz ) i + (VxWz − VzWx ) j + (VyWx − VxWy ) k
= U z (VxWz − VzWx ) − U y (VyWx − VxWy ) i + U x (VyWx − VxWy ) − U z (VzWy − VyWz ) j + U y (VzWy − VyWz ) − U x (VxWz − VzWx ) k
= (U xWx + U yWy + U zWz )(Vx i + Vy j + Vz k ) − (U xVx + U yVy + U zVz )(Wxi + Wy j + Wz k )
= ( U W) V − (U V) W
(Ans.)
( U V) (U V)
= (U zVy − U yVz ) i + (U xVz − U zVx ) j + (U yVx − U xVy ) k (U zVy − U yVz ) i + (U xVz − U zVx ) j + (U yVx − U xV y ) k
= (U zVy − U yVz ) + (U xVz − U zVx ) + (U yVx − U xVy )
2
2
2
= (U zVy ) + (U yVz ) + (U xVz ) + (U zVx ) + (U yVx ) + (U xVy ) − 2 U zVyU yVz + U xVzU zVx + U yVxU xVy
2
2
2
2
2
= (U x 2 + U y 2 + U z 2 ) (Vx 2 + Vy 2 + Vz 2 ) − (U xVx + U yVy + U zVz )
2
2
= ( U U )( V V ) − ( U V )( U V )
(Ans.)
( U − V) ( U + V)
= (U x − Vx ) i + (U y − Vy ) j + (U z − Vz ) k (U x + Vx ) i + (U y + Vy ) j + (U z + Vz ) k
= 2 (U zVy − U yVz ) i + 2 (U xVz − U zVx ) j + 2 (U yVx − U xVy ) k
= 2U V
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(Ans.)
( V )
= i+
j + k (Vx i + Vy j + Vz k )
z
x y
Vx Vy Vz
=
+
+
x
y
z
V Vy Vz
= x +
+
+ Vy
+ Vz
+ Vx
y
z x
y
z
x
= V + V
( V )
(Ans.)
= i+
j + k (Vxi + Vy j + Vz k )
z
x y
Vy Vz Vz Vx Vx Vy
=
−
−
−
i +
k
j+
y x
z y
x
z
Vy Vz Vx Vz Vy Vx
=
−
−
−
(Vx − Vz ) j + (Vy − Vx ) k
i +
k + (Vz − Vy ) i +
j+
y z
x x
y x
y
z
z
= V + ( ) V
(Ans.)
( U ) V + U ( V )
U x U y U z
V
Vx
V
=
+
+
i + y j+ z
(Vx i + Vy j + Vz k ) + (U x i + U y j + U z k )
y
z
y
z
x
x
U xVx U yVy U zVz
=
+
+
x
y
z
= i+
j + k (U xVxi + U yVy j + U zVz k )
z
x y
= ( UV )
k
(Ans.)
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-34-
PROBLEM SOLUTIONS
CHAPTER 2. FUNDAMENTAL EQUATIONS OF COMPRESSIBLE VISCOUS FLOW
2-1
Derive conservation of mass using the elemental control volume shown at right.
C.V. mass loss equals flow-out − flow-in:
(a) flow through front and back sides:
−vdzdr + v + ( v ) d dzdr
(b) flow through left and right sides:
−vr rddz + vr r + ( vr r ) dr ddz
r
(c) flow through bottom and top sides:
−vz rddr + vz + ( vz ) dz rddr
z
(d) mass loss within the clement: − ( r d dr dz )
t
The sum of (a,b,c) must equal (d) to conserve mass. Since ( r,θ,z ) are independent of time, we
may cancel the differentials ( dr d dz ) and rewrite the final result:
1
1
( rvr ) +
( v ) + ( vz ) + = 0
r r
r
z
t
(Ans.)
This is the (compressible) Equation of Continuity in cylindrical polar coordinates.
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-1-
2-2
Simplify the result of (2-1) above to the case of steady plane compressible flow in polar
coordinates and derive a suitable stream function for this case.
For plane flow,
/z = 0 and vz = 0; for steady flow, /t = 0. The continuity equation above in
Prob. 2-1 reduces to the following relation, with its stream-function equivalence:
( rvr ) + (v ) = 0 = + −
r
r r
By comparing terms, we see that the velocities are related to stream function as follows:
vr =
1
1
; v = −
r
r
(Ans.)
This is the steady plane polar-coordinate compressible stream function.
2-3
Simplify the result of (2-1) above to the case of steady plane compressible flow in
axisymmetric coordinates and derive a suitable stream function for this case.
For axisymmetric flow, / = 0 and v = 0; for steady flow, /t = 0. The continuity equation
above in Prob. 2-1 reduces to the following relation, with its stream-function equivalence:
( rvr ) + ( rvz ) = 0 = − +
r
z
r z z r
By comparing terms, we see that velocities are related to stream function as follows:
vr = −
1
1
; vz =
r z
r r
(Ans.)
This is one form of the steady plane axisymmetric compressible stream function.
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-2-
2-4
Use the Navier-Stokes equation to derive the “weak” form of the Bernoulli relation.
For steady, incompressible flow with zero viscosity, the Navier-Stokes equation reduces to
(
)
( V ) V = −p + g, where ( V ) V = V 2 /2 + V
where the latter identity follows from Eq. (1-10) of the text. Now “dot” this entire equation with
a differential arc length dr, giving the intermediate result
(
)
1
2
p + V /2 − g dr = ( V ) dr
For any direction dr, the left hand side consists of three exact differentials. The right hand side
vanishes for the particular direction dr along a streamline of the flow, since a streamline is
everywhere perpendicular to the cross-product of velocity and vorticity. For this case, then, the
dot product above becomes
( )
2
dp d V
+
+ g dz = 0,
2
or:
p/ + V2 /2 + g z = constant
(Ans.)
This is the “weak” form of the steady, incompressible Bernoulli equation [that is, it does not
account for heat transfer or shaft work] and assumes that the coordinate z is “up”. The “constant”
may vary in value from streamline to streamline.
2-5
Show that the incompressible energy equation (2-40) reduces, for zero and k, to the
“strong” form of Bernoulli’s equation.
If we simply set and k equal to zero in Eq. (40), we obtain the oversimplified result
De/Dt = 0, or e = constant as a particle moves through the flow. This is a “weak” isothermaltype result, implying no interaction between work and heat transfer in a flow. If, instead, we
combine energy and momentum, using (2-19) to substitute for the stress tensor in (2-40), we
obtain the result
ij
u i
Dp
DV
= −p div ( V ) −
− V
+ V g + viscous terms
x j
Dt
Dt
Substituting into Eq. (2-40) and setting , k, and div ( V ) equal to zero, we obtain
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-3-
(
)
D
e + p/ + V 2 /2 − g r = 0,
Dt
or:
e + p/ + V2/2 + g z = constant along a streamline
(Ans.)
where, again, z is “up”. This is the “strong” form of Bernoulli’s equation and is related to the
steady-flow-energy-equation of thermodynamics. It holds even for compressible flow as long as
viscous and heat-conduction effects are negligible.
2-6
Consider the proposed incompressible axisymmetric flow field vz = C ( R 2 − r 2 ), vr = 0
in the region 0 z L,0 r R, where C is a constant. Neglect gravity. (a) Determine if this is
an exact solution to the Navier-Stokes equation. (b) What might it represent? (c) If an
axisymmetric stream function ( r, z ) exists for this flow, find its form.
Solution: Substitute the given flow field into the radial and axial momentum Eqs. (B-6, 8):
r – momentum:
0=−
p
+0
r
z – momentum:
0=−
p
2
2
+
r C R − r
z r r r
(
) + z 2 C ( R 2 − r 2 )
2
The first of these tells us that p is independent of r and only varies axially. The second relation
yields a constant axial pressure gradient, p/z = −4C. (a) The given flow field is indeed an
exact solution. (b) It represents Poiseuille flow in a tube, Sect. 3-3 of the text. (c) The solution is
exact and a stream function ( r, z ) exists, from Appendix B or Prob. 2-3:
(
)
= 0;
= rvz = C R 2 r − r 3 .
z
r
R2r r 4
Integrate: = C
− + const
2
4
Ans. (c)
2-7
Investigate the stream function = Cxy, with C 0, to determine if it is a realistic
solution for (a) inviscid; or (b) viscous, incompressible flow.
The stream function, by definition, satisfies continuity exactly, with velocity components
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-4-
u=
= Cx;
y
v=−
= −Cy
x
(1)
There are no shear strains, but the normal strain rates are finite: xx = C, yy = −C, (hence there
are finite viscous normal stresses). The sum of these strain rates is zero, hence the flow is
incompressible. Note that the vorticity is also zero:
z =
v u
− =0
x y
Thus the flow is irrotational, and a velocity potential also exists. We conclude that the stream
function = Cxy represent a valid potential-flow field.
When plotted in the x-y plane for various values of C, the result is the pattern of streamlines
shown below. We could interpret this as flow (a) between two intersecting
streams; (b) of a stream against a plane wall; or (c) of flow around a 90° corner. Patterns (b) and
(b) would not be realistic for viscous flow, because the “walls” are not no-slip lines of zero
velocity.
Since the flow is irrotational, the viscous momentum term ( V) vanishes identically
[see Eq. (2-107) in the text] and, if we neglect gravity for convenience, the Navier-Stokes
equations yield the pressure gradients
2
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-5-
p
= − C2 x;
x
p
= − C2 y
y
Integration of these simple relations yields the exact pressure distribution in the fluid:
(
)
1
p = − C2 x 2 + y2 + constant
2
(Ans.)
But this is precisely Bernoulli’s equation, which is what we expect when an irrotational flow is
tested directly in the Navier-Stokes equation.
2-8
Investigate the incompressible stream function y = C(x y − y /3), where C 0.
Follow the same procedure as Prob. 2-7 above. The velocity components are
2
u=
(
)
= C x 2 − y2 ;
y
v=−
3
= −2Cxy
x
from which we may compute that the vorticity, z = v/x − u/y, is identically zero and the
flow is irrotational. The streamlines of this exact potential flow are plotted below.
Only the (symmetric) upper half plane is shown for convenience. We could interpret this flow as
(a) flow caused by three intersecting streams; (b) flow against a 120° corner; or (c) flow around a
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-6-
60° corner. Patterns (b) and (b) would not be realistic for viscous flow, because the “walls” are
not no-slip lines of zero velocity.
Since the flow is irrotational, the viscous momentum term ( V) vanishes identically
[see Eq. (2-107) in the text] and, if we neglect gravity for convenience, the Navier-Stokes
equations yield the pressure gradients
2
(
)
p
= −2C2 x 3 + xy 2 ;
x
(
p
= −2C2 y3 + x 2 y
y
)
Integration of these simple relations yields the exact pressure distribution in the fluid:
(
)
(
)
1
1
p = − C2 x 4 + 2x 2 y 2 + y 4 + constant = − u 2 + v 2 + constant
2
2
(Ans.)
But this is precisely Bernoulli’s equation, which is what we expect when an irrotational flow is
tested directly in the Navier-Stokes equation.
2-9
Analyze the following plane polar unsteady incompressible flow:
vr = 0
r 2
C
v = 1 − exp −
4t
r
with C and v constant and gravity neglected. This flow satisfies continuity exactly. If we
substitute these velocities into the θ-momentum equation in Appendix B, assuming radial
symmetry or p/ = 0, we find that it is also satisfied exactly. [The pressure distribution p ( r, t )
could then be found from the radial momentum equation but is not shown here.] Thus the given
distribution is indeed an exact solution to the Navier-Stokes equations.
This problem was discussed earlier, without proof of its exactness, in Prob. 1-6 on page 3. The
instantaneous vorticity profiles,
z =
r2
1
C
( rv ) = exp −
r r
2 t
4 t
and velocity profiles are plotted for various times on page 3.
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-7-
Using from Eq. (2-46), prove the inequalities in Eq. (2-47).
2-10
We are to find the conditions under which the dissipation function,
(
)
= 2 a 2 + b2 + c2 + f 2 + g 2 + h 2 + ( a + b + c )
2
is positive, where we adopt the simpler notation ( a, b, c, f , g, h ) for the strain rates
( xx , yy , zz , xy , yz , zx ). The first thing to notice is that only the coefficient is concerned
with the shear strain rates ( f ,g, h ) . It is easy to visualize pure shear flow with zero normal
strain, e.g. Couette flow or fully-developed duct flow. For this case, we require positive
(
)
= f 2 + g 2 + h 2 , with no contribution from . It follows that cannot be positive in such
flows unless
0
(Ans. a)
This is the first condition to be proved.
With thus guaranteed positive, it follows that the normal-strain-rate terms will also
have to stay positive to keep positive in cases where there is no shear strain (such as flow
through normal shock waves). Thus we require
(
)
( a + b + c ) + 2 a 2 + b 2 + c 2 0,
2
or:
/ + R 0,
where
R=
(
2 a 2 + b2 + c2
)
( a + b + c )2
This is equivalent to requiring that / ( −R min ) . To minimize R, we simply set
R/a = R/b = R/c = 0 and solve for a = b = c. Then R min = 2/3, and thus
−2/3
(Ans. b)
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-8-
This was the second condition to be proved.
2-11
Non-dimensionalize the differential equation of compressible irrotational flow:
2
t 2
+
(
u 2 + v2
t
) + u2 − a 2 2 + v2 − a 2 2 + 2uv 2 = 0
(
) x2 (
) y2
xy
where a is the speed of sound of the gas.
We only need a velocity scale U and a length scale L, since mass units are not present.
Appropriate nondimensional variables are:
* =
UL
u*, v*, a* =
u, v, a
U
x*, y* =
x, y
L
t* =
Ut
L
Substituting these variables into the basic differential equation and dividing out the leading
coefficient, we obtain the following nondimensional differential equation:
(
)
2
2
2
2
2* u* + v*
2*
2
2 *
2
2 *
+
+
u*
−
a*
+
v*
−
a*
+
2u*v*
=0
t*
x*y*
t*2
x*2
y*2
(
)
(
)
Examining this relation, we see that no dimensionless parameters appear (Ans.) [Actually,
correlating a* with temperature and velocity would in fact lead to two important parameters, the
Mach number and the specific heat ratio.]
2-12
Repeat the nondimensionalization of the Navier-Stokes equation for slow viscous flow by
noting that the pressure should scale as ( U/L ) . What happens if Re 1 ?
To proceed, it is only necessary to modify dimensionless pressure in Eq. (2-83) to
p* = ( p − po ) L/ ( U ) . Let us neglect gravity and buoyancy and consider only the Navier-Stokes
equation with constant and :
DV
= −p + 2 V
Dt
Introduce the variables from Eq. (2-83), including the new p*, and clean up by dividing out the
coefficient of the viscous term. The result is the following dimensionless equation:
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-9-
Re
DV*
= −*p* + *2V*,
Dt*
where
Re = UL/
The only parameter is the Reynolds number, which occurs only in the “inertia” or acceleration
term.
If the Reynolds number is very small, Re 1 , then inertia is negligible, and the balance
of fluid forces is between pressure and viscosity:
p = 2V
Re 1
if
(Ans.)
This is the creeping-flow or Stokes-flow approximation. It linearizes the Navier-Stokes equation
and enables many slow viscous flows to be solved analytically [see Section 3.9].
2-13 Nondimensionalize the equations of motion for free convection near a hot vertical plate,
using dimensionless variables which have no direct velocity scale (“U”):
u* =
uL
v* =
vL
where L is a length scale and
x* =
x
L
y* =
u
L
T* =
T − T1
To − T1
is kinematic viscosity.
The equations of motion are given in the statement of Prob. 2-13. Introducing these
variables into the continuity equation gives u*/x* +v*/y* = 0, with no parameters
appearing. Substitute these variables into the momentum equation and divide by the leading
coefficient. The result is the dimensionless momentum equation:
u*
u*
u*
2 u* 2 u*
+ v*
= GrL T* +
+
,
x*
y*
x*2 y*2
where
GrL =
g ( To − T1 ) L3
2
The single parameter, which occurs only in the buoyancy term, is the Grashof Number.
Substitute the same variables into the energy equation and divide by the leading
coefficient. The result is the dimensionless energy equation:
u*
T*
T* 1 2T* 2T*
+ v*
=
+
,
x*
y* Pr x*2 y*2
where
Pr =
cp
k
Here a second parameter arises, associated with the conduction term: the Prandtl Number.
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-10-
By merely nondimensionalizing the equations of motion, we conclude, correctly, that free
convection problems are dependent upon both the Grashof and Prandtl numbers.
2-14 Make a control-volume analysis of laminar flow in the pipe entrance as shown and find a
formula for the friction drag on the pipe walls.
(
)
2
2
Let the exit velocity be parabolic: u = C ro − r .
The proper control volume includes the entrance and exit and passes just inside the tube
walls, exposing the wall-friction drag force acting to the left on the fluid.
First relate “C” to the inlet by using the integral continuity relation:
(
V dA = 0 = −Uoro +
CV
or:
)
(
)
C ro2 − r 2 2r dr = Uo −ro2 + Cro4 /2 ,
2
exit
C = 2Uo /ro2
With C known, the drag is computed by applying the integral x-momentum equation:
Fx = ( po − px ) ro2 − Drag =
u ( V dA ) = −Uo2ro2 +
CV
exit
(
)
2
C ro2 − r 2 2r dr
The value of the last integral on the right is
C2ro6/3 = Uo2ro2/3
since
C = 2Uo2 ro2
The force equation above may thus be solved for the desired drag force:
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-11-
1
Drag = ro2 po − px − Uo2
3
2-15
(Ans.)
Illustrate “boundary-layer” behavior with Prandtl’s model differential equation:
d2u
dy
2
+
du
+ u = 0, with u ( 0 ) = 0 and u ( ) bounded
dy
assuming that 1. For finite , we assume a solution u = exp ( my ) , leading to
m2 + m + 1 = 0,
or:
m1,2 =
1
−1
2
Then the general solution has the form
(1 − 4 )
u = A exp(m1y) + B exp(m2 y). To satisfy the initial
condition u(0) = 0, we must have B = −A. Then our desired “boundary layer” type solution is
given by
u = C e m1y − e m 2 y
For 0, both m1,2 are negative and thus the solution is bounded at large y. For very small ,
we may approximate m1 = −1 and m2 = −1/.
If we neglect the second derivative term entirely ( = 0, corresponding to “inviscid” or
“slip” conditions), the basic differential equation reduces to
du
+ y = 0,
dy
with solution u = Ce− y
This is equivalent to the first term of the boundary-layer solution. It is impossible in this
“inviscid” case to satisfy the “no-slip” initial condition, u ( 0) = 0.
These two solutions, inviscid and boundary-layer, are compared in the figure on p. 28.
The boundary-layer solutions satisfy the no-slip condition and, for small ε, merge into the
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-12-
inviscid solution at larger y. The “boundary-layer” is very thin. This mathematical behavior can
be generalized into the asymptotic expansion methods of Section 4-11.
2-16
Consider the plane, incompressible, cartesian stream function in the region 0 y
b
c
= ax + by + e−cy
where ( a, b, c ) are positive constants. (a) Determine if this is an exact solution to the continuity
and Navier-Stokes equations if gravity and pressure gradient are neglected. (b) What are the
dimensions of ( a, b, c ) ? (c) If y = 0 represents a wall, does the no-slip condition hold there?
(d) Is there any vorticity in the flow field? If so, what is its form?
Solution: (a) First use the stream function ( x, y ) to determine the velocity components:
u=
= b − be−cy;
y
Check continuity:
v=−
= −a
x
u v
+ = 0+0
x y
OK, satisfied
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-13-
The Navier-Stokes y-momentum equation simply gives 0 0, while x-momentum yields
2u 2u
u
u
− cy
u + v = 0 − acbe
= 2 + 2 = −c 2be −cy
x y
y
x
(
)
Thus Navier-Stokes is satisfied if
(
)
a = c /
Ans. (a)
(b) The units of a and b are velocity, m/s. The units of c are inverse length,
(c) At y = 0, the “wall”, u = 0 (no-slip) and
v = −a
m −1.
Ans. (b)
(wall suction).
Ans. (c)
(d) This is a boundary-layer flow with suction, and there is vorticity, highest at the wall:
z =
2-17
v u
−
= 0 − cbe −cy
x y
Ans. (d)
Use Eq. (2-86) for density in the gravity term of the Navier-Stokes equation (2-29b) and
0 in the acceleration term and let = constant. For free convection, with U replaced by
/ ( 0 L ) and the hydrostatic assumption, p = 0 g = − 0g k, nondimensionalize Navier-Stokes
let
and show that the Grashof number appears.
Solution: With these assumptions, the Navier-Stokes equation for free-convection becomes
0
DV
0 (T − T0 ) gk + 2 V
Dt
Now introduce the dimensionless variables from Eqs. (2-81), with the single change
V* = 0 LV / , and collect terms, divide out the leading coefficient, and rearrange:
2
3
DV* 0 g (Tw − T0 ) L
T *k + *2 V* = GrT * k + *2 V*
2
Dt*
The Grashof number appears by itself, as the coefficient of the T * term. If we nondimensionalized the energy equation also, the Prandtl number would appear there.
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Hill LLC.
-14-
2-18 Flow through a well designed contraction or nozzle is nearly frictionless, as shown for
example in White (2003), Sec. 6.12. Suppose that water at 20°C flows through a horizontal
nozzle at a weight flow of 50 N/s. If entrance and exit diameters are 8 cm and 3 cm, respectively,
and the exit pressure is 1 atm, estimate the entrance pressure from Bernoulli’s equation.
3
Solution: For water at 20°C, = 998 kg/m . First find the inlet velocity from the weight flow:
w = 50
N
kg
m
m
2
= g D12V1 = 998 3 9.81 2 ( 0.08m ) V1, solve for V1 = 1.02
s
4
s
m
s 4
Now use one-dimensional continuity to estimate V2 in the exit and substitute into Bernoulli:
2
D1 V1 = D22V2 ,
4
4
p1 +
2
V12 = p1 +
2
or:
2
D1
m
m
0.08m
V2 = V1 =
1.02 = 7.23
s
s
0.03m
D2
998
998
(1.02 )2 = p2 + V22 = 101350Pa +
( 7.23)2 , solve p1 = 127, 000 Pa Ans.
2
2
2
2-19 Show, using Gauss’ theorem [Kreyzig (1999), Sect. 9.8] that the control volume mass
relation, Eq. (2-111), leads directly to the partial differential equation of continuity, Eq. (2-6).
Solution: Gauss’ theorem relates an area integral to a volume integral over the same region:
For any vector Q,
Qd ( vol ) = Q dA
vol
area
For continuity, Eq. (2-111), the relevant vector Q = V. Also, for a fixed control volume, we can
slip the time derivative inside the volume integral of Eq. (2-111) and then use the theorem:
d
d ( vol ) =
d ( vol )
dt CV
t
CV
Hence Eq. (2-111) becomes
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Hill LLC.
-15-
d ( vol ) + V dA = + V d ( vol ) = 0
t
t
CV
CS
CV
Since the elemental volume d ( vol ) is arbitrary, it follows that the terms inside the braces [] are
identically equal to zero. These terms are in the fact the equation of continuity, Eq. (2-6).
NOTE: A similar derivation for a moving control volume is given by Panton (1995), p. 88.
2-20 In discussing incompressible flow with constant , Eq. (2-30), we cavalierly said, “many
terms vanish” from Eq. (2-29a). Be less cavalier and show that the many viscous terms in Eqs.
2
(2-29a) do indeed reduce to the single vector term V, in three dimensions.
Solution: The writer goes ahead and laboriously works out each viscous term for constant
viscosity :
u u v u w v u v
2 + + + +
+ 2 + +
x
x y y x z z x y y x y x
+
w v w u w w v
+ + 2
+ + + =
+
z y z z
z x z x y y z
2u 2 v 2 w
u v w
+
+
+
2 + 2 + 2 + . . . exactly similar terms in v and w
x x y z
y
z
x
The original terms split into two groups, the first of which contains V = 0 and other is
2
multiplied by 2 ( u, v, w ) . Add up these latter three terms and you get V.
Ans.
2-21 In deriving the basic equations of motion in this chapter, we skipped over the partial
differential equation of angular momentum. Did we forget? Do some reading, perhaps in Lai et
al. (1995) or Malvern (1997), and explain the significance of the angular momentum differential
equation.
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Hill LLC.
-16-
Solution: In heavy-duty continuum mechanics, differential elements are allowed to have
concentrated couples or moments per unit volume. Let us denote such a vector moment by T .
Take Fig. 2-1 of the text and specialize to moments about the z axis only, as shown in the sketch
at below.
Summation of moments about the central z axis gives
( yx dx dz ) dy/2 + yx + ( yx /y ) dy ( dx dz ) dy/2
− ( xy dy dz ) dx/2 − xy + ( xy /x ) dx ( dy dz ) dx/2 = Tz dx dy dz
Neglecting 4th-order differentials, we obtain ( yx − xy ) dx dy dz = Tz dx dy dz, or:
( yx − xy ) = Tz . Similarly, for the other two axes, ( zy − yz ) = Tx and ( xz − zx ) = Ty .
So the concentrated couple causes shear-stress differences.
If T = 0, ij = ij , the stress tensor is symmetric.
2-22
Normalize the Navier–Stokes equation with constant properties, specifically,
V
p
+ ( V ) V = −
+ 2 V + g
t
(a) by setting:
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Hill LLC.
-17-
x
L
y* =
y
L
z* =
z
L
* = L
t * = t
V* =
V
U
p* =
p
P
g = − gk
x* =
(b) Define the dimensionless parameters that appear in the following nondimensional form:
V*
1
1
St
+ ( V* * ) V* = − Eu* p* + *2 V* − k
t
Re
Fr
Solution: Rearrange the dimensional variables
x = x* L
y = y* L
1
V = V*U
t=
t*
z = z*L
p = p* P
=
1 *
L
g = − gk
Use the rearranged variables to normalize the Navier–Stokes equation.
*
P * * *2 * gL
L V
*
*
*
p + 2 V − 2 k
* + ( V ) V = −
2
U t
UL
U
U
The dimensionless parameters are Strouhal number St = L , Euler number Eu =
U
Reynolds number Re =
(Ans.)
P
,
U 2
UL2
U2
, and Froude number Fr =
.
gL
(Ans.)
2-23 In order to better analyze the boundary-layer structure over a flat plate with a characteristic
length L , the traditional variables are rescaled using
x* =
x
y
u
p
, y* = , u* = , * =
, p* =
L
L
U
U
U 2
2u 2u
where
in the axial
1 and denotes the boundary-layer thickness. Show that
x 2
y 2
L
momentum equation written for the boundary-layer region. For steady-state motion with constant
fluid properties, you may start with the axial momentum equation given by:
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Hill LLC.
-18-
u
2u 2u
u
u
p
+
=−
+ 2 + 2
x
y
x
y
x
Solution: Rearrange the variables and use them in the axial momentum equation
u*U 2 u* *U 2 u*
U 2 p*
U 2u *
U 2u *
+
=
−
+
+
L x*
L y*
L x*
L2 x*2
2 L2 y*2
2
u * * u *
p* 1 2u * L 2u *
u * + * = − * + *2 +
*2
x
y
x Re x
y
where, Re =
*
UL
For steady-state motion with constant fluid properties on a flat plate parallel to the direction of flow,
there is no axial pressure gradient. Therefore, the equation reduces to
2
u * * u * 1 2u * L 2u *
u * +
=
+
x
y* Re x*2 y*2
*
Since it is assumed that the boundary layer is thin, i.e.,
L
2
L will be large.
1 then
Since each of the derivatives is assumed to be order one through proper scaling, then the second term
2u 2u
in parentheses must be much larger than the first term, 2
(Ans.)
x
y 2
2-24
Given the dimensionally scaled expressions of the basic fluid dynamical forces:
3
Gravitational force: Fg = mg L g
Viscous force: F = A = ( dV dy ) A UL
Steady inertial force: Fi = mVdV dx U L
2 2
Unsteady inertial force: Fu = mdV dt L U
3
2
Pressure force: Fp = pA pL
2 2
Compressibility force: Fc = ( dp d ) A c L
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Hill LLC.
-19-
Surface tension force:
F = D L
where the velocity gradient is approximated using dV dy = U L , verify the following characteristic
ratios:
Inertia and viscosity:
Fi UL
=
: Osborne Reynolds (British)
F
Inertia and compressibility:
Inertia and gravity:
Fi U 2
=
= Fr : William Froude (British)
Fg gL
Fi U 2 L
=
= We : Moritz Gustav Weber (German)
F
Inertia and surface tension:
Pressure and inertia:
Fi U 2
=
= Ma 2 : Ernst Mach (German)
Fc c 2
Fp
Fi
Unsteadiness and inertia:
p
= Eu : Leonhard Euler (Swiss)
U 2
Fu L
=
= St : Vincenc Strouhal (Czech physicist)
Fi U
Unsteadiness and viscosity:
Fu L2
=
= Sk 2 : Sir George Gabriel Stokes (Irish)
F
Viscosity and surface tension:
F
F
=
U
= Ca : Capillary Number
Solution: Inertia and viscosity:
Fi
mVdV dx
U 2 L2 UL
: Osborne Reynolds (British)
=
=
F ( dV dy ) A
UL
(Ans.)
Inertia and compressibility:
Fi
mVdV dx
U 2 L2 U 2
=
= 2 = Ma 2 : Ernst Mach (German)
Fc ( dp d ) A c 2 L2
c
(Ans.)
Inertia and gravity:
Fi mVdV dx U 2 L2 U 2
=
=
= Fr : William Froude (British)
Fg
mg
L3 g
gL
(Ans.)
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Hill LLC.
-20-
Inertia and surface tension:
Fi mVdV dx U 2 L2 U 2 L
=
=
= We : Moritz Gustav Weber (German)
F
D
L
(Ans.)
Pressure and inertia:
Fp
Fi
=
pA
pL2
p
=
= Eu : Leonhard Euler (Swiss)
2 2
mVdV dx U L
U 2
(Ans.)
Unsteadiness and inertia:
Fu
mdV dt
L3U L
=
=
= St : Vincenc Strouhal (Czech physicist)
Fi mVdV dx U 2 L2
U
(Ans.)
Unsteadiness and viscosity:
Fu
mdV dt
L3U L2
=
=
= Sk 2 : Sir George Gabriel Stokes (Irish)
F ( dV dy ) A
UL
(Ans.)
Viscosity and surface tension:
F
F
=
2-25
( dV
dy ) A UL U
=
= Ca : Capillary Number
D
L
(Ans.)
In 1851, Stokes introduced an expression for the drag force, FD = 3VD , in the context of
1
2
2
laminar flow over a smooth sphere of diameter D . Express the drag coefficient CD = FD V A
as a function of the Reynolds number. Recall that A should be the projected area of the sphere.
1 2 D 2
24
24
=
Solution: CD = 3VD V =
2 ( VD ) Re
2
(Ans.)
2-26 In normalizing the basic equations of motion in Sec. 2-9.1, we have assumed a characteristic
time constant that scales with the unsteady period, i.e., the reciprocal of the frequency, = 1 , when
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-21-
we set t * = t . Show that by taking a different t = tU L , for which = L U represents the time
for a fluid particle to travel a distance L at the characteristic mean-flow velocity,
*
(a) the continuity and momentum equations reduce to
*
*
* *
Continuity: * + ( V ) = 0
t
*
Navier–Stokes:
T
DV*
1
1
= − *k − * p* + * * *V* + ( *V* )
*
Dt
Fr
Re
(b) the energy equation reduces to
DT*
Dp*
1
Ec
= Ec * +
* ( k **T* ) + *
Energy:
*
Dt
Dt Re Pr
Re
*
Solution: From Sec. 2-9.1, the continuity equation is given by St
*
+ * ( * V * ) = 0 .
*
t
Here, Strouhal number St = L . From the question, = 1 . Then the equation becomes
U
L *
+ * ( * V * ) = 0
*
U t
If a new = L U
*
*
* *
is chosen, then the continuity equation reduces to * + ( V ) = 0 . (Ans.)
t
From Sec. 2-9.1, the Navier–Stokes equation is given by
* St
T
V *
1 *
1 *
+ ( V* * ) V* = −
k − * p* +
* *V* + ( *V* )
*
t
Fr
Re
Here, Strouhal number St = L . From the question, = L U . Then the equation reduces to
U
*
T
V *
1 *
1 *
*
*
*
* DV
+
V
V
=
=−
k − * p* +
* *V* + ( *V* )
(
)
*
*
Dt
Fr
Re
t
*
(Ans.)
From Sec. 2-9.1, the energy equation is given by
* St
T*
Dp*
1
Ec *
*
*
*
+
V
T
=
Ec
+
* ( k **T* ) +
(
)
*
*
t
Dt
Re Pr
Re
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Hill LLC.
-22-
Here, Strouhal number St = L . From the question, = L U . Then the equation reduces to
U
*
T*
Dp*
1
Ec *
*
*
*
* DT
+
V
T
=
=
Ec
+
* ( k **T* ) +
(
)
*
*
*
Dt
Dt
Re Pr
Re
t
*
(Ans.)
2-27 Renormalize the basic equations of motion in Sec. 2-9.1 assuming an incompressible fluid
with negligible dissipation function as well as constant viscosity and thermal conductivity,
= 0
and k = k0 . The normalization will correspond to
xi* =
xi
L
t * = t
* =
0
V* =
T* =
V
U
p* =
T − T0
Tw − T0
p
p0
=
=0
cp
c
V = 0
* = L
(a) Show that the continuity and momentum equations can be simplified into
Continuity: St
*
+ * ( * V * ) = 0
*
t
V*
1 *
1 *2 *
Navier–Stokes: * St * + V* *V* = −
k − Eu* p* +
V
Fr
Re
t
(b) Show that the energy equation reduces to
T*
− 1 p*
1
Energy: * St * + ( V* * ) T* =
St * + ( V* * ) p* +
*2T*
t
t
Re
Pr
(c) Under what circumstances does the Euler number become important?
(d) What happens to this set of equations if the density
= 0
is assumed to be constant?
Solution: Continuity equation is given by + ( V ) = 0 .
t
*
*
Using = 0 , V = UV , =
* U 0 *
*
t*
( * V* ) = 0 .
, and t = gives 0 * +
t
L
L
*
U 0
Divide both sides by
and using St = L gives St * +
t
L
U
*
( V ) = 0.
*
*
(Ans.)
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Hill LLC.
-23-
Navier–Stokes equation for incompressible flow
DV
V
= + ( V ) V = g − p + 2 V
Dt
t
Using normalized variables in the above equation gives
0 * U
p
U
V * U 2
+
V* * ) V* = − 0 gk0 * − 0 * p* + 02 *2 V*
(
*
t
L
L
L
Dividing by
0U 2
L V *
gk L
p
0
gives *
+ ( V* * ) V* = − 02 * − 0 2 * p* +
*2 V*
*
L
U
0U
0UL
U t
V *
1 *
1 *2 *
Or, * St * + ( V* * ) V* = −
− Eu* p* +
V
Fr
Re
t
(Ans.)
The energy equation for incompressible flow with negligible dissipation function is
cp
DT
Dp
= T
+ ( k T )
Dt
Dt
T
p
+ ( V ) T = T + ( V ) p + ( kT )
t
t
Or, c p
Using normalized variables in the above equation gives
T − T T * U
0 *c p w 0 * + ( V* * ) T * =
L
T0 t
*
T −T
p U
k T −T
w 0 T * p0 * + ( V* * ) p* + 02 w 0 *2T *
L
t
L T0
T0
T −T U
Dividing by 0c p w 0 gives
T0 L
L T *
T * p0 L L p*
k0
*
*
*
+
V
T
=
+ ( V* * ) p* +
*2T *
(
)
*
*
U
t
c
U
U
t
c
UL
0 p
0 p
*
Using the expression for coefficient of thermal expansion and specific-heat ratio
* St
T *
− 1 p*
1
*
*
*
+
V
T
=
St * + ( V* * ) p* +
*2T *
(
)
*
t
t
Re
Pr
(Ans.)
For such boundary conditions when the inlet and outlet flows are approximately uniform and
parallel, the viscous normal stresses become locally negligible and the only important boundary
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Hill LLC.
-24-
work is due to pressure forces at the inlet and outlet. Then, due to the presence of significant
pressure difference, the Euler number becomes important.
(Ans.)
If density
= 0 is assumed to be constant, the 2 terms in the set of equations become zero. (Ans.)
2-28 In deep water, the average speed V of a free-surface wave can be expressed as a function of
the density, , gravity, g , depth, D , and spatial wavelength, . Obtain a dimensionless
representation of the free-surface wave speed as a function of these controlling parameters.
Solution: Apply the Buckingham procedure:
Number of parameters n = 5:
Primary dimensions r = 3:
V D
Where
L
M
L
L3
t
g
L
t2
Repeat parameters m = 3:
V
M
D
g
L t
L
D
g
Then n − m = 2 dimensionless groups will set up one dimensionless equation.
a
b
V
L
M
c L
Summing exponents, 1 = g a b DcV = 2 3 ( L ) = M 0 L0t 0 . Hence, 1 =
.
gD
t L
t
L
M
c M
Summing exponents, 2 = g a b Dc = 2 3 ( L ) 2 = M 0 L0t 0 . Hence, 2 =
.
g D2
t L
t
a
By checking using F
The relation between
L
b
t as primary dimensions, give 1 = 1 and 2 = 1 .
1 and 2
is
1 = f ( 2 ) or, V = gD f
2
gD
(Ans.)
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Hill LLC.
-25-
2-29 A model sphere of diameter D is to be immersed in a water tunnel running at a speed of V
in order to obtain data that can be used to predict the drag force on a weather balloon that is traveling
in air at 1.5 m/s . For a model diameter of 50 mm , the measurement of the drag force on the small
sphere yields 3.78 N . Assuming similarity conditions have been established between the spherical
lab-scale model and the air balloon, what should be the water tunnel speed if the diameter of the
balloon is 3 m ? Estimate the drag force on the full-scale balloon.
Solution: For water at 20o C temperature, take
For water at 20o C temperature, take
= 998 kg/m3 and
= 0.001 kg/m s .
= 1.2255 kg/m3 and = 1.78 10−5 kg/m s .
The balloon velocity follows dynamic similarity, which requires identical Reynolds numbers
Remodel =
998 (1.5)( 0.05)
1.2255 (Vballoon )( 3)
VD
=
= 74850 = Reprototype =
model
0.001
1.78 10−5
−1
Therefore, Vballoon 3.624 10 m/s .
(Ans.)
Then, the two spheres will have identical drag coefficients.
CD ,model =
F
V 2 D 2
=
model
3.78
998 (1.5) ( 0.05)
2
Therefore, Fballoon 9.754 10
−1
2
= 0.6733 = CD ,prototype =
N.
Fballoon
1.2255 ( 3.624 10−1 ) ( 3)
2
2
(Ans.)
2-30 Surface tension is responsible for causing capillary waves to develop on a liquid-free surface
with an average speed, V , that is dependent on the fluid density, , surface tension, , and spatial
wavelength, . Obtain the nondimensional relations linking the speed of the capillary waves to these
three independent quantities.
Solution: Number of parameters n = 4 :
Primary dimensions r = 3:
V
Where
L
t
M
t2
M
V
L t
M
L
L3
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Hill LLC.
-26-
Repeat parameters m = 3:
Then n − m = 1 dimensionless group will set up one dimensionless equation.
a
c
L M
bM
Summing exponents, 1 = V V = 2 ( L ) 3 = M 0 L0t 0 . Hence, 1 = V
.
t t
L
a
By checking using F
b
c
t as primary dimensions, give 1 = 1 .
L
=C.
The functional relationship is 1 = V
Then, the nondimensional speed of capillary waves is V = C
.
(Ans.)
2-31 The lifting force F on an air-launched rocket depends on its length L , velocity V , diameter
D , angle of attack , density , dynamic viscosity , and speed of sound c . Using the
Buckingham Pi theorem, determine the dimensionless groups or Pi parameters needed to
characterize the functional relation between the lifting force and the quantities that affect it.
Solution: Apply the Buckingham procedure:
Number of parameters n = 8:
F
Primary dimensions r = 3:
F
L
Where
ML
2 L
t
L
V
M
L t
V
D
L
t
L
1
M
L3
Repeat parameters m = 3:
V
M
Lt
D
c
c
L
t
D
Then n − m = 5 dimensionless groups will set up one dimensionless equation.
a
b
F
c ML
M L
1 = V D F = 3 ( L ) 2 = M 0 L0t 0 . Hence, 1 =
V 2 D 2
L t
t
a
b
c
a
(Ans.)
b
c
L
M L
2 = aV b Dc L = 3 ( L ) ( L ) = M 0 L0t 0 . Hence, 2 =
D
L t
(Ans.)
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-27-
cM
M L
3 = aV b Dc = 3 ( L ) = M 0 L0t 0 . Hence, 3 =
VD
L t
Lt
a
b
a
b
a
b
(Ans.)
c L
cV
M L
4 = V D c = 3 ( L ) = M 0 L0t 0 . Hence, 4 = 2
D
L t
t
(Ans.)
a
b
c
c
M L
5 = V D = 3 ( L ) (1) = M 0 L0t 0 . Hence,
L t
a
b
c
5 =
(Ans.)
2-32 A solid propellant rocket chamber is often modeled as a porous tube with sidewall injection
(Majdalani 1999). The oscillatory Stokes boundary layer that develops inside a porous tube with
sidewall injection (due to pressure oscillations) is affected by variations in several parameters. These
include the injection velocity U , the frequency of oscillations , the kinematic viscosity , the
hydraulic radius of the chamber R , the chamber length L , and the axial location within the chamber
X . The dependence of the Stokes boundary-layer thickness on these parameters may be expressed
as = f (U , ,, R, L, X ) . Using scaling principles, show that the penetration depth of the unsteady
rotational wave, y p = R , may be related to the following dimensionless parameters
yp =
R X
= F Mj , St , ,
R
L L
where Mj =
U3
L
and St =
2
R
U
Note that the square, square-root, or reciprocal of a dimensionless parameter as well as the product
of two dimensionless parameters, such as X R and R L , can also produce a nondimensional group.
Solution: Number of parameters n = 7 :
Primary dimensions r = 3:
M
Where
L
L X
L L
U
R
L
t
1
t
L2
t
L
Repeat parameters m = 2 :
U
U
R L X
L t
R
Then n − m = 5 dimensionless groups will set up one dimensionless equation.
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Hill LLC.
-28-
The dimensionless groups in this case can be easily identified through inspection (check using F ,
L , t as primary dimensions is not really needed here).
3
X = U
R
, 4
, 5 = L
1 = , 2 = , 3 =
2
R
U
L
R
L
The relation between
1 , 2 , 3 , 4 , and 5 is
3
1 = f ( 2 , 3 , 4 , 5 ) or, y p = = f R , X , U2 , L = f Mj, St , R , X
L L R U
R
2-33
L L
(Ans.)
The fluid hammer or hydraulic shock can occur when a valve is closed rapidly in a fluid
supply line. Given a circular tube, the pressure impulse p = p1 − p2 , also known as the strength of
the fluid hammer, depends on the fluid density, , the speed of sound in the fluid, c , the diameter
of the tube, D , the dynamic viscosity, , and the mean-flow velocity, V . Using the Buckingham
Pi theorem, identify the dimensionless groups that will minimize the number of test cases needed to
fully characterize the pressure impulse sensitivity to these five independent parameters. Note that
the dimensionless pressure impulse is traditionally referenced to the speed of sound and that the
Reynolds number based on the speed of sound is known as the acoustic Reynolds number.
Solution: Number of parameters n = 6 :
Primary dimensions r = 3:
p
Where
M
2
Lt
M
L3
M
c
L
t
Repeat parameters m = 3:
D
c
D
V
L t
M
Lt
L
p
D
V
L
t
V
Then n − m = 3 dimensionless groups.
The dimensionless groups are as follows:
a
b
p
c M
M L
1 = aV b Dc p = 3 ( L ) 2 = M 0 L0t 0 . Hence, 1 =
V 2
L t
Lt
(Ans.)
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Hill LLC.
-29-
a
b
c L
c
M L
2 = aV b Dc c = 3 ( L ) = M 0 L0t 0 . Hence, 2 =
V
L
t
t
(Ans.)
cM
M L
3 = V D = 3 ( L ) = M 0 L0t 0 . Hence, 3 =
VD
L t
Lt
(Ans.)
a
a
b
b
c
2-34 The drag force acting on a submarine can be taken to be a function of the water density, ,
speed, V , viscosity, , and displaced volume, . Using a 1: 50 scaled-down model of the
submarine in a laboratory-size water tunnel, start by specifying the dimensionless parameters needed
to guide your drag force measurements. Assuming a leveling of the drag force coefficient at
increasing speeds, provide a best guess of the drag force on a full-scale prototype traveling at
27 knots if a value of 13 N is obtained in a water tunnel running at 10 knots .
Solution: Number of parameters n = 5:
Primary dimensions r = 3:
F
Where
ML
2
t
V
M
L3
L
t
Repeat parameters m = 3:
M
M
Lt
L
F
V
t
L3
V
Then n − m = 2 dimensionless parameters.
The dimensionless parameters are as follows:
a
b
F
c ML
M L
1 = aV b c F = 3 ( L3 ) 2 = M 0 L0t 0 . Hence, 1 =
V 2 2 3
L t
t
(Ans.)
cM
M L
2 = V = 3 ( L3 ) = M 0 L0t 0 . Hence, 2 =
V 1 3
L t
Lt
a
a
b
b
c
(Ans.)
From second dimensionless parameter,
p
m
=
13
pVp p
mVmm1 3
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Hill LLC.
-30-
13
13
13
p p Vm m
10 knots 1
10 knots 1
=
Therefore,
= (1)
=
m m Vp
27
knots
50
27 knots 50
p
From first dimensionless parameter,
Fp
pVp p
2
23
=
where p and
Fm
mVm 2m 2 3
m
subscripts in these equations represent the prototype and the model, respectively.
Therefore,
V
Fp = Fm p p
m Vm
2
p
m
23
13
2
10 knots 1 27 knots 50
= 13 N
27 knots 50 10 knots 1
23
= 129.3095 knots
(Ans.)
2-35
Experimental measurements of the pressure drop across a sudden contraction in a circular
p = p1 − p2 can be a function of the density, , dynamic viscosity, , meanflow velocity, V , and both large and small diameters, D1 and D2 . Identify the dimensionless groups
tube suggest that
that will minimize the number of test cases needed to fully characterize the pressure dependence on
these five independent parameters.
Solution: Number of parameters
Primary dimensions r = 3:
M
n = 6 : p D1 D2 V
L t
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Hill LLC.
-31-
p
Where
M
2
Lt
M
L3
Repeat parameters
D1
D2
L
M
Lt
L
V
L
t
m = 3: D1 V
Then n − m = 3 dimensionless groups.
The dimensionless groups are as follows:
a
b
p
c M
M L
1 = aV b D1c p = 3 ( L ) 2 = M 0 L0t 0 . Hence, 1 =
V 2
L t
Lt
a
(Ans.)
b
D2
c
M L
2 = V D D2 = 3 ( L ) ( L ) = M 0 L0t 0 . Hence, 2 =
D1
L t
(Ans.)
cM
M L
3 = aV b D1c = 3 ( L ) = M 0 L0t 0 . Hence, 3 =
VD
L t
Lt
(Ans.)
a
b
c
1
a
b
2-36 Water at 15 C flows through a 0.1 m internal diameter galvanized iron pipe ( = 0.00015 m
) at a mass flow rate, m , of 15 kg/s . We are told that:
The total pipe length, which is L = 150 m , runs in a horizontal plane.
The pipe contains one check valve ( Le D = 150 ) and four 90 elbows ( Le
The properties of water are:
D = 30 each).
= 1000 kg/m3 , = 0.00114 N s/m2 , and g = 9.81 m/s2 .
(a) Find the Reynolds number. Is the flow turbulent?
(b) Find the Darcy friction factor f .
(c) Find the major head loss,
hM .
(d) Find the minor head loss,
hm .
(e) What percentage of the total head loss is due to minor losses?
(f) Find the pressure drop across this pipe.
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Hill LLC.
-32-
Solution: We know that m = VA . Thus, V =
m
=
A
15 kg/s
= 1.9099 m/s .
2
3
2
(1000 kg/m ) 4 ( 0.1) m
VD (1000 kg/m ) (1.9099 m/s )( 0.05 m )
Re =
=
83767
( 0.00114 N s/m2 )
3
Darcy friction factor is calculated from
(
3500 ) . (Turbulent)
d
1
2.51
−2.0log
+
3.7 Re f
f
; f 0.0241 .
(150 m ) (1.9099 m/s ) = 6.7209 m
L V2
= ( 0.0241)
Major head loss hM = f
.
D 2g
( 0.1 m ) ( 2 9.81 m/s2 )
(Ans.)
(Ans.)
2
Sum of loss coefficients
(Ans.)
K = 0.0241 (150 + 4 30) = 6.507 .
(1.9099 m/s ) = 1.3146 m
V2
= ( 6.507 )
Minor head loss hm = ( K )
.
2g
( 2 9.81 m/s2 )
2
(Ans.)
Total head loss hL = hM + hm = 6.7209 m + 1.3146 m = 8.0355 m .
Therefore, % contribution of
hm to hL is %contribution =
hm
1.3146 m
100 =
100 = 16.36% . (Ans.)
hL
8.0355 m
Pressure drop p = 1 V 2 K = 1 (1000 kg/m3 ) (1.9099 m/s )2 ( 6.507 ) = 11.8679 kPa .
2
2
(Ans.)
Consider a water pipe of length 100 m , diameter of 0.1541 m ( 6 inches ), roughness of
4.6 10−5 m (DN 150 Schedule 40 Steel), viscosity of 1.08 10−5 m2 /s , specific weight of
8630 N/m3 , and an inlet pressure of 120 kPa at Point 1. In the presence of minor losses, determine:
2-37
(a) The maximum permissible volumetric flow rate (in m3 /s ) and average velocity at Point 2 that
will ensure that the pressure drop in the pipe does not exceed 60 kPa .
(b) The total head loss in meters.
(c) The percentage of the total head loss that corresponds to minor losses.
We are told that the pipe runs in a horizontal plane and that it contains an open butterfly valve (
Le D = 45 ) and two standard 90° elbows ( Le D = 30 each).
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Hill LLC.
-33-
Solution: Pressure drop across the pipe is p = 60 kPa .
Cross-sectional area A =
Density =
D2
4
=
( 0.1541 m )
2
4
= 0.01865 m 2 .
8630 N/m3
=
= 879.7146 kg/m3 .
g 9.81 m/s 2
Dynamic viscosity = = ( 879.7146 kg/m3 )(1.08 10−5 m 2 /s ) = 0.0095 N s/m 2 .
Velocity in terms of friction factor and pressure drop is given by
V=
2 Dp
=
f
2 ( 0.1541 m )( 60000 Pa )
(100 m ) (879.7146 kg/m ) ( f )
3
=
Initial guess for friction factor f = 0.02 gives V =
0.210205
f
0.210205
0.210205
=
= 3.242 m/s .
f
0.02
3
VD (879.7146 kg/m ) ( 3.242 m/s )( 0.1541 m )
=
46262 .
For f = 0.02 , Re =
( 0.0095 N s/m2 )
Using Haaland formula, friction factor can be estimated from
1.11
6.9
6.9 D 1.11
4.6 10−5 m 0.1541 m
1
= −1.8log
+
+
= −1.8log
3.7
f
46262
Re 3.7
Thus, f = 0.021946 .
An improved guess of f = 0.022168 is estimated through an iterative approach for convergence.
The respective velocity is V = 3.0793 m/s .
Then, volumetric flow rate is Q = VA = ( 3.0793 m/s ) ( 0.01865 m2 ) = 0.05743 m3 /s .
Sum of loss coefficients
(Ans.)
K = 0.022168 ( 45 + 2 30) = 2.32764 .
Total head loss hL = hM + hm = f
Therefore, % contribution of
L V2
V2
+ ( K )
= 8.0772 m .
D 2g
2g
hm to hL is %contribution =
(Ans.)
hm
1.1249 m
100 =
100 = 13.9269% .
hL
8.0772 m
(Ans.)
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Hill LLC.
-34-
2-38
A 200 m long cast iron pipe has a roughness of = 0.00025 m and a diameter of 0.05 m .
3
It carries water with a density of 1000 kg/m and a kinematic viscosity of 10−6 m2 /s . The gauge
pressure in the water main is 1500 kPa . The supply line will require the installation of eight elbows
( Le D = 30 each), two standard tees ( Le D = 60 each), and one square-edged entrance ( K = 0.5
), in the total length of 200 m . The gage pressure required at the discharge point is 500 kPa . Find
the discharge flow rate Q .
Solution: Pressure drop across the pipe is p = p1 − p2 = 1500 kPa − 500 kPa = 1000 kPa .
Cross-sectional area A =
D2
4
=
( 0.05 m )
4
2
= 0.0019635 m 2 .
Dynamic viscosity = = (1000 kg/m3 )(10−6 m2 /s ) = 0.001 N s/m 2 .
Velocity in terms of friction factor and pressure drop is given by
V=
2 ( 0.05 m )(1000000 Pa )
2 Dp
0.5
=
=
3
f
f
( 200 m ) (1000 kg/m ) ( f )
Initial guess for friction factor f = 0.02 gives V =
0.5
0.5
=
= 5 m/s .
f
0.02
3
VD (1000 kg/m ) ( 5 m/s )( 0.05 m )
=
= 250000 .
For f = 0.02 , Re =
( 0.001 N s/m2 )
Using Haaland formula, friction factor can be estimated from
1.11
6.9 D 1.11
6.9
1
0.00025 m 0.05 m
= −1.8log +
+
= −1.8log
3.7
f
Re 3.7
250000
Thus, f = 0.030775 .
An improved guess of f = 0.03086 is estimated through an iterative approach for convergence.
The respective velocity is V = 4.0252 m/s .
Then, volumetric flow rate is Q = VA = ( 4.0252 m/s ) ( 0.0019635 m2 ) = 0.0079 m3 /s .
(Ans.)
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Hill LLC.
-35-
2-39 Water at 15 C flows through a galvanized iron pipe with a roughness of = 0.00015 m , at
a mass flow rate, m , of 20 kg/s . We are told that:
The total pipe length, which is L = 200 m , runs in a horizontal plane.
The pipe contains one check valve ( Le
The properties of water are:
D = 55 ), and four
90 elbows ( Le
D = 30 each).
= 1000 kg/m3 , = 0.00114 N s/m2 , and g = 9.81 m/s2 .
The total pressure drop across the pipe is 169 kPa . What is the diameter of the pipe?
Solution: We know that Q =
m
20 kg/s
=
= 0.02 m3 /s .
3
1000 kg/m
Assume a pipe diameter of D = 0.25 m .
4 (1000 kg/m3 )( 0.02 m3 /s )
VD 4 Q
=
=
89350 .
Reynolds number in terms of Q is Re =
D ( 0.00114 N s/m2 ) ( 0.25 m )
Using Haaland formula, friction factor can be estimated from
6.9 D 1.11
6.9 0.00015 m 0.25 m 1.11
1
= −1.8log +
+
= −1.8log
3.7
f
Re 3.7
89350
Thus, f = 0.020758 .
Pressure drop in terms of Q is p = f
L V 2 8 fL Q 2
= 2 5 .
D 2
D
1
2 5
3
3
8 fL Q 2 8 ( 0.020758)( 200 m ) (1000 kg/m )( 0.02 m /s )
Then, D = 2
.
=
2 (169000 Pa )
p
1
5
An improved guess of D = 0.02614 m is estimated through iteration for convergence.
(Ans.)
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Hill LLC.
-36-
2-40 Water with a density of = 1000 kg/m negotiates a 180 turn through an elbow as shown
below. As the water enters the elbow, its gage pressure is 200 kPa . The pressure at the outlet section
3
is
atmospheric.
Assuming
uniform
properties
everywhere,
we
have
A1 = 0.002 m 2 ,
A2 = 0.0006 m 2 , and U1 = 2 m / s . Determine the horizontal force that is needed to prevent the
elbow from separating. Are the bolts in tension or compression?
Solution: Momentum flux in
x
direction for the elbow Fx = FSx + FBx =
u dV +CS u VdA
t CV
assuming steady, uniform, and incompressible flow with atmospheric pressure at exit.
Also assume that the horizontal force Rx to hold the elbow at its place is from left to right.
Therefore, Rx = − p1g A1 − (U12 A1 + U 22 A2 ) .
From continuity,
U1 A1 = U 2 A2 .
0.002 m2 )
(
A1
= ( 2 m/s )
= 6.6667 m/s .
Then, U 2 = U1
A2
( 0.0006 m2 )
Then,
(
Rx = − ( 200000 kg/m s 2 )( 0.002 m 2 ) − (1000 kg/m3 ) ( 2 m/s ) ( 0.002 m 2 ) + ( 6.6667 m/s ) ( 0.0006 m 2 )
Rx = −434.6669 N
2
2
(Ans.)
The negative sign indicates that the horizontal force needed to prevent the elbow from getting
separated is acting from right to left. Therefore, the bolts are in compression.
(Ans.)
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Hill LLC.
-37-
)
2-41
Consider a two-dimensional planar channel with one inlet and two outlets that include a
reducing bend. The velocity distribution at the inlet varies linearly from zero to U1 . The velocity at
the two outlets is uniform and the motion may be taken to be incompressible and steady. Determine
(a) the maximum velocity at Section 1 using the integral, control-volume approach and (b) the
reaction forces needed to support the channel.
Solution: The control-volume around sections 1, 2, and 3 yields Q1 = Q2 + Q3 ; in other words,
U1 A1 = U 2 A2 + U 3 A3
Thus, U1 =
U 2 A2 + U 3 A3
=
A1
2
+ ( 6 m/s ) ( 0.1 m )
4
= 0.6667 m/s
2
( 0.6 m )
4
( 2 m/s ) ( 0.3 m )
4
Considering water having a density of
2
= 998 kg/m3 flowing through the channel at
(Ans.)
20 C ,
the horizontal reaction force is Rx = m2u2 + m3u3 − m1u1 = ( Q2 )U 2 + ( Q3 )U3 cos 45 − ( Q1 )U1 .
Therefore,
Rx = 356.2828 N .
The vertical reaction force is Ry = m3u3 = ( Q3 )U3 sin 45 = 199.5299 N .
(Ans.)
(Ans.)
2-42 Consider the steady incompressible motion of a fluid across an axisymmetric tube of length
L and radius R = 3 cm .
(a) Determine the uniform velocity at the inlet, U , if the velocity profile at the outlet can be
r2
approximated using u ( r ) = U max cos 2 .
2R
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Hill LLC.
-38-
(b) Evaluate your result for U max = 10 m/s and water with a density of
= 1000 kg/m3 .
(c) Determine the difference in the fluid momentum flux between the inlet and outlet sections of
the tube.
Solution: For the given steady incompressible fluid motion inside the tube
−min + mout = 0
Then, with = constant , we have −Qin + Qout = 0 .
R
r2
−U R 2 + U max cos 2 2 rdr = 0
2R
0
R
r2 r2
2
2
−U R + 2 R U max cos 2 d 2 = 0
2R 2R
0
R
r 2
−U R + 2 R U max sin 2 = 0
2R 0
2
U=
2
2
(Ans.)
U max
For U max = 10 m/s , the inlet velocity is U = 3.1416 m/s .
(Ans.)
(
)
Difference in fluid momentum flux is mout u ( r ) − minU = Qout u ( r ) − QinU .
r2 r2
d
− U 2 R 2
2
2
2R 2R
R
2
( Qout u ( r ) − QinU ) = 2 U max
R 2 cos 2
0
= 2 U
2
max
R
1 r r
4
2
R sin 2 + 2 − U max
R2
4 R 4R 0
2
2
2
2
2
= 2 U max
R2 −
4
(Ans.)
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Hill LLC.
-39-
2-43 Consider a contraction between two pipes in a hydraulic system, which is known as a
reducer. With the pressure and velocity measurements available across the reducer, and given the
empty volume and mass of the reducer (r , mr ) , determine the reaction forces that must be
supplied by the surrounding system to hold the reducer. You may assume that gravity is downward.
Solution: From conservation of mass, we have U1 A1 = U 2 A2 .
Therefore, U 2 = U1
A1
D2
= U1 12 .
A2
D2
From conservation of momentum in x direction, we have
F = R + F
x
surface, x
+ Fbody, x = −mU
1 1 + m2U 2
Since there is no body force (weight) in the x direction,
Rx + Fsurface, x = Rx + p1 A1 − p2 A2 = U1 A1 (U 2 − U1 )
2 2 D12
2
2
Or, Rx = U1 A1 (U 2 − U1 ) − p1 A1 + p2 A2 = U1 D1 2 − 1 − p1D1 + p2 D2
4
D2
(Ans.)
From conservation of momentum in y direction, we have
F = R
y
+ Fsurface, y + Fbody , y = 0
Since there is no surface force (pressure) in the y direction,
Ry + Fbody , y = Ry − mr g − r g = 0
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Hill LLC.
-40-
Or, Ry = g ( mr + r ) .
2-44
(Ans.)
A two-dimensional velocity field is given by
u=−
Ay
Ax
v= 2 2
2 ,
x +y
x +y
2
where A is a constant. Does the field satisfy continuity?
Solution: The continuity equation should satisfy
Here,
u v
+ =0.
x y
u v
Ay Ax
2 xyA
2 xyA
+ = − 2
+ 2
=
−
=0.
2
2
2
x y x x + y y x + y ( x 2 + y 2 ) ( x 2 + y 2 )2
Therefore, the field satisfies continuity.
(Ans.)
2-45 For a certain incompressible, three-dimensional flow field, the velocity components in the
x and y directions are given by
u = x 2 + y 2 + z 2
v = xy + yz + z
(a) Determine the simplest velocity component
rate is zero.
w
in the z direction so that the volume dilatation
(b) Determine the components of the vorticity vector. Is this an irrotational flow field?
Solution: The volume dilatation rate is given by V =
Stated that
u v w
+ + .
x y z
w
u v w
u v
= − − = − ( 3x + z ) .
+ +
= 0 . Therefore,
z
x y
x y z
Then, the velocity component in the
z
z2
direction is w = − 3xz + .
2
(Ans.)
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Hill LLC.
-41-
w=
1 w v u w v u
1
− i +
−
k == − ( y + 1) i + 5 z j − y k 0
j+ −
2 y z z x x y
2
Thus, the flow is rotational.
(Ans.)
2-46 For a two-dimensional flow field, the velocity is defined by V = ( x 2 − y 2 ) i − 2 xyj . Determine
whether this motion is compressible or not.
Solution:
Given V = ( x 2 − y 2 ) i − 2 xyj
(Eq-1)
Generally, the two-dimensional fluid flow is given by V = ui + vj
(Eq-2)
where u and v are velocity components.
Comparing (Eq-1) and (Eq-2), we get
u = x2 − y 2
v = −2xy
Partially differentiating the above equations with respect to x and y, respectively,
u
= 2x
x
u
= −2 x
y
For the flow to be incompressible,
u u
+ =0
x y
2x − 2x = 0
Hence, the flow is incompressible.
2-47 For an incompressible, two-dimensional flow field, the velocity component in the y direction
is given by
v = 3xy + x 2 y
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Hill LLC.
-42-
Determine the velocity component u in the x direction so that the volume dilatation rate is zero.
Solution:
Given:
v = 3xy + x 2 y
The flow is incompressible, and for zero volumetric rate in a 2D flow,
u v
+ =0
x y
(Eq-1)
Since
v
= 3x − x 2
y
From (Eq-1), we have
u
= −3x + x 2
x
(Eq-2)
(Eq-2) can be integrated with respect to x as follows:
du = − 3xdx + x dx + f ( y)
2
u=
−3 2 x3
x + + f ( y)
2
3
(Ans)
where f(y) is the undetermined function of y.
2-48 The three components of the velocity in a flow field are given by
u = x 2 + y 2 + z 2
2
v = xy + yz + z
1
w = −3xz − z 2 + 4
2
Determine the divergence (volumetric dilatation rate) and interpret the results. Determine an
expression for the vorticity vector. Is this an irrotational flow field?
Solution:
The three components of the velocity in a flow field are given by
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Hill LLC.
-43-
u = x 2 + y 2 + z 2
2
v = xy + yz + z
1
w = −3xz − z 2 + 4
2
For 3D flow, we find w.
w
u
v
= −3z
= 2x ,
=y ,
x
x
x
v
u
w
= x+z ,
= 2y ,
=0
y
y
y
w
v
u
= −3x − z
= y + 2z ,
= 2z ,
z
z
z
Rotation component along x direction is
Wx =
1 w v 1
y
− = ( − y − 2z ) = − + z
2 y z 2
2
Rotation component along the y direction is
1 u w 1
5z
Wy = − = ( 2 z − ( −3z ) ) =
2 z x 2
2
Rotation component along the z direction is
1 v u 1
−y
Wz = − = ( y − 2 y ) =
2 x y 2
2
Rotation vector is given by
y 5z y
W = − + z iˆ + ˆj − kˆ
2 2 2
Since no component of the rotation vector is zero, this is rotational flow field.
Volumetric dilation rate is given by
v = v =
u v w
+ +
x y z
2z
= 2 x + x + z + ( −3x ) − = 0
2
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Hill LLC.
-44-
There is no change in the volume of the fluid element as it moves from one location to the other.
2-49 The velocity vector in a certain flow field is prescribed by
V( x, y, t ) = 2xti − 2 ytj
where (x, y) have units of feet and t has units of seconds. Determine expressions for the local
(unsteady) and convective (spatial) acceleration terms. Evaluate the magnitude and direction of
the velocity and acceleration at the point x = y = 2 ft at time t = 0 s.
Solution:
Given, velocity vector is
V( x, y, t ) = 2xti − 2 ytj
Comparing the above equation with v = uiˆ + vjˆ + wkˆ ,
u = 2 xt , v = −2 yt , w = 0
Acceleration is given by a = axiˆ + a y ˆj + az kˆ .
ax =
du
u
u
u
u
=u
+v +w +
dt
x
y
z
t
Convective part
Local part
dv
v
v
v v
=u +v +w +
dt
x
y
z t
dw
w
w
w w
az =
=u
+v
+w
+
dt
x
y
z t
ay =
(Eq-1)
Local acceleration point can be calculated as follows:
u = 2 xt
u
= 2x
t
v = 2 yt
v
= 2y
t
w=0
w
=0
t
Local acceleration = 2 xˆi + 2 yˆj
Convective part can be calculated as follows:
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Hill LLC.
-45-
u
= 0 , u = 0
u = 2 xt , u = 2t ,
x
v = 2 yt ,
y
z
v
v
= 2t , v = 0
=0 ,
y
x
z
So, combining the local and convective parts of acceleration in (Eq-1),
ax =
u
u
u
u
+u +v +w
t
x
y
z
= 2 x + ( 2 xt ) 2t
ax = 2 x + 4 xt 2
ay =
v
v
v
v
+u +v +w
t
x
y
z
= 2 y + ( 2 yt )( 2t )
ay = 2 y + 4 yt 2
2
2
And convective acceleration = 4 xt ˆi + 4 yt ˆj .
Net acceleration = ax ˆi + a y ˆj = ( 2 x + 4 xt 2 ) ˆi + ( 2 y + 4 yt 2 ) ˆj .
Velocity x = y = 2ft, t = 0.
V ( x, y, t ) = 2 xtˆi − 2 ytˆj
Substituting, x = y = 2 and t = 0,
V=0
Acceleration at x = 2, y = 2, and t = 0 is
a = ( 2 x + 4 xt 2 ) ˆi + ( 2 y + 4 yt 2 ) ˆj
a = ( 4 + 0 ) ˆi + ( 4 + 0 ) ˆj=4iˆ + 4jˆ
The magnitude of the acceleration = a = 42 + 42 = 4 2 ft/s 2
(Ans)
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Hill LLC.
-46-
The direction of the acceleration is
tan =
ay
ax
=
4
=1
4
= tan −1 (1) = 45
(Ans)
2-50 For steady flow, show that the curl of the inviscid momentum equation reduces to
(V ω) = 0 , where = V . Start with the steady expression of Euler’s equation, namely,
(V )V =
1
(V V) − V ( V) = −p / + g
2
Solution:
Steady expression of Euler’s equation is
(V )V =
1
(V V) − V ( V) = −p / + g
2
where = V
V2
−p
( V ) V =
+g
− V =
2
2
2
where V = V = V V
By taking the curl of Euler’s equation, we get the velocity equation.
(Euler’s equation)
V2
( V ) V =
− (V )
2
p
= − + g
(Eq-1)
Applying the identity scalar = 0 , the pressure term vanishes, provided that the density is
uniform.
p
V2
+ g = 0 and
2
=0
Therefore, (Eq-1) remains as
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Hill LLC.
-47-
( V ) V = − ( V )
For inviscid fluid, V = 0
( V ) = 0
(Ans)
2-51 For planar flow in the xy plane, show that the vorticity has only one nonvanishing component
in the z direction, specifically, ω = z k . Show that the vorticity, when written in terms of the
incompressible stream function, becomes
2 2
+ 2
2
y
x
z = −
, where u =
y
v=−
x
Solution:
For 2D planar consideration,
u v
+ =0
x y
2u 2u
u
u
u
1 p
+u
+v
= gx −
+ 2 + 2
t
x
y
x
y
x
And
2u 2u
v
v
v
1 p
+u +v
= gy −
+ 2 + 2
t
x
y
y
y
x
The pressure and gravity can be eliminated by cross-differentiation, that is, by taking the curl of
2D vectorial momentum equation.
The result is
2z 2z
z
z
+ u z +
=
+
2
t
x
y
y 2
x
v u
where z = −
x y
(Eq-1)
(Eq-2)
So the vorticity has only one nonvanishing component in the z direction, specifically
ω = z k .
For incompressible stream functions,
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Hill LLC.
-48-
u=
y
v=−
(given)
x
Therefore, (Eq-2) becomes
2 2
− 2
2
y
x
z = −
2 2
z = − 2 + 2
y
x
(Ans)
2-52 For steady, inviscid, planar flow in the xy plane, (a) show that (V ω) = 0 reduces to
z z
−
= 0 , where u = y
y x
x y
(b) Show that any solution of the form
v=−
x
z = f ( ) will satisfy this condition, including the linear
relation z = C , where C is a constant.
2
Solution:
(a) For steady, inviscid, planar flow in the xy plane,
(V ) = 0
Problem 2-50 shows that steady Euler equation reduces to (V ) = 0 for inviscid
fluid, and problem 2-51 shows that for 2D planar consideration, it can be written as
2z 2z
z
z
+ u z +
=
+
2
t
x
y
y 2
x
(Eq-1)
v u
where z = −
x y
2 2
Also z = − 2 + 2
y
x
2 = − 2
(Eq-2)
Substituting (Eq-2) in (Eq-1), we get
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Hill LLC.
-49-
2
2
2
)+
)−
) = 4
(
(
(
t
y x
x y
(Eq-3)
Considering the given assumptions,
2 ) = 0 (steady)
(
t
v 4 = 0 (inviscid)
(eq-4)
Substituting (Eq-4) in (Eq-3), we get
2
2
)−
(
( ) = 0
y x
x y
z
z
−
=0
y x
x y
(Ans)
(b) Eq-2 shows that 2 = − .
2
Due to symmetry of all the above equations, it can be easily shown that the resulting
vorticity transport equation is fully satisfied by a rotational flow so long as its vorticity
can be expressed as
z = f ( )
including the linear relation z = C .
2
2-53 For planar flow in the xy plane, any solution of the form z = C can be shown to satisfy
the steady, inviscid vorticity transport equation. When this relation is inserted into the vorticity
expression, one gets
2
2 2
+ 2 + C 2 = 0 , where u =
2
y
x
y
v=−
x
(a) Apply the classification test given by Eq. (2-68) to determine the character of the equation.
(b) How many boundary conditions will be needed in order to solve this equation?
Solution:
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Hill LLC.
-50-
z = C 2
(Eq-1)
When (Eq-1) is inserted into the vorticity expression, we get
2 2
+ 2 + C 2 = 0
2
x
y
(Eq-2)
(a) Applying classification test given by (Eq 2-68), to find the character of the equation.
Compare (Eq-2) with respect to (Eq 2-67), then we get
A = 0, C = 0, B = 0
B2 − 4 AC = 0
Equation is parabolic.
(b) Equation is parabolic, boundary conditions must be closed in one direction. If the
boundary condiations remain open at the other end of the other direction it becomes a
mixed initial and boundary-value problem.
2-54 The Taylor profile, which has been used to describe the bulk gaseous motion in twodimensional rocket chambers, corresponds to a planar profile in porous channels that bears
symmetry with respect to the chamber’s midsection plane at y = 0. Assuming unit wall injection
and unit chamber half height, the domain may be defined over 0 x and 0 y 1 , as shown
in Fig. 2-54. Using the stream-function vorticity approach, one may take z = C 2 to secure the
vorticity transport equation and then eliminate the velocities in favor of the stream function
everywhere. The resulting vorticity equation in terms of the stream function becomes
2 2
+ 2 + C 2 = 0
2
x
y
where u =
y
=−
x
(a) Using the method of separation of variables with ( x, y) = X ( x)Y ( y) , find the general form
of the solution.
(b) Express the problem’s four fundamental boundary conditions in terms of the stream function
knowing that:
u ( x,1) = 0
( x,1) = −1
( x, 0) = 0
u (0, y ) = 0
no axial speed at the wall as the fluid enters only normally
normal speed matches the unit injection value at the wall
symmetry about the midsection plane prevents flow crossing
no axial inflow at the chamber headwall
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Hill LLC.
-51-
(c) Find the solution that satisfies these boundary conditions.
(d) Determine the velocities associated with this solution and verify that they satisfy the four
boundary conditions stated in Part (b).
FIGURE P2-54
Solution:
(a) By inserting the product solution back into the vorticity equation, dividing through by
( x, y) = X ( x)Y ( y) , and rearranging, we get
Y
X
+ C2 = −
= 2
Y
X
where is the separation constant. We take = 0 because any other finite value can be shown
to be unphysical, as it would lead to either trigonometric or exponential variations in the axial
direction. By solving each of the separated ODEs individually, we obtain the general expression
( x, y ) = ( C1 x + C2 ) C3 cos ( Cy ) + C4 sin ( Cy )
Ans. (a)
(b) Next, we use the velocity-stream function definitions to relate the velocities to :
u ( x, y ) =
and ( x, y ) = −
y
x
We then convey the 4 boundary conditions to the stream function derivatives:
( x,1)
( x,1)
u ( x,1) = y = 0
y = 0
( x,1)
( x,1)
( x,1) = − x = −1
x = 1
or
( x, 0 ) = − ( x, 0) = 0
( x, 0) = 0
x
x
(0, y )
(0, y )
u ( 0, y ) =
=0
=0
y
y
Ans. (b)
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Hill LLC.
-52-
(c) Given the general form of the stream function, we can first evaluate
y = C ( C1 x + C2 ) −C3 sin ( Cy ) + C4 cos ( Cy )
= C C cos ( Cy ) + C sin ( Cy )
1 3
4
x
Direct substitution into the 4 boundary conditions yields:
( C1 x + C2 ) C C4 cos ( C ) − C3 sin ( C ) = 0
C1 C3 cos ( C ) + C4 sin ( C ) = 1
C1C3 = 0
CC2 C4 cos ( Cy ) − C3 sin ( Cy ) = 0
This set can now be solved simultaneously. The last boundary condition returns C2 = 0
especially that any other choice leads to an unphysical outcome. The third boundary condition
leads to C3 = 0 since the choice of C1 = 0 is no longer viable as it eliminates the axial
dependence altogether. At this juncture, the first boundary condition reduces to the problem’s
solvability condition, namely,
CC1C4 x cos C 0; x
To avoid a trivial outcome, it may be readily shown that the constants C , C1 , and C4 cannot be
permitted to vanish. This leaves us with
cos C = 0 or C = (2n +1) / 2; n = 0,1,2,
Realizing that the cases of n = 1,2, lead to an increasing number of flow reversal regions, we
select n = 0 or C = / 2 for a unidirectional motion. The stream function becomes:
( x, y ) = C1C4 x sin ( 12 y )
This enables us to set C4 = 1 with no loss of generality (i.e., whenever the product of two
undetermined constants appears jointly as yet another undetermined constant, one of the original
constants can be set equal to unity; alternatively, their product can be defined as a new
undetermined constant that supersedes the original two constants). Lastly, the second boundary
C =1
condition enables us to retrieve C1C4 sin ( / 2) = 1 or 1
. We thus arrive at the celebrated
profile due to Taylor (1956):
Taylor Profile:
( x, y ) = x sin ( 12 y )
Ans. (c)
As a testament to his ingenuity, G. I. Taylor was able to guess this solution with very little
analysis.
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Hill LLC.
-53-
(d) To verify that the above solution can indeed satisfy our boundary conditions, we first derive
the velocities directly from the stream function. We get
u ( x, y ) = 12 x cos ( 12 y )
and ( x, y ) = − sin ( 12 y )
These velocities can be evaluated at the appropriate locations at the wall, midsection plane, and
headwall. We find
u ( x,1) = 12 x cos 12 (1) = 0
( x,1) = − sin 12 (1) = −1
1
( x, 0 ) = − sin 2 ( 0 ) = 0
1
1
u ( 0, y ) = 2 ( 0 ) cos ( 2 y ) = 0
Ans. (d)
The solution clearly reproduces the 4 boundary conditions that are prescribed for this problem.
2-55 For axisymmetric flow in the rz plane, show that the vorticity has only one nonvanishing
component in the θ direction, specifically, ω = e . Show that the vorticity, when written in
terms of the incompressible stream function, becomes
1 2 1 1 2
= −
+
−
, where vr = − 1
r z 2 r 2 r r r 2
r z
vz =
1
r r
Solution:
The vorticity equation can be equally simplified for axis symmetric flows.
The radial and axial velocities can be related to the incompressible Stoke’s steam function using
the given relations
vr = −
1
and vz = 1
r r
r z
Since the flow is confined to the rz plane, the vorticity reduces to a single component in the
direction.
v v
ω = e = r − z e
z r
ω = e =
(Eq-1)
−1 2 1 2
−
+ 2 e
r r 2 r r
z
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Hill LLC.
-54-
Therefore, =
−1 2 1 1 2
+
−
.
r r 2 r 2 r r 2 z 2
(Ans)
2-56 For steady, inviscid, axisymmetric flow in the rz plane, (a) show that the vorticity transport
equation (V ω) = 0 reduces to
( / r ) ( / r )
−
= 0 , where vr = − 1
r
z
z
r
r z
(b) Show that any solution of the form
= rg ( )
vz =
1
r r
satisfies this condition, including the linear
relation = C r , where C is a constant.
2
Solution:
(a) We know that the vorticity transport equation, i.e., (V ω) = 0 , can be obtained by
taking curl as
D
= ( − )V + v2
Dt
(Eq-1)
As we know that z = − in 2D planar flow, we stumble upon the stokes operator D 2 , which
is a minor variant of the Laplacian in cylindrical coordinates. Hence
2
1
2
−1 2
+ 2
=
D with D = 2 −
r r r z
r
2
(Eq-2)
As for vorticity transport equation, since ( ) = 0 for all axis-symmetric flows bearing only
one vorticity component, (Eq-1) becomes
−
t
z r r
2 =
2
2 1
2 1 D ;
+
=
−
=
−
v
−
r2
r2 r
r z r
2 1 2
+
+
r 2 r r z 2
(Eq-3)
Expanding the expressions, (Eq-3) can be written as
D 2
t r
D 2 D 2
+
2 −
2 =
r z r z r r
1 4 4
4 2 3 3 3 2 3
= v 4 + 4 +2 2 2 − 2 3 +
−
+
z
r z r r r z 2 r 3 r 2 r 4 r
r r
(Eq-4)
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Hill LLC.
-55-
For limiting our attention with steady flow and inviscid flow (i.e., v = 0) with Re>>1, we get
D 2 D 2
−
=0
r z r 2 z r r 2
( / r ) ( / r )
−
=0
r
z
z
z
(b) Any rotational motion with
equation identically.
(Ans)
= rf ( ) can be shown to satisfy the vorticity transport
By using = C r , several solutions have been advanced and shown viable for
Reynolds number as low as 200.
2
2-57 For axisymmetric flow in the rz plane, any solution of the form = C r can be shown
to satisfy the steady, inviscid vorticity transport equation. When this relation is inserted into the
vorticity expression, one gets
2
2 1 2
−
+ 2 + C 2 r 2 = 0 where vr = − 1
2
r
r r z
r z
vz =
1
r r
(a) Apply the classification test given by Eq. (2-68) to determine the character of the equation.
(b) How many boundary conditions will be needed to solve this equation?
(c) How should you define B(ψ) and H(ψ) in the Bragg–Hawthorne Eq. (2-30e) to reproduce the
same partial differential equation considered here?
Solution:
(a) We have = C r
2
(Eq-1)
When inserted into the vorticity expression, we have
2 1 2
−
+ 2 + C 2 r 2 = 0
2
r 2 r z
(Eq-2)
Comparing this with (Eq 2-67), we have
B2 − 4 AC = 0
Therefore, the equation is parabolic (i.e., no real characteristic solution).
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Hill LLC.
-56-
(b) Since the equation is parabolic, boundary conditions must be closed in one direction but
remain open at one end of the other direction. So it becomes a mixed initial- and
boundary-value problem.
(c) Here the considered equation is
2 1 2
2 −
+ 2 + C 2 r 2 = 0
r
r r z
And Bragg–Hawthorne equation is (Eq 2-30e)
2 1 2 r 2 dH ( )
dB( )
2 −
+ 2 =
−B
r
r r z
d
d
So, we can get B(ψ) and H(ψ) to reproduce the same partial differential equation as
r 2 dH ( ) BdB ( )
−
= −C 2 r 2
d
d
(Ans)
2-58 The Taylor–Culick profile, which has been extensively used to describe the bulk gaseous
motion in axisymmetric rocket chambers (Culick 1966), corresponds to a self-similar
axisymmetric profile in porous tubes. Assuming a unit wall injection and a unit chamber radius,
the domain may be defined over 0 r 1 and 0 z , as shown in Fig. 2-58. Using the streamfunction vorticity approach, one may take = rC 2 to secure the vorticity transport equation
and then eliminate the velocities in favor of the Stokes stream function everywhere. The resulting
vorticity equation in terms of the Stokes stream function becomes
2 1 2
−
+ 2 + C 2 r 2 = 0
2
r r z
r
where
r = −
1
r z
z =
1
r r
(a) Using the method of separation of variables with (r, z) = R(r)Z ( z) , find the general form of
the solution.
(b) Express the problem’s four fundamental boundary conditions in terms of the stream function
knowing that:
no axial speed at the wall as the fluid enters only radially
z (1, z ) = 0
(1, z ) = −1
radial speed matches the unit injection value at the wall
r
symmetry about the centerline prevents flow crossing
r (0, z ) = 0
z (r , 0) = 0
no axial inflow at the chamber headwall
(c) Find the solution that satisfies these boundary conditions.
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Hill LLC.
-57-
(d) Determine the velocities associated with this solution and verify that they satisfy the four
boundary conditions stated in Part (b).
FIGURE P2-58
Solution:
(a) By inserting the product solution back into the vorticity equation, dividing through by
(r, z) = R(r)Z ( z) , and rearranging, we get:
R 1 R
Z
−
+ C 2 r 2 = − = 2
R r R
Z
where is the separation constant. We take = 0 because any other finite value can be shown
to be unphysical, as it would lead to either trigonometric or exponential variations in the axial
direction. By solving each of the remaining ODEs individually, we obtain
( r , z ) = ( C1 z + C2 ) C3 cos Cr 2 + C4 sin Cr 2
Ans. (a)
(b) Next, we use the Stokes stream function definitions to relate the velocities to :
1
1
r ( r , z ) = −
and z (r , z ) =
r r
r z
We then convey the 4 boundary conditions to the stream function derivatives, thus leading to a
well-posed problem in which both the governing equation and its boundary conditions are cast
under the same umbrella in terms of :
1 (1, z )
(1, z )
=0
z (1, z ) = 1 r
=0
r
(1, z ) = − 1 (1, z ) = −1
(1, z ) = 1
r
1 z
z
or
(r , z )
lim 1 (r , z ) = 0
z
=0
r →0 r
r (0, z ) = lim
z
r →0
r
(r , 0)
1 (r , 0)
=0
=0
r
z (r , 0) = r
r
It should be noted that the third boundary condition will be valid if, and only if, the limit is zero
despite the singularity in the denominator as r → 0 . This can occur if the ratio itself is
indeterminate to begin with, implying that the numerator should also vanish. Then, by applying
L’Hôpital’s rule, we must have
(r , z )
2 (r , z )
=0
lim
= 0 and lim
r →0
r →0
zr
z
(
)
(
)
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Hill LLC.
-58-
So finally, the fundamental constraints on the stream function become:
(1, z )
r = 0
(1, z ) = 1
z
2
lim (r , z ) = 0 and lim (r , z ) = 0
r →0 z
r →0
zr
(r , 0)
=0
r
(c) Given the general form of the Stokes stream function, we can write
Ans. (b)
2
2
r = 2C ( C1 z + C2 ) r −C3 sin Cr + C4 cos Cr
= C1 C3 cos Cr 2 + C4 sin Cr 2
z
2
= 2rC1C −C3 sin Cr 2 + C4 cos Cr 2
zr
Direct substitution into the main boundary conditions in Part (b) yields:
2C ( C1 z + C2 ) −C3 sin ( C ) + C4 cos ( C ) = 0
C ( C1 z + C2 ) ( C4 cos C − C3 sin C ) = 0
C1 C3 cos ( C ) + C4 sin ( C ) = 1
C1 ( C3 cos C + C4 sin C ) = 1
or
C C = 0
C1 C3 cos Cr 2 + C4 sin Cr 2 = 0
lim
1 3
r →0
CC r C cos Cr 2 − C sin Cr 2 = 0
2
2
3
2 4
2CC2 r −C3 sin Cr + C4 cos Cr = 0
Note that there is no need to impose the auxiliary constraint that accompanies the third boundary
condition,
lim 2 / zr = lim 2rC1C −C3 sin Cr 2 + C4 cos Cr 2 = 0
r →0
r →0
The latter proves to be self-satisfied due to the algebraic nature of the general solution obtained in
Part (a).
(
)
(
(
)
(
(
)
(
(
)
(
)
)
(
)
(
)
)
( )
(
)
(
)
)
(
)
( )
The last boundary condition returns C2 = 0 especially that any other choice leads to an unphysical
outcome. The third boundary condition leads to C3 = 0 since the choice of C1 = 0 is no longer
viable (it eliminates the axial dependence altogether). At this juncture, the first boundary condition
reduces to the problem’s solvability condition, namely,
CC1C4 z cos C 0; z
To avoid a trivial outcome, it may be readily shown that the constants C , C1 , and C4 cannot be
allowed to vanish. This leaves us with
cos(C) = 0 or C = (2n +1) / 2; n = 0,1, 2,
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Hill LLC.
-59-
Realizing that the cases of n = 1, 2, lead to an increasing number of flow reversal regions, we
select n = 0 or C = / 2 for which the flow remains unidirectional. The stream function becomes:
( r , z ) = C1C4 z sin ( 12 r 2 )
This enables us to set C4 = 1 with no loss of generality (i.e., whenever the product of two
undetermined constants appears jointly as yet another undetermined constant, one of the original
constants can be set equal to unity; alternatively, their product can be defined as a new
undetermined constant that supersedes the original two constants). Lastly, the second boundary
condition enables us to retrieve C1C4 sin ( / 2) = 1 or C1 = 1 . We thus arrive at the celebrated
profile due to Taylor (1956) and Culick (1966), known colloquially as the Taylor–Culick profile:
Taylor–Culick Profile:
Ans. (c)
( r , z ) = z sin ( 12 r 2 )
As a testament to his ingenuity, G. I Taylor was able to guess this solution with very little analysis.
Conversely, F.E.C. Culick derived it 10 years later with the intent of achieving a rotational mean
flow model of the bulk gaseous motion for an internal burning solid rocket motor with a circular
grain perforation.
(d) To verify that the above solution can indeed satisfy our imposed boundary conditions, we first
derive the velocities directly from the stream function:
r ( r , z ) = − sin ( 12 r 2 ) and z ( r , z ) = z cos ( 12 r 2 )
1
r
Next, we can evaluate these velocities at the appropriate locations at the wall, centerline, and
headwall. We find
(1, z ) = z cos 1 (1)2 = 0
2
z
1
2
r (1, z ) = − sin 12 (1) = −1
1
Ans. (d)
r cos 12 r 2
( 0, z ) = lim − 1 sin 1 r 2 = 0 = lim −
=0
2
0 r →0
r →0
r
1
r
r , 0 = 0 cos 1 r 2 = 0
2
z( ) ( )
Clearly, the analytical solution that we obtain is observant of the 4 principal boundary conditions
that are prescribed for this problem.
(
(
)
(
)
)
2-59 In order to derive an exact solution for the velocity field in a wall-bounded cyclonic flow
evolving in a cylindrical chamber (Fig. P2-58), Vyas and Majdalani (2006) assumed that the flow
is steady, incompressible, inviscid, and axisymmetric with respect to the centerline. Because of
the presence of swirl, a three-component velocity profile is used, namely,
V(r , z ) = r er + eθ + z ez . In this problem, a and b(= a / 2) denote the outer and outlet radii
of the cyclonic chamber, U stands for the circumferential tangential velocity at r = a, Ai = Qi / U
denotes the injection area for a volume flow rate of Qi , and
= Ai / (2 aL) represents a
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-60-
dimensionless offset swirl parameter that gauges the relative importance of the axial-to-tangential
speed ratio. The boundary conditions for this problem can be chosen to be
(r = a) = U
( z = 0) = 0
z
r (r = 0) = 0
(r = a) = 0
r
Q = 2 b (r , L) rdr
0 z
i
tangential speed matches imposed injection value
no axial flow penetration at chamber headwall
no flow crossing at centerline
no radial flow penetration at chamber sidewall
conservation of volume flowrate, Qi = UAi at z = L
(a) Guided by experimental data and numerical simulations, we may assume
= (r ) to be an
axially invariant function. Under this assumption, the azimuthal momentum equation in
cylindrical coordinates, namely,
r
1 p
+
+ z + r = −
r
r
z
r
r
decouples from its axial and radial counterparts. Recalling that the motion is axisymmetric
(i.e., there are no variations in ), show that the solution to the azimuthal momentum equation
given above reduces to
+
r
r
r
=0
or
1 (r )
=0
r r
(b) Realizing that the solution to Part (a) constitutes a statement of conservation of angular
momentum, r = B = const, solve for B and
(r ) using the first boundary condition,
(a) = U .
(c) Express the remaining four boundary conditions as functions of the Stokes stream function
using
1
1
and z =
r = −
r z
r r
(d) Because the vorticity transport equation ( V ω ) = 0 reduces to
( / r ) ( / r )
−
=0
r
z
z
r
use partial differentiation to prove that any solution of the form
= rg ( ) satisfies this
2
condition, including the linear relation = C r , where C is a constant.
(e) Substitute the Stokes stream function into the tangential vorticity to show that
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-61-
=
r z 1 1 2 1 2
−
=
−
−
z r r 2 r r r 2 r z 2
2
(f) Substitute = C r , which satisfies the vorticity transport equation, into the vorticity
equation to obtain
2 1 2
−
+ 2 + C 2 r 2 = 0
2
r r z
r
(g) Using the method of separation of variables with (r, z) = R(r)Z ( z) , find the general form of
the solution to the resulting partial differential equation.
(h) Find the particular solution that satisfies the four boundary conditions formulated in Part (c).
(i) Show that the velocities associated with this profile are:
a
r2
a
z
r2
0 r a
r = −U sin 2 ; = U ; z = 2U cos 2 ;
r a
r
a
a 0 z L
(j) Verify that the velocities satisfy the problem’s boundary conditions.
(k) Show that the same governing equation may be restored from the Bragg–Hawthorne Eq. (22 2
30e) where B = const and H ( ) = H0 − 12 C may be readily substituted. The form of the
latter may be easily deduced from a pressure analysis along a streamline.
FIGURE P2-59
Solution:
(a) Let us consider the azimuthal momentum equation:
r
+
+ z
r
r
z
axisymmetry
axial invariance
+
r
r
=−
1 p
r
axisymmetry
Since = (r ) , the tangential velocity has no dependence on (axisymmetry) or z (axial
invariance). Therefore, the second and third terms on the left-hand side vanish. Moreover, the
overarching condition of axisymmetry allows no variations in the azimuthal direction of any
property, including the pressure. As a result, the right-hand side vanishes; we are left with
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-62-
+ =0
r
r
r
or
The latter can be grouped conveniently using
( r )
r
=r
+
r
and so
The azimuthal momentum equation becomes
+
r
r
r
=0
or
1
+ = 0
r r
1 ( r )
=
+
r r
r r
1 ( r )
+ =
=0
r
r r r
Ans. (a)
(b) A single integration of Part (a) in the radial direction leads to r = B , where B must be
determined from one of the boundary conditions. We use the first condition, (a ) = U , to deduce
that B = aU .
Ans. (b)
(c) Next, we may express the remaining 4 boundary conditions as a function of the Stokes stream
function. We get
(r , 0)
1 (r , 0)
=0
=0
r
z (r , 0) = r
r
2
(0) = lim − 1 = 0
lim
=
0
and
lim
=0
r
r →0
r →0 zr
r z
r →0 z
or
Ans. (c)
(a, z )
1 (a, z )
r (a ) = −
=
0
=0
z
a
z
b (r , L)
b
Q UA
1 (r , L)
dr = i = i = UaL
2
rdr
=
Q
i
r
2 2
0
0 r r
It should be noted that the second boundary condition will be valid if, and only if, the limit is zero
despite the singularity in the denominator as r → 0 . This can occur if the ratio itself is
indeterminate to begin with, implying that the numerator must also vanish. Then, by applying
L’Hôpital’s rule, we are able to identify a total of two physical requirements:
=0
lim
= 0 and lim
r → 0 z r
r →0 z
2
Moreover, the last boundary condition can be further simplified into:
b
(r , L) dr = (b, L) − (0, L) = (b, L)
0
r
(d) Due to axisymmetry, and since the vorticity vector reduces to a single component ω = e ,
the vorticity transport equation collapses into
1
1
( V ω ) = e r + e z −
er +
e z ( e ) = 0
z r z
r r
r
By evaluating the second cross product on the right-hand side, we get
ez −
er = 0
er + e z −
z r z
r r
r
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Then, by evaluating the first cross product on the left-hand side, we collect a single azimuthal
vector component
r r z − z r r e = 0
To make further headway, we may expand the derivatives and write
+
−
−
=0
r r z z r r
r z r
r r z
Due to the cancellation of the mixed partial derivatives, we are left with
( / r )
r
z
−
( / r )
z
r
=0
or
g g
−
=0
r z z r
Ans. (d)
where g ( ) / r . At this stage, it can be easily shown that any g ( ) will satisfy the resulting
vorticity transport equation (VTE). For example, one may use the chain rule to express the partial
derivatives of g ( ) by putting
g dg
=
r d r
and
g dg
=
z d z
Direct substitution into the VTE yields
dg dg
−
=0
d r z d z r
Clearly, the above relation is true irrespective of g ( ) , i.e., as long as the ratio
Ans. (d)
/ r can be
2
written in terms of the stream function. This includes any linear relation of the form / r = C .
When the latter is substituted into the VTE, one recovers
2
2
C
−
C
=0
r z
z r
Ans. (d)
which satisfies the VTE identically.
(e) By combining the azimuthal vorticity definition and the Stokes stream function definitions, we
obtain the vorticity equation, which relates the vorticity directly to the stream function:
1 1 1 1 2 1 2
= r − z = −
−
−
Ans. (e)
−
=
z
r z r z r r r r 2 r r r 2 r z 2
2
(f) By substituting the linear relation = C r into the vorticity equation, we arrive at
1 1 2 1 2
C r = 2
−
−
r r r r 2 r z 2
2
Multiplying by
r
and rearranging, the vorticity equation becomes
2 1 2
−
+ 2 + C 2 r 2 = 0
2
r r z
r
Ans. (f)
(g) By inserting the product solution back into the resulting partial differential equation, dividing
through by (r, z) = R(r)Z ( z) , and rearranging, we get
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R 1 R
Z
−
+ C 2r 2 = −
= 2
R r R
Z
where is the separation constant. We take = 0 because any other finite value can be shown
to be unphysical, as it would lead to either trigonometric or exponential variations in the axial
direction. By solving each of the remaining ODEs individually, and consolidating the results, we
arrive at
( r , z ) = ( C1 z + C2 ) C3 cos ( Cr 2 ) + C4 sin ( Cr 2 )
Ans. (g)
(h) Before applying the boundary conditions obtained in Part (c), it is useful to express
2
2
r = 2C ( C1 z + C2 ) r −C3 sin Cr + C4 cos Cr
= C1 C3 cos Cr 2 + C4 sin Cr 2
z
2
= 2rC1C −C3 sin Cr 2 + C4 cos Cr 2
zr
Then the 4 boundary conditions become
2CC r −C sin Cr 2 + C cos Cr 2 = 0
2
3
4
C1C3 = 0
C C cos Ca 2 + C sin Ca 2 = 0
4
1 3
b
Q
2C ( C1L + C2 ) r −C3 sin Cr 2 + C4 cos Cr 2 dr = i
2
0
Note that the auxiliary constraint is not included because it is satisfied identically, namely,
(
(
)
)
(
)
)
(
(
(
)
)
(
(
(
(
)
)
)
)
(
)
(
)
( )
2
= lim 2rC1C −C3 sin Cr 2 + C4 cos Cr 2 = 0
r →0 zr
r →0
lim
( )
The first boundary condition returns C2 = 0 since any other choice leads to an unphysical
outcome. The second boundary condition leads to C3 = 0 since the choice of C1 = 0 is no longer
viable as it eliminates the axial dependence altogether. At this juncture, the third boundary
condition reduces to the problem’s solvability condition, namely,
(
)
C1C4 sin Ca 2 = 0
To avoid a trivial outcome, it may be readily shown that the constants C , C1 , and C4 cannot be
permitted to vanish. This leaves us with
sin(Ca 2 ) = 0 or Ca 2 = (2n + 2) / 2; n = 0,1, 2,
We select n = 0 or C = / a2 for which only one flow reversal will occur, as expected of the
bidirectional motion of a cyclonic flow field. Cases of n = 1, 2, are not considered here as they
give rise to an increasing number of flow reversal regions. For n = 0 , the stream function becomes:
( r , z ) = C1 z sin ( r 2 / a 2 )
where C4 = 1 is set with no loss of generality (i.e., whenever the product of two undetermined
constants appears jointly as yet another undetermined constant, one of the original constants can
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-65-
be set equal to unity). Lastly, the mass conservation condition may be used to retrieve C1 . We
obtain
b
2
r 2 2 r
Q
Qi
C1L cos 2 2 dr = i , C1L sin b 2 = Qi and so C1 =
2
2 L sin ( b 2 / a 2 )
0
a a
a 2
Alternatively, one may use the equivalent form given in Part (c) to express, more straightforwardly,
b
Qi
(r , L) dr = (b, L) = C L sin b2 / a 2 = Qi and so
C1 =
1
0
r
2
2 L sin ( b 2 / a 2 )
(
)
At the outset, the stream function becomes:
(
(
2
2
Qi sin r / a
( r, z ) =
z
2 L sin b2 / a 2
)
)
Ans. (h)
Note that for the special case of b = a / 2 , we get sin ( b2 / a 2 ) = 1 ; and since Ai = 2 aL and
Qi = UAi , the resulting expression can be further simplified into
( r, z ) =
r2
r2
2 aLU
z sin 2 = aUz sin 2
2 L
a
a
Ans. (h)
The resulting profile was first presented by Vyas and Majdalani (2006); it is sometimes referred
to as the quasi complex-lamellar model of a wall-bounded cyclonic motion.
(i) The radial and axial velocities may be deduced directly from the stream function; we get
r2
r2
1
1
a
z
r = −
= 2U cos 2 Ans. (i)
= − U sin 2 and z =
r r
r z
r
a
a
a
As for , we may recall the angular momentum conservation principle leading to r = B = aU
from Part (b). This implies
= U
a
r
Ans. (i)
(j) It is always a good habit to verify that the solution satisfies the originally prescribed boundary
conditions. To do so, we check one-by-one:
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(r = a ) = Ua = U
a
2
z ( z = 0) = 2U 0 cos r = 0
2
a
a
r2
2
r
cos
2
1 r 2 0
a = 0
− aU sin 2 = = lim − aU
r (r = 0) = lim
r →0
r
1
a 0 r →0
(a) 2
− aU
sin 2 = 0
r (r = a) =
a
a
2
2
b
Qi = 2 2U L cos r rdr = 2 aLU sin b = 2 aLU
2
2
0
a
a
a
Ans. (j)
Note that the continuation of the limit in the third boundary condition is carried out using
L’Hôpital’s rule. This confirms that the velocity profile fully satisfies the problem’s boundary
conditions.
(k) The Bragg–Hawthorne equation is defined as
dH ( )
dB ( )
2 1 2
−
+ 2 = r2
−B
2
r r z
d
d
r
Since B = Ua is constant, the last term on the right-hand side vanishes. Additionally, since we are
given a prescribed variation of the form H ( ) = H0 − 12 C 2 2 , the first term on the right-hand side
becomes dH ( ) / d = −C 2 . Forthwith, the Bragg–Hawthorne equation reduces to
2 1 2
−
+ 2 + C 2 r 2 = 0
2
r r z
r
The resulting expression is identical to the vorticity equation obtained in Part (f).
Ans. (k)
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CHAPTER 3. SOLUTIONS OF THE NEWTONIAN
VISCOUS FLOW EQUATIONS
3-1
Solve for constant-pressure Couette flow between parallel plates, as shown at right, for a
non-newtonian fluid, = K ( du/dy ) , n 1. Compare with the newtonian solution.
n
Since the boundary conditions are independent of x, we assume that velocity varies only with y.
The non-newtonian version of Eq. (3-4) is
0=−
dp
+
= 0+
dx y
y
since the pressure is constant. Thus we integrate to
= K ( du / dy ) = constant, or
n
du
= constant, or: u = a + b y
dy
Introducing no-slip boundary conditions at each wall, u ( −h ) = 0, u ( +h ) = V, we obtain
u=
U y
1 +
2 h
(Ans.)
This is exactly the same linear velocity distribution as Eq. (3-6), Fig. 3-2a, for newtonian flow.
The result is independent of the particular behavior of the non-newtonian fluid.
3-2
Solve for u ( r ) in annular Couette flow, Fig. 3-3, when both cylinders are moving. With
u = u ( r ) only and pressure constant, the analysis follows Section 3-2.2:
1 d du
r = 0, or: u = C1 ln ( r ) + C2
r dr dr
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-1-
The no-slip boundary conditions are u ( ro ) = Uo and u ( r1 ) = U1, which enable us to find the two
constants:
C1 =
U1 − Uo
,
ln ( r1 /ro )
C2 = Uo − C1 ln ( ro )
Substituting these constants into the solution above yields the desired velocity distribution:
u = Uo
ln ( r/ro )
ln ( r1 /r )
+ U1
ln ( r1 /ro )
ln ( r1 /ro )
(Ans.)
This is exactly the sum of the separate solutions for moving inner or outer cylinder, Eqs. (3-18)
and (3-19). This superposition is possible because the Navier-Stokes equations are linear for this
particular flow.
We are asked to plot u ( r ) for (a) U1 = Uo ; (b) U1 = − Uo ; and (c) U1 = 2Uo . The results are
shown below for the particular geometry r1 = 2ro . Also shown, for comparison, is the case where
the upper cylinder is not moving.
3-3
For axial Couette flow with only the inner cylinder moving. Eq. (3-18), find the
temperature distribution T ( r ) if the cylinder walls are at To and T1 , respectively.
If T = T(r) only and there are no radial or circumferential velocities, the energy equation from
Appendix B, Eq. (B-9), reduces to
0=
2
ln ( r1 /r )
k d dT
du
r
+
from Eq. (3-18)
, where u = U o
r dr dr
ln ( r1 /ro )
dr
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Hill LLC.
-2-
Introducing u(r), separating the variables, and integrating once, we obtain
ln ( r ) C
Uo2
dT
=−
+ , where C = constant
2
dr
r
k ln ( r1 /ro ) r
Integrating again, noting that ln ( r ) /r = ln 2 ( r ) /2, we obtain
2
U o2 ln ( r/ro )
T=−
+ C ln ( r ) + D
2 k ln 2 ( rr /ro )
Evaluating C and D from the boundary conditions T ( ro ) = To and T ( r1 ) = T1, we obtain the
desired result, after rearranging and cleaning it up a bit:
2
U o2 ln ( r/ro ) U o2 ln ( r/ro )
T = To + T1 − To +
−
2 k ln ( r1 /ro ) 2 k ln 2 ( r1 /ro )
(Ans.)
When plotted, these look very similar to the “channel” temperature profiles in Fig. 3-2b.
3-4
A long thin rod of radius R is pulled axially at speed U through fluid with constant ( , ) .
Find the velocity distribution u z = u ( r ) .
The analysis is similar to that of Section 3-2.2:
1 d du
r = 0, or: u = C1 ln ( r ) + C2
r dr dr
Application of the boundary conditions immediately leads to a paradox:
u ( ro ) = U = C1 ln ( R ) + C2 ; u ( ) = 0 = C1 ln ( ) + C2
(?)
It is impossible to find steady-flow constants if the rod moves through an infinite expanse of
fluid. Physically, it would require the finite-diameter rod to deliver an infinite amount of kinetic
energy to the fluid with only a finite wall shear stress.
3-5
A cylinder of radius R rotates at angular rate in an infinite expanse of fluid. Find the
velocity and pressure distributions and compare with an inviscid “potential” vortex.
The velocity follows from the -momentum equation in Section 3-2.3:
d2u
dr
2
+
d u
= 0,
dr r
or : u = C1r +
C2
r
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-3-
Introducing the boundary conditions, we obtain
u ( R ) = R = C1R +
C2
C
R 2
; u ( ) = C1 ( ) + 2 , or: u =
R
r
()
(Ans.)
This is the solution given by Eq. (3-25). It does indeed correspond to a potential vortex. The
pressure is found from the r-momentum equation:
dp u 2 2 R 4
2 R 4
=
=
, or: p = −
+C
dr
r
r3
2r 2
The constant C = po + 2R 2 / 2 is found from the single condition p = po at r = R. The final
expression for the pressure distribution is thus
(
1
p = po + 2R 2 1 − R 2 /r 2
2
)
(Ans.)
This is exactly what one would find by using Bernoulli’s equation for a potential vortex.
3-6
Find p ( r ) between rotating cylinders using the velocity distribution from Eq. (3-22).
With velocity known, the pressure is found from the r-momentum equation (3-20b):
r r ( − 1 )
r /r − r /r
dp u 2
b
=
, where u = ar + , and a = 1 1 o o o 1 , b = o 1 o
dr
r
r
r1 /ro − ro /r1
r1 /ro − ro /r1
(1)
Introducing u 2 and integrating, we find the pressure:
1
b2
p = a 2 r 2 + 2ab ln ( r ) − 2 + C
2r
2
The constant C = po − [a 2 ro2 /2 + 2ab ln ( ro ) − b 2 / 2ro2 ] is found from the single condition p = po
at r = ro . The final result may be rearranged in the form
1 1
1
p = po + a 2 r 2 − ro2 + 4ab ln ( r/ro ) + b 2 2 − 2
r
2
o r
(
)
(Ans.)
This is not at all what one would obtain by using the (invalid) Bernoulli equation.
3-7
Using one-dimensional integral analysis, derive the differential equation for the
oscillation amplitude X ( t ) of the U-tube given in the figure.
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Hill LLC.
-4-
For one-dimensional incompressible flow at constant area, it follows from the continuity relation,
AV = constant, that the fluid velocity
V=
dX
= fcn ( t ) only
dt
The fluid thus moves as a single “slug” of mass L long, with forces applied due to gravity (the
excess weight when the fluid is displaced) and wall friction. Newton’s law yields
F =
or:
−gro2
( 2X ) − w 2ro L
d2X
2w 2g
+
X=0
ro
L
dt
2
+
2
2 d X
= ro L 2 ,
dt
The problem is closed if we relate wall shear to velocity through the friction factor:
d 2X
dt
2
+
2 w
Cf dX dX 2g
|
|
+ X = 0, where Cf =
2
ro dt dt
L
( dX/dt )
(Ans.)
where the absolute value is needed to make the friction force reverse when the velocity reverses.
Given an initial displacement X ( 0) , we can solve for a damped oscillation X ( t ) . For laminar
flow, we could approximate C f by the pipe-flow value 16/ReD , as in Eq. (3-40). The
differential equation then becomes
d2X
dt 2
+
r2
2 dX 2g
+
X = 0, where t* = o
t* dt
L
4v
(1)
The quantity t* is a characteristic damping time of a laminar U-tube oscillation. If we displace
the fluid to a fixed initial point X o and release it, the solution of (1) above is
X = Xoe−t/t* cos ( n t ) , where n = ( 2g/L )
1/2
(Ans.)
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-5-
The natural frequency is the same as that of a (frictionless) pendulum of length (L/2), and the
oscillation damps to 1/2 amplitude in a time t = 0.693 t*. For turbulent flow, the basic
differential equation is nonlinear and a numerical solution is required.
3-8
Air at 20°C and 1 atm is driven between parallel plates 1 cm apart by a pressure gradient
and by the upper plate’s moving at 20 cm/s.
(a) Find the volume flow per unit width if the pressure gradient is –0.3 Pa/m.
The velocity distribution is given by Eq. (3-42):
u=
U y h 2 dp y 2
1 + +
− 1 −
2 h 2 dx h 2
The volume flow per unit depth is found by integrating this across the fluid between plates:
+h
Q=
−h
u dy = U h +
2h 3 dp
−
3 dx
(1)
We are given h = 1/2 cm, U = 20 cm/s, and dp/dx = −30 Pa/m. For air at 20C, =
1.81E-5 Pa-s. Equation (1) above may thus be evaluated for this case:
2 ( 0.005 m ) ( 0.3 Pa/m )
Q = ( 0.2 m/s )( 0.005 m ) +
= 0.00100 + 0.00138
3 (1.81E-5 Pa-s )
3
= 0.00238 m2 / 2 = 2380 cm3 /s/m
(Ans. a)
(b) Find the pressure gradient for which the shear stress at the lower plate is zero.
This is the ‘separation criterion’ given by Eq. (3-43) – [see Fig. 3-8, dashed line]:
dp 2U 2 (1.81E-5 Pa -s )( 0.2 m /s )
=
=
= + 0.0724 Pa/m
dx ( 2h )2
( 0.01 m )2
(Ans. b)
It takes a positive pressure gradient to cause flow separation.
3-9
Derive the solution u ( y, z ) for flow through an elliptical duct, Fig. 3-9, by solving
Eq. (3-30). Begin with an guessed quadratic solution, u = A + By 2 + Cz 2 , and work your way
through to the exact solution.
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-6-
Solution: The guessed solution seems OK since 2u = constant, as required by Eq. (3-30) for
fully-developed duct flow. Since u must be zero on the ellipse walls of semi-axes a and b, as
seen in Fig. 3-9, the constants ( A, B, C ) must be related to the ellipse formula. That is,
y2 z2
u = K 1 − 2 − 2
a
b
where K = constant
Substitute this u, which has no-slip (u = 0) at the walls, into the momentum equation (3-30):
2u
2u
1 dp a 2b2
−2 −2 1 dp
+
=K 2 + 2 =
, solve K =
−
2 dx a 2 + b2
y 2 z 2
b dx
a
1 dp a 2b 2
y2 z2
−
Thus, finally, for an ellipse, u =
1 −
−
2 dx a 2 + b 2 a 2 b 2
Ans.
3-10 Air at 20°C and 1 atm flows at an average velocity of 1.7 m/s through a rectangular 1-cm
by 4-cm duct. Estimate the pressure drop in Pa/m by (a) an exact calculation; and by (b) the
hydraulic diameter approximation.
Solution: First establish the hydraulic diameter: Dh = 4A/P =
4[(0.01m)(0.04m)/[2(0.01m+ 0.04m)] = 0.016 m
For air at 20°C and 1 atm, = 1.205 kg/m3 and = 1.81E-5 kg /( m-s ) . Then calculate
ReDh = VDh / = (1.205)(1.7 )( 0.016) / (1.81E-5) = 1810 2000, thus the flow is laminar.
(a) Exact analysis: Eq. (3-48) should apply to this rectangular section. Evaluate the flow:
Q=
4ba 3 dp 192 a tanh ( ib/2a )
− 1 −
3 dx 5 b 1,3,5,...
i5
3
4 ( 0.005 )( 0.02 ) dp 192 ( 0.02 ) 0.3737 0.8269 0.9614
=
−
1
−
+
+
+
...
3 (1.81E-5 ) dx 5 ( 0.005 ) 15
35
55
= 0.0001554 ( −dp/dx )
But Q/A = V, which is known. Dividing Q by the duct area gives the desired result:
Q/A = V = 1.7 m/s =
0.0001554 ( −dp/dx )
dp
= 0.3885 −
( 0.01)( 0.04 )
dx
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-7-
or:
dp
= − 4.38 Pa/m
dx
(Ans. a)
This answer could also be obtained by using Cf Re = 18.2 from Figure 3-13.
(b) Approximate: Simply assume that Cf 16/Re as if the duct were circular:
Cf
16
1
2
= 0.00884, or: avg = Cf V 2 = ( 0.00884 )( 0.5 )(1.205 )(1.7 )
1810
2
= 0.0154 Pa
Then, for fully developed flow, the relation between wall shear and pressure drop is
dp 4avg 4 ( 0.0154 Pa )
=
= 3.85 Pa/m
− =
Dh
0.016 m
dx
(Ans. b)
For this case, the hydraulic-diameter approximation predicts about 12% low.
3-11
Consider swirling motion superimposed upon axial flow and find the velocities.
We assume, for developed flow in a circular pipe, that
vr = 0
v = v ( r, t )
vz = vz ( r )
Substitute these into the complete incompressible equations of motion in cylindrical coordinates,
Appendix B. The continuity equation is satisfied identically, and the z-momentum equation
becomes
vz p
r
=
r r r z
This is independent of the circumferential motion v and thus may be solved directly, leading
simply to Poiseuille flow in a duct, Section 3-3.1.
The -momentum equation reduces to
v v v
=
r
t
r r r
This is uncoupled from axial motion v z and may be solved independently, subject to the swirlflow boundary conditions. Finally, with v known, the pressure may be found from the
r-momentum equation, p/r = v2 /r. Further discussion of superposition of swirl flows is given
in the text by Langlois (1964).
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-8-
3-12 If light oil viscosity (0.02 to 0.1 Pa-s) is measured by passing 1 m3 /h of fluid through an
annulus 30 cm long, with inner and outer radii of 9 mm and 10 mm, find p .
For an annulus, the hydraulic diameter is Dh = 2 ( a − b ) = 2 ( 0.010 − 0.009) = 0.002 m. The
annular area is (a 2 − b2 ) = 5.97E-5 m2 . Then the average velocity is V = Q/A 4.65 m/s.
Estimating the density of light oils to be about 900 kg/m3 , the Reynolds numbers are
ReDh =
VDh ( 900 )( 4.65)( 0.002 )
=
= 80 to 400
0.02 to 0.10
Thus the flow is laminar, and Eq. (3-51) applies:
(
)
2
2 2
a
−
b
p 4 4
Q=
a −b −
8 L
ln ( a/b )
Introducing the given numerical values, we obtain the pressure-drop estimate:
2
2
0.01) − ( 0.009 )
(
1
p
4
4
m3 /h =
( 0.01) − ( 0.009 ) −
3600
8 ( 0.02 to 0.10 )(.3)
ln ( 0.01/0.009 )
or:
p 335,000 to 1,680,000 Pa 3 to 17 atm
2
(Ans.)
For such a large pressure drop, a mechanical (e.g. Bourdon tube) gage might be recommended.
3-13 Lubricating oil at 20°C flows at 2 m/s into a tube, D = 3 cm, L = 2 m, with a wall
temperature of 10°C. Estimate (a) the heat transfer rate at x = 10 cm; and (b) the exit
temperature of the oil. The oil properties are
= 890 kg/m3 = 0.8 Pa-s k = 0.15
W
m-K
c p = 1800
J
kg-K
The Reynolds number is Re = VD/ = (890)( 2.0)( 0.03) / ( 0.8) = 67, hence the flow is laminar.
The Prandtl number is cp /k = ( 0.8)(1800 ) / ( 0.15 ) = 9600 ( oils ) .
(a) At x = 10 cm, the local Graetz number is
x* =
x
0.1
=
= 5.20E-6
D Re Pr ( 0.03)( 67 )( 9600 )
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Hill LLC.
-9-
At this very small value, we estimate the local Nusselt number from Eq. (3-83):
Nu x 1.076 ( 5.20E-6 )
−1/3
− 1.064 66.1 =
qw D
k T
This is close to the entrance, so we assume that T still equals 10°C. Then we calculate
q w |x =10cm = 66.1
( 0.15)(10) = 3060 W/m2
k T
= 66.1
D
( 0.03)
(Ans. a)
(b) At the exit, L* = L/ ( D Re Pr ) = 2.0/ ( 0.03)( 67 )( 9600 ) = 0.000104. This is too small to
make the chart in Fig. 3-15. Therefore we use Eq. (3-93) as an estimate:
Nu m 3.66 +
0.075/ ( 0.000104 )
1 + 0.05/ ( 0.000104 )
2/3
= 34.2 = −
1
*
ln Tm
( L )
4L
*
where the latter equality follows from Eq. (3-90). We may solve for
*
Tm
( L ) = 0.986 =
Tw − Tm 10 − Tm
=
, or: Tm ( L ) = 19.86C
Tw − To 10 − 20
(Ans. b)
The fluid has hardly cooled down at all over a two-meter length! This poor convection rate is
characteristic of oil flow under laminar conditions.
3-14
For plane flow with circular streamlines, solve the vorticity equation,
2 1
= v 2 +
r
t
r r
for the decay of a line vortex of strength o initially concentrated at the origin. The boundary
conditions are symmetry at the origin and no motion at infinity:
( 0, t ) = ( , t ) = 0
r
This is a classic problem in linear PDE theory and has an exact analogy in heat conduction
(a line heat pulse at the origin). The solution is given by Carslaw and Jaeger (1959):
r2 1
o
=
exp −
=
rv
4t r r ( )
4t
where o = V ds is the initial circulation about any closed curve enclosing the origin.
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-10-
The vorticity may be integrated to yield the instantaneous circumferential velocity:
v =
r
r 2
o
1
r
dr
=
1
−
exp
−
r 0
2 r
4t
These distributions were discussed earlier, without derivation, in Prob. 1-6, and both the velocity
and vorticity distributions are plotted on page 3.
3-15 A wide liquid film of thickness h flows due to gravity down the inclined plane shown
below. Find the (laminar) velocity distribution.
Since the boundary conditions are independent of x, we may assume u = u ( y ) , v = w = 0,
p/x = 0, and the momentum equation reduces to
d2u
dy2
= − g sin = constant, or: u =
−g sin 2
y + Ay + B
2
One boundary condition is no-slip at the wall:
u ( 0) = 0 = B
The second condition is hinted at in the problem statement: negligible shear at the free surface
due to weak interaction with the constant-pressure, low density atmosphere:
xy ( y = h ) = 0 =
or:
u
−g sin
|y = h =
( 2h ) + A,
y
2
A = gh sin /
Substituting for A and B, we find the complete solution for laminar developed film flow:
u=
ρgsinθ
y ( 2h − y )
2
(Ans.)
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Hill LLC.
-11-
This may be integrated to find the volume flow per unit width of film:
h
Q = u dy =
0
ρg h 3sinθ
3μ
(Ans.)
The flow rate varies as h3, so the rate of draining is highly dependent upon the film thickness.
Note the similarity of this (parabolic) solution with pressure-driven Poiseuille flow between
parallel plates (2h) apart, Eq. (3-44).
3-16 A laminar film drains at constant thickness down a vertical rod, as shown. Find formulas
for the velocity distribution and the flow rate.
Assume, due to the constant-pressure atmosphere outside the film, that p/z = 0, and, for a fully
developed film, u z = u ( r ) only. The z-momentum equation, Appendix B, reduces to
0 = g +
d du
r
r dr dr
This second-order ordinary differential equation may readily be integrated twice:
u=−
g r 2
+ A ln ( r ) + B
4
The boundary conditions are no-slip at the wall and no shear stress (weak atmospheric
interaction - see the top of page 44):
No-slip at wall:
g a 2
u ( a ) = 0 gives B =
− A ln ( a )
4
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-12-
No surface shear:
du
|y=b = 0 gives B = ρgb2 /2μ
dr
With A and B known, the final expression for the film velocity profile is
g b2
u=
4
2
2
r r a
2ln − +
a b b
(Ans.)
Considerable integration skill (not shown) is needed to evaluate the film flow rate:
b
Q = u ( r ) 2 r dr =
a
g a 4 4
b
4 ln ( ) + 42 − 1 − 34 , = 1
8
a
(Ans.)
For thin films approximating a flat wall, 1.0 b/a 1.2, Q increases (approximately) as the cube
of the film thickness (b-a) - see the solution to Prob. 3-15 above. For b/a 1.2, Q increases even
faster than the cube of the thickness, as film area increases with radius.
3-17 By extension of Prob. 3-15, consider a double layer of immiscible fluids 1 and 2, flowing
steadily down an inclined plane, as in Fig. P3-17. The atmosphere exerts no shear stress on the
surface and is at constant pressure. Find the laminar velocity distribution in the two layers.
Fig. P3-17
Solution: We solve for the velocity distribution in each layer and patch them together at y = h1.
Assuming u1,2 = fcn( y ) only, the layers satisfy similar momentum equations:
x − momentum:
0 = 1 g sin + i
y − momentum:
0=−
d 2ui
dy 2
pi
+ i g cos ,
y
,
i = 1, 2
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-13-
The pressure gradient pi /x = 0 because the surface pressure is constant. Thus the pressure is
hydrostatic in i g cos and is uncoupled from velocity. Forget pressure, find velocity:
u1 = −
1g sin 2
g sin 2
y + C1 y + D1 ; u2 = − 2
y + C2 y + D2
21
2 2
(1a, b)
By inspection, D1 = 0 to satisfy the no-slip condition at the bottom, y = 0. The other three
boundary conditions are negligible shear at the surface (weakly interacting atmosphere) and
matching velocities and shears in each layer at the interface, y = h1.
Surface shear: τ 2 0 = 2
Interface velocity: u1 = −
g sin
du2
| y =h1+h2 = 2 − 2
( h + h ) + 2C2
dy
2 1 2
1g sin 2
g sin 2
h1 + C1h1 = u2 = − 2
h1 + C2h1 + D2
21
2 2
g sin
Interface shear: 1 = 1 − 1
1
2 g sin
h1 + 1C1 = 2 = 2 −
h1 + 2C1
2
Solve simultaneously for the three constants:
g sin
C2 = 2
( h1 + h2 ) ;
2
C1 =
g sin
1
( 1h1 + 2h2 )
h h h h
D2 = h1g sin 1 1 + 2 2 − 2 1 − 2 2
1
2 2
2
21
Substitute these constants into Eqs. (1a, b ) above. (It’s too much typing to repeat here. Au)
To illustrate this flow, an example is shown below. The two layers are of equal thickness,
h1 = h2 = 0.5, and the densities are the same. Only the effect of upper-layer viscosity is shown. If
2 = 1, as in the center curve, the velocity profile is a continuous parabola, as calculated earlier
in Prob. 3-15. If 2 = 51, the lower curve, the outer velocity is less and two parabolas meet with
different slopes at the interface, y1 = 0.5. Similarly, if 2 = 0.21, the upper layer goes faster, as
shown, and again the slopes are different at y1 = 0.5.
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-14-
Two-layer flow down an inclined plane, for the special case 2 = 1 and h 2 = h1 .
3-18 The film surface in Prob. 3-15 is known to become unstable at a critical Reynolds number
Re* = hu max /v = Q (10) , which depends on , g, , , and surface tension T. .
Make a dimensional analysis of the correlation Re* = fcn ( , g, , , T ) , noting the dimensions
of each in the Mass-Length-Time system:
Re* and = 0; g = [L/T 2 ]; [] = [M/L3 ]; = M/LT ; T = [M/T 2 ]
We expect two dimensionless groups (“Pi’s”). One of them obviously is itself. The second
one, assumed proportional to , could be written as a power-group:
2 = ga bT c = M0L0T0 , from which we find a = −1, b = −4, and c = +3.
Dimensional analysis thus predicts that two Pi groups influence the Reynolds number:
T 3
Re* = fcn ,
g 4
(Ans.)
The last parameter, which depends only upon gravity and fluid parameters, is a form of the
Weber number for capillary motions. In his review of plane film motion, Fulford (1964) points
out that Weber number has only a small effect on film instability, so that the simple correlation
Re * 12 csc ( ) is a reasonable approximation for a draining plane film.
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-15-
3-19
Derive Eq. (3-96) for start-up of flow in a circular pipe.
This is a good exercise for the student. The first difficulty is that the basic differential equation
(3-94) contains the inhomogeneity ( −dp/dx ) , which must be removed before separation of
variables can work. To do this, define velocity V such that
(
)
dp r 2
V = u − u max 1 − r *2 , where r = r/ro and u max = − o
dx 4
We thus work with the difference between u and the final Poiseuille-flow profile at t = .
Substitution into Eq. (3-94) gives the following homogeneous PDE for V:
V 2 V 1 V
=
+
, where r* = r/ro and t* = vt/ro2
2
t* r*
r* r*
V (1,t*) = 0
subject to:
(no-slip)
(1)
(2)
2
V ( r*, 0 ) = u max 1 − r*
(3)
The solutions to (1) subject to condition (2) are all of the form V = J o ( λ n r*) exp(− n 2 t*),
where n are the roots of the Bessel function J o (see values in Table 3-3). The general solution
of the linear PDE (1) is thus a Fourier-Bessel series of such terms:
(
V = Cn J o ( n r *) exp − n2 t *
n =1
)
(4)
To evaluate C n , we use condition (3) and the Bessel function orthogonality condition:
1
xJo ( i x ) Jo ( jx ) dx
=
0
and
= 0
1
J1 ( i )
2
if i = j
if i j
We thus multiply Eq. (3) by r* J o ( j r*) and integrate the whole equation from 0 to 1. The
orthogonality condition enables us to pick off the constant one by one, with the result
Cn = −
8 u max
3n J1 ( n )
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-16-
With V thus known, we return to the original variable u and the desired solution:
u
u max
= 1 − r*2 −
8 J o ( n r*)
3
n =1 n J1
( n )
(
exp − 2n t*
)
Some typical profiles of u ( r*, t *) are shown in the text in Fig. 3-16. The profiles for very early
t* are shown at right (as requested in the problem) and are seen to consist of a linearly
accelerating core with thin growing boundary layers.
3-20 Consider air at 20°C and 1 atm initially at rest between fixed parallel plates h = 2 cm
apart. If the lower plate moves at speed Uo = 30 cm/s at time t = 0, evaluate u at the center.
For air at 20°C and 1 atm, v 1.50E-5 m2 /s. The Reynolds number of the flow is
Re = Uo h/v = (0.30)(0.02)/ (1.50E-5) = 400, hence the flow is laminar. The analysis of laminar
start-up between suddenly moved plates is given in Section 3-5.2, with velocity profiles plotted
in Fig. 3-20. If we insert the center point, y = h/2, into this formula, we obtain the velocity
prediction
u/Uo = 0.5 −
2 1
vt
nπ
exp −n 22 t* sin , t* = 2
n =1 n
2
h
(
)
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Hill LLC.
-17-
(a) Compute the midpoint velocity after 2 seconds. The corresponding dimensionless time is
t* = (1.50E-5)( 2.0) / ( 0.02) = 0.075. The formula above predicts
2
u midpoint
Uo
0.1966, or: u midpoint = ( 0.1966 )( 30 ) = 5.9 cm/s
(Ans. a)
(b) At what time will the midpoint velocity be 14 cm/s? This corresponds to u/U o =
14/30 = 0.4667. If we evaluate the formula above, searching for this ratio, we will find that
( 0.299)( 0.02) = 8.0 s
u
= 0.4667 occurs at t* = 0.299, or t =
Uo
(1.50 E-5)
2
(Ans. b)
One may check these results (approximately) by reading the midpoint values in Fig. 3-20.
3-21 Derive simplified formulas for the volume flow and
pumping power of a Couette pump (see figure in problem
statement). Illustrate for SAE 30 oil.
For small clearance,
( a − b ) a,
we assume a linear
velocity distribution, as shown at right. Let the annular
region have depth “L” into the paper. Then the flow area is
A = ( a − b ) L, with average velocity ( V/2) . Thus a
simplified expression for volume flow rate is
1
Q = Vavg A = b ( a − b ) L
2
(Ans. a)
As shown in the figure, the wall shear stress is approximately = V/ ( a − b ) , so the total torque
imparted to the fluid by the inner cylinder is T = ( 2 bL ) b. Thus the power needed to drive
this flow is
P = T = 2b2 L =
2 2 b3L
(a − b)
(Ans. b)
As a numerical example, take SAE-30 oil at 20°C, with a = 10 cm, b = 9 cm, L = 1 m, and a
rotation rate of 900 rpm. From Appendix A, Fig. A-2, read = 0.29 kg/m-s for SAE-30 oil. The
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-18-
angular velocity is w = ( 900 rev/min )( 2π rad/rev ) / ( 60 s/min ) = 62.83 rad/s. From formula (a)
above, we compute a volume flow rate
1
Q b ( a − b ) L = ( 0.5)( 62.83)( 0.09 )( 0.10 − 0.09 )(1.0 )
2
= 0.0283 m3 /s
(100 m3 /hr )
(Ans.)
From formula (b) above, we compute the power required:
2 ( 0.29 )( 62.83) ( 0.09 ) (1.0 )
= 520 W
( 0.10 − 0.09 )
2
P=
3
(Ans.)
Although it’s a lot of power for a small flow rate, the values seem reasonable for heavy oil.
3-22
Determine if the “Taylor vortex’’ is a solution to the Navier-Stokes equations:
v = ( const )
r2
exp
−
t 2
t
r
The vorticity is computed as in the “Oseen vortex” result, Prob. 3-14, page 34:
=
r2
1 v 2 ( const ) r 2
r
=
1
−
exp
−
r r r
t 2 t
t
By direct substitution into the -momentum equation (Appendix B) and the vorticity-decay
equation (top of page 43), we may verify after considerable algebra that the Taylor vortex is an
exact solution. The velocity and vorticity profile shapes are shown on the next page. By
comparison to the Oseen vortex profiles on page 3, the Taylor profiles are much flatter, less
peaked, and decay by dropping quickly near the axis with a reversal in vorticity away from the
axis.
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-19-
3-23
Consider purely radial outflow between parallel disks, as shown below. Assume
u r = fcn ( r, z ) with constant ( , ) and p = p ( r ) only. Solve as far as possible.
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Hill LLC.
-20-
The continuity equation (Appendix Eq.B-3) reduces to
( ru r ) = 0, or: ru r = f ( z )
r
Substitution of u r = f /r into the r-momentum relation (Appendix Eq.B-6) gives
u r
or:
1 u r
u r
p
= − +
r
r
r
r r r
−
f 2
r3
=−
2
ur ur
−
+
,
2
z 2
r
dp d 2f
+
dr r dz 2
(1)
to be solved for f ( z ) . (The first two terms in the brackets [] on the right cancel for the particular
distribution u r = f /r.) At any given section r, Eq. (1) is a nonlinear ordinary differential equation
which can be solved for f ( z ) , subject to two boundary conditions:
f = 0 at z = L;
No-slip:
Symmetry:
df
= 0 at z = 0
dz
The parameters involved in Eq. (1) are made clearer if we integrate the equation in the radial
direction between r = r1 and r = r2 [see page 114 of Bird et al. (1960)]. The result is
r2 d 2f
f 2 1 1
−
=
p
+
ln
2
2 r22 r12
r1 dz
(2)
where p = p1 − p2 . For “creeping” flow, the inertia term on the left hand side is negligible, and
the solution can be found immediately [Bird et al. (1960), page 114]:
Very low Reynolds number: f = ru f
p L2 z 2
1 −
2 ln ( r2 /r1 ) L2
(3)
This parabolic distribution is the exact analogy to Poiseuille flow in a channel. At higher Reynolds
numbers, Eq. (2) is nonlinear, and we may nondimensionalize it using Eq. (3) as a guide:
d 2
dz *2
= −K 2 − 2, where =
and
K=
(
pL2f
z
, z* = ,
2 ln ( r2 /r1 )
L
)
r12 − r22 p L4
4 2 ln 2 ( r2 /r1 )
(5)
The single parameter K is proportional to the Reynolds number of the flow. The boundary
conditions become ( 1) = 0 and d/dz*( 0 ) = 0. Equation (5) may be solved numerically by,
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-21-
e.g., the Runge-Kutta method. For a given K, the problem is to find the proper value of f ( 0)
which causes f ( l ) to be zero. The following solutions were found by the writer:
K=
f ( 0) =
0
0.2
0.4
0.6
0.7
1.0
1.0861
1.2135
1.4575
1.8569
Interesting, no solutions could be found for K 0.75 (approximately) - the reader is invited to
investigate this difficulty and explain it to the writer. The five tabulated solutions for velocity
profile (z*) are plotted below - all of them look approximately parabolic.
3-24
A porous cylinder of radius R exudes fluid radially at velocity U o into an unbounded
fluid of constant ( , ) . Find the velocity and pressure distributions if p = po at r = R.
Let u r = u ( r ) only, with u = u z = 0. The continuity equation is easily solved:
1d
R
( r u ) = 0, or: r u = constant, or: u = Uo
r dr
r
(Ans. a)
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Hill LLC.
-22-
This is the desired velocity distribution. The pressure follows from the r-momentum equation
(Appendix Eq. B-6) if we neglect gravity:
u
du
dp
d 1 d
= − +
( ru )
dr
dr
dr r dr
Introducing u ( r ) from solution (a) above, we find that the viscous term vanishes identically (a
consequence of u being proportional to 1/r and therefore a “potential” type of distribution) and
we obtain the result
Uo2R 2
dp Uo2 R 2
=
,
or:
p
=
−
+ constant
dr
r3
2 r2
The constant is evaluated from the condition p = po at r = R. The final result is
R 2
1
p = po + Uo2 1 −
2
r
(Ans. b)
For this particular case, the pressure relation is equivalent to Bernoulli’s equation.
3-25
A cylinder of radius ro rotates at surface vorticity o with a wall suction velocity of
( −vw ) at the surface. Find the vorticity distribution assuming that / = 0 and vz = 0.
With pressure and velocities dependent only upon r, the continuity equation (B-3) becomes
1
( rvr ) = 0,
r r
or: vr = −
ro v w
r
since vr |r =ro = −vw
With the radial velocity thus known, the θ-momentum equation,
vr
dv v r v v d dv v
+
= r
−
dr
r
r dr dr r
can be solved for tangential velocity. The general solution is
v = A r −1 + Br1−Re ,
where Re = v w ro / is the wall Reynolds number
The first term is the ‘potential’ vortex of a solid cylinder. The second term is a combined effect
of viscosity and wall suction and is unbounded if Re 1, which is unrealistic. Further, the flow
circulation, = 2 r v , is proportional to r 2-Re and is unbounded if Re 2, also unrealistic.
We conclude that, for physical realism.
B=0
if
Re 2
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-23-
The vorticity in the flow is obtained by differentiation:
=
1d
( r v ) = B ( 2 − Re ) r − Re
r dr
Utilizing the boundary condition for surface vorticity, = o at r = ro , we obtain the constant
B = o roRe / ( 2 − Re ) if Re 2. The final desired solution for vorticity is thus
r
= o o
r
Re
(Ans.)
With B known, the constant A follows from the condition v = roo at r = ro . The final solution
for velocity is in two parts, depending upon the wall Reynolds number:
v =
o ro2
Re-2
1 − Re + ( ro /r )
r ( 2 − Re )
=
3-26
o ro2 /r
if
Re 2
if
Re 2
For laminar flow in a porous tube, similar to the channel example in Section 3-6.3, use
the similarity variable = ( r/ro )
2
and find the proper stream function and its differential
equation and boundary conditions.
For uniform wall normal velocity, the stream function must have the form = ( a + bx ) g ( ) ,
and the constants are found by volume-flow balance between 0 and x:
1
= ro2 u ( 0 ) − v w ro x g ( )
2
(Ans.)
The velocity components are obtained by differentiation:
vr = −
vx =
1 v w g
=
r x
2v x dg
1
= u ( 0) − w
r r
ro d
If these velocities are substituted into the r- and z-momentum relations, Appendix Eqns. (B-6)
and (B-8), it is found that the pressure satisfies the relation
2p
=0
x
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Hill LLC.
-24-
exactly analogous to the porous channel in Section 3-6.3. Cross-differentiation to eliminate the
pressure thus gives the following differential equation for the stream function:
2 g + 4g+ Re ( gg − g g) , where Re = vw ro /
The boundary conditions at r = 0 are vr = vz /r = 0, which translates to
As − 0:
g
= 0 and g = 0
The boundary conditions to the wall, r = ro , are v r = v w and v z = 0, or:
g (l) = 1
and
g ( l ) = 0
This problem was first developed by Yuan and Finkelstein (1956), and a complete array of
numerical solutions - including vexing non-uniqueness and non-existence difficulties - is
presented by White (1962).
3-27 Modify the Ekman spiral solution for turbulent flow by using an “eddy viscosity”
correlation suggested by Clauser (1956):
veddy 0.04 D ( o /)
1/2
From Section 3-7.2, use the Ekman solution for surface velocity and penetration depth:
Vo =
o /
( 2v sin)
(1)
D = v/ ( ω sin )
(2)
For our particular case, Vwind = 6 m/s, air = 1.205 kg/m3 , and latitude 41°N, and = 2/
(86400 s ) = 7.27E-5 s−1. We have the surface-wind-shear estimate, Eq. (3-141):
2
o 0.002 air Vwind
= ( 0.002)(1.205)( 6.0) = 0.0868 Pa
2
Dividing this by seawater density gives the so-called “friction velocity”, v*;
v* =
( o /) = ( 0.0868/1025)1/2 = 0.00920 m/s
Substituting Clauser’s v into Eq. (2) above gives the penetration depth directly:
D = 2
( 0.04)( 0.00920) = 76 m
0.04 v*
= 2
ω sin
( 7.27E-5)(sin41°)
(Ans. a)
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Hill LLC.
-25-
We may then compute the eddy viscosity as v 0.028 m 2 /s from Clauser’s formula. The
surface seawater velocity may finally be computed from Eq. (1) above:
2
0.0092 )
(
Vo =
1/2
( 2 )( 7.27E-5 ) sin ( 41 )( 0.028 )
= 0.052 m/s
(Ans. b)
These are realistic results: surface velocity a few centimeters per second and penetration depth of
the order of 100 meters (varying with the wind velocity).
3-28
Repeat the Ekman-flow analysis of Section 3-7.2 for shallow water of depth h.
We are to solve the same differential equation with one changed boundary condition:
v w − 2i sin w = 0,
subject to:
w ( −h ) = 0,
w = u + iv, i =
( −1)
w ( 0) = i K, K = o /
The (linear) differential equation has a hyperbolic-function general solution:
w = Acosh ( bz ) + Bsinh ( bz ) , b = 2isin /v
1/2
= (1 + i ) sin /v
1/2
From the boundary conditions, we find that A = Btanh ( bh ) , B = iK/b. The desired solution for
(general) velocity and surface velocity is thus
w=
iK
iK
[tanh(bh) cosh(bz) + sinh(bz)], and w(0) =
tanh(bh) = u o + i vo
b
b
We use the identity tanh(x + iy) = [sinh(2x) + isin(2y)]/[cosh(2x) + cos(2y)] to untangle real and
imaginary parts of the surface velocity:
uo =
1
1
K
sinh 2h − sin 2h , vo = [sinh2βh + sin2βh], Q =
Q
Q
2 cosh 2h + cos 2h
where = /D. The ratio of these two is the tangent of the desired surface flow angle:
tan(θ) = u o /vo =
sinh(2πh/D) − sin(2πh/D)
sinh(2πh/D) + sin( 2πh/D)
(Ans.)
A plot of versus h/D is shown on the next page. The angle rises slowly from zero through an
interesting overshoot of 47° at h/D = 0.63 to level off at 45° in “deep” water. From the figure (or
the formula) we may estimate that = 20 at h/D = 0.24
(Ans.).
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-26-
3-29 Air at 20°C and 1 atm flows at U = 1 m/s across a cylinder of diameter 2 cm with wall
temperature of 40°C. Estimate (a) ; (b) w ; and (c) q w at the stagnation point.
Use the properties of air at an average temperature of 30°C. From Appendix A, the kinematic
viscosity is approximately = 1.60E-5 m2 /s. We can estimate B from inviscid theory, Eq. (1-3):
at the surface, U = 2Usin ( x/R ) , hence dU/dx at x = 0 equals B = 2U /R = 2 (1 m/s ) /
( 0.01 m ) = 200 s−1. From the analysis of Section 3-8.1.1.
(a) the shear layer thickness at the stagnation point is
1/2
1.60E-5 m2s
= 2.4 ( /B) = 2.4
−1
200s
= 0.00068 m = 0.68 mm
(Ans. a)
(b) From Eq. (3-159), wall shear is proportional to x, hence at the stagnation point, x = 0,
τw = 0
(Ans. b)
(c) for the heat transfer, estimate, at 30°C, from Appendix A, that k 0.0263 W/ ( m-K ) . The
Prandtl number is 0.71. From Section 3-8.1.4, the heat transfer rate is
q w −k ( T − Tw ) G ( Pr )
( B/ )
= − ( 0.0263)( 20 − 40 ) [0.570 ( 0.71)
0.4
1/2
200
]
1.60E-5
920 W/m2
(Ans. c)
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-27-
3-30
Solve Bodewadt’s “rotating disk” problem, with a fixed disk and a rotating stream.
This problem is very similar to the Kármán rotating-disk problem, with the same variables
( F,G, H, P ) as in Section 3-8.2 as functions of z*. There is one important modification (what
some students might call a “trick”): in order for the radial momentum equation to balance at
z* = with a rotating stream, there must be a positive nonzero radial pressure gradient. In
dimensionless form, this radial equation becomes
F = − G2 + F2 + FH + 1
This guarantees that, in the farfield, where G ( ) = 1, a rotating stream, the derivatives F' and
F will vanish and the radial and axial motion will decay to zero. The other three differential
equations are just the same as in Section 3-8.2:
H = −2F; G = 2FG + HG; P = 2FH − 2F
The boundary conditions change on G only, to make the wall fixed and the stream rotating:
F ( 0) = P ( 0) = H ( 0) = G ( 0) = 0;
F ( ) = 0; G ( ) = 1
This problem was first solved by hand by U. T. Bodewadt (Z. angew. Math. Mech., vol. 20,
1940, p. 241). Accurate numerical solutions were later given by Rogers and Lance (1960), who
also considered cases where both the disk and the freestream rotated (they concluded that no-slip
solutions were possible only if both rotated in the same direction).
Similar to Section 3-8.2, one must find the proper initial conditions F ( 0) and G ( 0 ) which
make the radial velocity F vanish and the tangential velocity G approach unity at infinity. After
considerable searching, we find the correct conditions:
F ( 0) = −0.94197
G ( 0) = +0.77289
(Ans.)
The velocity and pressure profiles are shown plotted below - they oscillate slowly before
decaying to their freestream values. [We used the Runge-Kutta routine in Appendix C.]
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-28-
3-31 Evaluate Kármán’s problem for a numerical case: one side of a disk rotating at 1200 rpm
in air at 20°C and 1 atm. Estimate (a) flow rate; (b) torque; and (c) power.
For air at 20°C and 1 atm, = 1.205 kg/m3 , = 1.81E-5 kg/ ( m-s ) , and = 1.50E-5 m2 /s. The
angular velocity is = (1200)( 2/60) = 40 = 125.7 rad/s.
(a) One way (the hard way) to evaluate flow rate is to integrate the radial velocity at the edge of
the disk, r = R:
Q = v r ( r = R ) 2 R dz = 2 R
2
( v) Fdz*
0
0
However, the last integral, which equals 0.441, is not shown or tabulated in the text. More
straightforward, the flow rate should also equal the fluid ‘pumped’ axially toward the disk:
Q = R 2 | vz ( ) |, where | vz ( ) | = 0.883
or:
(
)(
( ),
)
1/2
2
Q = ( 0.25 m ) ( 0.883) 1.5E-5 m2 /s 125.7 s−1
= 0.0075 m3 /s (Ans. a)
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-29-
(b) The torque on one side of the disk is given by Eq. (3-190):
M=
=
R 4Go
2
( 3 )
1/2
(1.205)( 0.25)4 ( 0.61592) (1.50E-5)(125.7 )3 = 0.025 N-m
2
(Ans. b)
(c) the power required to drive one side of the disk is
P = M = ( 0.025)(125.7 ) = 3.1 W
(Ans. c)
It should be pointed out that these numbers slightly exceed the expectations for laminar flow,
that is, the Reynolds number Re = R 2 /v is about 524,000.
3-32
Solve Jeffery-Hamel wedge flow for creeping flow, Re = 0 but 0 .
Equation (3-195) in Section 3-8.3 reduces for Re = 0 to: f + 4 2 f = 0. This is a third-order
linear differential equation which has the general solution
f = A1 sin ( 2) + A2 cos ( 2) + A3
Using the boundary conditions, f ( 0 ) = 1 and f ( l ) = 0, we find the constants:
A1 = 0; A2 = 1/ (1 − cos2 ) ; A3 = 1 − A2
Thus the desired solution for creeping Jeffery-Hamel wedge flow is
f=
cos2αη − cos2α
1
= 1 + csc2α sin − 2 − 1
1 − cos2α
2
2
(Ans.)
Some representative velocity profiles are plotted below. The case ( = 0) is the Poiseuille
parabola for channel flow, and the case ( = 90) is the separation point. For 90, separation
or backflow must occur in a diverging flow even at zero Reynolds number.
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-30-
3-33 Consider the following exact incompressible Navier-Stokes stream function in spherical
polar coordinates. Plot some streamlines and velocity profiles.
2v r sin 2θ
( r, ) =
,
1 + a − cosθ
a = constant
(1)
The radial velocity is computed from the stream function:
2v 2cosθ (1 + a − cosθ ) − sin 2θ
ur = 2
=
2
r sinθ
r (1 + a − cosθ )
/
(2)
This flow represents a laminar round jet issuing from the origin. The streamlines (Eq. 1) and jet
velocity profiles (Eq. 2 for constant r and unit centerline magnitude) are shown below for
a = 0.01. The jet becomes progressively thinner as a decreases [not shown].
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-31-
The jet width is defined as the point where u r = 0.01 u max . From Eq. 2 we compute that the
maximum velocity ( = 0 ) equals 2v/ ( ra ) . Then the jet velocity ratio is
ur
u max
a 2 cosθ (1 + a − cosθ ) − s in 2θ
= 0.01 at =
=
( )
2
2 (1 + a − cosθ )
That is, the jet “thickness” is a constant value of , or varies linearly with r. Some computed
values of the “1%-velocity” angle corresponding to jet thickness are as follows:
a
= 0.001 0.003 0.006 0.010
( ) =
7.5°
12.8°
17.6° 21.8°
The jet mass flow is found by integration over the jet profile at a given value of r:
( )
m=
u r dA =
jet
0
f ( , a )
2 r ( r d ) = ( const )( r )
r
( )
f ( , a ) d = ( const ) r
0
Thus the (laminar) mass flow increases linearly with r along the jet axis due to “entrainment”
from the ambient fluid outside the jet. [See Eq. (4-204) for details.]
3-34 A sphere of specific gravity 7.8 is dropped in oil of specific gravity 0.88 and viscosity
= 0.15 Pa-s. Estimate its terminal velocity for D = (a) 0.1 mm. (b) 1.0 mm, and (c) 10.0 mm.
At terminal velocity, the sphere’s net weight in oil equals its drag:
Wnet = sph − g D3 = Drag = CD V 2 D2
6
2
4
(
)
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Hill LLC.
-32-
or:
(
)
4 D g sph / − 1
V=
3CD
1/2
, where CD = fcn ( Re ) , Re =
VD
For creeping (Stokes) motion, we have CD = 24/Re, or V = W/ (3D ) , which could serve as
a first estimate. At higher Reynolds numbers, we could use Eq. (3-225):
C D ( sphere ) =
24
6
+
+ 0.4
Re 1 + Re
(a) D = 0.1 mm: The smallest sphere is most likely to be Stokes flow. Then estimate
( 7.8 − 0.88)( 998) ( 9.81)( /6 )( 0.0001)
W
V=
=
3 D
3 ( 0.0001)( 0.15 )
3
= 0.00025 m /s
(Ans. a)
This answer is accurate, as we may verify by checking the Reynolds number:
Re =
VD ( 0.88)( 998) ( 0.00025)( 0.0001)
=
= 0.00015 1
0.15
(Stokes flow)
(b) D = 1.0 mm: Use the Stokes-flow estimate V = W/ ( 3D ) to get a first guess of
V 0.025 m/s. Cheek the Reynolds number, Re = VD/ = 0.15, which is 1:
V 0.25 m/s
(Ans. b)
(c) D = 10 mm: Here the Stokes estimate is V = 2.51 m/s, with Re = 147, which is not small.
Therefore re-compute CD = 1.02 and re-estimate V from our formula above:
(
)
4Dg sph / − 1
V
3 CD
1/2
4 ( 0.01)( 9.81)( 7.8/0.88 − 1)
=
3 (1.02 )
1/2
= 1.00 m/s
This is a big change, so we need to iterate the formulas until they converge to
Re = 46, CD = 1.70,
3-35
V = 0.78 m /s
(Ans. c)
Verify Eq. (3-230), for the Stokes drag of a droplet, with a complete analysis.
We are to solve the Stokes stream function differential equation,
2
2 1 2 cot
2 + 2 2 − 2
= 0
r
r
r
(1)
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Hill LLC.
-33-
for ( r, ) subject to zero radial velocity, equal tangential velocity, and equal shear stress at the
droplet surface, r = a. By analogy with the sphere solution in Section 3-9.2, the stream function
yields to separation of variables, = Usin 2 f (r), and substitution into Eq. (1) above gives a
fourth-order linear “equidimensional” equation:
f −
4
r
2
f +
8
r
3
f−
8
r4
f =0 ,
with the general solution
f=
A
+ Br + Cr 2 + Dr 4
r
(2)
This result is valid for both inside and outside flows. Let subscripts “i” and “o” denote the inside
and outside flows, respective. The eight constants ( A, B, C, D )i,o are found by laborious
systematic application of the boundary conditions.
First, at large r, the outside flow must approach the uniform stream, o = (1/2 ) Ur 2 sin 2 , which
requires that
Co = 1/2
Do = 0
Second, the inside flow must have a finite velocity at the origin, r = 0, which requires that
Ai = Bi = 0
(
)
2
Third, u r = ( / ) / r sinθ must equal zero at r = a on both sides. This requires
Outside:
1
Ao = −Boa 2 − a 3 ;
2
Inside: Ci = −Dia 2
This leaves only two free constants, B o and Di , which are found by (a) equating the tangential
velocities, ( −/r ) / ( r sinθ ) , at the interface:
3
u o ( r = a ) = u i ( r = a ) , or: Bo = Dia 3 + a
4
Finally, equate the shear stresses at the interface:
o
u o
u
a ( 2o + 3i )
( r = a ) = i i ( r − a ) , or: Bo =
r
r
4 ( o + i )
(Ans.)
This is the desired solution for the inner and outer stream functions. The problem is completed
by evaluating the drag from the integrals in Eq. (2-219), with the result
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-34-
F = 8o U Bo = 6 a o U
1 + 2o /3i
1 + o /i
which was the liquid-droplet Stokes drag result to be proved.
3-36
Repeat Prob. 3-23, for radial outflow between parallel disks, for creeping flow.
Referring back to Prob. 3-23, continuity requires that ru r = fcn ( z ) only, and, if inertia or
convective acceleration is neglected (creeping flow), the r-momentum equation reduces to
dp u d 2
constant
=
r
u
=
(
)
r
dr r dz 2
r
where the latter equality follows from separation of variables. This may be integrated once with
respect to r and twice with respect to z, giving the velocity as a function of pressure:
ur =
L2 ( p1 − p 2 )
1 − z 2 /L2
2 r ln ( r2 /r1 )
(Ans.)
for any two points “1” and “2” between the disks. Thus, for creeping motion in this ‘slowly
varying’ duct, by analogy with lubrication theory, the local velocity profile is everywhere
parabolic, and the velocity level varies inversely with r to satisfy continuity.
The above result may be rewritten to show the logarithmic variation of pressure:
p = p1 −
2
2
L
r
r1
( ru max )1 ln
(Ans.)
The pressure drops logarithmically in the radial direction, at a rate proportional to the “level” of
flow ( ru max ) , i.e., the flow rate entering from the center of the disks.
3-37
Analyze creeping flow between parallel disks. The lower disk ( z = 0) is fixed, and the
upper disk ( z = L ) rotates at angular rate . Find the velocity and pressure distributions.
This is, in principle, a three-dimensional flow but with radial symmetry, i.e. / = 0. The
boundary conditions on velocity are
At z = 0: u r = u = u z = 0;
at z = L: u r = u z = 0, u = r
With inertia (acceleration) negligible in creeping flow, the -momentum equation (Eq. B-7 of
Appendix B) reduces, for p/ = 0 and zero gravity, to
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Hill LLC.
-35-
2 v =
1 v 2 v v
r
+
= 2
r r r 2
r
z
The hint u = r f (z) is a good one and cracks the puzzle, reducing this equation to f = 0, or
f (z) = A + Bz. The no-slip boundary conditions above yield A = 0 and B = /L. The desired
solution for creeping circumferential flow between the disks is thus
v = r
z
L
(Ans.)
The flow simply rotates in linear fashion, like Couette flow between plates. Since this particular
tangential velocity does not influence continuity, r-momentum, or z-momentum, the remainder
of the flow reduces, with no-slip between disks, to the trivial results
vr = vz = 0;
p = constant
(Ans.)
For creeping flow, no secondary motion is generated. As Reynolds number increases,
acceleration causes nonzero v r and v z to arise, like the Kármán rotating-disk solution.
3-38
Set-up the method of separation of variables for finding u ( y, z ) for a rectangular duct,
Fig. 3-9 and Eq. (3-48), by analyzing Eq. (3-30). First note that the separation will not work until
one defines a new variable U = u − F , where 2 F = (1/ )( dp /dx ) , so that 2U = 0. Then
separate U into z and y parts and find the form of each part. Show how an infinite series would
be required to satisfy the boundary conditions, but do not determine the series coefficients.
Solution: First find the function F. From the figure at right,
(
)
2
2
a suitable guess is F = K y − a , whence 2 F = 2 K =
(1/ )( dp/dx ) . If U = u − F , then 2U = 0.
Assume that U can be written as a separation of variables:
U ( y, z ) = Y ( y ) Z ( z ) , whence 2U = Y Z + Y Z = 0, or:
Y
Z
= − = constant = 0
Y
Z
We chose 0 so that Y would be trigonometric and Z would be exponential. Thus
U ( y, z ) = A cos ( y) + B sin( y)C cosh( z) + D sinh( z)
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-36-
The boundary conditions are no-slip, u = U + F = 0 at the duct walls, and symmetry, zero spatial
slopes along the center lines of the rectangle:
U ( a, z ) = 0;
(
)
U
U
( 0, z ) = 0; U ( y, b ) = − F ( y, b ) = − K y 2 − a 2 ;
( y, 0 ) = 0
y
z
The second of these conditions requires that B 0, the fourth requires that D 0. Now absorb C
into A and we are down to
U = A cos ( y ) cosh ( z ) .
Satisfying the first boundary condition above requires that cos ( a ) = 0, or:
a=
2
,
3 5
,
, ...
2
2
or
=
i
,
2a
i = 1,3,5, 7, ......
The third (and final) boundary condition requires that U equal a quadratic function along the
upper walls of the rectangle:
(
)
U ( y, b ) = − K y 2 − a 2 = A cos ( y ) cosh ( b )
?
This looks impossible, but since Laplace’s equation is linear, we may take U to be the sum of an
infinite series of cosine terms:
U ( y, z ) =
i
Ai cos
2a
i =1,3,5...
i
y cosh z
2a
Then we can find the Fourier coefficients Ai such that the above boundary condition for
U ( y, b ) is satisfied. We asked you not to carry this part out. The complete solution procedure is
given in the classic text by Rosenhead (1963), Laminar Boundary Layers, and we put the final
solution for u ( y, z ) in the text as Eq. (3-48).
3-39 Find the pressure distribution for lubrication flow [Section 3-9.8.1] in a parabolically
varying gap:
x2
h ( x ) = hL + ( h0 − hL ) 1 − 2
L
The relevant relation to be solved is Reynolds’ equation (3-23g). Using dimensionless variables
( )
x* = x/L, h* = h/h 0 , and p* = ph 02 / ( UL ) , Eq. (3-239) becomes
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-37-
d 3 dp*
dh *
,
h*
=6
dx *
dx *
dx *
with h* =
hL hL
2
+ 1 −
(1 − x *)
h0 h0
If we let “ p ” be zero for convenience, the boundary conditions on pressure are:
p* = 0 at x* = 0 and at x* = 1
Because h* is cubed and thus leads to the sixth power of x*, it is very laborious to solve for p *
analytically. We therefore program the differential equation above for a computer, using
Subroutine RUNGE from Appendix C. Some numerical values of slope are:
h L /h 0 =
dp*/dx *(0) =
0.9
0.7
0.5
0.4
0.3
–0.417 –1.357 –2.459 –3.078 –3.743
The pressure profiles from these results are plotted below
For this type of gap - with a steep change initially, decreasing to a shallow change at the end the pressure drops in the gap and there is a strong possibility of cavitation for this direction of
flow.
3-40 Set up Stokes’ first problem of the suddenly moved wall (Fig. 3-18a) on a digital
computer by the explicit finite-difference method of Eq. (3-247).
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Hill LLC.
-38-
Let us use dimensionless variables, y* = y/L, u* = u/U, and t* = vt/L2 , so the basic differential
equation and its explicit model are
(
)
t *
u* 2 u*
j
j
j+1
j
u
=
u
+
u
+
1
−
2
u
,
=
=
,
with
model:
(
)
n
n
n
+
1
n
−
1
t* y*2
( y*)2
(1)
The initial conditions at t = 0 are u1 = 1, u n 1 = 0 everywhere. The boundary conditions are
u1 = 1 and u n = = 0. For stability we must have 0.5. For convenience take y* = 1.0 and let
t* = 0.25, that is, = 0.25. Some numerical results are shown below:
t*
0.000
0.250
0.500
0.750
1.000
1.250
1.500
1.750
2.000
2.250
2.500
2.750
3.000
3.250
3.500
3.750
4.000
u1
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
u2
0.000
0.250
0.375
0.453
0.508
0.549
0.581
0.607
0.629
0.648
0.664
0.678
0.690
0.701
0.711
0.720
0.728
u3
0.000
0.000
0.063
0.125
0.180
0.227
0.267
0.302
0.332
0.359
0.383
0.405
0.424
0.442
0.458
0.473
0.487
u4
0.000
0.000
0.000
0.016
0.039
0.065
0.092
0.118
0.143
0.167
0.189
0.210
0.230
0.248
0.265
0.281
0.296
u5
0.000
0.000
0.000
0.000
0.004
0.012
0.022
0.035
0.049
0.064
0.078
0.093
0.108
0.122
0.136
0.150
0.163
u6
0.000
0.000
0.000
0.000
0.000
0.001
0.003
0.007
0.013
0.019
0.027
0.035
0.043
0.052
0.061
0.071
0.080
u7
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.001
0.002
0.004
0.007
0.011
0.015
0.019
0.024
0.029
0.035
u8
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.001
0.001
0.003
0.004
0.006
0.008
0.011
0.014
u9
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.001
0.002
0.002
0.003
0.005
u10
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.001
0.001
0.001
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Hill LLC.
-39-
When plotted versus = y*/ t*, we get a nearly unique error-function curve (Eq.3-108):
3-41 Repeat Prob. 3-40 using the implicit method, Eq. (3-249). We follow the ideas of
Prob. 3-40 except that we use the implicit model:
−u nj++11 + (1 + 2) u nj+1 − u nj+−11 = u nj , = t*/ ( y*)
2
There is no limitation on , but now we have a set of tri-diagonal simultaneous equations to
solve. Again let y* = 1.0 for convenience. We may use any time step without instability. Let us
choose t* = 1.0 (four times as large as Prob. 3-40), or = 1.0. The computed results are
tabulated below:
t*
0.000
1.000
2.000
3.000
4.000
u1
1.000
1.000
1.000
1.000
1.000
u2
0.000
0.382
0.553
0.642
0.696
u3
0.000
0.146
0.276
0.374
0.445
u4
0.000
0.056
0.130
0.203
0.267
u5
0.000
0.021
0.059
0.105
0.151
u6
0.000
0.008
0.026
0.052
0.082
u7
0.000
0.003
0.011
0.025
0.043
u8
0.000
0.001
0.005
0.012
0.022
u9
0.000
0.000
0.002
0.005
0.011
u10
0.000
0.000
0.001
0.002
0.005
Compared to the exact (error-function) solution, the accuracy is about 4%. We obtained
1.5% accuracy in Prob. 3-40 (see table on page 71) but used a time step 4 times smaller.
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-40-
3-42 Set up Stokes’ second problem, the steadily oscillating wall, Fig. 3-18b. Use the explicit
finite-difference model. We can follow the same procedure as Prob. 3-40, with variables
y* = y/L, u* = u/U, and t* = vt/L2 . For the wall velocity oscillation, take
u* ( 0, t*) = cos ( ω*t*) , where * = L2 /v
(1)
For simplicity, take y* = 0.5 (to get the needed spatial definition) and * = 1.0. To keep
less than 0.5, use the value = 0.25 from Prob. 3-40, which requires that t* = 0.0625. Run the
same explicit model as in Prob. 3-40:
(
)
u nj+1 = u nj +1 + u nj −1 + (1 − 2 ) u nj , = 0.25, u1 = cos ( t*)
From the print-out (not shown), we have plotted some velocity profiles on the next page (p. 60).
Initial transient effects do not die out until t* 10, so we have only plotted ‘steady-state’ profiles for
every unit value of t* = 10 to 20. Note the striking similarity with the exact profiles in Fig. 3-18b.
The numerical accuracy is about ±1% for these particular step sizes y* = 0.5 and t* = 0.0625.
3-43
Repeat Prob. 3-42 (Stokes’ 2nd problem) using an implicit method.
By analogy with Prob. 3-41, by using the implicit model (see top of page •••) we can choose,
e.g., a step size four times larger, that is, t* = 0.25, with y* = 0.5 and = 1.0, and of course
let u*( 0,t*) = cos ( t*) . We get good accuracy (about ±4%) over the entire cycle of varying wall
motion. The plot of instantaneous profiles would look almost exactly as above, so we will not
repeat those results.
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Hill LLC.
-41-
3-44
The solutions f n ( r*) to Eq. (3-76) are called Graetz functions, but they are not tabulated.
Set up a numerical solution of Eq. (3-76), perhaps using the Runge-Kutta subroutine of
Appendix C, and solve iteratively for the first three functions f1-2-3 and their eigenvalues 1-2-3.
Compare with Table 3-1, but no fair using that table for your initial guesses for eigenvalues.
Solution: The basic differential equation (3-76) should be solved for the highest derivative:
f n = −
(
)
1
f n − n2 1 − r 2 f n
r
With initial conditions fn ( 0) = 1 and fn (1) = 0. Because r = 1 is the centerline, we also know
from symmetry that f n ( 0) = 0, and, further, the value f n ( r ) /r |r =0 = 0 also. Thus we can start
at r = 0 and integrate numerically outward, for a small r, varying n until f n (1) = 0. We can
use Subroutine RUNGE or any effective numerical method. The writer used an Excel
spreadsheet. Only certain discrete eigenvalues n work, and we won’t cheat by selecting them
from Table 3-1 of the text! For example, for the first eigenvalue, we can guess 0 = 2.0, 2.5 and
3.0 and compute f1 (1) = 0.368, 0.104, − 0.141. The correct value of λ0 , for which f1 (1) 0,
lies between 2.5 and 3.0, and we can interpolate to find guessed λ0 2.7 or, even more closely,
2.704, just like Table 3-1. In a similar manner, we can home in on λ1 6.68 and λ 2 10.67.
The three eigenfunctions are shown in the chart below. They look like Bessel functions, but no,
they are Graetz functions.
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-42-
3-45 Extend Prob. 1-21, where we only found the Knudsen number, Kn 0.17, by using all
the data given there. (a) Find the Reynolds number and see if it is less than 2000 (laminar flow).
(b) Estimate the required pressure gradient in Pa/m. (c) Estimate the flow rate in mm3 /s.
Solution: From Prob. 1-21, for oxygen at 1200 Pa and 293 K, = 0.0158 kg/m3 and
= 2.0E-5 kg/m-s. The average velocity is 10 cm/s, and D = 35 μm. Thus the Reynolds number
is small and definitely laminar:
(
)
3
VD 0.0158 kg /m ( 0.1m/s )( 0.000053m )
=
= 0.0028
(a) Re D =
2.0E -5 kg / ( m s )
Ans. (a)
With velocity and diameter known, the volume flow rate follows immediately:
Q = AV =
4
( 0.000035m )2 ( 0.1m/s ) = 9.6E − 11m3 /s = 0.096
mm3
s
Ans. (c)
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Hill LLC.
-43-
With Kn = 0.17 0.01 from Prob. 1-21, we know slip flow is important, so Eq. (3-40) applies:
( 0.0000175m ) dp
m3 ro4 dp
Q = 9.6 E -11
=
− (1 + 8Kn ) =
− 1 + 8 ( 0.17 )
s
8 dx
8 ( 0.00002 kg /m s ) dx
4
dp
Solve for − = 22,000 Pa /m
dx
3-46
Ans. (b)
Starting from the axial momentum equation, derive Eqs. (3-40) for slip flow in tubes.
Solution: From Sect. 3-3.1 for the no-slip laminar pipe flow, the general solution of the fullydeveloped-flow equation, 2u = K = (1/ )( dp /dx ) , is u = Kr 2 /4 + C1 ln ( r ) + C2 . As before, we
must have C1 = 0 to enforce centerline symmetry and avoid a logarithmic singularity. Then we
find C 2 from the slip condition, noting that K is negative:
2
− ( dp /dx )
du
K 2
Kr Kr
|w = − o = o + C2 , or: C2 = −
ro + 2ro , K =
dr
4
4
2
− ( dp /dx ) 2 2
ro − r + 2ro
Thus the final solution is: u =
Ans.
4
(
uw = |
(
)
)
3-47 For the geometry of Fig. 3-1 assume a constant pressure gradient with both walls fixed.
Solve continuity and x-momentum for laminar slip flow between the plates. Find the velocity
distribution and the volume flow rate per unit depth. Does the Knudsen number appear?
Solution: The appropriate fully-developed x-momentum equation is Eq. (3-41):
d 2u
dy
Integrate twice:
2
=
1 dp
=K 0
dx
du
K
= Ky + C1; u = y 2 + C1 y + C2
dy
2
For symmetry, du /dy = 0 at y = 0, hence C1 = 0. To find C2 , apply the slip condition at y − h:
u y −h =
du
K 2
h + C2 = −
=
2
dy y −h
( − Kh ) ,
or: C2 = −
Kh
(h + 2
2
)
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Hill LLC.
-44-
The final slip-flow velocity distribution is thus
u=
( −dp /dx )
2
( h2 + 2 h − y 2 )
Ans.
The volume flow rate, with slip, does indeed contain the Knudsen number.
+h
udy =
Q=
−h
2 ( −dp /dx ) 3
h (1 + 3Kn ) , Kn = Knudsen number =
3
h
Ans.
3-48 Hadjiconstantinou (2003) has updated a second-order slip theory by Cercignani (2000) to
give new numerical coefficients for the wall-slip velocity, to be compared with Eq. (1-91):
uw 1.11
u
|w − 0.61
y
2
2u
y 2
|w
Repeat Prob. 3-52 with this formulation to solve continuity and x-momentum for laminar slip
flow between the plates. Find the velocity distribution and the volume flow rate per unit depth.
Does the Knudsen number appear?
Solution: From Prob. 3-52 we can get this far without no-slip:
d 2u
dy
Integrate twice:
2
=
1 dp
=K 0
dx
du
= Ky + C1 ;
dy
u=
K 2
y + C1 y + C2
2
Symmetry, du /dy = 0 at y = 0, hence C1 = 0. To find C2 , apply the new slip condition at
y = h:
u y =h
du
K
= h 2 + C2 = 1.11 −
− 0.61
2
dy y =h
Solve for
2d
u
2
dy y =h
2
h2
C2 = − K + 1.11 h + 0.61
2
2
The final modified slip-velocity distribution is:
u=
( −dp /dx )
2
( h2 + 2.22 h + 1.22 2 − y 2 )
Ans.
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Hill LLC.
-45-
The volume flow rate contains the Knudsen number twice:
+h
Q=
u dy =
−h
2 ( −dp /dx ) 3
h 1 + 3.33Kn + 1.83 kn 2 , Kn = Knudsen number =
3
h
(
)
Ans.
3-49 Glycerin, with a density of 1260 kg/m3 and viscosity of 1.5 kg/m-s, flows at 1 m/s past a
small smooth sphere. (a) What sphere diameter will cause the Reynolds number to be exactly
unity? For the sphere size in part (a), according to Stokes-flow theory, what is the gage pressure
(relative to freestream pressure) (b) at the front stagnation point; (c) at the rear stagnation point?
(d) What is the surface shear stress at these two points?
Solution: Numerical problems are a refreshing relief from heavy theory. Calculate Re D :
(
)
3
VD 1260 kg /m (1.0m /s ) D
Re D = 1.0 =
=
, solve for D = 0.00119m
1.5 kg /m − s
Ans. (a)
(b, c) From Eq. (3-215), with p = 0 pascals gage, the fluid gage pressure at the surface is
3 (1.5kg /m − s )(1.0 m/s )
3U
cos = −
cos = −3780cos
2a
0.00119m
2a
2
2
= −3780cos180 = +3780Pa Ans. (b); prear = −3780 cos 0 = −3780 Pa Ans. (c)
p |r =a = −
p front
3 aU
2
cos = −
(d) From Eq. (3-216), the surface shear rθ is zero at both the front (180 ) and back ( 0 ) .
3-50 Hill and Power (1956) proved that the creeping-flow (Stokes) drag of a solid object is
greater than the drag of any inscribed shape but less than the drag of any circumscribed shape.
Verify this result for the spheroid of Fig. 3-36 by comparing it to inscribed and circumscribed
spheres. Do the relative drag forces differ markedly or only by a few per cent?
Solution: The spheroid, shown as a heavy line above, has a length 2a and a diameter 2b. The
drag estimates for the spheroid are given by Eqs. (3-221):
Parallel to 2a:
C1 6 ( 4 + a/b ) /5; Normal to 2a: Cn 6 (3 + 2a/b ) /5
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Hill LLC.
-46-
For simplicity, let a b, the results would simply be reciprocal for a b. The force ratios may
be calculated as follows:
Ft −circumscribed
6 U a
5a /b
=
=
Ft −actual
6 ( 4 + a /b ) /5 Ub 4 + a /b
Ft −inscribed
6 U b
5
=
=
Ft −actual
6 ( 4 + a /b ) /5 Ub 4 + a /b
Fn−circumscribed
6 U a
5a /b
=
=
Fn−actual
6 ( 3 + 2a /b ) /5 Ub 3 + 2a /b
Fn−inscribed
6 U b
5
=
=
Fn−actual
6 ( 3 + 2a /b ) /5 Ub 3 + 2a /b
When plotted below in the curve-fit range, 1 a/b 5, we see that both “circumscribed” forces
are greater and both “inscribed” forces are less than the actual forces. The ratios are significantly
different from unity and vary by a factor of 2 to 3.
3-51 The elemental creeping-flow solution, = r (ln r )sin , is called an Oseenlet. Is this a
solution for plane flow, Eq. (3-206), or axisymmetric flow, Eq. (3-211), or both? How might
Oseenlets be used in analysis of more complex creeping flows?
Solution: Well, let’s first try axisymmetric creeping flow, Eq. (3-211):
( D ) = 0?, where D
2 2
2
=
2
r
2
+
1 2
r
2
2
−
cot
r 2
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Hill LLC.
-47-
With much labor, work out D 2 ( r ln r sin ) =
sin ln r sin cos2 ln r
−
−
r
r
sin r
With even more labor, D 2 D 2 ( r ln r sin ) = Six terms, nowhere near zero!
No, this particular Oseenlet is not a solution for axisymmetric creeping flow.
Then let us try plane creeping flow, Eq. (3-206):
(
)
4 = 2 2 = 0? where 2 =
1
r
r r r
2
1
+
r 2 2
1
1
Work out 2 ( r ln r sin ) = sin , then 2 sin = 0
r
r
Yes, this particular Oseenlet is indeed a solution for plane creeping flow.
Since the creeping-flow differential equation is linear, it follows that more complicated
creeping flows could be constructed as a superposition of Oseenlets.
3-52 Using a numerical method such as Subroutine RUNGE of Appendix C, solve the
axisymmetric stagnation flow Eq. (3-165) for the function F ( ) . Use an iterative scheme to
determine the proper value of F ( 0 ) .
Solution: From Eqs. (3-165) and (3-152), the axisymmetric equation and conditions are:
F + 2 F F + 1 − F 2 = 0
F ( 0) = F (0) = 0 ; F ( ) = 1
The challenge is to find the initial value F ( 0) which achieves F ( ) → 1. We could estimate
(guess) that this initial value is of order unity. The solutions are not as well behaved as the
Blasius equation – selecting F ( 0) too low causes F ( ) to keep falling far below unity, and
selecting F ( 0) too high causes F ( ) to keep rising far above unity. Anyway, we can make a
few guesses for F ( 0) and select = 4.0 as “infinity”. The equation above was solved
numerically, and here are some results:
F ( 0)
1.00
1.25
1.50
F ( 4.0 ) –0.8036 0.6686 1.9645
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Hill LLC.
-48-
So we see the correct solution F ( 4.0 ) 1.0 is somewhere between F ( 0 ) = 1.25 and 1.50, at
about 1.31. We could interpolate parabolically (giving F ( 0) 1.311 ) or we could just run off
two more solutions to narrowly bracket that region:
F ( 0)
1.30
1.32
F ( 4.0 ) 0.9369 1.0427
Now we can interpolate linearly to F ( 0) 1.3119, the same as Table 3-4 of the text. Ans.
And we can recompute the desired axisymmetric stagnation flow as in Fig. 3-25 of the text.
3-53 The rotating disk of Fig. 3-28 acts rather like a centrifugal pump. Consider a disk 60 cm in
diameter, rotating at 120 rev/min in air at 20°C and 1 atm. Is the flow at the disk tip laminar,
transitional, or turbulent? Using Kármán’s theory of Sect. 3-8.2, estimate the volume flow of air,
in cm3 /s, pumped outward by one side of the disk.
Solution: For air at 20°C and 1 atm, take = 1.205 kg/m3 and = 1.81E-5 kg/ ( m-s ) . Convert
= (120 rev/min )( 2π/60) = 12.57 rad/s. Then the disk Reynolds number, as defined by
Eq. (3-191), is
ro2 (1.205kg /m3 ) (12.57rad /s )( 0.3m )
=
= 75300 300, 000 ( transition )
1.8E − 5 kg /m − s
2
Re =
The Reynolds number is low, the flow is laminar.
Ans.
There are two ways to find the volume flow (l) integrate the radial outward flow proportional to
F ( z*) ; or ( 2) calculate the inward axial flow proportional to H ( z*) . The latter is easier, using
the approach velocity from Eq. (3-188):
vz ( ) = −0.8838 v = −0.8838 (12.57 )(1.81E − 5 /1.205) = −0.01214 m/s
The negative sign means flow toward the disk. The total volume flow is thus
2
Q = vz ( ) Adisk = ( 0.01214m/s ) ( 0.3m ) = 0.00343 m3 /s = 3430cm 3 /s
Ans.
There may be edge effects, since Kármán’s solution is for an infinite disk. But they cannot be
large, because the boundary layer thickness is only = 5.4 ( v / )
1/2
5.9 mm.
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-49-
3-54 Consider the fully developed flow in a long axisymmetric annulus, which is driven by a
pressure gradient.
(a) Using an appropriate coordinate system, make the necessary assumptions that will simplify
the incompressible Navier–Stokes equations.
(b) Write the differential equation of motion for this problem.
(c) Specify a judicious assortment of boundary conditions.
(d) Solve for the steady-state velocity profile u(r) for b ≤ r ≤ a.
Solution: Consider a strip of width dx and radius dr.
The flow is steady, incompressible, and fully developed.
The forces on the free body in the direction of flow must be zero.
Using the momentum and conservation of mass equation, we have
P
T
P ( 2 r ) dx − P + x dx 2 rdx + T ( 2 r ) dx − T + r dr 2 ( r + dr ) dx = 0
P
T
T
− x dx 2 rdr − r dr 2 rdx − T ( 2 r ) dxdr + r dr ( 2 dx ) dr = 0
Dividing the above equation throughout by the volume of the element 2 rdrdx and neglecting
the last term that is a higher-order differential, we obtain
P 1
T
+ T+
=0
x r
r
P 1
+
(Tr ) = 0
x r r
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-50-
P
1
=−
(Tr )
x
r r
P
(Tr ) = −r
r
x
Integrating the above equation, we obtain
r 2 P
Tr = −
+ C1
2 x
T =−
r P C1
+
2 x r
We know that T = −
V
.
r
Therefore, we have
−
V
r P C1
=−
+
r
2 x r
V
r P C1
=
−
r 2 x r
V=
r 2 P C1
− ln r + C2
4 x
Applying boundary conditions at r = Ri and r = Ro, we obtain
1 P 2 2 Ri2 − Ri2 Ri
V =−
Ri − r −
ln
4 x
Ri r
ln
Ro
(Ans.)
3-55 A fluid of constant thickness h flows steadily down a flat surface that is inclined at an angle
θ, as shown below.
(a) Make the necessary assumptions to reduce the Cartesian Navier–Stokes equations.
(b) Write the differential equation of motion for this problem.
(c) Specify a judicious assortment of boundary conditions.
(d) Solve for the steady-state velocity profile u(y) for 0 ≤ y ≤ h.
(e) Calculate the volumetric flow rate per unit width.
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Hill LLC.
-51-
Solution: τ is expanded in a Taylor series about the center of
the differential control volume shown in the
figure.
t = +
d dy
dy 2
and
b = +
d dy
−
dy 2
The boundary conditions are as follows:
For no slip at y = 0,
u=0
For no shear stress at y = h,
du
=0
dy
Applying the x component of the momentum equation to the differential control volume, we
obtain
FSx + FBx = updV + upV dA
Assuming steady and fully developed flow (i.e., u = u(y) and τ = τ(y)) and that there is no
variation of pressure in the x direction, we obtain
d dy
d dy
FSx + FBx = 0 = +
dx dz + +
− dx dz + g sin dx dy dz
dy 2
dy 2
Thus,
d
= − g sin .
dy
Integrating the above equation, we obtain
= − ( g sin ) y + C1
We know that
τ = 0 at y = h
Therefore,
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C1 = ( g sin ) h
and
du g sin
=
hy
dy
Integrating again, we obtain
u=
g sin
y2
hy
−
+ C2
2
Also,
u = 0 at y = 0
Therefore,
C2 = 0
and
g sin
y2
u=
hy −
2
(Ans. d)
The volumetric flow rate per unit width can be determined as follows:
h
g sin h
y2
g sin hy 2 y 3
Q = udy =
hy
−
dy
=
−
0
2
6 0
2
0
h
Thus, Q = h3 g sin / 3 .
(Ans. e)
3-56 Consider the flow configuration of two different fluid bodies that are separated by a rigid
interfacial surface that can be assumed to be infinitesimally thin. The heights and dynamic
viscosities of the two bodies are specified on the sketch. The upper body is bounded by a moving
surface that translates in the positive x direction at a uniform speed V. What should be the axial
speed of the interfacial surface in the absence of a pressure gradient in the streamwise direction?
Solution: At the interface,
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-53-
Z1 = Z2
and
V1 = V2
Therefore,
V=U
1V1
h1
=
2V2
h2
1U
F = Z1 A + Z 2 A =
U=
D1
=
2U
D2
1 0 2V
h1
+
h2
1
D1
= 0+
=
2
D2
2U
D2
FD2
2
(Ans.)
3-57 Consider fully developed laminar flow in the annulus between two concentric pipes as
shown below. The outer pipe is stationary, and the inner pipe moves in the x direction at a
constant speed V. Assume the axial pressure gradient is zero (dp/dx = 0). From the differential
form of the conservation laws, obtain a general expression for the velocity profile, u(r), in terms
of two integration constants, C1 and C2. Apply the no-slip boundary conditions to solve for C1
and C2. Give the final expression for u(r).
Solution: Applying the x component of momentum equation for the given annulus, we obtain
FSx + FBx = updV + upV dA
and
rx =
du
=
dr
The axial pressure gradient is zero. For a fully developed laminar flow, we have
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d dr
dr
d dr
dr
FSx = +
2 r + dx − −
2 r − dx = 0
dr 2
2
dr 2
2
Neglecting the products of differentials, the above equation reduces to
+r
d
=0
dr
or
d ( r )
dr
=0
Therefore, r = C1 or =
C1
.
r
We know that
=
du
dr
Therefore,
du C1
=
dr r
and
u=
C1
ln r + C2
Applying no-slip boundary conditions to the above equation gives us the following:
At u r = r = Vo ,
Vo =
At u r = r = 0 ,
0=
i
o
C1
C1
ln ri + C2
ln ro + C2
C1
or
C2 = −
C2 = −
Vo ln ro
ln ( ri / ro )
ln ro
Solving for C1 and C2, we obtain
C1 =
Vo
ln ( ri / ro )
and
(Ans.)
3-58 Consider a fully developed laminar flow in the annular space formed by the two concentric
cylinders shown in Prob. 3.57, but with a stationary inner cylinder and a nonzero pressure
gradient (dp/dx ≠ 0). Let ro = R and ri = kR, where 0 < k < 1 is a constant.
(a) Using the Navier–Stokes equations and proper boundary conditions, show that the velocity
profile and volume flow rate can be written as
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-55-
2
R 2 dp r
1− k 2
r
u (r ) = −
ln
1 − +
4 dx R ln (1/ k ) R
(1 − k )
R 4 dp
Q=−
1− k 4 −
8 dx
ln (1/ k )
and
2
2
(b) Obtain an expression for the location of the maximum velocity as a function of k.
(c) Obtain an expression for the average velocity in the annulus, u .
(d) By taking the limiting case of k → 0, compare your expressions to those corresponding to the
Poiseuille flow in a circular pipe.
Solution:
The differential control volume for nonzero dp/dx is
u=
r 2 dp C1
+ ln r + C2
4 dx
Applying boundary conditions to the above equation, we obtain the following:
R 2 dp C1
+ ln R + C2 = 0 .
At u r =r ,
i
4 dx
At u r =r ,
o
k 2 R 2 dp C1
+ ln kR + C2 = 0 .
4 dx
Solving for C1 and C2, we obtain
(
)
2
R 2 dp 1 − k
C1 = −
4 dx ln (1/ k )
and
(
)
2
R 2 dp R 2 dp 1 − k
C2 = −
+
ln ( R )
4 dx 4 dx ln (1/ k )
Substituting C1 and C2 in the expression of u, we obtain
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-56-
(
)
(
)
2
2
r 2 dp R 2 dp 1 − k
R 2 dp R 2 dp 1 − k
u=
−
ln r −
+
ln ( R )
4 dx 4 dx ln (1/ k )
4 dx 4 dx ln (1/ k )
Further simplification of this expression gives
u (r ) = −
2
R 2 dp r
1− k 2
r
1
−
+
ln
4 dx R ln (1/ k ) R
(Ans. a)
The volume flow rate is given by
R
R
kR
kR
Q = udR = u 2 r dr = 2 ur dr
Using the previously obtained expression for u(r), we obtain
R 2 dp R
r3
1− k 2
r
Q = 2 −
r
−
+
r ln dr
2
ln (1/ k )
R
4 dx kR R
R 4 dp 1 r r
1 − k 2 r r r
−
+
ln d
2 dx k R R ln (1/ k ) R R R
3
Q=−
1
R 4 dp 1 r 1 r
1 − k 2 r 1 r 1
Q=−
ln −
− +
2 dx 2 R 4 R ln (1/ k ) R 2 R 4
k
2
4
2
Further simplification of this expression gives
(1 − k )
R 4 dp
Q=−
1− k 4 −
8 dx
ln (1/ k )
2
2
(Ans. a)
(b) The condition for maximum velocity is
=
du
=0
dr
Using the expression of u from part (a), we obtain
0=−
R 2 dp 2r
1 − k 2 1
− 2 +
4 dx R
ln (1/ k ) r
The maximum velocity is located at α = r / R. Therefore,
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-57-
1/2
1
2 R 1 − k 2 1
1− k 2
or,
=
0=− 2 +
R
ln (1/ k ) R
2
ln (1/ k )
(Ans. b)
(c) The average velocity in the annulus is given by
u=
Q
A
The area is
r r
A = 2 r dr = 2 R 2 d = R 2 1 − k 2
R R
k
1
(
)
Therefore,
(1 − k )
R 4 dp
−
1− k 4 −
8 dx
ln (1/ k )
Q
u= =
A
2
R2 1 − k 2
(
)
2
(
)
2
2
= − R dp 1 + k 2 − 1 − k
8 dx
ln (1/ k )
(Ans. c)
(d) Considering the limiting case of k → 0, we have
Q=u =−
R 4 dp
8 dx
Thus, the results are congruent with the Poiseuille flow in a circular pipe.
(Ans. d)
3-59 Consider the problem associated with the spinning porous tube depicted in Fig. 3-28 of Sec.
3-6.4. By substituting the similarity transformations given by Eq. (3-182) into Eq. (3-181), derive
Eq. (3-183). Subsequently, recognizing that υθ = g(r) is a function of a single variable, derive Eq.
1
(3-184). Finally, introduce = r 2 into Eq. (3-184) and differentiate once to obtain Eq. (32
185). Verify that the four boundary conditions for this problem translate into
1
1. f = 1 from ur(1, z) = −1 (uniform radial injection at the wall)
2
1
2. f = 0 = 0 from uz(1, z) = 0 (zero axial speed at the injecting wall)
2
3. f(0) = 0 from ur(0, z) = 0 (no radial crossflow at the centerline)
4.
2 f ( 0 ) = 0 from ∂uz(0, z)/∂r = 0 (symmetric axial velocity at the centerline).
Solution:
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The spinning angular rate is ωs, and the tangential speed at the wall is vs.
For a constant radial injection speed at the porous wall, we have
vr ( r = ro ) = −v 0
ro = Radius of the tube
r=
r*
z*
v*
; z = ; vr =
ro
ro
v
v =
v*
v*
P*
; vz = z ; P = 2
v
v
v
For injection, we have
Re =
v ro
0
v
The steady, incompressible, and normalized Navier–Stokes equation simplifies to
1 ( rvr ) vz
+
=0
r r
z
vr
vr
v v 2
P 1 2vr 1 vr 2vr vr
+ vz r − = −
+
+
+
−
r
z r
r Re r 2 r z r 2 r 2
vr
v
v v v
1 2v 1 v 2v v
+ vz + r =
+
+
−
r
z
r
Re r 2 r r z 2 r 2
vz
vz
P 1 2vz 1 vz 2vz
vr
+ vz
=−
+
+
+
r
z
z Re r 2 r r z 2
The axisymmetric conditions
= 0 have been enforced.
A stream function that varies linearly in the flow direction is introduced.
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-59-
( r, z ) = zf ( r )
vr = −
vz =
f
r
zf
r
The pressure can be eliminated by cross-differentiation.
z
1
1
1
1 2 v
2
2 ff − ff − f +
rf − f + f = v
r r
r
Re
r r
z
Also, v = g ( r ) .
ff −
1
1
1 2
ff − f 2 +
rf − f + f = r C
r
Re
r
(1)
We know that
=
r2
2
Differentiating equation (1) with respect to η, we obtain
d 3 f df d 2 f 2 d 4 f
d3 f
−
+2 3 =0
f
+
3
d d 2 Re d 4
d
d
f(η) is subject to the following boundary conditions:
1
1. f = 1 from ur(1, z) = −1 (uniform radial injection at the wall)
2
1
2. f = 0 = 0 from uz(1, z) = 0 (zero axial speed at the injecting wall)
2
3. f(0) = 0 from ur(0, z) = 0 (no radial crossflow at the centerline)
4.
2 f ( 0 ) = 0 from ∂uz(0, z)/∂r = 0 (symmetric axial velocity at the centerline).
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CHAPTER 4. LAMINAR BOUNDARY LAYERS
Repeat the momentum-integral analysis of the flat plate in Sec. 4-1 for Pohlhausen’s
4-1
3
u 3 y 1 y
= − , where δ(x) is the boundary-layer or disturbance
U 2 2
thickness as pictured in Fig. P4-1. Is this profile any more (or less) realistic than the
approximation of Eq. (4-11)? For the above profile, compute (a) ( / x ) Re x , (b) ( * / x ) Re x
cubic polynomial profile
, (c) ( / x ) Re x , (d ) ( C f / x ) Re x , and (e) ( CD / x ) Re x .
Solution: We know that momentum thickness is approximated by
1
(
)(
)
u
u
39
y
3
3
, =
1 − dy 1.5 − 0.5 1 − 1.5 + 0.5 d =
U U
280
0
0
=
In a similar manner, we can find that * = (1 − u/U ) dy 3/8. [Note H = */ = 2.69 ]
Substitution into the integral relation (4-10) gives
Cf =
or:
2w
U
2
=
2 (1.5 U/ )
U
2
=2
d
d 39
=2
,
dx
dx 280
d = 140 dx/ (13 U )
Integrating, assuming = 0 at x = 0, yields
2
280 x
,
13 U
or
4.64
x
Re x
(Ans. c)
The other two thicknesses follow from = .1393 and * = 0.375 :
Re x = 0.646
x
*
Re x 1.740
x
(Ans. a, b)
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-1-
The errors, compared to the Blasius solution, are (–2.7%) and (+1.1%), respectively. The skin
friction follows by substituting ( x ) into the approximation w = 1.5U/. Thus
Cf Re x 0.646
CD ReL 1.293
(Ans. d, e)
The error is (−2.7%) for each. This is a better accuracy than the 10% errors of the parabolic profile
(see the Table on page 221), probably because the cubic profile is more realistic.
4.2
Repeat the integral flat-plate heat-transfer analysis of Sect. 4-1.7 by using the parabolic
velocity from Eq. (4-11) and the quartic temperature profile
2y 2y3 y 4
T − Te = ( Tw − Te ) 1 −
+ 3 − 4
T T
T
This is a realistic temperature profile, satisfying the five boundary conditions T ( 0) = Tw ,
2T/y 2 ( 0 ) = 0, T ( ) = Te , T/y ( ) = 0, and 2T/y 2 ( ) = 0. Substitution into the integral
energy equation, Eq. (4-21), gives
2k ( Tw − Te ) d r
cp U(Tw − Te )(2 − 22 )(1 − 2 + 23 − 4 )T d
qw =
=
dx 0
2 2 3
d
=
− ,
where = T 1
cp U ( Tw − Te )
dx
15 42
Now assume that = constant (flat plate with heating starting at the leading edge) and substitute
( d/dx ) 15v/U from page 220 (the integral momentum analysis). After cancellation of T and
U, we obtain
3 −
5 4 1
= ,
28
Pr
or:
=
T
= Pr −1/3
(Ans.)
where we neglect the quartic term in the last approximation. This is the same as Eq. (4-26), which
uses a parabolic temperature profile. The heat transfer estimate is
Nu x =
qw x
x 2kT
2
2
=
Re1/
Pr1/ 3
x
k ( Tw − Te ) kT 4.64
(Ans.)
This is similar to Eq. (4-27), except the constant ( 2/4.64) = 0.431 is about 30% higher than the
accurate estimate of 0.332 of Pohlhausen (1921). This is probably because we have mixed a crude
(parabolic) velocity with an accurate (quartic) temperature profile.
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-2-
4-3
Repeat the momentum analysis of C f from Sect. 4-1.4 with the better profile
y
u = U sin
2
(1)
Whereas the parabolic profile from Eq. (4-11) only satisfies three conditions, u ( 0) = 0, u ( ) = U,
and u/y ( ) = 0, the sine-wave profile (1) above also satisfies the zero-pressure-gradient
condition 2 u/y 2 ( 0 ) = 0 appropriate for flat-plate flow - as was also true for the cubic profile in
Prob. 4-1 on p. 86 of this Manual. The wall shear estimate is
w =
U
u
| y =0
y
2
Similarly, the momentum thickness estimate is
1
4−
= sin 1 − sin d =
= 0.1366
2
2
2
0
Then the momentum integral relation yields
Cf =
2 ( U/2 )
U 2
d
d
=2
= 2 ( 0.1366 ) ,
dx
dx
1/ 2
or:
Re x =
x
0.1366
= 4.80
The accuracy is good. Putting ( x ) back into the C f estimate gives
Cf Re x =
/2
= 0.3276
4.80
(Ans.)
The error is only (–1.4%), but we can’t usually expect such accuracy in integral theories.
4-4
Air at 20°C and 1 atm flows past a smooth flat plate as in Fig. P4-4. A pitot stagnation
tube, placed 2 mm from the wall, develops a water manometer head h = 21 mm. Use this
information with the Blasius solution, Table 4-1, to estimate the position x of the pitot tube. Check
to see if the flow is laminar.
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-3-
Fig. P4-4
Solution: For air at 20°C and 1 atm, take = 1.205 kg/m3 and = 1.81E-5 kg/(m-s).
Estimate the velocity u at y = 2 mm from the pitot stagnation tube formula:
p = ( water − air ) gh = (998 − 1.205 kg/m3 )(9.81 m/s 2 )(0.021 m) = 205 Pa
u = 2p/air = 2(205 Pa)/(1.205 kg/m3 ) = 18.46 m/s
Since u U = 20 m/s, the pitot tube is inside the boundary layer. The velocity ratio is
u /U = 18.46/20 = 0.923 = f ( )
We can interpolate in Table 4-1 of the text to find that this ratio corresponds approximately to
2.55. Everything is known in the definition of except the position x:
= 2.55 = y
(1.205)( 20 )
U
= ( 0.002m )
2 x
2 (1.81E − 5) x
Solve for x 0.41m = 41cm
Ans.
Check to see if the flow is laminar, so that the Blasius flat-plate theory holds:
Re x =
Ux (1.205)( 20 )( 0.41)
=
545, 000
1.81E − 5
For a smooth plate and a quiet freestream, transition to turbulence occurs for Re x 106. Thus we
conclude that the flow at x = 41 cm is indeed laminar.
4-5
Repeat the unheated-starting-length analysis on p. 224 of the text by keeping the term
( /5 ) . Solve numerically in the range 1 x 5 for Pr = 1 and x
3
o
= 1.
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-4-
The relevant differential equation is on p.
and, for Pr = 1 and 2 = 30vx/U, becomes
2 6 3 d 3 3 12
= 0.8, with = 0 at x = x o
4 −
+ − =
5 dx
5 15
This is first-order and nonlinear and easily solved numerically by, e.g., the Runge-Kutta routines
of Appendix C. If we neglect the second terms in each parentheses, we obtain the simple text
−1/3
3/4
solution = 1 − ( x o /x )
. The numerical and approximate solutions are shown below and
are seen to be in good agreement - which would be better if Pr > 1.
4-6
Develop an iterative solution of the Blasius equation (4-45) which will start with a
reasonable guess, say, f ( 0) = 0.3, and converge to the exact solution, f ( 0) = 0.4696.
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-5-
Solution: The Blasius equation is beautifully behaved, always yielding a constant value of f *( )
at large , so interpolation is quite easy. Note, for example, the following results:
f ( 0) =
0.30
0.35
0.40
0.45
0.50
f ( ) =
0.74176
0.82204
0.85857
0.97198
1.042705
One can almost interpolate linearly for the proper value f ( 0) = 0.4696 to give f ( ) = 1.0.
4-7
Consider a long flat plate emerging from a wall at velocity U, as in Fig. P4-7.
There is no freestream. Show that the Blasius Eq. (4-45) holds for this ease, with f ( 0 ) = 0,
f ( 0) = 1, and f ( ) = 0. Solve the equation numerically and show that Cf 0.444/Re1/x 2 . Also
evaluate v ( ) and discuss. [Hint: Note that f ( 0 ) is negative.]
Fig. P4-7
Solution: This is a moving flat plate, with zero pressure gradient and a boundary layer that grows
starting at x = 0. Therefore the Blasius equation is valid with boundary conditions reversed on the
streamwise ( u ) velocities:
f + f f = 0
with
f ( 0) = 0, f ( 0) = 1, f ( ) = 0
The correct initial condition which matches these conditions is f ( 0) −0.6276. The profiles of
f ( ) , f ( ) , and f ( ) are calculated (from Subroutine RUNGE or other routine) and are
plotted on the next page.
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-6-
Stream function ( f ), velocity ( f ) and shear stress ( f ) near a moving flat plate.
The wall shear stress is similar in form to the fixed-plate, moving-stream relation, Eq. (4-52):
w =
Uf ( 0 )
u
| y =0 =
;
y
2vx /U
Cf =
2 |w |
U
2
=
0.6276/ 2 0.4438
=
Ux /
Re x
Ans.
The vertical velocity at the outer edge of the boundary layer is
v () =
vU
vU
vU
( f − f ) | → =
( −1.142 ) = −0.808
2x
2x
x
Ans.
The negative sign means that the outer (still) fluid is entraining fluid into the boundary layer.
4-8
Develop a numerical solution to iterate the Falkner-Skan equation (4-71) for any value of
0. Start with a reasonable but incorrect value of f ( 0 ) .
Solution: Not as well behaved as Blasius, but still OK. We illustrate for = +0.3, “” = 5:
f ( 0) =
0.6
0.7
0.8
0.9
1.0
f ( ) =
0.27076
0.70748
1.09402
1.44830
1.77985
We may interpolate to f ( 0) = 0.77476 for this value = 0.3, as given in Table 4-2 of the text.
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-7-
4-9
The Blasius Eq. (4-45) must be iterated to find the value of f ( 0) which cause f ( ) to
equal 1.0. Panton (1996) suggests the following scheme to avoid iteration: Define
f ( ) = F ( ) ,
where is a constant
(a) Show that the function F also satisfies the Blasius equation. (b) If we arbitrarily set an initial
condition F ( 0) = 1.0, explain how can immediately be found without iteration. (c) Even with
found, explain why the solution for F is still awkward for filling out Table 4-1. (d) If an
integration (not required of you) then yields F ( ) = 1.6552, what is the proper value of ?
(e) Show that the value of from part (d) leads to the result f ( 0) = 0.4696.
Solution: (a) Introducing f = 2 F ( ) etc., we find indeed that F + FF = 0.
Ans. (a)
(b) From the transformation, f ( ) = 2 F ( ) = 1, compute = 1/F ( )
Ans. (b)
1/2
.
(c) The numerical solution, in even increments of , will not be spaced nicely in .
Ans. (c)
(d) Given the writer’s result F ( ) = 1.6552, compute = 1/1.6552
Ans. (d)
1/2
= 0.7773.
(e) With known, f ( 0) = 3 F ( 0) = ( 0.7773) (1.0) = 0.4696
3
4-10
Ans. (e)
Show that Clauser’s parameter, ( */w ) ( dp/dx ) , is a constant for the Falkner-Skan flows.
What is its value for the separating-flow condition?
Knowing that, for Falkner-Skan flow, = y ( m + 1) U/2vx
1/2
, and f ( ) = u/U, we may
compute
1/2
2 x
u
* = 1 − dy =
U
( m + 1) U
0
1
* = (1 − f ) d
*,
0
( m + 1) U
u
|y = 0 = U f ( 0 )
y
2 x
1/2
w =
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-8-
Then, since U = Kx m , the pressure gradient is
dp
dU
mU2
2 2m−1
= − U
= − m K x
=−
dx
dx
x
[Bernoulli relation]
Combining these three, we obtain the Clauser parameter for Falkner-Skan flow:
Clauser Parameter =
* dp
2m *
=−
w dx
( m + 1) f ( 0 )
(Ans.)
These values are independent of x, hence the Clauser parameter is a constant for a given wedge
flow. We may take , etc., from Table 4-2, page 243. For separating flow, = −0.19884,
m = −0.09043, since f ( 0 ) = 0, the Clauser parameter = +.
(Ans.)
When the (different) similarity variable = y ( IUI/vx )
equation reduces to
1/2
4-11
f +
(
)
m +1
f f + m f 2 − 1 = 0
2
is used, the Falkner-Skan
(1)
subject to the boundary conditions of Eq. (4-72): f ( 0) = f ( 0) = 0 and f ( ) = 1. Find the solution
for the special case flow U = −K/x = ( −K ) x −1.
This case is for m = −1, so the differential equation becomes
1/2
f − f + 1 = 0,
2
K/x
with = y
x
=
y
x
( K/ )
But this is exactly Eq. (4-88) for point-sink flow, and the solution is given by Eq. (4-89):
f=
u
= 3 tanh 2
+ 1.146 − 2
U
2
(Ans.)
Other details of this solution are given on p. 250 of the text.
4-12 The thin equilateral triangle below is subjected to airflow at 20°C and 1 atm. Estimate the
(laminar) plate drag in newtons.
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-9-
Assume that the local wall shear stress varies only with distance x from the leading edge, from
laminar flat-plate theory. Then the strip dA in the figure would have a uniform shear stress given
by Eq. (4-52), with f ( 0) = 0.4696:
1/2
w = 0.332
x
U3/2
For a triangle of side length a, the strip has an area dA = Ldx (1 − x/L ) , where L = a sin60° and
a = 2m. Then the total shear force on the plate is the integral of local strip force:
F = w dA ( 2 sides ) = 2 ( 0.332 )( )
1/2
L
U3/2 =
0
(
= 0.664 U3
= 0.885 ( )
)
1/2
1/2
1/2
2Lx
−
L−x
x
dx
2 3/2 L
x |0
3
( UL)3/2
(Ans.)
For our particular case, U = 12 m/s, L = 2sin 60 = 1.732 m, and for air = 1.2 kg/m3 and
= 1.8E-5 kg/m s. The numerical value of the laminar drag is thus
F = 0.885 (1.2 )(1.8E-5 )
1/2
(12 )(1.732 )
3/2
= 0.39 N
(Ans.)
We should check the Reynolds number to see if the flow is truly laminar: ReL = UL/ =
(1.2)(12)(1.732) / (1.8E-5) 462,000 (OK ). Transition should occur at
ReL 1E6.
4-13 Model a flow straightener as an array of thin-walled square ducts as in the figure below.
Estimate the pressure drop across an N N bundle of such ducts, using laminar flat-plate
boundary-layer theory.
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-10-
Each square cell has four surfaces, and each ( L a ) surface experiences a shear force given by
laminar boundary-layer theory, Eq. (4-53):
CD =
1.328
,
ReL
or
F = 0.664 ( L )
1/2
for one ( L a ) surface
U3/2a
An N N array of boxes has N 2 cells of four surfaces each, or a total shear force
Ftotal array = 2.656 N2 ( L )
1/2
U3/2a
A freebody around the entire array shows that this shear force has to be balanced by a pressure
drop across the array, multiplied by the array frontal area:
parray =
Ftotal
( Na )
2
=
2.656
( L )1/2 U3/2
a
(Ans.)
The result is independent of the array size N. This pressure drop is modest unless the cell width a
is very small. For example, for air flow at U = 10 m/s and a cell size a = 1 cm and L = 10 cm, we
may estimate p 12 Pa.
4-14 Develop a numerical solution for laminar flat-plate flow with wall suction or blowing and
compare your results with Fig. 4-15.
We are to solve the Blasius Eq. (4-45), f + f f = 0, subject to the blowing/suction conditions of
Eqs. (4-84,85):
f ( 0) = 0
f () = 1
f ( 0 ) = − 2v*w 0
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-11-
The required FORTRAN statements are the same as for the Blasius problem (bottom of p. 233 of
the text). The numerical problem is to find the unknown initial condition f ( 0) = Y1 ( x = 0) which
causes the farfield velocity profile to approach f ( ) = 1. For a particular example, let us take
blowing, v*w = 0.4, or f ( 0 ) = −0.4 2 = −0.5657. From Fig. 4-15(a) we see that “infinity”
extends out to about “ = 8”. The initial slope f ( 0 ) is clearly much smaller than the Blasius (no
blowing) value of 0.4696. Let us guess an array of initial values and tabulate f ( ) :
f ( 0) =
0.05
0.10
0.15
0.20
Y2 (8) = f ( ) =
0.7618
0.9531
1.1077
1.2423
The step size was = 0.05, adequate for 5-digit accuracy. The solutions for f ( ) are well
behaved and nearly equally spaced. Linear interpolation indicates f ( 0) = 0.1152 will cause
f ( ) = 1.0. A slightly more accurate value is
v*w = ( v w /U ) Re x = 0.4 :
Blowing:
f ( 0 ) = 0.1143
(Ans.)
With this initial value, the velocity profile f ( ) is exactly as shown in Fig. 4-15(a) for v*w = 0.4.
To check this, note that, for Pr = 1, from Eq. (4-60), the local heat transfer should be
Nu x / Re x = f ( 0 ) / 2 0.0810. This checks well with Fig. 4-15(b).
4-15 Derive a relation for local skin friction versus Reynolds number for point-sink flow.
Fig. 4-16, p. 250 of the text. Compare with the Falkner-Skan results if possible.
This is a favorable gradient, U = −K/x, and the velocity profile solution is Eq. (4-89):
u
= f ( ) = 3 tanh 2
+ 1.146 − 2
U
2
Differentiating this, we find that the wall shear is given by
W =
u
y
y =0
=
U K
f ( 0) ,
x v
where f ( 0 ) =
6
tanh (1.146 ) sech 2 (1.146 )
2
= 2/ 3 = 1.1547
Then the skin friction coefficient is
Cf =
2w
U
2
= 2.3094 ( v/K )
1/ 2
=
2.3094
Re x
(Ans.)
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-12-
where the last relation follows from the substitution K = |U|x. It is difficult to compare this result
with the Falkner-Skan solutions because we cannot easily take the limit, as m approaches (–1), of,
say, the listed values in Table 4-2, p. 243. The result does compare well with a comparably strong
favorable gradient, stagnation flow, Eq. (3-160), p. 158.
4-16 Develop a numerical solution for a laminar mixing layer between parallel streams and
compare with Fig. 4-17 for any value of k = ( 22 ) / ( 11 ) .
We must simultaneously solve two Blasius profiles and match them at = 0:
0: f1 + f1f1 = 0;
0: f 2 + f 2f 2 = 0
At = 0:
f1 = f 2 = 0;
f1 = f 2 0;
At = :
f1 = 1.0;
At = −: f 2 = 0
f1 = f 2 k
(Assume)
For simplicity, we take the lower fluid to be at rest far below the interface ( U2 = 0 ) . As a sample
case, we take k = 10, which is plotted in Fig. 4-17(c).
The numerical solution of this problem is an order of magnitude more difficult than the standard
Blasius profile. We must match the (unknown) interface velocity and the (unknown) interface
shear stress, at the same time ensuring that both the upper and lower velocity profiles approach
their asymptotic values of 1.0 and 0.0, respectively. And we must integrate one profile to ( + )
and the other to ( − ) . For k = 10, Fig. 4-17(c) indicates that the interface velocity f ( 0) 0.35,
which we may take as a first guess. Now guess a value of f 2 ( 0 ) and, using f1 ( 0) = f 2 ( 0) 10,
integrate the upper profile and see if the final velocity f1 ( ) = 1.0? If not, try another value of
f 2 ( 0 ) . When the upper velocity finally does approach 1.0, then integrate the lower profile to see
if it approaches f1 ( − ) = 0.0? It probably will not because our initial value for f ( 0 ) was only a
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-13-
guess. Change this guess and try again. Take “infinity” to correspond to a value = 10, which
is well off the chart in Fig. 4-17(c). Tabulate some results as follows:
f ( 0 ) = 0.35: f1 ( +10 ) = 1.0 at f 2 ( 0 ) = 0.12503, but
f 2 ( −10 ) = +0.0199 0
f ( 0 ) = 0.34: f1 ( +10 ) = 1.0 at f 2 ( 0 ) = 0.12623, but
f 2 ( −10 ) = −0.0067 0
Our first guess for f ( 0 ) was very good, so we didn’t have to iterate much. Clearly we are close
to the correct answer, and we may interpolate linearly to obtain
k = 10: f ( 0) 0.3425,
f 2 ( 0) = 0.12593,
f1 (0 ) = 0.39823
(Ans.)
The velocity profiles are in close agreement with those plotted in Fig. 4-17(c).
4-17
Air at 20°C and 1 atm issues from a slot and forms a plane laminar jet. At x = 50 cm,
u max = 20 cm/s. Estimate the jet (a) width; (b) mass flow; and (c) Reynolds number.
For air take = 1.2 kg/m3 and = 1.8E-5 kg/m s. Knowledge of u max and its position allows us
to compute the jet momentum J from Eq. (4-104):
1/3
u max
J2
= 0.2 m/s = 0.4543
x
1/3
J2
= 0.4543
(1.2 ) j (1.8E-5 )( 0.5 )
,
J = 0.00096 kg/s 2
or:
Then the jet width estimate follows from Eq. (4-106):
1/3
x 2 2
b = 21.8
J
1/3
( 0.5)2 (1.8E-5 )2
= 21.8
( 0.00096 )(1.2 )
= 0.090 m
(Ans. a)
= 0.0072 kg/s/m
(Ans. b)
The jet mass flow per unit depth is given by Eq. (4-107):
m = ( 36Jx )
1/3
= 36 ( 0.00096 )(1.2 )(1.8E-5 )( 0.5 )
1/3
The jet Reynolds number could be expressed in three essentially equivalent ways:
Re jet
m
= = 400;
1/3
Jx
or: 2
= 120;
or:
u max b
= 1200
(Ans. c)
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-14-
4-18
Air at 20°C and 1 atm flows at 1 m/s past a slender two-dimensional body of length
L = 30 cm and CD = 0.05 based on ‘plan’ area (bL). At 3 m downstream of the trailing edge,
estimate (a) wake velocity defect; (b) wake thickness; and (c) Reynolds number.
For air take = 1.2 kg/m3 and = 1.8E-5 kg/m s. The body-length Reynolds number is
ReL = UL/ = (1.2)(1.0)( 0.3) / (1.8E-5) = 20,000. For x = 3 m, the centerline wake defect
velocity is computed by applying Eq. (4-112) at y = 0:
1/2
1/2
1/2
1/2
u1 ( y = 0 )
ReL L
20000 0.3
= CD
=
0.05
(
)
= 0.315,
U
16 3.0
16 x
or:
u1 ( y = 0) = u max 0.32 m/s
(Ans. a)
[This is perhaps pushing wake theory a bit, since u is not truly small compared to U.] By analogy
with jet theory, the wake half-thickness could be defined as the point where the defect velocity
drops to 1% of its maximum velocity, from the gaussian profile:
Uy 2
exp −
= 0.01,
4xv
or:
4.605 ( 4xv )
=
U
1/2
or:
y1% = 0.0144 m,
y1%
4.605 ( 4 )( 3.0 )(1.8E-5/1.2 )
=
1.0
Width b = 2y1% 0.029 m
1/2
(Ans. b)
Finally, the wake Reynolds number could be expressed in terms of wake width and the maximum
defect velocity:
Re wake
u max b (1.2 )( 0.32 )( 0.029 )
=
= 600
1.8E-5
(Ans. c)
There are alternate Reynolds numbers, such as m/, similar to jet flow, Eq. (4-107).
4-19 Derive the steady-flow two-dimensional momentum-integral relation, Eq. (4-120 or 121),
including wall suction or blowing, by applying conservation laws to the control volume shown.
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-15-
First apply conservation of mass, starting at the bottom left and working clockwise:
0
0
0
( V n )dA = 0 = − ubdy − Ubd + ubdy + d ubdy − vw b dx
CS
Rearranging this gives a net continuity relation to be used later as a substitution:
U
d d
=
u dy − v w
dx dx 0
In a similar manner, the x-momentum relation, again starting at the bottom left point, is
dp
Fx = pb + p + 2 b d − ( p + dp )( + d ) − w b dx = − u 2b dy − U 2b d
0
+ u b dy + d u 2 b dy − v w ( 0 ) b dx
2
0
0
where the last (vanishing) term is included only to show that there is zero x-momentum coming in
at the wall because u = 0 there. Clean up and drop second order terms dpd, etc:
dp d
d
w = − − u 2dy + U 2
dx dx 0
dx
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-16-
Now eliminate ( d/dx ) using the continuity relation above:
w = −
d
d
dp
u 2dy + U u dy − Uv w −
dx 0
dx 0
dx
This is one common form of the momentum-integral relation [see, e.g., Kays and Crawford (1980),
p. 48.] To put this equation in the form of Eq. (4-121), we use the Bernoulli relation and the
definitions of momentum and displacement thicknesses:
dp
dU
= −U
;
dx
dx
u dy = U ( − *) ;
2
2
u dy = U u dy − U
0
0
0
After these three substitutions, we obtain Eq. (4-121) for steady flow ( /t = 0):
w
U
4-20
2
=
d
1 dU v w
+ ( 2 + *)
−
dx
U dx U
(Ans.)
Solve Eq. (4-121) for flat-plate flow with nonzero blowing, v w 0, with u/U given by
Eq. (4-11) and both and 1/v w proportional to x1/2 . Compare to Fig.4-15(b).
The velocity profile is a simple (and unrealistic for large blowing) parabola
u
y y2
=2 − 2 ,
U
with
2
15
and
w =
2U
For flat-plate flow with dU/dx = 0, substituting these formulas into Eq. (4-121) gives
2v 2 d v w
=
−
U 15 dx U
Now assume (correctly) a square-root relationship, to which the blowing parameter v*w in
Fig. 4-15(b) can be related:
Let
= B x1/ 2
and
v w = K x −1/ 2 ,
whence v*w =
vw
U
Re x =
K
( vU )
(1)
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-17-
Substitution into the momentum integral relation above gives a quadratic equation in B:
B2 −
15K
30v
B−
=0
U
U
Note that the no-blowing ( K = 0) solution is Bo2 = 30/U, or o ( 30x/U ) , as in Eq. (4-14).
Then the solution to the quadratic equation above can be written in the form
1/2
(
B = Bo 1 + 1.875 v*2w
)
1/2
+ (1.875)
1/2
v*w ,
where we have substituted for K = v*w
Bo = ( 30 v/U )
1/2
where
( U )
|noblowing
from Eq. (1) above. Since w = 2U/ and
= Bx1/2 , the skin friction on the plate may be written in the following final form:
Cf =
Cfo
(
)
1 + 1.875v*2 + (1.875 )1/2 v*
w
w
Using the Reynolds analogy for
Pr = 1,
,
Cfo =
0.73
Re x
(Ans.)
this may also be written in the form
Nu x / Re x = Cf /2 = 0.365/ , where denotes the bracketed factor above. The agreement
with Fig. 4.15(c) for suction is excellent, as shown by the following calculated values:
v*w =
Nu x /Re x =
0.0
–1.0
–2.0
–3.0
–4.0
–5.0
0.365
1.12
2.06
3.04
4.03
5.02
Note the good agreement with the suction values in Fig. 4-15. The reason is that the u ( y ) profiles
remain approximately parabolic when suction is applied [see Fig. 4-15(a)]. For blowing, however,
the agreement of the approximate theory is more qualitative:
v*w =
Nu x /Re x =
0.0
0.1
0.2
0.4
0.6
0.8
0.365
0.318
0.278
0.216
0.173
0.142
Blow-off is not predicted, and the friction/heat-transfer estimates are too high. This is because the
blowing profiles become S-shaped and the parabolic formula is not accurate.
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-18-
4-21
Improve Prob. 4-20 by developing a parametric polynomial profile u ( y ) which accounts
for blowing and suction similar to Fig. 4-15(a). Do not solve or integrate for Cf .
We need profiles which develop the S-shape when there is wall blowing [we did OK with the
simple parabola in Prob. 4-20 for suction cases]. At a minimum, we add one more boundary
condition at the wall, obtained from the x-momentum boundary-layer equation: non-zero curvature
when there is blowing or suction:
At y = 0, u = 0, but v w 0, thus: v w
u
2u
|wall = 2 |wall
y
y
(1)
We therefore try a cubic polynomial for the velocity profile:
u
= a + b 2 + c 3 ,
U
= y/
This already satisfies the no-slip condition. We further require that it merge smoothly with the
freestream, plus condition (1) above:
At y = :
u = U, or:
a + b + c = 1
At y = :
u/y = 0, or:
a + 2b + 3c = 0
At y = 0, Condition (1) above:
Re*b = 2c,
Re* = v w /
The parameter Re* is directly related to the “blowing parameter” v*w from Fig. 4-15:
Re* =
vw vw
Const
=
Re x = ( Const ) v*w if
=
U
x
x
Re x
Solving, we obtain a = 6/ ( 4 + Re*) , b = 3Re*/ ( 4 + Re*) , and c = −2 (1 + Re*) / ( 4 + Re*) . The
desired polynomial profile is thus listed and plotted as follows:
u
6
3Re* 2 2 + 2Re* 3
+
−
U 4 + Re*
4 + Re*
4 + Re*
(Ans.)
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-19-
We see below that the profiles are realistic for Re* −2.0, but they do not “blow off”.
4-22
Modify Prob. 4-20 by using the same parabolic profile, Eq. (4-11), but let the wall velocity
v w 0 (suction) be constant. Solve for Cf ( x ) and ( x ) and compare with Fig. 4.21 and
Eq. (2-123) of the text.
This problem begins the same way as Prob. 4-20 on p. 79 of this Manual. The parabolic velocity
profile leads to a differential equation for ( x ) :
2
2 d v w
=
−
U 15 dx U
The difference here is that we assume constant suction velocity, v w 0, not a square-root
variation. We may solve for ( x ) by separating the variables:
x
d
15
( 2 /U ) + ( vw /U ) = 2 dx
0
0
where we have assumed that = 0 at x = 0. Both integrals are easy to evaluate, and the result may
be rearranged as follows:
−
2 v w 15xv w
ln 1 +
=
v w
2
2U
If we rewrite this in terms of the variables plotted in Fig. 4-21, we have a better result:
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-20-
( −vw ) − ln 1 − ( −vw ) = 15 ,
2v
2v
4
where
=
( −vw )
U
(Ans.)
Re x
Meanwhile, the skin-friction estimate comes from differentiation of the parabolic profile:
w =
2U
4
4
, hence Cf
=
CQ ,
U / ( − v w ) /
CQ =
−vw
U
(Ans.)
Thus, with ( −vw ) / known as a function of from the first answer, C f may be computed and
is directly related to the “suction coefficient” ( −v w ) /U. Some numerical values from these integral
estimates may be tabulated as follows:
( −vw ) / =
0.1
0.3
0.5
1.0
1.5
1.8
1.9
2.0
=
0.027
0.083
0.143
0.318
0.570
0.854
1.052
Cf (U/ − v w ) =
40.0
13.33
8.00
4.00
2.67
2.22
2.11
2.0
The thicknesses are a little small, e.g., at
0.6, ( −v w ) / 1.54 whereas Fig. 4.21 indicates
a value of about 2.0. At z = , ( −vw ) / = 2.0, whereas Eq. (2-232) for ‘asymptotic’ (exponential
profile) suction indicates a value of 4.6.
The skin friction estimate is realistic and, as expected, smaller than the CD values
plotted in Fig. 4-21. For example, ReL = 106 and (− v w )/U = 0.001 are equivalent = 1.0, for
which C f from the tabulation above would equal about 0.0022, whereas CD from Fig. 4-21 is
about 0.0025, which is quite reasonable agreement.
4-23 Apply the method of Thwaites (Sect. 4-6.6) to laminar boundary flow on a circular
cylinder. Compute local wall friction and compare with Fig. 4-24(b).
We select the inviscid-flow theory for freestream velocity: U(x) = 2Uo sin(x/a), where a is
cylinder radius and x starts at the front stagnation point. [See Fig. 4.24(a) for the geometry.]
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-21-
The momentum thickness from Thwaites’ method is given by Eq. (4-138):
=
2
x
0.45
U
U = 2Uo sin ( x/a )
U dx,
5
6
0
Thwaites’ factor equals (2 /) ( dU/dx ) , where dU/dx = ( 2Uo /a ) cos ( x/a ) . After substitution
into the momentum-thickness relation above, we obtain, denoting = x/a,
=
0.45 cos
sin
6
sin
5
d, where
0
= 15 8 − cos (3sin
1
4
)
+ 4sin 2 + 8
Then, with known, the wall shear stress is given by Eq. (4-139):
w
Ux)
S() ,
(x)
Finally, we are asked to plot ( Cf /2)
where S = ( + 0.09 )
0.62
( )
( Ux/ ) ,
where Cf /2 = w / U 2 . Putting all this together
from the above formulas results in the following desired expression:
1/2
sin 5
1
Cf Re x = S ( )
2
0.45 F
,
where
F = sin 5 d
(Ans.)
0
where = ( x/a ) is the local angle in radians. Some numerical values from this formula may be
listed as follows for various positions along the cylinder:
=
0°
30°
45°
60°
75°
90°
95°
100°
103.1°
=
0.075
0.072
0.068
0.059
0.041
0.0
–0.025
–0.060
–0.090
0.5Cf Re =
1.195
1.148
1.084
0.984
0.830
0.575
0.444
0.256
0.0
The skin friction values are very close to Terrill’s digital computer results plotted in
Figure 4-24(b). Thwaites predicts separation at = 103.1, quite close to Terrill’s value, 104.5°.
(b) We could also calculate using the measured velocity distribution of Eq. (4-144):
3
5
U U 1.814 ( x /a ) − 0.271( x /a ) − 0.0471( x /a )
2
4
dU /dx (U /a ) 1.814 − 0.813 ( x /a ) − 0.2355 ( x /a )
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-22-
Carrying out Thwaites’ integral (4-138) for ( 2 / ), we can find Thwaites’ parameter ( x ) :
( x) =
2 dU
v dx
=
0.45
(U /U )
6
d (U /U )
d ( x /a )
x /a
5
(U /U ) d ( x /a )
0
The results are shown in the following plot. Separation is at x /a = 1.372 radians = 78.6°.
4-24 Apply Thwaites’ method to the test cases in Table 4-5 of the text. Have each student take
a different case. Compute and plot Cf Re x and compute the separation point.
The cases U = 1 − x and U = sin ( x ) have already been covered, but Table 4-5 still contains
9 other cases to try, with (presumably accurate) separation points listed by two different methods.
We illustrate here with only one case:
Tani (1949 ) : U = 1 − x 2 ;
U5 = 1 − 5x 2 + 10x 4 − 10x 6 + 5x 8 − x10
We have to integrate this for Thwaites method, Eq. (4-138), to obtain and :
1/2
0.45vF)
(
=
U3
and = −
0.9xF
U6
7
5x 3
5x 9 x11
5 10x
where F = x −
+ 2x −
+
−
3
7
9
11
The separation point, = −0.09, occurs at x = 0.268.
The local skin friction is given by
Cf =
2 w
U
2
,
where w
(Ans.)
[see Table 4-5]
U
0.62
S ( ) , and S = ( + 0.09 )
For the present case, this may be rearranged and evaluated as follows:
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-23-
Cf Re x =
(
2S 1 − x 2
)
3/2
( 0.45 G )1/2
,
where G = 1 −
5x 2
10x 6 5x 8 x10 F
+ 2x 4 −
+
−
=
3
7
9
11 x
This wall friction distribution is plotted below and is very close to the exact (digital computer)
solution for this problem.
4-25 Consider a two-dimensional flat-walled diffuser as in Fig. P4-25. Assume incompressible
flow with a one-dimensional freestream velocity U ( x ) and entrance velocity U o ( x ) . The
entrance height is W and the constant depth into the paper is b.
Fig. P4-25
Using Thwaites’ method, find an expression for the angle at which separation will occur at
x = L. What is the value of if L = 1.5 W ?
Solution: By one-dimensional continuity, the average velocity U ( x ) is given by
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-24-
U obU = U b (W + 2 x tan ) ,
or:
U=
Uo
1 + ( 2 x tan ) /W
Thwaites’ method requires us to compute the position where ( x ) = −0.09, where
x
dU 2 dU 0.45 5
=
=
U dx
dx v
dx U 6 0
with U ( x ) given by the expression above.
As it happens, this velocity U ( x ) is the same as Görtler’s third case in Table 4-5, (1 + x ) (“x”
here is not the same quantity). Thus we can save ourselves a laborious calculation and use Görtler’s
Thwaites-method result:
−1
= −0.09 ( separation)
if
2 x tan
0.079W
0.158, or xsep = Lsep
W
tan sep
If L = 1.5W , then tan sep = 0.079/1.5 = 0.0527,
or
sep 3.0.
Ans
Laminar flow has much less resistance to separation than turbulent flow.
4-26 Apply the explicit model, Sect. 4-7.1, to laminar boundary layer flow over a circular
cylinder, with U ( x ) given by either Eq. (4-143) or (4-144). Compute and plot Cf .
In the explicit model, we step forward directly to u m +1,n using Eq. (4-146), then compute
v m +1,n from Eq. (4-148). Step sizes are limited by Eqs. (4-147):
x u min y 2 / ( 2 )
y 2 /v max |
We can use either the inviscid or the actual freestream. U ( x ) - see p.
of the text. Actual
numbers for u, v are immaterial, and C f will scale, when completed, with Re x . For example,
we could take Uo = 1, D = 1, and v = 1E-5, in arbitrary units, simulating a Reynolds number
U o D/v = 1E5. The computed results will, if stability limits are satisfied, be very close to those
plotted in Fig. 4-24(b), p. 270, for either potential flow or actual flow. The computed separation
point will be somewhat inaccurate unless very small step sizes x are used as we approach
separation.
4-27
Repeat Prob. 4-26 using the implicit finite-difference model of Section 4-7.2.
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-25-
Here we use Eq. (4-149) instead of (4-146) and solve simultaneously for u m +1,n either by iteration
or the TDMA [Sect. 4-7.2.1], after which v m +1,n are again computed by Eq. (4-148). There are no
stability limitations, but for numerical accuracy the step size y should be no larger than about
5% of the local boundary layer thickness, and the marching step x should be less than about 1°
of arc, say, x* = 0.01. Computed skin friction will be in good agreement with Fig. 4-24(b) except
possibly near separation, where smaller x* should be used as we approach this singular point.
4-28 Investigate the Crank-Nicolson (1947) finite-difference method as implemented by, e.g.,
Blottner (1970). What are its advantages and disadvantages in boundary layers?
The Crank-Nicolson method splits the difference between an implicit and an explicit model for the
(second-order) viscous term in the x-momentum equation:
2u
y2
=
1 u m,n +1 − 2u m,n + u m,n −1 1 u m+1,n +1 − 2u m+1,n + u m+1,n −1
+
2
2
y2
y2
where the last term is implicit, involving velocities at the next station ( m + 1) . We thus have to
solve simultaneously for these new values. However, since the scheme is only fifty-per-cent
implicit, so to speak, the iteration for the new values of u m +1,n will be faster and the TDMA will
not usually be necessary. The Crank-Nicolson model is unconditionally stable for linear systems,
e.g. one- or two-dimensional linear diffusion. Eqs. (3-246) or (3-250), and it is generally quite
stable for weakly nonlinear systems such as the laminar boundary layer equations. See Blottner
(1970) for further details.
4-29 Apply the explicit finite-difference method of Sect. 4-7.1 to one of the laminar-flow test
cases in Table 4-5. Compute CfRex and compare with Table 4-5 at separation.
The explicit model works well for all these cases in Table 4-5, as long as the stability limits of Eqs.
(4-147) are satisfied. As an example, we show computations for the same case used in Prob. 4-24
on page 85 of this Manual: U = 1 − x 2 (Tani’s first case). The computed skin friction is compared
below with the Thwaites’ results from Prob. 4-24. The step size x should be smaller near
separation to match the “exact” xsep = 0.271.
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-26-
4-30
Sherman (1990) gives a CFD solution for laminar flow due to a freestream U o approaching
a parabolic cylinder, as in Fig. P4-30. The cylinder surface is defined by y /R = ( 2 x /R )
1/ 2
(
R is the cylinder nose radius. The arc length s along the surface is defined by ds = R 1 + 2
(
where = y /R. From potential theory, the surface velocity is U = U o / 1 + 2
)
1/2
, where
)
1/ 2
d ,
. (a) Show that
the surface velocity approaches the stream velocity U o as one moves up the surface. (b) Using
Thwaites’ method, estimate the distance s /R along the surface where ( U ) is within 10% of
the flat plate value of 0.22.
Fig. P4-30
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-27-
Solution: First plot the surface velocity. It is easiest to use = y /R as the abscissa, as follows:
We see that U U o , or zero-pressure-gradient flat-plate behavior, at about y /R 6. Ans. (a)
(
(b) Apply Thwaites’ method by first evaluating s /R = 1 + 2
(
2 /v = 0.45/U 6
)
1/2
) U 5ds and thence = ( 2 /v ) ( dU /ds ) and plot
d , after which we compute
w /U
We see that w /U is within 10% of 0.22 at about s /R 9, or y /R 4.
( + 0.09)
0.62
.
Ans. (b)
4-31 Apply the Smith-Spalding thermal-integral method of Eq. (4-161) for Pr = 1 to the
Howarth distribution, U = Uo (1 − x/L ) , for constant wall temperature. Compute and plot local
Nusselt number and also the Reynolds analogy factor, Cf /2C h .
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-28-
The skin friction for Thwaites’ method can be computed from the results of the sample Howarth
analysis on pp. 268–269 of the text, including Eq. (4-142):
1/ 2
( x ) = −0.075 (1 − x/L )
−6
− 1 ;
Cf
x/L
Re x = 2S ( )
−
5
0.075 (1 − x/L ) − (1 − x/L )
where S = ( + 0.09) . The computed values of Cf Re x vary from 0.664 at x = 0 to zero at
separation, x/L = 0.123, as shown in Fig. 4-26, p. ••• of the text. Meanwhile, the heat transfer is
estimated from the Smith-Spalding formula, Eq. (4-161), for Pr = 1:
0.62
0.330 ( 2.95 )
Nu L
0.332
=
=
1/2
1/2
x
ReL
(1 − x/L )−2.95 − 1
2.95
1.95 dx
(1 − x/L )
(1 − x/L ) L
0
1/2
(Ans.)
Since Nu = Ch RePr, we may convert this, for Pr = 1, to Ch Re x by the substitution
Ch
1/ 2
Nu L Uo x
Re x =
=
Re x
ReL U L
Nu x
where
U/Uo = 1 −
x
L
(Ans.)
The values of Cf Re x are plotted below and decrease slightly, from 0.332 at x = 0 to 0.311 at
x/L = 0.123 (separation). The Reynolds analogy factor ( Cf /2Ch ) is also shown below.
4-32
Modify Prob. 4-31 by using an explicit or implicit finite-difference method.
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-29-
Prob. 4-31, a Howarth velocity distribution, U = Uo (1 − x/L ) with an isothermal wall, was
programmed by the implicit finite-difference model of Eq. (4-165), p. 282 of the text. The
following numerical values were used, for simplicity: Uo = 1, L = 1, Tw − T = 1, x = 0.001,
(
)
y = 0.00003, and = 1E-5. Computation was carried out until x/L 0.12, where separation
occurred. Two hundred y-steps were used, and, after the velocities and temperatures were
computed, the wall heat flux q w = −k ( T/y )w was evaluated and non-dimensionalized in the
form ( hx/k ) /
( Ux/ ).
The results are plotted below for Pr = 1.
We see that the Nusselt numbers predicted by the simple Smith/Spalding quadrature method from
Prob. 4-31 are somewhat high. The finite-difference results are more realistic - the value of 0.22
at separation agrees with the Falkner-Skan result on p. 248.
4-33 Apply the Smith-Spalding method, Eq. (4-161) with Pr = 0.72, to flow past an isothermal
circular cylinder, Eq. (4-167), and compare results with Fig. 4-29.
By substituting U ( x ) from Eq. (4-167) into the Smith-Spalding method, Eq. (4-1671), and using
the constants a and b from Table 4-6 for Pr = 0.72, we obtain the following estimate for local
Nusselt number around the cylinder:
Nu R
0.296 1.82z − 0.4z3
ReR
(
)
−2.88 z
(1.82z − 0.4z
0
)
3 1.88
dz
−1/2
, z = x/R
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-30-
No closed form integral is known to this writer, but the integration is readily performed by, e.g.,
the Runge-Kutta Subroutine of Appendix C. However, the values plotted in Fig. 4-29 are based on
diameter D as the length scale. Hence we need to multiply the above results by 2. That is,
Nu D
=
ReD
( 2 ) NuReR
R
The results are nearly identical to the dotted line plotted as “Smith-Spalding” in Fig. 4-29. Some
particular values may be tabulated as follows (for air, Pr = 0.72 ):
=
Nu/ Re =
0°
10°
20°
30°
40°
50°
60°
70°
80°
0.958
0.953
0.936
0.907
0.865
0.811
0.743
0.659
0.558
4-34 In Eq. (4-144) for Rea = 9500, U /U 1.814 x /a near the front of the cylinder. For air at
20°C and 1 atm, this corresponds approximately to a = 5 cm and U = 2.85 m/s. Using the
Falkner-Skan theory, Tables 4-2 and 4-3, and a temperature difference (Tw − T ) = 12C, estimate
(a) the momentum thickness in mm; and (b) the heat transfer rate in W/m2 at the front of this
cylinder.
Solution: For air at 20°C and 1 atm, take = 1.205 kg/m3 , = 1.81E-5 kg/( m-s ) , Pr = 0.71, and
k = 0.026 W/( m-K ) . Calculate v = / = 1.50E-5 m2 /s.
(a) Use Table 4-2 for m = = 1 (stagnation flow) to estimate the momentum thickness.
* = 0.29235 =
( m + 1)U = (1 + 1)(1.814U /a )
2v x
2v x
va
= 0.29235
Solve for 0.29235
1.814U
(1.50E − 5m2 /s ) ( 0.05m ) = 0.00011m = 0.11mm
1.814 ( 2.85 m /s )
An alternate computation by Thwaites’ method would yield 0.10 mm.
Ans. (a)
(b) For the heat transfer, use Table 4-3 or Eq. (4-79) for m = = 1 (stagnation flow). Both
formulations give the same numerical result:
Nu x =
hx
0.4 1.814U x x
= 0.57 Pr 0.4 Re1/2
x = 0.57 ( 0.71)
, or:
k
a
v
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-31-
h = ( 0.497 ) k
Then
1.814 ( 2.85m /s )
1.814U
W
= ( 0.497 ) 0.026
av
m−K
( 0.05m ) (1.5E − 5m
2
/s
)
= 34
W
W
qw = h (Tw − T ) = 34 2 (12 K ) 410 2
m
m K
W
m2 K
Ans. (b)
Note that h is proportional to a −1/2 , that is, small cylinders have higher heat transfer.
4-35
(
)
3 3
For flat-plate flow with a wall temperature difference Tw − To = To 1 − x /L , find the
point x at which the local wall heat transfer changes sign.
This is a short exercise in understanding of the superposition technique of Sect. 4-8.4. If T is a
polynomial, the local heat transfer is given by Eqs. (4-172) and (4-173). In this particular case, all
polynomial coefficients are zero except a o and a 3 . Thus, from Eq. (4-173), we obtain
(
)
q w = ( const ) To 1 − 2.2091 x 3 /L3 ,
or:
4-36
qw = 0
at
x/L = 2.2091−1/3 = 0.768
(Ans.)
Modify Prob. 4-35 by using an explicit or implicit finite-difference method.
We may verify the integral-theory method of Prob. 4-35 (taken from Sect. 4-8.4) by programming
the problem of Blasius flow with a cubic wall temperature difference. This writer used the implicit
method of Eq. (4-165) with the following conditions: U = 1,
L = 1, Te = 0,
Tw = 1 − x3 , v = 1E-5, x = 0.005, and y = 0.0001. The computed results are plotted below in
the form of local Nu x / Re x , which would equal 0.332 everywhere if the wall temperature were
constant. We see that the integral theory is very accurate.
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-32-
4-37
Analyze laminar flow in the entrance region between parallel plates ( 2H ) apart, if the
entrance velocity is uniform, u = Uo . Approximate the flow as a potential ‘core’, where
u = U ( x ) , plus parabolic-shaped boundary layers on each wall [Sparrow (1955)].
The geometry is shown on the previous page. Boundary layers grow out from each wall:
u ( x, y )
y y2
=2 − 2 ,
U(x)
= (x)
where y is measured from the wall. The boundary layer merges at y = with a flat inviscid core
of velocity U ( x ) . At any x, the volume flow must match the entrance flow:
Continuity:
2
2HUo = 2 ( H − ) U + 2 U ,
3
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-33-
or:
= 3H (1 − Uo /U )
(1)
Meanwhile, the x-momentum equation is given by boundary layer theory, Eq. (4-120):
d
dU
w / = u ( U − u ) dy +
( U − u ) dy
dx 0
dx 0
Substituting for u ( x, y ) from the parabolic approximation above, we obtain
2U 2 2 d 9
dU
=
U
+ U
15
dx 15
dx
Now we eliminate using Eq. (1) above and rewrite our differential equation for U ( x ) in terms
of the dimensionless variables = U/Uo and x* = x/H:
16 7 d
10
=
= constant
9 − + 2
2
dx*
3U
H
o
The initial condition (at the entrance) is = 1.0 at x* = 0. The solution is
x/ ( 2H )
U
Uo
3 U
= 9
− 16 ln
−
7
−
2
U o ( 2H ) / 40 U o
U
Uo
(Ans.)
Note that the left hand side matches the grouping ( x/D) / ReD plotted for “exact” entrance flow in
Fig. 4-33, p. 288 of the text. This (approximate) equation of Sparrow (1955) is plotted on the next
page of this Manual and may be compared with Fig. 4-33. The predicted “entrance length”,
x = Xe , occurs when U = 1.5 Uo :
Xe /2H
0.02594
2HUo /
(Ans.)
Figure 4-33 indicates that this estimate may be low - at least, based on the pressure drop parameter
K in Fig. 4-33, fully-developed flow seems to be reached at a somewhat higher value,
( Xe /2H) /Re2H = 0.04.
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-34-
4-38 For potential freestream flow past a sphere, use the Rott-Crabtree method, Eq. (4-191), to
predict the separation point for a laminar boundary layer.
The Rott-Crabtree method consists of the quadrature (4-191) for a given U ( x ) and ro ( x ) :
2 =
0.45
ro2 U6
x
ro U dx,
2
5
x
U ( x ) = 1.5Uo sin ,
a
for
0
x
ro = a sin
a
after which the shape factor = ( /)(dU/dx). Combining all the above relations, we obtain the
laminar shape factor for potential flow past a sphere:
2
=
0.45cos
sin
8
sin
7
d, where
0
= 35 16 − cos (5sin
1
6
)
+ 6sin 4 + 8sin 2 + 16
We may evaluate this relation for various angles and tabulate the results as follows:
=
30°
60°
75°
90°
95°
100°
103°
103.6°
=
0.0545
0.0459
0.0330
0.0
–0.0220
–0.0553
–0.0836
–0.0900
Thus, separation is predicted at = 103.6.
(Ans.)
If these values of are used to compute wall shear stress from Eq. (4-139), the result is the solid
curve for ‘potential flow’ in Fig. 4-35, page 295 of the text.
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-35-
4-39 In the spirit of Eq. (4-146) for two-dimensional flow, develop an explicit finite difference
model for the thick axisymmetric-flow momentum relation, Eq. (4-195b). Use the same mesh as
shown in Fig. 4-25, with yn being the radial coordinate, but do not analyze the axisymmetric
continuity Eq. (4-195a). Do the parameters and still appear? Note that there is no pressure
gradient, U m+1 = U m .
Solution: We are to model the momentum equation (4-195b):
u
u
u
v
u
+v
=
( a + y ) , where
x
y a + y y
y
a
is the body radius,
After substituting in the finite-difference approximations for the derivatives and solving
laboriously for the explicit value of um +1,n , the writer found the following result:
2a + yn + yn+1
4a + 2 yn + yn+1 + yn−1
um+1,n
− um,n+1 + 1 −
um , n
2 a + 2 yn
2 a + 2 yn
2a + yn + yn−1
+
+ um,n−1
2 a + 2 yn
(
)
2
Here = vx / um,n y and = vm,n x / ( 2um,n y ) have the same meanings as in Eq. (4-146).
Compare the above formulation with Eq. (4-146) for two-dimensional momentum, which does not
contain the complicated parenthesis terms which multiply in the above equation.
4-40 Air at 20°C and 1 atm issues from a circular hole to form a round laminar jet. At 20 cm
downstream of the hole the maximum jet velocity is 35 cm/s. Estimate (a) the “1%” jet thickness;
(b) the jet mass flow; and (c) an appropriate Reynolds number.
3
2
For air at 20°C take = 1.2 kg/m , = 1.8E-6 kg/m-s, and v = 1.5E-5 m /s. With u max and
known, we may compute jet momentum J at this section from Eq. (4-207):
u max =
3J
, or
8x
J=
x
8
kg
m
kg-m
1.8E-5
( 0.2 m ) 0.35 = 1.06 E-5 2
3
m-s
s
s
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-36-
With J known, we evaluate the (dimensionless) constant C from Eq. (4-206):
1/2
3J
C=
2
16v
1/2
3 (1.06 E-5)
=
16 (1.2 )(1.5E-5)2
= 48.3
The “1%” jet thickness occurs when the profile shape function in Eq. (4-207) equals 0.01:
C22
1 +
4
−2
= 0.01,
C
= 3,
2
or
or:
=
r
6
= 1% = 0.124
48.3
x
Jet thickness b = 2r1% = 2 ( 0.124)( 0.2 m) = 0.050 m
(Ans. a)
The jet thus spreads outward at a half-angle of tan −1 ( 0.124 ) 7.1. The jet mass flow at this
station is given by Eq. (4-208):
m = 8x = 8 (1.8E-5)( 0.2) = 9.05 E-5 kg/s
(Ans. b)
Finally the local jet Reynolds number could be expressed in at least two ways:
Re =
J
v2
= 39100;
or:
Re =
u max b ( 0.35)( 0.050 )
=
= 1160
v
1.5E-5
(Ans. c)
4-41 Air at 20°C and 1 atm flows at 1 m/s past a slender body of revolution, L = 15 cm, whose
drag coefficient is 0.008 based on area L2 . Assuming laminar flow at 3 m downstream of the
trailing edge, estimate (a) the maximum wake velocity defect; (b) the “l%” wake thickness; and
(c) the wake-thickness Reynolds number.
For air at 20°C take = 1.2 kg/m , = 1.8E-6 kg/m-s, and = 1.5E-5 m2 /s. With x, L, and
CD known, the maximum wake defect velocity follows from Eq. (4-211):
3
U2 L2
(1.0) ( 0.15) = 0.159 m/s = 16 cm/s
= CD o = ( 0.008)
8x
8 (1.5E-5)( 3.0 )
2
u max
2
(Ans. a)
The 1% thickness occurs when the Gaussian profile in Eq. (4-210) equals 0.01:
U r2
exp − o = 0.01, or:
4x
or:
Wake thickness
4.605 ( 4 )( 3.0 )(1.5E-5 )
=
(1.0 )
1/ 2
r1%
b = 2r1% = 0.0576 m = 58 cm
= 0.0288 m,
(Ans. b)
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-37-
Finally, an appropriate wake Reynolds number is
Re =
u max b ( 0.159 )( 0.0576 )
=
= 610
1.5E-5
(Ans. c)
4-42 Given a streamlined airfoil as pictured in Fig. P4-42a, axial velocity profiles may be
measured at both the upstream and downstream sections labeled 1 and 2 using a streamlinebounded control volume (i.e., one that forms a streamtube). Assuming equal pressure around the
control volume and symmetry with respect to the midsection plane, determine (a) the half height
of the upstream station, H , and (b) the drag coefficient on this airfoil if the vertical dimension at
the downstream station is Y = φc , where c is the cord. The downstream velocity is given by
1
1
u = U 1 − cos y / Y . Hint: You may take advantage of symmetry by selecting a control
2
2
volume as pictured in Fig. P4-42b.
Solution:
u
* = 1 − dy
0
U
1
2
y
dy
2
* = 1 − cos
0
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-38-
0
0
* = dy −
y
1
cos
dy
2
2
1 2 y
= −
sin
2 0
2
*
* = − sin − 0
2
* = − sin
2
* = −
1 ( − 1)
= 1 − =
* =
( −)
(Ans.)
4-43 Consider the steady, incompressible, and two-dimensional flow past a circular cylinder of
diameter d, as pictured in Fig. P4-43. The upstream velocity U may be taken to be uniform while
the downstream velocity may be set to be
y
; +0 y +
u ( y) sin
= 2
U
1;
+ y +
where both U and d are known. Calculate (a) δ and (b) the drag coefficient CD.
Solution:
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-39-
Consider the conservation of mass in integral form:
( )
d
dv +CS v n dA = 0
dt CV
For steady-state conditions,
d
dv = 0
dt CV
(v n)dA = 0
CS
a.
(v n)dA + (v n)dA + (v n)dA + (v n)dA = 0
AB
BC
CD
(eq-1)
AD
(v n)dA = (v n)dA = 0
BC
AD
Since streamlines are the upper boundary and lower boundary of the control volume,
along surfaces BC and AD, the velocity vector v and normal vector to the surface are
perpendicular such that v n =| v || n | cos90 = 0 .
(v n )dA = − U (d 1)
AB
Here v and n are in opposite directions.
v n =| v | +n∣ cos180 = −v = −U
(v n )dA =
CD
0
y
u ( y )(dy 1) = U sin
dy
0
2
y
− cos 2
= U
2
0
2
( v n )dA = U ( − ( 0 − 1) )
CD
=
2 U
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-40-
Therefore, (Eq-1) becomes
− Ud +
Ud =
=
2 U
=0
2 U
2
d
(Ans)
b. Linear momentum equation in integral form is
F =
d
vdv + CS v(v n)dA
dt CV
Under steady-state conditions,
d
vdv = 0.
dt CV
ΣF = v (v n )dA
CS
Assuming only drag force is the external force acting,
D = v(v n )dA
CS
D=
(v n)dA + (v n)dA + (v n)dA + (v n)dA
AB
BC
CD
AD
For similar reasoning made during conservation of mass equation simplification
(v n)dA = (v n)dA = 0
BC
AD
v (v n )dA = − U ( d 1)
2
AB
= − U 2 d
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-41-
v (v n)dA =
0
(u( y)) 2 (dy 1)
CD
y
= 0 U 2 sin 2 dy
2
1
y
= U 2 0 1 − cos 2
dy
2
2
=
y
U 2 0 1 − cos
dy
2
y
sin
= U2 y −
2
0
U 2
=
2
U 2
D = − U d +
2
2
= − U 2 d +
2
2
dU 2
= U 2 d −1 +
4
2
Drag on the body = U d 1 −
4
2 U 2 d 1 −
D
4
=
Drag coefficient =
1
U 2 d
U 2 ( d 1)
2
4
= 4 −
d
(Ans)
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-42-
4-44
Plot the rotating-disk solution, Table 3-5, in the form of a hodograph of ‘streamwise’ flow
( r− v ) versus ‘transverse’ flow vr . Compare with Fig. 4-42(c).
The rotating disk is a fully three-dimensional (laminar) boundary layer. As seen in Fig. 3-28,
p. 156 of the text, the ‘secondary’ flow is radially outward, and the ‘main’ flow is circumferential.
To view this flow relative to the wall, we subtract out the moving-wall velocity ( r ) . Thus we
plot dimensionless streamwise velocity, 1 − G ( z*) , versus secondary or transverse velocity,
F ( z*) . We could do this from the values listed in Table 3-5, p. 159. Or, if your PC, like
mine, has a plotting program, you could just re-run the Runge-Kutta solution,
Eqs. (3-186) of the text, and make a running plot of Y ( 3) versus 1 − Y ( 5 ) . In this latter manner
the hodograph below was prepared. It clearly resembles ordinary unidirectional skewing, as in
Figs. 4-42(a,c) on p. 311 of the text.
4-45 Verify that the free-convection similarity variables of Eqs. (4-255) do indeed lead to the
coupled ordinary differential equations of Eqs. (3-256).
This is a standard exercise in variable-substitution, using the chain rule. It is very good. I think,
for students to perform such an exercise at least once in their graduate training, since similarity
solutions do crop up in many areas of applied mechanics.
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-43-
4-46 Develop a numerical solution to the vertical-plate free-convection relations, Eqs. (4-256)
and (4-257), for a particular Prandtl number not listed in Table 4-10.
This is a challenging numerical-integration exercise, since there are two unknown initial
conditions, f ( 0 ) and Θ ( 0 ) . The table is a big help, as otherwise we would have to make very
crude guesses of these two values.
As an example, let us take the case Pr = 8. The table indicates that the proper values of
f ( 0) and Θ ( 0) are about +0.44 and –1.09, respectively. Equations (4-256) may be modelled by
the Runge-Kutta Subroutine of Appendix C, letting Y1 = f , Y2 = f , Y3 = f ,
and Y5 = Θ. The proper FORTRAN relations are then
F (1)
=
−3* Y ( 3) * Y (1) + 2* Y ( 2 ) * Y ( 2 ) − Y ( 5 )
F( 2)
=
Y (1)
F ( 3)
=
Y ( 2)
F( 4)
=
−3* Pr* Y ( 3) * Y ( 4 )
F ( 5)
=
Y ( 4)
Y4 = Θ,
with known initial conditions Y ( 2) = Y ( 3) = 0 and Y ( 5) = 1. Let us guess that f ( 0 ) is
somewhere between 0.43 and 0.45, and that ( 0 ) is between –1.08 and –1.10. From Fig. 4-50,
page 326 of the text, we see that “infinity” is at about = 6. Based on these four guesses, we run
off four sets of solutions and check to see whether the values of f ( ) or Y2 ( 6 ) and ( )
or Y5 ( 6 ) are near zero. We list the following results:
f ( 0)
( 0 )
f ( 6)
( 6)
0.43
–1.08
–0.0466
–0.0055
0.43
–1.10
+0.0756
–0.0203
0.45
–1.08
–0.1373
+0.0327
0.45
–1.10
–0.0103
+0.0174
We may interpolate linearly to find that f ( 6) ( 6 ) 0 and f ( 0 ) = 0.438 and ( 0) = −1.094.
We then repeat the procedure using a narrower range of guesses. The process converges this second
time to an accurate estimate of the solution:
Pr = 8.0:
f ( 0 ) = 0.4386
( 0 ) = −1.0957
Nu/Gr1/ 4 = 0.775
(Ans.)
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-44-
The solutions for velocity ( f ) and temperature ( ) are plotted on the next page.
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-45-
4-47 A vertical isothermal plate 40 cm high and 30 cm wide is immersed in air at 20°C and 1 atm.
Each side of the plate is to dissipate 100 W of heat to the air. Calculate (a) the wall temperature (in
°C ) and (b) the Grashof number. Is the flow laminar? Hint: Assuming an average wall-fluid (film)
temperature of 100°C, evaluate the necessary thermophysical properties, and repeat until the
properties no longer change.
With T unknown, we cannot compute the Grashof number, but let us assume the flow is laminar
and proceed (checking later for transition). Make a guess that the average fluid temperature is
50°C—from Table A-2, k 0.0276 W/m-K and = 1.8E − 5m 2 /s .
For air, Pr 0.71, the laminar flow relation (4-260) in the text predicts
Nu x = 0.354 Grx1/4 , or Nu L,mean =
4
Nu x = 0.472 GrL1/4
3
For air, an ideal gas, = 1/T ( K ) . So we know everything except T, and the above relation
becomes, for qw = 100 W ,
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-46-
Nu L =
qw L
(100 W )( 0.4 m )
=
A w kT ( 0.4 )( 0.3) m 2 ( 0.0276 W/m-K ) T
1/4
g
= 0.472 2 L3T
v
(
)(
2
−1
9.81 m/s 1/323 K
= 0.472
2
1.8E-5 m 2 /s
(
)
)
1/4
3
( 0.4 m )
Solve this for T5/4 517, or T 148C. A better estimate of average fluid temperature is thus
(1/2) ( T + Tw ) 94C (367 K ) , and we may correct the fluid properties to k 0.0306 W/m-K,
v 2.23E-5 m2 /s, = 1/367 K−1. Repeating the calculation gives little change, and we may take
the desired result to be
T 150C,
or:
Tw = 170 °C
(Ans.)
3
2
9
Finally, check the Grashof number, gL T/ = 5.2E8, which is less than 10 ; hence the flow
is presumed to be laminar and our estimate is accurate.
4-48 A horizontal pipe of diameter 5 cm is immersed in air at 20°C and 1 atm. If the cylinder
surface is at 300°C, estimate the heat loss in watts per meter of pipe length.
Estimate the fluid properties at the average temperature between the wall and the ambient
air: T ( avg ) = (1/2)( 20 + 300 ) 160 C = 433 K. From Table A-2, p. 575 of the text,
−1
k 0.0350 W/m-K and 2.96E-5 m2 /s. Also, = 1/T = 1/433 K . The Rayleigh number of
the cylinder is thus
Ra D =
( 9.81)(1/433)
3
3
( 0.05 ) ( 300 − 20 )( 0.71) = 643000
D
T
Pr
=
2
2
( 2.96E-5 )
g
The relevant heat-transfer correlation for free convection to horizontal cylinders is Eq. (4-265),
p. of the text:
0.387 ( 643000 )
1/6
Nu1/2
L
or:
= 0.60 +
9/16 8/27
1 + ( 0.559/0.71)
= 3.58,
or:
Nu L = 12.85 =
hD
,
k
h = 12.85 ( 0.035 W/m-K ) / ( 0.05 m ) = 8.99 W/m 2 -K
The total heat loss from cylinder to air is thus estimated as
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-47-
(
)
q w = h ( DL ) T = 8.99 W/m 2 -K ( 0.05 m )(1.0 )( 300 − 20 C ) = 400 W / m
(Ans.)
4-49 A model two-dimensional airfoil has the following theoretical potential-flow surface
velocities on its upper surface at a small angle of attack:
x /C
0.0
0.025
0.05
0.1
0.2
0.3
0.4
0.6
0.8
1.0
V /U
0.0
0.97
1.23
1.28
1.29
1.29
1.24
1.14
0.99
0.82
The stream velocity U is varied to set the Reynolds number. The chord length C is 30 cm. The
fluid is air at 20°C and l atm. Assume that x is a good approximation to the arc length along the
upper surface. Using any laminar-boundary-layer method of your choice, find the predicted
separation point, if any, for (a) ReC = 1E6; and (b) ReC = 4E5.
Solution: For air at 20°C and 1 atm, = 1.205 kg/m and = 1.81E-5 kg/(m-s). We need to
integrate V 5 versus x, but the data are sparse and highly variable, as shown below:
3
We either need a terrific curve-fit, or we can just plow along. Thwaites’ method requires
2
v
0.45
V6
x
5
V dx and then find where =
0
2 dV
= −0.09 (separation)
v dx
The writer plowed along and produced the reasonably accurate plot of versus x on the next page.
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-48-
Separation is predicted by Thwaites’ method at approximately x/C 0.45. Ans. (a, b)
The problem called for calculated separation “for (a) ReC = 1E6; and (b) ReC = 4E5. ”
However, for a laminar boundary layer, the separation point is independent of ReC !
The reader is invited to prove this as Problem 4-51.
4-50 Air, at about 1 atm and 20°C, flows through a 12 cm square duct at 0.4 m3 /s as in
Fig. P4-50. Two hundred thin flat plates of chord-length 1 cm are stretched across the duct at
random positions. Their wakes do not interfere with each other. How much additional pressure
drop do these plates contribute to the duct flow loss?
Fig. P4-50
Solution: For air at 20°C and 1 atm, = 1.205 kg/m and = 1.81E-5 kg/(m-s). First find the
average velocity through the duct from one-dimensional continuity:
3
V=
Q 0.4 m3 /s
m
=
=
27.8
A ( 0.12m )2
s
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-49-
Check the Reynolds number: ReDh = (1.205)( 27.8)( 0.12)(1.81E-5) 220,000. Thus the duct flow
is turbulent, and we will assume (why not?) that each tiny plate sees this average approach velocity
V = 27.8 m/s. The plate length Reynolds number is only Re L = 18, 500, low enough that the flow
over the plates is laminar and not tripped by the stream turbulence. Calculate the drag of one plate
from the Blasius formula:
Re L = 18500, CD =
F1 plate = CD
1.328
1.328
=
= 0.00977
Re L
18500
2
2
1.205
V bL ( 2sides ) = 0.00977
( 27.8) ( 0.12m )( 0.01m ) (2) = 0.0109 N
2
2
The total drag of 200 plates is thus approximately Ftotal = ( 200)( 0.0109N ) = 2.18 N. A freebody
of the plates shows that this force can only be balanced by an overall pressure drop across the array
of plates:
p plates =
4-51
Ftotal
2.18 N
=
150 Pa
Aduct ( 0.12m )2
Ans.
Nondimensionalize Thwaites’ method, using U o and L as reference values, for laminar
boundary layer flow with an external potential flow distribution U ( x ) . Show that the predicted
separation point is independent of the Reynolds number U o L /v. Apply this conclusion to discuss
the expected results for Prob. 4-49.
Solution: For typing ease, let V = U /U o and let = x /L. Then Thwaites’ method calls for the
calculation of the Thwaites parameter ( x ) , with separation as follows:
0.45 x
= 2 /v ( dU /dx ) =
U
6
dU
5
U
dx
= −0.09
dx
0
(separation)
Now introduce the dimensionless variables and rewrite:
0.45 5 5
d (U oV )
0.45 5 dV
= 6 6 U oV Ld
= −0.09 = 6 V d
U oV 0
d ( L )
V 0
d
The final expression depends only on the dimensionless velocity distribution and is entirely
independent of U o , L, and v, that is, it is independent of Reynolds Number! See Problem 4-49
for a numerical example.
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-50-
4-52
A conical diffuser of initial radius R expands at a uniform angle , as in Fig. P4-52.
The flow enters at uniform velocity U o . Assuming a one-dimensional freestream, use any laminar
boundary layer method of your choosing to estimate the angle for which flow separation occurs
at x = 2 R.
Fig. P4-52
Solution: The writer is going to choose the Rott-Crabtree axisymmetric modification, Equation (4191), of Thwaites’ integral method. The cone radius is ro ( x ) , and we estimate the velocity U ( x )
from one-dimensional continuity and compute momentum thickness:
x
2 0.45 2 5
R2
2 tan
2 6 ro U dx, U = U o 2 , ro = R (1 + Kx ) , K =
v ro U 0
R
ro
Substitute for U and ro in the 2 equation and integrate, with the result
2 0.45 (1 + Kx )
−7
=
1 − (1 + Kx ) ,
v
7U o K
10
where K = 2 tan
R
Now evaluate the velocity gradient:
−2 K U o
dU
d
−2
=
U o (1 + Kx ) =
(1 + Kx )3
dx dx
Then the Thwaites’ parameter ( x ) is given by
2 dU
0.9
2 tan
7
=
=−
1 + Kx ) − 1 , K =
(
v dx
7
R
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-51-
Separation occurs at = −0.09, or (1 + Kx )7 = 1.7, or Kx = 0.0788. If, as specified for this
problem, x = 2R, then Kx = 4 tan = 0.0788, tan = 0.0197, θ = 1.13°.
Ans.
This is smaller than the (laminar) result for the flat diffuser in Prob. 4-25, because the conical
geometry causes U ( x ) to drop about twice as fast.
4-53 Show that the point-sink boundary-layer solution of Eqs. (4-86) and (4-89) may be
interpreted through Thwaites’ method as a constant value of Thwaites’ parameter, = 9/80.
(a) How does this value of compare to that for plane stagnation flow? (b) Use Eq. (4-89) to
derive an expression for skin friction C f as a function of Rex.
Solution: From Fig. 4-16, sink flow is a strong acceleration, U = ( − K /x ) . Evaluate the Thwaites’
parameter ( x ) . We have to integrate from x to
to avoid the x = 0 singularity:
0.45 5 dU 0.45x6 − K 5 d
=
=
U dx
=
( − K /x )
6
5
v dx U 6 0
dx K
x x
dx
2 dU
0.45 x6 K 5 K 0.45 9
=
=
= = 0.1125 = constant
6
4 2
4
80
K 4 x x
Ans.
The comparable Thwaites estimate for plane stagnation flow is stag 0.075, or 33% less.
4-54 Show that the exact Falkner-Skan solutions for two-dimensional stagnation flow are
equivalent to Thwaites-method parameters of = 0.0855, S = 0.360, and H = 2.216. How do
these values compare with the use of Thwaites’ method for the freestream U = Bx ?
Solution: For stagnation flow, = m = 1, the dimensionless wall shear, displacement thickness,
and momentum thickness are given in Table 4-2: fo = 1.23259, * = 0.64790, and * = 0.29235.
We calculate the desired Thwaites-type parameters if U = Bx, dU /dx = B:
* =
m +1 U
v
= 0.29235, U = Bx, m = 1, = 0.29235
,
2 vx
B
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-52-
=
Thus
w =
2 dU
2
= ( 0.29235) = 0.0855
v dx
Ans.
u
m +1 U
|w = U fo
, S = w = (1.23259 )( 0.29235 ) = 0.360 Ans.
y
2 vx
U
H=
* * 0.64790
=
=
= 2.216
* 0.29235
Ans.
Use Thwaites’ method, Eq. (4-138), to estimate for stagnation flow, U = Bx, dU /dx = B:
x
x
2 0.45 5
0.45
0.45
2 dU 0.45
6 U dx = 6 6 B5 x5 dx =
, =
=
= 0.075
v
6B
v dx
6
U 0
B x 0
Ans.
By comparison, from Thwaites’ correlation, Table 4-4, for 0.075 (12% low from FalknerSkan), S 0.327 (9% low), and H 2.36 (6% high).
4-55 When a jet is formed normal to a wall, it is called a wall jet, as shown in Fig. P4-55. The
jet origin can be thought of as a source of momentum, and further down the wall the velocity
profiles are similar. Glauert (1956) showed that both plane and axisymmetric wall jets satisfy the
same ordinary differential equation for laminar flow:
2
df
+f
+ 2 = 0
3
2
d
d
d
d3 f
d2 f
Fig. P4-55
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-53-
where y, f , f u, and f . The boundary layer thickness is proportional to x3/4 .
(a) Verify Glauert’s choice of boundary conditions: f ( 0) = f ( 0) = f ( ) = 0. (b) To enforce the
normalized condition f ( ) = 1, Glauert showed that f ( 0) = 2/9. For this value, using any
convenient method, numerically integrate the similarity equation out to = 8 and plot both
f ( ) and f ( ).
Solution: Read Glauert’s paper to see how this interesting similarity solution was deduced. There
is no freestream, the jet discharges into still air, so f u has to be zero at both the wall and at
infinity. Letting f ( 0) = 0 simply sets the wall stream function to zero. A numerical solution for
f ( 0) = 2/9 (Glauert’s choice to make f ( ) = 1) is plotted as follows:
4-56 Consider the steady two-dimensional flow past a circular cylinder of diameter d, as
pictured in Fig. P4-56. The upstream velocity U may be taken to be uniform while the
downstream velocity is allowed to vary from U⧸2 to U according to the following symmetric
profile:
1
1
y
1
;
0
y
u ( y ) 1 + sin
u
(
)
1 + sin ; 0 1
= 2
= 2
or
, where
2
2
U
U
1;
y
1
1;
both U and d are known. Calculate (a) δ as a function of d and (b) the drag coefficient CD. Here,
ξ = y⧸δ.
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-54-
Solution:
*
a. = 0 1 −
u( y)
dy
U
1
y
= 0 dy − 0 1 + sin
dy
2
2
1
y
= [ y ]0 − 0 dy + 0 sin
dy
2
2
2
= − +
2
y
− cos 2
0
2
= − +
2
0
− cos 2 − − cos 2
2
= − + [0 + 1]
2
2
= − +
2
+ 4
+4
=
=
2
2
+4
2 d
=
=
2
+2
b. cD =
FD
=
1
Av 2 (2 + )
2
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-55-
4-57 Repeat the previous problem using the following polynomial profile at the downstream
section of the control volume:
− y −
1;
− −1
1;
2
u( y) 1 y
u ( ) 1
= 1 + 2 ; − y + or
= (1 + 2 ) ; −1 +1
U
U
2
2
+1 +
1;
1;
+ y +
Solution:
1; − y −
u ( y ) 1 y2
= 1 + 2 ; − y +
Given
U
2
1; + y +
From RTT,
dN
dt
=
system
ndv + CS n(v ˆ )dA
t CV
a. Using conservation of mass,
N = m, n = 1
0 = 0 + 2 0 udyw − 2 vd = 0
d =
0
udy
U
=
U
2U
0
y2
1
+
dy
2
1 1 y3
= [ y]0 + 2
2
3 0
1
1 4
= + =
2
3 2 3
2
=
3
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-56-
=
3d
2
(Ans)
b. Using conservation of linear momentum,
N = mv, n = v
Fx = 0 + 2 u 2 dy − 2 U 2 d
0
u 2 y 2 2
Fx = 2
1 + 2 dy − u 2 d
0 4
u2
y2 y4
= 2
1
+
2
+ 4 dy − u 2 d
2
0
4
u2
3 5
= 2 + 2 2 + 4 − u 2 d
3 5
4
1 2 1
= 2 u 2 1 + + − d
4 3 5
3
1
FD = − u 2 wd = cD u 2 wd
5
2
cD = 1.2
(Ans)
4-58 Consider the steady, incompressible, and two-dimensional flow past an ellipsoidal cylinder
with unit depth W in the z direction. The upstream velocity, U, may be taken to be uniform while
U
the downstream velocity varies from
until it reaches U according to the following profile:
2
− y −Y
1;
u( y) 1 y 4
= 1 + 4 ; −Y y +Y
U
2 Y
1;
+Y y +
− −1
1;
u ( ) 1
= (1 + 4 ) ; −1 +1
or
U
2
+1 +
1;
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-57-
where U is known and Y defines the edge of the boundary-layer region. Calculate (a) Y in terms
of H and (b) the drag coefficient CD using the major axis of the ellipse as the characteristic
length. You may use half of the domain shown below for simplicity.
Solution:
1; − y −
u ( y ) 1 y4
= 1 + 4 ; − y +
Given:
U
2
1; + y +
*
a. =
u( y)
1−
dy
0
u
1 1 y4
= 1 − +
dy
4
0
2 2
4
1 y
1
= d y − dy +
dy
4
0
0 2
0 2
1
1 1
= + [ − 0] + 4 5
0
2
2 5
= −
5
= − +
4
2 10
2 10
+
* = H =
10 − 5 + 3
=
10
5
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-58-
=
5H
3
(Ans)
b. cD =
2 FD
20 H
cD =
2
u w
276
(Ans)
4-59 The Blasius equation can be expressed as
f
+ f =0
f
where the primes denote differentiation with respect to = y / U / (2 x) . Given that the
Blasius constant is
= f (0) , show that an integrodifferential form of the solution consists of
f ( ) = exp − f (t )dt
0
Solution:
The Blasius solution is given by
f "'
+ f =0
f"
(eq-1)
The initial condition is given by,
f "( 0) = and =
y
u
2vx
From equation (1),
f "'
=−f
f"
Integrating both sides from 0 to
0
f
dt
=
0 − f (t )dt
f
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-59-
0
0
ln f " ( t ) = − f ( t )dt
ln f " ( ) − ln f " ( 0 ) = − f ( t )dt
0
f ( )
ln
= − f ( t )dt
0
f (0)
"
But f ( 0) =
f " ( )
ln
= − 0 f ( t )dt
Taking exponents on both sides,
(
f " ( )
= exp − 0 f ( t )dt
(
f " ( ) = exp − f ( t )dt
0
)
)
(Ans)
4-60 The Blasius solution arises in the context of laminar, incompressible, constant property,
boundary-layer flow over a flat plate, where the freestream velocity above the plate is U. As
usual, the boundary-layer equations for flow over a flat plate reduce to
u v
x + y = 0
2
u u + v u = u
x
y
y 2
with
u ( x, 0) = 0, u ( x, ) = U ,
and
u
y
=0
y =
An equally common form of the Blasius similarity variable does not carry the arbitrary “2” factor
under the radical. More specifically, the Blasius equation can be rederived using the following
transformations:
=
U
y, = xU f ,
x
and
u ( x, y ) df ( )
=
U
d
Show that the corresponding Blasius equation becomes f +
1
ff = 0 with the same boundary
2
conditions as before.
Given:
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-60-
Boundary layer equations:
u
du
dv
d2y
+ v = 2
dx
dy
dy
u v
+ =0
x y
(eq-1) =
(eq-2)
Boundary conditions are as follows:
At y = 0 , u = v = 0
At y = , u = v
The local free stream velocity u(x) at section x is scale factor for u, because the dimensionless
u(x) varies in x direction and is unity with all values of y at all sections.
The scale factor for y, denoted by g(x), is proportional to local boundary layer thickness so that
“u” itself varies between 0 and unity.
Now, if Blausius flow, it is possible to identify g(x) with boundary layer thickness , we know
that
=
L
1
Re L
In terms of “x,”
x
=
1
=
v x
u
v
u
u
y
That is,
= F
= f ( )
v
u
u
Where = y and
=
vx
v
u
v
y
y = x dy = x d
u
u
ux
u
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-61-
The stream function can be obtained in terms of velocity component
= udy = u F ( )
vx
d
u
= u vx F ( ) d
= uvx f ( ) + D
Also,
where D = constant
(Ans)
F ( ) d = F ( ) , and the constant of integration is zero if the stem function at solid
surface is equal to zero.
Now the velocity component and their derivates
u=
du d d
= u f −1 ( )
=
,
dy d dy
−d
1 1
1 y
−1
v=
= − v v
f + x f ( ) −
2 x ( )
2n v
dx
x
u
du
d
1
"
"
Therefore,
= u f ( )
= u f ( ) −
2
dx
dx
y 1
vx x
u
dy
d
1
= u f " ( )
= u f " ( )
v
du
dy
x
u
u
du
= v f (n)
dy
vx
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-62-
−
2
d v
= u
dy 2
v
1
f ( )
vx
x
u
2u v2
=
f ( n)
2
y
vx
(eq-3)
Substitute (eq-3) in (eq-1)
u2
u2
u2
f ( ) f ( ) + 0 f ( x) − f ( ) f (n) = f ( )
2 x
2x
x
u2
1 u2
=−
f ( ) f ( ) = f ( x)
2 x
x
2 f ( ) + f (n) f ( ) = 0
Where f ( ) =
4-61
f ( )d + C
u
u d + C
(Ans)
and =
y
vx
u
Use the Runge–Kutta method with shooting to solve the Blasius equation f + ff = 0
and thus capture the flat plate boundary layer using = y U / (2 x). Use the far-field value of
= 6 and a step size of = 0.02.
(a) Select f (0) = {0.3,0.5} as the first two guesses for the unknown initial condition. Report the
number of additional guesses required to converge on f () 1 within an absolute tolerance
of ò = 10−5.
(b) Plot f , f , and f versus . Tabulate f , f , and f for increments of = 0.2 (i.e., every
tenth data point). Feel free to check your results against Table 4-3.
(c) Use the 99% definition of the boundary-layer thickness to estimate the value of = that
corresponds to the “edge” of the boundary layer.
(d) Compute the parameters:
*
Rex ,
Rex , and C f Rex
x
x
Solution:
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-63-
A finite difference formulation of the Blasius equation may be implemented using a Runge–
Kutta solver with a shooting method. To begin, we use the following finite-difference
discretization scheme,
fi +2 = 3 fi +1 − 3 fi + fi −1 − fi ( fi +1 − 2 fi + fi −1 )
with
f − 2 fi + fi −1
f − 3 fi +1 + 3 fi − fi −1
f −f
fi = i +1 i
fi= i +1
and fi= i + 2
2
3
(a) If we take f (0) = {0.3,0.4} as the first two initial guesses, we find that both of these values
undershoot our desired convergence on f () 1. However, if we use a value of f (0) = 0.5 ,
we overshoot the initial “Blasius” slope needed for convergence. The use of f (0) = 0.5 to
bracket the initial guesses is therefore more practical. In fact, it is possible to determine how
many additional guesses are needed over an interval [a, b] for the bisection method to converge.
For the bisection method, the number of iterations leading to convergence corresponds to:
1
b−a
b − a ln ( b − a ) − ln(2 )
n = log 2
ln
=
=
ln 2
2 ln 2 2
To obtain a tolerance of = 10−5 when converging on f () 1 , we determine that our guesses
for f (0) must actually fall within a tolerance of = 10−7 . Thus, using a bracketing interval of
[0.4,0.5] , the total number of guesses needed to reach convergence may be estimated to be
0.5 − 0.4
= 3 + 18.93 22 guesses Ans. (a)
Total Number of Guesses = 3 + n = 3 + log 2
2 10−7
However, this number can be subjective as it depends on the numerical scheme used by each
individual.
(
)
(b) To check our results, we may compare the plots and tabulated data to Figure 4-10(a) and
Table 4-3. Our results agree with the tabulated values within 2 significant digits, as the solution
method employed implements a far coarser discretization than that produced in Table 4-3.
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-64-
0.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
2.2
2.4
2.6
2.8
3.0
3.2
3.4
3.6
3.8
4.0
4.2
4.4
4.6
4.8
5.0
5.2
5.4
5.6
5.8
6.0
f ( )
0.00000
0.00675
0.03204
0.07600
0.13843
0.21895
0.31694
0.43151
0.56148
0.70541
0.86165
1.02844
1.20395
1.38646
1.57436
1.76628
1.96105
2.15780
2.35585
2.55472
2.75409
2.95375
3.15358
3.35349
3.55345
3.75343
3.95342
4.15341
4.35341
4.55341
4.75340
f ( )
0.00000
0.08434
0.17789
0.27081
0.36231
0.45128
0.53641
0.61629
0.68952
0.75494
0.81170
0.85942
0.89820
0.92861
0.95158
0.96828
0.97995
0.98778
0.99282
0.99594
0.99779
0.99884
0.99941
0.99971
0.99986
0.99993
0.99997
0.99998
0.99999
0.99999
0.99999
f ( )
0.00000
0.46843
0.46688
0.46220
0.45281
0.43736
0.41501
0.38552
0.34945
0.30813
0.26358
0.21820
0.17443
0.13443
0.09974
0.07115
0.04876
0.03209
0.02026
0.01227
0.00713
0.00397
0.00212
0.00108
0.00053
0.00025
0.00011
0.00005
0.00002
0.00001
0.00000
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-65-
5
f
f'
f"
f (η) at large η
4.5
4
3.5
3
2.5
2
1.5
1
0.5
0
0
1
2
3
η
4
5
6
Ans. (b)
(c) Being limited to a discretization of = 0.02 , the closest value we are able to determine is
3.46
Ans. (c)
This value can be obtained numerically and checked graphically. The actual value to 16
significant digits is = 3.471886880405967 , which is given in Table 4-3; however, the relative
coarseness of the discretization size used here leads to a value of either 3.46 or 3.48, with 3.46
being closer to the true value of 3.47188688…
(d) Based on the present numerical calculations, U / (2 x) can be determined from the intercept of a 45˚ line, which is asymptotic to the curve of f ( ) at large . Using the slopeintercept for the asymptotic line and a data pair ( , f ) , one may deduce U / (2 x) . For
example, using (6.000,4.793) and a slope of m = 1 for a 45˚ line, we can write, for y = mx + b ,
6 = 4.793 + b and so b = 1.207
Then using the expression given by Eq. (4-68), we have
U
U
1.207 or
1.706
2 x
x
This leaves us with
1.706
Rex 1.706
or
Ans. (d)
x
x
Rex
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-66-
Using the present discretization scheme, and using f (0) = U / (2 x) 0.4686 as determined
from Part (a) –and consistently with Eq. (4-69),– we can perform a similar manipulation for the
momentum thickness. We get
Rex 0.4686 2
Rex 0.662
Ans. (d)
x
x
Finally, using Eq. (4-70), we are able to retrieve
0.662
Cf =
or C f Rex 0.662
Ans. (d)
x
Rex
These results generally agree with those obtained through Eqs. (4-68)–(4-70) and Table 4-1.
Their differences from the ‘true values’ serve as an important reminder of the appreciable
sensitivity of the solution to the discretization size. The so-called classic values associated with
the Blasius problem, given in Table 4-1, were originally obtained using a relatively coarse grid.
and so
4-62 Under standard temperature and pressure conditions, consider a laminar, incompressible
flow of air over a flat plate, as pictured in Fig. P4-1, where the velocity may be taken to be
linear, specifically, u( y) = Uy / . Using von Kármán’s momentum-integral approach, evaluate
(a) the fundamental boundary-layer properties, δ, δ*, θ and τw; (b) the drag force per unit width
( FD / b) as well as the drag coefficient, CD = 2FD / ( U 2 Lb ) . You may use a constant
freestream velocity, U = 10 m/s, b = 1 m, and L = 1 m.
Solution:
a. Given equation of velocity profile is
u( y) =
Uy
u( y) y
=
U
According to von Kármán momentum-integral equation,
d
=
2
U
dx
Momentum thickness
u
u
= 1 − dy
0 U
U
y
y
= 1 − dy
0
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-67-
=
0
y y 2
− dy
y 2 y3
= − 2 = − =
2 3 0 2 3 6
=
(Ans)
6
du
d Uy
U
w = = =
dy y =0
dy y =0
w
d
=
U 2 dx
U 1 d
=
u 2 6 dx
6 d
=
u
dx
6
U dx = d
6
2
x2 =
Ux
2
2 =
12
x2
( Re x )
2
12
Re x
x
x
2
=
=
(Ans)
3.4641
Re x
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-68-
6 2 2
x =
ux
2
2 =
12 x 12 8.90 10−4 1
=
u
1000 10
= 1.03344 10−3 m = 1.03344mm
=
6
=
1.03344
= 0.1722 mm
6
Disp. Thickness ( * ) is calculated as follows:
u
y
* = 1 − dy = 1 − dy
0
0
U
y2
= y − = − =
2 0 2 2
* =
2
* =
1.03344
= 0.51672 mm
2
w =
U 10 8.9 10−4
=
= 8.612 N / m 2
−3
1.03344 10
b. Drag force per unit width can be calculated as follows:
FD = wb dx
LV
FD
U L 10 8.9 10−4 1
=
dx =
=
= 8.612 N / m
b 0
1.03344 10−3
Drag coefficient can be calculated as follows:
cD =
2 FD
2 8.612
=
= 0.0001722
v 2 Lb 1000 102 1
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-69-
4-63 Under standard temperature and pressure conditions, consider a laminar, incompressible
flow over a flat plate where the velocity may be taken to be of the Pohlhausen type, specifically,
u / U = c0 + c1 + c2 2 + c3 3 + c4 4 ; y / . Evaluate (a) the fundamental boundary-layer
properties, δ, δ*, θ, and τw; (b) the drag force per unit width
( FD / b)
as well as the drag
coefficient, CD = 2 FD / ( U 2 Lb ) . Assume a uniform freestream velocity with no pressure
gradient.
a. For Karman–Pohlhausen approximation,
u
y
= C0 + c1 + c2 2 + c3 3 + c4 4 ; =
U
For boundary conditions,
i.
u(x, 0)=0
ii.
u( x, s) = u( x)
iii.
iv.
v.
u
( x, s ) = 0
y
2u
u ( x) d (u )
( x,0) = −
2
y
dx
2
u
( x, ) = 0
y 2
Karman–Pohlhausen approximation got the following result:
0 = C0
− = 2C2
1 = C0 + C1 + C2 + C3 + C4
0 = C1 + 2C2 + 3C3 + 4C4
0 = 2C2 + 6C3 + 12C4
So, C0 = 0
C1 = 2 +
C2 =
6
−
2
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-70-
C3 = −2 +
C4 = 1 −
2
6
And rewrite the equations as follows:
u
3
= 1 − (1 + )(1 − ) + s (1 − )3
U
6
u
= C5 ( ) + C6 ( ) , where −12 ( x) 12
U
1
*
So, = 1 −
0
3
= −
10 120
u
d
U
(from K–P approximation)
3 2C
= + 2
10 120
Similarly,
37
2
−
−
315 945 9072
=
37 2C2 4C22
=
+
−
315 945 9072
And =
=
U
2+
6
2C2
U
2−
Here, the answer is written in terms of C2.
6
(Ans)
Here are the results obtained by Pohlhansen approximation.
0.686
=
x
Re
U = constant,
du
=0
dx
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-71-
=
37
and
315
And =
* =
1.75
x
Re
1
0.686
U 2
2
Re
So, we have been asked
= nominal boundary thickness
= 0.99U
(Ans)
Here, U is the flow velocity.
* = displacement boundary thickness
* =
1.75 x
Re
(Ans)
Re = Reynolds number
= momentum boundary thickness
=
0.686 x
Re
(Ans)
And stress w is calculated as follows:
w =
1
0.686
U 2
2
Re
(Ans)
b. Drag force per unit width can be calculated as follows:
Let L be length of plate and b be its width. So, FD = CD
So,
FD
1
0.686
= CD U 2
L
b
2
Re
CD =
2 FD
U 2 Lb
CD =
0.686
Re
1
0.686
U 2
Lb
2
Re
(Ans)
(Ans)
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-72-
4-64 (a) Consider steady, incompressible, laminar flow over a flat plate at 20°C, 1 atm, and 0.1
m/s. Does water or air produce a thicker boundary layer at the trailing edge? Show your work
and any assumptions that you need to make.
(b) Consider the attached two-dimensional laminar motion over two streamlined airfoils as
pictured below.
If the only difference between the two airfoils is that Airfoil # 1 has twice the length of Airfoil #
2, what is the relation between their boundary-layer thicknesses at the trailing edge? You may
treat the two airfoils as two-dimensional flat plates. You may also resort to any suitable laminar
correlation.
Solution:
a. Given data is
At 20⁰C, 1 atm
vx = 0.1m/s
Let x = 1 m, where x is the length of the plate.
Given fluid is
Air: air = 1.214 kg/m3
air = 1.812 10−5 N-s/m 2
Water: water = 998 kg/m3
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-73-
water = 1.002 10−3 N-s/m 2
Let the velocity profile be linear, i.e.,
x
v
So, air = 3.46
air
u
y
= .
v s
1.812 10−5 1
= 3.46
1.214 0.1
air = 0.4224 m
(Ans)
Now for water,
Water = 3.46
1.002 10−3 1
998 0.1
water = 0.1096 m
So,
(Ans)
ain
0.4224
=
water 0.1096
air
= 3.85401
water
air = 3.86 water
(Ans)
b. First streamlined airfoil length = L
Second streamlined airfoil length = 2L
We know that
=c
x
v
i.e., x
1
x
= 1
2
x2
1
1
=
2
2
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-74-
2 = 21
(Ans)
4-65 For nearly a century, the boundary-layer problem for flow over a flat plate, which was first
reduced to a third-order nonlinear differential equation by Blasius (1908), could only be
solved numerically or through the use of infinite series. In seeking practical
alternatives, several polynomial approximations have been attempted, and these are
mainly attributed to Pohlhausen (1921). Because the polynomial profiles are intended
to capture the flow behavior inside the boundary layer only, the profiles are expressed
as piecewise solutions of the form:
2
3
4
y
y
y
y
u c0 + c + c2 + c3 + c4 ; 0 y
=
U
y
1;
To ensure smooth merging between the velocity in the boundary layer ( u ) and the outer,
freestream velocity (U ) , several boundary conditions have been suggested by Pohlhausen
(1921). These are:
(1) No slip condition at the wall: u ( x,0 ) = 0 .
(2) Matching the far-field velocity at the boundary-layer edge: u ( x, ) = U .
(3) Smooth merging with the far field, where the shear vanishes:
(4) Axial momentum balance at the wall:
2u
y 2
layer equation, and where the pressure gradient
=
y =0
u
y
= 0.
y −b
dp
, which is recovered from the boundarydx
dp
dU
in conformance with Euler’s far= − U
dx
dx
field equation.
(5) No further variations in the slope of the velocity at the boundary-layer edge (zero curvature):
2u
= 0 Note that in the case of a zero pressure gradient, which corresponds to the Blasius
y 2 y =
problem, condition # 4 reduces to
2u
y 2
= 0 . For this particular case, determine which of the
y =0
five boundary conditions (1 to 5) are satisfied by the following Pohlhausen profiles:
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-75-
u
y y2
=2 − 2
U
u 3y
y3
=
− 3
(b) Third-order, cubic polynomial:
U 2 2
(a) Second-order, quadratic polynomial:
(c) Fourth-order, quartic polynomial:
u
y
y3 y 4
= 2 −2 3 + 4
U
.
Hint: Although condition #5 is true as y → , it proves to be inaccurate at the edge of the
boundary layer, where the slope is still changing. Fortuitously, this condition only
affects the quartic Pohlhausen polynomial, which explains why it tends to be less
precise in predicting boundary-layer characteristics than its lower-order counterparts
(cf. Table 4-1).
Solution: The given profile is
2
3
u
y
y
y
y
= c0 + c1 + c2 + c3 + c4
U
4
The boundary conditions are as follows:
(i) y = 0, u = 0
(ii) y = , u = U
(iii) y = ,
u
=0
y
2u
(iv) y = 0, 2 = 0
y
2u
(v) y = , 2 = 0
y
Conditions (iv) and (v) are valid when the pressure gradient is zero, i.e.,
P
= 0.
x
The given Pohlhausen profile is
u
y y2
= 2 − 2 (Second-order, quadratic polynomial)
(a)
U
Therefore, we have
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-76-
u
y
y
= c0 + c1 + c2
U
2
Applying boundary conditions, we have
(i)
y = 0, u = 0
0 = c0 + c1 ( 0) + c2 ( 0)
c0 = 0
(1)
(ii)
y = 6, u = U
1 = 0 + c1 ( / ) + c2 ( / )2
1 = c1 + c2
(2)
(iii)
y =,
u
=0
y
u 1 c1
y
= + 2c2 2
y U
0=
c1
+ 2c2
0=
c1
+2
y
2
c
0 = 1 + 2c2 2
c2
c1 + 2c2 = 0
(3)
Using Eq (2) and Eq (3), we obtain
c1 = 2, c2 = −1
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-77-
Therefore, the profile is
u
y y
= 2 −
U
2
This profile satisfies the first three boundary conditions.
P
0.
x
In this case,
(b)
(Ans. a)
u 3y
y3
=
− 3 (Third-order, cubic polynomial)
U 2 2
Therefore, we have
2
u
y
y
y
= c0 + c1 + c2 + c3
U
3
Applying boundary conditions, we have
(i)
y = 0, u = 0
0 = c0 + c1 ( 0) + c2 ( 0) + c3 ( 0)
c0 = 0
(ii)
(4)
y = , u = U
1 = 0 + c1 ( / ) + c2 ( / )2 + c3 ( / )3
1 = c1 + c2 + c3
(5)
(iii)
y =,
u
=0
y
u 1
c1
y
y2
=
+
2
c
+
3
c
2
3
2
3
y U
0=
c1
+ 2c2
2
+
3
c
3
2
3
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-78-
0=
c1
+2
c2
+3
c3
c1 + 2c2 + 3c3 = 0
(6)
(iv)
y = 0,
1
U
2u
=0
y 2
2u
c3 y
c2
2 = 0+2 2 +6 3
y
0=2
c2
2
c2 = 0
+0
(7)
Using equations (4) and (5), we obtain
3
−1
c1 = , c3 =
2
2
Therefore, the profile is
u
3 y 1 y
= −
U 2 2
3
This profile satisfies the first four boundary conditions.
(c)
(Ans. b)
u
y
y3 y 4
= 2 − 2 3 + 4 (Fourth-order, quartic polynomial)
U
Therefore, we have
2
3
u
y
y
y
y
= c0 + c1 + c2 + c3 + c4
U
4
Applying boundary conditions, we have
(i)
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-79-
y = 0, u = 0
c0 = 0
(ii)
(8)
y = , u = U
1 = c1 + c2 + c3 + c4
(9)
(iii)
y =,
u
=0
y
u 1
c1
y
y2
y3
= 0 + + 2c2 2 + 3c3 3 + 4c4 4
y U
0=
c1
+2
c2
+3
c3
+4
c4
c1 + 2c2 + 3c3 + 4c4 = 0
(10)
(iv)
y = 0,
1
U
2u
=0
y 2
2u
c3 y
c2
c4 y 2
=
0
+
2
+
6
+
12
2
2
3
4
y
0=2
c2
2
c2 = 0
+0+0
(11)
(v)
y =,
1
U
2u
=0
y 2
2u
c3 y
c2
c4 y 2
=
2
+
6
+
12
2
2
3
4
y
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-80-
0=2
2
c2
2
c2
+6
+6
c3
2
2
c3
+ 12
3
+ 12
c4
2
c4 2
4
=0
c2 + 3c3 + 6c4 = 0
Using equation (11), we have
c3 + 2c4 = 0
(12)
Using equations (9), (10), and (12), we obtain
c1 = 1, c3 = −2, c4 = 1
Therefore, the profile is
3
u
y
y y
= − 2 +
U
4
This profile satisfies all five boundary conditions.
(Ans. c)
4-66 Consider the quartic Pohlhausen profile for the laminar, incompressible flow over a flat
plate with a norzero pressure gradient:
u
y
= c0 + c1 + c2 2 + c3 3 + c4 4 ; 0 1 where =
U
(a) Apply Pohlhausen’s five boundary conditions to show that the unknown coefficients must be
1
1
1
1
c0 = 0, c1 = 2 + Λ, c2 = − Λ, c3 = Λ − 2, and c4 = 1 − Λ .
6
2
6
2
2 dp 2 dU
where Λ = −
=
U dx
dx
Here, Λ represents the Pohlhausen pressure gradient parameter. Simplify your solution into the
form:
1
u / U = 2 − 2 3 + 4 + Λ (1 − )3
6
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-81-
(b) For what value of Λ will flow separation occur? Recall that separation occurs when
u / y = 0 at y = 0 .
(c) Plot u / U versus for Λ = −12, −6,0, +6, +12 .
Solution:
u
= c0 + c1 + c2 2 + c3 3 + c4 4
U
( x, y ) =
y
The boundary conditions are as follows:
(i) u ( x,0 ) = 0
(ii) u ( x, ) = u ( x )
(iii)
u
( x, ) = 0
y
(iv)
−u ( x ) du ( x )
2u
x, 0 ) =
2 (
y
U
dx
(v)
2u
( x, ) = 0
y 2
The Kármán–Pohlhausen approximation yields the following results:
0 = c1
−Λ = 2c2
1 = c1 + c2 + c3 + c4 + c0
0 = c1 + 2c2 + 3c3 + 4c4
0 = 2c2 + 6c3 + 12c4
c0 = 0
Λ
6
Λ
c2 = −
2
Λ
c3 = −2 +
2
c1 = 2 +
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-82-
c4 = 1 −
Λ
6
The profile is rewritten as follows:
u ( y)
U
u ( y)
U
= 1 − (1 + )(1 − ) +
3
Λ
3
(1 − )
6
(Ans. a)
= f ( ) +ΛG ( ) , where −12 Λ ( x ) 12
Therefore,
0
1
= 1 −
*
u
U
Λ
3 2c3
3
d = −
= +
10 120
10 120
Similarly, we have
37 2c3
37
4c32
Λ
Λ2
−
−
=
+
−
315 945 9072
315 945 9072
=
and
w = u
U Λ
U 2c
2+ = u 2−
8
6
8
6
Here are the results obtained by Pohlhausen approximation.
2 = 0.4 x
ux
U
0.686
x
=
Re
U = constant
dU
=0
dx
=
37
315
* =
1.75 x
Re
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-83-
=
1
0.686
U 2
2
Re
δ = nominal boundary thickness = 0.99U
δ* = displacement boundary thickness =
= momentum thickness =
w = shear stress =
1.75x
Re
0.686
x
Re
1
0.686
U 2
2
Re
Let L be the length of the plate and b be its width.
1
0.686
U 2
Lb
2
Re
FD
1
0.686
= CD U 2
L
b
2
Re
FD = CD
0.686
1
2 U 2
Lb
2
2 FD
Re
= 0.686
CD =
=
2
2
U Lb
U Lb
Re
4-67
(Ans.)
Consider the quartic Pohlhausen profile for the laminar, incompressible flow over a flat
plate at zero incidence and with no pressure gradient. Using dimensionless variables,
we have:
F ( ) =
y
u 2 − 2 3 + 4 ; 0 1
where =
=
U 1;
1
(a) Show that the dimensionless displacement thickness, momentum thickness, and shape factor
lead to:
* 1 u
=
= 1−
0 U
*
(b)
u
*
d = 0.3 = =
0U
1
u
1 −
U
*
and
H
=
= 2.554
d
=
0.1175
Show that the Kármán–Pohlhausen momentum-integral equation reduces to
w
d U
dF
= U 2 *
=
s; s
= dimensionless velocity slope at the wall.
dx
d −0
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-84-
(c) After separating variables and integrating from ( 0 ) = 0 to ( x ) , show that the
boundary-layer thickness can be written as a function of the local Reynolds number,
Rex = Ux / , specifically,
x
=
a
; a = 5.836
Rex
(d) Recalling that the friction coefficient is C f = 2 w / ( U 2 ) = 2 *d / dx , and that
= ax1/2 / U / , show that
d ( a / 2 )
=
and C f =
dx
Rex
b
; b = a * = 0.6855
Rex
(e) Recalling that the displacement thickness can be deduced from / x = / x , show that
*
*
c
; c = a * = 1.751
Rex
=
x
*
(f) Recalling that the momentum thickness can be deduced from / x = * / x , show that
x
=
d
; d = a * = 0.6855
Rex
v
u
= − 0y dy , integrate the axial profile to show that the transverse
U
x U
velocity can be written as
(g) Using continuity,
v 2 3 4 4 5 ( a / 2 ) 2 3 4 4 5 2.9178
= − +
= − +
U
2
5 Rex
2
5 Rex
(h) Show that the maximum transverse velocity occurs at the edge of the boundary layer where it
can be obtained from
v
U
=
max
v
U
=
y −
( 3a / 20 ) = 0.8753
Rex
Rex
(i) Verify that the errors in predicting / x and C f are 17 and 3.2 percent relative to the classic
Blasius results.
Solution: The given quartic Pohlhausen profile is
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-85-
F ( ) =
u 2 − 2 3 + 4 ; 0 1
=
U 1;
1
where
=
y
0 1 0
* = 1 −
0
u
U
y
1 0 y
3
4
y
y y
3
4
d
=
1
−
2
−
2
+
d
=
1
−
2
−
2
+
0
0 dy
(
)
3
4
y
y
y
2 4
5
* = dy − 2 dy − 2 3 dy + 4 dy = − − 3 + 4
0
0
0
5
0
2
3
* = − − + =
2 5 10
* 3
=
= 0.3
10
*
= = 0.3
*
=
=
0
0
2
u
u u
u
1 − dy =0 − dy
U U
U U
2
u
u
dy − dy
0 U
U
y
3
y3 y 4
=
− 2 − 2 3 + 4 dy
10 0
2
= 0.1175
* =
1 u
u
= 1 − d = 0.1175
0
U U
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-86-
H=
* *
0.3
=
=
= 2.554
0.1175
(Ans. a)
(b) According to the von Karman equation, we have
w
=2
x
1
U 2
2
2
w2 = 2
U
dx
w
d
u
=
. where w = u
2
U
dx
y
y =0
u
2 y 2 y3 y 4
=
− 3 + 4
U
du 1
2 6 y2 3y4
=
− 3 + 4
dy V
du
1
2
du 2U
=
=
dy
dy y =0 V
w = u
2
=
2U
2uU d
=
U2 dx
2u
d
=
U dx
2udx = u d
2 x = U d
2 x = U *
2
2
where
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-87-
* =
= * d = *d
=
U
s
(Ans. b)
(c) For the given profile, we have
w
d
=
2
U dx
1 2uU
U 2
d 2
=
dx 15
u
dx = d
U
15
d =
15u
dx
U
Integrating the above equation, we obtain
2
2
=
15ux
+C
U
When x = 0, = 0 , C = 0.
x
30
5.836
a
=
=
Re x
Re x
Re x
=
(d) c f =
w
1
U 2
2
=2
(Ans. c)
d
dx
2 w
d
= 2 *
2
U
dx
where
* =
d
d
= *
=*
dx
dx
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-88-
1/2
d ax
w2 =*
U
dx U
v
ax1/2
U / v
=
d =
1 −1/2
x
d
2
dx
=
dx
U
v
a
Using the value of
cf =
a
2
xU
v
d
in the above equation, we have
dx
w
a *
b
* ( a / 2)
=
c
=
=
f
2
U
xU
Re x
Re x
(Ans. d)
v
(e) Now,
*
x
= *
x
*
x
*
x
*
x
= *
a
a
, where =
x
Re
Re x
=
c
, where c = a * = 1.751
Re x
=
c
1.751
=
Re x
Re x
(f) * =
* =
(Ans. e)
x
a
d
0.6855
=* =*
=
=
x
x
x
Re x
Re x
Re x
(Ans. f)
(g) The given profile is
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-89-
u
= 2 − 2 3 + 4
U
=
0 1
y
Using continuity equation, we have
v
u
= − dy
U
x 0 U
y
y
y
y
y
v − y
y 3 y 4 − y
y3
y4
=
2
−
2
+
dy
=
2
dy
−
2
dy
+
0 3 0 4 dy
U x 0
3 4 x 0
2
4
5
− 2 y 2 y
y
=
− 3 + 4
x 2 4
5
It is given that
x
=
a
ax
=
Re x
Re x
Therefore,
we
v − y 2
y4
y 5 − y 2
y4
y5
=
− 3 3 + 5 5 =
−
+
2a x
5a x x ax
2a 3 x 3
5a 5 x 5
U x ax
3/2
5/2
Re
Re3/2
Re5/2
V x V x
x
V x
x
x
u
u
u
v − y 2
=
U x a
v y2
=
U a
V 1
u
V
y 4 V
− 3
x 2a u
1
y 4 V
−
u 2 x3/2 2a3 u
3/2
3/2
1
y 5 V
+
x3/ 2 5a5 u
3
y 5 V
+
2 x5/2 5a5 u
5/2
5/2
have
1
x5/ 2
5 1
2 x7 / 2
v 1 y 2 ax 1 3 y 4 a3 x3 1 1 y 5 a5 x5 1
=
−
+
U 2 a x 2 4 a3 3 x 4 2 a5 5 x6
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-90-
v 2 3 4 4 5 ( a / 2)
= − +
U
2
5 Re x
(1)
v 2 3 4 4 5 2.9178
= − +
U
2
5 Re x
(Ans. g)
(h) Differentiating with respect to y gives
2
3 y 4 4 y5 ( a / 2) 2 y
y3
y 4 ( a / 2)
d y
=
−
+
=
−
6
+
4
2
2 4 5 5 Re x 2
4
5 Re x
dy
d u
dy U
To obtain the stationary point, set
(2)
u
= 0.
y U
2y
y3
y4
−
6
+
4
=0
y 2
4
5
2
−
2
16 y 3
5
8 y3
3
8 y3
18 y 2
4
−
+
18 y 2
−
9 y2
−
9 y2
2
4
16 y 3
5
=−
=0
2
2
= −1
+1 = 0
3 2
8 3 − 9 2 + 1 = 0
This equation is satisfied when = 1 .
Therefore,
y
=1 y =
v
v
=
U max U y =
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-91-
v y 2 3 y 4 4 y5 ( a / 2) 2 3 4 4 5 ( a / 2) 3 4 ( a / 2)
= −
+
= −
+
= 1 − +
U 2 2 4 5 5 Re x 2 2 4 5 5 Re x 2 5 Re x
v 3a 1
0.8753
=
=
U 20 Re x
Re x
(Ans. h)
−
x Quartic Pohlhausen x Blasius
% error in =
100%
x
x Blasius
=
5.836 − 5
100%=16.61% 17%
5
% error in c f =
(c )
f
Quartic Pohlhausen
(c )
f
=
− (c f )
(Ans. i)
Blasius
100%
Blasius
0.6855 − 0.664
100% = 3.162% 3.2%
0.664
(Ans. i)
4-68 Consider the quartic Majdalani–Xuan profile for the laminar, incompressible flow over a
flat
plate:
1 4
5
3
y
u − + ; 0 1
where =
[Majdalani and Xuan (2020)]
F ( ) = = 3
3
U
1
1;
(a) Show that the dimensionless displacement thickness, momentum thickness, and
shape factor lead to:
* 1 u
1u u
*
*
= = 1 − d = 0.3500, = = 1 − d = 0.1337, and H = = 2.618
0 U
0U U
*
(b) Show that
a = Rex / x = 4.993, b = C f Rex = Rex / x = 0.6676, and c = * Rex / x = 1.748 .
(c) Calculate the total skin friction force and the drag coefficient for a plate of length L. Express
ReL = UL /
your
result
in
terms
of
(d) Using the continuity equation, integrate the axial velocity profile to show that
v 2 3 4 4 5 ( a / 2 ) 2 3 4 4 5 2.9178
= − +
= − +
U
2
5 Rex
2
5 Rex
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-92-
(e) Show that the maximum transverse velocity occurs at the edge of the boundary layer where it
can be obtained from
v
U
=
max
v
U
=
y −
( 7a / 40 ) = 0.8738
Rex
Rex
(f) Verify that the errors in predicting / x and C f are 1.7 . and 0.53 percent relative to the
classic
Blasius
results
in
Table
4-4.
(g) Show that the quartic Majdalani–Xuan profile satisfies all of the fundamental
boundary conditions except for the vanishing curvature at the edge of the boundary
layer. Explain why better predictions are obtained when this condition is relaxed.
Hint: Although the zero-curvature boundary condition is true as y → , it is inaccurate at the
edge of the boundary layer, where the velocity slope continues to change. By relaxing
this condition, Majdalani and Xuan ( 2020 ) are able to derive a simple approximation
that is more accurate than Pohlhausen’s quartic polynomial by one order of magnitude.
Consequently, the Majdalani–Xuan polynomial profile may be viewed as a practical
equivalent to the Blasius solution in several engineering applications.
Solution: The given profile is
1 4
5
3
u − + ; 0 1
F ( ) = = 3
3
U
1
1;
where
=
y
0 1 0
u
= 1 −
0
U
*
y
1 0 y
5 y y 3 1 y 4
1 4
5
3
d = 0 1 − − + d = 0 1 − − + dy
3
3
3 3
3
y
5 y
5 2 4
1 y4
5
* = dy − dy − 3 dy + 4 dy = −
− 3+
4
0
0
3 0
3 0
6 4 15
5 21
− +
=
6 4 15 60
* = −
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-93-
* 21
=
= 0.3500 = *
60
=
=
0
0
2
u
u u
u
1 − dy =0 − dy
U U
U U
2
u
u
dy − dy
0 U
U
5 y y3 1 y 4
5 y y3 1 y 4
=
− 3+
dy
−
0 3 − 3 + 3 4 dy
3
34
0
2
* =
= 0.1337
H=
* *
0.35
=
=
= 2.618
0.1337
(Ans. a)
(b) For the given profile, we have
w
1
U 2
2
u
du
dy
1
U 2
2
=
d
dx
=
d
dx
5 U
3 = d 11
1
U 2 dx 31250
2
u
31250 u
d
=
33 U dx
d =
31250 u
dx
33 U
Integrating the above equation, we obtain
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-94-
2
2
=
31250 ux
+C
33 U
When x = 0, = 0 , C = 0.
x
Re x
1
43.52
=
a=
= 4.993
x
Re x
Re x
62500
33
=
4.993 ( 0.1337 ) 0.6676
Re x
a *
b
cf =
=
=
=
b = c f Re x =
x
Re x
Re x
Re x
Re x
Now
*
x
*
x
*
x
= *
x
= *
=
a
Re x
c
Re x
where
*a = c
c = 0.35 4.993
c = 1.748
c=
* Re x
x
= 1.748
(Ans. b)
(d) The given profile is
u 5
1
= − 3 + 4
U 3
3
=
a=
y
Re x
x
a = 4.993
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-95-
Using continuity equation, we have
v
5 y y3 1 y 4
= −
−
+
U
x 0 3 3 3 4
y
dy
y
y 3
y
− 5 y
y
1 y4
=
dy − 3 dy + 4 dy
x 3 0
30
0
=
− 5 y 2
y4
y5
−
+
x 3 2 4 3 15 4
It is given that
x
=
a
=
Re x
ax
Re x
Therefore, we have
2
y 5 Re5/2
v − 5 y Re x y 4 Re3/2
x
x
=
−
+
U x 6 ax
4a 3 x 3
15a 5 x5
v − 5 y 2
=
U x 6 a
v 5 y2
=
U 6a
V 1
u
y 4 V
− 3
x 4a u
V −1
3/2
1
y 5 V
+
x3/2 15a 5 u
5/2
1
x5/2
y 4 V −3 y 5 V
−
+
u 2 x3/2 4a3 u 2 x5/2 15a5 u
v 2 3 4 4 5 ( a / 2)
= − +
U
2
3 Re x
3/2
5/2
−5
7/2
2 x
(1)
u y 2 3 y 4 4 y5 ( a / 2)
= −
+
U 2 2 4 3 5 Re x
(Ans. d)
Differentiating with respect to y gives
d u 2y
y3
y 4 ( a / 2)
= 2 −6 4 +4 5
dy U
Re x
(2)
7 ( a / 2 ) 7a 1
0.8738
v
v
=
=
= =
Re x
U max U y = 20 Re x 40 Re x
(Ans. e)
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-96-
−
x Quartic M − x x Blasius
% error in =
100%
x
x Blasius
=
1.748 − 1.72
100%=1.7%
1.72
% error in c f =
=
(c )
f
Quartic M − x
(c f
)
(Ans. f)
− (c f
)
Blasius
100%
Blasius
0.668 − 0.664
100% = 0.53%
0.664
(Ans. f)
4-69 Consider the continuous Majdalani–Xuan exponential profile for the laminar,
incompressible flow over a flat plate:
y
− s (1+ 12 s + 2 )
u
=
1
−
e
; =
U
[Majdalani and Xuan (2020)]
Re
a
x
s =
=
1.6304, 0
x
2
2
(a) Show that the dimensionless displacement thickness, momentum thickness, and shape factor
lead to:
u
u
u
* = 1 − d = 0.35068 , * =
1 − d = 0.13559 , and H = 2.5862
0 U
0
U
U
(b) Show that
Rex / x = 4.9039 , C f Rex = Rex / x = 0.66494 , and * Rex / x = 1.7197
(c) Verify that the errors in predicting / x , C f , and * / x are negligible, being, respectively,
0.12%, 0.13%, and 0.06% relative to the computed Blasius solution in Table 4-4.
(d) In addition to the fundamental boundary conditions that are satisfied by this profile, verify
that the continuous Majdalani–Xuan exponential profile captures the behavior of the Blasius
solution very robustly: (1) it exhibits the same slope as the exact Blasius solution at the wall,
(2) it equals, as it should, 0.99U at y = , (3) it matches the slope of the Blasius velocity
profile at the edge of the boundary layer with reasonable accuracy, and (4) it approximates
the curvature of the exact Blasius solution at y = , where the Blasius curvature does not
vanish.
(e) Contrary to the piecewise approximations that are confined to the 0 1 range, this profile
remains valid beyond the edge of the boundary layer, specifically as → , with u / U → 1
asymptotically, as it should, thus allowing for the smooth and seamless merging of the
viscous solution with the far-field freestream. Explain why this simple analytic profile is
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practically equivalent to the “exact” Blasius solution given the small truncation error entailed
in the derivation of the Blasius equation itself.
(f) By writing = y U / (2 x ) = C , and recalling that = a x / U , show that the
conversion constant between the Blasius similarity variable and
is simply
C = a / 2 3.47 .
(g) Plot this solution for 0 , where = 1.5 corresponds to 5.2 written in terms of
the Blasius variable. Add to your plot the Blasius derivative function df / d , which
represents the normalized axial velocity.
Solution:
(a) We are asked to determine the dimensionless displacement and momentum thicknesses as
well as the shape factor for the given velocity profile. Using the defining expressions for * ,
*, and H , we have:
− s (1+ 1 s + 2 )
u
2
d 0.35068
* = 1 − d = 0 e
*
0 U
2.5862
and H =
Ans. (a)
*
2
2
1
1
− s (1+ 2 s + )
− s (1+ 2 s + )
*
= 1− e
e
d 0.13559
0
The above integrals may be evaluated numerically using an available symbolic program or
integrator (e.g., Mathematica, Matlab, Maple, Python, C, Fortran, Basic, etc.). An example of a
Mathematica code is provided below.
(
)
s:=1.63040
u[x]:=1-Exp[-s*x*(1+(1/2)*s*x+x^2)]
EtaStar:=NIntegrate[1-u[x],{x,0,Infinity}]
ThetaStar:=NIntegrate[u[x]*(1-u[x]),{x,0,Infinity}]
H:=EtaStar/ThetaStar
a:=Sqrt[2*s/ThetaStar]
b:=Sqrt[2*ThetaStar*s]
c:=a*ThetaStar*H
ErrorDeltax := (Abs[(a-4.90999)]/4.90999)*100
ErrorCf:=(Abs[(b-0.66411)]/0.66411)*100
ErrorDeltaxStar:=(Abs[(c-1.72079)]/1.72079)*100
(b) Given the properties in Part (a) and noting that s 1.6304 stands for the slope of F at the
wall, we can calculate
2s
Rex =
4.9039
a =
x
*
b = C f Rex =
Ans. (b)
Rex = 2 * s 0.66494
x
*
Rex = a * H 1.7197
c =
x
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(c) We can compute the relative errors using the robustly computed or ‘exact’ Blasius values
located in Table 4-4 as benchmarks. We get
Rex / x − 4.90999
error:
100% 0.124%
x
4.90999
C f Rex − 0.66411
C
error:
100% 0.125%
Ans. (c)
f
0.66411
*
* Rex / x − 1.72079
error:
100% 0.064%
x
1.72079
These coincide with the reported relative errors and show that the relative errors are virtually
insignificant.
(d) We are asked to verify 4 conditions, namely,
1. F ( 0 ) s
2. F (1) = 0.99
3. F (1) matches the slope of the Blasius velocity with reasonable accuracy
4. F (1) approximates the curvature at y =
First, we may proceed by defining the function and its derivatives using
1
2
F ( ) = 1 − exp − s 1 + s +
2
3 2s 2
2
2
−s
F ( ) = 3 s + s + s exp − s −
2
2 2
F ( ) = − s 9 3 s + 6 2 s 2 + s s 2 + 6 + 2 s 2 − 3 exp − 3 s − s − s
2
Checking these conditions one-by-one, we find
F ( 0 ) = s identically;
(
)
(
) (
)
1
F (1) = 1 − exp − s 1 + s 0.99 ;
2
s2
2
F (1) = s + 4s exp − − 2s 0.0932 which mimics Blasius’s value of 0.0904 in Table 4 2
2;
−s 2
3
2
F (1) = − s s + 8s + 15s − 6 exp
− 2s −0.729 which mimics the -0.709 Blasius value.
2
Ans. (d)
(
)
(
)
(e) The Blasius equation is accompanied by a small truncation error that depends on the size of
the flow Reynolds number. This is due to its original derivation requiring the neglect of the
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second-order diffusion term in the streamwise direction. We also recall that the Blasius solution
is derived assuming laminar conditions, thus ensuring that the Reynolds number remains of
modest size. As a result, the truncation error associated with the Blasius equation itself may be
viewed to be of order Re −1 = O(10−5 ) . As shown in Table 4-4, since the largest L2 error
associated with the present analytical solution is O (10 −4 ) , it makes its predictions practically
equivalent to the Blasius solution obtained numerically, especially that the latter carries a finite
truncation error that depends on the discretization scheme employed by the user.
Ans. (e)
(f) The following derivation enables us to specify the conversion constant between the Blasius
similarity variable and , i.e., the fractional distance spanning the viscous boundary layer:
= C , y
y
U
= C
, and so
2 x
C
a
4.9039
U
=
3.47 Ans. (f)
=
or C =
2 x a x / U
2
2
Since = ax / Rex = a x / U from Part (b).
(g) The results shown below resemble those of Fig. 4-13 and confirm that the Majdalani–Xuan
and Blasius solutions remain imperceptible, not only within the viscous layer, but also in the far
field as → . This may be viewed as being significant due to the simplicity of the Majdalani–
Xuan profile. In fact, a feature article on this subject has appeared in the Physics of Fluids where
the systematic approach leading to a family of exponentially-decaying formulations by
Majdalani and Xuan (2020) is explained.
1
0.8
0.6
u
U
Blasius (1908)
Majdalani–Xuan (2020)
0.4
0.2
0
0
0.2
0.4
0.6
0.8
y/
1
1.2
1.4
1.6
Ans. (g)
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4-70 Consider the sinusoidal Schlichting profile for the laminar, incompressible flow over a flat
plate:
u
y
= sin
; 0 y
U
2
[Schlichting (1979)]
(a) Using continuity, v = − 0y u dy , integrate the axial profile to show that the transverse
x
U
U
velocity can be written as
Re x
v y y y a
= cos +
sin − 1
;a =
U 2 2
x
2 Re x
(b) Plot u / U and v / U versus y / and show that the maximum transverse velocity occurs at
the edge of the boundary layer where it can be obtained from
v
U
=
max
v
U
=
y =
− 2 ( a / 2 ) 0.871264
=
Re x
Re x
(c) Evaluate at (v / U )max at x = 0.6 m and = 0.006 m . Hint: First show that, irrespective of
the Reynolds number,
v
U
=
max
−2
2 x
Solution: Given that
u
y
= sin
; 0 y
U
2
(a)
v v
yu
y 2
y
= = − dy = − 0y sin
− 1
dy = cos
0
U U
x U
x
x
2
2
y
v
y
2ax
=
cos
− 1
2ax
Re x
U x
Re x
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-101-
v
y V 2a ux
= cos
− 1
U x
2ax ux
V
y Re x
= − sin
2ax
=
− y y Re x
+ cos
2x
2ax
a
− 1
Re x
y
y y a
sin
+ cos
− 1
2 x 2 2 Re x
y
y Re x y a
= sin
+ cos
− 1
2 Re x
2 x 2 a
v y
a
y
y
=
sin
+ cos
− 1
U 2
2
2 Re x
(b)
(Ans. a)
v y
a
y
y
=
sin
+ cos
− 1
U 2
2
2 Re x
Differentiating the above equation with respect to y, we obtain
2
d v
y
y
= − sin + sin + y cos
dy U
2 2 2
2 2
2
2
d v
y a
= y cos
dy U 2
2 Re x
To obtain the stationary points, set
a
Re x
(1)
d v
=0.
dy U
2
y a
=0
y cos
2
2
Re
x
y = 0,
Differentiating equation (1) with respect to y, we obtain
3
2
d2 v
y
y a
=
−
y
sin
+
cos
2
dy U 2 y
2 2
2 Re x
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Substituting y = , we obtain
3
a
d2 v
=
−
y
+
0
dy 2 U 2 y
Re x
3
d2 v
a
0
= y
2
dy U 2 y Re x
Therefore, v is maximum at y = .
U
v
v
=
U max U y =
= cos
2
sin
+
2
2
a
a
− 2 ( a / 2)
= 0 + − 1
=
− 1
2 Re x
Re x
Re x
(Ans. b)
(Ans. b)
(c) Given that a =
Rex
x
−2 −2
v
=
=
2 x 2 x
U max
At x = 0.6 m and = 0.006 m,
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v
− 2 0.006 0.18169
=
=
100
U max 2 0.6
4-71
0.182%
(Ans. c)
Consider the cubic Pohlhausen profile for the laminar, incompressible flow over a flat
plate:
u 3 y y3
=
−
;0 y
U 2 2 3
[Pohlhausen 1921]
(a) Using continuity, integrate the axial profile to show that the transverse velocity can be written
as
v
3 y2 y4 a
3 y2 y4
= 2 2 − 4
= 2 2 − 4
U 16
Re x 16
x
(b) Plot u / U and v / U versus y / and show that the maximum transverse velocity occurs at
the edge of the boundary layer where it can be obtained from
v
U
=
max
U
=
y −
3a
0.8702
=
16 Re x
Re x
(c) Evaluate at (v / U )max at x = 0.7 m and = 0.007 m .
Solution: Given that
u 3 y y3
=
−
;0 y
U 2 2 3
(a) Using continuity, we have
v
yu
=−
dy
U
x 0 U
v
y 3y
y3
3y2
y4
= − 0
−
−
dy = −
U
x 2 2 3
x 4 8 3
Also,
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a=
Re x
3 =
x
=
ax
=
Re x
ax
=
V x
u
u ax
V x
a 3 x3/2
V
u
3/2
3/2
3/2
2 V
4 V
4 V
y
3
y
3y
2
4
v
u − u = 1 y 2 V −
u = 3 2 y − y a
=−
U
x 4a x
8a 3 x 3/2 8 x 3/2
u
16a 3 x 5/2
16 2 4 Re x
v
3 y2 y4
= 2 2 − 4
U 16
x
(Ans. a)
2
4
(b) v = 3 2 y 2 − y 4 a
U 16
Re x
Differentiating the above equation with respect to y, we obtain
d v 3 4 y 4 y4 a
−
=
dy U 16 2 4 Re x
To obtain the stationary points, set
(1)
d v
=0.
dy U
3 4 y 4 y4 a
−
=0
16 2 4 Re x
4 y 4 y4
−
=0
2 4
4 y2
y2
1
−
2 2
=0
4 y2
=0 y =0
2
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1−
y2
y2
=
0
= 1 y =
2
2
y = 0, , −
Differentiating equation (1) with respect to y, we obtain
d 2 v 3 4 16 y 3 a
− 4
=
dy 2 U 16 2
Re x
Substituting y = , we obtain
d 2 v 3 4 16 a
−
0
=
dy 2 U 16 2 Re x
Therefore, v is maximum at y = .
U
v
v
=
U max U y =
=
3 2 4 a
3
=
2 2 − 4
16
Re x 16
a
0.8702
=
Re x
Re x
(Ans. b)
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(Ans. b)
v
3 a
(c) =
U max 16 Re x
a=
Re x
a
=
x
Re x x
3
v
=
U max 16 x
At x = 0.7 m and = 0.007 m,
3 0.007
v
=
= 0.188%
U max 16 0.7
(Ans. c)
4-72 Consider the continuous, one-parameter, tangent hyperbolic Yun profile for laminar flow
over a flat plate:
( )
u
y
= tanh s ; = , s = 1.6304, 0
U
[Yun (2010)]
(a) Show that the dimensionless displacement thickness, momentum thickness, and shape factor
lead to:
0
* = 1 −
u
U
u
u
*
d = 0.4251, = 1 −
U U
0
d = 0.1882 and H = 2.2589
(b) Show that
Re x / x = 4.1624, C f Re x = Re x / x = 0.7834, and * Re x / x = 1.7696
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-107-
(c) Verify that the errors in predicting / x and C f are 15.2 and 18 percent relative to the
computed Blasius values in Table 4 − 4 .
Solution: The given profile is
( )
u
y
= tanh s ; = , s = 1.6304, 0
U
=
y
0 0
y
0 y
The possible condition is
0 y
* = 1 −
0
u
U
(
( ))
y
d = 0 1 − tanh s d = 0 1 − tanh s dy
*
0.9748
* = − ln cosh s − ln1 = − 0.9748 − 0 = 1 −
= 0.4251
1.6304
s
s
=
=
0
0
2
u
u
u
u
1 − d =0 − dy
U U
U U
2
y
y
tanh s dy − tanh s dy
0
Integrating and solving gives
* =
= 0.1882
* * 0.4251
H= =
=
= 2.2587
0.1882
c fx =
w
1
U 2
2
=
(Ans. a)
u
y
d
d
=
1
dx
U 2 dx
2
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=
y
U tanh s
y
y
y
y
d
= tanh s − tan 2 h s dy
1
dx 0
U 2
2
Applying the boundary condition x = 0, = 0 , we obtain
Re x = 4.1624 = a
x
cf =
a *
*
=
Re x b = a
x
Re x
cf =
b
0 1882 4 1624
=
Re x
Re x
c f Re x = 0.7834
Also,
*
x
= *
x
= *
a
Re x
*a = c c = 0.4251 4.1624 = 1.7696
*
x
=
c
1.7696
=
Re x
Re x
*
Re x = 1.7696
x
(Ans. b)
−
x Yun x Blasius
% error in =
100%
x
x Blasius
=
4.1624 − 4.90999
100% = 15.22%
4.90999
(Ans. c)
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-109-
% error in c f =
=
(c )
f
Yun
− (c f )
(c f )
Blasius
100%
Blasius
0.783 − 0.66411
100% = 18%
0.66411
(Ans. c)
4-73 Consider the continuous, two-parameter tangent hyperbolic Savaş profile for laminar
flow over a flat plate:
u
y
5/3 3/5
= tanh ( ) ; = , = s = 1.6304, 0 [Savaş (2012)]
U
(a) Show that the dimensionless displacement thickness, momentum thickness, and shape factor
lead to:
u
u
u
* = 1 − d = 0.3509 , * =
1 − d = 0.1363 , and H = 2.5738
0
0
U U
U
(b) Show that
Rex / x = 4.8906 , C f Rex = Rex / x = 0.6667 , and * Rex / x = 1.7161
(c) Verify that the errors in predicting / x and C f are 0.40% and 0.40% relative to the
computed Blasius solution in Table 4-4.
Solution:
(a) We are asked to find the dimensionless displacement and momentum thicknesses as well as
the shape factor for the given velocity profile. Using the defining expressions for * , *, and
H , we have:
u
5/3 3/5
* = 1 − d = 1 − tanh ( ) d 0.3509
0
*
0 U
2.5738 Ans. (a)
and H =
*
5/3 3/5
5/3 3/5
*
tanh ( ) 1 − tanh ( ) d 0.1363
=
0
The above integrals may be evaluated numerically using an available symbolic program or
integrator. An example of a Mathematica code is provided below.
s:=1.6304
u[x]:=(Tanh[(s*x)^(5/3)])^(3/5)
EtaStar:=NIntegrate[1-u[x],{x,0,Infinity}]
ThetaStar:=NIntegrate[u[x]*(1-u[x]),{x,0,Infinity}]
H:=EtaStar/ThetaStar
a:=Sqrt[2*s/ThetaStar]
b:=Sqrt[2*ThetaStar*s]
c:=a*ThetaStar*H
ErrorDeltax := (Abs[(a-4.90999)]/4.90999)*100
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ErrorCf:=(Abs[(b-0.66411)]/0.66411)*100
(b) Given the properties in Part (a) and noting that = s 1.6304 stands for the slope of F at
the wall, we can write
a = Re / x = 2s / * 4.8906
x
*
Ans. (b)
b = C f Rex = Rex / x = 2 s 0.6667
*
c = * Rex / x = a H 1.7161
(c) We can compute the relative errors using the robustly computed or ‘exact’ Blasius values
located in Table 4-4 as benchmarks. We get
Rex / x − 4.90999
error:
100% 0.40%
x
4.90999
Ans. (c)
C
Re
−
0.66411
f
x
100% 0.40%
C f error:
0.66411
These coincide with the reported relative errors in Table 4-4.
4-74 Consider the continuous, one-parameter Moeini–Chamani profile for laminar flow over a
flat plate:
u
y
= erf ( ); = , = 1.59261, 0 [Moeini and Chamani (2017)]
U
(a) Show that the dimensionless displacement thickness, momentum thickness, and shape factor
lead to:
u
u
u
* = 1 − d = 0.3543 , * =
1 − d = 0.1467 , and H = 2.4142
0
0
U U
U
(b) Show that
Rex / x = 4.9491 , C f Rex = Rex / x = 0.7262 , and * Rex / x = 1.7532
(c) Verify that the errors in predicting / x and C f are 0.8% and 9.4% relative to the computed
Blasius solution in Table 4-4.
Solution:
(a) We are asked to find the dimensionless displacement and momentum thicknesses as well as
the shape factor for the given velocity profile. Using the defining expressions for * , *, and
H , we have:
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u
* = 1 − d = 0 1 − erf ( ) d 0.3543
0 U
* u u
=
Ans. (a)
1 − d = 0 erf ( ) 1 − erf ( ) d 0.1467
U
U
0
*
2.4142
H =
*
The above integrals may be evaluated numerically using an available symbolic program or
integrator (e.g., Mathematica, Matlab, Maple, Python, C, Fortran, etc.). An example of a
Mathematica code is provided below.
s:=1.7971
u[x]:=Erf[1.59621*x]]
EtaStar:=NIntegrate[1-u[x],{x,0,Infinity}]
ThetaStar:=NIntegrate[u[x]*(1-u[x]),{x,0,Infinity}]
H:=EtaStar/ThetaStar
a:=Sqrt[2*s/ThetaStar]
b:=Sqrt[2*ThetaStar*s]
c:=a*ThetaStar*H
ErrorDeltax := (Abs[(a-4.90999)]/4.90999)*100
ErrorCf:=(Abs[(b-0.66411)]/0.66411)*100
(b) Given the properties in Part (a), we must find the value of s for this particular profile. The
parameter s is defined as
dF ( 0 )
s=
d
In view of the Moeini–Chamani error function, we have
2
dF
dF (0) 2 −( 0 )2
d
2
=
=
e
1.7971
erf ( ) = e− and so s =
d d
d
This value of s can be used in the remaining analysis. For example, we readily find
a = Re / x = 2s / * 4.9491
x
*
Ans. (b)
b = C f Rex = Rex / x = 2 s 0.7262
*
c = * Rex / x = a H 1.7532
(c) We can compute the relative errors using the robustly computed or ‘exact’ Blasius values
located in Table 4-4 as benchmarks. We obtain
Rex / x − 4.90999
error:
100% 0.80%
x
4.90999
Ans. (c)
C f Rex − 0.66411
100% 9.4%
C f error:
0.66411
These coincide with the reported relative errors.
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-112-
4-75 Because of the popularity of approximate profiles in laminar boundary-layer studies, let
us consider a quartic mean flow in the case of no pressure gradient:
u
y
= c0 + s + c2 2 + c3 3 + c4 4 ; =
0 1
U
where represents the fractional distance within the laminar boundary layer, and s c1 = F (0)
is the axial velocity slope at the wall, also known as the Blasius constant or connection
parameter.
F ( ) =
(a) Show that the particular profile that will satisfy Pohlhausen’s four essential conditions
described in Sec. 4–1 must be:
F ( ) = s + ( 4 − 3s ) 3 + ( 2s − 3) 4
Hint: These conditions translate into F (0) = F (1) = F (0) = 0 and F (1) = 1 .
(b) Show that the normalized boundary-layer properties, * , * , and H , can be written as a
direct function of s :
1
*
3
* = 0 (1 − F ) d = 20 ( 4 − s )
1
4 13s 19 s 2
*
=
F
(1
−
F
)
d
=
+
−
0
35
210
630
189 ( 4 − s )
* *
=
=
H
* 144 + 78s − 38s 2
( normalized displacement thickness )
( normalized momentum thickness )
( momentum shape factor )
(c) Making use of C f = 2d / dx , show that the wall shear stress and skin friction coefficient can
be written as
w =
U
s
and
Cf =
4 13s 19s 2 d
2 s
= 2 +
−
U
35 210 630 dx
(d) Use separation of variables to identify
630s
dx
2
72 + 39s − 19s U
(e) After integrating from (0) = 0 to ( x) , show that the boundary-layer thickness and its
d =
derivative can be written as functions of the local Reynolds number, Rex = Ux / , and the
Blasius constant s :
x
=
a
1260s
; a=
72 + 39s − 19s 2
Rex
and
d ( a / 2)
=
dx
Rex
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(f) Using the definition of the friction coefficient, C f = 2 w / ( U 2 ) , show that
(
b
s
; b=
72 + 39 s − 19 s 2
315
Rex
Cf =
)
(g) Recalling that the displacement thickness may be specified as * = * , show that
*
x
=
c
3a
; c = (4 − s)
20
Rex
(h) Recalling that the momentum thickness may be deduced from = * , show that
d
=
; d = a * = b
x
Rex
(i) Using the continuity equation = −
y
udy , show that the transverse velocity component
x 0
becomes
3
2
s
a
= 2 + ( 4 − 3s ) 4 + ( 2s − 3) 5
U 4
8
5
Rex
(j) Show that the maximum value of the transverse velocity occurs at the edge of the boundary
layer where
=
U max U
=
y =
e
567 s
; e = (1 − 14 s )
360 + 195s − 95s 2
Rex
(k) Calculate (a, b, c, d , e) for the two values of s = 2 and 5 / 3 , which correspond to the
quartic Pohlhausen and Majdalani–Xuan profiles.
Blasius values in Table 4-1.
Compare your results to the classic
Solution:
(a) In order to determine the 4 unknown polynomial coefficients, we revert back to Pohlhausen’s
essential boundary conditions and apply them systematically to the generic quartic polynomial:
u
F ( ) = = c0 + s + c2 2 + c3 3 + c4 4
U
Condition 1: F (0) = 0 (no slip condition at the wall)
c0 + s (0) + c2 (0) 2 + c3 (0)3 + c4 (0) 4 = 0
or
c0 = 0
Condition 4: F (0) = 0 (zero pressure gradient in the far field)
F ( ) = 2c2 + 6c3 + 12c4 2 ,
2c2 + 6c3 (0) + 12c4 (0) 2 = 0
and so
c2 = 0
This leaves us with only 2 unknowns.
Condition 2: F (1) = 1 (matching the freestream velocity at the boundary-layer edge)
s + c3 + c4 = 1 and so c4 = 1 − s − c3
Condition 3: F (1) = 0 (negligible shear at the boundary-layer edge)
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F ( ) = s + 2c2 + 3c3 2 + 4c4 3 ,
F (1) = s + 3c3 + 4c4 = 0
and so
c4 = − 14 s − 34 c3
Equating the last two expressions for c4 , we get c4 = − 14 s − 34 c3 = 1 − s − c3 ; this enables us to
deduce
c3 = 4 − 3s and c4 = 2 s − 3
We thus arrive at
F ( ) = s + ( 4 − 3s ) 3 + ( 2s − 3) 4
Ans. (a)
(b) The normalized displacement thickness can be computed from:
1
1
*
* =
= (1 − F )d = 1 − s − (4 − 3s ) 3 − (2 s − 3) 4 d
0
0
1
2
4
5
s
3s 2s 3
3 3s
= − s − (4 − 3s ) − (2s − 3) = 1 − − 1 + − + − 0 = −
2
4
5 0
2
4 5 5
5 20
3
(4 − s)
20
The normalized momentum thickness can be computed from:
* =
or
*=
Ans. (b)
1
= F (1 − F )d
0
1
= s + (4 − 3s) 3 + (2 s − 3) 4 1 − s − (4 − 3s) 3 − (2 s − 3) 4 d
0
1
= ( s − s 2 2 + 4 3 − 3s 3 − 3 4 − 6 s 4 + 6 s 2 4 + 6 s 5 − 4 s 2 5 − 16 6 + 24 s 6
0
−9s 2 6 + 24 7 − 34s 7 + 12s 2 7 − 9 8 + 12s 8 − 4s 2 8 )d
4 13s 19s 2
+
−
35 210 630
The momentum shape factor can be computed from:
3
3
(4 − s)
(4 − s)
189 ( 4 − s )
*
20
20
H=
=
=
or H =
2
2
4 13s 19 s
144 + 78s − 38s
144 + 78s − 38s 2
+
−
35 210 630
1260
or
*=
Ans. (b)
Ans. (b)
(c) Here we derive two different forms of the skin friction coefficient starting with the
fundamental shear stress definition:
u
u
U U
U F
U
and so w = s
Ans. (c)
w =
=
=
y y =0
y
=0
y =0
Therefore, based on the shear stress definition, we have, on the one hand,
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U
s
w
2 s
Cf = 1
=1
=
2
2
U
2 U
2 U
2 s
U
On the other hand, using Kármán’s momentum-integral equation, we have
d ( * )
4 13s 19s 2 d
d
d
Cf = 2
=2
= 2 *
= 2 +
−
dx
dx
dx
35 210 630 dx
and so
Cf =
Ans. (c)
(d) Equating the two forms of C f found in Part (c), we get:
s
s
4 13s 19s 2 d
2 s
U
U
dx = d
dx = d and so
= 2 +
−
or
2
72
4
+
39
13
s − 19 s 2
s
19
s
U
35
210
63
0
dx
+
−
35 210 630
630
This can be rearranged into
630s
Ans. (d)
d =
dx
2
72 + 39s − 19s U
(e) In order to express the boundary layer as a function of the local Reynolds number, the result
in Part (d) can be integrated from the leading edge of the plate to any location x . This can be
accomplished by taking
x
2
630 s
x
630s
0 d = 0 72 + 39s −19s2 U dx or 2 = 72 + 39s − 19s 2 U
Solving for the displacement thickness, we find
1260s
1260s
x
1260s
x
=
=
=
2
2
2
2
x
72 + 39s − 19s Ux
72 + 39s − 19s U
72 + 39s − 19s Ux
and so
x
=
1260s
72 + 39s − 19s 2
1
a
1260s
=
;a =
72 + 39s − 19s 2
Rex
Rex
As for the spatial gradient of the disturbance thickness, we obtain
( 12 a )
d
d
d
1260s
x 1
1260s
and
so
=
=
=
dx dx 72 + 39s − 19s 2 U 2 72 + 39s − 19s 2 Ux
dx
Rex
Ans. (e)
Ans. (e)
(f) We start with the outcome of Part (c):
4 13s 19 s 2 d
4 13s 19 s 2 1
1260 s
Cf = 2 +
−
=
2
−
+
2
35 210 630 dx
35 210 630 2 72 + 39 s − 19 s Ux
72 + 39 s − 19 s 2
1260 s
=
2
630
72 + 39 s − 19 s Ux
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or
Cf =
s
1
b
s
72 + 39s − 19s 2 )
=
;b =
72 + 39s − 19s 2 )
(
(
315
315
Rex
Rex
(g) Starting with * = * , we may simply divide both side by x and simplify. We get
3a
(4 − s)
*
c
3a
* * a 3
20
or
=
;c = (4 − s)
= =
( 4 − s ) =
Re 20
x
20
x
x
Rex
Rex
x
Ans. (f)
Ans. (g)
(h) Starting with = * , we may simply divide both side by x and simplify. We get
a
a *
= * =
* =
Re
x x
Rex
x
Ans. (h)
Specifically, we have
72 + 39 − 19s 2
a 4 13s 19s 2 1
1260s
=
−
=
+
x Rex 35 210 630 Rex 72 + 39s − 19s 2
630
Collecting terms and simplifying, we retrieve
b
d
s
1 s
=
=
;d =
72 + 39s − 19s 2
=
72 + 39s − 19s 2 ) or
(
x Rex 315
x
3
1
5
Rex
Rex
Ans. (h)
(
)
(i) To obtain the normal velocity component, the continuity equation may be integrated, starting
y
with = − udy :
x 0
y
3
4
y u
y
y
y
=−
dy = − s + (4 − 3s) + (2s − 3) dy
x
U
x 0 U
0
2
4
5
s y 4 − 3s y 2s − 3 y
= − +
+
3
4
x
2 4 5
2
4
5
d s y
d 4 − 3s y
d 2s − 3 y
= − −
−
3
−
4
dx 4
dx 5
dx 2
4
d s 2 3
4
5
=
+ ( 4 − 3s ) + ( 2s − 3)
4
5
dx 2
=
(a / 2) s 2 3
4
4
5
2 + 4 ( 4 − 3s ) + 5 ( 2s − 3)
Rex
Finally, we have
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3
2
s
a
= 2 + ( 4 − 3s ) 4 + ( 2 s − 3) 5
U 4
8
5
Rex
Ans. (i)
(j) To show that the maximum transverse velocity occurs at the boundary-layer edge, we set
d ( / U )
=0
d
Substituting the results of Part (i) and differentiating, we get
d ( / U ) s
3
a
= + ( 4 − 3s ) 3 + 2 ( 2s − 3) 4
d
2
2
Rex
s 3
a
= + ( 4 − 3s ) 2 + 2 ( 2s − 3) 3
=0
2 2
Rex
Accordingly, the extrema will correspond to the roots of [ 12 s + 32 ( 4 − 3s ) 2 + 2 ( 2s − 3) 3 ] = 0 .
We get
3 11s 2 − 16 s − s
(always negative, unphysical)
−
2
8
s
−
12
(
)
0
(minimum of where = 0)
=
(maximum of within 0 1)
1
2
s + 3 11s − 16 s
(negative or out of range for 1 s 2)
2 ( 8s − 12 )
Interestingly, for s = 2 , we recover a double root at = 1, namely, when both the 3rd and 4th
roots predict the maximum to occur at the edge of the boundary layer. For s 2 , the initial slope
starts gravitating away from the true Blasius slope of 1.63; as such, it starts affecting the physical
behavior of the transverse velocity and should not be considered.
Having determined the location of the peak transverse velocity, it is now possible to evaluate its
magnitude. We have
=
U max U =1
Forthwith, the maximum transverse velocity may be calculated at the edge of the boundary layer
by taking:
3 9
4
6
1260s
1
1
= s+ − s+ s−
2
U =1 4
2 8
5
5 72 + 39 s − 19s
Rex
1260 s
12
= (1 − 14 s )
2
40
72 + 39s − 19s
or
=
U max
3 2
1
1260s
1
= (1 − 4 s )
10 72 + 39s − 19s 2
Rex
e
567 s
; e = (1 − 14 s )
360 + 195s − 95s 2
Rex
1
Rex
Ans. (j)
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(k) An evaluation of these parameters and their errors relative to the Blasius values are provided
in the table below.
Profile
a=
x
Re x
b = C f Re x = d =
x
Re x
c=
*
x
Re x
e=
Re x
U max
5.83558515
16.72%
0.685449684
3.16%
1.750675545
1.80%
0.8753
1.73%
− 3 + 13 4
4.993399337
0.13%
0.667547918
0.53%
1.747689767
1.63%
0.8738
1.55%
Blasius (1908)
5
0.664
1.72
0.8604
2 − 3 + 4
5
3
Clearly, the Majdalani–Xuan profile outperforms Pohlhausen’s quartic polynomial in predicting
the classic Blasius values, especially in its expressions for the boundary-layer thickness and skin
friction coefficient. For example, instead of a 17% error in predicting the boundary-layer
thickness, the use of the Majdalani–Xuan profile leads to a rather negligible error which, at
0.13%, proves to be two orders of magnitude smaller than that associated with Pohlhausen’s
quartic polynomial.
4-76 Because of the popularity of approximate solutions in laminar boundary-layer studies, let
us consider a quartic mean-flow profile in the presence of a pressure gradient:
u
y
= c0 + s + c2 2 + c3 3 + c4 4 ; =
0 1
U
where represents the fractional distance within the laminar boundary layer, and s c1 = F (0)
is the axial velocity slope at the wall which can be written as function of the zero-pressure slope
s0 and a pressure gradient correction:
F ( ) =
s = s0 + s1
(a) Show that the particular profile that will satisfy Pohlhausen’s four essential conditions with a
non-vanishing pressure gradient [Sec. 4–1] must be:
F ( ) = s − 12 2 + ( 4 − 3s + ) 3 + ( 2s − 3 − 12 ) 4 ; = −
2 dp 2 dU
=
U dx
dx
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Hint: These translate into F (0) = F (1) = 0 , F (1) = 1 and F (0) = − , where is
Pohlhausen’s pressure parameter.
(b) Show that flow separation will occur for = − s0 / s1 . Recall that separation occurs when
u / y = 0 at y = 0.
(c) Show that the normalized boundary-layer properties, * , * , and H , can be written as a
function of s and :
1
( 36 − 9s + )
( normalized displacement thickness )
60
19s 2 ( − 12 )( 24 + ) s (156 + 17 )
*
=−
−
+
630
2520
2520
s (156 + 17 ) − ( − 12 )( 24 + ) − 76 s 2
=
( normalized momentum thickness )
2520
42 ( 9s − 36 − )
H=
( momentum shape factor )
2
76s + ( − 12 )( 24 + ) − s (156 + 17 )
(d) Making use of C f = 2 * d / dx + 2(2 + H ) * (U / U ) , where U = dU / dx , show that the
* =
wall shear stress and skin friction coefficient can be written as,
U
2
w =
s and so C f = 1 w 2 =
s = 2 * d / dx + 2(2 + H ) * (U / U )
U
U
2
or
U
3 − (17 s + 9 ) 2 + ( 76 s 2 + 33s − 1044 ) + 1260 s
d 2s
=
−
2
2
+
H
=
−
4
(
)
dx U *
2 + (12 − 17 s ) + 76s 2 − 156s − 288
(e) For the case of Hiemenz flow where the velocity outside the boundary layer is given by
U = ax and V = −ay in Sec. 3–8.1, the analytic solution leads to a constant boundary-layer
thickness, show that = ( 2 / )U = const.
(f) Using Pohlhausen’s quartic profile with s = 2 + / 6 , determine and the corresponding
skin friction coefficient, w / U / (U ) , for the Hiemenz flow; compare your predictions
to those obtained from the numerical solution provided in Table 3-4. Show that:
Pohlhausen = 7.052
and
Hiemenz = 5.760 and
w
s
=
= 1.196
U U
versus
the
numerical
value
of
w
= 1.233
U U
The latter appears as “ F (0) ” in Table 3-4.
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(g) Using Majdalani–Xuan’s quartic profile with s = 5 / 3 + 0.2075 , determine and the
corresponding skin friction coefficient, w / U / (U ) , for the Hiemenz flow; compare
your predictions to those in Part (f ). Show that:
Majdalani-Xuan = 5.761 matches the Hiemenz value and
w
s
=
= 1.192 is off by
U U
3.3%.
(h) Use Thwaites’ approach to predict the value of and the corresponding skin friction
coefficient w / U / (U ) by applying Eqs. (4-139) and (4-140). Compare the skin friction
coefficient predicted by Thwaites’ approach to the one obtained from the numerical solution
provided in Table 3-4. Show that:
w
6
= S ( )
= 1.195
U U
0.45
(i) Compute the normalized displacement and momentum thicknesses as well as the shape factor
using Pohlhausen’s quartic profile for the Hiemenz flow. Show that: * = 0.241, * = 0.105,
= 0.075 ,
S ( ) = 0.3272 , and
and H =2.308 .
(j) Compute the normalized displacement and momentum thicknesses as well as the shape factor
using Majdalani–Xuan’s quartic profile for the Hiemenz flow. Show that: * = 0.267,
* = 0.115, and H = 2.318 .
Hint: The reference Hiemenz solutions obtained by numerical integration are * = 0.270,
* = 0.122, and H = 2.219.
Solution:
(a) In order to determine the 4 unknown polynomial coefficients, we revert back to Pohlhausen’s
essential boundary conditions and apply them systematically to the generic quartic polynomial:
u
F ( ) = = c0 + s + c2 2 + c3 3 + c4 4
U
Condition 1: F (0) = 0 (no slip condition at the wall)
c0 + s (0) + c2 (0) 2 + c3 (0)3 + c4 (0) 4 = 0
or
c0 = 0
Condition 2: F (1) = 1 (matching the freestream velocity at the boundary-layer edge)
s + c2 + c3 + c4 = 1 and so c2 = 1 − s − c3 − c4
Condition 3: F (1) = 0 (negligible shear at the boundary-layer edge)
F ( ) = s + 2c2 + 3c3 2 + 4c4 3 , F (1) = s + 2c2 + 3c3 + 4c4 = 0
Substituting the expression of c2 found through Condition 2, we get c3 = s − 2 − 2c4 .
Condition 4: F (0) = − (pressure gradient in the far field)
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2
This value of c2 may be readily substituted into the expression found through Condition 2 and
F ( ) = 2c2 + 6c3 + 12c4 2 , F (0) = 2c2 = −
and so
c2 = −
retrieve c3 = 1 − s + 2 − c4 . This reduces the problem to two equations with two unknowns for c3
and c4 . Combining this expression for c3 with the expression found through Condition 3, we
have
s − 2 − 2c4 + c4 = 1 − s +
or c4 = 2s − 3 − 12
2
This enables us to deduce
c3 = s − 2 − 2c4 = s − 2 − 2 ( 2s − 3 − 12 ) and so c3 = 4 − 3s +
We thus arrive at
F ( ) = s − 12 2 + (4 − 3s + ) 3 + (2s − 3 − 12 ) 4
Ans. (a)
(b) At separation, the flow detaches. This means that there is no change in velocity between the
wall and the flow directly above it (zero shear stress). This condition translates into F (0) = 0 .
Starting with
F ( ) = s − + 3(4 − 3s + ) 2 + 4(2 s − 3 − 12 ) 3
setting F (0) = 0 leaves us with
F ( ) = s − (0) + 3(4 − 3s + )(0) 2 + 4(2 s − 3 − 12 )(0)3 or s = s0 + s1 = 0
We deduce
s
Ans. (b)
=− 0
s1
(c) The normalized displacement thickness can be computed from:
1
* 1
* =
= (1 − F )d = 1 − s + 12 2 − (4 − 3s + ) 3 − (2 s − 3 − 12 ) 4 d
0
0
1
2 1 3
4
5
1
= − s + − (4 − 3s + ) − (2s − 3 − 2 )
2 6
4
5 0
s
3s 2s 3
3 3s
= 1 − + −1 + − − + + − 0 = − +
2 6
4 4 5 5 10
5 20 60
1
* = ( 36 − 9s + )
60
or
Ans. (c)
The normalized momentum thickness can be computed from:
*=
1
= F (1 − F )d
0
1
= s − 12 2 + (4 − 3s + ) 3 + (2 s − 3 − 12 ) 4 1 − s + 12 2 − (4 − 3s + ) 3 − (2 s − 3 − 12 ) 4 d
0
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*=
or
s (156 + 17 ) − ( − 12 )( 24 + ) − 76 s 2
2520
Ans. (c)
The momentum shape factor can be computed from:
1
( 36 − 9s + )
42 ( 36 − 9s + )
*
60
H=
=
=
2
s (156 + 17 ) − ( − 12 )( 24 + ) − 76s
s (156 + 17 ) − ( − 12 )( 24 + ) − 76s 2
2520
42 ( 9 s − 36 − )
or
Ans. (c)
H=
2
76s − s (156 + 17 ) + ( − 12 )( 24 + )
(d) Here we derive two different forms of the skin friction coefficient starting with the
fundamental shear stress definition:
u
u
U U
U F
U
Ans. (d)
w =
=
=
=
s
y y =0
y
=0
y =0
Based on the shear stress definition, we now have, on the one hand,
U
s
w
2 s
2 s
and so C f =
Ans. (d)
Cf = 1
=1
=
2
2
U
U
2 U
2 U
On the other hand, using Kármán’s momentum-integral equation, we have
2
2s
d
2 s
2 d
U
=U
+
2
2
+
H
and so
C f = 2 *
+ 2 ( 2 + H ) * =
) U
(
*
dx
U
dx
U
To make further headway, we recall that from the definition of Pohlhausen’s pressure parameter,
we have
2
U
2
d d
d 2 2 d
= U and so
=
=
=
dx U dx U dx dx
When these groupings are employed, the momentum-integral equation becomes
2s
d 2s
d
= U + 2 ( 2 + H ) or U =
− 2(2 + H )
*
dx U *
dx U
Next, we may substitute H and * from Part (c) into the right-hand side to obtain:
U
and so
d
==
dx U
2s
s (156 +17 ) −( −12 )( 24 + ) −76 s 2
2520
42 ( 9s − 36 − )
− 2 2 +
2
76s − s (156 + 17 ) + ( − 12 )( 24 + )
3 − (17 s + 9 ) 2 + ( 76s 2 + 33s − 1044 ) + 1260s
d
U = −4
dx U
2 + (12 − 17 s ) + 76s 2 − 156s − 288
Ans. (d)
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(e) For the Heimenz flow, both and U = a are constant (because U = ax ). This leaves us
with
2 dU
2
=
=a
= const.
Ans. (e)
dx
Since a is constant as well as , the pressure parameter must also be invariant.
(f) Since is constant for the Hiemenz flow, and U = a is independent of x , the left-hand side
term in Part (d) becomes:
d
U =0
dx U
We now have, referring to the right-hand side of Part (d):
3 − (17 s + 9 ) 2 + ( 76 s 2 + 33s − 1044 ) + 1260 s
−4
=0
2 + (12 − 17 s ) + 76 s 2 − 156 s − 288
or
3 − (17 s + 9 ) 2 + ( 76s 2 + 33s − 1044 ) + 1260s = 0
At this stage, we may insert s = 2 + / 6 for Pohlhausen’s quartic profile to obtain
2
3 − 17 ( 2 + 6 ) + 9 2 + 76 ( 2 + 6 ) + 33 ( 2 + 6 ) − 1044 + 1260 ( 2 + 6 ) = 0
5 3 79 2
+ − 464 + 2520 = 0
or
18
6
Based on the resulting cubic polynomial, three roots may be deduced, namely,
1 = −72.26, 2 = 17.803, and 3 = 7.052
Both 1 and 2 fall outside the range of acceptable , which is −12 12 . The only
acceptable outcome is, therefore,
Pohlhausen = 7.052
Ans. (f)
This value entails a 22.43% error compared to the numerical value of Hiemenz = 5.760 . As for
the skin friction coefficient, we get
s
2+ 6
(C f ) Pohlhausen = w
=
=
= 1.1957
Ans. (f)
U U
This value entails a 3.00% error compared to the numerical value of (C f ) Hiemenz = 1.233 , which
appears as F (0) in Table 3-4.
(g) We now repeat the same analysis using s = 5 / 3 + 0.2075 for Majdalani–Xuan’s quartic
profile. Substitution into 3 − (17 s + 9 ) 2 + ( 76s 2 + 33s − 1044 ) + 1260s = 0 yields
3 − 17 ( 53 + 0.2075 ) + 9 2
2
+ 76 ( 53 + 0.2075 ) + 33 ( 53 + 0.2075 ) − 1044 + 1260 ( 53 + 0.2075 ) = 0
3
2
or
0.744775 + 22.08083 − 516.43888 + 2100 = 0
Based on the resulting cubic polynomial, three roots may be deduced, namely,
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1 = −46.04, 2 = 10.63 and 3 = 5.761
Both 1 and 2 fall outside the range of acceptable , which is −8.03 8.03 . The only
acceptable outcome is, therefore,
Ans. (g)
Majdalani-Xuan = 5.761
This value entails a 0.017% error compared to the numerical value of Hiemenz = 5.760 . As for
the skin friction coefficient, we get
5
+ 0.2075
v
s
(C f ) Majdalani-Xuan = w
=
=3
= 1.1924
Ans. (g)
U U
This value entails a 3.3% error compared to the numerical value of (C f ) Hiemenz = 1.233 , which
appears as F (0) in Table 3-4.
Clearly, by using the Majdalani–Xuan profile in lieu of Pohlhausen’s, we gain a staggering 3
orders of magnitude improvement in predicting the pressure parameter for the Hiemenz flow.
(h) Based on Thwaites’ method, the momentum thickness may be written as a function of the
mean flow using Eq. (4-159),
0.45 x 5
2 =
U dx
U 6 0
2
2U 2 dU
dU .
=
Then based on Thwaites’ parameter, =
, we can retrieve dx =
dx
Substituting back into Thwaites’ integral, we have
x
0.45 5 2
=
U
dU ,
6
U 0
2
=
0.45 x 5
U dU
U 6 0
and so
=
0.45 U 6 0.45 3
=
=
U6 6
6
40
= 0.075
or
As for Thwaites’ shear correlation, which is given by Eq. (4-161), it becomes
0.62
0.62
S ( ) = ( + 0.09 ) = ( 0.075 + 0.09 )
S ( ) = 0.3272
and so
Ans. (h)
Ans. (h)
This enables us to deduce the wall shear stress using Eq. (4-160), namely,
U
w =
S ( )
The skin friction coefficient follows. We get
U
S ( )
= S ( )
Cf = w
=
U U
U
U
U
However, since = 2U / , , we have / U = 2 / . Inserting this ratio under the radical leaves
us with
S ( ) 2 S ( ) 0.3272
and so C f = 1.195
Ans. (h)
Cf =
=
=
0.075
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(i) Before computing the normalized displacement and momentum thicknesses for the Hiemenz
flow, we recall that, from Part (c), we have
1
* = ( 36 − 9 s + )
60
−76s 2 + s (156 + 17 ) − ( − 12 )( 24 + )
*
=
2520
42 ( 9 s − 36 − )
H =
2
76s − s (156 + 17 ) + ( − 12 )( 24 + )
For Pohlhausen’s profile, Pohlhausen = 7.052 and s = 2 + 6 . Backward substitution yields:
1
or * = 0.241
* = 36 − 9 ( 2 + 7.052
Ans. (i)
6 ) + 7.052
60
Next, we calculate
2
7.052
−76 ( 2 + 7.052
6 ) + ( 2 + 6 ) (156 + 177.052 ) − 7.052 ( 7.052 − 12 )( 24 + 7.052 )
*
=
2520
* = 0.105
or
Ans. (i)
At this point, the shape factor may be readily evaluated. We get
42 9 ( 2 + 7.052
6 ) − 36 − 7.052
H=
7.052 2
7.052
76 ( 2 + 6 ) − ( 2 + 6 ) (156 + 17 7.052 ) + ( 7.052 − 12 )( 24 + 7.052 )
H = 2.308
and so
Ans. (i)
(j) For the Majdalani–Xuan profile, Majdalani-Xuan = 5.761 and s = + 0.2075 . Backward
5
3
substitution yields:
* =
1
(36 − 9 ( 53 + 0.2075 5.761) + )
60
or
* = 0.267
Ans. (j)
Next, we calculate
2
−76 ( 53 + 0.2075 5.761) + ( 53 + 0.2075 5.761) (156 + 17 ) − ( − 12 )( 24 + )
*=
2520
* = 0.115
or
Ans. (j)
Finally, the shape factor may be readily evaluated. We obtain
42 9 ( 53 + 0.2075 5.761) − 36 − 5.761
H=
2
76 ( 53 + 0.2075 5.761) − ( 53 + 0.2075 5.761) (156 + 17 5.761) + ( 5.761 − 12 )( 24 + 5.761)
and so
H = 2.318
Ans. (j)
The errors associated with the Majdalani–Xuan profile in predicting the key boundary-layer
characteristics may be determined relative to the reference Hiemenz values obtained by
numerical integration. These are * = 0.270 and * = 0.122 . In comparison, the Majdalani–
Xuan characteristics entail errors of 1.1% and 5.7%, respectively. These constitute one order of
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magnitude improvements over the same two properties computed using Pohlhausen’s quartic
polynomial and yielding, respectively, relative errors of 11% and 14%.
4-77 Consider a shape-preserving mean-flow profile in the laminar boundary layer over a flat
plate with no pressure gradient:
u
y
=
= F ( );
0 1
U
As usual, let s = F (0) denote the axial velocity slope at the wall. For an arbitrary profile F ( ) ,
the normalized boundary-layer properties, * = * / , * = / , and H = * / *, consist of
pure constants that can be determined from their defining integrals once F ( ) is specified.
(a) Recalling that C f = 2d / dx , show that the wall shear stress and skin friction coefficient can
be reduced to:
2 s
(definition)
U
U
w =
s and C f =
2 * d (momentum-integral relation)
dx
(b) By equating the defining expression for C f and its momentum-integral relation to , show
that
d =
s
dx
U *
(c) By integrating from the plate’s leading edge to any station x , show that ( x) and its
derivative can be expressed in terms of Rex = Ux / , * , and the slope s :
x
=
a
2s
; a=
*
Rex
and
d ( a / 2)
=
dx
Rex
(d) Using the defining expression for the friction coefficient, C f = 2 w / ( U 2 ) , show that
Cf =
b
; b = 2 s *
Rex
(e) Prove the two identities ab = 2s and b / a = * .
(f) Recalling that the displacement and momentum thicknesses may be specified as * = * and
= * , show that
*
x
=
c
; c = a *
Rex
and
x
=
d
; d = a * = b
Rex
(g) Using CD , plate = 2 ( L) / L from Eq. (4-10) and ( x) above, show that
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2a *
ReL
(h) Evaluate (a, b, c) for the Majdalani–Xuan quartic profile and compare your results to the
CD, plate =
classic Blasius values.
Solution:
(a) From the definition of the wall shear stress, we have, on the one hand,
u
u
U U
U F
U
w =
=
=
=
s
y y =0
y
=0
y =0
The corresponding skin friction coefficient returns
U
s 2 s
w
Cf = 1
=1 2=
2
U
2 U
2 U
On the other hand, using Kármán’s momentum-integral equation, we have
U
d
d
C f = 2 *
+ 2 ( 2 + H ) *
= 2 *
dx
dx
U
Ans. (a)
Ans. (a)
Ans. (a)
(b) Equating the two expressions of C f from Part (a), we get, in general,
2 *
d 2 s
=
dx U
and so
d =
s
dx
U *
Ans. (b)
(c) Using the result of Part (b), both sides of the equation may be integrated from the leading
edge of the plate to any position x . We get:
x
2
s dx , = xs and so = 2s x or = 2s
d
=
0
2 U *
x
* U
* Ux
0 U *
This shows that in general, the disturbance thickness coefficient can be easily predicted using
x
=
a
2s
;a=
*
Rex
Ans. (c)
(d) Using the second expression for the skin friction coefficient from Part (a), we have
1 2s
d
2s *
*
C f = 2 *
= 2 * 2 =
dx
Rex
Rex
and so
Cf =
b
; b = 2s *
Rex
Ans. (d)
(e) It is straightforward to show that
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b
2s *
2s * *
=
=
= *
a
2s
2s
*
(f) Using the displacement thickness definition and dividing both sides by x , we have
*
a *
* = * and so
= * =
x
x
Rex
ab =
2s
2s * = 2s
*
and
*
hence
x
=
c
; c = a *
Rex
Ans. (e)
Ans. (f)
Similarly, using the momentum thickness definition and dividing both sides by x , we have
= * and so
hence
x
=
* a *
= =
x x
Rex
d
; d = a * = b
Rex
Ans. (f)
(g) Using CD , plate = 2 ( L) / L from Eq. (4-10) and ( x) from Part (f) at x = L , we have:
2
2 a *L
CD, plate = ( L) =
L
L ReL
Simplifying, we get
2a *
Ans. (g)
CD, plate =
ReL
(h) Before evaluating the key boundary-layer properties for the Majdalani–Xuan profile, we
recall from the foregoing analysis that:
2s
a=
; b = 2s * ; c = a* ; ab = 2s
*
For the Majdalani–Xuan profile, the slope is set at s = 5 / 3 . This means
3
4 13s 19 s 2
−
= 0.1337
* = ( 4 − s ) = 0.35 and * = +
35 210 630
20
By substituting these values into the above expressions, we find:
a = 4.993; b = 0.668; c = 1.748
Ans. (h)
For the classic Blasius (1908) solution:
a = 5; b = 0.664; c = 1.72
The boundary-layer predictions based on the Majdalani–Xuan profile deviate from the classic
Blasius values for a, b and c by 0.13%, 0.53% and 1.6%, respectively. These percentages
depend on the use of double precision in computing all properties directly from their defining
expressions. If 10 significant digits are retained before computing the errors in Table 4-1, one
would obtain the following results:
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Table 4-1 Boundary-layer predictions from five piecewise analytic profiles with their errors relative to the
classic Blasius values showing 10 significant digits
u
= F ( )
U
2 − 2
3
2
− 12 3
* =
sin ( 12 )
5
3
− 3 + 13 4
Blasius (1908)
* =
H=
*
x
Re x
Cf
Re x
*
x
Re x
L2 error
0.333333333 0.133333333 2.500000000 5.477225575 0.730296743 1.825741858 0.020
3.1%
0.25%
3.5%
9.5%
10%
6.1%
0.375000000 0.139285714 2.692307692 4.640954808 0.646418705 1.740358053 0.034
9.0%
2 − 2 3 + 4
*
4.7%
4.0%
7.2%
2.6%
1.2%
0.300000000 0.117460317 2.554054054 5.835585150 0.685449684 1.750675545 0.054
13%
12%
1.4%
17%
3.2%
1.8%
0.363380227 0.136619772 2.659792366 4.795326227 0.655136377 1.742526735 0.021
5.6%
2.7%
2.7%
4.1%
1.3%
1.3%
0.350000000 0.133686067 2.618073878 4.993399337 0.667547918 1.747689767 0.008
1.7%
0.52%
1.1%
0.13%
0.53%
1.6%
0.344
0.133
2.590
5.000
0.664
1.720
n/a
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CHAPTER 5. THE STABILITY OF LAMINAR FLOWS
P5-1
While holding ( g , 1, 2 , J
)
constant, show that the right hand side of Eq. (5-9) has a
/2
minimum at the wavenumber − g ( 1 − 2 ) /J . Find experimental data somewhere and
estimate this “critical” wavelength and velocity difference for air blowing over gasoline.
Solution The equation (5-9) to be studied is repeated here:
(U1 − U 2 )
2
g ( 1 − 2 ) + 2J ( 1 + 2 )
(5-9)
1 2
Instead of >, we will use an equality ( = ) to find the minimum value of (U1 − U 2 ) which we will
2
(
)
call Y for short. The equation is of the form Y = a + b 2 / . Differentiating with respect to ,
we find that dY /d = 0 when 2 = a/b. Thus our immediate result:
2 |min velocity =
g ( 1 − 2 ) ( 1 + 2 ) / ( 12 ) = g ( 1 − 2 )
J
( 1 + 2 ) / ( 12 )
Ans.
J
2
Plot (U1 − U 2 ) versus dimensionless wavenumber J / g ( 1 − 2 )
1/2
The curve is rather flat, but the minimum is at = g ( 1 − 2 ) /J
1/2
as follows:
, as predicted.
(b) For air blowing over gasoline, 1 680 kg/m3 , 2 1.2 kg/m3 , J 0.022 N/m. We may
calculate a minimum at 550 m−1 = 2 / , or a wavelength 0.011 m. This corresponds to
a minimum value of (U1 − U 2 ) 4.5 m/s.
Ans. (b)
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-1-
P5-2
Show that, if the upper and lower velocities in Fig. 5-2 are negligible, and if surface
tension is neglected, a disturbance of the interface will propagate at the phase speed
g ( 1 − 2 )
2 ( 1 + 2 )
c=
where is the wavelength of the disturbance. Discuss what might happen if 1 2 . Estimate
this propagation speed for an air-water interface when the wavelength is 3 meters.
Solution: If surface tension and the velocities are negligible, then we are studying the effect of
gravity on an interface between two still fluids of different density. The expression for wave
frequency, from Eq. (5-8), reduces to
−
g ( 1 − 2 )
( 1 + 2 )
If an interfacial wave is produced, its propagation speed is
C=
T
=
2 /T
=
2 /
Thus, for the present case of an interfacial travelling wave,
C=
g ( 1 − 2 )
g ( 1 − 2 )
=i
=i
( 1 + 2 )
2 ( 1 + 2 )
Ans.
This is the speed of a deep-water wave, far from any upper or lower boundaries. For air layered
over fresh water, 1 998 kg/m3 , 2 1.2 kg/m3. If = 3 m, the propagation speed is
C 9.81( 3m )( 998 − 1.2 ) / 2 ( 998 + 1.2 )
1/2
= 2.16 m/s.
Ans.
If 1 2 , then the argument of the square root is positive and the wave is unstable. Presumably
the two layers will overturn so that the heavy fluid goes to the bottom.
Ans.
P5-3
Derive a linearized disturbance equation for the celebrated van der Pol equation:
d2X
dt
2
(
+ X = 0,
) dX
dt
+ C X2 − 1
C = constant
(1)
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-2-
Assume that X o (t) is a known solution to (1). Substitute the proposed solution X o + into (1),
where ( t ) is a (small) disturbance. Write out all the resulting terms.
Solution:
(
)
(
)
Xo" + " + C Xo2 − 1 Xo' + C Xo2 − 1 + 2 ' + 2 Xo CXo ' + 2CX o 2 +
+ CXo ' 2 + C2 ' + Xo + = 0
where the prime ( ') denotes differentiation with respect to time. The sum of the boldface terms
vanishes, because X o is a known solution to (1). What remains is the disturbance equation, which
we linearize by neglecting power and products of e and e' . What then remains is the desired
linearized disturbance equation:
(
)
''+ C X o2 − 1 '+ (1 + 2CX o X o ' ) = 0
(2) (Ans.)
This is a second order linear differential equation with variable (oscillatory) coefficients. The
original solution Xo ( t ) is stable if the solutions to (2) do not grow indefinitely with time. It helps
(
)
a little to modify (2) by defining a new variable Z = exp (1/2 ) C Xo2 − 1 dt , resulting in a
differential equation without a first-derivative term:
Z" + f ( t ) Z = 0,
where f ( t ) = 1 + CXo Xo ' −
(
)
1 2 2
C Xo − 1
4
(4)
The coefficient f ( t ) is periodic, thus (4) is related to the Mathieu Equation.
We will not proceed any further with the analysis. It has been proven in the literature that
solutions of the van der Pol equation (1) are stable, that is, bounded, for any positive value of the
constant C.
P5-4
Verify the Orr-Sommerfeld equation (5-18) in the text, page 347
Solution: Begin with the linearized two-dimensional (parallel-flow) equations (5-15a,b,c):
iu + v ' = 0
(
iu ( U − c ) + U ' v = −ip/ + v u "− 2u
(
iv ( U − c ) = −p ' / + v v"− 2 v
)
)
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-3-
Eliminate p by differentiating the second relation, multiplying the third relation by ( i ) , and
subtracting. The result is:
(
v ( U − c ) 2 + U"− i3v = −i ( U − c ) u '+ v u '''− 2 2 u '
)
Now eliminate u by substituting u = ( −v'/i ) , u ' = ( −v"/i ) ,etc., from the first relation. The
result is the desired Orr-Sommerfeld equation which was to be proved:
( U − c ) ( v"− 2 v ) − U"v +
P5-5
(
)
iv
v ''''− 22 v ''+ 4 v = 0
(Ans.)
Consider Rayleigh’s inviscid stability equation (5-21). The perturbation amplitude v ( y )
is complex, and, for temporal stability analysis, is real and c is complex. Show that Rayleigh’s
equation may be split into real and imaginary parts, as follows:
U " (U − cr )
U " ci
vr +
vr "− 2 +
=0
2
2
(U − cr ) + ci (U − cr )2 + ci2
U " (U − cr )
U " cr
vi "− 2 +
v
−
=0
i
2
2
2
2
U
−
c
+
c
U
−
c
+
c
(
(
r)
i
r)
i
Explain, in words, a possible method for solving this system numerically for a given U ( y ) .
Solution: This is a good exercise for students to split a linear differential equation into real and
imaginary parts. The numerical scheme would typically begin at large y, where vr and vi are
exponential in y, and integrate inwards, varying cr and ci to match the wall conditions.
P5-6
For an approximate quartic polynomial flat-plate velocity distribution,
u/ U = 2− 23 + 4 ,
= y/
solve the inviscid Orr-Sommerfeld equation, find the eigenvalues, and plot ci versus . Nondimensionalize with respect to and U.
Solution: The Orr-Sommerfeld relation becomes
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-4-
2
d 2 v*
2 d U*
U*
−
c*
−
*
v*
=
v*,
(
) 2
2
d
d
where v* =
v
c
, c* = , * =
U
U
and where U* = 2− 23 + 4 , d2 U*/d2 = −12+ 122
This quartic polynomial is a reasonable approximation to U( y ) but is not very accurate for
U"( y ) , since it does not satisfy U"' ( 0 ) = 0 as required for flat plate flow. [One might try, as an
alternate profile, a 5th or 6th order polynomial to meet this wall condition.]
As suggested in the problem statement, we start the integration out at = 2, where
v* = exp ( −*) ,
and integrate inward numerically to satisfy the wall conditions
v*( 0) = v*' ( 0) = 0. We guess values of c* = c*r + ic*i . When passing through the ‘critical’ layer,
where U* = c*, there will be an unavoidable blip in the integrated curve, which we avoid by not
letting the denominator |U* − c*| drop below, say, 0.001. The integrated solution then passes
smoothly through this singularity.
The (inviscid) eigenvalues and eigenfunctions are found for a given by varying ( cr ,ci )
until we satisfy v ( 0) = v' ( 0) = 0. Here we show only the solutions for the neutral condition,
ci = 0. The eigenvalues vary with wavenumber as follows:
=
0.5
1.0
1.5
2.0
2.5
c r /U o
=
0.190
0.332
0.469
0.620
0.762
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-5-
The neutral eigenfunctions v* ( y*) for this case are shown in the following graph.
They are not especially accurate compared to solutions using the exact Blasius functions for U ( y )
and U" ( y ) - see, for example, the text by Betchov and Criminale (1967), pp. 46–50.
P5-7 For two-dimensional stagnation flow, U = Kx, estimate the position Re x where instability
first occurs.
Solution: From Table 5-1, p. 354 of the text, stagnation flow ( = 1.0 ) becomes unstable at
Re = 5636. Meanwhile, from the Falkner-Skan solutions in Table 4-2, p. 243, the dimensionless
momentum thickness in stagnation flow is θ* = 0.29235. This gives the prediction
1/ 2
m +1 U
2 vx
= 0.29235, for m = 1, or:
0.29235
=
x ( Ux/v )1/ 2
Solve this for Re and set it equal to the critical value:
Re = 0.29235 Re1/2
x = 5636, or
Re x,crit = 3.72 E8
(Ans.)
This is a large Reynolds number, illustrating the extreme (but not infinite) stability of the
favorable-gradient stagnation-flow velocity profile.
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-6-
We can check this result against the Wazzan et al. (1981) shape factor correlation in
Fig. 5-31, page 381. The shape factor for stagnation flow is, from Table 5-1, H = */ =
12490/5636 = 2.22. Entering Fig. 5-31 at H = 2.22, we can read Re x,crit = 3 E8, in good
agreement with our estimate above.
P5-8 For the separating Falkner-Skan wedge flow, = −0.19884, estimate the position Re x
where instability first occurs.
Solution: From Table 5-1, p. 354 of the text, separating flow ( = −0.19884) becomes unstable at
Re 17. Meanwhile, from the Falkner-Skan solutions in Table 4-2, p. •••, the dimensionless
momentum thickness in separating flow is θ* = 0.58544. This means that
1/2
m +1 U
2 vx
= 0.58544 for m = −0.09043, or:
0.8681
=
x Re1/2
x
Solve this for the momentum-thickness Reynolds number and set it equal to the critical value:
Re = 0.8681 Re1/2
x = 17, or: Re x,crit 380
(Ans.)
This is a remarkably small value of the critical Reynolds number, illustrating the extreme
instability of the S-shaped separating-flow velocity profile.
We cannot check this result very well against the correlation of Wazzan et al. (1981) in
Fig. 5-31, page 381, because H in that figure only goes up to 3.1. The shape factor for FalknerSkan separating flow is, from Table 4-2, H = 2.35885/0.58544 = 4.0, which is off the bottom right
of Fig. 5-31. However, our value of Re x,crit = 380 seems to be a reasonable extrapolation of Fig.
5-31.
P5-9 For the Howarth freestream velocity U = Uo (1 − x/L ) , estimate the point ( x/L ) where
boundary layer instability first occurs, if Uo L/v = 1 E6.
Solution: This is not a similarity solution, so we need estimates of ( x ) and H ( x ) along the wall,
assuming that instability occurs before boundary-layer separation. We could use Thwaites’ method
from Sect, 4-6.7, pp. 265–268 of the text, for which
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-7-
2 = −
vL
,
Uo
−0.075 (1 − x/L )
where
Thus: Re = ( − ReL )
1/2
= 1000 ( − )
1/2
−6
− 1
since ReL = 106
One computes Re ( x ) and, with ( x ) known from Table 4-4 or Eq. (4-141) on p. 267, one
computes H ( x ) and consults the shape factor correlation, Fig. 5-12 on p. 360 to see if instability
has been reached. For example:
At x/L = 0.02: = −0.0097, Re = 98, H = 2.63, Re = 257
Check this against H = 2.63 in Fig. 5-12, for which Re*,crit = 400. Thus we have not yet reached
the point of instability. Move a bit further downstream:
At x/L = 0.03: = −0.015,
Re = 123,
H = 2.65, Re* = 325
At x/L = 0.04: = −0.021,
Re = 144,
H = 2.68, Re* = 386
For H = 2.65, from Fig. 5-12, read Re*,crit 350 - perhaps just a bit shy of instability.
For H = 2.68, from Fig. 5-12, read Re*,crit 300 - perhaps just a bit downstream of instability.
Our best estimate, then, reading a very small-scale correlation curve, is:
Howarth flow:
Uo L/v = 106 :
( x/L )instability = 0.035
(Ans.)
This result is equivalent to a point Rex = (1E6)( 0.035) = 35,000. We can check this estimate from
the Wazzan correlation in Fig. 5-31: for H = 2.65, read Re x,crit = 35, 000, which is excellent
agreement for this somewhat uncertain prediction.
P5-10 Generalize Prob. 5-9 to calculate and plot the instability point ( x/L )crit as a function of
nominal Reynolds number ReL = Uo L/v.
Solution: We follow the same procedure as Prob. 5-9, systematically varying the Reynolds
number:
= −0.075 (1 − x/L )
−6
− 1 , Re =
( − ReL ) , H = H ( ) , Re = H Re , Rex = ReL ( x/L )
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-8-
We then read Fig. 5-12 to see if Re is critical for the local H, and check by reading Fig. 5-31 to
see if the local Re x is critical for that value of H. The results may be plotted as follows for a wide
range of nominal Reynolds numbers:
The value x/L = 0.035 for ReL = 1E6 is taken from Prob. 5-9 above. We see that the point of
instability always occurs before the point of boundary-layer separation ( x/L = 0.120) for a very
wide range 1E4 Uo L/v 1E8. The two predictions, Figs. 5-12 and 5-31, give similar results.
P5-11 For potential freestream flow across a circular cylinder, U = 2Uo sin ( x/a ) , with
Re D = 1E6, estimate the position ( x/a )crit where instability first occurs.
Solution: This is a variation of Prob. 5-9, with a different freestream velocity distribution. We
need to estimate ( x ) , ( x ) , H ( x ) , and Re ( x ) for cylinder flow using, say, Thwaites’
method. The results of this prediction were given in Prob. 4-23 on page 107 of this Manual:
2 =
v a ()
0.45cos
x
, where =
8 − cos 3sin 4 + 4sin 2 + 8 , =
6
2Uo cos
a
15sin
(
)
We compute H ( ) from Table 4-4 or Eq. (4-141) on p. 267, and we also compute
Re = sin ( ReD / cos ) , Re = H Re , Rex = ReD sin
1/2
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Then we check Fig. 5-12 to see if Re is critical for this value of H and ReD . A sample
calculation is as follows for ReD = 1E6:
= 60: = +0.059, Re = 210, H = 2.41, Re = 507, Rex = 9.1E5
For H = 2.41, Fig. 5-12 predicts that Re,crit = 25, 000, so we have not yet reached instability.
Also, for H = 2.41, Fig. 5-31 predicts that Re x,crit 8E6, so this correlation also indicates that we
are too early to be unstable. Try a position further along:
= 80: = +0.031, Re = 420, H = 2.49, Re = 1050, Re x = 1.4 E6
For H = 2.49, we read from Fig. 5-12 that Re,crit = 1200, so we are before the point of instability.
For H = 2.49, we read from Fig. 5-31 that Re x,crit = 8 E5, just past the point of instability. This
seems to bracket our prediction well enough:
Cylinder Potential Flo w:
Uo D/v = 106 :
crit = 80, or:
( x/a )crit = 1.4
(Ans.)
P5-12 Using the guidance of Probs. P5-5 and P5-6 and any numerical method of your choosing,
solve the Rayleigh equation (5-21) for a simplified boundary layer flow, U = tanh ( y ) , 0 y ,
with a typical value of 0.2 to 0.3. Do you expect any inviscid instability? If time permits, plot
some computed values of cr versus . [HINT: Begin at large y 4 with the exponential solution
to Eq. (5-6) and integrate backwards to the wall.]
Solution: This assignment is quite similar to Prob. 5-6: a simple function for U ( y ) , numerical
integration backward from large y, and a search for neutral eigenvalues ( , cr ) with ci = 0. Here
are some neutral pairs computed by the writer, plotted on the next page:
=
0.5
1.0
1.5
cr =
0.30
0.41
0.55
There is no point of inflection in tanh ( y ) , except y = 0, thus there is no inviscid instability.
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P5-13 Using the guidance of Probs. P5-5 and P5-6 and any numerical method of your choosing,
solve the Rayleigh equation (5-21) for the Blasius boundary layer flow, which you should generate
from Eqs. (4-45) and (4-46). Do you expect any inviscid instability? Select a value of in the
range 0.2 to 0.3. [HINT: Begin at large 5 with the exponential approximation of Eq. (5-6) and
integrate backwards to the wall.]
Solution: First we have to generate the Blasius solution numerically, using a small enough step
size, say, = 0.05, to enable us to compute an accurate neutral Rayleigh solution. The Rayleigh
procedure follows the examples set in Problems 5-6 and 5-12. We need U f and U f and
then solve Eq. (5-21), splitting it up, real and imaginary, as in Prob.
5-5. There is no inviscid instability because the Blasius profile only has a point of inflection at
= 0. The neutral eigenvalues ( ci = 0) found by the writer are as follows:
=
0.3
0.5
1.0
cr =
0.42
0.54
0.83
The eigenfunctions vr ( ) are shown on the next page.
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-11-
Neutral inviscid eigenfunctions vr ( ) for the Blasius velocity profile.
P5-14 For stagnation-point flow, U = Kx, estimate the position Re x where transition occurs,
using the method of Michel, Eq. (5-38). Why is Granville’s method, Section 5-5.1.1,
inappropriate?
Solution: Michel’s method requires an estimate of Re ( x ) , which we could take from the
Falkner-Skan solution of Table 4-2, p. 243 of the text:
1/2
* = 0.29235,
vx
or: = 0.29235
U
= 0.29235 ( v/K )
1/2
The same formula was used in Prob. 5-7, page 143 of this Manual. This relation is equivalent to
Re = 0.29235 Re x . Thus Michel’s correlation predicts that
10
2.9
0.4
Re = 0.29235 Re1/2
x 2.9 Re x , or: Re x,trans =
9.2 E9
0.29235
(Ans.)
This is a very large transition Reynolds number, showing the extreme stability of stagnation flow,
which has a strong favorable pressure gradient.
An alternate estimate, always possible whether the flow is “similar” or not, is to use
Thwaites’ method, Eq. (4-138), to estimate the momentum thickness for the given U ( x ) :
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-12-
x
2
0.45
=
v ( Kx )6
( Kx ) dx =
0
5
0.075
,
K
1/2
or:
v
0.274
K
This estimate for is about 6% less than the Falkner-Skan result and leads to the result
10
2.9
Rex,trans. =
= 1.8 E10
0.274
(Ans.)
[The 6% discrepancy is greatly magnified after taking the tenth power.] Both estimates are
acceptable to this writer. The method of Granville, Eq. (5-37), is inappropriate because it is limited
to a Thwaites parameter 0.4. From our Thwaites calculation above, stagnation flow has
= 0.075, or too high for Granville’s correlation. Should we choose to extrapolate Granville’s
formula to such a high , it would predict Re,tr = 42100, or Re x,tr 2.3 E10, about a factor of
2 too large.
P5-15 For the separating Falkner-Skan flow, = −0.19884, use any appropriate correlation to
estimate the point of transition, Re x,tr . Compare with Fig. 5-32.
Solution: We may use Michel’s method, which requires an estimate of Re ( x ) , which we could
take from the Falkner-Skan solution of Table 4-2, p. 243 of the text:
1/2
m +1 U
2 vx
= 0.58544 for m = −0.09043, or:
0.8681
=
x Re1/2
x
Michel’s method, Eq. (5-38), then predicts
10
2.9
0.4
Re = 0.8681 Re1/2
x 2.9 Re x , or: Re x,tr
= 1.73 E5
0.8681
(Ans.)
This estimate is in agreement with Fig. 5-32. Other methods could be used as well, as illustrated
in Fig. 5-32.
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-13-
P5-16 For the Howarth flow, U = Uo (1 − x/L ) , use Michel’s method, Eq. (5-38), to estimate the
transition point if Uo L/v = 4E6. Neglect freestream turbulence.
Solution: For Michel’s method we need momentum thickness ( x ) , which was computed in Prob.
5-9 on page 144 of this Manual:
2 = −
Thus
vL
−6
, where = −0.075 (1 − x/L ) − 1
Uo
Re = U/v = ( U/Uo ) ( − ReL )
1/2
for this particular flow. Noting that
x/L =
until it is
( Ux/v ) / ( ReL ) ( U/Uo ) , we may evaluate Michel’s correlation, Re = 2.9 Re0.4
x
satisfied. Some numerical results, for Uo L/v = 2 E6, are as follows:
x/L =
0.02
0.04
0.06
0.08
0.10
0.12
Rex =
39200
76800
112800
147200
180000
211200
Re =
136
196
244
287
327
366
2.9Re0.4
x = 199
261
304
339
367
391 (?)
We see that, at the separation point, ( x/L ) 0.120, the momentum-thickness Reynolds number is
still not quite high enough for transition. Therefore, for ReL = 2E6, this flow does not undergo
transition before separation is reached.
If we increase Re L , we do get transition before separation, as listed below:
Re L =
2 E6
3 E6
4 E6
6 E6
8 E6
1 E7
( x/L )tr =
none
0.131
0.119
0.102
0.091
0.081
P5-17 Generalize Prob. 5-16 into a parametric computer plot of ( x/L ) tr versus Re L .
Solution: All we do is follow the same procedure of Prob. 5-16 above with different values of
Re L . Some results are tabulated above, and we may plot a broad spectrum of results as follows:
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-14-
P5-18 For (potential) cylinder flow, U ( x ) = 2Uo sin ( x/a ) , use Michel’s method, Equation
(5-38), to estimate the transition point if Uo D/v = 2 E6. Neglect freestream turbulence.
Solution: Let us reproduce the Thwaites’ result for momentum thickness in potential-stream
cylinder flow from Prob. 5-11, page 145 of this Manual:
2 =
v a ()
0.45cos
x
, where =
8 − cos 3sin 4 + 4sin 2 + 8 , =
6
2Uo cos
a
15sin
(
)
where a = D/2 is the cylinder radius. Thus Re = U/v = sin ( ReD /cos)
1/2
for this particular
flow. Noting that x/D = ( Ux/v ) / ( ReD ) ( U/U o ) , we may evaluate Michel’s correlation,
Re = 2.9 Re0.4
x , continuously (as increases) until it is satisfied. Some numerical results for our
case, ReD = 2 E6, may be listed as follows:
=
50°
60°
70°
80°
90°
100°
110°
120°
2.75E6
3.14E6
3.44E6
3.61E6
3.63E6
Rex =
1.34E6
Re =
345
420
501
589
693
820
990
1237
2.9 Re0.4
x = 818
924
1016
1092
1151
1194
1217
1220
1.81E6 2.30E6
We see that the Michel criterion is satisfied just before 120°, say, tr 119.5°
(Ans.).
However, the laminar separation point, from Prob. 4-23, page 107 of this Manual, is at
103.1. Therefore our estimate of 119.5° is unrealistic according to the limitations of boundary
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-15-
layer theory. Furthermore, it is known from experiment (see, e.g., Fig. 3-38a) that transition in
cylinder flow actually occurs at about ReD 2.5E5, or much less than the value 2E6 studied here.
Thus it appears to this writer that Michel’s method may be unrealistic in general for estimating
the transition point in strong favorable gradients.
Meanwhile, the method of Wazzan et al, Fig. 5-31, indicates that transition occurs at about
tr = 92, which is quite reasonable when compared to experiment. Wazzan’s method is used in
Prob. 5-19 below.
P5-19 Air at 20°C and 1 atm flows quietly toward a wedge of half-angle 36°, resulting in a powerlaw freestream and a laminar boundary layer along the surface. Use Wazzan’s method, Eq. (5-42),
to estimate the transition Reynolds number Rex,tr .
Solution: This problem is dimensionless, and it doesn’t matter if the fluid is air or water or
whatever. For wedge flow, Fig. 4-10, the half-angle is 36 = /2, or = 0.4, where the power
exponent in U = Cx m is m = 1 4. The shape factor H is constant, and we can either use Thwaites’
method or simply interpolate in Table 4-4 to estimate H 2.34. (a) Wazzan’s transition
correlation, Eq. (5-42), gives log10 ( Rex,tr ) 8.03, or Rex,tr 1.1E8.
Ans. (a)
(b) The Thwaites/Michel approach, Eqs. (4-138) and (5-38), yields
2
v
0.45
( )
1/4 6
K6 x
x
K
0
5
( x1/4 )
5
dx =
0.45 3/4 0.2 x
0.2
x =
; =
;
2.25K
U
x
Re x
Re = 0.2 Re x = 2.9 Re0.4
x ,tr , Solve for Re x ,tr 1.3E8
Ans. (b)
The two estimates are remarkably close but, as seen in Fig. 5-32 of the text, the two theories spread
apart as the pressure gradient becomes more favorable, 0.4.
P5-20 Repeat Prob. 5-14 (stagnation flow) for a freestream turbulence level T = 1%.
Solution: We must change methods here, since the Michel and Wazzan correlations are valid only
for negligible freestream turbulence. Equation (5-47) is quite simple, and predicts
Re x,tr 1.4E6 ± 40%, reasonable, but it is only suggested for high turbulence, T 1%.
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-16-
Dunham’s correlation, Eq. (5-46) or Fig. 5-34(b), is not valid for 0.04, whereas
stagnation flow has 0.075. This leaves only the van Driest & Blumer semi-empirical theory,
Eq. (5-45), p. 385 of the text. For stagnation flow, m = 1.0 and = 2.4 from Table 4-2, page 243.
Equation (5-45) then becomes, for T = (1%) = 0.01:
1690
Re1/2
x
or:
= 0.312 (1 + 0.11)
−0.578
+ 1.6 ( 2.4 ) Re1/2
x ( 0.01) ,
2
2
Z2 + 320 Z − 1.834E6 = 0, Z = Re1/2
x , with solutions Z = 1203 and −1524
We reject the negative solution, Z = −1524, as spurious. Therefore the predicted transition
Reynolds number in stagnation flow, for a turbulence level of 1%, is
Re1/2
x = 1203,
or:
Re x,tr = 1.45 E6
(Ans.)
This is approximately verified by the plotted value in Fig. (5-34a) at m = 1 and T = 1%. This is a
factor of 6000 less than the value predicted for T = 0% in Prob. 5-14.
P5-21 Repeat Prob. 5-15 (separating flow) for a freestream turbulence level T = 1%.
Solution: Since 0.04 for this case, both the Van Driest & Blumer (Fig. 5-34a) and Dunham (Fig.
5-34b) correlations are valid. The Michel, Granville, and Wazzan methods are not, since they assume
negligible freestream turbulence. For the Van Driest and Blumer formula. Equation (5-45), take
m = −0.09043 for separating flow and read = 4.9 from Table 4-4. Then, for T = (1%) = 0.01,
Eq. (5-45) predicts
1690
Re1/2
x
or:
( 0.312 )( −0.09043 + 0.11)
−0.528
+ 1.6 ( 4.9 ) Re1/2
x ( 0.01) ,
2
2
Z2 + 648 Z − 440000 = 0, z = Re1/2
x , with solutions Z = +414 and Z = −1062
We reject the negative solution, Z = −1062, as spurious. Therefore the predicted transition
Reynolds number in separating flow, for a turbulence level of 1%, is
Re1/2
x = 414,
or:
Re x,tr = 172, 000
(Ans.)
This value is plotted (approximately) in Fig. 5-34(a).
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-17-
For the Dunham correlation, Eq. (5-46), we take T = 0.01 and we need an estimate for
the value of at separation. For Falkner-Skan flow, m = −0.09043, we find from Table 4-4, page
267 of the text, that
2 dU ( 0.58544 )2 ( 2x ) mU
=
=
= −0.0682 for m = −0.09043
v dx ( m + 1) U
x
We may then evaluate Dunham’s correlation:
680
− 80 0.01
Re,tr = 0.27 + 0.73e ( )( ) 550 +
= 447
1 + 100 ( 0.01) − 21( −0.0682 )
Meanwhile, from Prob. 5-15, for separating flow, Re = 0.8681 Rex . Therefore
2
447
Rex,tr =
= 266, 000
0.8681
(Ans. - Dunham)
Considering the uncertainty of transition-point estimation, this result is quite comparable to the
Van Driest & Blumer value of 172,000 predicted above. Note - exasperatingly - that these values
for T = 1% are about the same as the ( T = 0% ) Michel prediction in Prob. 5-15! This result casts
further doubt upon the accuracy of the Michel correlation.
P5-22 Repeat Prob. 5-16 (Howarth flow) for a freestream turbulence level T = 1%.
Solution: Since, strictly speaking, the Van Driest & Blumer method, Eq. (5-45), applies only to
similar (Falkner-Skan) flows, we should not apply it to Howarth (nonsimilar) flow. That leaves
only the Dunham correlation. Eq. (5-46), which for T = 0.01 becomes
680
Re,tr = 0.598 550 +
2 − 21 tr
We follow the same procedure as in Prob. 5-16, this time monitoring ( x ) and using the above
relation to compute Re,tr for comparison with the local Re until they are equal. Some example
results for Re L = 2 E6 are as follows:
x/L =
=
Re =
0.02
0.04
0.06
0.08
0.10
0.12
–0.0097
–0.0208
–0.0337
–0.0487
–0.0661
–0.0865
136
196
244
287
327
366
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-18-
Re,tr =
513
487
479
463
449
435
Thus, even at separation ( x/L = 0.12) for a turbulence level of 1%. Dunham’s method does not
predict transition for Howarth flow at ReL = 2 E6 - which was also the case with Michel’s method
at T = 0% from Prob. 5-16.
If we try some higher values of Re L , we do find transition before the separation point, as
follows:
Re L =
( x/L )tr
2 E6
3 E6
4 E6
6 E6
8 E6
N/A
0.116
0.096
0.0723
0.0583
1 E7
0.0489
These transition points are earlier than those predicted for T = 0% by Michel’s method in
Prob. 5-17 - see the plot at the bottom of page 150 of this Manual.
P5-23 Repeat Prob. 5-18 (cylinder flow) for a freestream turbulence level T = 1%.
Solution: As discussed in Prob. 5-22 above, the Van Driest & Blumer method is not strictly valid,
since cylinder flow is not ‘similar’. We therefore use the Dunham correlation, Eq. (5-46):
For T = (1% ) = 0.01:
Re,tr
680
0.598 550 +
2 − 21 tr
We follow the same procedure as in Prob. 5-18, this time monitoring ( x ) and using the above
relation to compute Re,tr for comparison with the local Re until they are equal. Some example
results for ReD = 2 E6 are as follows:
=
40°
50°
60°
70°
80°
90°
100°
Re =
274
345
420
501
589
693
820
Re,tr =
1081
977
862
743
631
532
453
We see that transition is predicted somewhere between 80° and 90°. If we use smaller increments,
we will find the point of equality of Re at 82.1°
(Ans.).
The effect of Reynolds number for this same turbulence level T = 1% is as follows:
Re L =
1 E6
2 E6
3 E6
4 E6
6 E6
8 E6
1 E7
tr =
86.4°
82.1°
75.9°
71.5°
65.3°
60.8°
57.3°
These occur (plausibly) earlier than the predictions for T = 0% by Michel’s method in
Prob. 5.18, p. 151 of this Manual. They seem to be reasonable predictions.
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-19-
P5-24
For a pipe flow started from rest with acceleration a, as in Fig. 5-37b, the momentum
thickness initially grows according to the formula 0.35 ( vt )
1/2
. Apply this relation to Michel’s
steady-flow transition correlation, Eq. (5-38), assuming that “V” and “x” are given by constantacceleration formulas. Show that the result is a constant value of the dimensionless transition time
(
ttr a 2 /v
)
1/3
. Why does the diameter D not appear?
Solution: By “constant-acceleration formulas”, we simply mean to take “V” = at and
“x” = (1/2 ) at 2 . Substitute into Michel’s transition correlation, Eq. (5-38):
Re
V ( at ) ( 0.35
=
=
v
vt
v
x
v
1/2
a 2t 3
0.35
Clean up:
v
a 2t 3
2.9
2v
1/3
Solve:
) 2.9 Re0.4 = 2.9 Vx 0.4 = 2.9 ( at ) ( at 2 /2)
a2
t tr
v
0.4
2 1/3
a
, or: 0.35 ttr
v
v
3/2
0.4
1/3
2.9 a 2
= 0.4 ttr
2 v
6/5
10/3
2.9
= 0.4
2 ( 0.35 )
457
Ans.
This simple estimate, which is not too bad, is given as a straight line in Fig. 5-37b of the text. Here
there is no diameter effect on ttr , because the momentum thickness estimate, 0.35 ( vt )
1/2
, is
based on unsteady boundary layer theory, with no constraint or outer wall at the outer edge. The
data in Fig. 5-37b show that there is indeed an effect of diameter.
P5-25 Air at 20°C and 1 atm flow at U = 12 m/s past a smooth flat plate. A 1-mm diameter wire
trip is placed across the plate at x = 1 m. If freestream turbulence is negligible, at what position
x tr will transition occur? What wire location x will cause earliest transition?
Solution: For air at 20°C and 1 atm, take v = 1.5 E-5 m2 /s. Then, at x = 1 m, Re x = Ux/v = 8 E5,
at which point the displacement thickness is, from laminar flat-plate theory,
* =
1.72 x
Re1/2
x
=
1.72 (1.0 )
( 800,000 )1/2
0.00192 m = 1.92 mm
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-20-
Thus, at this position, a wire of diameter k = 1 mm has a ratio k/δ* = 1/1.92 0.52, which will
certainly cause early transition. From Fig. 5-36(a), read Re(rough)/Re ( smooth ) 0.40. We need an
estimate for Retr ( smooth ) : from Fig. 5-33, for the data of Schubauer and Skramstad, read, at zero
turbulence, Retr ( smooth ) = 3 E6. Then our estimate for the transition point location with the wire
trip is
Re x,tr = 0.4 ( 3 E6 ) = 1.2 E6 =
Ux (12 ) x
=
, or:
v 1.5 E-5
x tr = 1.5 m
(Ans.)
The wire causes transition at about 50 cm further downstream, whereas without the wire transition
would not occur until x = 3.75 m.
Since Re ( smooth ) = 3 E6 is a basic parameter in this problem, the earliest transition occurs
when the wire causes the lowest possible value of Re ( rough ) . From Fig. 5-36(a), the lowest
measured data give Re ( rough ) /Re ( smooth ) = 0.1 at k/* = 0.8. From flat-plate theory, this
corresponds to
k 0.1( 3 E6 )
k Re1/2
k
x
= 0.8 =
*
1.72 x
1.72 x
1/2
, or x = 398 k = 0.4 m
(Ans.)
A 1-mm wire placed at x = 0.4 m would cause transition almost immediately downstream.
P5-26 Repeat Prob. 5-25 for a freestream turbulence level T = 1 %.
Solution: This analysis is identical to Prob. 5-25 except that the reference value Re tr,smooth must
be modified for T = 1%. Fig. 5-33 does not extend out to T = 1%, therefore use the Van Driest &
Blumer correlation, Eq. (5-44), which is based upon the same data:
Re1/2
x,tr =
2
−1 + 1 + 132500 ( 0.01)
39.2 ( 0.01)
2
708, or:
Rex,tr ( T = 1% ) = 500, 000
Other parameters follow from Prob. 5-25: at x = 1 m and k = 1 mm, Rex = 8 E5. Thus the wire
is too late, transition has already occurred at Rex = 5 E5, or at the position x 0.625 m (Ans.).
At this high level of freestream turbulence, a trip wire is not needed.
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-21-
P5-27 Find out more about the narrow white band in the chaotic region of Fig. 5-38, for
3.825 r 3.865, by making computer iterations in that region for, say, r 0.0005. Plot the
resulting iterates only in this expanded region and comment on the resulting remarkable pattern
which appears.
Solution: The scale of Fig. 5-38 is so large that it hints little of the beauty of the narrow white area.
However, plotting the iterates in the narrow region 3.825 r 3.865 discussed above immediately
reveals a beautiful pattern, as shown below [one needs a personal computer with point-plotting
ability to do this - this writer uses the PLOT routine that is part of the TRUE BASIC software].
The “chaos” stops at about r = 3.827, and the iterates then regroup into three single lines, each of
which bifurcates at about r = 3.842 into a pattern resembling the original “period-doubling”
pattern of Fig. 5-38. In fact, each pattern - labelled #1, #2, and #3 in the above figure - is quite
similar in shape to the original pattern.
A greatly enlarged version of pattern #3, which lies in the region 0.95 x 0.97, is shown
below. We see that it closely resembles the original chaotic large-scale pattern of
Fig. 5-38. Thus chaotic phenomena often contain - hidden in the chaos - microscopic and
submicroscopic images - presumably occurring multiply at smaller and smaller scales - of the
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McGraw Hill LLC.
-22-
chaos. There is a very good discussion of this phenomenon in the popular-science book by James
Gleich, Chaos - Making a New Science, Viking Press, New York, 1987.
P5-28 Repeat Prob. P5-19 if the freestream has a turbulence level of 4%. Find the estimated
transition Reynolds number Rex,tr by two different methods and compare.
Solution: Recall that Prob. 5-19 involves flow past a wedge of half-angle 36°, with a power-law
freestream and a laminar boundary layer along the surface. We showed there that U = Kx1/2 ,
with = 0.4. That problem shows that, by Thwaites’ method, 2 /v 0.2 x /U and 0.05, a bit
too high to use Dunham’s freestream-turbulence correlation, Eq. (5-46).
(a) If used as an estimate, Dunham’s formula would yield too high a transition value:
Re ,tr 217, Re x,tr ( Re ,tr ) /0.2 = 234, 000
2
Ans. (a) – Dunham
(b) As a second estimate, for m = 1 4, = 0.4, Table 4-2 indicates 2.8, and we can estimate
from the van Driest and Blumer correlation, Eq. (5-45), that
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McGraw Hill LLC.
-23-
1690
Re1/2
x,tr
= 0.312 ( 0.25 + 0.11)
−0.528
+ 1.6 ( 2.8) Re1/2
x,tr ( 0.04 ) , solve Re x,tr 77, 000
2
2
Ans. (b)
(c) The simple estimate of Schmid and Henningson, Eq. (5-47), is quick and reasonable:
T ( Re x,tr )
1/2
1200, T = 4%,
solve
Re x,tr 90, 000
Ans. (c)
This is the limit of our prediction skills. A reasonable transition estimate is Rex,tr 100,000.
P5-29 The famous neutral curve of Taylor (1923), for Couette flow between rotating cylinders,
is in Fig 5-22b. The region above the curve is simply labeled unstable. Some amazingly diverse
flow regimes lie in this region, as shown in a wonderful chart by Andereck, Liu, and Swinney
(1986). Report to the class on this chart and its many unstable flow patterns.
Solution: The writer’s best effort at scanning the chart is shown below. The writer lacks words to
discuss the nineteen different unstable flow regimes in the chart. Good luck!
Flow regimes observed by Andereck et al. (1986) between two rotating cylinders. The Reynolds
numbers are Reinner = ri2i /v and Reouter = ro2o /v.
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McGraw Hill LLC.
-24-
P5-30 For two-dimensional inviscid flow, the vorticity equation (2-116) may be written as
+u
+ v = 0,
t
x
y
Defining a basic flow
( ,U ,V ,)
where
=
v u
−
= − 2
x y
and disturbances
( , u, v, ) ,
derive a linearized
disturbance equation for this flow. Then assume normal modes as travelling waves:
( , u, v, ) = ˆ ( y ) , uˆ ( y ) , vˆ ( y ) , ˆ ( y ) exp i ( x − ct )
Derive the disturbance equations and, if possible, combine them to obtain a single differential
equation for a single disturbance amplitude.
Solution: The instructions above are pretty clear. Substitution of (basic flow + disturbances) into
the vorticity equation yields
Ω '
Ω
'
'
Ω
'
'
+
+U
+U
+ u'
+ u'
+V
+V
+ v'
+ v'
=0
t
t
x
x
dx
x
y
y
y
y
Ω + ' = −2 −2 '
The boldface terms are basic flows and cancel themselves. The products of fluctuations are
neglected, with an arrow through them. The disturbance differential equations are thus
'
'
'
+U
+ u'
+V
+ v'
0
t
x
x
y
y
v' u'
' = − 2 ' =
−
x y
These can be solved, as suggested in the problem statement, by assuming travelling waves. After
substitution, all terms contain exp i ( x − ct ) , which can be cancelled out. Thus we have the
following disturbance equations:
ˆ ( −i c ) + ˆ ( iU ) + uˆ
ˆ = − ( i ) ˆ −
2
dˆ
+V
+ vˆ
=0
x
dy
y
d 2ˆ
dy
2
= i vˆ −
duˆ
dy
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McGraw Hill LLC.
-25-
To make this work, the “basic” coefficients (U , V , /x, /y ) have to vary with y only. In
fact, it makes sense to neglect the term uˆ ( /x ) as being unlikely.
To boil this down to one single equation in one single variable, clearly we could substitute
for in terms of . We could also use the definition of stream function to eliminate the two
velocities:
u=
, v=−
,
y
x
hence uˆ =
dˆ
dy
and
vˆ = −iˆ
Making these substitutions, we obtain the final third-order ordinary differential equation for the
amplitude of the stream function:
d 2ˆ
d 3ˆ
dˆ
i (U − c ) 2 − 2ˆ − V 3 − 2
− iˆ
=0
dy
dy
dy
y
Ans.
We would have to know, in advance, the (y) variations of the coefficients U, V, and .
P5-31 Use the airfoil surface-velocity data of Prob. P4.49 and Michel’s method, Eq. (5-38), to
estimate the position of transition to turbulence if Rec = 2 106. Assume air at 20°C and 1 atm. If
you note an ambiguity in the results, please criticize them.
Solution: Recall the data from Prob P4.49: A model two-dimensional airfoil has these potentialflow surface velocities on its upper surface at a small angle of attack:
x /C
0.0
0.025
0.05
0.1
0.2
0.3
0.4
0.6
0.8
1.0
V /U
0.0
0.97
1.23
1.28
1.29
1.29
1.24
1.14
0.99
0.82
The stream velocity U is varied to set the Reynolds number. The chord length C is 30 cm. The
fluid is air at 20°C and 1 atm. Assume that x is a good approximation to the arc length.
For air at 20°C and 1 atm, = 1.205 kg/m3 and = 1.81E-5 kg/(m-s). Thus, if
Rec = 2 106 , the appropriate velocity is U = Re v /C = ( 2E6)( 0.000015) / ( 0.3) = 100 m/s.
Problem 4-49 shows how Thwaites’ method was used to estimate ( x ) and ( x ) . One can then
calculate Re ( x ) and compare it with Michel’s transition line, 2.9 Re0.4
x . Here is a graph:
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McGraw Hill LLC.
-26-
According to this estimate, transition occurs at Rex 1,080,000, or x /C 0.54. The ambiguity
is that this is past the point of separation ( x /C 0.45) predicted in Prob. 4-49!
P5-32 Consider the cubic flat-plate velocity distribution:
u 3
1
y
= − 3
=
U 2
2
Solve the Orr–Sommerfeld Eq. (5-23) numerically for the inviscid case with = 0 . Begin at
− y
and integrate inwardly to satisfy the wall conditions = = 0 at
y = 2 assuming that e
y = = 0. Assuming temporal amplification, find some (damped) eigenvalues and plot ci versus
in dimensionless form.
Solution:
We are asked to numerically solve the Orr-Sommerfeld equation for the inviscid case with = 0 .
Based on Eq. (5-29), we can derive a non-dimensional Rayleigh equation of a form similar to Eq.
(5-27). The nondimensional Orr-Sommerfeld equation can be written as
i
U * − c* − 2 − U * +
− 22 + 4 = 0
Re
with
u
c
U
=
U* =
c* =
= Re =
U
U
U
This step aids in the formulation by casting the problem into a nondimensional form. Next, we let
= 0. This can be thought of as letting Re → . The result is the Rayleigh equation that stands
at the basis of inviscid-stability theory:
U * − c* − 2 − U * = 0
(
)(
)
(
(
)(
)
)
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McGraw Hill LLC.
-27-
Next, a computational strategy needs to be chosen given several traditional techniques to solve the
stability equation. For example, one could rely on a shooting method as traditionally done in this
case. However, the problem is formulated as a boundary-value problem, and the shooting method
will require solving an initial value problem multiple times until convergence is achieved. Since
our goal is to retrieve the eigenvalues, we can rewrite the above expression as an eigenvalue
problem. For more insight, please refer to the Matlab code provided as part of the solution for P535 below. Note that this formulation can be extended from the inviscid Rayleigh equation to the
full Orr-Sommerfeld equation rather straightforwardly.
To proceed, we may reformulate our continuous differential eigenproblem from
L1 = c* L2
to a discrete, generalized eigenproblem of the form
M = c* N
where M and N represent finite-difference operator matrices that only depend on the
nondimensional wave number and discretization scheme. The nondimensional Rayleigh stability
equation can then be separated into two matrices, namely,
U * − U *2 + U * = c* − 2
)
(
( )
(
)
Now that we have our assumed form, we may discretize . Using a second-order finite difference
scheme, we may put
d 2 j +1 − 2 j + j −1
=
+ O 2
2
2
d
Our continuous eigenvalue problem can thus be discretized. We get
B j j −1 + Aj + B j j +1 = c* Bˆ j −1 + Aˆ j + Bˆ j +1
(
(
)
)
where
A = − 2 2U * − 2U * − U * 2
j
*
B j = U
2
2
Aˆ = − 2 +
Bˆ = 2
From this set-up, two tridiagonal matrices can be computed for M and N . To solve the general
eigenvalue problem, we can multiply both sides by N −1 and retrieve
N −1M = c*
This problem can be simply solved using linear algebra and a code of your choice for the spectrum
of eigenvalues.
(
)
As we go through this exercise, it may be useful to recall the properties of the Rayleigh stability
equation (for an additional reference, see Drazin, Introduction to Hydrodynamic Stability). In the
Rayleigh equation, if a complex eigenvalue exists, so does its complex conjugate pair. Therefore,
to each damped stable mode, there exists a corresponding amplified, unstable mode which may be
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McGraw Hill LLC.
-28-
attributed to the time reversibility of the problem. In other words, the only way that an inviscid
flow may be stable is if all modes are neutrally stable. Although such behavior may at first seem
paradoxical; it should be rather expected for a non-dissipative system. A Hamiltonian system with
one degree of freedom has centers for stable equilibrium points and saddles for unstable
equilibrium points in the phase plane analysis. For this reason, the spectral plot will contain both
positive and negative complex conjugate pairs, if modes are not all neutrally stable.
For this profile, all eigenvalues predicted from the Rayleigh equation are found to be neutrally
stable. Specifically for = 0 , we obtain ci* 0 . This should not be surprising because the velocity
profile has an inflection point at the outer edge of the viscous boundary-layer region, i.e., not inside
the viscous domain.
In fact, the absence of unstable modes can be further investigated by solving the complete OrrSommerfeld equation using an artificially large Reynolds number. By way of illustration, a plot of
the eigenvalues for a Reynolds number of 1,000,000 is reproduced below for a nondimensional
wave number ranging from 1 to 10 and a discretized coordinate that is based on 20 points.
We see that this Orr-Sommerfeld solution with very small viscosity is stable within the domain
given that ci* 0 .
Before leaving this problem, it may be useful to remark that the presence of eigenvalues is quite
sensitive to the discretization level. In our framework, we are approximating a continuous
eigenvalue problem with a discrete formulation. In any such analysis, it is important to ensure that
the discretization is sufficient by repeatedly increasing the number of points until further changes
in the eigenvalues become negligible.
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McGraw Hill LLC.
-29-
P5-33 Consider the sinusoidal flat-plate velocity distribution:
u
y
= sin
=
U
2
Solve the Orr–Sommerfeld Eq. Error! Reference source not found. numerically for the inviscid
case with = 0 . Begin at y = 2 assuming that e− y and integrate inwardly to satisfy the wall
conditions = = 0 at y = = 0. Assuming temporal amplification, find some (damped)
eigenvalues and plot ci versus in dimensionless form.
Solution:
We are asked to numerically solve the Orr-Sommerfeld equation for the inviscid case with = 0 .
Based on Eq. (5-29), we can derive a non-dimensional Rayleigh equation of a form similar to Eq.
(5-27). The nondimensional Orr-Sommerfeld equation can be written as
i
U * − c* − 2 − U * +
− 22 + 4 = 0
Re
with
u
c
U
=
U* =
c* =
= Re =
U
U
U
This step aids in the formulation by casting the problem into a nondimensional form. Next, we let
= 0. This can be thought of as letting Re → . The result is the Rayleigh equation that stands
at the basis of inviscid-stability theory:
U * − c* − 2 − U * = 0
(
)(
)
(
(
)(
)
)
Next, a computational strategy needs to be chosen given several traditional techniques to solve the
stability equation. For example, one could rely on a shooting method as traditionally done in this
case. However, the problem is formulated as a boundary-value problem, and the shooting method
will require solving an initial value problem multiple times until convergence is achieved. Since
our goal is to retrieve the eigenvalues, we can rewrite the above expression as an eigenvalue
problem. For more insight, please refer to the Matlab code provided as part of the solution for P535 below. Note that this formulation can be extended from the inviscid Rayleigh equation to the
full Orr-Sommerfeld equation rather straightforwardly.
To proceed, we may reformulate our continuous differential eigenproblem from
L1 = c* L2
to a discrete, generalized eigenproblem of the form
M = c* N
where M and N represent finite-difference operator matrices that only depend on the
nondimensional wave number and discretization scheme. The nondimensional Rayleigh stability
equation can then be separated into two matrices, namely,
U * − U *2 + U * = c* − 2
( )
(
)
(
)
Now that we have our assumed form, we may discretize . Using a second-order finite difference
scheme, we may put
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McGraw Hill LLC.
-30-
d 2 j +1 − 2 j + j −1
=
+ O 2
2
2
d
Our continuous eigenvalue problem can thus be discretized. We get
B j j −1 + Aj + B j j +1 = c* Bˆ j −1 + Aˆ j + Bˆ j +1
(
(
)
)
where
A = − 2 2U * − 2U * − U * 2
j
B j = U *
2
2
Aˆ = − 2 +
Bˆ = 2
From this set-up, two tridiagonal matrices can be computed for M and N . To solve the general
eigenvalue problem, we can multiply both sides by N −1 and retrieve
N −1M = c*
This problem can be simply solved using linear algebra and a code of your choice for the spectrum
of eigenvalues.
(
)
As we go through this exercise, it may be useful to recall the properties of the Rayleigh stability
equation (for an additional reference, see Drazin, Introduction to Hydrodynamic Stability). In the
Rayleigh equation, if a complex eigenvalue exists, so does its complex conjugate pair. Therefore,
to each damped stable mode, there is a corresponding amplified, unstable mode which may be
attributed to the time reversibility of the problem. In other words, the only way that an inviscid
flow may be stable is if all modes are neutrally stable. Although such behavior may at first seem
paradoxical; it should be rather expected for a non-dissipative system. A Hamiltonian system with
one degree of freedom has centers for stable equilibrium points and saddles for unstable
equilibrium points in the phase plane analysis. For this reason, the spectral plot will contain both
positive and negative complex conjugate pairs, if modes are not all neutrally stable.
For this profile, all eigenvalues predicted from the Rayleigh equation are found to be neutrally
stable. Specifically for = 0 , we obtain ci* 0 . This should not be surprising because the velocity
profile has an inflection point at the outer edge of the viscous boundary-layer region, i.e., not inside
the viscous domain.
In fact, the absence of unstable modes can be further investigated by solving the complete OrrSommerfeld equation using an artificially large Reynolds number. By way of illustration, a plot of
the eigenvalues for a Reynolds number of 1,000,000 is reproduced below for a nondimensional
wave number ranging from 1 to 10 and a discretized coordinate that is based on 20 points.
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McGraw Hill LLC.
-31-
We see that this Orr-Sommerfeld solution with very small viscosity is unstable at the
nondimensional wave numbers of = 4 and 5 where ci* 0 .
Before leaving this problem, it may be useful to remark that the presence of eigenvalues is quite
sensitive to the discretization level. In our framework, we are approximating a continuous
eigenvalue problem with a discrete formulation. In any such analysis, it is important to ensure that
the discretization is sufficient by repeatedly increasing the number of points until further changes
in the eigenvalues become negligible.
P5-34 Consider the Majdalani–Xuan polynomial profile, which represents a surprisingly simple
and precise quartic velocity approximation to the Blasius boundary-layer problem for flow over a
flat plate with no pressure gradient:
u 5
1
y
= − 3 + 4
=
U 3
3
Solve the Orr–Sommerfeld Eq. Error! Reference source not found. numerically for the inviscid
case with = 0 . Begin at y = 2 assuming that e− y and integrate inwardly to satisfy the wall
conditions = = 0 at y = = 0. Assuming temporal amplification, find some (damped)
eigenvalues and plot ci versus in dimensionless form.
Solution:
We are asked to numerically solve the Orr-Sommerfeld equation for the inviscid case with = 0 .
Based on Eq. (5-29), we can derive a non-dimensional Rayleigh equation of a form similar to Eq.
(5-27). The nondimensional Orr-Sommerfeld equation can be written as
i
U * − c* − 2 − U * +
− 22 + 4 = 0
Re
(
)(
)
(
)
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McGraw Hill LLC.
-32-
with
u
c
U
c* =
= Re =
U
U
U
This step aids in the formulation by casting the problem into a nondimensional form. Next, we let
= 0. This can be thought of as letting Re → . The result is the Rayleigh equation that stands
at the basis of inviscid-stability theory:
U * − c* − 2 − U * = 0
=
U* =
(
)(
)
Next, a computational strategy needs to be chosen given several traditional techniques to solve the
stability equation. For example, one could rely on a shooting method as traditionally done in this
case. However, the problem is formulated as a boundary-value problem, and the shooting method
will require solving an initial value problem multiple times until convergence is achieved. Since
our goal is to retrieve the eigenvalues, we can rewrite the above expression as an eigenvalue
problem. For more insight, please refer to the Matlab code provided as part of the solution for P535 below. Note that this formulation can be extended from the inviscid Rayleigh equation to the
full Orr-Sommerfeld equation rather straightforwardly.
To proceed, we may reformulate our continuous differential eigenproblem from
L1 = c* L2
to a discrete, generalized eigenproblem of the form
M = c* N
where M and N represent finite-difference operator matrices that only depend on the
nondimensional wave number and discretization scheme. The nondimensional Rayleigh stability
equation can then be separated into two matrices, namely,
U * − U *2 + U * = c* − 2
)
(
( )
(
)
Now that we have our assumed form, we may discretize . Using a second-order finite difference
scheme, we may put
d 2 j +1 − 2 j + j −1
=
+ O 2
2
2
d
Our continuous eigenvalue problem can thus be discretized. We get
B j j −1 + Aj + B j j +1 = c* Bˆ j −1 + Aˆ j + Bˆ j +1
(
(
)
)
where
A = − 2 2U * − 2U * − U * 2
j
*
B j = U
2
2
Aˆ = − 2 +
Bˆ = 2
From this set-up, two tridiagonal matrices can be computed for M and N . To solve the general
eigenvalue problem, we can multiply both sides by N −1 and retrieve
(
)
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McGraw Hill LLC.
-33-
N −1M = c*
This problem can be simply solved using linear algebra and a code of your choice for the spectrum
of eigenvalues.
As we go through this exercise, it may be useful to recall the properties of the Rayleigh stability
equation (for an additional reference, see Drazin, Introduction to Hydrodynamic Stability). In the
Rayleigh equation, if a complex eigenvalue exists, so does its complex conjugate pair. Therefore,
to each damped stable mode, there is a corresponding amplified, unstable mode which may be
attributed to the time reversibility of the problem. In other words, the only way that an inviscid
flow may be stable is if all modes are neutrally stable. Although such behavior may at first seem
paradoxical; it should be rather expected for a non-dissipative system. A Hamiltonian system with
one degree of freedom has centers for stable equilibrium points and saddles for unstable
equilibrium points in the phase plane analysis. For this reason, the spectral plot will contain both
positive and negative complex conjugate pairs, if modes are not all neutrally stable.
For this profile, all eigenvalues predicted from the Rayleigh equation are found to be neutrally
stable. Specifically for = 0 , we obtain ci* 0 . This should not be surprising because the velocity
profile has an inflection point at the outer edge of the viscous boundary-layer region, i.e., not inside
the viscous domain.
In fact, the absence of unstable modes can be further investigated by solving the complete OrrSommerfeld equation using an artificially large Reynolds number. By way of illustration, a plot of
the eigenvalues for a Reynolds number of 1,000,000 is reproduced below for a nondimensional
wave number ranging from 1 to 10 and a discretized coordinate that is based on 50 points.
We see that this Orr-Sommerfeld solution with very small viscosity is stable within the domain
given that ci* 0 .
Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of
McGraw Hill LLC.
-34-
Before leaving this problem, it may be useful to remark that the presence of eigenvalues is quite
sensitive to the discretization level. In our framework, we are approximating a continuous
eigenvalue problem with a discrete formulation. In any such analysis, it is important to ensure that
the discretization is sufficient by repeatedly increasing the number of points until further changes
in the eigenvalues become negligible.
P5-35 Consider the Majdalani–Xuan exponential profile, which represents a very accurate
velocity approximation to the Blasius boundary-layer problem for flow over a flat plate with no
pressure gradient:
u
y
= 1 − exp − s 1 + 12 s + 2
=
U
and s = 1.630398038629397 is the Blasius slope at the wall, also known as the Blasius constant
or connection parameter. Solve the Orr–Sommerfeld Eq. Error! Reference source not found.
numerically for the inviscid case with = 0 . Begin at y = 2 assuming that e− y and integrate
inwardly to satisfy the wall conditions = = 0 at y = = 0. Assuming temporal amplification,
find some (damped) eigenvalues and plot ci versus in dimensionless form.
(
)
Solution:
We are asked to numerically solve the Orr-Sommerfeld equation for the inviscid case with = 0 .
Based on Eq. (5-29), we can derive a non-dimensional Rayleigh equation of a form similar to Eq.
(5-27). The nondimensional Orr-Sommerfeld equation can be written as
(U
*
)(
)
− c* − 2 − U * +
with
(
)
i
− 22 + 4 = 0
Re
u
c
U
c* =
= Re =
U
U
U
This step aids in the formulation by casting the problem into a nondimensional form. Next, we let
= 0. This can be thought of as letting Re → . The result is the Rayleigh equation that stands
at the basis of inviscid-stability theory:
=
U* =
(U
*
)(
)
− c* − 2 − U * = 0
Next, a computational strategy needs to be chosen given several traditional techniques to solve the
stability equation. For example, one could rely on a shooting method as traditionally done in this
case. However, the problem is formulated as a boundary-value problem, and the shooting method
will require solving an initial value problem multiple times until convergence is achieved. Since
our goal is to retrieve the eigenvalues, we can rewrite the above expression as an eigenvalue
problem. For more insight, please refer to the attached Matlab code. Note that this formulation can
be extended from the inviscid Rayleigh equation to the full Orr-Sommerfeld equation rather
straightforwardly.
Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of
McGraw Hill LLC.
-35-
To proceed, we may reformulate our continuous differential eigenproblem from
L1 = c* L2
to a discrete, generalized eigenproblem of the form
M = c* N
where M and N represent finite-difference operator matrices that only depend on the
nondimensional wave number and discretization scheme. The nondimensional Rayleigh stability
equation can then be separated into two matrices, namely,
)
(
(U * ) − U *2 + U * = c* ( − 2 )
Now that we have our assumed form, we may discretize . Using a second-order finite difference
scheme, we may put
d 2 j +1 − 2 j + j −1
=
+ O 2
2
2
d
Our continuous eigenvalue problem can thus be discretized. We get
(
)
(
B j j −1 + Aj + B j j +1 = c* Bˆ j −1 + Aˆ j + Bˆ j +1
)
where
A = − 2 2U * − 2U * − U * 2
j
B j = U *
2
2
Aˆ = − 2 +
Bˆ = 2
(
)
From this set-up, two tridiagonal matrices can be computed for M and N . To solve the general
eigenvalue problem, we can multiply both sides by N −1 and retrieve
N −1M = c*
This problem can be simply solved using linear algebra and any code of your choice for the
spectrum of eigenvalues.
As we go through this exercise, it may be useful to recall the properties of the Rayleigh stability
equation (for an additional reference, see Drazin, Introduction to Hydrodynamic Stability). In the
Rayleigh equation, if a complex eigenvalue exists, so does its complex conjugate pair. Therefore,
to each damped stable mode, there is a corresponding amplified, unstable mode which may be
attributed to the time reversibility of the problem. In other words, the only way that an inviscid
flow may be stable is if all modes are neutrally stable. Although such behavior may at first seem
paradoxical; it should be rather expected for a non-dissipative system. A Hamiltonian system with
one degree of freedom has centers for stable equilibrium points and saddles for unstable
equilibrium points in the phase plane analysis. For this reason, the spectral plot will contain both
positive and negative complex conjugate pairs, if modes are not all neutrally stable.
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-36-
The following plot is for 200 points in the domain and a nondimensional that is discretized
from 1 to 50 with 50 points.
Based on the solution of the Rayleigh equation, both stable and unstable eigenvalue pairs are
detected, albeit with very small growth rates. This behavior seems to indicate that the exponential
profile is very weakly unstable in the absence of viscosity.
Sample Code:
clear; clc;
%% Orr-Sommerfeld equation solver
clear; clc;
format long
% Mean Flow Profile
% % P5-32
u = @(y) (3/2)*y - (1/2)*y.^3;
upp = @(y) -3*y;
% % P5-33
% u = @(y) sin(pi()*y/2);
% upp = @(y) ((-pi()^2)/4)*sin(pi()*y/2);
% % P5-34
% u = @(y) (5/3)*y - y.^3 + (1/3)*y.^4;
% upp = @(y) 4*y.^2 - 6*y;
% % P5-35
% s = 1.630398038629397;
% u = @(y) 1 - exp(-s*y*(1+(1/2)*s*y+y.^2));
% upp = @(y) -y*(9*y.^3 *s + 6*y.^2 * s^2 + y*s*(s^2 + 6)+2*(s^2 3))*s*exp(-y.^3 *s-(y.^2 * s^2)/2 - y*s);
% Parameters
alpha = linspace(1,30,30);
Re = 1000000;
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-37-
for k = 1:length(alpha)
yCoord = linspace(0,2,50);
dy = yCoord(2) - yCoord(1);
P = @(y) u(y) - 2*alpha(k)*1i/Re;
Q = @(y) (alpha(k)^3)*1i/Re -1*alpha(k)^2 * u(y) - upp(y);
A = @(y) 6*1i/(alpha(k)*Re) + dy^2 * (dy^2 * Q(y) - 2*P(y));
B = @(y) -4*1i/(alpha(k)*Re) + dy^2 * P(y);
C = 1i/(alpha(k)*Re);
Ahat = -1*dy^2 * (2+alpha(k)^2 * dy^2);
Bhat = dy^2;
SizeMatrix = length(yCoord)-1;
M = zeros(SizeMatrix, SizeMatrix);
N = zeros(SizeMatrix, SizeMatrix);
for i = 2:SizeMatrix-1
N(i,i) = Ahat;
N(i,i+1) = Bhat;
N(i+1,i) = Bhat;
end
N(1,1) = Ahat;
N(1,2) = Bhat;
N(end,end-1) = Bhat;
N(end,end) = Ahat;
for i = 3:SizeMatrix-2
M(i,i) = A(yCoord(i+1));
M(i,i+1) = B(yCoord(i+1));
M(i+1,i) = B(yCoord(i+1));
M(i,i+2) = C;
M(i+2,i) = C;
end
M(1,1) = C+ A(yCoord(2));
M(1,2) = B(yCoord(2));
M(1,3) = C;
M(2,1) = B(yCoord(3));
M(2,2) = A(yCoord(3));
M(2,3) = B(yCoord(3));
M(2,4) = C;
M(end,end) = C + A(yCoord(end));
M(end,end-1) = B(yCoord(end));
M(end,end-2) = C;
M(end-1,end) = B(yCoord(end-1));
M(end-1,end-1) = A(yCoord(end-1));
M(end-1,end-2) = B(yCoord(end-1));
M(end-1,end-3) = C;
EigenValues{k} = eig(N^(-1)*M);
EigenValuesNonzero{k} = find(imag(eig(N^(-1)*M)));
end
alphaPlot = [];
for i = 1:length(alpha)
if EigenValuesNonzero{i} ~=0
alphaPlot = [alphaPlot; alpha(i)];
EigenP = EigenValues{i};
Index = EigenValuesNonzero{i};
for j = 1:length(Index)
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-38-
EigenPlot(j) = imag(EigenP(Index(j)));
figure(1)
scatter(alpha(i),EigenPlot(j))
hold on
end
end
end
title('Problem 5-34')
xlabel('Dimensionless wave number')
ylabel('Dimensionless imaginary wave speed')
hold off
%% Rayleigh equation (inviscid stability)
clear; clc;
format long
% Mean Flow Profile
% % P5-32
u = @(y) (3/2)*y - (1/2)*y.^3;
upp = @(y) -3*y;
% % P5-33
% u = @(y) sin(pi()*y/2);
% upp = @(y) ((-pi()^2)/4)*sin(pi()*y/2);
% % P5-34
% u = @(y) (5/3)*y - y.^3 + (1/3)*y.^4;
% upp = @(y) 4*y.^2 - 6*y;
% % P5-35
% s = 1.630398038629397;
% u = @(y) 1 - exp(-s*y*(1+(1/2)*s*y+y.^2));
% upp = @(y) -y*(9*y.^3 *s + 6*y.^2 * s^2 + y*s*(s^2 + 6)+2*(s^2 3))*s*exp(-y.^3 *s-(y.^2 * s^2)/2 - y*s);
% Parameter
alpha = linspace(1,100,500);
for k = 1:length(alpha)
yCoord = linspace(0,2,500);
dy = yCoord(2) - yCoord(1);
P = @(y) u(y) ;
Q = @(y) -1*alpha(k)^2 * u(y) - upp(y);
A = @(y) 1 * (dy^2 * Q(y) - 2*P(y));
B = @(y)
P(y);
C = 0;
Ahat = -1 * (2+alpha(k)^2 * dy^2);
Bhat = 1;
SizeMatrix = length(yCoord)-1;
M = zeros(SizeMatrix, SizeMatrix);
N = zeros(SizeMatrix, SizeMatrix);
for i = 2:SizeMatrix-1
N(i,i) = Ahat;
N(i,i+1) = Bhat;
N(i+1,i) = Bhat;
end
N(1,1) = Ahat;
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-39-
N(1,2) = Bhat;
N(end,end-1) = Bhat;
N(end,end) = Ahat;
for i = 3:SizeMatrix-2
M(i,i) = A(yCoord(i+1));
M(i,i+1) = B(yCoord(i+1));
M(i+1,i) = B(yCoord(i+1));
M(i,i+2) = C;
M(i+2,i) = C;
end
M(1,1) = C+ A(yCoord(2));
M(1,2) = B(yCoord(2));
M(1,3) = C;
M(2,1) = B(yCoord(3));
M(2,2) = A(yCoord(3));
M(2,3) = B(yCoord(3));
M(2,4) = C;
M(end,end) = C + A(yCoord(end));
M(end,end-1) = B(yCoord(end));
M(end,end-2) = C;
M(end-1,end) = B(yCoord(end-1));
M(end-1,end-1) = A(yCoord(end-1));
M(end-1,end-2) = B(yCoord(end-1));
M(end-1,end-3) = C;
%
%
end
EigenValues{k} = eig(N^(-1)*M);
EigenValuesNonzero{k} = find(imag(eig(N^(-1)*M)));
Index = EigenValuesNonzero{1};
Values = EigenValues{1};
alphaPlot = [];
for i = 1:length(alpha)
if EigenValuesNonzero{i} ~=0
alphaPlot = [alphaPlot; alpha(i)];
EigenP = EigenValues{i};
Index = EigenValuesNonzero{i};
for j = 1:length(Index)
EigenPlot(j) = imag(EigenP(Index(j)));
figure(1)
scatter(alpha(i),EigenPlot(j))
hold on
end
end
end
title('Problem 5-35')
xlabel('Dimensionless wave number')
ylabel('Dimensionless imaginary wave speed')
hold off
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McGraw Hill LLC.
-40-
P5-36 Determine whether the following profiles are prone to instability in the context of an
inviscid, incompressible, external fluid domain ( 0 ≤ y ≤ ∞ ):
(a) U = 1 + y 2 ; (b) U = y3 ; (c) U = y 4 ;(d )U = e− y ; (e) U = e− y ;( f )U = e− y cos( y);( g )U = tanh( y)
2
2
Solution: The general stability criteria are
i.
ii.
iii.
Stable: U " 0
Stable: U " 0
Stable: U " = 0 and also U "(U − U s ) 0
iv.
Unstable: U " = 0 and also U "(U − U s ) 0
U s = U ( ys ) , where ys is a point at which U " = 0 .
Assumption: Assuming that the profile is an inviscid, incompressible, external fluid domain ( 0 ≤
y ≤ ∞ ).
(a) U = 1 + y
2
U ' = 2y
U"= 2
U"0
It is stable.
(b) U = y
(Ans. a)
3
U ' = 3y2
U " = 6y
U " y 0 0
It is stable.
U " y = 0 = 0 and U "(U − U s ) = 0
Therefore, at y = 0 this profile is prone to instability.
(c) U = y
(Ans. b)
4
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-41-
U ' = 4 y3
U " = 12 y 2
U " y 0 0 . It is stable.
U " y = 0 = 0 and U "(U − U s ) = 0
Therefore, at y = 0 this profile is prone to instability.
(Ans. c)
(d) U = e− y
U ' = −e− y
U " = e− y
U " y 0 0
It is stable as e0 = 1 .
(e) U = e
(Ans. d)
− y2
U ' = −2 y e− y
2
U " = 4 y 2e− y − 2e− y
2
2
U " = e− y ( 4 y 2 − 2 )
2
U " y = 0 0 . It is stable.
U " y 0 0 . It is stable.
(Ans. e)
(f) U = e− y cos ( y )
2
U ' = −e − y sin ( y ) + cos ( y ) e − y
2
2
( −2 y ) = −e− y ( sin ( y ) + 2 y cos ( y ) )
2
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U " = e− y ( 4 y 2 cos ( y ) + 4 y sin ( y ) − cos ( y ) )
2
U " y = 0 0 . It is stable.
U " y 0 0 . It is stable.
(Ans. f)
(g) U = tanh ( y )
U ' = sech 2 ( y )
U " = 2sech ( y ) sech ( y ) tanh ( y ) = 2sech 2 ( y ) tanh ( y )
U " y 0 0 . It is stable.
U " y = 0 = 0 and U "(U − U s ) = 0
Therefore, at y = 0 this profile is prone to instability.
(Ans. g)
P5-37 Given an inviscid motion U(y), show that if σ represents an eigenvalue of the Rayleigh
equation given by Eq. (5-27) then so does its complex conjugate.
U"
+ 2 v = 0 (Eq. 5-27)
U −c
Solution: Rayleigh equation: v "−
Boundary layer: U ( 0) = U ( ) = 0
U "( y; c ) is a non-trivial solution of Rayleigh’s equation. Integrating Eq. (5-27) from 0 to ∞ gives
0
( v'
2
+ 2 v
2
) dy +
0
U"
2
v dy = 0
(U − c)
c represents an eigenvalue. Therefore
*
U " 2 U " (U − c ) 2 U " (U − Cr + ici ) 2
v =
v =
v
2
2
U −c
U −c
U −c
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-43-
where ci is the complex conjugate of c.
So we can write the imaginary part as
ci
0
U"
U −c
v dy = 0
2
2
For stable flow, ci = 0 or for instability ci 0 . Hence proved.
(Ans)
P5-38 Compare the benefits and deficiencies of the Orr–Sommerfeld equation to the en method in
the treatment of boundary-layer instability.
Solution: Benefits of the Orr–Sommerfeld equation:
i.
ii.
iii.
It is used for linear stability theory for parallel flow problems.
It governs small perturbations in the velocity component normal to the wall in parallel shear
flow.
Boundary layer responds to external forcing.
Benefits of the en method:
i.
ii.
iii.
It is a useful tool for both linear and nonlinear stability theory.
It links the transition location to a certain degree of amplification.
Small deviations may accumulate during the integration over a large streamwise distance.
Deficiencies of the Orr–Sommerfeld equation:
i.
ii.
iii.
It cannot predict where breakdown of laminar boundary layer will occur.
Factors not taken into consideration in the equation:
a. The normal velocity of the mean flow profile
b. Upstream conditions
c. Wall and freestream disturbances
d. Nonlinear mode of interaction for large amplitude disturbances and mean flow
distortion
Solutions are lengthy and sophisticated.
Deficiencies of the en method:
i.
ii.
iii.
Boundary layer responding to the external forcing has not been considered.
Amplitude A0 is same for all frequencies.
Initial disturbance predicted by linear theory will have a factor of en.
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P5-39 Discuss the implications of Squire’s theorem on the stability of two-dimensional planar
motions that are subject to three-dimensional disturbances under (a) inviscid and (b) viscous
conditions.
Solution:
(a) Inviscid case:
Squire’s theorem implies that to each unstable three-dimensional disturbance there
corresponds a more unstable two-dimensional one. It is sufficient to look at 2D
perturbations governed by the Rayleigh equation.
Proof:
If c = f ( ) is the solution of the two-dimensional problem of Eqs. (5-27) and (5-28), then
c = f ( ) is the solution of the equivalent two-dimensional problem.
1/2
By Squire’s transformation, c = f ( 2 + 2 ) is the solution of the three-dimensional
problem, where is the wave number.
Thus, to each unstable three-dimensional disturbance with growth rate ci , there
corresponds a two-dimensional disturbance with growth rate ci which is more unstable
since if = 0 .
(Ans.a)
(b) Viscous case:
Squire’s theorem implies that if a 3D mode ei x+i y becomes unstable at a particular value
of Re, then there is an equivalent 2D mode which is unstable at a smaller value of Re. It
therefore suffices to consider only 2D modes.
Proof:
Consider the 3D perturbation qˆ = q ( y ) exp i ( x + y − ct ) and a simplified form of the
eigenvalue problem.
i u + v + i = 0
(Eq. 1)
(c + iU )u + vu = −i p +
(c + i u )v = − p +
1 n
u − 2u − 2u )
(
Re
1
v − 2v − 2v )
(
Re
(c + i ) = −i p +
1
− 2 − 2 )
(
Re
(Eq. 2)
(Eq. 3)
(Eq. 4)
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-45-
Variables
u=
u and p can be defined as
( u + ) and p = p
2 = ( 2 + 2 )
(Eq. 1) becomes
i u + v ' = 0
(Eq. 5)
(Eq.2) + (Eq.4) we obtain,
(c + i u ) + vU = −i 2 p +
R=
1
( u − 2v )
Re
Re
c
, c=
(c + iU )u + vU = −i p +
(c + iU )V = − p +
1
( u − 2u )
R
1
( v − 2v )
R
(Eq. 6)
(Eq. 7)
Comparing the above equations to the first four equations, we find that for 3D problems,
finding c in terms of and R is mathematically identical to the 2D problem obtained by
setting = 0 .
Then c = −i , c = −i
If Re is finite, then the 3D mode ( , ) will become unstable at some critical Reynolds
Rc
.
By mathematical similarity, the 2D mode ( ,0 ) will go unstable at Re = Rc . Since
number Re = Rc , corresponding to R = Rc =
Rc Rc , an equivalent 2D mode will go unstable at a lower Reynolds number. Thus, 2D
modes are the first to go unstable as Re increases, and it is sufficient to consider 2D only.
P5-40 Rayleigh’s criterion for rotational instability states that “an inviscid rotating flow is
unstable if the square of its circulation decreases outward.” Using cylindrical polar coordinates (r,
θ, z), consider the motion of an inviscid rotating field prescribed by V(r ) = 0,V (r ),0 over a
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-46-
finite interval ri ≤ r ≤ ro, where the vorticity retains a single nonzero component in the z direction,
namely, ω = ( 0,0, z ) . (a) Show that
1 d
z =
( rV ) ;
r dr
dz
0;
dr
and
d ( r 2V2 )
dr
0
Is the flow stable or unstable?
(b) Assuming Couette flow [Eq. (3-22)] between two long cylinders where the subscripts i and o
denote the inner and outer surfaces rotating at angular speeds Ωi and Ωo, show that a condition
leading to instability corresponds to
i ri 2 o ro2
From this condition, what can you say about the stability of the motion if the outer cylinder is
stationary and the inner cylinder is rotating? Conversely, what happens if the inner cylinder is
stationary and the outer cylinder is rotating? Finally, what happens anytime the two cylinders rotate
in opposite directions?
Solution:
(a) The vorticity is = v .
Here v = ( 0, v (r ),0) = v ( r ) and = ( 0,0, z ) are in cylindrical coordinates.
Now, we know that in cylindrical coordinates,
1 vz v
vr vz
v =
−
−
rˆ +
z
z r
r
ˆ 1 ( rv ) vr
−
+
r r
zˆ
Here vr = 0, v = v ( r ) and vz = 0 .
Also,
v
= 0 (as v is a function of r only).
z
Hence only
v =
z =
ẑ component remains.
1 ( rv )
zˆ
r r
1 d ( rv )
r dr
(Ans)
Hence proved.
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-47-
dv dv v
1
z = v + r = +
r
dr dr r
d z d 2 v v
=
− 2 0
dr
dr 2
r
(Ans)
Hence proved.
d ( r 2 v2 )
dr
= 2rv2 + r 2 2v
dv
0
dr
This solution for v holds good only for the finite interval ri ≤ r ≤ ro.
The flow is stable.
(b) (i) When the outer cylinder is stationary and the inner cylinder is rotating, then 0 = 0 .
Then the flow is always unstable.
(ii) When the outer cylinder is rotating and the inner cylinder is stationary, then the flow is
linearly stable. Hence, this is the basis for the modern viscometer, which works by avoiding
the vortical structure and obtains an accurate measurement of the viscosity of fluids.
(iii) For two cylinders rotating in opposite directions, the region close to the inner cylinder
is unstable and the region close to the outer cylinder is stable.
According to Taylor, “In the inner region, the square of the circulation decreases outward,
so that centrifugal force tends to make the flow unstable. In the outer region the square of
the circulation increases so that centrifugal force tends to make the flow stable.”
P5-41 Consider the following time-dependent convective equation for u(x, t) ,
u
u
+c
= u
t
x
where c and σ are positive constants. (a) Show that any disturbance of the form u(x, t) = εeσt f (x −
ct) represents a valid solution. (b) By specifying disturbances to the zeroth-order solution u0 = 0
using the normal mode form u(x, t) = εeσt exp [iα(x − ct)] , determine the condition for stability.
(c) Assuming u(x, t) = εeσt sech2(x − ct) , illustrate the behavior of this disturbance graphically. (d)
How does the solution behave as t → ∞ at a fixed point in space? (e) After a very long time, if u
becomes unbounded at a fixed point in space, it will signal the presence of an absolute instability.
Show that this disturbance becomes absolutely unstable for σ > 2c.
Solution:
(a) Disturbance of the form
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-48-
u = e t f ( x − ct )
u
= e t f ( x − ct ) − c e t f ( x − ct )
t
u
= u − c e t f ( x − ct )
t
u
= e t f ( x − ct )
x
Substituting the partial derivatives in the following time-dependent convection equation
u
u
+c
= u , we get
t
x
u − c e t f ( x − ct ) + c e t f ( x − ct ) = u
t
Hence, the disturbance in the form u = e f ( x − ct ) is a valid solution.
(Ans.a)
(b) For obtaining the zeroth-order solution using the normal mode form, we can write the timedependent convection equation as
u u
+c = u
t
x
u = e t exp[i ( x − ct )]
u
= t e t exp[i ( x − ct )] − i c e exp[i ( x − ct )]
t
u
= ( − i c ) u
t
u
= i e t exp[i ( x − ct )]
x
Hence the time-dependent convection equation becomes
− i c u + ci cu = u
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-49-
u − i cu + ci cu = u
u ( −1) = i cu ( − 1)
= ic
Therefore, the condition for stability is
= ic .
(Ans.b)
(c) Behavior of the disturbance u ( x, t ) = e t sech 2 ( x − ct ) is shown in the following graphs.
Fig. (a)
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-50-
Fig. (b)
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-51-
Fig. (c)
(d) The solution behaves asymptotically stable as t → at fixed point in space.
(Ans.d)
(e) This disturbance becomes absolutely unstable for a 2c as shown in fig. (c). It states that
small perturbation grows above a given threshold at a fixed point of the flow.
(Ans.e)
P5-42 Consider the following time-dependent convection-diffusion equation for u(x, t),
u
u 2u
+c
=
+ ( − 0 ) u
t
x x 2
where c, σ, and σ0 are positive constants. (a) Show that the flow will be stable so long as σ < σ0.
(b) If, instead, we have σ > σ0, find the range of c that will give rise to convectively unstable
disturbances as well as the range that will trigger a case of absolute instability.
Solution: Let us consider the normal modes of the form
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-52-
u( x, t ) = Aei ( x −t )
u
= −i Aei ( x −t ) = −iu
t
u
= i Aei ( x −t ) = i u
x
2u
= − 2 Aei ( x −t ) = − 2u
x 2
Substituting the partial differentiations in the equation,
−iu + ci u + 2u − ( − 0 ) u = 0
−i + i c + 2 − ( − 0 ) = 0
− c + i 2 − i ( − 0 ) = 0
= f ( )
f ( ) = c − i 2 + i ( − 0 )
f ( ) = c − 2i
f (0) = c
f (0) = i ( − 0 )
− 0 = −if (0)
(a) The flow is asymptotically stable when 0 because all modes decay exponentially.
(b) The flow is convectively unstable if 0 and c 0 because the modes which grow
exponentially also propagate.
The flow is absolutely unstable if 0 and c = 0 .
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P5-43 Using cylindrical polar coordinates (r, , z) in conjunction with a velocity vector
V = ( vr , v , vz ) and pressure P, the continuity and momentum equations for inviscid,
incompressible motion reduce to
vr
v v v
v v 2
P
+ vr r + r + vz r − = −
r
r
z
r
r
t
1
1 v v
( rvr ) + + z = 0
r r
r z
and
v v v
v v v
v
+ vr + + vz + r
r r
z
r
t
v
v v v
v
z + vr z + z + vz
r r
z
t
1 P
=−
r
P
=−
z
Let us now consider the case of a rotating axisymmetric flow between two concentric cylinders,
where the velocity field can be prescribed by V = V0 + v(x, t ) with 0 1 . The basic motion
in this problem may be captured by a strictly swirling (azimuthal) velocity component of the form,
V0 (r ) = [0,V (r ),0] . (a) Start by writing the disturbances as v(x, t ) = (uˆ, vˆ, wˆ ) = (u, v, w)ei z + cit and
show that the radial disturbance must satisfy a linear, second-order, differential equation, namely,
d 2u 1 du u 2 B
1 d 2 2
+
− 2 − + 2 u = 0, where B = 3
(r V ) .
2
dr
r dr r
ci
r dr
(b) A journal bearing may be modeled as being composed of an inner cylinder of radius ri and an
outer cylinder of radius ro ri . Assuming that the steady viscous motion in the cylindrical annulus
−1
is expressible as V (r ) = C1r + C2 r , determine the appropriate constants and then proceed to
deduce B.
(c) By fixing the longitudinal wavenumber α, and assuming that B can be negative, discuss the
stability characteristics of this motion.
Solution:
i z + ci t
i z + ci t
i z + ci t
(a) Vr = ue
, V = V (r ) + òve
, Vz = e
V
vx
Vr
= ci uei z + cit , = ci vei z + cit ,
= ciei z + cit
t
t
t
Vr
du i z + cit V V (r ) Vz
,
,
= iei z + cit
=
e
=
r
z
dr
r
dr
For linearly stable, higher-order terms are neglected.
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From continuity equation:
Vr Vr Vz
+ +
=0
r r z
du i z + ci t uei z + ci t
e
+
+ i ei z + ci t = 0
dr
r
(Eq. 1)
From momentum balance,
Vr V2
1 P
−
=−
t
r
r
ci uei z + cit
( v(r ) + ve
−
ci u −
=0
r
ci uei z + c t −
i
)
i z + ci t 2
2V (r ) ei z + ci t
=0
r
2V (r )v
=0
r
(Eq. 2)
V
V V V
1 P
+ Vr + r = −
t
r
r
r
i z + ci t
ci e
ci e
i z + ci t
ci + u
)
V (r ) + ei z + cit
dV (r )
i z + cit
+ ue
=0
dr
r
i z + ci t
dV (r ) e
+
dr
+ ue
+ ue
(
i z + ci t
dV (r ) uV (r )
+
=0
dr
r
i z + ci t
r
uV ( r )
=0
(Eq. 3)
Vz
P
=−
t
z
ci ei z + cit = 0
ci = 0
(Eq – 4)
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-55-
Substituting = 0 in (Eq. 1), we get
du u
+ =0
dr r
(Eq. 5)
Differentiating (Eq. 5) with respect to r, we get
d 2u 1 du u
+
−
=0
dr 2 r dr r 2
(Eq. 6)
From (Eq. 3), we get
v=−
dV (r ) u
u dr + r V (r )
1
ci
Substituting in (Eq. 2), we get
ci u +
2 V (r ) dV (r ) u
u
+ V (r ) = 0
ci r dr
r
Multiplying the above equation with
ci
, we get
dV (r ) 2V 2 (r )
2
V
(
r
)
+
=0
dr
r
2u +
u
rci2
2u +
u
dV (r )
2V (r )r 2
+ 2V 2 (r )r = 0
rc
dr
2u +
u 1 d 2
( r V (r ) ) = 0
ci2 r 3 dr
3 2
i
2 B
+ 2
ci
u = 0
where B =
(Eq. 7)
1 d 2
( r V ( r )) .
r 3 dr
Subtracting (Eq. 7) from (Eq. 6), we get
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d 2u 1 du u 2 B
+
− − + 2
dr 2 r dr r 2
ci
u = 0
(Ans. a)
Hence proved.
−1
(b) V (r ) = C1r + C2 r
Boundary conditions are
At r = ri , = i = C1 +
C2
ri 2
At r = r0 , = o = C1 +
C2
ro2
C C
i − 0 = 22 − 22
r0
ri
− 0 2 2
C2 = 2i
r r
2 i 0
r0 − ri
Let =
r
o
, R = i 1
ro
i
1−
C2 = i ri 2
2
1 − R
Then C1 becomes
− R2
C1 = i
2
1− R
− R2
1− 1
V (r ) = i
r + i ri 2
2
2
1 − R r
1− R
B=
1 d 2
( r v (r ) )
r 3 dr
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B=
C
1 d 2
r C1r + 2
3
r dr
r
B=
1 d
C1r 3 + C2 r
3
r dr
B=
1
3r 2C1 + C2
3
r
B=
1
r3
2 − R2
1−
+ i ri 2
3r i
2
2
1 − R
1− R
ri 2
i 3
2
B=
−
R
+
(1 − )
)
2 (
3
r
1 − R r
(c) B 0
R2
Hence, the motion becomes unstable.
Therefore, the stability criterion is
1 d 2
( r V (r ) ) 0 .
r 3 dr
P5-44 Assuming strictly two-dimensional flow conditions with a spanwise wave number of n = 0
in the z direction and a basic motion that is prescribed by a two-component velocity vector of the
form V0 ( x, y) = [U ( x, y),V ( x, y),0] , (a) show that the linearized Navier–Stokes equations with
biglobal stability disturbance modes, given by Eqs. (5-82)–(5-85), can be readily reduced into
u v
+
=0
x y
(continuity)
2u 2u
U
U
u
u 1 p
u+
v +U
+V
+
= 2 + 2 ( x -mom. )
x
y
x
y x
x y
2v 2v
V
V
v
v 1 p
−i v +
u+
v +U
+V
+
= 2 + 2 ( y -mom. )
x
y
x
y y
x v
−i u +
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where pˆ ( x, y, z, t ) = p( x, y)e−i t and v( x, y, z, t ) = (uˆ, vˆ,0) = [u( x, y), v( x, y),0]e−i t .
(b) Using a disturbance stream function that satisfies the linearized continuity equation identically,
eliminate u and υ from the remaining momentum equations by substituting
y
u=
and
v=−
x
Show that the resulting momentum equations can be rearranged into
3
1 p
U U
2
2
3
= i
−
+
−U
− V 2 + 2 + 3 ( x -mom.)
x
y x y y x
xy
y
x y y
3
1 p
V V
2
2
3
= −i
−
+
+U 2 +V
− 3 +
( y -mom.)
y
x x y y x
x
xy
xy 2
x
(c) The pressure can be eliminated by taking the partial derivative of the x-momentum pressure
gradient with respect to y and setting it equal to the partial derivative of the y-pressure gradient
with respect to x. Bearing in mind that the basic flow motion represented by (U, V) must satisfy
continuity, show that the two momentum equations can be consolidated into a single fourth-order
partial differential equation for the disturbance stream function viz.
2 2 2V 2U 2U 2V
i 2 + 2 =
−
+
−
y xy y 2 x xy x 2 y
x
3
3
3
3 4 4
4
+U 3 +
+ V 3 + 2 − 4 + 4 + 2 2 2
xy 2
x y x
y
x y
x
y
Solution: Assumptions:
The flow is two dimensional ( = 0 ) .
i.
ii.
iii.
Spanwise wave number n = 0 .
Two-component velocity vector is of the form V0 ( x, y) = [U ( x, y),V ( x, y),0] .
(a) Navier–Stokes equation with biglobal stability disturbance modes is given by
u v
+
+ inw = 0 (continuity)
x y
−i u +
(5-82)
2u 2u
U
U
U
u
u
1 p
u+
v+
w +U
+V
+ inWu +
= 2 + 2 − n 2u ( x-mom.) (5-83)
x
y
z
x
y
x
y
x
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-59-
2v 2v
V
V
V
v
v
1 p
u+
v+
w +U
+V
+ inWv +
= 2 + 2 − n 2v ( y -mom.) (5-84)
x
y
z
x
y
y
y
x
2
2
w w
W
W
W
w
w
p
−i w +
u+
v+
w +U
+V
+ inWw + in = 2 + 2 − n 2 w ( z -mom.) (5-85)
x
y
z
x
y
y
x
−i v +
Substituting the assumptions in the above equations,
the continuity equation becomes
u v
+
= 0 (continuity) (Eq. 1)
x y
x-mom and y-mom equations become
2u 2u
U
U
u
u 1 P
−i u +
u+
v +U
+V
+
= 2 + 2 ( x-mom) (Eq. 2)
x
y
x
y x
y
x
−i v +
2v 2v
V
V
v
v 1 P
u+
v +V
+V
+
= 2 + 2 ( y -mom) (Eq. 3)
x
y
x
y y
y
x
where pˆ ( x, y, z, t ) = p( x, y)e−i t and v( x, y, z, t ) = (uˆ, vˆ,0) = [u( x, y), v( x, y),0]e−i t .
Hence proved.
(Ans. a)
(b) Disturbance stream function is given by
u=
y
u 2
2u
3
=
2 = 2
x xy
x
x y
u 2
2u 3
= 2 2= 3
y y
y
y
v=−
x
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-60-
v
2
2v
3
=− 2 2 =− 3
x
x
x
x
v
2
2v
3
=−
2 =−
y
xy
y
xy 2
Substituting the partial derivatives of u and v in Eqs. (1) and (2), we get
−i
U
+
y
x y
U
+
y
−
x
2
2
+
U
+
v
2
xy
y
1 P
3
3
+
=
+
2
3
x
x y y
2 2
3
1 P
U U
3
= i
−
− v 2 + 2 + 3 ( x-mom) (Eq. 4)
+
−U
x
y
x y y x
xy y
x y y
Hence proved.
(Ans. b)
2
2 1 P
3
3
V V
−i −
= 3 +
+
+
+
−
+U 2 +V −
xy 2
x x y y x
x
xy x
x
1 P
V
= −i
−
y
y x y
2
2
U
+
+
V
+
V
2
y x
xy
x
3
3
−
+
3
( y -mom) (Eq. 5)
xy 2
x
Hence proved.
(Ans. b)
(c) To eliminate pressure term, (Eq. 4) is partially differentiated with respect to y, and we get
4
1 2 P
2 U 2 V
3
2
4
= i
−
+
−
U
−
V
+
+
x 2y 2 y 4
xdy
y 2 x y 2 y 2 x
xy 2
y 3
(Eq. 6)
(Eq. 5) is partially differentiated with respect to x, then we get
4
1 2 p
2 2V V 2
3
2
4
= −i
−
+
+
V
+
V
−
+
4
xdy
x 2 x 2 y y x 2
x3
x 2y
x 2y 2
x
(Eq. 7)
Equating (Eq. 6) and (Eq. 7), we get
2 2 2V 2U 2U 2V
i 2 + 2 =
−
+
−
y xy y 2 x xy x 2 y
x
3
3
3
3 4 4
4
+U 3 +
+ V 3 + 2 − 4 + 4 + 2 2 2
xy 2
x y x
y
x y
x
y
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-61-
Hence proved.
(Ans. c)
5-45 The Taylor profile, which has been used to describe the ideal gaseous motion in slab-type
rocket motors (Taylor 1956), corresponds to a two-dimensional self-similar profile in a porous
channel [Griffond and Casalis (2001)]. When the channel half height is taken to be unity, the basic
flow stream function may be written as 0 ( x, y) = x sin( 12 y) for 0 x X N ,0 y 1, where
X N represents the length of the porous channel (see Fig. P5-45).
FIGURE P5-45
Assuming a wall injection speed of unity, the basic flow velocity may be expressed as
V0 ( x, y) = U ( x, y)i + V ( y) j with U = 12 x cos( 12 y) and V = − sin( 12 y ) . As shown in the
problem above, the linearized Navier–Stokes equations with two-dimensional biglobal stability
disturbance modes (and n = 0 in the z direction) can be consolidated into a single, fourth-order,
partial differential equation for the disturbance stream function , where ˆ = ( x, y )e
V ( y) is independent of x , the stream function formulation can be further reduced into
2 2
i 2 + 2
y
x
− i t
. Since
3
2U 2U
3
=
−
+
+
U
+
3
y 2 x xy y
xy 2
x
3
4 4
3
4
+ V 3 + 2 − 4 + 4 + 2 2 2
x y
y
x y
y
x
The basic flow boundary conditions require uniform wall-normal injection, no slip at the headwall,
and symmetry with respect to the midsection plane, y = 0 , specifically:
0
0
( x,1) = 1;
( x,1) = 0
(wall-normal injection)
y
x
0
0
(0, y ) =
(0, y ) = 0
(no slip at headwall)
x
y
2 0
0 ( x, 0) =
( x, 0) = 0
(V = U / y = 0 at y = 0)
y 2
x
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(a) Recalling that the instantaneous motion is recoverable from = 0 +ˆ , and that disturbances
are not allowed to alter the basic flow conditions at the boundaries, show that the disturbances
must satisfy:
( x,1) =
( x,1) = 0
(wall-normal injection)
y
x
(0, y ) =
(0, y) = 0
(no slip at headwall)
y
x
2
( x, 0) = 2 ( x, 0) = 0
(midsection symmetry)
y
x
You may also impose the outflow condition suggested by Theofilis, Duck and Owen (2004) by
setting a simple extrapolation relation at the domain’s exit plane, x = X N , namely,
( X N , y) =
X N − X N −2
X − XN
( X N −1 , y) + N −1
( X N −2 , y)
X N −1 − X N −2
X N −1 − X N −2
(b) Using the Taylor basic flow profile and a numerical solver of your choice, find the eigenvalues
for n = 0, X N = 8 , and = 1/ 900 = 1.1110−3m2 /s .
(c) Optional: Compare your findings to those of Griffond and Casalis (2001), “On the nonparallel
stability of the injection induced two-dimensional Taylor flow.”
Solution:
(a) First, the two components of the mean flow are expressible by
U = 12 x cos( 12 y) and V = − sin( 12 y )
We also know that that the instantaneous velocity and flow streamfunction are related through
and V = −
U=
y
x
Since the instantaneous motion is recoverable from = 0 +ˆ = 0 + ( x, y)e−i t , we may
differentiate with respect to x and y to obtain
0 ˆ 0 −i t
0 ˆ 0 − i t
and =
=
+
=
+
e
+
=
+
e
y
y
y
y
y
x
x
x
x
x
For a porous channel with uniform wall injection, the fundamental boundary conditions on the
instantaneous field are the same as those on the mean flow (because the “disturbances are not
allowed to alter the basic flow conditions at the boundaries”). We have, for any time t ,
(1) A uniform injection and no-slip condition at the sidewall, U ( x,1, t ) = 0 and V ( x,1, t ) = −1
(2) A prescribed zero injection pattern at the headwall, U (0, y, t ) =V (0, y, t ) = 0
U
( x, 0, t ) = V ( x, 0, t ) = 0
y
By conveying these boundary conditions to the streamfunction, we get
Sidewall:
(3) A vanishing speed at the centerline,
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0
( x,1, t ) =
( x,1) +
( x,1)e−i t
x
x
x
and
0
( x,1, t ) =
( x,1) +
( x,1)e − i t
y
y
y
or
−V ( x,1, t ) = 1 +
U ( x,1, t ) = 0 +
( x,1)e −i t 1 = 1 +
( x,1)e −i t ; t
( x,1) = 0
x
x
x
( x,1)e −i t 0 = 0 +
( x,1)e −i t ; t
( x,1) = 0
y
y
y
Headwall:
0
0
(0, y, t ) =
(0, y) +
(0, y)e−i t and
(0, y, t ) =
(0, y ) +
(0, y )e − i t
y
y
y
x
x
x
or
−V (0, y, t ) = 1 +
(0, y)e −i t 0 = 0 +
(0, y)e −i t ; t
(0, y) = 0
x
x
x
U (0, y, t ) = 0 +
(0, y)e −i t 0 = 0 +
(0, y)e −i t ; t
(0, y) = 0
y
y
y
Centerline:
2 0
2
2
0
−i t
( x, 0, t ) =
( x, 0) + 2 ( x, 0)e−i t
and
( x, 0, t ) =
( x, 0) +
( x, 0)e
2
2
y
y
y
x
x
x
or
−V ( x, 0, t ) = 1 +
( x, 0)e −i t 1 = 1 +
( x, 0)e −i t ; t
( x, 0) = 0
x
x
x
U
2
2
2
( x, 0, t ) = 0 + 2 ( x, 0)e −i t 0 = 0 + 2 ( x, 0)e −i t ; t
( x, 0) = 0
y
y
y
y 2
(b) Eigenvalues for n = 0, X N = 8 , and = 1/ 900 = 1.1110−3m2 /s may be obtained using a QZ
solver with Chebychev discretization and N=50 with application to the incompressible Taylor
mean flow profile as the basic flow solution. Results are illustrated below.
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-64-
Here, the hollow squares (□) represent eigenmodes obtained from the present incompressible
stability solver.
(c) Based on Griffond and Casalis (2001), “On the nonparallel stability of the injection induced
two-dimensional Taylor flow,” the stability approach is based on the Local-Non-Parallel (LNP)
approach. Although the LNP approach is one dimensional, it takes into account the fact that
changes in the axial direction are slower than changes in the normal direction. No such assumptions
are made using the biglobal approach, which is truly two dimensional. Predictions based on the
biglobal approach are therefore expected to outperform in accuracy those of the LNP approach.
There are several differences between the two methodologies. Griffond and Casalis (2001) solve
a local stability problem, investigating how a location in space responds to small disturbances.
However, the biglobal approach is a global method that accounts for the entire flow at once.
Therefore, a specific point in the solution domain can be predicted to be locally unstable, yet
globally stable. Additionally, since the LNP approach is essentially one dimensional, it assumes
the form of the spatial growth or decay in the remaining spatial dimensions. For the present case,
Griffond and Casalis (2001) assume that
( x, y, t ) = ( y ) xr +ii exp ( −i t )
where represents the real temporal frequency and = r + i i combines the spatial growth rate
r and the spatial eigenfrequency i ; on the other hand, the biglobal approach employs the
following ansatz:
( x, y, t ) = ( x, y ) exp i ( x − r t ) exp ( it )
where is the spatial mode number in the axial direction and = r + i i stands for the complex
eigenmode.
In short, the LNP approach represents a one-dimensional local or spatial approach, whereas the
biglobal approach represents a global temporal approach.
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P5-46 Assuming strictly axisymmetric flow conditions with an azimuthal wave number of m = 0
and a basic motion that is prescribed by a two-component velocity vector of the form
V0 (r, z) = U r (r ),0,U z (r, z) , (a) show that the linearized Navier–Stokes equations with biglobal
stability disturbance modes, which are given by Eqs. (5-86)–(5-89), can be readily reduced into
ur ur uz
+ +
= 0 (continuity)
r
r
z
−i ur + U r
2u 1 ur ur 2ur
ur
U r
u
1 p
+ ur
+ U z r − 2r +
− 2 + 2 =−
(r -mom.)
r
r
z
r r r
z
r
r
−i uz + U r
2u 1 uz 2uz
uz
U z
u
U z
+ ur
+ U z z + uz
− 2z +
+ 2
r
r
z
z
r r
z
r
1 p
( z -mom.)
=−
z
where pˆ (r , , z, t ) = p(r , z )e −i t and v(r , , z , t ) = ( uˆr , 0, uˆ z ) = ur (r , z ), 0, u z (r , z ) e −i t .
(b) Using a disturbance stream function that satisfies the linearized continuity equation identically,
eliminate ur and uz from the remaining radial and axial momentum equations by substituting
ur = −
1
r z
and
uz =
1
r r
(c) The pressure can be eliminated by taking the partial derivative of the radial pressure gradient
with respect to z and setting it equal to the partial derivative of the axial pressure gradient with
respect to r. Show that the problem reduces to a single fourth-order partial differential equation for
the disturbance stream function, namely, the one advanced by Chedevergne et al. (2006):
2 1 2
i 2 −
+
r r z 2
r
3 U r U z 2U r 2
3 2U r 1 U r 2U z 1 U z
=
U
+
+
−
+
U
+
−
+
−
r
z
r 3 r
z
r r 2
z 3 r 2 r r
rz r z
3 U r U r U z 2 1 U z 2U z U z 2
3
+U r
+
−
+
+
+
−
−
+
U
z
rz 2 r
r
z z 2 r r
r 2 z
r r z
r 2 z
4 2 3 3 2 3
4
2 3
4
− 4 −
+
−
+2 2 2 −
+
r r 3 r 2 r 2 r 3 r
r z
r r z 2 z 4
r
Solution: Assumptions:
i.
Axisymmetric flow condition ( u = 0 )
ii.
Azimuthal wave number ( m = 0)
iii.
Two-component velocity vector is of the form V0 (r, z) = U r (r ),0,U z (r, z) .
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r
(a) Navier–Stokes equations with biglobal stability disturbance modes is given by
u u
ur ur
+ + im + z = 0
r
r
r
z
−i ur + U r
(continuity) (5-86)
U u
u U r 2U u
ur
U r
u
U r 1 p
+ ur
+ im r +
−
+ U z r + uz
+
r
r
r
r
r
z
z
r
2u 1 ur
u
2u
2im
= 2r +
− (1 + m2 ) 2r − 2 u + 2r
r r
r
r
z
r
−i u + U r
(r -mom.) (5-87)
u
U
U u
u U U r u urU
u
U
m
+ ur
+ im +
+
+
+ U z + uz
+i
p
r
r
r
r
r
r
z
z
r
2u
u
2u
1 u
2im
= 2 +
− (1 + m2 ) 2 + 2 ur + 2 ( -mom.) (5-88)
r r
r
r
z
r
−i u z + U r
U u
u U z
u z
U z
u
U z 1 p
+ ur
+ im z +
+ U z z + uz
+
r
r
r
r
z
z
z
2u 1 uz m2
2u
= 2z +
− 2 uz + 2z ( z -mom.) (5-89)
r r
r
z
r
Substituting the assumptions in the above equations, the continuity equation becomes
ur ur uz
+ +
=0
r
r
z
(continuity) (Eq. 1)
r-momentum and z-momentum equations become
−i ur +
2u 1 ur ur 2ur
U r ur
U r
u
1 p
+ ur
+ U z r − 2r +
− 2 + 2 =−
(r − mom)
r
r
z
r r r
z
r
r
(Eq. 2)
−i uz + U r
d 2u 1 uz 2uz
uz
U z
u
U z
1 P
+ ur
+ U z z + uz
− 2z +
+ 2 =−
( z − mom)
r
r
dz
z
r r
z
z
r
(Eq. 3)
Hence proved.
(Ans. a)
(b) Disturbance stream functions are as follows:
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-67-
ur = −
1
r z
and
uz =
1
r r
Let us consider the r-momentum equation:
ur = −
1
r z
ur
1 1 2
= 2
−
r
r z r r z
2 ur
1 2
2 1 d 2
1 2
2 2
2 1
=
−
−
+
=
−
−
r 2
r 2 r z r 3 z r r 2 z r 2 rdz r 2 rdz r 3 z r r 2 z
ur = −
1
r z
ur
1 2
=−
z
r z 2
2ur
1 3
=−
2
dz
r z 3
Substituting the partial derivatives of ur in (Eq. 2), we get r-momentum as follows:
−
1 1 2 1 U r
1 P
1
1 2
= i
+ Ur 2
−
−
−
U
z
r
r z
r z 2
r z r rz r z r
2 2
2 1 3
1 1 2
1 1 3
− 2
− 3
−
+
−
+
−
2
3
2
3
3
r rz r z r r z r z r rdz r z r z
−
1 1 2 1 ur
1 P
1
1 2
= i
+ Ur 2
−
−
−
U
z
r
r z
r z 2
r z r rx r z r
1 2 1 2
1
− 2
−
−
(r − mom)
2
3
r rz r r z r z
(Eq. 4)
Let us consider the z-momentum equation:
1
uz =
r r
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-68-
u z
1 1 2
=− 2
+
r
r r r r 2
2u z
1 2
2 1 3
1 2
2
2 d 2 1 3
=− 2
+
+
−
=
−
+
r
r r 2 r 3 r r r 3 r 2 r 2
r 3 r r 2 r 2 r r 3
uz =
1
r r
u z 1 2
=
z
r r z
2u z 1 3
=
z 2
r r z 2
Substituting the partial derivatives of uz in (Eq. 3), we get the z-momentum equation as
follows:
−
1 1 2
1 P
1
= −i
+ U r − 2
+
2
z
r r
r r r r
1 U z
1 2 1 uz
−
+
U
z
r z r
+
r rz r r z
2 2 2 1 3 1 1 2 1 3
− 3
− 2
+
−
+
+
2
r r 3 r 3 r r 2 r 2 r rdz 2
r r r r
−
1 P
1
= −i
+ Vr
z
r r
1 1 2 1 U z
1 2 1 uz
−
+
−
+
U
z
r 2 r r r 2 r z r
+
r ry r z
1 1 2 1 3 1 3
− 3
− 2
+
+
( z − mom) (Eq. 5)
2
r r 3 r rdz 2
r r r r
Hence proved.
(Ans. b)
(c) To eliminate pressure term, (Eq. 4) is partially differentiated with respect to z, we get
−
1 2 P
1 2 U r 2 U r 3 1 2 ur uz 3 1 2 uz
= i
+
−
−
−
−
rdz
r z 2 r 2 z 2
r rz 2 r z 2 r r z 3 r z 2 z
1 3
1 4
1 4
− 2
−
−
2
2
2
4
r rz
r r z
r z
r 2 P
2 U r 2
3 2 U r
3 2 U z
−
= i 2 +
−Ur
−
−Uz 3 − 2
rz
z
r z 2
rz 2 z 2 r
z
z z
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1 3
4
4
−
−
−
2
r 2z 2 z 4
r rdz
(Eq. 6)
(Eq. 5) is partially differentiated with respect to r, we get
−
1 2 P
= −i
rz
1 2 1
1 2 2 1 3 1 2
−
+
U
+ 3
+
−
r − 2
r r 2
2
r r 2
r r r r 3 r 2 r 2
r r
1 1 2 U r 1 U z 2 U z 1 1 2U z
+ − 2
+
−
+
−
2
r r zr
r r 2 z
z r r 2
r r r r r
1 3
1 2 1 2 U z U z
+U z
−
+
+
2
2
z
r r z r rz r rz r
+
1 2 1
r r 2 − r 2 r
1 2 3 1 3 2 2 1 4 1 3 1 4
1 2U z
1 2
− 3
−
−
+
+
−
+
−
2
r r rz
r 4 r r 2 r 3 r 3 r 2 r r 4 r 2 r 3 r dr 2 dz 2 r 2 r z 2
r r
2 1
2 2 2 3 U r 2 1
r 2 P
−
= −i 2 −
+ U r −
+ 2
+
+
−
2
r z
r r
r r r 3 r r 2 r r
r
r r
3
U z 2 1 U z 2U z
1 2 2 U z
−
+
−
+Uz 2 −
+
r zr r r z
z r 2
r z r rdz rdz r
3 2 3 2 3 4
U z 2 1 2U z
4
1 3
+
−
+
− 2
− 3
−
+
+
−
2
z r 2 r r r rdz
r r r 2 r 3 r 4 r 2z 2 r 2 r z 2
r r
(Eq. 7)
Equating (Eq. 6) and (Eq. 7), we get
2 1 2
i 2 −
+
r r z 2
r
3 U r U z 2U r 2
3 2U r 1 U r 2U z 1 U z
=
U
+
+
−
+
U
+
−
+
−
r
z
r 3 r
z
r r 2
z 3 r 2 r r
rz r z r
3 U r U r U z 2 1 U z 2U z U z 2
3
+U r
+−
+
+
− 2
−
+Uz 2
2 +
2
rz r
r
z z
r z
r r z
r z
r r
4 2 3 3 2 3
4
2 3
4
− 4 −
+
−
+
2
−
+
r r 3 r 2 r 2 r 3 r
r 2z 2 r r z 2 z 4
r
Hence proved.
(Ans. c)
P5-47 The Taylor–Culick profile, which has been repeatedly used to describe the ideal gaseous
motion in solid rocket motors (Culick 1966), corresponds to an axisymmetric, self-similar profile
in a porous tube. When the tube radius is taken to be unity, the basic flow stream function may be
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-70-
written as 0 = z sin( 12 r 2 ) for 0 r 1 and 0 z Z N , where Z N represents the length of the
porous tube (see Fig. P5-47a).
FIGURE P5-47a
Assuming a wall injection speed of unity, the basic flow velocity may be expressed as
V0 (r , z ) = U r er + U z e z with U r = −r −1 sin( 12 r 2 ) and U z = z cos( 12 r 2 ) . As shown in the
problem above, the linearized Navier–Stokes equations with biglobal stability disturbance modes
can be reduced into a single, fourth-order, partial differential equation for the disturbance stream
− i t
function , where ˆ = (r , z )e . Following Chedevergne et al. (2012), the basic flow boundary
conditions require uniform wall-normal injection, no slip at the headwall, and axisymmetry with
respect to r = 0 , specifically:
0
0
(wall-normal injection)
r (1, z ) = 0; z (1, z ) = 1
0
0
(r , 0) =
( r , 0) = 0
(no slip at headwall)
z
r
3 0
0
0
(0, z ) =
(0, z ) = 0 (U r = U z / r = 0)
3 (0, z ) =
r
z
r
= +ˆ
0
(a) Recalling that the instantaneous motion is recoverable from
, and that disturbances
must be prevented from altering the basic flow conditions at the boundaries, show that the
disturbances must satisfy:
(wall-normal injection)
r (1, z ) = z (1, z ) = 0
(r , 0) =
(r , 0) = 0
(no slip at headwall)
z
r
3
(0, z ) =
(0, z ) = 0
(axisymmetry)
3 (0, z ) =
r
z
r
You may also impose the outflow condition suggested by Theofilis, Duck and Owen (2004) by
setting a simple extrapolation relation at the domain’s exit plane, z = Z N , namely,
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-71-
(r , Z N ) =
Z N − Z N −2
Z −Z
(r, Z N −1 ) + N −1 N (r, Z N −2 )
Z N −1 − Z N −2
Z N −1 − Z N −2
(b) Using the Taylor–Culick basic flow and a numerical solver of your choice, show that one of
the eigenvalues for
= 40.4 − 9.16i.
m = 0, Z N = 8
2
−4 2
and = 1/1975 m /s 5 10 m /s
is approximately
(c) Obtain the “boomerang” set of eigenvalues for Z N = 8 and Z N = 10 using = 5 10−4 m2 /s .
(d) Compare your results to those of Chedevergne et al. (2012) provided in Fig. P5-47b and explain
the differences.
FIGURE P5-47b
Solution:
(a) First, the two components of the mean flow are expressible by
U r = −r −1 sin( 12 r 2 ) and U z = z cos( 12 r 2 )
We also know that that the instantaneous velocity and flow streamfunction are related through
1
1
and U z =
Ur = −
r r
r z
Since the instantaneous motion is recoverable from = 0 +ˆ = 0 + ( x, y)e−i t , we may
differentiate with respect to z and r ,
0 ˆ 0 −i t
0 ˆ 0 −i t
and
=
=
+
+
=
=
+
+
e
e
r
z
r
z
r
z
r
z
r
z
For a porous tube with uniform wall injection, the fundamental boundary conditions on the
instantaneous field are the same as those on the mean flow (because the “disturbances must be
prevented from altering the basic flow conditions at the boundaries”). We have, for any time t ,
(1) A uniform injection and no-slip condition at the sidewall, U z (1, z , t ) = 0 and U r (1, z , t ) = −1
(2) A prescribed zero injection pattern at the headwall, U z (r , 0, t ) = U r (r , 0, t ) = 0
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-72-
2U z
(0, z, t ) = U z (0, z, t ) = U r (0, z, t ) = 0
y 2
By conveying these boundary conditions to the Stokes streamfunction, we get
Sidewall:
0
0
(1, z, t ) =
(1, z, t ) =
(1, z ) +
(1, z ) +
(1, z )e−i t and
(1, z )e−i t
r
z
r
z
r
z
or
−rU r (1, z ) = 1 +
(1, z )e −i t 1 = 1 +
(1, z )e −i t ; t
(1, z ) = 0
z
z
z
(3) A vanishing speed at the axis of symmetry,
rU z (1, z ) = 0 +
(1, z )e −i t 0 = 0 +
(1, z )e −i t ; t
(1, z ) = 0
r
r
r
Headwall:
0
0
(r , 0, t ) =
(r , 0, t ) =
(r , 0) +
(r , 0) +
(r , 0)e−i t and
(r , 0)e−i t
r
z
r
z
r
z
or
−rU r (r , 0) = 1 +
(r , 0)e −i t 0 = 0 +
(r , 0)e −i t ; t
( r , 0) = 0
z
z
z
rU z (r , 0) = 0 +
(r , 0)e −i t 0 = 0 +
(r , 0)e −i t ; t
( r , 0) = 0
r
r
r
Centerline:
3 0
3
3
0
(0,
z
,
t
)
=
(0,
z
)
+
(0, z )e −i t
(0, z, t ) =
(0, z ) +
(0, z )e−i t and
3
3
3
r
r
r
z
z
z
or
−rU r (0, z ) = 1 +
(0, z )e −i t 1 = 1 +
(0, z )e −i t ; t
(0, z) = 0
z
z
z
r 3 2U z
3
3
3
− i t
−i t
(0,
z
)
=
0
+
(0,
z
)
e
0
=
0
+
(0,
z
)
e
;
t
(0, z ) = 0
2 r 2
r 3
r 3
r 3
(b) Using the Taylor–Culick basic flow and a numerical solver, one can show that among
the eigenvalues for m = 0, Z N = 8 and = 5 10−4m2 /s is approximately = 40.4 − 9.16i.
(c) The “boomerang” set of eigenvalues for Z N = 8 and Z N = 10 may be obtained using
= 5 10−4m2 /s and a numerical solver Results are illustrated below.
In the figure below, we compare the biglobal stability frequency spectra for a porous tube driven
by the Taylor–Culick mean flow motion at a crossflow Reynolds number of Re = 1975 , m = 0 ,
r = 1 , and using (a) ZN = 8 and (b) 10. Here, both (□) and (■) reflect eigenmodes obtained from
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-73-
the present incompressible stability solver whereas (▲) represent eigenmodes obtained from the
incompressible biglobal stability solver developed by Chedevergne et al. (2012).
(d) One may note a slight leftward shift of the eigenmodes obtained from the
incompressible biglobal stability solver in comparison to the results obtained by
Chedevergne et al. (2012). The leftward shift may be ascribed to the use of a slightly
different formulation. The additional modes shown here are due to the use of the QZ
algorithm and a larger discretization stencil, thus leading to a broader spectrum of modes
than obtained previously through the Arnoldi algorithm used by Chedevergne et al. (2012);
the latter has the capability of pinpointing modes that are in the neighborhood of a userspecified value.
P5-48 Identify the operators that appear in the biglobal stability disturbance equations using
cylindrical polar coordinates, which are given by Eqs. (5-94)–(5-97), in order to provide the source
coefficients for the eigenvalue problem represented by Aij f i = Bij f i , where σ and fi denote the
eigenvalues and corresponding eigenvectors, while Aij and Bij refer to
Ac ,ur
Ar ,ur
Aij =
A
, ur
Az ,u
r
Ac ,u
Ac ,u2
Ar ,u
Ar ,uz
A ,u
A ,u2
Az ,u
Az ,uz
Ac , p
Ar , p
A , p
Az , p
and
Bc ,ur
Br ,ur
Bij =
B
, ur
Bz ,u
r
Bc ,u
Bc ,uz
Br ,u
Br ,uz
B ,u
B ,uz
Bz ,u
Bz ,uz
Bc , p
Br , p
B , p
Bz , p
Show that
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-74-
U U
U
im
U r
A ,ur =
+
− 2 2
+
+ im + U z
1 Ar ,ur = U r
r
r
r
r r
r
z
Ac ,ur = r + r
2
2
2
U
1 U U r
1 1+ m
+
+
+ U z + im
− 2 +
− 2 + 2 A ,u = U r
A = im
r r
r
z
r
r r
r
z
c ,u
r
r
2
2
2
1 1+ m
1 U r 2U
im
− 2 +
− 2 + 2
A
=
−
+
2
r ,u
r r
r
z
Ac ,uz =
r
r
r
r2
z
U
U r
1
im
Ac , p = 0
Ar ,uz =
Ar , p =
A ,uz =
A , p =
z
r
z
r
And
U z
1 U z
Az ,ur =
Az ,u =
r
r
U
U z
+Uz +
+ im
Az ,uz = U r
r
z z
r
2
2
2
1 m
1
− 2 + 2 Az , p =
− 2 +
r r r
z
z
r
and that the only nonvanishing values of Bij are Br ,ur = B ,u = Bz ,uz = i .
Solution: Cylindrical coordinates are regrouped into operators and written as follows:
1
im
+ ur + u + uz = 0 (continuity) (5-94)
r r
r
z
2 1 1 + m2 2
U
U r
U
+
+
im
+
U
−
− 2 + 2 ur
r
2+
z
r
z
r r
r
z
r
r r
1
1 U r 2U 2 im
U
+
−
+ 2 u + r u z +
p = (i )ur
r
r
z
r
r
(r − mom.) (5-95)
2 1 1 + m2 2
imU 1 U U r
U U 2 im
+
−
u
+
U
+
+
+
−
− 2 + 2 + U z u
r
2+
2 r
r
r
r
r
r
r r
r
z
z
r
r
r
U
+
z
U z
r
1 U z
ur +
r
im
p = (i )u
uz +
r
( − mom.) (5-96)
2 1 m2 2
U
U z
u
+
U
+
im
+
U
+
−
−
+
2+
uz
z
r
r
z z
r r r 2 z 2
r
r
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-75-
1
+
p = (i )u z
z
( z -mom ) (5-97)
Eigenvalue problem is represented by Aij f i = Bij f i , where σ and fi denote the eigenvalues and
corresponding eigenvectors,
while Aij and Bij refer to
Ac ,ur
Ar ,ur
Aij =
A
, ur
Az ,u
r
Ac ,u
Ac ,u2
Ar ,u
Ar ,uz
A ,u
A ,u2
Az ,u
Az ,uz
Ac , p
Ar , p
A , p
Az , p
and
Bc ,ur
Br ,ur
Bij =
B
, ur
Bz ,u
r
Bc ,u
Bc ,uz
Br ,u
Br ,uz
B ,u
B ,uz
Bz ,u
Bz ,uz
Bc , p
Br , p
B , p
Bz , p
Equations (5-94)–(5-97) are rewritten with operator matrices elements, and we get
( A ) u + ( A ) u + ( A ) u
c ,ur
r
c ,u
c ,uz
= 0 (Eq. 1)
z
( A ) u + ( A , ) u + ( A ) u + ( A ) P = ( B ) u
r ,ur
r
r u
r ,u z
z
r ,P
r ,ur
r
(Eq. 2)
( A )u + ( A )u + ( A )u + ( A ) P = (B )u
(Eq. 3)
( A )u + ( A )u + ( A )u + ( A ) p = ( B )u
(Eq. 4)
,ur
r
z ,ur
r
,u
z ,u
,u z
z ,u z
z
z
,P
z, p
,u
z ,u z
z
Comparing the continuity equation (5-94) with (Eq. 1),
1
Ac,ur = +
dr r
Ac ,u =
im
r
Ac ,uz =
z
Ac , P = 0
Bc ,ur = Bc ,u = Bc ,u z = Bc , p = 0
Comparing the r-momentum equation (5-95) with (Eq. 2),
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-76-
Ar ,ur = U r
2 1 1 + m2 2
U
U r
+
+ im + U z − 2 +
− 2 + 2
r r
r
z
r
z
r r r
Ar ,u =
1 U r 2U 2 im
−
+ 2
r
r
r
Ar ,u z =
U r
z
Ar , p =
1
r
Br ,ur = i
Br ,u = Br ,uz = Br , p = 0
Comparing the -momentum equation (5-96) with (Eq. 3),
A ,ur =
U U 2 im
+
− 2
r
r
r
A ,u = U r
2 1 1 + m2
imU 1 U U r
+
+
+
− 2 2 +
− 2 + 2 +Uz
r
r
r
r
r r
r
z
z
r
A ,uz =
U
z
A ,P =
im
r
B ,u = i
B ,ur = B ,uz = B , p = 0
Comparing the z-momentum equation (5-97) with (Eq. 4),
Az ,ur =
U z
r
Az ,u =
1 U z
r
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Az ,uz = U r
Az , P =
2 1 m2 2
U
U z
+ im + U z +
− 2 +
− 2 + 2
r
r
z z
z
r r r r
1
z
Bz ,uz = i
Bz ,ur = Bz ,u = Bz , p = 0
Hence proved.
(Ans)
P5-49 Using Eq. (5-106), provide the linear coordinate transformations that are needed to map an
axisymmetric problem defined in the (r, z) domain from the 0 ≤ r ≤ 1 and 0 ≤ z ≤ ZN intervals to
the Chebyshev domain of [−1, 1].
Solution: Coordinate transformation is necessary to redistribute the collocation points within an
interval for the purpose of giving high resolution to regions of very rapid change.
Using Eq. (5-106), map an axisymmetric problem for r ( a, b ) and z ( c, d ) . The two auxiliary
variables are and .
r=
b
a
( + 1) − ( − 1)
2
2
=
2r − ( a + b )
b−a
z=
d
c
( + 1) − ( − 1)
2
2
=
2z − ( c + d )
d −c
and
2
=
r b − a
2
=
z d − c
Domain that consists 0 ≤ r ≤ 1 and 0 ≤ z ≤ ZN may be reconditioned as follows:
r=
1
=2
( + 1) = 2r − 1
2
r
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and
z=
ZN
2z
2
=
( + 1) = − 1
2
ZN
z Z N
(Ans)
P5-50 Using Eq. (5-106), provide the linear coordinate transformations that are needed to map a
cylindrical polar problem defined in the (r, θ, z) domain from the 0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π, and 0 ≤ z ≤
ZN intervals to the Chebyshev domain of [−1, 1].
Solution: Using Eq. (5-106), map a cylindrical polar problem for r, θ, and z where r ( a, b ) ,
( c, d ) , and z ( e, f ) . The three auxiliary variables are , , and .
b
a
r = ( + 1) − ( − 1)
2
2
d
c
= ( + 1) − ( − 1)
2
2
f
e
z = ( + 1) − ( + 1)
2
2
2r − ( a + b)
b−a
2 − (c + d )
=
d −c
2 z − (e + f )
=
f −e
=
2
=
r b − a
2
and
=
d − c
2
=
z f − e
Domain that consists of 0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π, and 0 ≤ z ≤ ZN intervals may be reconditioned as
follows:
1
r = ( + 1)
2
= ( + 1)
Z
z = N ( + 1)
2
= 2r − 1
= −1
=
2z
−1
ZN
=2
r
2
1
=
and
2
=
z Z N
(Ans)
P5-51 Identify the block matrices that must be used to convert the biglobal stability disturbance
equations in cylindrical polar coordinates, based on Eqs. (5-94)–(5-97), to an eigenvalue problem
of the form Aij f i = Bij f i , where σ and fi denote the eigenvalues and corresponding eigenvectors.
The block matrices appear as
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Ac ,ur
Ar ,ur
Aij =
A
, ur
Az ,u
r
Ac ,u
Ac ,uz
Ar ,u
Ar ,uz
A ,u
A ,uz
Az ,u
Az ,uz
Ac , p
Ar , p
A , p
Az , p
and
Bc ,ur
Br ,ur
Bij =
B
,ur
Bz ,u
r
Bc ,u
Bc ,uz
Br ,u
Br ,u2
B ,u
B ,uz
Bz ,u
Bz ,uz
Bc , p
Br , p
B , p
Bz , p
Show that the only nonvanishing values of Bij are Br ,ur = B ,u = Bz ,uz = iI N and that
im
U U
A ,ur = r + r − 2 r 2
ii
ii
A = (U ) D r + (U ) D z + U r + im U
r ,u r
r ii
N
z ii
N
r ii
1
r
1 U
r
A ,u = (U r )ii DNr +
Ac ,ur = DN + r
r ii
r 2 DNr 1 + m 2
ii
z 2
− ( DN ) +
− 2 + ( DN )
im
r
rii
+ U r + im U + U D z
ii
A
=
c ,u
( z )ii N
rii
r ii
r
U
im
1 U r
Ar ,u =
−2
+ 2 2
Ac ,u2 = DNz
r 2 D r 1 + m2
2
r ii
rii
r
− ( DN ) + N − 2 + ( DNz )
rii
rii
Ac , p = 0
1 r
U r
A
=
A
=
D
r
,
u
r
,
p
N
2
z ii
A ,u = U A , p = im
t
rii
z ii
dU z
1 U z
Az ,ur =
Az ,u =
r ii
r ii
U
U z
+ im
and Az ,uz = (U r )ii DNr + (U z )ii DNz +
r ii
z
2
2
1
m2
1
− ( DNr ) + DNr − 2 + ( DNz ) Azp = DNz
rii
rii
Solution: Partial derivatives with respect to r and z will be expressed in pseudo-spectral
differentiation matrix as follows:
= DNr ,
= DNz
r
z
When applying domain mapping and differentiation matrices to the answer for Problem 5-48, we
can easily extract the block matrices element as follows:
For continuity equation:
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Ac ,ur = DNr +
Ac ,u =
1
rii
im
rii
Ac,uz = DNz
Ac , P = 0
Bc ,ur = Bc ,u = Bc ,u z = Bc , p = 0
For r-momentum equation:
2
r 2 DNr 1 + m2
u imU
Ar ,ur = (U r )ii DNr + (U z )ii DNz + r +
−
− 2 + ( DNz )
( DN ) +
r ii
rii
rii
r
im
1 ur 2U
Ar ,u =
−
+ 2 2
r ii
rii
r
1
U
Ar ,uz = r , Ar , P = DNr
z ii
Br ,ur = iI N
Br ,u = Br ,uz = Br , p = 0
For θ-momentum equation:
U U 2 im
A ,ur = + − 2
r ii
rii
r
2
r 2 DNr 1 + m2
U u 1 U
z
A ,u = (U r )ii DNr + im + r +
+
U
D
−
− 2 + ( DNz )
( DN ) +
z N
r
r ii r ii
rii
rii
U
A ,uz =
y ii
A , p =
im
rii
B ,u = iI N
B ,ur = B ,uz = B , p = 0
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For z-momentum equation:
U
Az ,ur = z
r ii
1 U z
Az ,u =
r ii
Az ,uz
2
r 2 DNr m2
imU U z
z
= ( ur )ii D +
+
− 2 + ( DNz )
+ (U z )ii DN − ( DN ) +
z ii
rii
rii
r
Az , P =
r
N
1
DNz
Bz ,uz = iI N
Bz ,ur = Bz ,u = Bx , p = 0
Hence proved.
(Ans)
P5-52 Reduce the block matrices of Eqs. (5-94)–(5-97) that you just derived assuming
axisymmetric disturbances with no azimuthal variations and m = 0.
Solution: Equations for axisymmetric disturbance with no azimuthal variation and m = 0 were
shown in the answer for problem 5-46.
The block matrices appear as follows:
Ac ,ur
Aij = Ar ,ur
Az ,ur
As we know
Ac ,uz
Ar ,uz
Az ,uz
Bc ,ur
Ac , p
Ar , p and Bij = Br ,ur
Az , p
Bz ,ur
Bc ,uz
Br ,uz
Bz ,uz
Bc , p
Br , p
Bz , p
= DNr ,
= DNz
r
z
For continuity equation:
Ac ,ur = DNr +
1
rii
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Ac,uz = DNz
Ac , p = 0
Bc ,ur = Bc ,uz = Bc , p = 0
For r-momentum equation:
2
2
Dr
1
U
Ar ,ur = (U r )ii DNr + r − ( DNr ) + N + ( DNz ) − 2 + (U z )ii − DNz
rii
rii
r ii
U
Ar ,uz = r
z ii
Ar , P =
1
DNr
Br ,ur = iI N
Br ,uz = Br , p = 0
For z-momentum equation:
U
Az ,ur = z
r ii
2
2
Dr
U
Az ,uz = (U r )ii DNr + z − DNr ) + N + ( DNz ) + (U z )ii DNz
rii
z ii
Az , P =
1
DNz
Bz ,uz = iI N
Bz ,ur = Bz , p = 0
Hence proved.
(Ans)
P5-53 Consider a problem with a simple axisymmetric viscous-flow configuration, namely, a
cylindrical fluid domain that extends over 0 r 1 and 0 z Z N . Assume infinite impedance
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(hardwall) boundaries everywhere along r = 0, r = 1, and z = 0 , and that an outflow condition
exists at z = Z N , where a zero normal stress requirement based on Eq. (5-114) translates into
u
uz ur
u
+
= 0;
= 0; 2 z − p = 0
r
z
z
z
Show that the boundary conditions on the ( ur , u , uz , p ) disturbance eigenfunctions can be written
as
u z
u (0, z ) = 0
(0, z ) = 0
ur (0, z ) = 0
r
u (1, z ) = 0
u z (1, z ) = 0
ur (1, z ) = 0
u (r , 0) = 0
u (r , 0) = 0
u z (r , 0) = 0
r
ur
p ( r, Z N )
u
u
u z
( r, Z N ) = − z ( r, Z N ) ( r, Z N ) = 0
( r, Z N ) =
r
z
z
2
z
Solution:
p(0, z )
=0
r
−−
−−
−−
(centerline)
(sidewall)
(headwall)
(exit plane)
u
uz ur
u
+
= 0;
= 0; 2 z − p = 0
r
z
z
z
It is a simple axisymmetric flow. Therefore, at the centerline the boundary conditions are
ur ( 0, x ) = u ( 0, x ) = 0,
uz
( 0, z ) = 0
r
uz
=P
z
2u z P
=
zdr r
P
(0, z ) = 0
r
In headwall and sidewall no slip boundary conditions are applied.
ur (1,, z ) = u (1, z ) = u z (1, z ) = 0
(sidewall)
ur (r , 0) = u (r , 0) = ux (r , 0) = 0
(Headwall)
In exit plane outflow conditions are valid.
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-84-
uz
u
=− r
r
dz
ur
u
( r, zN ) = − z ( r, zN )
x
r
u
( r, zN ) = 0
z
u z
p
=
z
p ( r, zN )
uz
( r, zN ) =
z
2
Hence proved.
(Ans)
P5-54 Using the cylindrical polar formulation of P5-51–P5-53 and a numerical solver of your
2
1
choice, investigate the stability of the Taylor–Culick profile corresponding to 0 = z sin( 2 r )
for
0 r 1, 0 z Z N
.
Here
V0 (r , z ) = U r er + U z e z
with
U r = −r −1 sin( 12 r 2 )
and
U z = z cos( 12 r 2 ) . Using m = 0 and = 5 10−4m2 /s find the spectrum of eigenvalues for
Z = 10
ZN = 8
and N
. Compare your results to those of Chedevergne et al. (2012) provided in Fig.
P5-47b.
Solution:
This problem follows the solution to P5-53. In the figure below, we compare the biglobal stability
frequency spectra for a porous tube driven by the Taylor–Culick mean flow motion at a crossflow
Reynolds number of Re = 2000 , m = 0 , r = 1 , and using aspect ratios of (a) Z N = 8 and (b)
Z N = 10 . We assume infinite impedance (hardwall) boundaries everywhere along r = 0 , r = 1
and z = 0 . An outflow condition exists at z = Z N , as described in P5-53. Here, both (□) and (■)
reflect eigenmodes obtained from the present incompressible stability solver whereas (▲)
represent eigenmodes obtained from the incompressible biglobal stability solver developed by
Chedevergne et al. (2012).
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-85-
P5-55 Using the cylindrical polar formulation of P5-51–P5-53, investigate the stability of the
Cecil–Majdalani profile, which extends the Taylor–Culick motion to spinning rocket motors and
porous tubes (see Fig. P5-55). As shown by Cecil and Majdalani (2017) in their AIAA awardwinning paper, the basic motion can be described using V0 (r , z ) = U r er + U er + U z e z with
U r = −r −1 sin( 12 r 2 ) , U = r −1 1 − exp ( − 14 r 2 Re ) , and U z = z cos( 12 r 2 ) for 0 r 1 ,
0 2 , and 0 z Z N . Here = Vs / Vw represents the tangential speed of the spinning
chamber relative to the wall injection speed, and Re = Vwa / stands for the injection Reynolds
number based on the wall injection speed Vw , chamber radius a, and kinematic viscosity . When
the injection speed and chamber radius are taken to be unity, the Reynolds number becomes
equivalent to the reciprocal of the kinematic viscosity, Re = 1/ and the Hertzian frequency of
rotation may be deduced from f s = Vs / (2 a).
Using m = 0 and = 5 10−4m2 /s , find the
spectrum of eigenvalues for Z N = 8 and Z N = 10 . By comparing your results to those of
Chedevergne et al. (2012) in Fig. P5-47b. what can you say about the effect of axial rotation on
the stability of the basic motion? You may use the boundary conditions described in P5-53.
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FIGURE P5-55
Solution:
Using an incompressible stability solver with m = 0 and = 5 10−4m2 /s , and the boundary
conditions described in P5-53, we obtain the spectrum of eigenvalues for Z N = 8 and Z N = 10 .
By comparing our results to those of Chedevergne et al. (2012) in Fig. P5-47b. we see that the
effect of axial rotation reduces the stability of the basic motion.
Biglobal stability frequency spectra for a porous tube driven by the Cecil–Majdalani mean flow
motion at a crossflow Reynolds number of Re = 1975 , m = 0 , r = 1 , and using (a) XN = 8 and (b)
10.
Here, both hollow and solid square symbols, (□) and (■), represent eigenmodes obtained from the
present incompressible stability solver. The difference between them is that the solid squares, (■),
illustrate the characteristic ‘boomerang’ pattern of eigenvalues associated with the Taylor–Culick
mean flow motion. The difference here is that the eigenmodes appear to be slightly more
hydrodynamically unstable in comparison to the Taylor–Culick mean flow motion studied by
Chedevergne et al. (2012). The slight increase in the stability growth rate i in the present case
may be attributed to the additional swirling component that accompanies the Cecil–Majdalani
profile.
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-87-
P5-56 Identify the operators that appear in the compressible biglobal stability disturbance
equations in cylindrical polar coordinates, which are given by Eqs. (5-155)-(5-160), in order to
provide the source coefficients for the eigenvalue problem represented by Aij fi = Bij fi , where
and fi denote the eigenvalues and corresponding eigenvectors, while Aij and Bij refer to
Ac ,
Ar ,
Aij = A ,
Az ,
Ae,
Ac ,ur
Ac,u
Ac ,uz
Ar ,ur
Ar ,u
Ar ,uz
A ,ur
A ,u
A ,uz
Az ,ur
Az ,u
Az ,uz
Ae,ur
Ae,u
Ae,uz
Ac ,T
Ar ,T
A ,T
Az ,T
Ae,T
and
Bc ,
Br ,
Bij = B ,
Bz ,
Be,
Bc ,ur
Bc,u
Bc ,uz
Br ,ur
Br ,u
Br ,uz
B ,ur
B ,u
B ,uz
Bz ,ur
Bz ,u
Bz ,uz
Be,ur
Be,u
Be,uz
Bc ,T
Br ,T
B ,T
Bz ,T
Be,T
Show that
U
U r U r 1 U U z
Ac , = r + r + r + z + U r r + im r + U z z
U r
U r U U r
U r U2
T0
+U r
+
+Uz
−
+ T0
+ R
Ar , =
r
r
z
r
r
r
t
U U U
U U rU R
T0
U
+U r
+
+Uz
+
(density terms)
A , =
+ r imT0 +
t
r
r
z
r
U z U U z
U z
U
T
+
+Uz
+ R 0 + T0
Az , = z +U r
r
r
z
z
t
z
T
T U T
T
U
Ae, = c p 0 + cv U r 0 + 0 + U z 0 − RT0 U r
+ U z + im
t
r
r
z
z
r
r
0
1
Ac ,ur = r + 0 r + r
4 2 1 1 2 m2
U
U r
Ar ,ur = 0 U r
+ im + U z +
−
− +
−
2+
r
z r
r r r 2 z 2 r 2
r
3 r
(radial terms)
A = U + U − im 7 +
0
, ur
r
3r r r
r
T P
U z 1
2
−
+
Ae,ur = 0 c p 0 − 0 − r
Az ,ur = 0
r
3 r z r z
r
r
U
1 0
im
im 7
Ar ,u = 0 r − 2U − −
Ac,u = r + 0 r
r
3r r r
2 1 1 2 4m 2
imU + U r
1 U
A
=
U
+
+
U
+
−
− +
−
,u
2+
(tangential terms)
0 r
z
r
z r
r r r 2 z 2 3r 2
r
r
0 c p T0 1 P0
Az ,u = 0 U z − im
Ae,u =
−
−
r
3r z
r r
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0
U
U r 2
im
+ 0
Ar ,uz = 0
−
A ,uz = 0
−
Ac ,uz =
z
z
z
3 r z
z
3r z
2
2
2
U
U z
1 m
4
(axial terms)
+ im + U z +
− 2 +
− 2 +
Az ,uz = 0 U r
r
r
z
z
r r r
3 z 2
r
T P
Ae,uz = 0 c p 0 − 0 − z
z
z
and
R
Ar ,T = R 0 + 0
A ,T = im 0 + 0
Az ,T = R 0 + 0
Ac ,T = 0
r
r
z
r
z
2
2
2
A = c U + im U + U − R U 0 + U 0 + U 0 − k + 1 − m +
2
0 v r
z
z
r r
e,T
r
z
z
r
r r r 2 z 2
r
r
(temperature terms)
where the dissipation function may be reconstructed from
4 U U 1 U U z 1 2 U
U r U z
= 2 r − r −
−
+
+ Ur −
−
3 r
r r
z r r r
z
r
r
im U
U r
1 U
U
+2 + r − U + z +
r
z z
r
r r
U
4 2 U
1 U
U r U z im
1 1 U z U
= + Ur −
−
+ 2 + r − U − +
+
3 r
z r
r
z
r r r
r
r
z
U z U r 1 U z U im
U r U r 1 U
4 U
= 2 z −
−
−
+ 2
+
+
+
3 z
r
r r z
z r r
z r
r
and that the only nonvanishing values of Bij are Bc, = i, Br ,ur = B ,u = Bz ,uz = i 0 , Be,T = i 0cv , and
z
Be, = −iRT0 .
Solution: We are asked to determine the operators that appear in the block matrices for the stability
analysis. To start, we first expand the given equation:
Aij fi = Bij fi
Ac ,
Ar ,
A
,
Az ,
Ae,
Ac ,ur
Ac ,u
Ac ,u z
Ar ,ur
Ar ,u
Ar ,uz
A ,ur
A ,u
A ,uz
Az ,ur
Az ,u
Az ,uz
Ae,ur
Ae,u
Ae,uz
Ac ,T
Bc ,
Ar ,T u
Br ,
r
A ,T u = B ,
Bz ,
Az ,T u z
Be,
Ae,T T
Bc ,ur
Bc ,u
Bc ,u z
Br ,ur
Br ,u
Br ,u z
B ,ur
B ,u
B ,uz
Bz ,ur
Bz ,u
Bz ,uz
Be,ur
Be,u
Be,uz
Bc ,T
Br ,T u
r
B ,T u
Bz ,T u z
Be,T T
After expanding out, we can collect the following 5 equations:
Ac , + Ac ,ur ur + Ac ,u u + Ac ,uz u z + Ac ,T T = ( Bc , + Bc ,ur ur + Bc ,u u + Bc ,u z u z + Bc ,T T )
(
Ar , + Ar ,ur ur + Ar ,u u + Ar ,u z u z + Ar ,T T = Br , + Br ,ur ur + Br ,u u + Br ,u z u z + Br ,T T
)
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-89-
(
T = (B
T = (B
A , + A ,ur ur + A ,u u + A ,uz u z + A ,T T = B , + B ,ur ur + B ,u u + B ,uz u z + B ,T T
Az , + Az ,ur ur + Az ,u u + Az ,uz u z + Az ,T
z,
Ae, + Ae,ur ur + Ae,u u + Ae,uz u z + Ae,T
e,
+ Bz ,ur ur + Bz ,u u + Bz ,uz u z + Bz ,T T )
)
+ Be,ur ur + Be,u u + Be,u z uz + Be,T T )
It may be useful at this juncture to clarify the convention used to define the subscripts. While the
first subscript denotes the type of equation to which a term belongs (e.g., continuity, r -, -, z momentum, or energy), the second subscript denotes the primitive variable associated with that
term. For example, the coefficient for the density in the continuity equation is labeled as Ac , .
Bearing this in mind, we proceed by considering the continuity equation [Eq. (5-155)]:
0
im
U r U r 1 U U z
1
r + r + r + z + U r r + r U + U z z + r + 0 r + r ur
1
+ 0 + im0 u + 0 + 0 u z = i
(continuity)
r
z
z
By applying this labeling convention, we identify the following
U
U r U r 1 U U z
1
Ac ,ur = 0 + 0 +
Ac , =
+
+
+
+ Ur
+ im + U z
r
r r
z
r
r
z
r
r r
1 0
im
Ac ,u =
Ac ,uz = 0 + 0
+ 0
Ac,T = 0 Bc, = i
r
z
z
r
This specification can be easily seen by comparing the continuity equation to the expanded matrix
equation. We are simply matching these two equations and identifying their coefficients. The same
strategy can be extended to the momentum and energy equations.
The r -momentum equation [Eq. (5-156)] is provided below:
U r
U r U U r
U r U2
T0
+ Ur
+
+Uz
−
+ T0
+ R
r
r
z
r
r
r
t
4 2 1
U
U r
1 2 m2
U
+
im
+
U
+
−
+
−
+
− 2 ur
2
0 r
z
2
2
r
z
r
r r r z
r
3 r
r
U r 2
U
0
im 7
+ 0 r − 2U −
−
u
+
−
+ 0 T = i 0 ur ( r -mom.)
uz + R
0
z
3 r z
r
3r r r
r
r
By matching the coefficients against the expanded matrix form, we are able to identify
U
U r U U r
U r U2
T0
Ar , = r + U r
+
+Uz
−
+ T0
+ R
r
r
z
r
r
r
t
4 2 1 1 2 m2
U
U r
Ar ,ur = 0 U r
+ im + U z +
−
− +
−
2+
r
z r
r r r 2 z 2 r 2
r
3 r
U
U r 2
im 7
Ar ,u = 0 r − 2U −
−
Ar ,T = R 0 + 0 Br ,ur = i 0
A
=
−
r ,u z
0
r
r
z
3 rz
3r r r
r
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The same may be repeated for the -momentum equation [(Eq. 5-157)], specifically:
U
U U im 7
U U U
U U rU R
T0
t + U r r + r + U z z + r + r imT0 + + 0 r + r − 3r r + r ur
2 1 1 2 4m2
imU + U r
1 U
+ 0 U r
+
+Uz +
− 2 +
− +
−
u
r
z r
r r r 2 z 2 3r 2
r
r
im
R
U
+ 0 −
u z + im0 + 0 T = i 0 u
( -mom.)
z
3r z
r
As before, by matching the coefficients against the expanded matrix form, we retrieve
U U U
U U rU
U
A , = + U r
+
+Uz
+
r
r
z
r
t
U im 7
U
A ,ur = 0 + −
+
r 3r r r
r
imU + U r
1 U
A ,u = 0 U r +
+Uz +
r
z r
r
A ,uz = 0
U im
−
z
3r z
A ,T =
T0
R
+ r imT0 +
2 1 1 2 4m 2
−
− +
−
2+
r r r 2 z 2 3r 2
r
R
im 0 + 0
r
B ,u = i 0
The z -momentum equation [Eq. (5-158)] follows suit. We have:
U z
U z U U z
U z
T0
t + U r r + r + U z z + R z + T0 z
2
U z 1
0 U z im
+ 0
−
+
−
u
ur +
r
3 r z rz
3r z
r
2 1 m2 4 2
im
U z
+ 0 U r
+ U + U z +
−
−
+
2+
uz
z
z
r r r 2 3 z 2
r r
r
+ R 0 + 0 T = i 0 u z
z
z
( z -mom.)
As such, we extract the following coefficients
U z U U z
U z
U
T
Az , = z + U r
+
+Uz
+ R 0 + T0
r
r
z
z
t
z
2
0 U z im
U z 1
−
Az ,ur = 0
−
+
Az ,u =
r
3r z
r
3 r z r z
im
U z
Az ,uz = 0 U r
+ U + U z +
z z
r r
2 1 m2 4 2
−
−
+
2+
r r r 2 3 z 2
r
Az ,T = R 0 + 0 Bz ,uz = i 0
z
z
Lastly, we may express the energy equation [(Eq. 5-159)] as follows:
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T0
T0
im
T0 U T0
c p t + cv U r r + r + U z z − RT0 U r r + U z z + r U
0 c p T0 1 P0
T P
T P
+ 0 c p 0 − 0 − r ur +
−
− u + 0 c p 0 − 0 − z uz
r
r
z
z
r r
2 1 m2 2
U 0
im
+ 0cv U r
+ U + U z − R U r 0 + U z 0 +
−
k
−
+
2+
T
z
r
z
r
r r r 2 z 2
r r
r
= i ( 0cvT − RT0 )
(energy)
The corresponding coefficients are
T0
T
im
T U T
+ cv U r 0 + 0 + U z 0 − RT0 U r + U z + U
t
r
r
z
z r
r
0c p T0 1 P0
T0 P0
T0 P0
Ae,ur = 0 c p
−
− r Ae,u =
−
− z
−
− Ae,uz = 0 c p
r
r
z
z
r r
U 2 1 m2 2
im
Ae,T = 0 cv U r + U + U z − R U r 0 + U z 0 + 0 − k 2 +
−
+
z
r
z
r r
r r r 2 z 2
r r
Be,T = i 0cv Be, = −iRT0
Ae, = c p
As for the dissipation functions given by Eq. (5-160), they may be expressed as
U
U r U z
4 U U 1 U U z 1 2 U
= 2 r − r −
−
+
+2 r −
−
3 r
r r
z r r r
r
r
z
im U
1 U r U U z U r
+2 +
−
+
+
r
r r
z z
r r
r
U
U
U r U z
4im 2 U
1 U r U 1 1 U z U
+2 r −
−
+ 2 +
−
+
− +
3r r
r
r
z
r
r r r r
z
r
U z U r im 1 U z U
z 4 U z U r U r 1 U
= 2
−
−
−
+ 2
+
+
+
3 z
r
r r z
z r r r
z
r
=
z
P5-57 Identify the block matrices that must be used to convert the compressible biglobal stability
disturbance equations in cylindrical polar coordinates, based on Eqs. (5-155)-(5-160), to an
eigenvalue problem of the form Aij fi = Bij fi , where and fi denote the eigenvalues and
corresponding eigenvectors. The block matrices appear as:
Ac ,
Ar ,
Aij = A ,
Az ,
Ae,
Ac ,ur
Ac,u
Ac ,uz
Ar ,ur
Ar ,u
Ar ,uz
A ,ur
A ,u
A ,uz
Az ,ur
Az ,u
Az ,uz
Ae,ur
Ae,u
Ae,uz
Ac ,T
Ar ,T
A ,T
Az ,T
Ae,T
and
Bc ,
Br ,
Bij = B ,
Bz ,
Be,
Bc ,ur
Bc,u
Bc ,uz
Br ,ur
Br ,u
Br ,uz
B ,ur
B ,u
B ,uz
Bz ,ur
Bz ,u
Bz ,uz
Be,ur
Be,u
Be,uz
Bc ,T
Br ,T
B ,T
Bz ,T
Be,T
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Show that the only nonvanishing values of Bij are Bc, = iI N , Br ,ur = B ,u = Bz ,uz = i 0 I N ,
Be,T = i 0cv I N , Be, = −iRT0 I N and that
Ac ,
Ar ,
A ,
A
z,
Ae,
U
1 U U z
U U
= r + r +
+
+ im + (U r )ii DNr + (U z )ii DNz
r r
z
r ii
r
U
T0
U r U U r
U r U2
= r +U r
+
+Uz
−
+ (T0 )ii DNr
+ R
r
r
z
r ii
r ii
t
U U U
U U rU
R
U
T
= +U r
+
+Uz
+
+ im (T0 )ii + 0
r
r
z
r ii rii
t
ii
T
U z U U z
U z
U z
=
+U r
+
+Uz
+ R 0 + (T0 )ii DNz
r
r
z ii
t
z ii
T U T
T
T
U
= c p 0 + cv U r 0 + 0 + U z 0 − R (T0 )ii (U r )ii DNr + (U z )ii DNz + im
r
r
z ii
t ii
r ii
(density)
Ac ,ur
A
r ,ur
A ,ur
Az ,u
r
1
= 0 + ( 0 )ii DNr +
rii
r ii
U
4
U
= ( 0 )ii (U r )ii DNr + im + (U z )ii DNz + r − DNr
r ii
r ii
3
( )
2
+
1 r 1
DN − 2 + DNz
rii
rii
( )
2
−
m 2
rii2
U
im 7 1
U
= ( 0 )ii + − 2 + DNr
r ii
3 rii rii
r
z
DN
T P
U z
= ( 0 )ii
+ DNr DNz
Ae,ur = ( 0 )ii c p 0 − 0 − r
−
r ii 3 rii
r ii r ii
(radial terms)
Ac ,u
A
,u
Az ,u
im 1
7
1 U r 2U
Ar ,u = ( 0 )ii
−
− DNr − 2
r ii
3 rii
rii
r
im (U )ii + (U r )ii
1 U
= ( 0 )ii (U r )ii DNr +
+ (U z )ii DNz +
rii
rii ii
2
2
Dr
1
4m 2
− DNr + N − 2 + DNz − 2
rii rii
3rii
=
1 0
+ im 0
rii
ii
( )
=
( 0 )ii U z
rii
(tangential)
( )
im z
− 3r DN
ii
ii
Ae,u = c p
( 0 )ii T0
rii
1 P0
− r −
ii
ii
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Ac ,u z
A ,uz
A
z ,u z
U
= 0 + ( 0 )ii DNizi
Ar ,u z = ( 0 )ii r − DNr DNz
z ii 3
z ii
im z
U
= ( 0 )ii −
DN
3rii
z ii
U
= ( 0 )ii (U r )ii DNr + im
r
r
2
D
m2 4
− DNr + N − 2 +
rii
3
rii
( )
(axial terms)
U z
z
+ (U z )ii DN + z
ii
ii
2
T P
DNz
Ae,u z = 0 c p 0 − 0 − z
z
z ii
( )
and
Ac ,T
A
,T
Ae,T
=0
=
Ar ,T = R 0 + ( 0 )ii DNr
r ii
R
im0 + 0
rii
ii
Az ,T = R 0 + ( 0 )ii DNz
z ii
U
= ( 0 )ii cv (U r )ii DNr + im + (U z )ii DNz
r ii
r 2 DNr m 2
U 0
− R U r 0 + U z 0 +
−
k
− 2 + DNz
DN +
r
z
r ii
rii
rii
( )
(temperature terms)
( )
2
where the dissipation function may be reconstructed from
U
U r U z 1
4 U U 1 U U z r 2 U
= 2 r − r −
−
DN +
+2 r −
−
3 r
r r
z ii
r
r
z ii rii
r
4 2 U
U
1 U
1
U r U z im
=
+ Ur −
−
+ 2 + r − U DNr −
3 r
z ii rii
r
rii
ii
r
r
r
1 U z U z
+
DN
+
z ii
r
U
z 4 U z U r U r 1 U z
U r r 1 U z U im
= 2
−
−
−
DN + 2 z +
DN +
+
3 z
r
r r ii
z ii
z ii rii
r
r
Solution: This problem involves converting our previous matrix formulation to a form that can be
conveniently applied to a grid. To start, any primitive or mean flow variable will be evaluated at
the ii location, which stands for the ( r , z ) grid location, as we assume an axisymmetric
configuration for our biglobal formulation. Next, all derivatives are converted to derivative
operators according to the following terminology:
= DNr
r
= DNz
z
2
= = DNr
2
r r
r
( )
2
2
= = DNz
2
z z
z
( )
2
By applying these notations to the above set of matrix equations, we obtain:
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Ac ,
Ar ,
A ,
A
z,
Ae,
U
1 U U z
U U
= r + r +
+
+ im + (U r )ii DNr + (U z )ii DNz
r r
z
r ii
r
U
T0
U r U U r
U r U2
= r +U r
+
+Uz
−
+ (T0 )ii DNr
+ R
r
r
z
r ii
r ii
t
U U U
U U rU
R
U
T
= +U r
+
+Uz
+
+ im (T0 )ii + 0
r
r
z
r ii rii
t
ii
T
U z U U z
U z
U z
=
+U r
+
+Uz
+ R 0 + (T0 )ii DNz
r
r
z ii
t
z ii
T U T
T
T
U
= c p 0 + cv U r 0 + 0 + U z 0 − R (T0 )ii (U r )ii DNr + (U z )ii DNz + im
r
r
z ii
t ii
r ii
(density)
Ac ,ur
A
r ,ur
A ,ur
Az ,u
r
1
= 0 + ( 0 )ii DNr +
rii
r ii
U
4
U
= ( 0 )ii (U r )ii DNr + im + (U z )ii DNz + r − DNr
r ii
r ii
3
( )
2
+
1 r 1
DN − 2 + DNz
rii
rii
( )
2
−
m 2
rii2
U
im 7 1
U
= ( 0 )ii + − 2 + DNr
r ii
3 rii rii
r
z
DN
T P
U z
= ( 0 )ii
+ DNr DNz
Ae,ur = ( 0 )ii c p 0 − 0 − r
−
r ii 3 rii
r ii r ii
(radial terms)
Ac ,u
A
,u
Az ,u
im 1
7
1 U r 2U
Ar ,u = ( 0 )ii
−
− DNr − 2
r ii
3 rii
rii
r
im (U )ii + (U r )ii
1 U
= ( 0 )ii (U r )ii DNr +
+ (U z )ii DNz +
rii
rii ii
2
2
Dr
1
4m 2
− DNr + N − 2 + DNz − 2
rii rii
3rii
=
1 0
+ im 0
rii
ii
( )
=
( 0 )ii U z
rii
(tangential)
( )
im z
− 3r DN
ii
ii
Ae,u = c p
( 0 )ii T0
rii
1 P0
− r −
ii
ii
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Ac ,u z
A ,uz
A
z ,u z
U
= 0 + ( 0 )ii DNz
Ar ,u z = ( 0 )ii r − DNr DNz
z ii 3
z ii
im z
U
= ( 0 )ii −
DN
3rii
z ii
U
= ( 0 )ii (U r )ii DNr + im
r
r
2
D
m2 4
− DNr + N − 2 +
rii
3
rii
( )
U z
z
+ (U z )ii DN + z
ii
ii
2
T P
DNz
Ae,u z = 0 c p 0 − 0 − z
z
z ii
(axial terms)
( )
and
Ac ,T
A
,T
Ae,T
=0
=
Ar ,T = R 0 + ( 0 )ii DNr
r ii
R
im0 + 0
rii
ii
Az ,T = R 0 + ( 0 )ii DNz
z ii
U
= ( 0 )ii cv (U r )ii DNr + im + (U z )ii DNz
r ii
r 2 DNr m 2
U 0
− R U r 0 + U z 0 +
−
k
− 2 + DNz
DN +
r
z
r ii
rii
rii
( )
(temperature terms)
( )
2
As for the dissipation function, its components can be expressed as:
U
U r U z 1
4 U U 1 U U z r 2 U
= 2 r − r −
−
DN +
+2 r −
−
3 r
r r
z ii
r
r
z ii rii
r
4 2 U
U
1 U
1 1 U z U z
U r U z im
=
+ Ur −
−
+ 2 + r − U DNr − +
+
DN
3 r
z ii rii
r
rii r
z ii
ii
r
r
U
z 4 U z U r U r 1 U z
U r r 1 U z U im
= 2
−
−
−
DN + 2 z +
DN +
+
3 z
r
r r ii
z ii
z ii rii
r
r
r
Lastly, the right-hand side block matrix has to be converted. Recalling that the derivatives are a
Kronecker product, the B terms have to be arranged to ensure that the equations are properly
configured as per Eq. (5-108). The right-hand side is multiplied by the identity matrix I N to
reproduce the correct distribution. This is accomplished by taking:
Bc , = iI N Br ,ur = i 0 I N B ,u = i 0 I N Bz ,uz = i 0 I N Be,T = i 0cv I N Be, = −iRT0 I N
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CHAPTER 6. INCOMPRESSIBLE TURBULENT MEAN FLOW
6-1
By direct substitution of fluctuating variables, Eqs. (6-6), into the x-momentum relation
and time-averaging using (6-7), develop the x-component of the three-dimensional “Reynolds”
equation (6-12) and reduce to a two-dimensional turbulent boundary layer.
The key to this development is to write the convective acceleration in Eq. (6-11) in the
“conservative” form mentioned just below. For the x direction, we obtain
(V ) u =
( uu ) + ( vu ) + ( wu )
x
y
z
Then introduce u = u + u, etc. into the full x-momentum relation, Eq. (6-11):
( u + u) + ρ ( uu + uu + u2 ) + ρ ( uv + uv + uv + uv )
t
x
y
+ ρ ( uw + uw + uw + uw ) = − ( p + p ) + ρg x + μ 2 ( u + u )
z
x
ρ
(1)
Now take the time-average of Eq. (1), using the averaging relations, Eq. (6-7), leaving:
u
u
u
u
ρ +u
+v
+w
+ ( u u ) + ( u v ) + ( uw )
x
y
z x
y
z
t
p
= − + ρg x + μ 2 u
x
(Ans. 3-dimensional)
Since the first four terms on the left equal ρ ( du/dt ) , this relation is, as desired, the
x-component of the mean (turbulent) momentum equation, Eq. (6-12).
For a steady two-dimensional boundary layer, /t = 0, /z = 0, w = 0, and the
y-derivative (across the layer) is the dominant one. The relation above reduces to
u
u
dp
2u
ρ u
+v
+ ( u v ) = −
+ ρg x + μ 2
y y
dx
y
x
(Ans. 2-d T.B.L.)
where the pressure gradient could be replaced by dp/dx = −ρU ( dU/dx ) in the freestream.
This is the desired result and is equivalent to Eq. (6-21b) in the text, page 411.
6-2
Using the Navier-Stokes equation in cylindrical coordinates (Appendix B), develop a
three-dimensional Reynolds-stress equation for z-momentum and simplify it to axial steady
turbulent boundary flow along me outside of a circular cylinder.
This is somewhat more complicated than Prob. 6-1 above, because of the added complexity of
the cylindrical geometry. The axial convective acceleration can be written in a “conservative”
form, using the continuity relation, Eq. (B-3), as follows:
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-1-
( V ) vz = vr
vz vθ vz
v 1
1
+
+ vz z =
( rvr vz ) +
( vθ v z ) + ( v z v z )
r
r θ
z r r
r θ
z
The full Navier-Stokes z-momentum equation is given by Eq. (B-8):
p
v
ρ z + ( V ) vz = − + ρg z + μ2 vz ,
z
t
where 2 =
1 1
2
r
+
+
r r r r 2 θ 2 z 2
Introduce fluctuating variables into this relation, using ( vr , vθ ,vz ) = ( u,v,w ) for convenience
in the notation:
ρ
ρ
( w + w) +
( ruw + ruw + ruw + ruw ) +
( vw + vw + vw + vw )
t
r r
r θ
+ ρ ( ww + 2ww + w w ) = − ( p + p ) + ρg z + μ 2 ( w + w )
z
x
ρ
Now take the time-average of this relation, using the averaging rules, Eqs. (6-7), leaving:
w v w
w 1
1
w
ρ
+u
+
+w
+
( ruw ) +
( vw ) + ( w w )
r r θ
z r r
r θ
z
t
p
= − + ρg z + μ 2 ( w )
(Ans. 3-dimensional)
z
This is the desired 3-D result. For steady axisymmetric flow along the outside wall of a circular
cylinder, /t = 0, /θ = 0, vθ = “v” = 0, and the dominant derivative is /r across the
boundary layer. Further, if boundary-layer displacement is (presumed) small, there is no
pressure gradient. Further neglect gravity along the cylinder. The 3-D equation above
reduces to
w 1
μ w
w
ρ w
+u
+
( ruw)
r
r r r
z
r r r
(Ans. 2-d T.B.L.)
where, as mentioned, u represents v r and w represents v z . We may rewrite this in terms of a
cylindrical boundary-layer “turbulent shear stress”.
τ = τlam + τ tub , where τlam = μ
w
and τ tub = −ρuw
r
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The axisymmetric cylinder boundary-layer flow then may be written as
w 1
w
p w
+u
(r τ)
(Ans. 2-dim. T.B.L.)
=
r r r
z
This has exactly the same form as laminar axisymmetric cylinder boundary-layer flow from
Eq. (4-195b), page 299 of the text.
6-3
Consider turbulent flow past an isothermal flat plate of length L and width b, with
constant ( ρ,μ,cp ,k ) . Assume Pr = 1, that is, δ u = δ T , with uniform stream velocity U and
stream temperature Te at the plate leading edge. At the trailing edge, x = L, the mean velocity
and temperature may be approximated by one-seventh power-law profiles:
u T − Tw y
U Te − Tw δ
With no further information, estimate the drag force and total heat transfer of one side of the
plate, in terms of the boundary layer thickness δ and other flow parameters.
1/7
The flat-plate control-volume analysis of Section 4-l is appropriate for turbulent flow also, as
long as the mean profiles are known. The drag force is given by Eq. (4-6), page 218 of the text:
δ
F = ρu ( U − u ) b dy = ρδbU
0
1
2
η (1 − η )dη = 72 ρδbU
1/7
7
1/7
2
(Ans. a)
0
In like manner, the total heat transfer is given, for either laminar or turbulent flow, by the flatplate thermal integral analysis, Eq. (4-19), neglecting kinetic energy changes:
δ
1
dQ
= ρcp u ( T − Te ) b dy = ρδc p U ( Tw − Te ) η1/7 (1 − η1/7 )dη
dt 0
0
or:
dQ 7
= ρδbcp U ( Tw − Te )
dt 72
(Ans. b)
The analysis would be complete if we had an accurate estimate for the growth of the boundary
layer thickness, δ(x). For example, use of Prandtl’s flat plate estimate from
Eq. (6-72), δ/L = 0.37/ Re L1/5 , gives the dimensionless drag force formula
CD =
2 ( 7/72 )( 0.37 ) 0.072
2F
(turbulent flat-plate flow)
2
ρU bL
Re1/5
Re1/5
L
L
which is popular in the literature, e.g., Kays & Crawford (1980), page 175. Similarly, the
Stanton number could be approximated as
CH =
0.37 ( 7/72 ) 0.036
dQ/dt
=
=
bLρUcp
Re1/5
Re1/5
L
L
This formula (which assumes Pr = 1 ) is given on p. 213 of Kays & Crawford (1980).
6-4
The first experiment of Clauser (1954) used air at 24°C and 1 atm. Velocity data at the
first station, x = 6.92 ft, are given below. Taking δ = 3.5 in. and dU/dx = 1.06 sec−1 at this
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-3-
station, analyze the data to find (a) the inner law and wall shear stress; (b) the outer law and
Clauser’s parameter β and Coles’ parameter P ; and (c) the logarithmic overlap.
y, in
u, ft/s
y, in
u, ft/s
0.1
16.14
0.8
22.88
0.15
17.02
0.9
23.70
0.2
17.54
1.0
24.38
0.25
18.16
1.25
26.51
0.3
18.69
1.5
28.21
0.4
19.60
2.0
31.22
0.5
20.49
2.5
32.27
0.6
21.24
3.0
32.44
0.7
22.03
3.5
32.50
For air at 24C and 1 atm, take ρ = 0.00230 slug/ft 3 and v = 0.000165 ft 2 /s. A preliminary plot
(not shown) of u versus log(y) reveals that the first five or six points form a straight line - the
logarithmic overlap. We may thus test the log-law relation,
u 1 yv*
ln
+ B, with κ = 0.41 and B 5.0
v* κ v
for the first seven data points y and u(y) to estimate the value of friction velocity v*. The results
are shown in the following table:
y ( in.) =
v* ( ft/s ) =
y+ =
0.1
0.15
0.2
0.25
0.3
0.4
0.5
1.091
1.080
1.068
1.078
1.072
1.080
1.094
55
82
109
136
164
218
273
All these points lie in the usual logarithmic overlap range 30 y+ 300. Therefore, at this
station, we conclude that the friction velocity and wall shear stress are
v* = 1.08 ft/s τ w = ρv *2 = ( 0.00230 )(1.08) = 0.0027 psf
2
(Ans. a)
The first data point is not close enough to the wall to demonstrate an inner law.
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-4-
To establish Clauser’s parameter β, we need to estimate δ*. Numerical integration (by
the trapezoidal rule) of the given velocity data give the estimate δ* = 0.59 in. Then
β=
δ* dp
δ*
dU ( 0.59/12 )( 32.50 )(1.06 )
=
−ρU
= 1.45 (Ans. b)
=
2
2
τ w dx ρv*
dx
(1.08)
Coles and Hirst (1968) give β 1.358 at this station for their data reduction.
To establish Coles’ parameter P , we need to fit the wake law, Eq. (6-47) to the
velocity-profile data. Well, the fit is good in the intermediate region, 0.5 in < y < 2 in, but the
edge of the boundary layer is overpredicted with such a fit. This is seen in the following graph
of the data in u+ versus log(y+) coordinates:
The wake law for P 2.4 (Ans. b) fits very well nearer the wall but overshoots the last three
data points. The last point, y = δ = 3.5 in, is best fit by P 1.4. Coles and Hirst (1968) do
not give an estimate of P for this station.
The graph on above shows that the logarithmic overlap fits the data very well.
However, Coles’ law does not fit the entire profile but only its interior portion. Coles reported
similar edge discrepancies for other experimental boundary layer profiles.
With P uncertain, we cannot estimate * accurately from the wake-law formula (6-48).
With a best-fit P 2.4, the formula gives * = 0.96 inches, or about 50% too high.
6-5
Analyze the velocity data of Prob. 6-4 to achieve a power-law overlap approximation, as
in Eq. (6-50). Find and C and compare with Eq. (6.50). To save work, use Eq. (6-69) to
estimate the momentum thickness. At this station the wall shear stress is approximately
0.0027 lb/ft 2 . (HINT: Use only the first 10 data points, the rest are well into the wake region.)
Solution: For air at 24C, take = 0.00230 slug/ft 3 and v = 0.000165 ft 2 /s. The data from Prob.
6-4 may be plotted as follows in log-log coordinates. The initial slope is about 0.16.
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-5-
Based on the first 10 data points, the log-law is approximately u + = 7.8 ( y + )
0.16
, that is, C 7.8
and 0.16. From Eq. (6-69), approximately / = 7/72 , or = 7 ( 3.5 in ) /72 = 0.34 inches.
Thus Re = U /v = ( 32.5 ft/s )( 0.34/12 ft ) / ( 0.000165 ft 2 /s ) 5600. Then Fig. 6-14 and
Eq. (6.50) predict
C 3 + 0.62 ln ( 5600) 8.3 ( 6% high ) and = 1.24/ln (5600 ) 0.14 (10% low )
The given data τ w = 0.0027 lb/ft 2 is not needed here but could be found from the power-law.
6-6
For developed turbulent pipe flow, assuming that the log-law is va1id across the entire
tube, find a formula for centerline velocity as a function of Darcy friction factor .
The desired formula is hidden within Eq. (6-53), which computes average pipe-flow velocity
by integrating the log-law across the entire pipe:
3
1 av*
u av = v* ln
+B−
v
2κ
κ
( 6-53)
The terms in boldface are, in fact, equal to u max /v* (neglecting the slight ‘wake’ at the
centerline). Thus we may rearrange the above equation as follows:
u max = u av +
u max
3
3 v*
3
v*, or:
= 1+
= 1+
2κ
u av
2κ u av
2κ 8
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where the last substitution follows from the definition of Darcy friction factor,
2
= 4Cf = 8v*2 /u av
. Taking κ = 0.41, we have the final desired relation:
u max
1 + 1.29
u av
(Ans.)
6-7
Water at 20C flows through a smooth 3-cm-diameter pipe at 30 m3 /hr. Assuming
developed flow, estimate (a) the wall shear stress; (b) the pressure drop per meter; and (c) the
centerline velocity. What is the laminar-flow rate? What flow rate gives τ w = 100 Pa?
For water at 20C take ρ = 998 kg/m3 and μ = 0.001 kg/m-s. The average velocity in the
pipe is
3
Q ( 30/3600 ) m /s
u av = =
= 11.79 m/s
A ( π/4 )( 0.03 m )2
whence Re D =
ρu av D ( 998 )(11.79 )( 0.03)
=
= 353, 000
μ
( 0.001)
The flow is clearly turbulent ( Re 2000) . Since the walls are ‘smooth’, the Darcy friction
factor is given by Eq. (6-54) for developed smooth-wall pipe flow:
(
)
1
= 2.0 log10 ReD − 0.8, or = 0.01403 if ReD = 353, 000
The wall shear stress is then given by
1 2
2
1
τ w = ρu av
= ( 0.01403) ( 998)(11.79 ) = 243 Pa
8
8
( Ans. a )
The pressure drop in developed flow is given by
p = τ w
4 (1.0 )
4Δx
= ( 243)
32, 400 Pa/m
D
0.03
( Ans. b )
Finally, the centerline velocity is estimated from our formula derived in Prob. 6-6 above:
(
)
u max u av 1 + 1.29 = (11.79 ) 1 + 1.29
( 0.01403) = 13.6 m/s ( Ans. c )
The flow rate would have to be much less for the flow to be laminar:
ReD =
ρu av D 998u av ( 0.03)
=
2000 if u av 0.0668 m/s,
μ
0.001
or: Q transition = u av A = ( 0.0668 )( π/4 )( 0.03) = 4.72 E-5 m3 /s = 0.17
2
m3
hr
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-7-
Finally, finding the flow rate for which wall shear equals 100 Pa is vexing, because we know
neither the velocity nor the friction factor and thus have to iterate. However. Eq. (6-55) gives
us the estimate τ w proportional to Q7/4 . Thus a good first guess would be
Q100Pa
100
= Q243Pa
243
4/7
0.6, or Q 0.6 ( 30 ) 18 m3 /hr
Try this value with the numerical procedure above:
Q = 18
m3
m
; u av = 7.074 ; Re D = 211800; = 0.01546; τ w = 96.5 Pa
hr
s
Not bad. [The correct power on τ w = CQ n should have been n 1.81. ] Anyway, we can
interpolate between these two shear stresses to find that
τ w = 100 Pa if
6-8
Re D = 216000, or: Q = 0.00510
m3
m3
= 18.4
s
hr
The overlap region of Clauser’s velocity profile in Prob. 6-4 may be fit by a power-law
estimate y + 7.5 ( u + ) . Use this result to estimate the wall shear stress in lbf /ft 2 and the shape
1/ 6
factor H.
Solution: This is close to, but slightly different from, the power-law found in Prob. 6-5,
u + = 7.8 ( y+ ) . Find the wall shear stress by writing this in terms of v * :
0.16
1/ (1+ )
uv *a
u
yv *
=C
, or: v* =
a
v*
v
Cy
with C = 7.5 and =
1
6
Apply this formula to the first 10 points in the velocity table in Prob. 6-4. The results are
v* = 1.11 0.04 ft/s
Ans.
τ w = v *2 = 0.0028 0.0002 lb f /ft 2
Problem 6-5 stated that the wall shear stress was approximately 0.0027 lbf/ft2, or 4% lower.
For the shape factor estimate, we need to compute displacement and momentum
thicknesses. Write the power law in the form u /U = 1/6 , = y / . Then
1
1
* = (1 − ) d = ;
7
0
1/6
1
= 1/6 (1 −1/6 ) d =
0
3
28
The ratio of these two is the desired power-law shape factor estimate:
H=
* (1/7 )
4
=
= 1.33
( 3/28) 3
Ans.
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-8-
6-9
Analyze developed turbulent flow through a smooth-wall square duct, using the loglaw, the double symmetry, and a suitable assumption about distribution of wall shear stress.
Compare your result for friction factor with the hydraulic-radius concept.
By double symmetry, we need only consider one-eighth of the duct cross-section, shown as the
shaded area below. Although the flow in this region is three-dimensional (secondary circulation
toward the corners), the mean velocity u in the duct-axis direction is well represented by the
log-law, Eq. (6-38a), based on distance “y” from the wall, as sketched in the figure.
In general, however, the wall shear stress varies with z along the wall, being zero in the corners
and rather flat near z = a/2.
The total volume flow rate in the duct would be 8 times the flow rate in the shaded area:
z
1 yv*
Q = 8 dz v* ( z ) ln
+ B dy
v
κ
0
0
a/2
If, as a first approximation, we assume constant v* ( z ) around the cross-section, the integration
is easily performed:
u av =
1 a+
Q
3
av*
=
v*
+ B − , where a + =
av ln
2
a
2κ
v
κ 2
Now, by definition, the friction factor for any duct is such that v*av /u av =
( / 8 ).
Therefore the above relation may be rewritten as a friction factor formula:
(8/ )
1
ln Rea
κ
( /32 ) + B −
3
,
2κ
Rea =
u av a
v
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-9-
For κ = 0.41 and B = 5.0, this may be written in numerical form as
(
)
1
1.99 log10 Rea − 1.02
But these are exactly the same numbers which result for turbulent pipe flow - see the middle of
page 426 of the text. This simple theory thus predicts that a square duct friction is equivalent
to pipe friction if Rea = ReD , that is, if “a” is the effective diameter.
In fact, the ‘hydraulic diameter’ of a square duct does equal its side length, a:
D h ( square ) =
4A 4a 2
=
=a
P
4a
Therefore, a first-order use of the log-law in square-duct flow leads to the formula
(
)
1
2.0 log10 Rea − 0.8, a = side length
(Ans.)
If we modified the theory to have a variable v*(z) which satisfies the physical condition that
τw = 0 ( or v* = 0) in the duct corners, the friction formula might be improved. However, we
1/7
can’t guess how v*(z) varies: perhaps v*/v*max = ( 2z/a ) could be tried? Meanwhile, the
‘effective diameter’ concept of page 428 is known to work well.
6-10 In the overlap layer, turbulent shear is dominant, and the effect of viscosity is small.
Suppose that we neglect and replace Eq. (6-32) by the approximate gradient relationship
du
fcn ( y , w , )
dy
Show, by dimensional analysis, that this leads directly to the logarithmic overlap law (6-38a).
Solution: There are four variables and their dimensions in the {MLT} system are:
Variable
du /dy
y
w
Dimension
T −1
L
ML−1T −2
ML−3
With 4 variables and 3 primary dimensions, we expect only one Pi group, and it is:
P1 =
du y
du
v*
v*
= constant, or:
= constant
= C1
dy w
dy
y
y
Integrate:
u = C1v *ln ( y ) + C2
If the constants are arranged so that the logarithm has a dimensionless argument, we obtain
u /v* = u + =
1
ln ( y + ) + B
κ
6-11 Use the log-law to analyze developed turbulent flow in a smooth annulus of inner radius
ri and outer radius ro. Compare the friction result with hydraulic-diameter theory.
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The (turbulent) velocity profile in the annular region would look like this:
In general, the outer-wall shear stress is somewhat less than the inner-wall shear, and the radial
position rm of the maximum velocity (the zero-shear line) is somewhat toward the inner wall
rather than in the exact middle of the annular region. This is seen by making a force balance
on the outer region and comparing to the inner region, as follows:
Inner region: p π ( rm2 − ri2 ) = τi 2πri Δx
Outer region: p π ( ro2 − rm2 ) = τ o 2πro Δx
Divide one by the other to get the ratio of inner to outer shear stress:
2
2
τo ri ( ro − rm )
=
τi ro ( rm2 − ri2 )
(1)
Meanwhile, we can write a log-law profile for each region, using wall coordinates y i and yo,
respectively, as shown in the figure on (previous) page 175 of this Manual:
Inner region:
ui
1 y v*
ln i i + B, v*i =
v*i κ
v
( τi /ρ )
Outer region:
uo
1 y v*
ln o o + B, v*o =
v*o κ
v
( τo /ρ )
These two must meet at equal (maximum) velocity at the point of zero shear:
1 ( rm − ri ) v*i
1 ( r − r ) v*o
+ B = v*o ln o m
+ B
At r = rm: u max = v*i ln
v
v
κ
κ
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-11-
(2)
Equations (1) and (2) above can be solved simultaneously to relate the shear stress ratio to the
position of the maximum velocity. Obviously the algebra is quite complex, however. The
volume flow rate can be computed by integrating over each region:
rm
ro
ri
rm
Q = u i 2πrdr + u o 2πrdr
(3)
where u i and u o are given by their respective log-laws. The average velocity and hydraulic
diameter are given by
Q
V=
2
π ( ro − ri2 )
Dh =
4π ( ro2 − ri2 )
2π ( ro + ri )
= 2 ( ro − ri )
from which we may form the Reynolds number Re = VDh /v. Finally, the pressure drop is
directly related to inner and outer shear stress by an overall force balance:
pπ ( ro2 − ri2 ) = 2π ( τ o ro + τi ri ) x
(4)
These relations may be combined to find the overall pressure drop as a function of flow rate
for any particular annulus size. The dimensionless pressure drop can be taken as the (overall)
Darcy friction factor, :
=
p 2D h
x ρV 2
(5)
For a given ratio ( ro /ri ) , would vary with Re. The above relations constitute a complete
“log-law theory” of developed turbulent flow in an annulus.
As a first approximation, assume the maximum velocity to be in the middle of the
annular region, rm = ( ri + ro ) /2. Then Eq. (2) above predicts that
v*i = v*o = v*;
τi = τ o
This cannot be exactly true because it violates the force balance, Eq. (1) above. Proceeding
anyway, the volume flow for this approximation becomes
1 hv*
1
Q = 2πv*h ( ro + ri ) ln
+ B− ,
κ
κ v
1
1
or:
V/v* = ln ( h + ) + B −
κ
κ
where h =
1
( ro − ri ) ,
2
Since h = Dh /4, this latter relation is equivalent to a friction-factor/Reynolds-number equation:
(8/ ) =
or:
1 1
1
ln ReDh Λ/8 + B −
κ 4
κ
(
)
1/ = 1.99log10 ReDh Λ − 1.19
(6)
(Ans.)
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-12-
We may compare this to the hydraulic-diameter approximation from pipe theory, page 426 of
the text:
(
1
1.99 log10 Re Dh
)
Λ − 1.02
(7)
The difference is slight. Some numerical values can be listed as follows:
Re =
( Eq. 6 ) =
1E4
0.0353
1E5
0.0200
1E6
0.0127
1E7
0.00875
1E8
0.00635
( Eq. 7 ) =
Difference =
0.0334
0.0192
0.0123
0.00850
0.00620
6%
4%
3%
3%
2%
Jones and Leung (1981) give a more accurate way to compute annulus friction.
6-l2 Modify the flat-plate integral analysis of Sec. 6-6.1 by combining Eq. (6-67) with
Eq. (6-71) rather than (6-69). Numerical integration may be necessary. Compare this friction
estimate with Eq. (6-70).
We are to combine the momentum relation (6-67) with the wall-wake correlation (6-71):
Ct =
2
dθ
=2 ;
2
dx
θ 3.54 22.21
− 2 ;
δ
Reδ = κ ( − 7.2 )
where the last relation is the wall-wake profile evaluated at y = δ (see p. 429 of text). These
relations constitute a first-order differential equation relating δ ( x ) to ( x ) , which we may
integrate numerically. One possible formulation would be
dF
1
= 2,
dRex
where F = exp κ ( − 7.2 ) ( 3.54 − 22.21/ )
With “F” known at the next station, we iterate to find the new value of . Later, if desired, we
can compute Reδ from the algebraic relation above. To solve this differential equation, we need
an initial value for at an initial Reynolds number. Strictly speaking, = 0 at Re x = 0 , but
this blows up in our equation. Therefore we start at the extremely low (non-turbulent) value
= 7.2 and assume that Re x = 0 there. The calculations then proceed very smoothly. The
results may be tabulated and compared with the standard power-law approximation, Eq. (6-70):
Re x =
1E5
3E5
1E6
3E6
1E7
3E7
1E8
Cf =
0.00533
0.00437
0.00358
0.00301
0.00252
0.00216
0.00185
0.00521
0.00446
0.00375
0.00321
0.00270
0.00231
0.00194
Cf ( 6-70 ) =
The differences are only a few percent, therefore we don’t bother with this more complicated
algebraic formulation.
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-13-
6-13
Devise a scheme to compute Cf ( x ) and δ ( x ) in the turbulent region of flat-plate flow
when there is an initial region of laminar flow up to a transition value Retr.
One could equate total thickness δ or momentum thickness θ at the transition point, thus
defining an “effective” origin x 0 for the ensuing turbulent boundary layer:
6/7
1/2
6/7
Equate δ : 5.0 Re1/2
tr = 0.16 Re XD ; Equate θ: 0.664 Re tr = 0.01555 Re XD
Both give about the same estimate of the “virtual” turbulent origin x 0 , but equating θ is more
realistic. Downstream of Retr, local boundary layer parameters are computed based on the
effective local Reynolds number
Re x,eff = Re x − Re tr + Re XD
For example,
δ
0.16
= 1/7
x eff Re x,eff
and
Cf =
0.027
Re1/7
x,eff
Schlichting (l979, page 641) suggests modifying our turbulent drag formulas as follows:
CD ( L x tr ) =
0.031 A
−
, where A depends upon Retr
Re1/7
ReL
L
One computes “A” by equating the above formula to the laminar drag at transition,
1/2
C D = 1.328 / Re1/2
tr . For example, if Retr = 500,000 ( or 1E6 ) , then A = 1400 ( or 2980 ) .
Let us compare these two approaches for an assumed value of Re1/2
tr = 500, 000. From
equating θ, the effective origin is Rexo = 168000. List some tabulated values of friction by the
two methods at various Reynolds numbers:
Re x =
5E5
7E5
1E6
2E6
4E6
1E7
1E8
Re x,eff =
1.68E5
3.68E5
6.68E5
1.668E6
3.668E6 9.668E6
9.967E7
Cf =
0.00484
0.00433
0.00397
0.00349
0.00312
0.00271
0.00194
0.00395
0.00375
0.00340
0.00308
0.00270
0.00194
Fully-turbulent flow:
Cf =
0.00414
The wall shear, at first higher, approaches the fully-turbulent results far downstream.
Re x =
δ/x =
5E5
7E5
1E6
2E6
4E6
1E7
1E8
0.00964
0.0135
0.0157
0.0172
0.0169
0.0155
0.0115
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-14-
Fully-turbulent flow:
δ/x =
0.0245
0.0234
0.0222
0.0201
0.0182
0.0160
0.0115
At first much thinner, δ ( x ) also approaches the fully-turbulent value downstream.
6-14 Water at 20C and 1 atm flows at 6 m/s past a smooth flat plate 1 m long and 60 cm
wide. Estimate (a) the trailing edge boundary layer thickness, (b) the trailing edge wall shear
stress, and (c) the drag of one side of the plate, if Re tr = 1 E6.
For water at 20C and 1 atm, take ρ = 998 kg/m3 and μ = 0.001 kg/m-s. Then
ReL =
ρUL ( 998)( 6.0 )(1.0 )
=
= 5,988,000
μ
( 0.001)
Thus the first one-sixth of the plate flow is laminar, and the rest turbulent. If we ignore the
laminar portion and assume fully turbulent flow, then the power-law Eqs. (6-70) give
δ ( L) =
0.16 L ( 0.16 )(1.0 )
δ
=
0.0172 m, δ* = = 0.0022 m
1/7
1/7
ReL
8
( 5988000 )
Cf ( L ) =
0.027
= 0.0029, or:
Re1/7
L
7
CD = Cf ( L ) = 0.00339, or:
6
2
1
τ w ( L ) = ( 0.0029 ) ( 998 )( 6 ) = 52 Pa
2
1
Drag = ( 0.00339 ) ( 998 ) ( 6 2 )( 0.6 m 2 ) 36.5 N
2
We don’t have a ‘log-law’ estimate for δ ( L ) , but Eq. (6-78) predicts Cf ( L ) 0.00278
or τw ( L ) = 50 Pa, and Eq. (6-81) predicts CD = 0.00320, or Drag 34.4 N.
If we account for the laminar leading-edge now, using the methods of Prob. 6-13 above,
for Re x,tr = 1E6, we equate momentum thicknesses there to obtain Re xo = 36600,
or x ( origin ) 0.161 m. Then Leff = 0.839 m, and the thickness formula predicts
ReL,eff =
ρULeff
0.16Leff 0.16 ( 0.839 )
= 5025000, or δ ( L ) =
=
0.0148 m
1/7
μ
Re1/7
( 5025000 )
L,eff
Similarly, the trailing edge shear stress is estimated by
Cf ( L ) =
0.027
2
1
0.00298, or: τ w ( L ) = ( 0.00298) ( 998)( 6 ) = 53.5 Pa
1/7
Re L,eff
2
Finally, using Schlichting’s idea for drag force with a laminar region (see p. 141 above),
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-15-
CD =
0.031 A
0.031
1440
−
−
= 0.003095,
1/7
1/7
ReL
ReL ( 5988000 )
5988000
1
Drag = ( 0.003095) ( 998) ( 62 )( 0.6 m 2 ) 33.4 N
2
or:
These latter ‘laminar-leading-edge’ results would be the preferred engineering estimates.
6-15
Repeat Prob. 6-14 if the plate average roughness is 0.1 mm. Estimate B at x = L.
From Prob. 6-14 above, ReL = 5.99E6. With roughness height k = 0.0001 m, L/k = 10,000,
and it is difficult to estimate if the flow is “fully rough”. From Eq. (6-61), the criterion for fullyrough flow would be
k+ =
kv* kU v* Rek
=
=
60, or: Rek 60λ = 1200 or so
v
v U
λ
For our case, Rek = ρUk/v = ( 998)( 6)( 0.0001) / ( 0.001) 599, thus the flow is in the
“intermediate roughness” regime and (the complicated) Eq. (6-82) should apply.
Let us first concentrate on the wall shear stress at the trailing edge. We substitute
Rex = 5.99E6 and ( x/k ) = 10,000 in Eq. (6-82) and iterate for λ. The result is
λ ( L ) 22.07 =
or:
U
6
, v* =
= 0.272 m/s,
v* ( L )
22.07
τ w ( L ) = ρv*2 = ( 998 )( 0.272 ) = 74 Pa
2
(Ans. b)
Then k + ( L ) = kv*/v = 27.1 (intermediate roughness) and Eq. (6-62) predicts that
ΔB ( L )
1
1
ln (1 + 0.3k + ) =
ln 1 + 03 ( 27.1) 5.4
κ
0.41
(Ans. d)
The law-of-the-wall with roughness, Eq. (6-60) then predicts (if we neglect any ‘wake’),
998 ( δ )( 0.272 )
U 1 δv*
1
ln
ln
+ 5.0 − 5.4
+ B − B = 22.07 =
v* κ v
0.41
0.001
or:
δ ( L ) = 0.37 m (Ans. a)
This is more than twice as thick as the smooth-wall boundary layer in Prob. 6-14.
To compute the drag - for which there is no simple formula since the plate has
intermediate roughness - we evaluate wall shear at various positions along the plate, using Eq.
(6-62). Some results may be tabulated as follows:
x ( m) =
0.01
0.05
0.1
0.2
0.4
0.6
0.8
1.0 m
τ w ( Pa ) =
201
136
117
101
88
81
77
74 Pa
These are very well fit by the power-law approximation
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-16-
τ w ( Pascals ) 75 x ( m )
−0.2
0.01 x 1.0 m
,
[If the entire plate had been “fully rough”, the wall shear would be correlated by Equation
(6-83) in the text, but that formula is about ±10% in error.] Integration of the above formula
gives an approximation for the drag on one side of the plate:
L
Drag = τ w b dx =
0
1m
−0.2
75x ( 0.6) dx =
0
75 ( 0.6 )
= 56 N
0.8
(Ans. c)
With 0.1-mm roughness, the plate has about 60% more drag compared to Prob. 6-14.
[Equation (6-84b) is not bad, either, predicting Drag = 53 N for “fully rough flow”.]
6-16 Derive Stevenson’s law-of-the-wall with suction or blowing, Eq. (6-86).
The derivation assumes (1) constant normal velocity v w near the wall; and (2) flat-plate flow,
that is, dp/dx = 0. Then Eq. (6-85) applies:
τ = τ w + ρv w u = μ t
du
du
, where μ t ρκ 2 y 2
dy
dy
and τ w = ρv*2
We use the mixing-length eddy-viscosity model, Eq. (6-88), without the van Driest damping
factor. Non-dimensionalize the above equation by dividing by τ w to obtain
+
w
+
1+ v u = κ
2
(y )
+ 2
2
du +
u + yv* + v w
+
, vw =
, where u = , y =
v*
v
v*
dy +
Separate the variables and integrate
du +
1 + v
+
w
1/ 2
u +
=
1/ 2
dy +
2
1
1 + v +w u + = ln ( y + ) + constant
, or:
+
+
κy
vw
κ
The “constant” was adjusted by Stevenson (1963) so that, in the limit as v w = 0, we recover
the no-blowing law-of-the-wall, Eq. (6-38a). Thus we split the “constant”:
1/ 2
2
1
1 + v+w u + ) − 1 ln ( y+ ) + B, κ = 0.41 and B = 5.0
+ (
κ
vw
(Ans.)
This is Stevenson’s law-of-the-wall with suction or blowing, Eq. (6-86).
6-17 Rewrite Stevenson’s law-of-the-wall with suction, Eq. (6-86), in the form of a ratio of
local friction Cf to the no-blowing value, Cfo and relate this to a “blowing parameter”,
B = ( 2vw ) / ( UeCf ) . Plot this in the range −0.5 B 2.0 and compare to the correlation
formula Cf / Cfo = ln (1 + B ) / B, from Kays and Crawford (1980, p. 181).
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-17-
First let us note that the quantity ( v +w u + ) in Eq. (6-86), when evaluated at y = δ, u = U, is none
other than the blowing parameter B:
vw U vw U2 vw
v U =
=
=
2 / Cf = B, the blowing parameter
v* v* U v*2 U
+
w
+
Further, the quality v +w itself is related to loca1 skin friction:
v+w =
vw vw U vw
1/2
=
=
( 2 / Cf )
v* U v* U
Then Eq. (6-86) - see the bottom of page 182 of this Manual - may be rewritten as
2 ( Cf /2 )
B
1/2
(1 + B )1/2 − 1 = 1 ln ( δ + ) + B = ( 2 / Cfo )1/2 if δ+ δo+
κ
where the last substitution follows since [ln(δ + ) / κ + B] exactly equals the no-blowing ratio
U/v* - we assume (without proof) that this dimensionless thickness is not affected by the
blowing. [A similar assumption is made by Kays and Crawford (1980, p. 181).] We apologize
that italic B means ‘blowing parameter’ while non-italic B means the log-law constant, 5.0.
Clean up the above equation and solve for the skin friction ratio:
2
1/2
2
Cf / Cfo = (1 + B ) − 1
B
to be compared to Kays & Crawford: Cf / Cfo =
ln (1 + B )
B
(1)
(2)
(Ans.)
It is interesting that the two formulas do not look alike at all, but they are very similar
numerically. [Kays and Crawford (1980, pp. 180–181) make a different assumption about the
constancy or eddy viscosity, rather than the constancy of boundary-layer thickness which we
assumed.]
The two formulas are plotted versus B for −0.5 B 2.0 below. The two agree with
each other and also with available suction/blowing flat-plate friction data.
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-18-
We see that suction ( B 0) increases friction (and heat transfer), while blowing ( B 0)
decreases friction (and heat transfer). This was also the case for laminar flat-plate flow (see
Fig. 4-l5b) - but turbulent flow does not “blow off” at finite B.
6-18 Water at 20C flows through a smooth permeable pipe of diameter 8 cm. The volume
flow rate is 0.06 m3/s. Estimate the wall shear stress, in pascals, if the wall velocity is
(a) 0.01 m/s blowing; (b) 0 m/s; and (c) 0.01 m/s suction. To avoid excessive iteration, assume
that the ratio of average to centerline velocity is 0.85.
Solution: For water at 20C, = 998 kg/m3 and = 0.0010 kg/m-s. First find the average
velocity and the (casually estimated) maximum velocity:
V=
Q
0.06 m3 /s
=
= 11.94 m /s
A π ( 0.04 m )2
;
Vmax
V
= 14.04 m/s
0.85
Now, for blowing and/or suction, apply Stevenson’s law, Eq. (6-86) at the centerline:
2
vw+
1/ 2
Rv *
1
+ Vmax
ln
1 + vw
− 1 =
+ 5,
v*
0.41
vw+ =
vw
v*
For y = R = 0.04 m and Vmax = 14.04 m/s, try vw = ( a ) + 0.01 m/s; ( b ) 0; and ( c ) − 0.01 m/s.
Solve for v*. EES is excellent for this type of iteration. The three results are:
(a)
vw = 0.01 m/s, v* = 0.418 m/s, w = v*2 = ( 998 )( 0.418 ) 174 Pa
Ans. (a)
(b)
vw = 0 m/s, v* = 0.483 m/s, w = v*2 = ( 998)( 0.483) 233 Pa
Ans. (b)
(c)
vw = −0.01 m/s, v* = 0.551 m/s, w = v*2 = ( 998)( 0.551) 303 Pa
Ans. (c)
2
2
2
These values are only approximate because V is not exactly equal to 85% of Vmax .
6-19 Use Stevenson’s relation (6-86) to develop a formula for wall friction in pipe flow with
blowing or suction. Apply your result to Prob. 6-7, water flowing at 30 m3 /hr in a 3-cmdiameter smooth pipe, modified so that the wall blowing rate is 3 cm/s.
Solution: Recall from Prob. 6-7 that uav = 11.79 m/s, D = 3 cm, = 998 kg/m 3 ,
and = 0.0010 kg/m-s. The Reynolds number is ReD = 353,000. One way is to proceed
exactly as in Prob. 6-18, that is, apply Stevenson’s law at the centerline, y = R:
2
vw+
1/2
Rv *
1
+ umax
ln
1 + vw
+ 5,
− 1 =
v*
0.41
vw+ =
vw
v*
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-19-
If the wall velocity vw is known, and if umax can be approximately related to uav , the on1y
unknown in this relation is the friction velocity v*. By definition, w = v *2 . The writer cannot
solve for v* because it appears in 4 places in the above correlation. However, one can iterate
and plot the variation of w with vw using, say, a spreadsheet. For our particular data from
Prob. 6-7 with vw = 0.03 m/s, we estimate umax 1.15uav = 13.6 m/s and obtain
2v * 0.03 m/s 13.6 m/s
1 +
0.03
v*
v*
1/ 2
( 998 )( 0.015 ) v *
1
− 1 =
ln
+5
0.001
0.41
By iteration or EES or whatever, we obtain the result v* = 0.333 m/s, and
w = v *2 = (998 kg/m3 ) ( 0.333 m/s ) = 111 Pa
2
Ans.
This is 54% less wall shear than the result of 243 Pa obtained for the impermeable pipe in Prob.
6-7. This is strong blowing, vw+ = +0.09, off the top of the chart in Fig. 6-22.
An alternate approach is to use the wall shear ratio developed in Eq. (6-172) of the
textbook:
w ( blowing )
w ( no blowing )
=
Cf
=
Cf0
e −1
, where =
2vw
umax C f 0
For a smooth impermeable pipe with ReD = 353,000, 0 = 0.014 from Eq. (6-54) and hence
C f 0 = 0 /4 = 0.0035. This gives = 2 ( 0.03 m/s ) / (13.6 m/s )( 0.0035 ) = 1.26. Thus
w ( blowing )
w ( no blowing )
=
e −1
=
1.26
1.26
=
= 0.50
e − 1 2.525
1.26
This is close to our estimate by the first (Stevenson) method. The second estimate is
w 0.50 w0 = 0.50 ( 243 Pa ) = 122 Pa
Ans.
6-20 As an alternative to Eq. (6-62), Bergstrom et al. (2002) suggest the following formula
for the downshift of the log-law (6-60) due to uniform surface roughness of height k:
B
1
ln ( k + ) − 3.5
κ
for
k + 4.2
First compare this correlation, with a sketch or graph, to Eq. (6-62). Then apply this correlation
to derive a formula for pipe friction factor similar to Eq. (6-64).
Solution: The two formulas are compared in the figure below. Bergstrom’s formula cannot be
used below k + = 4.2, for then it predicts negative (unrealistic) B.
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-20-
Since k + is independent of y, we simply subtract Bergstrom’s B formula from the smooth
pipe analysis of Eq. (6-53):
3
1
3
1 Rv *
1 Rv *
uav = v * ln
+ B − B −
+ B − ln k + + 3.5 −
= v * ln
v
2
v
2
where
Rv *
= Re D
,
v
32
uav
8
=
,
v*
and
k+ =
k
Re D
D
32
Use the traditional constants = 0.41 and B = 5.0, clean up, and use base-10 logarithms. As if
by magic, the Reynolds number cancels out, with the final result
1
D
2.0log10 + 1.7
k
Ans.
In other words, Bergstrom’s roughness formula, which does not level out towards zero as k +
decreases, predicts a fully rough wall friction for all Reynolds numbers. Its predictions, though,
are considerably smaller, by 10% to 25%, than the fully rough predictions of the Colebrook
formula (6-65) as Re D → .
6-21 Use numerical quadrature to evaluate and sketch Eq. (6-96), the van Driest turbulent
velocity profile, for zero pressure gradient. Compare with Eq. (6-41).
Solution: Both formulas are models of the wall-law and log-law combined:
y+
+
Van Driest: u =
0
Spalding:
2dy +
1 + 1 + 4 y + 1 − exp ( − y + / A)
2
2
2 1/2
y + = u + + e− B e Z − 1 − Z − Z 2 / 2 − Z 3 / 6 ,
( 6 − 96 )
Z = u+
( 6 − 41)
The latter is purely algebraic, if implicit in u + , while the former may be integrated by, say, the
Runge-Kutta Subroutine of Appendix C in the text or with a spreadsheet.
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-21-
The two formulas may be compared in the following chart:
There is little to choose between the two models as far as accuracy is concerned. However, the van
Driest model is more flexible, since it may be made to fit other conditions - pressure gradient,
blowing/suction, roughness, etc. - by changing the value of the clamping constant “A”.
6-22 Use the log-law, Eq. (6-38a), to analyze Couette flow between plates as shown. Since
there is no pressure gradient, the shear stress is constant in the fluid between the plates.
The profile is broken into an upper and a lower part, as shown below. The lower log-law begins
at u = 0 and rises to u = U/2 at the center, y = h/2. The upper part begins at u = U and drops
down to u = U/2 at the center.
In the both upper and lower regions, the log-law must satisfy the centerline condition
U/2 1 hv*
ln
+B
v* κ 2v
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-22-
This is a relation between shear stress, velocity, and plate separation distance. In
dimensionless friction-factor/Reynolds-number from, we may rewrite this as
1/2
Reh
4f
1 1
ln
κ 2
( Reh f ) + B,
where Reh =
τ h
Uh
and f = w
v
μU
This relation is implicit but easily computed. When Uh/v = 1E5, the result for f is
f = 51.1
(Ans.)
Other values can be listed as follows:
Uh/v =
τ w h/μU =
1E5
51.1
1E6
337
1E7
2375
1E8
17595
The velocity profile may be plotted by computing U+ = Re/ f and h + =
1E9
135270
( Re*f ).
Then plot
the log-law, Eq. (6-38a) as follows, in two parts:
Lower layer, 0 y h/2:
Upper layer, h/2 y h:
1 + y
ln h + B
κ h
1 h−y
U + − u + = ln h +
+B
κ
h
u+ =
The results are plotted as follows for various Reynolds numbers:
6-23
Use the log-law, Eq. (6-38a), to analyze flow about a cylinder of radius R rotating at
angular rate
ω
in an infinite fluid. Sketch Cf = 2τ w /ρω R versus Re = ωR 2 /v.
2
2
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The circumferential velocity profile vθ ( r ) must equal Rω at the surface, dropping off to zero
as r gets very large. However, the “log-law” does not have this asymptotic behavior and will
need a little help. Near the wall, we can write the log-law as follows:
Rω − vθ 1 ( r − R ) v *
ln
+B
v*
κ
v
(1)
However, we do not know where to stop this formula, that is, we do not know the thickness
“ δ ” where vθ 0. A crude but effective remedy is to assume that δ equals the thickness of a
flat-plate boundary layer of length L = 2πR. From Eq. (6-70), thus,
δ
0.16
2πR ( 2πR 2ω/v )1/ 7
(2)
Assume that, at this value of ( r − R ) , vθ = 0 in Eq. (1) above. That is,
1/2
Rω 2
1 δv*
= = ln
+ B,
v* Cf
κ v
δ from Eq. ( 2 ) above
This completes the analysis. For example, if Re = ωR /v = 1E5, Eq. (2) predicts that
2
0.1493 R, whence the log-law predicts Rω/v* = 21.01, or Cf = 2/ ( 21.01) = 0.00453.
We may tabulate a few friction factors as follows:
2
Re =
1E4
1E5
1E6
1E7
1E8
1E9
δ/R =
0.2074
0.1493
0.1074
0.00773
0.00556
0.00401
Cf =
0.00713
0.00453
0.00311
0.00225
0.00170
0.00133
The italic values are probably not really turbulent flow. A nice curve-fit to the remaining values
is a one-seventh power law for Re 1E5 :
Rotating cylinder turbulent flow: Cf 0.023
1/7
Re
(Ans.)
The writer knows of no other “theory” for this flow - even as crude as the one above. However,
friction on rotating cylinders has been measured by Theodorsen and Regier (1944), who
recommend the following correlation for the skin friction coefficient:
the text for an example), they recommend the following correlation for the skin friction
coefficient of a rotating cylinder in turbulent flow:
(
)
1
4.07 log10 Re Cf − 0.6
Cf
Theodorsen & Regier (1944 )
This experimental curve-fit differs by about ±10% from the power-law analysis found above.
The two may be plotted versus Reynolds number as follows:
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-24-
We see that the agreement is quite satisfactory, considering the simplifications made.
6-24 Consider the flat-plate flow of Prob. 6-14. Use any integral method from Section 6-8 to
compute τ w (x), δ*(x), and H(x) along the plate surface. Compare your results with traditional
algebraic flat-plate formulas. Begin at x = 4 cm with Cf = 0.005.
We choose Head’s entrainment method, Eqs. (6-123) and (6-124),
d
−0.6169
( UeθH1 ) = Ue F ( H1 ) , where F = 0.0306+ ( H 1 − 3.0 )
dx
H1 3.3 + 0.8234 ( H − 1.1)
−1.287
for H 1 1.6
(6-123)
(6-124)
to be combined with the Kármán integral relation, Cf = dθ/dx when U e = constant. We have
U e = 6 m/s, with v = 0.001/ 998 m 2 /s. Begin at x = 0.04 m with Cf = 0.005. Take H(0) 1.3
(flat-plate turbulent flow at low Reynolds number). Estimate θ(0) from power-law theory, Eqs.
1/ 7
(6-69,70): Rex = (998)(6)(0.04) /(0.001) = 239500, whence δ/x 0.16/Re x = 0.0273 and
θ/x (7 / 72)(d/x), or (0) 0.00011 m. With H(0) = 1.3, Eq. (6-124) above predicts
H1 (0) = 9.83. We are then ready to begin the integration of the Head-method differential
equations. The computed results Cf (x) and θ( x) can be compared with the power-law flatplate formulas:
Power-laws: Cf
0.027
7
Re11/
x
θ 0.0156
x Re1/x 7
(1, 2)
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-25-
A comparison may be given in the following graph:
The agreement is good enough for an integral method - I believe the Head-method skin friction
is too high because we started at too low a Reynolds number (239500), where the flow is barely
turbulent and the Head correlations are not well documented.
The computed shape factors in Head’s method may be tabulated as follows:
x ( m) =
0.08
0.20
0.30
0.40
0.50
0.60
0.80
1.00
H
H1
1.406
7.65
1.410
7.33
1.400
7.49
1.393
7.64
1.386
7.76
1.381
7.86
1.372
8.04
1.365
8.20
We don’t have an algebraic formula for H(x) for a flat plate: power-law theory assumes a
constant value of about H = 1.3. We could, if necessary, estimate H(x) using Cf (x) and
Coles wall-wake profile correlation, Eqs. (6-48), taking = 0.5. We won’t do that now. The
Kármán-based theory (p. 451 of the text) does this and gives similar good agreement.
6-25 Use any integral method from Section 6-8 to analyze the Howarth freestream
velocity distribution, U ( x ) = U0 (1 − x/L ) , to compute Cf ( x ) and the separation point for
ReL = U0 L/v = (a) 1 E6; (b) 1 E7 ; and (c) 1 E8. (Recall that xsep /L ( laminar ) = 0.120. ]
Let us again choose Head’s entrainment method. Eqs. (6-123) and (6-124), for our
computation. We need a starting point, which we take to be the value of x where Cf = 0.005
as computed from flat-plate theory (the initial stages of the Howarth distribution being at low,
nearly zero, β). At this initial x0 we take H0 1.4 and compute θ 0 by flat-plate theory, say,
Eqs. (6-69,70). Then we program the Kármán integral relation, Eqs. (6-28), and Head’s two
correlations (6-123,124). The results for local skin friction Cf ( x ) may be plotted as follows:
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-26-
We see that turbulent flow is much more resistant to separation, and the separation point is
about four times further downstream than for laminar flow and increases slightly with Reynolds
number. Other methods in the text give similar results. We compare here the Head separationpoint results and the Kármán-based theory on p. 451 of the text:
Re L =
( x/L)Head
( x/L)Karman
1 E6
1 E7
1 E8
0.44
0.49
0.53
0.40
0.45
0.50
They are not the same. Turbulent boundary layer prediction methods give varying results.
6-26 For potential flow past a cylinder, U ( x ) = 2U0 sin ( x/a ) , use any convenient turbulent
boundary method to compute the separation point for U0a / v = (a) I E6; (b) I E7; and
(c) I E8. Compare with the laminar separation point of = 105 (Fig. 4-24b).
We choose the Kármán-based method, p. 456 of the text, starting at a small value of x such that
Cf0 0.004 to 0.006 depending upon Reynolds number. Take H0 1.4 and θ0 given by flat-
plate theory, Eqs. (6-69) and (6-70). Then program this method and plot the local skin friction,
as follows:
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Head’s entrainment method (Sect. 6-8.1.2) was not used because, to this writer, it is difficult
to get started when there are no known initial conditions - the method is sensitive to H0 ,H10 ,
and θ 0 . It works fine when initial data is given - see Prob. 6-27 below.
The plot above indicates that, depending upon Reynolds number, the separation point
lies between 130 and 140, or about three to four times further down the back side of the
cylinder than for laminar flow. This is typical of the increased resistance to separation of a
turbulent boundary layer.
6-27 Use any convenient turbulent boundary layer method from Section 6-8 to compute skin
friction for the Clauser II experiment, Flow 2300 of Coles and Hirst (1968), using kinematic
viscosity of 0.000165 ft2/s. The measured freestream velocities are
x ( ft ) =
7.5
9.0
11.0
12.67
16.17
19.17
23.92
26.67
U ( ft/s ) =
26.1
24.8
23.5
22.8
21.3
20.2
18.9
18.1
These velocities, as expected of “near-equilibrium” turbulent boundary layers, fit a power-law
distribution with an accuracy of 1% :
U ( x ) 47.3x −0.29 ,
x in feet and U in ft/s
The sparse initial data, only Cf ( 0) and U ( x ) , suggests the method of Das (1988), page 458 of
the text, which keys on Cf ( x ) without thickness or shape factor information. However, for
those who would like to use, say, the Head entrainment (Sect. 6-8.1.2) or Kármán type
(Sect. 6-8.1.1) method, we add here the initial values of momentum thickness and shape factor
at x = 7.5 ft, as taken from Coles and Hirst (1968), p. 213: θ0 = 0.609 in, H 0 = 1.788. We have
programmed all three methods and plot the theoretical and actual skin friction distributions as
follows:
We see that Das’s method is in fairly good agreement, while the Head and Kármán methods
jump away from the initial point to new curves more or less parallel to the data. Almost all the
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-28-
thickness-based methods used at the 1968 Stanford Conference had the same trouble - see p.
545 of Kline et al. (1968).
The trouble seems to be that Clauser’s initial data are incompatible with the commonly
used correlations between friction, momentum thickness, and shape factor. For example, if
Cfo = 0.00127 and H0 = 1.788, the Ludwieg-Tillmann formula (6-117) predicts that
Reθo = 13340, or 66% higher than Clauser’s measured value of 8032 (Coles and Hirst 1968, p.
214). Das’s method does not use any input but Cfo and the wall-wake law.
6-28
Use any integral method from Section 6-8 to compute Cf ( x ) for the experiment of
Moses, Case 5 [Flow 4000 of Coles and Hirst (1968)]. Take v = 0.000166 ft 2 /s and
Cfo = 0.00471 at x = 0 ft.
The measured freestream velocities are as follows:
x ( ft ) =
0.0 0.162 0.323 0.484 0.646 0.865 1.058 1.303 1.516 1.734 1.979 2.198
U ( ft/s ) = 82.0 76.26 69.09 61.36 55.49 51.14 49.54 48.93 48.65 48.79 48.86 48.51
We thus have a “relaxing” flow, with an initial strong adverse gradient which levels off to nearconstant freestream velocity.
The Das method is available, or you may use a Kármán-type or Head entrainment
method, in which case we need initial values of momentum thickness and shape factor. From
Coles and Hirst (1968), page 385, Moses Flow 4000, θo = 0.0137 in and H o =1.62. We integrate
these methods for an assumed 5th-order polynomial curve fit to U(x):
U ( ft/s ) 82.17 − 47.409x − 2.6711x 2 + 29.209x 3 −14.105x 4 + 2.0027x5 , ( x in ft )
This fit has a correlation of 0.995 when compared to the velocity data above. The results of
these three theories are shown below and compared with Moses’ skin friction data:
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-29-
We see that agreement with experiment is not that great. Head’s method drops down to
separation, Cf 0, at x = 1 ft, which was a difficulty of many of the methods tested at the 1968
Stanford Conference. The Kármán-type and Das methods are better but not outstanding. This
experiment was hard to predict because of the ‘relaxing’ freestream.
6-29 Use the similarity concepts of Section 6-9 to derive power-law expressions for the
variation of total mass flux with x for (a) a plane jet; (b) a round jet: and (c) a plane mixing,
layer. Compare with laminar flow results.
In general, the total mass flow equals
ρu dA
over the cross-section of the flow. In all three
cases ( a,b,c ) above, u = Umax fcn ( η) , where η = σy/x. Our three results are:
+b
(a) Plane jet: m =
ρu dy = Const U
max
b
b
C1
C2 x = Cx1/2
x1/2
(Ans. a)
The related result for the laminar plane jet, Eq. 4-107, is mass flow proportional to x1/3.
+b
(b) Round jet: m = ρu 2πr dr = Const Umax b2
b
C1
C2 x 2 = Cx
x
(Ans. b)
The same result holds for the laminar round jet, Eq. 4-208, mass flow proportional to x.
+b
(c) Plane mixing layer: m =
ρu dy = C ( U
2
− U1 ) b = C1C2 x = Cx
(Ans. c)
−b
The result for the laminar mixing layer, Eq. 4-92, is mass flow proportional to x1/2 .
6-30 At a certain section of a developed turbulent plane water jet, the maximum velocity is
3 m/s and the mass flow is 800 kg/s per meter or width. At 2 m further downstream, estimate
(a) the jet width; (b) maximum velocity; and (c) mass flow.
Assuming developed flow, we use Eq. (6-151) to evaluate the mass flow:
+
m=
+
xη 2ρU x
2
− ρu dy = ρ− U maxsech ( η) d σ = σmax
For the plane jet, σ = 7.67, so we may evaluate the local position for the given mass flow:
x=
(800 kg/s/m )( 7.67 ) 1.025 m
mσ
=
2ρU max 2 ( 998 kg/m3 ) ( 3 m/s )
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-30-
Then at 2 m further downstream, or x = 3.025 m, we may use these formulas from Section
6-9.1.1 to evaluate the new flow conditions, as follows:
(a) Width b = x tan (13) = ( 3.025)( 0.23) = 0.694 m
(Ans. a)
1/2
(b) Maximum velocity Umax
(c) Mass flow =
6-31
1.025
= U x −1.025
3.025
1.75
m
s
(Ans. b)
2ρU max x 2 ( 998 )(1.75 )( 3.025 )
kg
=
= 1370
σ
7.67
s-m
(Ans. c)
Air at 20C and 1 atm issues at 0.001 kg/s from a 4-mm diameter orifice into still air.
At 1 m downstream, estimate (a) maximum velocity; (b) jet width; and (c)
μ t /μ.
For air take ρ 1.2 kg/m . Estimate the air velocity issuing from the orifice:
3
*
m
0.001
m
UO =
=
66.3
2
ρA (1.2 )( π/4 )( 0.004 )
s
Then an estimate of the jet momentum issuing from the orifice is
J=
ρu 2dA ρUo2 A0 = (1.2 )( 66.3)
orifice
2
π
kg-m
2
( 0.004 ) 0.0663 2
4
s
We then use this value of J to estimate, at x = 1 m, jet properties, using Sect. 6-9.1.2:
7.4 ( J/ρ )
=
x
1/2
At x = 1 m: U max
b1%
( 7.4 )( 0.0663/1.2 )
=
(1.0 )
1/2
= 1.74
xη1% (1.0 )( 2.993)
0.20 m
15.2
15.2
m
s
(Ans. a)
(Ans. b)
μ t KρU max b ( 0.018)(1.2 )(1.74 )( 0.20 )
=
=
= 410
μ
μ
0.000018
(Ans. c)
The mass flow at this station is approximately 4πρU max x 2 / (15.2 ) = 0.114 kg/s, or about one
2
hundred times greater than the orifice outlet mass flow.
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-31-
6-32 A long 5-m-diameter vertical cylinder is placed in the ocean where the current is 60
cm/s across the cylinder. At 1 km downstream of the cylinder, estimate (a) the wake width; and
(b) the wake velocity defect.
From Sect. 6-9.1 of the text, we have the following far-wake estimates for a cylinder:
y1/2 = 0.275 ( xθ )
u max = 1.75 ( θ/x )
1/2
1/2
where θ is the wake momentum thickness, assumed constant. The drag force on the cylinder is
related to θ as follows:
F = ρU 2θ = CD
1
ρU 2 D,
2
or: θ =
CD D
2
Now estimate the diameter Reynolds number of the cylinder in 20C seawater:
ReD = ρUD/μ = (1025)( 0.6)( 5.0) / ( 0.0011) 2.8 E6. From Fig. 3-38a, estimate CD = 0.5.
Then a crude estimate for cylinder wake momentum thickness is θ = ( 0.5)( 5) / 2 1.25 m.
Our wake thickness estimate at x = 1000 m is thus
y1/2 = 0.275 (1000 )(1.25)
1/2
This corresponds to a total wake thickness
u max = 1.75 U ( θ/x )
1/2
= 10 m
(Ans. a)
b1% 2.6y1/2 26 m. Similarly,
= 1.75 ( 0.6 ) 1.25 /1000
1/2
0.037
m
s
(Ans. b)
These are clearly very crude estimates, based on uncertain drag and wake correlations.
6-33
Evaluate the temperature law-of-the-wall, Eq. (6-164), numerically, using the van
Driest eddy viscosity (6-90) and assuming
Prt = 1.0. Compare with Eqs. (6-165,166).
We are to integrate Eq. (6-164) with Prt = 1.0:
y+
μ
dy +
T
, with t =
1/Pr + μ t /μ
μ 1+
0
+
ζ
, ζ = 2κ 2 y+ 2 1 − e − y
(1 + 2ζ )
+
/26
2
This quadrature is easily performed with Subroutine RUNGE of Appendix C. As we reach,
say, y+ = 1000, the values of T+ become logarithmic in y+ and have the form of Eq. (6-165).
We may list some computed values for various molecular Prandtl numbers:
Pr =
A ( Pr ) =
0.7
2.38
1.0
5.28
2.0
12.97
3.0
19.28
6.0
34.87
10.
52.19
30.
119.6
100.
291.6
300.
657.6
1000.
1608.
These agree reasonably well (±10% in the range 1.0 Pr 100 ) with the accepted correlations
2/3
for A ( Pr ) , Eqs. (6-166) and (6-167). They fit the formula A ( Pr ) 13Pr − 8.0 to within
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-32-
±10% except for the values at Pr = 300 and 1000. These theoretical values depend upon the
particular eddy-viscosity correlation chosen.
6-34 Air at 20C and 1 atm flows at 60 m/s past a smooth flat plate 1 m long and 60 cm
wide. If the plate surface temperature is 50C, estimate the total heat loss in watts from one
side of the plate.
For air at a mean temperature of 35C, take
ρ = 1.15 kg/m3 ,μ = 1.88 E-5 kg/m-s,
Pr = 0.71,cp = 1005 J/kg-K, and k = 0.0265 W/m-K. Then the plate Reynolds number is
ReL =
ρUL (1.15)( 60 )(1.0 )
=
= 3.67 E6
μ
1.88E-5
( turbulent flow )
Thus Sect. 6-10.3 applies for turbulent flow. The skin friction coefficient at x = L is
Cf =
0.455
= 0.00301
ln ( 0.06 )( 3.67E6 )
2
Then the trailing edge Stanton number is given by Eq. (6-168):
qw
0.001505
= 0.00168 =
2/3
1/2
ρUcp ΔT
1+13 ( 0.71) − 1 ( 0.001505)
or: q w ( x = L ) 0.00168 (1.15 )( 60 )(1005 )( 50 − 20 ) = 3486 W/m 2
Ch =
The mean heat transfer over the whole plate should be about 15% higher, or
q mean = 4010 W/m 2 . Then the heat loss from one side of the plate should be approximately
dQ
W
= q mean Aplate = 4010 2 (1.0 m )( 0.6 m ) 2400 W
dt
m
(Ans.)
This is a relatively modest rate (compared to water flow in the next problem) because air has
modest specific heat, low density, and low thermal conductivity.
6-35
Repeat Prob. 6-34 above for water flow at U = 6 m/s.
For water at a mean temperature of 35°C, use Appendix A to find ρ = 993 kg/m3 ,
μ = 7.19E-4 kg/m-s, k = 0.624 W/m-K, c p = 4177 J/kg-K, and Pr = 4.8. The plate Reynolds
number is
Re L =
ρUL ( 993)( 6.0 )(1.0 )
=
8.29 E6
μ
0.000719
( turbulent flow )
Thus Sect. 6-10.3 applies for turbulent flow. The skin friction coefficient at x = L is
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-33-
Cf =
0.455
= 0.000264
ln ( 0.06 )( 8.29E6 )
2
Then the trailing edge Stanton number is given by Eq. (6-168):
qw
0.00132
= 0.000706 =
2/3
1/2
ρUcp ΔT
1+13 ( 4.8) − 1 ( 0.00132 )
or: q w ( x = L ) = 0.000706 ( 993)( 6.0 )( 4177 )( 50-20 ) = 527, 000 W/m2
Ch ( x = L ) =
The mean heat transfer over the whole plate should be about 15% higher, or q mean = 6.06E6
W/m2. Then the heat loss from one side of the plate should be approximately
dQ
W
= q mean Aplate = 606,000 2 (1.0 m )( 0.6 m ) 364, 000 W
dt
m
(Ans.)
This is 150 times higher than the (faster) air flow in Prob. 6-34. The reason is that water has
very high specific heat and thermal conductivity, plus relatively low viscosity, making it one
of the best heat-convecting fluids known.
6-36 Modify Prob. 6-34 (airflow at U = 60 m/s ) is the plate is permeable, with a uniform
blowing velocity of 2 cm/s.
From Prob. 6-34 we know that ReL = 3.67E6 and Pr = 0.71. The appropriate blowing
correlation is Eq. (6-172):
Ch ( blowing )
vw / U
ζ
ζ , where ζ =
Ch ( no blowing ) e − 1
Ch ( no blowing )
(6-172)
We have already computed Ch ( no blowing ) = 0.00168 (see p. 200 of this Manual). Then the
blowing parameter is approximately ζ = ( 0.02 / 60.0 ) / ( 0.00168) = 0.20. Equation (6-168) then
predicts
Ch Ch ( blowing )
0.20
=
0.20
0.904, or:
Cho
0.00168
e −1
or:
Ch 0.00152 =
qw
,
ρUcp ΔT
q w ( x = L ) = ( 0.00152 )(1.15)( 60 )(1005)( 50 − 20 ) = 3160 W/m2
In this case, the blowing effect is slight (−10%) - corresponding to “F” = v w /U 0.000333 in
Figure 6-45 of the text. The mean heat transfer on the plate should be about 15% higher, or
q mean = 3620 W/m 2 . The heat loss from one side of the plate should be
dQ
cm
W
blowing = 3630 2 (1.0 m )( 0.6 m ) 2180 W
with 2
dt
s
m
(Ans.)
6-37 Water at 20C and 1 atm flows at 4 kg/s into a smooth tube of diameter 3 cm and length
2 m. If the wall temperature is 40C, estimate the average outlet water temperature.
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-34-
For water at a mean temperature of 30C, take ρ = 995 kg/m3 ,μ = 7.79E-5 kg/m-s,
k = 0.617 W/m-K, cp = 4177 J/kg-K, and Pr = 5.4. The tube Reynolds number is
ReD =
4 ( 4.0 )
4m
=
= 213,000
πμD π ( 7.97E-4 )( 0.03)
( turbulent flow )
From Eq. (6-65), the turbulent flow friction factor is
1
2.0log10 2130001/2 − 0.8, or: 0.0154
1/2
Then Eq. (6-169) is used for the (smooth) pipe Stanton number, with /8 0.00193:
h av
0.00193
=
0.000883
=
,
2/3
1/2
ρVav cp
1+13 ( 5.4 ) − 1 ( 0.00193)
W
m
m
or: h av = ( 0.000893)( 995 )( 5.69 )( 4177 ) = 20900 2
since Vav =
= 5.69
m -K
ρA
s
Ch =
The mean heat transfer must be compatible with an energy balance. Let T0 be the outlet water
temperature. Then the total heat transferred is
( T − 40 ) − ( 20 − 40 )
dQ
= mcp ( T0 − 20 ) = h av πDLΔTmean , where ΔTmean = 0
dt
T − 40
ln 0
20 − 40
(1)
That is, the concept of “logarithmic-mean temperature difference” holds for turbulent pipe flow
just as it did for laminar flow, page 123 of the text. Rather than solve Eq. (1) iteratively, we
rearrange it into the following explicit form:
h πDL
( 20900 ) π ( 0.03)( 2.0 )
T0 − 40
= exp − av
= exp −
= 0.790,
20 − 40
mc p
( 4.0 )( 4177 )
or:
Toutlet = 40 + ( 20 − 40 )( 0.790 ) = 24.2°C
(Ans.)
This corresponds to Tmean = 17.82C, or a total heat transfer dQ/dt 70100 W.
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-35-
6-38 Reconsider the liquid film flowing down an inclined plane from Prob. 3-15. This time
let the flow be turbulent and of constant depth h. Using the log-law for the velocity, (a) find an
expression for the bottom wall shear stress as related to , g, , , h, and the surface velocity
Us, (b) Find a formula for the flow rate Q per unit width and compare to laminar flow, Q h3 .
Solution: Let the width into the paper be b. Neglect air friction at the top surface. For steady
flow, the bottom shear force balances the x-directed weight:
Fig. P6-38
w Lb = gLbh sin , or:
w = gh sin
A turbulent boundary layer of thickness = h forms. From Eq. (6-69),
2 w
0.020
, or: w 0.01 5/6 1/6U 11/6 h−1/6
1/6
2
U
( Uh / )
Ans. (a)
6
Or we could solve for U [200 gh7/6 sin 1/6 −1/6 ]6/11. The flow rate Q would be Q Uhb.
7
7/11
18/11
Since U is proportional to h , Q is proportional to h
Ans. (b)
6-39 Use the method of Ambrok, Eq. (6-178), to estimate the overall heat transfer on an
isothermal circular cylinder at ReD = 1 E6 and Pr = 1.0. Compare your result with the
experimental value Nu D = 1400. Discuss the discrepancy, if any.
For Pr = 1.0, ro = ( 2-D flow ) , T = constant, and ρ = constant, Eq. (6-178) reduces to
Ch =
h (x)
0.0295v0.2
=
0.2
ρU ( x ) cp x
U ( x ) dx
0
For freestream velocity near the cylinder surface, we assume the inviscid relation,
U ( x ) = 2Uosin ( x/R ) , where U o is the approach velocity and R the cylinder radius. [We
know this formula does not work on the back side of the cylinder, but we have little better.]
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-36-
Ambrok’s formula above then becomes
Ch ( x ) =
0.0295v0.2
x
2U o sin ( x/R ) dx
0
0.2
=
0.0295 ( U 0 D/v )
1 − cos ( x/R )
0.2
0.2
Noting that C h is non-dimensionalized by U ( x ) , not U0 , we may, finally, rewrite this as
h (x)D
sin ( x/R )
0.059 Re0.8
D
0.2
k
1 − cos ( x/R )
(Ans.)
This result has at least two flaws: (1) near the leading edge, x/R 50, the flow should be
laminar; and (2) on the back side, x/R 120, the flow is separated and boundary layer theory
should not apply. Nevertheless, we may plot the local heat transfer as follows:
We have added some pseudo “data points”, based on experiments performed at somewhat
lower Reynolds numbers 2E5 [see, e.g., Giedt (1949)]. The actual flow would be laminar
near the front and separated near the back. The experimental Nusselt number Nu D is about
1400, while the integrated average of Eq. (6-178) in the graph above is about 2560, or 80%
larger. Unfortunately, the writer knows of no better theory than this for overall heat transfer in
flow past a blunt body.
6-40 Investigate modifying the finite-difference methods of Section 4-7 for a twodimensional turbulent boundary layer, using an eddy viscosity model. Comment.
The turbulent continuity equation is the same as for laminar flow, so the model is the same, Eq.
(4-148). If ( m, n ) denote the ( x, y ) directions, respectively, we have
vm+1,n vm+1,n-1 −
Δy
( u m+1,n − u m,n + u m+1,n−1 − u m,n−1 )
2Δx
(4-148)
where ( u, v ) denote time-mean velocities, of course. This formula presumes knowledge of the
u’s at the new station ( m + 1) . Meanwhile, the x-momentum equation has the same type of
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-37-
acceleration and pressure-gradient terms as for laminar flow, but the shear terms now contains
strongly variable nonlinear coefficients. That is,
u
u
u
dU
u
+v
=U
+ ( v + v t )
x
y
dx y
y
(6-21b)
The last term requires us to account for the y-variation of eddy viscosity in the model:
u 1 ( v + v t )n+1/2 ( u m,n+1 − u m,n ) ( v + v t )n −1/2 ( u m,n − u m,n −1 )
v
+
v
=
−
(
)
t
y
y y
y
y
That is, we either model the derivative of vt or we evaluate vt at ( n + 1/2) and ( n −1/2 ) ,
respectively, as shown above.
If we use an explicit model, that is, a modification of Eq. (4-146), the step size x will
be so limited that the computer-intensive turbulent-flow calculation will be too lengthy.
Therefore we adopt an implicit model modified from Eq. (4-149), as follows:
−α n +1/2 u m+1,n+1 + (1 − α n +1/2 − α n −1/2 ) u m+1,n − α n −1/2 u m+1,n −1 = u m,n + β ( u m,n+1 − u m,n −1 ) +
where α n +1/2 =
( v + v t )n +1/2 y2
u m,n x
and α n −1/2 =
( v + v t )n −1/2 y2
u m,n x
and β =
U 2m +1 − U m2
2u m,n
u m,n x
2u m,n y
Only the coefficients on the left change. We must evaluate the eddy viscosity at points midway
between n and n + 1, or “ n + 1/2 ”, and midway between n and n −1, or “ n − 1/2 ”. The model
can then be solved for the three unknown u’s with the TDMA, Section 4-7.2.1.
We may step along with this model with x limited only by numerical accuracy, that
is, changes in u and v should be no greater than about 5 per cent per step. The transverse step
size y is limited by the need for accuracy near the wall, where gradients are very large. Since
the mesh in this model is rectangular, we will need many y ’s to model the entire boundary
layer, yet keep the viscous sublayer accurate - 1000 y ’s are not unusual.
6-41 Apply your ‘turbulent finite difference’ technique of Prob. 6-40 to the computation of
Cf ( x ) for 0 x < L in fully turbulent flat-plate flow with ReL = 1 E6.
The model used is implicit and is described in Prob. 6-40 above. For flat-plate flow, we take
U = constant, or dU/dx = 0. It remains only to specify the step sizes and the eddy viscosity
model. Although the van Driest ‘damping’ model of Eq. (6-90) is probably the most widely
used in commercial boundary layer codes, it makes the local shear term proportional to the
square of local gradient u/y and is thus highly nonlinear. We can use that model, and perhaps
the students should be encouraged to do so. An alternate near-wall formula, which does not use
a mixing-length model, was proposed by G. L. Mellor at a symposium in 1968 (see p. 477 of
the 1st edition of the text):
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-38-
( κy )
μt
for the inner and overlap layers
μ ( κy+ )3 + 328.5
+ 4
Mellor (1968):
This model is easy to apply. We can then switch over in the outer or ‘wake’ layer to Clauser’s
model, Eqn. (6-93):
Clauser (1956):
μ t 0.016ρUδ*
in the outer layer
The cross-over point is where the two formulas are equal to each other, as in Fig. 6-23 of the
text. [No ‘intermittency’ correction was made to Clauser’s formula.] The formula requires an
estimate of local displacement thickness to complete the calculation.
Using the continuity and momentum models of Prob. 6-40, this Mellor/Clauser model
was applied to flat-plate flow with the following parameters (arbitrary units):
U = 1 unit; L=1 unit; v =1E-6 units; x = 0.01 unit; y = 0.00003 unit
The choice of x was strictly for accuracy (100 steps along the plate), since the implicit
(TDMA) method is unconditionally stable.
The choice of y was made for two reasons: (1) to ensure that at least three mesh points
would be in the viscous sublayer, and (2) to ensure that the thickest boundary layer (at x = L )
could be encompassed with about 600 total mesh points in the y direction.
The computations - over 600 y points and 100 x stations - took about 30 minutes
on a small personal computer, with a program written in the BASIC language. The computed
velocity profiles at three stations - x/L = 0.1,0.4, and 1.0 - are plotted in law-of-the-wall
coordinates as follows:
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-39-
We see that the profile shapes are excellent and closely resemble typical flat-plate flow profiles,
as in Fig. 6-9 of the text. The log-layer is very well represented for κ = 0.41.
The only complication in the computation was the need to know δ* in order to use
Causer’s outer-layer eddy viscosity. Since x was small, we simply evaluated δ* by
integrating the previous profile (at x − x ) and used it instead of iterating the local profile to
match eddy viscosity with actual displacement thickness. The error is very slight.
The following computed parameters can be listed at the three stations plotted above:
x/L =
Re x =
0.1
1 E5
0.4
4 E5
1.0
1 E6
δ99%/ x =
0.0255
0.0186
0.0155
Cf =
0.00505
0.00372
0.00326
=
0.55
0.75
0.75
δ/x ( Eq. 6-70 ) :
0.0309
0.0253
0.0222
Cf ( Eq. 6-70) :
0.00521
0.00428
0.00375
Cf ( Eq. 6-78) :
0.00601
0.00447
0.00376
The values of Coles’ parameter are slightly larger than the accepted value of 0.45 for
turbulent flat-plate flow. [This is an artifact of the Mellor eddy-viscosity formula.]
The computed boundary layer thickness is about 25% thinner than predicted by the
simple power-law formula of Eq. (6-70). [The power-law extrapolates along the log-layer and
overestimates the thickness.] Similarly, the computed skin friction is lower than the traditional
formulas, Eqs. (6-70) and (6-78) - here the problem may simply be an inaccuracy of the
Mellor/Clauser model for such low ‘turbulent’ Reynolds numbers.
6-42 Use the log-law, Eq. (6-38a), to analyze turbulent flow near a rotating disk of radius R
and angular velocity ω . Let the crossflow velocity w be given by a parabolic hodograph
assumption:
w u tan β (1 − y/δ ) , β 12
2
Compute local skin friction C f = 2τ w / ( ρω2 r 2 ) and moment coefficient C M = 4M/ρω2 R 5 versus
Re = ωR 2 /v and compare with Kármán’s formulas, CM = 0.146/Re1/5 and Cf = 0.053/Re1/5
(see Fig. (3-30)].
First let us estimate the skin friction by using the flat-plate formula, Eq. (6-68), which is based
upon the local log-law:
Cf =
0.020
,
Re1/5
δ
where Reδ =
( ωr ) δ = Re
v
R
r δ
RR
for this case
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-40-
Now we need to estimate local boundary layer thickness by, say, modifying Eq. (6-70):
δ
0.16
r ( ωr 2 /v )1/7
Putting these two together gives an estimate for local skin friction in terms of local r:
Cf =
0.020
Re η 0.16
R ( Re η2 )1/7
R
1/6
0.0271
( Re η )
2 1/7
,
where, η =
R
r
R
We will use this in computing the moment coefficient. At the point r = R, we have:
Cf ( r = R )
0.0271
Re1/7
R
(Ans. a)
We may compare this with two accepted correlations for this rotating disk problem:
Kármán (1921): Cf
0.053
; Theodorsen (1944 ) :
Re1/5
R
(
1
= −2.05 + 4.07 log10 Re Cf
Cf
)
Some computed results are given as follows:
Re :
Cf :
Our formula:
Kármán
Theodorsen
1 E5
1 E6
1 E7
1 E8
1 E9
0.00523
0.00530
0.00535
0.00377
0.00334
0.00333
0.00271
0.00211
0.00226
0.00195
0.00133
0.00162
0.00140
0.00084
0.00122
The present formula is in fairly good agreement - Theodorsen’s being the most accurate
correlation, since it is fitted to actual rotating-disk data.
To compute the moment on the disk, we integrate the torque caused by shear stress on
each circular strip of the disk:
R
M = τ w r ( 2πr dr )
-see Eq. ( 3-190 ) of the text
0
Now introduce τ w C f (1/2 ) ρ ( ωr )
2
from our power-law formula above and rewrite in
dimensionless form:
1
0.0271 4
0.072
η dη = 1/7
1/7 2/7
ReR η
Re
0
CM = 4π
(Ans. b)
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-41-
We may compare this result with von Kármán’s (1921) formula, CM 0.146 / Re1/5
R ;
Re :
CM :
Our formula:
Kármán
1 E5
1 E6
0.0139
0.0146
0.0100
0.00921
1 E7
1 E8
1 E9
0.00720
0.00581
0.00518
0.00367
0.00373
0.00231
The two agree for low Reynolds numbers but then our formula falls higher as Re increases.
Experimental data for rotating disks [Theodorsen and Regier (1944)] lie between these two
estimates.
Kármán’s formula came, not from wall friction estimates, but from the angular
momentum theorem applied to the disk velocity profiles. If y is normal to the disk,
δ
M=
( r V ) ρVn dA = ( ru ) ρw 2πr dy
CS
0
where the radial or ‘crossflow’ velocity w(y) was related to u(y) by the hodograph formula at
2
the beginning of this problem. [Actually Kármán used just (1 − y/δ ) rather than (1 − y/δ ) . ]
Kármán assumed a one-seventh power-law for the profile u(y):
1/7
u ( rω ) 1 − ( y/δ ) ,
plus
w u tanβ (1 − y/δ )
2
After carrying out the integration and using a one-fifth power-law estimate for boundary layer
thickness δ , the result is Kármán’s formula, CM 0.146 / Re1/5
R .
Goldstein (Proc. Cambridge Philo. Soc., v. 31, pt. 2, 1935, pp. 232–241) improved the
analysis somewhat by assuming that u(y) followed the logarithmic law, Eq. (6-38a), rather than
a one-seventh power-law.
6-43 Consider a two-dimensional flat-walled diffuser
as in Fig. P6-43. Assume incompressible flow with a
U ( x ) and
one-dimensional freestream velocity
entrance velocity U 0 at x = 0 . The entrance height is W
and the constant depth into the paper is b = 4W . Assume
turbulent flow at x = 0, with momentum thickness
/ W = 0.02, H ( 0) = 1.3, and U0W / v = 105. Using the
method of Head. Sect. 6-8.1.2, numerically estimate the
angle
for which separation will occur
at x = 1.5 W .
Solution: By one-dimensional continuity, the average velocity U ( x ) is given by
U 0bW = Ub (W + 2 x tan ) , or:
U=
U0
1 + ( 2 x tan ) / W
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-42-
We solve Head’s method for a given and the initial conditions until separation occurs. We
try a range of values of , and the results, for U 0W /v = 105 , are:
( x /W )separation
10
1.56
8
1.99
6
2.73
12
1.28
So, according to Head’s theory, separation occurs at 10.4. Ans. The plot is below:
6-44 Solve Prob. 6-43 instead by the Kármán-based method of Sect. 6-8.1.1. If time permits,
investigate three different Reynolds numbers, U 0W / v = 105 ,106 , and 107 . Assume turbulent
flow at x = 0, with momentum thickness / W = 0.02, H ( 0) = 1.3, and U 0W / v = 105. Do you
expect the separation angle to increase as Reynolds number increases?
Solution: Yes, I would make a guess that separation angle increases as Reynolds number
increases. The Kármán-oriented method is a bit algebraically heavy, since the momentum
integral (6-28) and skin friction correlation (6-120) are supplemented by a shape factor relation
(6-119a), a - correlation (6-121), and the definition of (6-122). In fact, the method has
difficulty matching the given initial conditions, because the adverse gradient dU /dx is very
strong for any significant diffuser angle , so is large and H cannot be as low as 1.3.
Anyway, the Kármán-based method predicts separation near 4, as in this table for the given
Reynolds number U 0W / v = 105.
( x / W )separation
3
3.4
4
1.6
5
0.8
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-43-
A plot of the (rather low) skin friction coefficients for = 4 is as follows:
The effect of Reynolds number slight, but, yes, an increase: For ReW = 106 , sep = 4.2. For
ReW = 106 , sep = 4.4. The Kármán-based method is poor for this diffuser.
6-45 Solve Prob. 6-43 again by Head’s method if the diffuser is an expanding cone with an
inlet pipe of diameter W. This time there is no “depth into the paper b”. For a given Reynolds
number, do you expect a cone to have a smaller separation angle than a flat-walled diffuser?
Solution: This time use axisymmetric one-dimensional continuity to estimate U ( x ) :
U0
4
W2 =U
4
(W + 2 x tan )
2
or: U
,
U0
1 + ( 2 x tan ) / W
2
Since U drops twice as fast as in the flat-walled diffuser, yes, indeed, I do expect a cone to have
a smaller separation angle for a given distance, in this case, x = 1.5L. Although the initial
boundary layer is rather thick, here assume that Head’s two-dimensional theory is satisfactory,
so that we don’t have to put together an axisymmetric thick-layer theory. Running an array of
diffuser angles gives the following separation lengths:
( x / W )separation
3
2.39
4
1.76
5
1.39
6
1.15
The desired length of L = 1.5W lies between four and five degrees. We interpolate.
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-44-
So, with Head’s theory, cone separation occurs at 4.6. Ans. The plot is below:
6-46 Sink flow, Fig. 4-16, if begun far out with a turbulent boundary layer, may relaminarize
as it approaches the origin. Show that the acceleration parameter K crit relates to sink flow and
also relates to the Reynolds number of the flow. According to K crit data, what Reynolds
number should cause sink flow to relaminarize?
Solution: For a two-dimensional sink, as shown in the
figure, the stream velocity takes the form U = −B / x ,
where B is a constant (see Fig. 4-16 of the text). From Eq.
(6-142) the dimensionless acceleration parameter is
defined as
K=
v dU
with K crit 3 10−6 for reverting to
U 2 dx
Fig. P6-46
laminar flow. Evaluate K for sink flow:
K sink =
v
( B / x)
2
d B vx 2 B v
= = constant
− =
dx x B 2 x 2 B
Thus, in sink flow, the acceleration parameter is constant. Now evaluate the local Reynolds
number and relate it to K:
Re x =
Ux B x B 1
=
= =
v
xv v K
Ans.
Interestingly, Re x is the inverse of K. The critical Reynolds number for relaminarization is
Re x ,crit =
1
1
=
330, 000
Kcrit 3E − 6
Ans.
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-45-
6-47 As an improvement to Stevenson’s inner velocity-law with suction and blowing,
Eq. (6-86), Oljaca and Sucec (1997) offer the following inner/overlap/outer velocity profile
formula based upon wall transpiration data compiled by Donald Coles in 1971:
u+ +
2
f+
vw+ 2 4
4 2 2
+
f
+
f
4
2
where =
1
ln ( y + ) + A
The function f 3 ( y / ) − 2 ( y / ) is the wake shape as in Eq. (6-46). The constant A 5
2
3
for moderate blowing or suction. Show how this profile differs from the Fig. 6-22 curves for
vw+ = 0.02 and also compare it with Fig. 6-44a for vw+ = 0.0038.
Solution: We are to compare the above formula with Stevenson’s formula (6-86);
1/2
2
1
1 + vw+u + ) − 1 = ln ( y + ) + B
+ (
vw
We take A = B = 5.0 and = 0.41 and = 0.5 (flat plate). Since the Oljaca/Sucec formula
requires a specific boundary layer thickness , we plot these only for the + 1000, as in the
Fig. 6-44a data. The results are shown in the following plot:
We see that the Oljaca/Sucec formula, in spite of its algebraic complexity, is simply the
Stevenson formula plus a wake. The data go much higher, i.e., is much larger.
6-48 Consider as a starting point the shear stress expression in the viscous sublayer, i.e.,
u
=
− u v = const
y
Make the necessary assumptions and simplifications leading to the linear relation u + = y + .
Solution:
Considering the starting point,
u
=
− u v = const
y
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-46-
The linear relationship is valid only in the viscous sublayer, where turbulent fluctuations (u′,
v′) are damped out. So we can assume
u
= Const
y
If we integrate the above relation, we get
=
dy = du
u= y
Now we use 𝜏 = 𝜌𝑢𝜏2 , where 𝑢𝜏 = velocity at the edge of the viscous sublayer.
u=
u2
u u
y = y
u
We can name the LHS as u + and RHS as y + , which are the nondimensional velocity and
length.
So, u + = y + .
(Ans)
6-49 Consider as a starting point the shear stress expression in the log layer, i.e.,
u
− u v = const
y
Make the necessary assumptions and simplifications, including the mixing-length
approximation, that result in the logarithmic relation u + = −1 ln y + + B .
=
Solution:
Continuing from the solution of problem 6-48,
+
− u + = f ( ) .
the outer layer profile is given as uinf
Here, =
and is the boundary layer thickness.
u = uinf at y =
This relation is obtained from the dimensional analysis.
For the inner layer,
u+ = g ( y+ )
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-47-
Let + =
UT
v
y + = +
Now adding the two equations above for outer and inner layer,
+
uinf
= f ( ) + g ( + )
Differentiating this equation with respect to + , we get
+ '
uinf
= g ' ( + )
where ′ means differentiation with respect to + .
Now differentiating this equation with respect to ,
0 = g ' ( + ) + + g '' ( + )
g ' ( y + ) + y ' g '' ( y + ) = 0
This can be written as
d + dg
y
=0
dy + dy +
Integrating with respect to
y + , we get
+ dg
= const
y
+
dy
This will be named 1 .
dy
1
= +
+
dy y
g=
1
ln ( y + ) + B
From the inner layer equation, we know that
g = u+ .
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-48-
1
u+ =
ln ( y + ) + B
u + = −1 ln ( y + ) + B
(Ans)
6-50 The average velocity for a turbulent pipe flow can be evaluated using the expression
uav =
Q
1
=
A a2
a
0
u (r )dA =
1
a2
a
0
u (r )2 r dr =
2
a2
a
0
u ( y )(a − y )dy
where the simple y = a − r coordinate transformation is used, with y being the dimensional
distance from the wall. Assuming that the law of the wall is accurate all the way to the wall,
i.e., neglecting the very thin viscous sublayer, substitute the log-law,
the average velocity definition to prove that
u + = −1 ln y + + B , into
uav 1 av*
3
= ln
+ B−
*
v
2
Hint: Before substituting u = ln y + B into the average velocity integral, express it first
as
u ( y ) 1 yv*
= ln
+ B
v*
+
−1
+
Solution:
uav =
=
Q
1
=
A a2
1
a2
uav =
a
0
2
a2
a
0
u (r )dA
u (r )2 dr
a
0
u ( y )(a − y )dy
(eq-1)
where y = a − r and y is the distance from the wall.
We know that u + = −1 ln ( y + ) + B .
Now in the dimensional form
u ( y)
*
=
y *
ln
+ B
1
u ( y) =
* y *
*
ln
+ B
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-49-
Substituting the value of u ( y ) in (Eq-1),
* *
*
ln y + B ( a − y ) dy
uavg =
2
a2
uavg =
2 *a a *
*
ln
y
dy
a 2 0 v
a
0
a
0
a
*
ln y + * B ( a − y )dy
0
v
C
B
A
To simplify, integrate A, B, and C separately.
a
*
2 *a a *
2 *a a
ln
y
dy
=
ln
y
−
ln
( ) 0 dy
A. 2
a 0 v
a 2 0
v
Using ln ( x ) dx = x ln ( x ) − x
2
= 2
a
a
*a
a
*
( y ln ( y ) − y )0 + y ln
y 0
2
= 2
a
*a
a *
a ln
− a
y
a
*
* a *
* a
ln y ydy = y ln ydy + y ln dy
B.
0
0 v
0
v
x 2 ln x x 2
−
Using x ln xdx =
2
4
* y2
* y 2 ln y y 2
B =
− + ln
2
4 0
v 2 0
a
a
* a 2 ln a a 2 a 2 *
B =
− + ln
2
4 2 v
B=
* a2 * a2
ln a −
2 v 4
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-50-
*
C. C = v ( B )
a
0
(a − y )dy
a
y2
= v ( B ) ay −
2 0
*
a 2 *( B)a 2
= B a2 − =
2
2
Now substituting the values of A, B, and C in eq-2, we get
uavg
=
2
= 2
a
2
a2
*a
a * * a 2 * a 2 *( B)a 2
a ln
− a − ln a − +
2
y 2 v 4
2 * a * * 2 1 * 1 * B 2
ln
a
− 1 − a ln a + a
2 v 4 2
1 v* 1 1 B
= 2v* ln a − 1 − +
2 v u 2
uavg
*
=
a *
3
ln
+ B−
v
2
1
(Ans)
6-51 The wall-friction factor in a pipe of radius a is defined as
Cf =
2 w
4 uav2
where Λ is the Darcy friction factor. (a) Show that the following identities hold:
Cf
uav
2
av*
(2a)uav
=
and
=
Re
, where ReD =
D
v*
Cf
8
(b) Substitute these identities into our previous result, namely,
uav 1 av*
3
= ln
+ B−
*
v
2
Then using κ = 0.41, B = 5.0, and log 10 (x) = ln(x)/ln10, derive Prandtl’s 1935 Darcy friction
factor formula for smooth pipes, namely,
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-51-
(
1/ = 1.99log10 Re D
)
− 1.02;
4C f
Note that Prandtl slightly adjusted the constants in this correlation to better fit pipe-friction
measurements at low Reynolds numbers, thus leading to Eq. (6-54).
Solution:
2 uav2 uav
2
=
= *
Cf
2 w
a.
Because * (shear velocity) =
so uav* =
2
Cf
v
Now, Re D
=
w
,
(Ans)
Cf
8
=
(2a)uav
1
v
2uav
(2a)uav
2
uav2 8
a *
=
because * = w
Hence,
b.
av*
= ReD
Cf
8
(Ans)
uav 1 av*
3
= ln
+ B−
*
v
2
2X 4 1
3
= ln Re D
+ B −
8 4
2
1
1
3
8=
ln Re D
+ 5 −
0.41
2 0.41
32
1
1
1
3
5
=
ln Re D
+
−
0.41 8
32
8 0.82 8
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-52-
(
)
1
= 1.99 log10 Re D − 1.02
(Ans)
6-52 One of the simplest ways to recapture the Reynolds stress is to time-average the
conservation form of the axial momentum equation in Cartesian coordinates. (a) Start by
expanding
(u 2 )
x
+
(uv)
1 dp
2u
=−
+ 2 (axial momentum in conservation form)
y
dx
y
Then using the continuity equation, show that the result is identical to the traditional form:
u
u
u
1 dp
2u
+v = −
+ 2 (steady, incompressible, axial momentum form)
x
y
dx
y
b) By substituting the Reynolds decomposed variables,
u = u + u , v = v + v , p = p + p , into
the conservation form, time-averaging, cancelling terms that average out, and assuming that
streamwise variations in
show that
u u are much slower than the cross-streamwise variations in u v ,
(
u
u
1 dp
2u
u
+v
=−
+ 2 +
−u v
x
y
dx
y y
)
Hint: The time-averaged products of mean and fluctuating quantities, such as
uu , uv , vu , and vv , all vanish identically. Similarly, the time-averaged fluctuating quantities
vanish unless they are squared or multiplied by other fluctuating quantities. Finally, the timeaveraged mean quantities are the mean quantities.
Solution:
(a)
(u 2 )
x
+
(uv)
1 dp
2u
=−
+v 2
y
dx
y
(Eq-1)
u 2 (uv)
u
v
v
+
= 2v + u + v
x
y
x
y
y
u
v
u
u
u + u + u + v
y
x
y
x
For incompressible flow, continuity equation can be written as
u v
+ = 0 (for 2D flow)
x y
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-53-
( v2 )
x
+
u v
(uv)
u
v
=u + +u
+v
y
x
y
x y
equal to 0
u 2 (uv)
u
v
+
=u +v
x
y
x
y
(Eq-2)
From (Eq-1) and (Eq-2),
u
u −1 dp
2v
u +v
=
+v 2
x
y dx
y
(Ans)
u = u + u
(b) v = v + v
p = p + p
For steady incompressible 2D flow, the continuity equation is
u v
+ = 0 . (Eq-3)
x y
u
u −1 dp
2u
+v
=
+ v 2 .(Eq-4)
Axial momentum equation is u
x
y dx
y
Substitute decomposed variables in (Eq-3) and (Eq-4).
u + u ) + ( v + v ) = 0
(
x
y
( u + u )
(Eq-5)
2
−1 d
(u + u ) + ( v + v ) ( u + u ) =
p
+
p
+
v
+
v
u + u )
(
)
(
)
2 (
x
y
dx
y
'
u
u
v
u (u ) + v
+u
v +u )+v
+v
+ v ( u + u )
(
x
y
x
x
y
y
2
1 d
=−
p + p ) + ( v + v ) (u + u )
(
dx
y
(Eq-6)
uv = ( u + u )( v + v ) equal to uv + uv + vu + uv .
All the terms linear in fluctuations variables are zero, i.e., u' = v' = 0 , but this is not
true for nonlinear terms.
uv = uv + u'v
So, the continuity equation for the decomposed variables is
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-54-
u v
+
=0
x y
From (Eq-6),
u
u
1 dp
2u
u
+v
+ (uv ) = −
+ v 2 (because all other linear terms become zero)
x
y y
dx
y
u
u
u
1 dp
2u
+v
=−
+ 2 +
−uv
x
y
dx
y y
( )
(Ans)
The equation is very similar to the original equation except the term
−uv which is
y
( )
rooted to Reynold’s stress. It introduces the coupling between the mean and
fluctuating part.
6-53 The average velocity for a turbulent channel flow can be evaluated using the expression
uav =
Q 1 h
=
u ( y )dy
A h 0
where h represents the distance from the wall to the midsection plane of the channel.
Assuming that the law of the wall is accurate all the way to the wall, i.e., neglecting the very
thin viscous sublayer, substitute the log-law, u = ln y + B , into the average velocity
+
−1
+
definition to prove that
uav 1 hv*
1
= ln
+ B−
*
v
Solution:
uav =
Q 1 h
=
u ( y )dy
A h 0
Given log-law, u = ln y + B
+
−1
+
( )
*
u + is the dimensionless velocity, i.e., velocity divided by the shear velocity
u+ =
u
*
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-55-
y + is the wall coordinate i.e., distance from the wall made dimensionless by shear velocity
( * ) and kinematic viscosity ( ) .
y *
y =
and * = w , where is the density of the flowing fluid and w is the wall
+
shear.
u
*
=
y *
ln
+ B
v
1
Given formula for average velocity = uav = 1 u ( y ) dy ,
h
h
u
*
u=
=
0
y *
ln
+ B
v
1
* y *
*
ln
+ B
v
uavg =
1 h * y *
ln
+ B * dy
h 0 v
uavg =
*
h
h
0
y *
B *
ln
dy
+
h
v
h
0
dy
Let’s assume
h
y *
I = ln
dy
0
v
h
y *
I = 1 ln
dy
0
v
Integrating by parts, we get
y *
d ln
y *
h v
I = ln
1dydy
1 dy − 0
v
dy
h
h
y *
1
*
I = ln
ydy
y − 0
y * v
v 0
v
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-56-
h
h
y *
I = ln
y − 0 dy
v 0
h
y *
h
I = ln
y − y 0
v 0
h *
I = h ln
−h
v
And
h
0
uav =
dy = h
*
h
h
0
y *
B *
ln
dy +
h
v
h
0
dy
h * B *
u*
uav = h ln
(h)
− h +
hk
v h
h * B
uav 1
* = h ln
− h + (h)
hk
v h
h * 1
1
B
h ln
h + (h)
−
hk
h
v hk
uav
=
uav
h * 1
= ln
− + B
v
*
*
1
uav 1 hv*
1
= ln
+ B−
*
v
(Ans)
6-54 Because a channel is not round, it is necessary to use the concept of a hydraulic diameter
on which the Reynolds number can be based. For a sufficiently wide channel of height 2h, (a)
show that the hydraulic diameter is Dh = 4 A / P . Next, (b) using ReDh = (4h)uav / , and
recalling that uav / v (8 / )
*
1/2
, show that
hv*
=
1 4huav
8
1
=
Re Dh
2 8 2
(c) In addition, by substituting κ = 0.41, B = 5.0, and log 10 (x) = ln (x)/ln10 into
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-57-
uav 1 hv*
1
= ln
+ B−
*
v
derive the Darcy friction factor formula for smooth channels, namely,
(
1/ = 2.0log10 Re Dh
)
− 1.19
Solution:
4 A 4 w 2h
=
P 2 ( w + 2h )
Since the weight of the channel is very large when compared to the height, the height can be
neglected w >>(2h).
(a) Dh =
Dh =
4 ( w 2h )
= 4h
2w
Hence, the hydraulic diameter of the channel is 4h.
(Ans)
1/2
8
(b) uav = v*
Write the relation for Reynold’s number
Re Dh =
( 4h ) uav
Substituting the expression for uav,
1
Re Dh
( 4h ) * 8 2
=
Now rearrange the equation
v*
=
Re Dh
4
8
h * Re Dh
=
8
2
h *
=
1
8 2
Re Dh
Thus, the equation is
hv*
=
1
8 2
Re Dh .
(Ans)
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-58-
(c) Write the expression for ratio of velocity:
uav
*
=
1
ln
h *
1
+B−
v
Substitute the values.
8
1
1
1
=
ln
Re Dh + 5 −
0.41 8 2
0.41
2 828
1
= 2.44 ln
Re Dh
8 2
+ 2.561
Use the relation, log 10 (x) = ln (x)⧸ln10.
2.828
1
= 2.44 ln(10) log
+ log Re Dh
8 2
(
(
)) ) + 2.561
)
2.828
= −5.92 + 5.62 log Re Dh + 2.561
(
1
= 2.0 log10 Re Dh
)
− 1.19
(Ans)
6-55 Determine the beginning, ending, and maximum entrance lengths in a circular duct of
diameter D = 0.03 m with air flowing between 1 m/s (beginning) and 50 m/s (ending). Repeat
the analysis in a square duct of 0.03 m height. For air, use a density of 1.2 kg/m3 and a
viscosity of 1.8 × 10−5 kg/(m ⋅ s) . Recall that the hydraulic diameter can be calculated using
Dh = 4A/P. Calculate the velocity leading to the maximum entrance lengths in both circular
and square ducts.
Solution:
Area of the circular duct =
D2
4
=
( 0.03)
2
4
Ac = 0.0007065 m 2
Perimeter of the circular duct = Pc = 2 r
0.03
Pc = 2
= 0.094 m
2
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-59-
Area of the square As = a 2 = ( 0.03) = 0.0009 m 2
2
Perimeter of the square = Ps = 4a = 4 0.03 = 0.12 m
For square, Dh =
4 As 4 0.0009
=
= 0.03m
Ps
0.12
Required length =
Area 0.0009
=
Dh
0.003
l = 0.03m
For circular duct, Dh =
4 Ac 4 0.0007065
=
Pc
0.094
Dh = 0.03m
Required length =
l=
Area
Dh
0.0007065
= 0.023m
0.03
So, the beginning length in both cases is same, i.e., l = 0.03m
(Ans)
But the ending length = 0.023 m.
(Ans)
6-56 Because of the popularity of power-law profiles in turbulent boundary layer studies,
Majdalani (2018) introduced a generic mean-flow pattern of the form:
u
F ( ) = 1/ q
Ue
where ξ ≡ y/δ represents the fractional distance within the turbulent boundary layer, and the
power-law exponent q = {5, 6, 7, 8, 9} can be used to prescribe the value that best captures
the motion under consideration.
(a) Based on Eq. (6-28), show that the normalized displacement and momentum thicknesses,
η* and θ*, as well as the shape factor H, can be expressed as direct functions of q, namely,
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-60-
* *
1
1
= (1 − F )d =
0
q +1
*
1
q
= 0 F (1 − F )d =
(q + 1)(q + 2)
* * q + 2
=
=
H
*
q
(normalized displacement thickness)
(normalized momentum thickness)
(momentum shape factor)
(b) Recognizing that most wall shear stress correlations, such as Eqs. (6-68) and (6-72), can
be specified as functions of Reδ, Majdalani wrote a generic expression of the form
1/ m
w = Cw u
uav
2
av
2Cw
Re1/ m
Cf =
or
where Cw and m are characteristics of the problem at hand. Assuming no pressure gradients,
apply the momentum-integral relation, given by Eq. (6-67), to obtain
1/ m
d Cw
=
dx * uav
(c) After separating variables and integrating from δ(0) = 0 to δ(x) , show that the boundary
layer thickness may be written as a function of the local Reynolds number, Rex = uav x/ν ,
namely,
x
m
C m + 1 m +1
; a w*
1
m
m +1
a
=
Rex
(d ) Recalling that the friction coefficient may be defined as C f = 2 w / ( uav2 ) , show that
1
m *Cwm m+1
b
Cf =
;
b
2
1
m +1
m +1
Re
x
1
(e) Recalling that the displacement thickness may be specified * = (1 − F )d , show that
0
*
x
m
=
1 Cw m + 1 m +1
; c
1
q +1 * m
m +1
c
Re x
1
(f) Recalling that the momentum thickness may be deduced from = F (1 − F )d , show
0
that
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McGraw Hill LLC.
-61-
x
m
=
q
Cw m + 1 m +1
; d
1
(q + 1)(q + 2) * m
m +1
d
Re x
(g) Using the continuity equation v = −
y
udy , integrate the axial velocity profile to show
x 0
that
1/ q
v
y y
m
=e ; e
Ue
x
(m + 1)(q + 1)
(h) Recognizing that Prandtl’s 1927 power-law model uses Cw = 0.0233, q = 7, and m = 4,
determine the characteristic coefficients (a, b, c, d, e) and compare your results to Eq. (6-72).
(i) Recognizing that the wall–wake law of Coles uses Cw = 0.010, q = 7, and m = 6, determine
the characteristic coefficients (a, b, c, d, e) and compare your results to Eq. (6-70).
Solution:
u
F ( ) = 1/ q , where ξ ≡ y/δ
Ue
(a)
d
dU e C f
+ (2 + H )
=
dx
U e dx
2
(Given Eq 6-28)
From momentum-integral analysis, we know that
Cf = 2
d
dx
Also, we know that
u+
1
2 y
ln ( x + ) + B +
f
k
k
Since equilibrium factor = 0 , = 0.45 .
Evaluation of the profile function (Eq. 6-47) at y = then gives
2
* =
c
f
Ue
1/2
=
1 *
2
ln
+ B+
k
k
1/2
Cf
2.44ln Re
+ 7.2
2
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-62-
By analogy with the following equation,
1 av*
3
uav = v * ln
+B−
2k
k
This relation represents a friction relation between Cf and Re .
It is laborious algebraically. So, C f = 0.020Re−1/6 or in terms of wall shear stress,
1/
= 0.010 u
uav
2
av
Substituting all the above equations, we get
0.020 Re−1/6 2
d 7 7 d Re
=
dx 72 36 d Re x
Integrating with knowledge that = 0 at x = 0 , we get
Re 0.16 Re6/7
x
x
=
0.16
0.027
and C f =
1/7
Re x
Re1/7
x
And with k = 0.41 and = 0.45
1/2
2
3 54 22.21
=
− 2 , where =
C
f
2
We know Prandtl in the year 1927 used = 0.0233 uaV
/ uav
1/4
C f 0.0466Re6−1/4 . Therefore,
x
=
0.058
0.37
and C f
.
1/5
Re1/5
Re x
x
1/q
Therefore,
u y
=
Ue
or
and s y
u
= ( )1/ q .
Ue
In the given Eq 6-28, we know that
0
* = 1 −
u
*
dy =
Ue
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McGraw Hill LLC.
-63-
1
u y
* = 1 − d
0
Ue
Therefore, displacement thickness becomes normalized displacement thickness as
* =
1
*
u
1
= 1 − d =
0
q +1
Ue
Also, =
0
(Ans)
u
u
*
1 − dy =
Ue Ue
u
u y
1− d
0U
Ue f
e
* =
1
Therefore, momentum thickness becomes normalized momentum thickness as
*
1 u
u
q
=
1 − d =
0U
Ue
( q + 1)( q + 2 )
e
(Ans)
Also, momentum shape factor is
* * q + 2
H=
=
=
*
q
(Ans)
1/ m
(b) w = Cw u
uav
or C f =
2
av
2CW
Re1/m
Assuming that there are no pressure gradients and applying momentum-integral, we
get
C
d
d
=*
= 1/w m
dx
dx Re
*
C
d Re
C
= 1/w m or Re1/ m d Re = * d Re x
d Re x Re
Using the initial condition (0) = 0 , we can integrate and get
Re
0
Re1/ m d Re =
Cw
*
Re x
0
d Re x
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-64-
Or Re =
uav
m
m
C m + 1 m+1
m +1
= w*
Re
x
m
Solving for ( x ) , rearranging, and simplifying, we get
m
C m + 1 m +1
= *
x m
1
Re
1
m +1
x
Differentiating, substituting, and simplifying, we get
1/ m
d Cw
=
dx * uav
(Ans)
m
C m + 1 m +1
= w*
x m
1
1
Rexm +1
1
2C
C f = 1/wm =
Re
*
m *Cwm m +1
1
m +1
Re m +1
2
x +1
m
*
1 Cw m + 1 m +1
=
=
And
x
x q +1 * m
1
1
Rexm +1
m
*
q
Cw m + 1 m +1
=
=
x
x (q + 1)(q + 2) * m
1
Re
1
m +1
x
These relations can be conveniently expressed as
x
a
=
(Ans. c)
1
m +1
x
Re
m
C m + 1 m+1
where a w*
m
Cf =
b
1
m +1
x
(Ans. d)
Re
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McGraw Hill LLC.
-65-
m C
where b 2
m +1
*
*
x
m
w
1
m +1
c
=
Re
(Ans. e)
1
m +1
x
m
1 Cw m + 1 m+1
where c
q +1 * m
And
x
=
d
Re
(Ans. f)
1
m +1
x
m
q
Cw m + 1 m+1
where d
(q + 1)(q + 2) * m
Note that the transverse velocity can be obtained from the continuity equation,
namely,
v
y u
y y
=−
dy = − dy , and so
Ue
x 0 U e
x 0
1/ q
1/ q
v
m
y y
=
U e (m + 1)(q + 1) x
y y 1/7
; Prandtls
10 x
=
1/7
3 y
28 x ; wall-wake law
(Ans. g)
The validity of the expressions for (a, b, c, d, e) can be conformed by taking Prandtl’s
Cw = 0.0233, q = 7, and m = 4 to reproduce, within the round-off error affecting Eq.
(6-72), the following correlations:
x
=
0.381
1
5
x
Re
Cf =
0.0371
0.0593 * 0.0477
and =
=
1
1
1
x
x
5
5
Rex5
Re x
Rex
(Ans. h)
The same may be repeated for the wall-wake law. Using Cw = 0.010, q = 7, and m = 6,
we may bypass intermediate calculations and recover
x
=
0.162
1
7
x
Re
Cf =
0.0271 * 0.0203
0.0158
=
and
=
1
1
1
x
x
7
7
Re x
Rex7
Re x
(Ans. i)
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