Solution Frank-White-Fluid Mechanics-7th ed Chapter4

Prévia do material em texto

Chapter 4  Differential Relations 
for a Fluid Particle 
P4.1 An idealized velocity field is given by the formula 
24 2 4tx t y xz  V i j k 
Is this flow field steady or unsteady? Is it two- or three-dimensional? At the point (x, y, z)  
(–1, 1, 0), compute (a) the acceleration vector and (b) any unit vector normal to the 
acceleration. 
Solution: (a) The flow is unsteady because time t appears explicitly in the components. 
(b) The flow is three-dimensional because all three velocity components are nonzero. 
(c) Evaluate, by laborious differentiation, the acceleration vector at (x, y, z)  (1, 1, 0). 
2 2
2 2 4
2 2
du u u u uu v w 4x 4tx(4t) 2t y(0) 4xz(0) 4x 16t x
dt t x y z
dv v v v vu v w 4ty 4tx(0) 2t y( 2t ) 4xz(0) 4ty 4t y
dt t x y z
dw w w w wu v w 0 4tx(4z) 2t y(0) 4xz(4x) 16txz 16x z
dt t x y z
   
   
   
   
   
   
         
            
         
 
2 4 2dor: (4x 16t x) ( 4ty 4t y) (16txz 16x z)
dt
      V i j k 
at (x, y, z)  (1, 1, 0), we obtain 2 3d 4(1 4t ) 4t(1 t ) 0 (c)
dt
Ans.     V i j k 
(d) At (–1, 1, 0) there are many unit vectors normal to dV/dt. One obvious one is k. Ans. 
 
P4.2 Flow through the converging nozzle 
in Fig. P4.2 can be approximated by the 
one-dimensional velocity distribution 
o
21 0 0xu V w
L
       
(a) Find a general expression for the fluid 
acceleration in the nozzle. (b) For the 
specific case Vo  10 ft/s and L  6 in, 
compute the acceleration, in g’s, at the 
entrance and at the exit. 
 
Fig. P4.2 
 
 Chapter 4  Differential Relations for a Fluid Particle 303
Solution: Here we have only the single ‘one-dimensional’ convective acceleration: 
221 . (a)oo
Vdu u xu V Ans
dt x L L


               
2
o2V x1
L L
 
22(10) 2 6 10 , 1 400(1 4 ),
6 /12 6 /12o
ft du xFor L and V x with x in feet
s dt
          
At x  0, du/dt  400 ft/s2 (12 g’s); at x  L  0.5 ft, du/dt  1200 ft/s2 (37 g’s). Ans. (b) 
 
P4.3 A two-dimensional velocity field is given by 
V  (x2 – y2  x)i – (2xy  y)j 
in arbitrary units. At (x, y)  (1, 2), compute (a) the accelerations ax and ay, (b) the 
velocity component in the direction   40, (c) the direction of maximum velocity, and 
(d) the direction of maximum acceleration. 
Solution: (a) Do each component of acceleration: 
2 2
x
2 2
y
du u uu v (x y x)(2x 1) ( 2xy y)( 2y) a
dt x y
dv v vu v (x y x)( 2y) ( 2xy y)( 2x 1) a
dt x y
 
 
 
 
          
           
 
At (x, y)  (1, 2), we obtain ax  18i and ay  26j Ans. (a) 
(b) At (x, y)  (1, 2), V  –2i – 6j. A unit vector along a 40 line would be n  cos40i  
sin40j. Then the velocity component along a 40 line is 
40V ( 2 6 ) (cos 40 sin 40 ) . (b)Ans         40V n i j i j 5.39 units 
(c) The maximum velocity is [(-2)2 + (-6)2]1/2 = 5.32 units, at an angle in the third quadrant, 
θ = 180° + arctan(-6/-2) = 180° + 71.6° = .251.6°. Ans. (c)
(d) The maximum acceleration is amax . [182  262]1/2 . 31.6 units at .55.3. Ans. (c, d) 
 
 
_______________________________________________________________________ 
�P4.4 A simple flow model for a two-dimensional converging nozzle is the distribution 
 
(a) Sketch a few streamlines in the region 0<x/L<1 and 0<y/L<1, using the method of 
Section 1.11. (b) Find expressions for the horizontal and vertical accelerations. 
0)1(  w
L
yUv
L
xUu oo
 Solutions Manual  Fluid Mechanics, Fifth Edition 
 
304 
(c) Where is the largest resultant acceleration and its numerical value? 
 
Solution: The streamlines are in the x-y plane and are found from the velocities: 
 
These may be plotted for various values of the dimensionless constant C, as shown: 
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
0 0.2 0.4 0.6 0.8 1
x/L
y/L
C = 1
C = 0.75
C = .5
C = 0.25
 
The streamlines converge and the velocity increases to the right. Ans.(a) 
(b) The accelerations are calculated from Eq. (4.2): 
).(:sstreamlinetheFinally 
)]/1)(/ln[(:,)/ln()/1ln(
Cancel
/)/1(
:integrateor
aAns
constantLxLyorconstLyLLxL
U
LyU
dy
LxU
dx
v
dy
u
dx
o
oo
x/L1
C
L
y


 
 Chapter 4  Differential Relations for a Fluid Particle 305
(c) Find the resultant of ax and ay from Ans.(b) above and introduce y/L from Ans.(a): 
2 2 2 2 21 2 /(1 ) , where / .( )x ya a a C x L Ans c           
We observe that the resultant acceleration increases with x and is greatest at x = L, where its 
numerical value is (Uo2/L) [4 + C2/4]1/2. 
 
 
P4.5 The velocity field near a stagnation point (see Example 1.10) may be written in 
the form 
 o o ou v and are constantsU x U y U LL L 
(a) Show that the acceleration vector is purely radial. (b) For the particular case L  1.5 m, 
if the acceleration at (x, y)  (1 m, 1 m) is 25 m/s2, what is the value of Uo? 
Solution: (a) For two-dimensional steady flow, the acceleration components are 
2
o o
o o 2
2
o o
o o 2
U Udu u u x yu v U U (0) x
dt x y L L L L
U Udv v v x yu v U (0) U y
dt x y L L L L
 
 
 
 
                 
                    
 
Therefore the resultant 2 2 2 2o o(U /L )(x y ) (U /L (purely radial) (a)Ans.  a i j )r 
(b) For the given resultant acceleration of 25 m/s2 at (x, y)  (1 m, 1 m), we obtain 
2 2
o o
o2 2 2
U Um ra 25 2 m, solve for U . (b)
s L (1.5 m)
Ans       m6.3
s
 
 
).()/)(/(0
)1(0)/)](/1([
2
2
bAns
L
y
L
ULULyU
y
vv
x
vua
L
x
L
ULULxU
y
uv
x
uua
o
ooy
o
oox






 Solutions Manual  Fluid Mechanics, Fifth Edition 
 
306 
 
 
 
P4.6 An incompressible plane flow has the velocity components u = 2y, v = 8x, w = 0. 
(a) Find the acceleration components. (b) Determine if the vector acceleration is radial. 
(c) Which two streamlines of this flow are straight lines? 
 
Solution: (a, b) With no z activity, we can stick to steady two-dimensional formulas: 
 
(2 )(0) (8 )(2) 16
(2 )(8) (8 )(0) 16 .( )
16 16 16( ) 16 , yes. .( )
x
y
x y
u ua u v y x x
x y
v va u v y x y Ans a
x y
a a x y x y Ans b
      
      
      a i j i j i j r Radial
 
(c) Find the streamlines in the manner suggested in Chapter 1, Eq. (1.41): 
 
2 2
, or : , 2 8
2 8
Integrate : 2 8 , or : 4
dx dy dx dy y dy x dx
u v y x
ydy x dx y x const
  
   
 
The streamlines are all curved except when the constant of integration is zero, for which 
there are two straight streamlines: y =  2 x . Ans.(c) 
 
 Chapter 4  Differential Relations for a Fluid Particle 307
 
P4.7 Consider a sphere of radius R immersed in a uniform stream Uo, as shown in 
Fig. P4.7. According to the theory of Chap. 8, the fluid velocity along streamline AB is given by 
3
o 3u 1
RU
x
     V i i 
 
Fig. P4.7 
Find (a) the position of maximum fluid acceleration along AB and (b) the time required 
for a fluid particle to travel from A to B. Note that x is negative along line AB. 
Solution: (a) Along this streamline, the fluid acceleration is one-dimensional: 
3 3 3 4 3 4 3 7
o o o
du uu U (1 R /x )( 3U R /x ) 3U R (x R x ) for x R
dt x


          
The maximum occurs where d(ax)/dx  0, or at x  –(7R3/4)1/3  –1.205R Ans. (a) 
(b) The time required to move along this path from A to B is computedfrom 
R t
3 3
o o3 3
4R 0
dx dxu U (1 R /x ), or: U dt,
dt 1 R /x


     
R
4R
2
1
o 2 2
R (x R) R 2x Ror: U t x ln tan
6 x Rx R 3 R 3


            
 
It takes an infinite time to actually reach the stagnation point, where the velocity is 
zero. Ans. (b) 
 
P4.8 When a valve is opened, fluid flows in the expansion duct of Fig. P4.8 according 
to the approximation 
1 tanh
2
x UtU
L L
    V i 
308 Solutions Manual  Fluid Mechanics, Fifth Edition 
 
Find (a) the fluid acceleration at (x, t)  (L, L/U) and (b) the time for which the fluid 
acceleration at x  L is zero. Why does the fluid acceleration become negative after 
condition (b)? 
 
Fig. P4.8 
 
Solution: This is a one-dimensional unsteady flow. The acceleration is 
2
x
u u x U Ut x U Uta u U 1 sech U 1 tanh
t x 2L L L 2L 2L L
 
 
                                 
 
2
2 2U x Ut 1 Ut(1 )[sech tanh ]
L 2L L 2 L
            
 
At (x, t)  (L, L/U), ax  (U2/L)(1/2)[sech2(1) – 0.5tanh2 (1)]  0.0650 U2/L Ans. (a) 
 
The acceleration becomes zero when 
 
 
 
The acceleration starts off positive, then goes through zero and turns negative as the 
negative convective acceleration overtakes the decaying positive local acceleration. 
 
 
P4.9 An idealized incompressible flow has the proposed three-dimensional velocity 
distribution 
V  4xy2i  f(y)j – zy2k 
 Chapter 4  Differential Relations for a Fluid Particle 309
Find the appropriate form of the function f(y) which satisfies the continuity relation. 
Solution: Simply substitute the given velocity components into the incompressible 
continuity equation: 
2 2 2 2(4 ) ( ) 4 0u v w f dfxy zy y y
x y z x y z dy
     
               
2 2: 3 . Integrate: ( ) ( 3 )dfor y f y y dy Ans.
dy
      3y constant 
 
P4.10 Consider the simple incompressible plane flow pattern u = U, v = V, and w = 0, 
where U and V are constants. (a) Convert these velocities into polar coordinate components, vr 
and v. [HINT: Make a sketch of the velocity components.] (b) Determine whether these new 
components satisfy the continuity equation in polar coordinates. 
Solution: This is harder than it looks. Make a sketch of each separate cartesian component: 
 
 
 
 
(a) We can resolve each figure into radial and circumferential components. For Figure (a), 
U has a radial component U cos and a circumferential component (-U sin). For Figure 
(b), V has a radial component V sin and a circumferential component V cos. Combine 
these into the result 
 cos sin ; sin cos .( )rv U V v U V Ans a        
(b) The original (cartesian) distribution, being constant velocity, obviously satisfied 
continuity. The new version, in polar coordinates, requires some effort. From Eq. (4.9) for 
incompressible flow, 
x
y 

x 
y

U
V 
(a) (b)
310 Solutions Manual  Fluid Mechanics, Fifth Edition 
 
 
1 1 1 1( ) ( ) [ ( cos sin )] ( sin cos )
1 1( cos sin ) ( cos sin ) Yes, satisfied. .( )
rr v v r U V U Vr r r r r r
U V U V Ans b
r r
     
   
           
      0
 
 
P4.11 Derive Eq. (4.12b) for cylindrical coordinates by considering the flux of an 
incompressible fluid in and out of the elemental control volume in Fig. 4.2. 
Solution: For the differential CV shown, 
out ind ol dm dm 0t
       
 
Fig. 4.2 
 
r r
z z
r z
drr d dr dz v r dz d ( v )dr(r dr)dz d v dz dr
t 2 r
dr dr( v )d dz dr v r d dr ( v ) r d dr
2 z 2
drv r dz d v dz dr v r d dr 0
2



       
       
    
       
             
       
 
Cancel (d drdz) and higher-order (4th-order) differentials such as (dr d  dz dr) and, 
finally, divide by r to obtain the final result: 
.Ans
        r z
1 1rv v v 0
t r r r z
          
 
 Chapter 4  Differential Relations for a Fluid Particle 311
 
 
 
 
 
 
P4.12 Spherical polar coordinates (r, , ) are defined in Fig. P4.12. The cartesian 
transformations are 
x  r sin cos 
y  r sin sin 
 z  r cos 
Do not show that the cartesian incompressible continuity relation (4.12a) can be 
transformed to the spherical polar form 
 
Fig. P4.12 
 
2
2
1 1 1( ) ( sin ) ( ) 0
sin sinr
r
r r rr  
            
What is the most general form of r when the flow is purely radial, that is,  and  
are zero? 
Solution: Note to instructors: Do not assign the derivation of this continuity 
relation, it takes years to achieve, the writer can’t do it successfully. The problem is 
only meant to acquaint students with spherical coordinates. 
2
r2
1If 0, then (r ) 0, so, in general, 
rr
Ans. 
       r 2
1 fcn ,
r
   
 
312 Solutions Manual  Fluid Mechanics, Fifth Edition 
 
 
 
P4.13 For an incompressible plane flow in polar coordinates, we are given 
 3 2cos sinrv r r   
Find the appropriate form of circumferential velocity for which continuity is satisfied. 
 
Solution: Substitute into continuity, Eq. (4.9), for incompressible flow: 
 
 
3 2
2
3 2
1 1 1 1( ) ( ) [ ( cos sin )] ,
1or : 4 cos 3 sin
Integrate : 4 sin 3 cos ( ) .
r
v
r v v r r r
r r r r r r
v
r r
r
v r r f r Ans



  
 
 
        
   
   
 
We can’t determine the form of the “constant of integration” f(r) without further 
information. 
 
P4.14 For incompressible polar-coordinate flow, what is the most general form of a 
purely circulatory motion,   (r, , t) and r  0, which satisfies continuity? 
Solution: If vr  0, the plane polar coordinate continuity equation reduces to: 
 (1/r) ∂vθ/∂θ = 0 , or: vθ = fcn(r, t) Ans. 
P4.15 What is the most general form of a purely radial polar-coordinate 
incompressible-flow pattern, r  r(r, , t) and   0, which satisfies continuity? 
Solution: If v  0, the plane polar coordinate continuity equation reduces to: 
r
1 (r ) 0, or: only
r r
v Ans.  r
1 fcn
r
  v 
 Chapter 4  Differential Relations for a Fluid Particle 313
________________________________________________________________________ 
 
P4.16 Consider the plane polar coordinate velocity distribution 
where C and K are constants. (a) Determine if the incompressible equation of continuity 
is satisfied. (b) By sketching some velocity vector directions, plot a single streamline for 
C = K. What might this flow field simulate? 
 
Solution: (a) Evaluate the incompressible continuity equation (4.12b) in polar coordinates: 
 
Incompressible continuity is indeed satisfied. (b) For C = K, we can plot a representative 
streamline by putting in some velocity vectors and sketching a line parallel to them: 
 
-0.2
0.0
0.2
0.4
0.6
0.8
1.0
1.2
-0.2 -0.1 0.0 0.1 0.2 0.3 0.4
O
 
0 zr vr
Kv
r
Cv 
).(000)(1)(1)(1)(1 aAns
r
K
rr
Cr
rr
v
r
vr
rr r





 
314 Solutions Manual  Fluid Mechanics, Fifth Edition 
 
The streamlines are logarithmic spirals moving out from the origin. [They have axisymmetry 
about O.] This simple distribution is often used to simulate a swirling flow such as a tornado. 
 
P4.17 An excellent approximation for the two-dimensional incompressiblelaminar boundary layer on the flat surface in Fig. P4.17 is 
 
3 4
1/ 2
3 42 2 for , where , constant( )
y y yu U y C x C         
 
(a) Assuming a no-slip condition at the wall, find an expression for the velocity component 
v(x, y) for y  . (b) Then find the maximum value of v at the station x = 1 m, for the 
particular case of airflow, when U = 3 m/s and  = 1.1 cm. 
Solution: (a) With u known, use the two-dimensional equation of continuity to find v: 
 
3 4
2 4 5
3 4 2 4 5
2 4 5 2 4 5
0
2 6 4( ) ,
3 2 3 2: 2 ( ) 2 ( ) .( )
2 4 5
y
v u y d y d y dU
y x dx dx dx
d y y y d y y yor v U dy U Ans a
dx dx
  
  
 
     
        
     
 
 (b) First evaluate C from the given data at x = 1 m: 
 
1/ 2 1/ 2
1/ 2 1/ 2
1/ 2
0.011 (1 ) , hence 0.011
1 1Or, alternately , ( )
2 2 2
m C m C m
d C x x
dx xx

   
  
   
Substitute this into Ans.(a) above and note that v rises monotonically with y to a maximum 
at the outer edge of the boundary layer, y =  . The maximum velocity v is thus 
Fig. P4.17 
 Chapter 4  Differential Relations for a Fluid Particle 315
 max
1 3 2 0.011 32 ( ) 2(3 )[ ]( ) .( )
2 4 5 2(1 ) 20
d m mv U Ans b
dx s m
     m0.0050
s
 
This is slightly smaller than the exact value of vmax from laminar boundary theory (Chap. 7). 
 
 
P4.18 A piston compresses gas in a cylinder by moving at constant speed V, as in 
Fig. P4.18. Let the gas density and length at t  0 be o and Lo, respectively. Let the gas 
velocity vary linearly from u  V at the piston face to u  0 at x  L. If the gas density varies 
only with time, find an expression for (t). 
V = constant
x = 0 x = L(t)
p(t)
x
 
Fig. P4.18 
Solution: The one-dimensional unsteady continuity equation reduces to 
o
d u x( u) , where u V 1 , L L Vt, (t) only
t x dt x L
        
           
o
t
oo
u V d dtEnter and separate variables: V
x L L Vt


 
      
           o o oThe solution is ln( / ) ln(1 Vt/L ), or: .Ans
o
o
L
L Vt
 
 
P4.19 An incompressible flow field has the cylindrical velocity components   Cr, z  
K(R2 – r2), r  0, where C and K are constants and r  R, z  L. Does this flow satisfy 
continuity? What might it represent physically? 
Solution: We check the incompressible continuity relation in cylindrical coordinates: 
z
r
v1 1 v(rv ) 0 0 0 0 .
r r r z
Ans         satisfied identically 
This flow also satisfies (cylindrical) momentum and could represent laminar flow inside a 
tube of radius R whose outer wall (r R) is rotating at uniform angular velocity. 
 
316 Solutions Manual  Fluid Mechanics, Fifth Edition 
 
P4.20 A two-dimensional incompressible velocity field has u  K(1 – e–ay), for x  L 
and 0  y  . What is the most general form of v(x, y) for which continuity is satisfied and 
v  vo at y  0? What are the proper dimensions for constants K and a? 
Solution: We can find the appropriate velocity v from two-dimensional continuity: 
ayu [K(1 e )] 0, or: fcn(x) only
y x x
v v    
       
oSince at y 0 for all x, then it must be that .v v Ans  o const v v 
The dimensions of K are {K}  {L/T} and the dimensions of a are {L–1}. Ans. 
 
P4.21 Air flows under steady, 
approximately one-dimensional conditions 
through the conical nozzle in Fig. P4.21. If 
the speed of sound is approximately 340 m/s, 
what is the minimum nozzle-diameter ratio 
De/Do for which we can safely neglect 
compressibility effects if Vo  (a) 10 m/s and 
(b) 30 m/s? 
Solution: If we apply one-dimensional 
continuity to this duct, 
 
Fig. P4.21 
2 2 2
o o o e e e o e e o o eV D V D , or V V (D /D ) if4 4
       
To avoid compressibility corrections, we require (Eq. 4.18) that Ma  0.3 or, in this case, 
the highest velocity (at the exit) should be Ve  0.3(340)  102 m/s. Then we compute 
1/2 1/2
e o min o e o o(D /D ) (V /V ) (V /102) if V 10 m/s . (a)Ans   0.31 
oif V 30 m/s . (b)Ans 0.54 
 
 
P4.22 In an axisymmetric flow, nothing varies with ; the only nonzero velocities 
are vr and vz (see Fig. 4.2 of the text). If the flow is steady and incompressible and vz = Bz, 
where B is constant, find the most general form of vr which satisfies continuity. 
 
 Chapter 4  Differential Relations for a Fluid Particle 317
Solution: With no  variation and no v, the equation of continuity (4.9) becomes 
 2
1 1( ) 0 ( ) ( ) ,
or : ( ) ; Integrate : ( )
2
( )Finally, .
2
z
r r
r r
r
v
r v r v Bz
r r z r r z
Br v B r r v r f z
r
B f zv r Ans
r
        
     
  
 
The “function of integration”, f(z), is arbitrary, at least until boundary conditions are set. 
__________________________________________________________________________ 
 
 
 
P4.23 A tank volume V contains gas at conditions (o, po, To). At time t  0 it is 
punctured by a small hole of area A. According to the theory of Chap. 9, the mass flow 
out of such a hole is approximately proportional to A and to the tank pressure. If the tank 
temperature is assumed constant and the gas is ideal, find an expression for the variation 
of density within the tank. 
Solution: This problem is a realistic approximation of the “blowdown” of a high-
pressure tank, where the exit mass flow is choked and thus proportional to tank pressure. 
For a control volume enclosing the tank and cutting through the exit jet, the mass relation is 
tank exit exit
d d(m ) m 0, or: ( ) m C p A, where C constant
dt dt
        
o
p(t) t
o
o p o
CRT Ap dpIntroduce and separate variables: dt
RT p
     
The solution is an exponential decay of tank density: p  po exp(–CRToAt/V ). Ans. 
 
P4.24 For incompressible laminar flow between parallel plates (see Fig. 4.12b), 
the flow is two-dimensional (v  0) if the walls are porous. A special case solution is 
2 2( ) ( )u A Bx h y   , where A and B are constants. (a) Find a general formula for 
velocity v if v = 0 at y = 0. (b) What is the value of the constant B if v = vw at y = +h? 
 
Solution: (a) Use the equation of continuity to find the velocity v: 
318 Solutions Manual  Fluid Mechanics, Fifth Edition 
 
 
 
2 2
3
2 2 2
3
2
( )( )
Integrate : ( ) ( ) ( )
3
If 0 0, then ( ) 0. ( ) .( )
3
       
    
    

v u B h y
y x
yv B h y dy B h y f x
yv at y f x v B h y Ans a
 
(b) Just simply introduce this boundary condition into the answer to part (a): 
 
3
3
3
3
( ) ( ) , .( )
3 2
w
w
vhv y h v B h hence B Ans b
h
     
 
P4.25 An incompressible flow in polar 
coordinates is given by 
2v cos 1r
bK
r
      
2v sin 1
bK
r
       
Does this field satisfy continuity? For 
consistency, what should the dimensions of 
constants K and b be? Sketch the surface 
where vr  0 and interpret. 
 
Fig. P4.25 
Solution: Substitute into plane polar coordinate continuity: 
r 2
v1 1 1 b 1 b(rv ) 0 K cos r K sin 1 0
r r r r r r r r
Satisfied      
                           
The dimensions of K must be velocity, {K}  {L/T}, and b must be area, {b}  {L2}. The 
surfaces where vr = 0 are the y-axis and the circle r = b, as shown above. The pattern 
represents inviscid flow of a uniform stream past a circular cylinder (Chap. 8). 
 
 
 
 
 
 Chapter 4  Differential Relations for a Fluid Particle319
 
P4.26 Curvilinear, or streamline, coordinates are defined in Fig. P4.26, where n is 
normal to the streamline in the plane of the radius of curvature R. Show that Euler’s 
frictionless momentum equation (4.36) in streamline coordinates becomes 
 
 
 / ( / ) (1/ ) ( / ) sV t V V s p s g          (1) 
 
 
 
2 1
n
V pV g
t R n
 
       (2) 
 
Fig. P4.26 
Further show that the integral of Eq. (1) with respect to s is none other than our old friend 
Bernoulli’s equation (3.76). 
Solution: This is a laborious derivation, really, the problem is only meant to 
acquaint the student with streamline coordinates. The second part is not too hard, 
though. Multiply the streamwise momentum equation by ds and integrate: 
s
2
V dp dp dpds V dV g ds gsin ds g dz
t
             
Integrate from 1 to 2: (Bernoulli) .Ans  
2 22 2
2 1
2 1
1 1
V V dpV ds g z z 0
t 2
       
 
 
 
P4.27 A frictionless, incompressible steady-flow field is given by 
V  2xyi – y2j 
in arbitrary units. Let the density be o  constant and neglect gravity. Find an expression 
for the pressure gradient in the x direction. 
320 Solutions Manual  Fluid Mechanics, Fifth Edition 
 
Solution: For this (gravity-free) velocity, the momentum equation is 
2
ou v p, or: [(2xy)(2y ) ( y )(2x 2y )] px y
   
          
V V i i j  
2 3
oSolve for p (2xy 2y ), or: .Ans   2opi j 2xyx  
  
 
P4.28 Consider the incompressible flow field of Prob. P4.6, with velocity components 
u = 2y, v = 8x, w = 0. Neglect gravity and assume constant viscosity. (a) Determine whether 
this flow satisfies the Navier-Stokes equations. (b) If so, find the pressure distribution p(x, y) if 
the pressure at the origin is po. 
Solution: In Prob. P4.6 we found the accelerations, so we can proceed to Navier-Stokes: 
 
2
2
( ) [0 (8 )(2)] 0 0; 16
( ) [(2 )(8) 0] 0 0; 16
x
y
u u p p pu v x g u x
x y x x x
v v p p pu v y g v y
x y y y y
    
    
                    
                   
 
Noting that 
2 /( ) 0 in both cases, we conclude .( )p x y Ans a    Yes, satisfies Navier - Stokes. 
(b) The pressure gradients are simple, so we may easily integrate: 
 
2 2
2 2
, or : 16 16 8 ( )
If (0,0) , 8 ( ) .( )o o
p pdp dx dy p x dx y dy x y const
x y
p p then p p x y Ans b
  

           
   
  
This is an exact solution, but it is not Bernoulli’s equation. The flow is rotational. 
 
P4.29 Consider a steady, two-dimensional, incompressible flow of a newtonian fluid 
with the velocity field u  –2xy, v  y2 – x2, and w  0. (a) Does this flow satisfy 
conservation of mass? (b) Find the pressure field p(x, y) if the pressure at point (x  0, y  0) 
is equal to pa. 
 Chapter 4  Differential Relations for a Fluid Particle 321
Solution: Evaluate and check the incompressible continuity equation: 
( )0 2 2 0 0 au v w y y
x y z
  
          Yes! Ans. 
(b) Find the pressure gradients from the Navier-Stokes x- and y-relations: 
2 2 2
2 2 2 , :
u u u p u u uu v w or
x y z x x y z
             
               
2 2 2 3[ 2 ( 2 ) ( )( 2 )] (0 0 0), : 2 ( )p pxy y y x x or xy x
x x
                 
and, similarly for the y-momentum relation, 
2 2 2
2 2 2 , :
v v v p v v vu v w or
x y z y x y z
             
               
2 2 2 3[ 2 ( 2 ) ( )(2 )] ( 2 2 0), : = 2 ( )p pxy x y x y or x y y
y y
                
The two gradients p/x and p/y may be integrated to find p(x, y): 
2 2 4
2 ( ), :
2 4y Const
p x y xp dx f y then differentiate
x
  
        
2 2 3 3 42 ( ) 2 ( ), 2 , : ( )
2
p df dfx y x y y whence y or f y y C
y dy dy
               
2 2 4 4: (2 ) ( , ) (0,0), :
2 a
Thus p x y x y C p at x y or       aC p 
Finally, the pressure field for this flow is given by 
(b)Ans. 2 2 4 4ap p (2x y x y )
     
 
 
P4.30 For the velocity distribution of Prob. P4.4, determine if (a) the equation of continuity 
and (b) the Navier-Stokes equation are satisfied. (c) If the latter is true, find the pressure 
distribution p(x,y) when the pressure at the origin equals po. Neglect gravity. 
Solution: Recall that we were given u = Uo(1+x/L) and v = -Uo y/L. (a) Test continuity: 
322 Solutions Manual  Fluid Mechanics, Fifth Edition 
 
(b) Now substitute these velocities into the x- and y- Navier-Stokes equations: 
Solve for the two pressure gradients and cross-differentiate to see if they agree: 
 
Thus, before finding p(x,y), we know this is an exact solution to Navier-Stokes. Ans.(b) 
(c) Integrate the two pressure gradients to find the pressure distribution: 
 
This is the same as Bernoulli’s equation, but that is a bit hard to see. 
 
 
 
 
 
P4.31 According to potential theory (Chap. 8) for the flow approaching a rounded two-
dimensional body, as in Fig. P4.31, the velocity approaching the stagnation point is given 
by u  U(1 – a2/x2), where a is the nose radius and U is the velocity far upstream. 
).(.satisfiedOK,0)()]1([ aAns
L
U
L
U
L
yU
yL
xU
xy
v
x
u oo
oo 




011))(()0()1(
011)0)(()1(
2
2










y
pv
y
p
L
U
L
yU
L
xU
y
vv
x
vu
x
pu
x
p
L
yU
L
U
L
xU
y
uv
x
uu
o
oo
o
o
o


bothfor0Check)1(
222




yx
p
L
y
L
U
y
p
L
x
L
U
x
p oo 
).()
22
(
2
,Then;)()
2
(
222
2222
cAnsp
L
y
L
xx
L
Up
const
L
y
L
Uf
dy
df
y
pyf
L
xx
L
Udx
x
pp
o
o
oo



 


 Chapter 4  Differential Relations for a Fluid Particle 323
Compute the value and position of the maximum viscous normal stress along this 
streamline. Is this also the position 
 
Fig. P4.31 
of maximum fluid deceleration? Evaluate the maximum viscous normal stress if the fluid 
is SAE 30 oil at 20°C, with U  2 m/s and a  6 cm. 
Solution: (a) Along this line of symmetry the convective deceleration is one-dimensional: 
2 2 2 4
2
x 2 3 3 5
u a 2a a aa u U 1 U 2U
x x x x x


                   
 
xdaThis has a maximum deceleration at 0, or at (5/3) a . (a)
dx
Ans   x 1.29a 
The value of maximum deceleration at this point is 2x,maxa 0.372U /a.  
(b) The viscous normal stress along this line is given by 
2
xx 3
u 2a U2 2 with a maximum . (b)
x x
Ans  
      max
4 U at x a
a
   
Thus maximum stress does not occur at the same position as maximum deceleration. For 
SAE 30 oil at 20°C, we obtain the numerical result 
max3
kg kg 4(0.29)(2.0)SAE 30 oil, 917 , 0.29 , . (b)
m s (0.06 m)m
Ans      39 Pa 
 
P4.32 The answer to Prob. 4.14 is   f(r) only. Do not reveal this to your friends if 
they are still working on Prob. 4.14. Show that this flow field is an exact solution to the 
Navier-Stokes equations (4.38) for only two special cases of the function f(r). Neglect 
gravity. Interpret these two cases physically. 
324 Solutions Manual  Fluid Mechanics, Fifth Edition 
 
Solution: Given v  f(r) and vr  vz  0, we need only satisfy the -momentum relation: 
2
r 2 2 2
v v v v v v1 p 1 1v r ,
r r r r r r r r
               
                
 
2
1 d df for: (0 0) 0 r 0 , or:
r dr dr r
                2
1 1f f f 0
r r
   
This is the ‘equidimensional’ ODE and always has a solution in the form of a power-law, 
f  Crn. The two relevant solutions for these particular coefficients are n = 1: 
f1  C1r (solid-body rotation); f2  C2/r (irrotational vortex) Ans. 
 
P4.33 Consider incompressible flow at a volume rate Q toward 
a drain at the vertex of a 45 wedge of width b, as in Fig. P4.33. 
Neglect gravity and friction and assume purely radial 
inflow. (a) Find an expression for vr(r). (b) Show that 
the viscous term in the r-momentum equation is zero. 
(c) Find the pressure distribution p(r) if p = po at r = R. 
Solution: (a) Assume one-dimensional, steady, radial inflow. Then, at any radius r, 
 4, .( )
( / 4)r
Q Q C Qv where C Ans a
area r b r b 
      
The velocity is negative because the flow is inward. (b) The r-momentum equation is not 
written out in Chapter 4; it is Eq. (D.5) of Appendix D. The viscous term is 
 
2
2 2 2 2
2 3 3 3
2 1 2( ) [ ( ) ]
1 ( / ) 1[ ( ( )) 0] [ ( ) ] ( ) .( )
r r r
r
v vv v v
v r
r r rr r r r
C C r C C C Cr Ans b
r r r r r r rr r r r
   
  
          
               0
 
 
r 

Q
 = /4 
Drain Fig. P4.33
 Chapter 4  Differential Relations for a Fluid Particle 325
(c) With the viscous term zero, the r-momentum equation reduces to 
 
2
2 3
2 2 2
1 1 12 2 2
2 2
2 2
( ) ( )( ) , or :
Integrate : ; , ,
2 2 2
4Finally, ( ) , .( )
2
r
r
o o
o
v CC C p pv
r r r rr r
C C Cp C at r R p p C C p
r R R
QC Cp p where C Ans c
bR r
 
  


        
         
   
 
The two terms in parentheses are the velocities-squared at r = R and r = r, respectively. In other 
words, it integrates to Bernoulli’s equation because the viscous term is zero (irrotational flow). 
 
 
P4.34 A proposed three-dimensional incompressible flow field has the following vector 
form: 
V  Kxi  Kyj – 2Kzk 
(a) Determine if this field is a valid solution to continuity and Navier-Stokes. (b) If g  –gk, 
find the pressure field p(x, y, z). (c) Is the flow irrotational? 
Solution: (a) Substitute this field into the three-dimensional incompressible continuity 
equation: 
.
( ) ( ) ( 2 )
2 0 (a)
u v w Kx Ky Kz
x y z x y z
K K K Ans.
     
          
    Yes, satisfied
 
(b) Substitute into the full incompressible Navier-Stokes equation (4.38). The laborious 
results are: 
2x momentum: ( 0 0) (0 0 0)pK x
x
         
2y momentum: (0 0) (0 0 0)pK y
y
         
z momentum: {0 0 ( 2 )( 2 )} ( ) (0 0 0)pKz K g
z
              
Integrate each equation for the pressure and collect terms. The result is 
326 Solutions Manual  Fluid Mechanics, Fifth Edition 
 
p  p(0,0,0) – gz – (/2)K2(x2  y2  4z2) Ans. (b) 
Note that the last term is identical to (/2)(u2  v2  w2), in other words, Bernoulli’s 
equation. 
(c) For irrotational flow, the curl of the velocity field must be zero: 
  V  i(0 – 0)  j(0 – 0)  k(0 – 0)  0 Yes, irrotational. Ans. (c) 
 
P4.35 From the Navier-Stokes equations 
for incompressible flow in polar coordinates 
(App. E for cylindrical coordinates), find 
the most general case of purely circulating 
motion (r), r  z  0, for flow with no 
slip between two fixed concentric cylinders, 
as in Fig. P4.35. 
Solution: The preliminary work for this 
r
No slip
r = b
r = a
υ (r)θ
 
Fig. P4.35 
problem is identical to Prob. 4.32 on an earlier page. That is, there are two possible 
solutions for purely circulating motion (r), hence 
2
1 1 2 1 2
Cv C r , subject to v (a) 0 C a C /a and v (b) 0 C b C /b
r  
        
This requires C1  C2  0, or v  0 (no steady motion possible between fixed walls) Ans. 
 
 
 
 
P4.36 A constant-thickness film of 
viscous liquid flows in laminar motion 
down a plate inclined at angle , as in Fig. 
P4.36. The velocity profile is 
u  Cy(2h – y) v  w  0 
Find the constant C in terms of the specific 
weight and viscosity and the angle . Find 
the volume flux Q per unit width in terms 
of these parameters. 
(y)
g y
x
h
θ
 
Fig. P4.36 
 
Solution: There is atmospheric pressure all along the surface at y  h, hence p/x  0. 
The x-momentum equation can easily be evaluated from the known velocity profile: 
2
x
u u pu v g u, or: 0 0 gsin + ( 2C)
x y x
         
            
Solve for . (a)Ansg sinC
2
 
 
The flow rate per unit width is found by integrating the velocity profile and using C: 
h h
3
0 0
2Q udy Cy(2h y)dy Ch per unit width . (b)
3
Ans      3gh sin3  
 
 
P4.37 A viscous liquid of constant 
density and viscosity falls due to gravity 
between two parallel plates a distance 2h 
apart, as in the figure. The flow is fully 
developed, that is, w  w(x) only. There 
are no pressure gradients, only gravity. Set 
up and solve the Navier-Stokes equation 
for the velocity profile w(x). 
Solution: Only the z-component of Navier-
Stokes is relevant: 
 
Fig. P4.37 
2
20 , : , ( ) ( ) 0 ( )
dw d w gg or w w h w h no-slip
dt dx
             
The solution is very similar to Eqs. (4.142) to (4.143) of the text: 
.w Ans 
2 2g (h x )
2
 
 
P4.38 Show that the incompressible flow distribution, in cylindrical coordinates, 
00  znr vrCvv 
328 Solutions Manual  Fluid Mechanics, Fifth Edition 
 
where C is a constant, (a) satisfies the Navier-Stokes equation for only two values of n. 
Neglect gravity. (b) Knowing that p = p(r) only, find the pressure distribution for each 
case, assuming that the pressure at r = R is po. What might these two cases represent? 
 
Solution: (a) The important direction here is the -momentum equation, Eq. (D.6): 
Cancel C and  and rn-2. These terms equal zero only if n2 = 1, or n = 1. Ans.(a). 
(b) Find the respective pressure distributions for n = 1 and n = -1. Use Eq. (D.5), which 
reduces simply to p/r = v2/r. Try this for each distribution, n = 1: 
 
Case 1, v = Cr, is solid-body rotation. Case 2, v = C/r, is an irrotational potential vortex. 
 
 
 
 
0)(])(1[
:or,]))((1[)0(0000
:or,)2(1
r
1 )(
22221
22
2
22
2












nnnn
n
n
r
CrrCnCrrnCr
rr
r
CrCr
r
r
rrr
vv
v
rr
vvp
r
vvv
t
v




 V
).()11(
2
:or;)/(:1,2
).()(
2
:or;)(:1,1
222
222
1
22
222
bAns
rR
Cppdr
r
rCdpnCase
bAnsRrCppdr
r
rCdpnCase
o
r
R
p
p
o
r
R
p
p
o
o






 Chapter 4  Differential Relations for a Fluid Particle 329
 
P4.39 Reconsider the angular-
momentum balance of Fig. 4.5 by adding 
a concentrated body couple Cz about the z 
axis [6]. Determine a relation between the 
body couple and shear stress for 
equilibrium. What are the proper 
dimensions for Cz? (Body couples are 
important in continuous media with 
microstructure, such as granular 
materials.) 
Solution: The couple Cz has to be 
per unit volume to make physical sense 
in Eq. (4.39): 
 
Fig. 4.5 
2
xy yx 2 2
xy yx z 2
1 1 1 ddx dy dx dy dz C dx dy dz dx dy dz(dx dy )
2 x 2 y 12 dt
     
        
 
Reduce to third order termsand cancel (dx dy dz): yx  xy  Cz Ans. 
The concentrated couple allows the stress tensor to have unsymmetrical shear stress terms. 
 
 
P4.40 For pressure-driven laminar flow between parallel plates (see Fig. 4.12b), 
the velocity components are u = U(1– y2/ h2), v = 0, and w = 0, where U is the centerline 
velocity. In the spirit of Ex. 4.6, find the temperature distribution T(y) for a constant wall 
temperature Tw. 
 
Solution: There are no variations with x or z, so the energy equation (4.53) reduces to 
 
2 2
2 2
2 2
2 22 3
2 2
12 2 4 4
0 ( ) ( ) ,
4 42: ( ) ( ) ; Integrate : ( )
3
  
 
       
      
p
T T u d T duc u k k
x y dyy dy
U Ud T Uy dT yor y C
k dydy h k h k h
 
 
330 Solutions Manual  Fluid Mechanics, Fifth Edition 
 
The condition T = Tw at h is equivalent to dT/dy = 0 at y = 0. Thus C1 = 0. Integrate again: 
 
2 2 24 4
2 2 24 4
4 4( ) ; at : ( ) ,
12 12 3
           w wU U Uy hT C y h T T C C T kk h k h 
The final solution for T(y) is, like Ex. 4.6, a quartic polynomial: 
 
2 4
4( ) (1 ) .3w
U yT y T Ans
k h
   
 
 
 
 
P4.41 As mentioned in Sec. 4.10, the velocity profile for laminar flow between two 
plates, as in Fig. P4.40, is 
max
2
4 ( ) 0u y h yu w
h
   
If the wall temperature is Tw at both walls, use 
 
Fig. P4.41 
the incompressible-flow energy equation (4.75) to solve for the temperature distribution 
T(y) between the walls for steady flow. 
Solution: Assume T  T(y) and use the energy equation with the known u(y): 
2 22 2
max
p p2 2 2
4udT d T du d Tc k , or: c (0) k (h 2y) , or:
dt dydy dy h
                
2 22 3
2 2 2 2max max
12 4 4
16 u 16 ud T dT 4y(h 4hy 4y ), Integrate: h y 2hy C
dy 3dy kh kh
             
 
 Chapter 4  Differential Relations for a Fluid Particle 331
Before integrating again, note that dT/dy  0 at y  h/2 (the symmetry condition), so 
C1  –h3/6. Now integrate once more: 
         
2 2 3 4
2max
1 24
16 u y y yT h 2h C y C
2 3 3kh
 
If T  Tw at y  0 and at y  h, then C2  Tw. The final solution is: 
.Ans
   
2 2 3 4
max
w 2 3 4
8 u y y 4y 2yT T
k 3h h 3h 3h
     
 
 
This is exactly the same solution as Problem P4.40 above, except that, here, the 
coordinate y is measured from the boo tom wall rather than the centerline. 
 
P4.42 Suppose that we wish to analyze the rotating, partly-full cylinder of Fig. 2.23 as a 
spin-up problem, starting from rest and continuing until solid-body-rotation is achieved. 
What are the appropriate boundary and initial conditions for this problem? 
Solution: Let V  V(r, z, t). The initial condition is: V(r, z, 0)  0. The boundary 
conditions are 
Along the side walls: v(R, z, t)  R, vr(R, z, t)  0, vz(R, z, t)  0. 
At the bottom, z  0: v(r, 0, t)  r, vr(r, 0, t)  0, vz(r, 0, t)  0. 
At the free surface, z   : p  patm, rz  z  0. 
 
P4.43 For the draining liquid film of Fig. 
P4.36, what are the appropriate boundary 
conditions (a) at the bottom y  0 and (b) at 
the surface y  h? 
 
Fig. P4.36 
Solution: The physically realistic conditions at the upper and lower surfaces are: 
(a) at the bottom, y  0, no-slip: u(0)  0 Ans. (a) 
u(b) At the surface, y h, no shear stress, 0, or . (b)
y
Ans  
u ( ) 0
y
h  
332 Solutions Manual  Fluid Mechanics, Fifth Edition 
 
 
 
 
 
P4.44 Suppose that we wish to analyze the sudden pipe-expansion flow of Fig. P3.59, 
using the full continuity and Navier-Stokes equations. What are the proper boundary 
conditions to handle this problem? 
Solution: First, at all walls, one would impose the no-slip condition: ur  uz  0 at all 
solid surfaces: at r  r1 in the small pipe, at r  r2 in the large pipe, and also on the flat-
faced surface between the two. 
 
Fig. P3.59 
Second, at some position upstream in the small pipe, the complete velocity 
distribution must be known: u1  u1(r) at z  z1. [Possibly the paraboloid of Prob. 4.34.] 
Third, to be strictly correct, at some position downstream in the large pipe, the 
complete velocity distribution must be known: u2  u2(r) at z  z2. In numerical 
(computer) studies, this is often simplified by using a “free outflow” condition,  u/ z  0. 
Finally, the pressure must be specified at either the inlet or the outlet section of the 
flow, usually at the upstream section: p  p1(r) at z  z1. 
 
P4.45 For the sluice gate problem of Example 3.10, list all the boundary conditions 
needed to solve this flow exactly by, say, Computational Fluid Dynamics (CFD). 
 
1 
2
3
3
2
4 
 Chapter 4  Differential Relations for a Fluid Particle 333
Solution: There are four different kinds of boundary conditions needed, as labeled. 
(1) Known velocity V1 upstream, and of course the depth y1 must be known. 
(2) Known pressure patm at both the upstream and downstream free surfaces. 
(3) No-slip (V = 0) all along the bottom and on the gate inner wall. 
(4) The downstream flow is complicated because we don’t know V2 or y2 and therefore 
cannot specify them. What CFD modelers do is to have an adjustable upper boundary 
and specify that the exit flow is “smooth”, or “zero gradient”, that is, V/x = 0. 
 
 
 
P4.46 Fluid from a large reservoir at temperature To flows into a circular pipe of radius 
R. The pipe walls are wound with an electric-resistance coil which delivers heat to the 
fluid at a rate qw (energy per unit wall area). If we wish to analyze this problem by using 
the full continuity, Navier-Stokes, and energy equations, what are the proper boundary 
conditions for the analysis? 
Solution: Letting z  0 be the pipe entrance, we can state inlet conditions: typically 
uz(r, 0)  U (a uniform inlet profile), ur(r, 0)  0, and T(r, 0)  To, also uniform. 
At the wall, r  R, the no-slip and known-heat-flux conditions hold: uz(R, z)  ur(R, z)  0 
and k( T/ r)  qw at (R, z) (assuming that qw is positive for heat flow in). 
At the exit, z  L, we would probably assume ‘free outflow’:  uz/ z   T/ z  0. 
Finally, we would need to know the pressure at one point, probably the inlet, z  0. 
 
 
 
 
 
 
 
 
 
 
 
P4.47 Given the incompressible flow V  3yi  2xj. Does this flow satisfy continuity? 
If so, find the stream function (x, y) and plot a few streamlines, with arrows. 
334 Solutions Manual  Fluid Mechanics, Fifth Edition 
 
 
 
 
Solution: With u  3y and v  2x, we 
may check  u/ x   v/ y  0  0  0, OK. 
Find the streamlines from u   / y  3y 
and v  –/x  2x. Integrate to find 
.Ans  2 23
2
y x 
Set   0, 1, 2, etc. and plot some 
streamlines at right: flow around corners of 
half-angles 39 and 51. 
39°
51°
x
y
 
Fig. P4.47 
 
P4.48 Consider the following two-dimensional incompressible flow, which clearly satisfies 
continuity: 
u  Uo  constant, v  Vo  constant 
Find the stream function (r, ) of this flow, that is, using polar coordinates. 
Solution: In cartesian coordinates the stream function is quite easy: 
u  / y  Uo and v  –/x  Vo or:   Uoy – Vox  constant 
But, in polar coordinates, y  rsin and x  rcos. Therefore the desired result is 
(r, )  Uor sin – Vor cos  constant Ans. 
 
 
P4.49 Investigate the stream function   
K(x2 – y2), K  constant. Plot the streamlines 
in the full xy plane, find any stagnation 
points, and interpret what the flow could 
represent. 
Solution: The velocities are given by 
u 2Ky; v 2Kx
y x
 
        
This is also stagnation flow, with the stream-lines turned 45 from Prob. 4.48. 
 
Fig. P4.49 
 
P4.50 In 1851, George Stokes (of Navier-Stokes fame) solved the problem of steady 
incompressible low-Reynolds-number flow past a sphere, using spherical polar coordinates 
(r, ) – [Ref. 5, page 168]. In these coordinates, the equation of continuity is 
 2( sin ) ( sin ) 0rr v r vr  
    
(a) Does a stream function exist for these coordinates? (b) If so, find its form. 
 
 
Solution: Two velocity components and two continuity terms. Yes,  exists! Ans.(a) 
(b) The stream function should be defined such that continuity takes the form 
 
 
2 2
2
0 , or :
1 1; .( )
sin sin
r
r r
v v Ans b
r r r

 
 
 
 
      
    
 
 
 
P4.51 The velocity profile for incompressible pressure-driven laminar flow between parallel 
plates (see Fig. 4.12b) has the form u = C(h2 – y2), where C is a constant. (a) Determine if a 
stream function exists. (b) If so, determine a formula for the stream function, 
 
Solution: (a) A stream function exists, for a single velocity component u, if u/x = 0, which 
it certainly is, since u is a function only of y. Yes,  exists. Ans.(a) 
(b) Finding the stream function is just a matter of direct integration: 
 
 3
2 2 2
0 , hence is a function only of
( ) ; Integrate : ( ) .( )
3
v y
x
yu C h y C h y constant Ans b
y
 
 
  
     
 
 
P4.52 A two-dimensional, incompressible, frictionless fluid is guided by wedge-shaped 
walls into a small slot at the origin, as in Fig. P4.52. The width into the paper is b, and the 
volume flow rate is Q. At any given distance r from the slot, the flow is radial inward, 
with constant velocity. Find an expression for the polar-coordinate stream function of this 
flow. 
 
Fig. P4.52 
Solution: We can find velocity from continuity: 
r
Q Q 1v from Eq. (4.101). Then
A ( /4)rb r

      
.Ans   4Q constant
b
θπ 
This is equivalent to the stream function for a line sink, Eq. (4.131). 
 
P4.53 For the fully developed laminar-pipe-flow solution of Eq. (4.137), find the 
axisymmetric stream function (r, z). Use this result to determine the average velocity 
V  Q/A in the pipe as a ratio of umax. 
Solution: The given velocity distribution, vz  umax(1 – r2/R2), vr  0, satisfies 
continuity, so a stream function does exist and is found as follows: 
2 4
2 2
z max max 2
1 r rv u (1 r /R ) , solve for u f(z), now use in
r r 2 4R
 
         
r
1 df v 0 0 , thus f(z) const, .
r z dz
Ans 
          
2 4
max 2
r ru
2 4R
  
We can find the flow rate and average velocity from the text for polar coordinates: 
2 4
2
1-2 2 1 0-R max max max2
R RQ 2 ( ), or: Q 2 u u (0 0) R u
2 24R
                 
 
 Chapter 4  Differential Relations for a Fluid Particle 337
2 2
avg pipe maxThen V Q/A [( /2)R u /( R )] .Ans    max1 u2 
 
338 Solutions Manual  Fluid Mechanics, Fifth Edition 
 
 
P4.54 An incompressible stream function 
is defined by 
2 3
2( , ) (3 )
Ux y x y y
L
   
where U and L are (positive) constants. 
Where in this chapter are the streamlines of 
this flow plotted? Use this stream function to 
find the volume flow Q passing through the 
 
Fig. E4.7 
rectangular surface whose corners are defined by (x, y, z)  (2L, 0, 0), (2L, 0, b), (0, L, b), 
and (0, L, 0). Show the direction of Q. 
Solution: This flow, with velocities u  / y  3U/L2(x2 – y2), and v  –/ x  
–6xyU/L2, is identical to Example 4.7 of the text, with “a”  3U/L2. The streamlines are 
plotted in Fig. E4.7. The volume flow per unit width between the points (2L, 0) and (0, L) is 
2 3
2 2
U UQ/b (2L, 0) (0, L) (0 0) [3(0) L L ] UL, or: .
L L
Ans        Q ULb 
Since  at the lower point (2L, 0) is larger than at the upper point (0, L), the flow through 
this diagonal plane is to the left, as per Fig. 4.9 of the text. 
 
P4.55 For the incompressible plane flow of Prob. P4.6, with velocity components 
u = 2y, v = 8x, w = 0, determine (a) if a stream function exists. (b) If so, determine the 
form of the stream function, and (c) plot a few representative streamlines. 
 
Solution: (a) Check to see is two-dimensional continuity is satisfied: 
 
 (2 ) (8 ) 0 0 0 .( )u v y x Ans a
x y x y
            Yes, ψ exists. 
(b) Find the stream function by relating velocities to derivatives of  : 
 2 22 ; 8 ; Integrate : 4 .( )u y v x y x constant Ans b
y x
            
 
 Chapter 4  Differential Relations for a Fluid Particle 339
(c) Plot a few streamlines, that is, plot y2 = 4x2 + C for various C. Here are the 
results: 
 
0
1
2
3
4
5
6
-3 -2 -1 0 1 2 3
y2 = 4x2 + C
 
We are showing only the upper half plane, which is the mirror image of the lower half. 
 
 
P4.56 Investigate the velocity potential 
 = Kxy, K = constant. Sketch the potential 
lines in the full xy plane, find any 
stagnation points, and sketch in by eye the 
orthogonal streamlines. What could the 
flow represent? 
Solution: The potential lines,   constant, 
are hyperbolas, as shown. The streamlines, 
 
Fig. P4.56 
sketched in as normal to the  lines, are also hyperbolas. The pattern represents plane 
stagnation flow (Prob. 4.49) turned at 45. 
 
P4.57 A two-dimensional incompressible flow field is defined by the velocity 
components 
2 2x y yu V v V
L L L
       
340 Solutions Manual  Fluid Mechanics, Fifth Edition 
 
where V and L are constants. If they exist, find the stream function and velocity potential. 
Solution: First check continuity and irrotationality: 
2 2 0
20 0
u v V V
x y L L
v u Vx
x y L
 
 
 
 
   
            


exists
does not exist
;
V k k
 
To find the stream function , use the definitions of u and v and integrate: 
2
2 , 2 ( )
2
x y xy yu V V f x
y L L L L
 
             
 
2 2Evaluate
Thus 0 and .
Vy df Vyv
x L dx L
df Ans
dx

     
      
xy yV const
L L
22 
 
P4.58 Show that the incompressible velocity potential in plane polar coordinates (r,) 
is such that 
1
r r r
     
Finally show that  as defined satisfies Laplace’s equation in polar coordinates for 
incompressible flow. 
Solution: Both of these things are quite true and easy to show from the definition of the 
gradient vector in polar coordinates. Ans. 
 
P4.59 Consider the two-dimensional incompressible velocity potential   xy  x2 – y2. 
(a) Is it true that 2  0, and, if so, what does this mean? (b) If it exists, find the stream 
function (x, y) of this flow. (c) Find the equation of the streamline which passes through 
(x, y)  (2, 1). 
Solution: (a) First check that 2  0, which means that incompressible continuity is 
satisfied. 
2 2
2
2 2 0 2 2 0x y
            Yes 
 Chapter 4  Differential Relations for a Fluid Particle 341
(b) Now use  to find u and v and then integrate to find . 
2
2 , hence 2 ( )
2
yu y x xy f x
x y
          
2
2 2 , hence ( )
2
df xv x y y f x const
y x dx
 
            
The final stream function is thus . (b)Ans      2 21 2
2
y x xy const 
(c) The streamline which passes through (x, y)  (2, 1) is found by setting   a constant: 
2 21 3 5( , ) (2, 1), (1 2 ) 2(2)(1) 4
2 2 2
At x y        
Thus the proper streamline is . (c)Ans      2 21 52
2 2
y x xy 
 
P4.60 Liquid drains from a small hole in 
a tank, as shown in Fig. P4.60, such that 
the velocity field set up is given by r  0, z 
 0,   R2/r, where z  H is the depth of 
the water far from the hole. Is this flow 
pattern rotational or irrotational? Find the 
depth zc of the water at the radius r  R. 
Solution: From Appendix D, the angular 
velocity is 
patm
z
r
z = 0
r = R
z = H
 
Fig. P4.60 
z
1 1(rv ) (v ) 0 (IRROTATIONAL)
r r r 
      
Incompressible continuity is valid for this flow, hence Bernoulli’s equation holds at the 
surface, where p  patm, both at infinity and at r  R: 
2 2
atm r atm r R c
1 1p V gH p V gz
2 2
         
2 2
r r RIntroduce V 0 and V R to obtain .2C
Rz H Ans
g
     
 
342 Solutions Manual  Fluid Mechanics, Fifth Edition 
 
 
 
P4.61 For the incompressible plane flow of Prob. P4.6, with velocity components u = 
2y, v = 8x, w = 0, determine (a) if a velocity potential exists. (b) If so, determine the form of 
the velocity potential, and (c) plot a few representative potential lines. 
 
Solution: (a) A velocity potential exists if the vorticity is zero. Here, for plane flow in (x, y) 
coordinates, we need only evaluate rotation around the z axis: 
 
 2 8 2 6 0 Rotational , does not exist. .( )z z
v u Ans a
x y
            
(b, c) There is no velocity potential – no plot, no formula. The flow has constant vorticity. 
 
P4.62 Show that the linear Couette flow 
between plates in Fig. 1.6 has a stream 
function but no velocity potential. Why is 
this so? 
Solution: Given u  Vy/h, v  0, check 
continuity: 
 
Fig. 1.6 
?u v 0 0 0 Find from
x y
       Satisfied therefore exists .  
Vyu , v 0 , solve for .
h y x
Ans        
2V y const
2h
 
Now check irrotationality: 
?
z
v u V2 0 0 0! .
x y h
Ans         Rotational, does not exist. 
 
 
 
 Chapter 4  Differential Relations for a Fluid Particle 343
P4.63 Find the two-dimensional velocity potential (r,) for the polar-coordinate flow 
pattern r  Q/r,   K/r, where Q and K are constants. 
Solution: Relate these velocity components to the polar-coordinate definition of  : 
r 
Q K 1v , v ; solve for .
r r r r
Ans
        Q ln(r) K const   
 
P4.64 Show that the velocity potential (r, z) in axisymmetric cylindrical coordinates 
(see Fig. 4.2 of the text) is defined by the formulas: 
r zr z
     
Further show that for incompressible flow this potential satisfies Laplace’s equation in (r, z) 
coordinates. 
Solution: Both of these things are quite true and are easy to show from their definitions. Ans. 
 
 
P4.65 A two-dimensional incompressible 
flow is defined by 
2 2 2 2
Ky Kxu
x y x y
    
where K  constant. Is this flow irrotational? 
If so, find its velocity potential, sketch a 
few potential lines, and interpret the flow 
pattern. 
 
Fig. P4.65 
Solution: Evaluate the angular velocity: 
2 2
z 2 2 2 2 2 2 2 2 2 2
v u K 2Kx K 2Ky2 .
x y x y (x y ) x y (x y )
Ans             0 (Irrotational) 
Introduce the definition of velocity potential and integrate to get (x, y): 
2 2 2 2
Ky Kxu ; v , solve for .
x yx y x y
Ans   
         
 1 yK tan K
x
 
The  lines are plotted above. They represent a counterclockwise line vortex. 
344 Solutions Manual  Fluid Mechanics, Fifth Edition 
 
 
 
 
P4.66 A plane polar-coordinate velocity potential is defined by 
cos constK K
r
   
Find the stream function for this flow, sketch some streamlines and potential lines, and 
interpret the flow pattern. 
 
Solution: Evaluate the velocities and thence find the stream function: 
 
Fig. P4.66 
r 2 2
Kcos 1 1 Ksinv ; v ,
r r r rr r
solve .Ans

     
   

        
  Ksin
r
 
The streamlines and potential lines are shown above. This pattern is a line doublet. 
 
 Chapter 4  Differential Relations for a Fluid Particle 345
 
P4.67 A stream function for a plane, 
irrotational, polar-coordinate flow is 
ln and constC K r C K    
Find the velocity potential for this flow. 
Sketch some streamlines and potential 
lines, and interpret the flow pattern. 
Solution: If this problem is given early 
enough (before Section 4.10 of the text), the 
 
Fig. 4.14 
students will discover this pattern for themselves. It is a line source plus a line vortex, a 
tornado-like flow, Eq. (4.134) and Fig. 4.14 of the text. Find the velocity potential: 
r
1 C K 1v ; v , solve .
r r r r r r
Ans
               C ln(r) K  
The streamlines and potential lines are plotted above for negative C (a line sink). 
 
P4.68 For the velocity distribution of Prob. P4.4, (a) determine if a velocity potential 
exists and, if it does, (b) find an expression for (x,y) and sketch the potential line which 
passes through the point (x, y) = (L/2, L/2). 
 
Solution: Recall the given flow, u = Uo(1+x/L) and v = Uo(y/L). (a) Calculate if the 
flow is irrotational. For plane flow, only one term of the curl(V) is needed: 
 
Therefore a velocity potential does exist. Ans.(a) (b) To find , integrate from u and v: 
 
0 )curl(Yes,;0002 

 V
y
u
x
v
z
346 Solutions Manual  Fluid Mechanics, Fifth Edition 
 
For a potential line to pass through (L/2, L/2), we must have /(Uo/L) = [1/2 + {(1/2)2 -
(1/2)2}/2] = ½. For convenience let the const = 0. Thus we are to plot this potential line: 
 
The result is plotted (red) in the graph below, along with the (blue)  line, which has the 
analytic form  = Uo(y + xy/L) = 3UoL/4. 
 
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
x/L
y/L
 /UoL  /UoL
 
 
 
2
2
2 2
2
; (1 ) ( ) ( )
2
0 , or :
2
Thus ( )( ) .( )
2
o o
o o
o
x xu Thus u dx U dx U x f y
x L L
df y yv U f U constant
y dy L L
x yxU L const Ans b
L L
 


       
        
  
 
2
22
22
1
L
yx
L
x
LUo

 Chapter 4  Differential Relations for a Fluid Particle 347
 
P4.69 A steady, two-dimensional flow has the following polar-coordinate velocity potential: 
 
where C and K are constants. Determine the stream function (r, ) for this flow. For extra 
credit, let C be a velocity scale U, let K = UL, and sketch what the flow might represent. 
 
Solution: Write out the  and  expressions for polar-coordinate velocities: 
 
Extra credit: Plot a typical streamline for C = U and K = UL: 
 
-0.2
0.0
0.2
0.4
0.6
0.8
1.0
1.2
-0.2 -0.1 0.0 0.1 0.2 0.3 0.4
O
 
All the streamlines are logarithmic spirals coming out from the origin in every direction. 
 
 
 
rKrC lncos  
.sin,)0(sin1
)(sinhence,1cos
AnsconstantKrChence
r
KC
r
v
rfKrC
rr
KC
r
vr











348 Solutions Manual  Fluid Mechanics, Fifth Edition 
 
 
P4.70 A CFD model of steady two-dimensional 
incompressible flow has printed out the values of 
stream function (x, y), in m2/s, at each of the 
four corners of a small 10cm-by-10cm cell, as 
shown in Fig. P4.70. Use these numbers to 
estimate the resultant velocity in the center of 
the cell and its angle  with respect to the x axis. 
 
Solution: Quick analysis:the  values are higher on the top than the bottom, therefore u is to 
the right. The  values are higher on the right than the left, therefore v is down. There are 
several ways to estimate the center velocities. One simple way is to compute average values of 
 on the sides: 
 
 
 
 
 
x = 1.5 m x = 1.6 m
y = 1.0 m
y =1.1 m
? 
V ? 
 = 1.9552 m2/s 2.0206
1.79781.7308 m2/s 
Fig. P4.70 
u
v
1.9879
1.9092
1.7643
1.8430 
 Chapter 4  Differential Relations for a Fluid Particle 349
Then ucenter  /y = (1.9879-1.7643 m2/s)/(0.1m) = 2.236 m/s to the right. And vcenter  
/x = (1.9092-1.8430 m2/s)/(0.1m) = 0.662 m/s down. The resultant and its angle are 
 
The  values in this problem are in fact taken 
from an exact solution, V = 2.3315 m/s,  = 16.505. 
 
P4.71 Consider the following two-dimensional function f(x, y): 
(a) Under what conditions, if any, on (A,B,C,D) can this function f be a steady, plane-flow 
velocity potential? (b) If you find a (x, y) to satisfy part (a), also find the associated stream 
function x, y), if any, for this flow. 
Solution: (a) If f is to be a plane-flow velocity potential, it must satisfy Laplace’s equation: 
(b) To find , use to get u and v and work backwards to get the stream function: 
 
 
.)
236.2
662.0(tan;)662.0()236.2( 122 AnsdownV o16.5m/s2.332  
V 
0where,223  ADxCyxBxAf
).(3ispotentialvelocityThe
0and3026
23
2
aAnsDxyAxA
CABifCBxAxf



).(3,Finally
,66
)(3,33
32
3222
bAnsconstyAyxA
constf
dx
dfAxy
x
Axy
y
v
xfyAyxA
y
yAxA
x
u










350 Solutions Manual  Fluid Mechanics, Fifth Edition 
 
P4.72 Water flows through a two-dimensional 
narrowing wedge at 9.96 gal/min per meter of 
width into the paper. If this inward flow is purely 
radial, find an expression, in SI units, for (a) the 
stream function, and (b) the velocity potential of the flow. 
Assume one-dimensional flow. The included angle of the wedge is 45. 
 
Solution: The wedge angle equals /4 radians. At any given position r, the inward flow equals 
 
 We have already been advised that v = 0. (a) Work from radial velocity to stream function: 
Note that r must be in meters. (b) Work from radial velocity to obtain velocity potential: 
 
 
r 
Q Drain 
Fig. P4.72 
3 3
4 /( )
( / 4)
4(9.96 / min/ )(6.309 5 / / )where 4( / ) / 0 00080


 
     
  
r
Q bQ Qv
A rb r
gal m E m s gpm mQ b .
s m
30.00080 / 1 ; Solve 0.00080 .( )  
     r
m s mv Ans a
r r
30.00080 ; Solve 0.00080 ln( ) .( )      r
m s mv r Ans b
r r
 Chapter 4  Differential Relations for a Fluid Particle 351
 
 
 
 
P4.73 A CFD model of steady two-dimensional 
incompressible flow has printed out the values of 
velocity potential (x, y), in m2/s at each of the 
four corners of a small 10cm-by10cm cell, as 
shown in Fig. P4.73. Use these numbers to 
estimate the resultant velocity in the center of 
the cell and its angle  with respect to the x axis. 
 
Solution: Quick analysis: the  values are lower on the left than the right, therefore u is to the 
right. The  values are lower on the top than the bottom, therefore v is down. There are several 
ways to estimate the center velocities. One simple way is to compute average values of  on 
the sides: 
 
 
x = 1.5 m x = 1.6 m
y = 1.0 m
y = 1.1 
? 
V ? 
= 4.8338 m2/s 5.0610
5.12364.9038 m2/s 
Fig. P4.73
352 Solutions Manual  Fluid Mechanics, Fifth Edition 
 
 
 
 
 
 
 
Then ucenter  /x = (5.0923-4.8688 m2/s)/(0.1m) = 2.235 m/s to the right. And vcenter  
|/y| = (5.0137-4.9474 m2/s)/(0.1m) = 0.663 m/s down. The resultant and its angle are 
 
The  values in this problem are in fact taken 
from an exact solution, V = 2.3315 m/s,  = 16.505 down. 
 
P4.74 Consider the two-dimensional incompressible polar-coordinate velocity potential 
 
where B is a constant and L is a constant length scale. (a) What are the dimensions of B? 
(b) Locate the only stagnation point in this flow field. (c) Prove that a stream function exists 
and then find the function (r, ). 
 
u
v
4.9474 m/s
5.0923
5.0137
4.8688 
.)
235.2
663.0(tan;)663.0()235.2( 122 AnsV o16.5m/s2.331  
V 

 LBrB  cos
 Chapter 4  Differential Relations for a Fluid Particle 353
Solution: (a) To give  its correct dimensions of {L2/T}, the constant B must have the 
dimensions of velocity, or {L/T}. Ans.(a) 
(b) Calculate velocities in polar coordinates: 
 
At first it doesn’t look as if we can find a stagnation point, but indeed there is one: 
As discussed later in Chap. 8, this is the velocity potential of a Rankine half-body. 
(c) With the velocities known, check the continuity equation: 
 
Continuity is satisfied. Find the stream function from the definition of (r, ): 
 
P4.75 Given the following steady axisymmetric stream function: 
 
valid in the region 0  r  R and 0  z  L. (a) What are the dimensions of the constant B? 
(b) Show whether this flow possesses a velocity potential and, if so, find it. (c) What might this 
flow represent? [HINT: Examine the axial velocity vz.] 
r
BLB
r
vB
r
vr 

 
  sin1;cos
).(0,0:180, bAns
L
BLBvvLr r   
satisfiedYes,0coscos01)(1 


r
B
r
Bv
r
vr
rr r



).(lnsin:Integrate
sin;cos1
cAnsconstrLBrB
r
BLB
r
vB
r
vr








constantsareandwhere,)
2
(
2 2
4
2 RB
R
rrB 
354 Solutions Manual  Fluid Mechanics, Fifth Edition 
 
 
Solution: (a) From the definition of (r, z) in Eqs. (4.105), the dimensions of  are {L3/T}. 
Thus B has velocity dimensions, {B} = {L/T}. Ans.(a) 
(b) To test for irrotationality, first find the velocity components from Eqs. (4.106): 
 
Now evaluate the curl of the velocity, which has only one possible non-zero component. From 
Appendix D, Eq. (D.11), 
 
(c) The interpretation of the flow follows immediately from the velocity components. The 
velocity profile is a paraboloid of revolution and represents Poiseuille pipe flow, Eq. (4.137). 
Ans.(c) 
 
*P4.76 A two-dimensional incompressible flow has the velocity potential 
 
where K and C are constants. In this discussion, avoid the origin, which is a singularity (infinite 
velocity). (a) Find the sole stagnation point of this flow, which is somewhere in the upper half 
plane. (b) Prove that a stream function exists and then find (x, y), using the hint that 
dx/(a2+x2) = (1/a)tan-1(x/a). 
 
)1()
2
42(
2
11;01 2
2
2
3
R
rB
R
rrB
rrr
v
zr
v zr 

 
).(.existnotdoes,0202 2 bAnsR
Br
r
v
z
v zr  Rotational


)ln()( 2222 yxCyxK 
 Chapter 4  Differential Relations for a Fluid Particle 355
Solution: (a) Find the velocity components and see where they both equal zero: 
 
For positive K and C, u cannot be zero anywhere except at x = 0. Then v = 0 if 
 
(b) First check the velocities to see if continuity is satisfied: 
 
The algebra is messy but, indeed, continuity is satisfied,  exists. Ans.(b) – part 1. Now 
integrate the velocity components to find the stream function : 
 
 
 
 
P4.77 Outside an inner, intense-activity circle of radius R, a tropical storm can be 
simulated by a polar-coordinate velocity potential (r, ) = Uo R , where Uo is the windvelocity at radius R. (a) Determine the velocity components outside r = R. (b) If, at R = 
2222
22;22
yx
CyKy
x
v
yx
CxKx
x
u 

 
).(0:,22 aAns
K
Cyandxatstagnationor
y
CKy 
0]
)(
422[]
)(
422[ 222
2
22222
2
22 


yx
Cy
yx
CK
yx
Cx
yx
CK
y
v
x
u
).()(tan22obtaintoIntegrate
2222
1
2222
bAnsconst
x
yCKxy
yx
CyKy
x
vand
yx
CxKx
y
u






356 Solutions Manual  Fluid Mechanics, Fifth Edition 
 
25 mi, the velocity is 100 mi/h and the pressure 99 kPa, calculate the velocity and pressure 
at r = 100 mi. 
Solution: (a) First, convert Uo = 100 mi/h = 44.7 m/s and R = 25 mi = 40,200 m. The 
velocities are calculated from , as requested in Prob. P4.58: 
 1( ) 0 ; ( ) .( )or o o
U R
v U R v U R Ans a
r r r 
      
 
Outside the “intense” region, the wind is simulated as a circulating “potential vortex” 
whose velocity drops off inversely as the radius. (b) The flow is irrotational, otherwise  
would not exist. Thus Bernoulli’s equation applies outside r = R, with no elevation change 
at the ocean surface. Take surface air density to be sea-level standard,  = 1.225 kg/m3. 
44.7At 4 100 , 11.2
4 4 4
o oU R U mr R mi v
R s      
2 2
1 1 2 2Bernoulli : , or :2 2
p V p V    
 2 2
2 10099,000 (1.225 / 2)(44.7) (1.225 / 2)(11.2) , Solve .( )mip p Pa Ans b    100,150
 
The pressure far from the storm is approximately sea-level standard pressure. 
 
 
 
P4.78 An incompressible, irrotational, two-dimensional flow has the following stream 
function in polar coordinates: 
 
 sin( ) , where and are constants.nAr n A n  
 
Find an expression for the velocity potential of this flow. 
 
 
 
 
 Chapter 4  Differential Relations for a Fluid Particle 357
 
Solution: Use  to find the velocity components, then integrate back to find . 
 
 
1
1
1 cos( ) ; Integrate : cos( ) ( )
1 1sin( ) [ sin( )]
Compare : 0 , and thus cos( ) constant .
n n
r
n n
n
v n Ar n Ar n f
r r
dfv n Ar n nAr n
r r r d
df Ar n Ans
d

    
   
 


      
         
  
 
 
 
 
 
 
P4.79 Study the combined effect of the 
two viscous flows in Fig. 4.16. That is, find 
u(y) when the upper plate moves at speed V 
and there is also a constant pressure gradient 
(dp/dx). Is superposition possible? If so, 
explain why. Plot representative velocity 
profiles for (a) zero, (b) positive, and 
(c) negative pressure gradients for the same 
upper-wall speed V. 
 
Fig. 4.16 
Solution: The combined solution is 
2 2
2
V y h dp yu 1 1
2 h 2 dx h
                
 
The superposition is quite valid because the 
convective acceleration is zero, hence what 
remains is linear: p  2V. Three 
representative velocity profiles are plotted 
at right for various (dp/dx). 
Fig. P4.79 
 
 
 
 
 
358 Solutions Manual  Fluid Mechanics, Fifth Edition 
 
P4.80 An oil film drains steadily down the side of a vertical wall, as shown. After an 
initial development at the top of the wall, the film becomes independent of z and of 
constant thickness. Assume that w  w(x) only that the atmosphere offers no shear 
resistance to the film. (a) Solve Navier-Stokes 
 
for w(x). (b) Suppose that film thickness and [ w/ x] at the wall are measured. Find an 
expression which relates  to this slope [ w/ x]. 
Solution: First, there is no pressure gradient  p/ z because of the constant-pressure 
atmosphere. The Navier-Stokes z-component is (d2w/dx2)  g, and the solution requires 
w  0 at x  0 and (dw/dx)  0 (no shear at the film edge) at x  . The solution is: 
. (a) NOTE: is negative (down)Ans wgxw x    2 )2 
The wall slope is dw/dx / , rearrange [ / | ] . (b)/wall wallg : g dw dx Ans        
 
 
 
 
P4.81 Modify the analysis of Fig. 4.17 to 
find the velocity v when the inner cylinder 
is fixed and the outer cylinder rotates at 
angular velocity o. May this solution be 
added to Eq. (4.146) to represent the flow 
caused when both inner and outer cylinders 
rotate? Explain your conclusion. 
Solution: We apply new boundary condi-
tions to Eq. (4.145) of the text: 
1 2
i 1 i 2 i
v C r C /r;
At r r , v 0 C r C /r


 
    
Fixed
Fluid: ρ, μ
ro
ri
υ
r
Ω θ
 
Fig. 4.17 
o o o 1 o 2 oAt r r , v r C r C /r     
1 2Solve for C and C . The final result: .Ans
 
  
i i
o o
o i i o
r/r r /rv r
r /r r /r
  
 Chapter 4  Differential Relations for a Fluid Particle 359
This solution may indeed be added to the inner-rotation solution, Eq. (4.146), because the 
convective acceleration is zero and hence the Navier-Stokes equation is linear. 
 
P4.82 A solid circular cylinder of radius 
R rotates at angular velocity  in a viscous 
incompressible fluid which is at rest far 
from the cylinder, as in Fig. P4.82. Make 
simplifying assumptions and derive the 
governing differential equation and boundary 
conditions for the velocity field v in the 
fluid. Do not solve unless you are obsessed 
with this problem. What is the steady-state 
flow field for this problem? 
 
Fig. P4.82 
Solution: We assume purely circulating motion: vz  vr  0 and /  0. Thus the 
remaining variables are v  fcn(r, t) and p  fcn(r, t). Continuity is satisfied identically, 
and the -momentum equation reduces to a partial differential equation for v: 
subject to v (R, t) R and v ( , t) 0 .Ans 
          
    
    2
v v v1 r
t r r r r
  
I am not obsessed with this problem so will not attempt to find a solution. However, at 
large times, or t  , the steady state solution is v  R2/r. Ans. 
 
 
 
P4.83 The flow pattern in bearing lubrication can be illustrated by Fig. P4.83, where a 
viscous oil (, ) is forced into the gap h(x) between a fixed slipper block and a wall 
moving at velocity U. If the gap is thin, ,h L it can be shown that the pressure and 
velocity distributions are of the form p  p(x), u  u(y),   w  0. Neglecting gravity, 
reduce the Navier-Stokes equations (4.38) to a single differential equation for u(y). What 
are the proper boundary conditions? Integrate and show that 
21 ( ) 1
2
dp yu y yh U
dx h
       
where h  h(x) may be an arbitrary slowly varying gap width. (For further information on 
lubrication theory, see Ref. 16.) 
360 Solutions Manual  Fluid Mechanics, Fifth Edition 
 
 
Fig. P4.83 
Solution: With u  u(y) and p  p(x) only in the gap, the x-momentum equation becomes 
2 2
2 2
du dp u d u 1 dp0 , or: constant
dt dx dxy dy
        
2
1 2
1 dp yIntegrate twice: u C y C , with u(0) U and u(h) 0
dx 2     
With C1 and C2 evaluated, the solution is exactly as listed in the problem statement: 
.Ans   
21 dp yu y yh) U 1
2 dx h
     
 
P4.84 Consider a viscous film of liquid 
draining uniformly down the side of a 
vertical rod of radius a, as in Fig. P4.84. 
At some distance down the rod the film 
will approach a terminal or fully 
developed draining flow of constant outer 
radius b, with z  z(r),   r  0. 
Assume that the atmosphere offers no 
shear resistance to the film motion. Derive 
a differential equation for z, state the 
proper boundary conditions, and solve for 
the film velocity distribution. How does 
the film radius b relate to the total film 
volume flow rate Q? 
 
Fig. P4.84 
 Chapter 4  Differential Relations for a Fluid Particle 361Solution: With vz  fcn(r) only, the Navier-Stokes z-momentum relation is 
2z
z
dv p0 g v ,
dt z
        
2
z
z 1 2
1 d dv g gror: r , Integrate twice: v C ln(r) C
r dr dr 4
 
 
          
zvThe proper B.C. are: u(a) 0 (no-slip) and (b) 0 (no free-surface shear stress)
r
   
 The final solution is .Ans   
2
2 2
z
gb r gv ln r a
2 a 4
       
b
z
a
The flow rate is Q v 2 r dr
bwhere .
a
Ans


 

     
4
4 2 4ga 3 1 4 4 ln ,
8
    
 
 
P4.85 A flat plate of essentially infinite 
width and breadth oscillates sinusoidally in its 
own plane beneath a viscous fluid, as in 
Fig. P4.85. The fluid is at rest far above the 
plate. Making as many simplifying assump-
tions as you can, set up the governing 
differential equation and boundary conditions 
for finding the velocity field u in the fluid. Do 
not solve (if you can solve it immediately, 
you might be able to get exempted from the 
balance of this course with credit). 
 
Fig. P4.85 
Solution: Assume u  u(y, t) and p/x  0. The x-momentum relation is 
2 2
x 2 2
u u u p u uu v g ,
t x y x x y
            
               
2
2
u uor: 0 0 0 0 0 , or, finally:
t y
   
              
subject to: .Ans     
2
o2
u u u(0, t) U sin t and u , t 0.
t y
       
 
P4.86 SAE 10 oil at 20C flows between parallel plates 8 mm apart, as in Fig. P4.86. A 
mercury manometer, with wall pressure taps 1 m apart, registers a 6-cm height, as shown. 
Estimate the flow rate of oil for this condition. 
Solution: Assuming laminar flow, this geometry fits Eqs. (4.143, 144) of the text: 
 
Fig. P4.86 
2
avg max
2 dp hV u , where h plate half-width 4mm
3 dx 3
       
For SAE 10W oil, take   870 kg/m3 and   0.104 kg/ms. The manometer reads 
p  (Hg – oil)gh  (13550 – 870)(9.81)(0.06)  7463 Pa for x  L  1m 
2 2p h 7463 Pa (0.004) mThen V 0.383 
x 3 1 m 3(0.104) s
       
The flow rate per unit width is Q VA (0.383)(0.008) .Ans  
3m0.00306 
s m 
NOTE: The Reynolds number, based upon plate half-width, is 16, laminar. 
 
P4.87 SAE 30W oil at 20°C flows 
through the 9-cm-diameter pipe in Fig. 
P4.87 at an average velocity of 4.3 m/s. (a) 
Verify that the flow is laminar. (b) 
Determine the volume flow rate in m3/h. 
(c) Calculate the expected reading h of the 
mercury manometer, in cm. 
Solution: (a) Check the Reynolds number. 
For SAE 30W oil, from Appendix A.3,   
891 kg/m3 and   0.29 kg/(ms). Then 
2.5 m
D = 9 cm
Hg
SAE 30W Oil
h
V
 
Fig. P4.87 
 Chapter 4  Differential Relations for a Fluid Particle 363
Red  Vd/  (891 kg/m3)(4.3 m/s)(0.09 m)/[0.29 kg/(ms)]  1190  2000 Laminar Ans. (a) 
(b) With average velocity known, the volume flow follows easily: 
Q  AV  [(/4)(0.09 m)2](4.3 m/s)(3600 s/h)  98.5 m3/h Ans. (b) 
(c) The manometer measures the pressure drop over a 2.5 m length of pipe. From 
Eq. (4.147), 
2 2(0.045 )4.3 , 12320 
8 2.5 8(0.29 / )
m p R p mV solve for p Pa
s L m kg m s
      
12320 ( ) (13550 891)(9.81) , Solvemano merc oilp gh h       h 0.099 m Ans. (c) 
 
P4.88 The viscous oil in Fig. P4.88 is set 
into steady motion by a concentric inner 
cylinder moving axially at velocity U 
inside a fixed outer cylinder. Assuming 
constant pressure and density and a purely 
axial fluid motion, solve Eqs. (4.38) for the 
fluid velocity distribution vz(r). What 
are the proper boundary conditions? 
Fig. P4.88 
Solution: If vz  fcn(r) only, the z-momentum equation (Appendix E) reduces to: 
2z z
z z
dv p d dvg v , or: 0 0 0 r
dt z r dr dr
   
            
The solution is vz  C1 ln(r)  C2, subject to vz(a)  U and vz(b)  0 
Solve for C1  U/ln(a/b) and C2  –C1 ln(b) 
The final solution is: .Ansz ln(r/b)v U ln(a/b) 
 
 
P4.89 Modify Prob. 4.88 so that the outer cylinder also moves to the left at constant 
speed V. Find the velocity distribution z(r). For what ratio V/U will the wall shear stress 
be the same at both cylinder surfaces? 
364 Solutions Manual  Fluid Mechanics, Fifth Edition 
 
 
Solution: We merely modify the boundary conditions for the known solution in 4.88: 
vz  C1 ln(r)  C2, subject to vz(a)  U and vz(b)  –V 
Solve for C1  (U  V)/ln(a/b) and C2  U – (U  V)ln(a)/ln(a/b) 
The final solution is =  z ln(r/a)v U (U V) ln(a/b) Ans. 
The shear stress   (U  V)/[r ln(a/b)] and is never equal at both walls for any ratio of 
V/U unless the clearance is vanishingly small, that is, unless a  b. Ans. 
 
P4.90 It is desired to pump ethanol at 20C through 25 meters of straight smooth 
tubing under laminar-flow conditions, Red = Vd/ < 2300. The available pressure drop 
is 10 kPa. (a) What is the maximum possible mass flow, in kg/h? (b) What is the 
appropriate diameter? 
Solution: For ethanol at 20C,  = 789 kg/m3 and  = 0.0012 kg/m-s. From Eq. (4.138), 
 
4 4
laminar 8 128
R p d pQ
L L
 
 
   
Clearly, flow increases with diameter, so maximum mass flow requires the maximum 
diameter consistent with the maximum Reynolds number. The Reynolds number may be 
written out: 
 
34
2
2 2
3 3
max
4 4Re ( )( ) 2300
128 32
2300(32) 2300(32)(0.0012) (25)Or : 3.36 7
(789) (10,000)
0.00695 .( )
d
d pVd Q d p
d d L L
Ld E m
p
Solve for d m mm Ans b
   
      


    
   
  7
 
The maximum mass flow is 
4
max max 3
(0.00695 ) 10,000(789 )[ ]( ) 0.0151 .( )
128(0.0012 / ) 25
kg m Pa kg kgm Q Ans a
kg m s m s hm


    54
 
 Chapter 4  Differential Relations for a Fluid Particle 365
Light liquids like ethanol stay laminar only for tiny diameters. To work the same problem 
with, say, SAE 30W oil,  = 0.29 kg/m-s, would result in dmax = 26 cm, or 37 times larger. 
The maximum oil mass flow would be nearly nine thousand times larger. 
 
P4.91 Consider 2-D incompressible steady Couette flow between parallel plates with 
the upper plate moving at speed V, as in Fig. 4.16a. Let the fluid be nonnewtonian, with 
stress given by 
 
,
2
c c c
xx yy xy yx
u v a u va a
x y y x
         
                    a and c are constants 
Make all the same assumptions as in the derivation of Eq. (4.140). (a) Find the velocity 
profile u(y). (b) How does the velocity profile for this case compare to that of a 
newtonian fluid? 
Solution: (a) Neglect gravity and pressure gradient. If u  u(y) and v  0 at both walls, 
then continuity specifies that v  0 everywhere. Start with the x-momentum equation: 
xyxx
x
u u pu v g
x y x x y
       
        
Many terms drop out because v  0 and xx and u/x  0 (because u does not vary with x). 
Thus we only have 
1 20, : ,2
c
xy d a du duor constant u C y C
y dy dy dy


           
 
The boundary conditions are no-slip at both walls: 
u(y  –h)  0  C1(–h)  C2; u(y  h)  V  C1(h)  C2, solve 1 2,2 2
V VC C
h
  
The final solution for the velocity profile is: 
. (a)Ansu y V V( ) y
2h 2
  
This is exactly the same as Eq. (4.140) for the newtonian fluid! Ans. (b) 
 
366 Solutions Manual  Fluid Mechanics, Fifth Edition 
 
 
 
 
 
 
P4.92 A tank of area Ao is draining in laminar flow through a pipe of diameter D and 
length L, as shown in Fig. P4.92. Neglecting the exit-jet kinetic energy and assuming the 
pipe flow is driven by the hydrostatic pressureat its entrance, derive a formula for the 
tank level h(t) if its initial level is ho. 
 
Fig. P4.92 
Solution: For laminar flow, the flow rate out is given by Eq. (4.147). A control volume 
mass balance shows that this flow out is balanced by a tank level decrease: 
4
where ( )
128out o
D p dhQ A p gh t
L dt
 
     
Thus we can separate the variables and integrate to find the tank level change: 
4
0
, : .
128
o
h t
oh
dh D g dt or Ans
h LA
 

       o o
D gh h t
LA
 
4
exp
128
 
 
 
P4.93 A number of straight 25-cm-long microtubes, of diameter d, are bundled together 
into a “honeycomb” whose total cross-sectional area is 0.0006 m2. The pressure drop 
from entrance to exit is 1.5 kPa. It is desired that the total volume flow rate be 1 m3/h of 
water at 20°C. (a) What is the appropriate microtube diameter? (b) How many 
microtubes are in the bundle? (c) What is the Reynolds number of each microtube? 
 Chapter 4  Differential Relations for a Fluid Particle 367
Solution: For water at 20°C,   998 kg/m3 and   0.001 kg/ms. Each microtube of 
diameter D sees the same pressure drop. If there are N tubes, 
3 4 4
41 (1500 ) 1.47 5 
3600 128 128(0.001 / )(0.25 )tube
m D p D PaQ NQ N N E N D
s L kg m s m
 

     
2
2
0.0006At the same time, /
( /4)bundle tube
mN A A
D  
Combine to find D2  2.47E6 m2 or D  0.00157 m and N  310 Ans.(a, b) 
With D known, compute V  Q/Abundle  Qtube/Atube  0.462 m/s and 
ReD  VD/  (998)(0.462)(0.00157)/(0.001)  724 (laminar) Ans. (c) 
 
P4.94 A long solid cylinder rotates steadily 
in a very viscous fluid, as in Fig. P4.94. 
Assuming laminar flow, solve the Navier-Stokes 
equation in polar coordinates to determine the 
resulting velocity distribution. The fluid is at rest 
far from the cylinder. [HINT: the cylinder does 
not induce any radial motion.] 
 
Solution: We already have the useful hint that vr = 0. Continuity then tells us that 
(1/r)v/ = 0, hence v does not vary with . Navier-Stokes then yields the flow. From Eq. 
D.6, the tangential momentum relation, with p/0 and v = f(r), we obtain Eq. (4.139): 
 
R 
r 

Fig. P4.94

368 Solutions Manual  Fluid Mechanics, Fifth Edition 
 
Rotating a cylinder in a large expanse of fluid sets up (eventually) a potential vortex flow. 
________________________________________________________________________ 
*P4.95 Two immiscible liquids of 
equal thickness h are being sheared 
between a fixed and a moving plate, 
as in Fig. P4.95. Gravity is neglected, 
and there is no variation with x. 
Find an expression for (a) the velocity at the 
interface; and (b) the shear stress in each fluid. Assume steady laminar flow. 
 
Solution: Treat this as a Ch. 4 problem (not Ch. 1), use continuity and Navier-Stokes: 
This tells us that there is no velocity v, hence we need only consider u(y) in Navier-Stokes: 
.,Finally;;,At
0hence,0,As
:Solution,)(1
2
2
2
2
1
2
12
Ans
r
RvRC
R
CRvRr
Cvr
r
CrCv
r
v
dr
dvr
dr
d
r






h
h
V 
Fixed 
x
y
1, 1 
2, 2 
Fig. P4.95
wallstheatslipnofor0thus;00:Continuity 


 constv
y
v
y
v
x
u
ybau
dy
ud
y
u
x
u
x
p
y
uv
x
uu







Thus
)0(000:or)()( 2
2
2,12
2
2
2
2,12,1 
 Chapter 4  Differential Relations for a Fluid Particle 369
The velocity profiles are linear in y but have a different slope in each layer. Let uI be the 
velocity at the interface. (a) The shear stress is the same in each layer: 
 
(b) In terms of the upper plate velocity, V, the shear stress is 
 
________________________________________________________________________ 
 
P4.96 Reconsider Prob. P1.44 and calculate (a) the inner shear stress and (b) the power 
required, if the exact laminar-flow formula, Eq. (4.140) is used. (c) Determine whether this 
flow pattern is stable. [HINT: The shear stress in (r, ) coordinates is not like plane flow.] 
 
Solution: The exact laminar-flow velocity is Eq. (4.140), and the shear stress is Eq. (D.9): 
 
 
Recall the data from Prob. P1.44: ri = 5 cm, ro = 6 cm, L = 120 cm,  = 0.86 kg/m-s (SAE 
50W oil), and i = 900 rev/min = 94.25 rad/s. At the inner cylinder, 
 
 
The moment and power required are 
 
).(forSolve
21
2
21 aAnsVuh
uV
h
u
I
II

 

).()(
21
21 bAns
h
V

 
2
( / ) ( / )
[ ]
( / ) ( / )
2
( ) [ ]
( / ) ( / )

   
  
   
o o
i i
o i i o
i i o
r
o i i o
r r r r
v r
r r r r
dv v r r
dr r r r r r r
2 2
2 94.25(0.05) 2(0.06)[ ] (0.86)[ ] .( )
( / ) ( / ) 0.06 / 0.05 0.05 / 0.06 (0.05)
     
i i o
inner
o i i o i
r r
Pa Ans a
r r r r r
531
2 22 (531) 2 (0.05) (1.20)
(94.25 / )(10.0 ) .( )
i i
i
M r L N m
Power M rad s N m watts Ans b
     
    
10.0
943
370 Solutions Manual  Fluid Mechanics, Fifth Edition 
 
 
The shear stress, moment, and power are all 31% larger than the approximate linear-
profile analysis of Prob. 1.44. 
(c) The stability of this flow is determined by Taylor’s criterion, Eq. (4.141): 
 
 
To finish this, we had to look up the density of SAE 50W oil,  = 902 kg/m3. 
_______________________________________________________________________ 
 
P4.97 For the pressure-gradient flow between two parallel plates of Fig. 4.12(b), reanalyze 
for the case of slip flow at both walls. Use the simple slip condition uwall = l (du/dy)wall, 
where l is the mean free path of the fluid. (a) Sketch the expected velocity profile. (b) Find 
an expression for the shear stress at each wall. (c) Find the volume flow between the plates. 
Solution: The velocity profile will be similar to Fig. 4.12b except for wall slip. Ans.(a) 
 
 
 
 
 
The no-slip analysis of Eq. (4.133) is correct up to the point of integration: 
The boundary conditions, however, give rise to new constants: 
y 
x 
u(y)u
u
p/x < 0 
Fig, P4.97 
y = +h 
y = -h
2
1 2
1
2
dp yu C y C
dx  
2
1 2
2
1 2
1At :
2
1 ( )At :
2
wall
wall
dp hy h u u C h C
dx
dp hy h u u C h C
dx
 
 
     
     
3 2 3 2
2 2
( ) (0.05)(0.06 0.05) (94.25) 490 1700 .( )
( / ) (0.86 / 902)
i o i ir r rTa Ans c 
      STABLE
 Chapter 4  Differential Relations for a Fluid Particle 371
Subtract these two and we find that C1 = 0, which is a symmetry condition. The slip 
velocity is 
Finally, the velocity profile with wall slip is given by 
 This is the shape that the writer attempted to sketch above in Ans.(a). 
(c) The volume flow between the plates, per unit width into the paper, is 
The added flow due to slip is proportional to twice the Knudsen number, 2Kn = 2  /h. 
______________________________________________________________________ 
 
 
P4.98 For the pressure-gradient flow in a circular tube, in Sect. 4.10, reanalyze for the 
case of slip flow at the wall. Use the simple slip condition vz, wall = l (dvz/dr)wall, where l is the 
mean free path of the fluid. (a) Sketch the expected velocity profile. (b) Find an expression for 
the shear stress at the wall. (c) Find the volume flow through the tube. 
 
 
Solution: (a) The velocity profile has equal slip u all around, as shown: 
 
2
2
2
2
1 1| ( ) ,
2
1whence ( ) ( 2 )
2
wall h
du dp dp hu h C
dy dx dx
dpC h h
dx
  

    
  
 
2 21 ( ) ( 2 ) .( )
2
dpu h h y Ans b
dx   
3
2 21 2 2[ ( )( 2 ) ( ) (1 ) .( )
2 3
h
h
dp h dpQ u dA h h y dy Ans c
dx dx h 


         
372 Solutions Manual  Fluid Mechanics, Fifth Edition 
 
 
 
 
 
 
 
(b) The analysis of the velocity is correct up to the first result after Eq. (4.136) of the text: 
Once again, C1 = 0 to avoid a logarithmic singularity at the centerline. The constant C2 is 
found from the slip boundary condition: 
The velocity profile thus is given by the slip-flow formula 
 
The volume flow, with slip, is given by 
The slip-flow correction factor is 4 times the Knudsen number, 4Kn = 4  /R. 
____________________________________________________________________________ 
r 
z 
vz(r)u
u
p/z < 0 
Fig, P4.98 
r = R
2
1 2ln( )4z
dp rv C r C
dz   
2
2
2
2
At : | ( ) ,
4 2
1whence ( ) ( 2 )
4
  

      
  
 

z
z r R
dvdp R dp Rr R v u C
dz dr dz
dpC R R
dz
2 21 ( ) ( 2 ) .( )
4z
dpv R R r Ans b
dz   
4
2 2
0
1[ ( ) ( 2 ) ]2 ( )(1 4 ) .( )
4 8
R
z
dp R dpQ v dA R R r r dr Ans c
dz dz R
          
 Chapter 4  Differential Relations for a Fluid Particle 373
FUNDAMENTALS OF ENGINEERING EXAM PROBLEMS: Answers 
Chapter 4 is not a favorite of the people who prepare the FE Exam. Probably not a single 
problem from this chapter will appear on the exam, but if some did, they might be like these: 
 
FE4.1 Given the steady, incompressible velocity distribution V  3xi  Cyj  0k, 
where C is a constant, if conservation of mass is satisfied, the value of C should be 
(a) 3 (b) 3/2 (c) 0 (d) –3/2 (e) –3 
 
FE4.2 Given the steady velocity distribution V  3xi  0j  Cyk, where C is a constant, 
if the flow is irrotational, the value of C should be 
(a) 3 (b) 3/2 (c) 0 (d) –3/2 (e) –3 
 
FE4.3 Given the steady, incompressible velocity distribution V  3xi  Cyj  0k, 
where C is a constant, the shear stress xx at the point (x, y, z) is given by 
(a) 3 (b) (3x  Cy) (c) 0 (d) C (e) (3  C)
 
FE4.4 Given the steady incompressible velocity distribution u = Ax, v = By, and w = 
Cxy, where (A, B, C) are constants. This flow satisfies the equation of continuity if A 
equals 
 (a) B , (b) B + C , (c) B – C , (d) – B , (e) –(B + C) 
 
FE4.5 For the velocity field in Prob. FE4.4, the convective acceleration in the x 
direction is 
 (a) A x2 , (b) A2 x , (c) B2 y , (d) B y2 , (e) C x2 y 
FE4.6 If, for laminar flow in a smooth straight tube, the tube diameter and length both 
double, while everything else remains the same, the volume flow rate will increase by a 
factor of 
 (a) 2 , (b) 4 , (c) 8 , (d) 12 , (e) 16 
 
 
374 Solutions Manual  Fluid Mechanics, Fifth Edition 
 
COMPREHENSIVE PROBLEMS 
C4.1 In a certain medical application, water at room temperature and pressure flows 
through a rectangular channel of length L  10 cm, width s  1 cm, and gap thickness 
b  0.3 mm. The volume flow is sinusoidal, with amplitude Qo  0.5 ml/s and frequency 
f  20 Hz, that is, Q  Qosin(2 f t). 
(a) Calculate the maximum Reynolds number Re  Vb/, based on maximum average 
velocity and gap thickness. Channel flow remains laminar for Re < 2000, otherwise it 
will be turbulent. Is this flow laminar or turbulent? 
(b) Assume quasi-steady flow, that is, solve as if the flow were steady at any given Q(t). 
Find an expression for streamwise velocity u as a function of y, , dp/dx, and b, where 
dp/dx is the pressure gradient required to drive the flow through the channel at flow rate Q. 
Also estimate the maximum magnitude of velocity component u. 
(c) Find an analytic expression for flow rate Q(t) as a function of dp/dx. 
(d) Estimate the wall shear stress w as a function of Q, f, , b, s, and time t. 
(e) Finally, use the given numbers to estimate the wall shear amplitude, wo, in Pa. 
Solution: (a) Maximum flow rate is the 
amplitude, Qo  0.5 ml/s, hence average 
velocity V  Q/A: 
30.5 6 / 0.167 /
(0.0003 )(0.01 )
Q E m sV m s
bs m m
   
max
(0.167)(0.0003)Re
(0.001/ 998)
 . (a)
Vb
Ans
 
  50 laminar
 
 
(b, c) The quasi-steady analysis is just like Eqs. (4.142144) of the text, with “h”  b/2: 
. (b, c)Ans
 
  
dp b dp b sb dpu y u Q u bs
dx dx dx
      
2 2 3
2
max max max
1 1 2, ,
2 4 2 4 3 12   
(d) Wall shear: ( ) . (d)wall
wall
du Ans
dy
     oQb dp Q f t
dx sb sb2 2
66 sin 2
2
  
(e) For our given numerical values, the amplitude of wall shear stress is: 
2 2
6 6(0.001)(0.5 6)
(0.01)(0.0003)
o
wo
Q E
sb
    3.3 Pa Ans. (e) 
 
 Chapter 4  Differential Relations for a Fluid Particle 375
 
C4.2 A belt moves upward at velocity V, 
dragging a film of viscous liquid of 
thickness h, as in Fig. C4.2. Near the belt, 
the film moves upward due to no-slip. At 
its outer edge, the film moves downward 
due to gravity. Assuming that the only non-
zero velocity is v(x), with zero shear stress 
at the outer film edge, derive a formula for 
(a) v(x); (b) the average velocity Vavg in 
the film; and (c) the wall velocity VC for 
which there is no net flow either up or 
down. 
(d) Sketch v(x) for case (c). 
Solution: (a) The assumption of parallel 
flow, u  w  0 and v  v(x), satisfies conti-
nuity and makes the x- and z-momentum 
equations irrelevant. We are left with the 
y-momentum equation: 
 
Fig. C4.2 
2 2 2
2 2 2
v v v v v v vu v w g
x y z y x y z
              
                
There is no convective acceleration, and the pressure gradient is negligible due to the free 
surface. We are left with a second-order linear differential equation for v(x): 
2 2
1 1 22 Integrate: Integrate again: 2
d v g dv g g xx C v C x C
dxdx
  
        
At the free surface, x  h,   (dv/dx)  0, hence C1  –gh/. At the wall, v  V  C2. 
The solution is 
. (a)Ansgh gv V x x    
2
2
 
(b) The average velocity is found by integrating the distribution v(x) across the film: 
2 3
0 0
1 1( ) . (b)
2 6
hh
avg
ghx gxv v x dx Vx Ans
h h
 
 
        
ghV  
2
3
 
(c) Since hvavg  Q per unit depth into the paper, there is no net up-or-down flow when 
. (c)AnsV gh   2 / 3 
376 Solutions Manual  Fluid Mechanics, Fifth Edition 
 
(d) A graph of case (c) is shown below. Ans. (d) 
 


