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= \u2212 \u2212 = =
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\u20131}. Ans.

4.21 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
\u3c0 \u3c0\u3c1 \u3c1 \u3c1 \u3c1= \u2248 \u2248
To avoid compressibility corrections, we require (Eq. 4.18) that Ma \u2264 0.3 or, in this case,
the highest velocity (at the exit) should be Ve \u2264 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

268 Solutions Manual \u2022 Fluid Mechanics, Fifth Edition

4.22 A flow field in the x-y plane is described by u = Uo = constant, v = Vo = constant.
Convert these velocities into plane polar coordinate velocities, vr and v\u3b8.

Solution: Each pair of components must
add to give the total velocity, as seen in the
sketch at right.
The geometry of the figure shows that
Ans.\u3b8
v U \u3b8 V \u3b8
v U \u3b8 V \u3b8
r o o
o o
cos sin ;
sin cos
= +
= \u2212 +

Fig. P4.22

4.23 A tank volume V contains gas at conditions (\u3c10, p0, T0). 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 \u201cblowdown\u201d 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 CpA, where C constant
dt dt
\u3c1\u3c5+ = = \u2212 = \u2212 =\ufffd \ufffd
o
p(t) t
o
o p o
CRT Ap dpIntroduce and separate variables: dt
RT p
\u3c1
\u3c5
= = \u2212\ufffd \ufffd
The solution is an exponential decay of tank density: p = po exp(\u2013CRToAt/\u3c5). Ans.

4.24 Reconsider Fig. P4.17 in the following general way. It is known that the boundary
layer thickness \u3b4(x) increases monotonically and that there is no slip at the wall (y = 0).
Further, u(x, y) merges smoothly with the outer stream flow, where u \u2248 U = constant
outside the layer. Use these facts to prove that (a) the component \u3c5(x, y) is positive
everywhere within the layer, (b) \u3c5 increases parabolically with y very near the wall, and
(c) \u3c5 is a maximum at y = \u3b4.
Solution: (a) First, if \u3b4 is continually increasing with x, then u is continually decreasing
with x in the boundary layer, that is, \u2202 u/\u2202 x < 0, hence \u2202 v/\u2202 y = \u2013\u2202 u/\u2202 x > 0 everywhere. It
follows that, if \u2202 v/\u2202 y > 0 and v = 0 at y = 0, then v(x, y) > 0 for all y \u2264 \u3b4. Ans. (a)
Chapter 4 \u2022 Differential Relations for a Fluid Particle 269
(b) At the wall, u must be approximately linear with y, if \u3c4w \u2265 0:
u df df v u dfNear wall: u y f(x), or y , where 0. Then y
x dx dx y x dx
\u2202 \u2202 \u2202
\u2202 \u2202 \u2202
\ufffd \ufffd
\u2248 = < = \u2212 = \ufffd \ufffd\ufffd \ufffd

2
0
df df yThus, near the wall, v y dy . (b)
dx dx 2
y
Ans\ufffd \ufffd \ufffd \ufffd\u2248 \u2248\ufffd \ufffd \ufffd \ufffd\ufffd \ufffd \ufffd \ufffd\ufffd Parabolic
(c) At y = \u3b4, u \u2192 U, then \u2202 u/\u2202 x \u2248 0 there and thus \u2202 v/\u2202 y \u2248 0, or v = vmax. Ans. (c)

4.25 An incompressible flow in polar
coordinates is given by
2v cos 1r
bK
r
\u3b8 \ufffd \ufffd= \u2212\ufffd \ufffd\ufffd \ufffd
2v sin 1
bK
r
\u3b8 \u3b8
\ufffd \ufffd
= \u2212 +\ufffd \ufffd\ufffd \ufffd

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\u3b8\u2202\u2202 \u2202 \u2202\u3b8 \u3b8\u2202 \u2202\u3b8 \u2202 \u2202\u3b8
? \ufffd \ufffd \ufffd \ufffd\ufffd \ufffd \ufffd \ufffd
+ = = \u2212 + \u2212 + =\ufffd \ufffd \ufffd \ufffd\ufffd \ufffd \ufffd \ufffd

\ufffd \ufffd \ufffd \ufffd

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 = \u221ab, as shown above. The pattern
represents inviscid flow of a uniform stream past a circular cylinder (Chap. 8).

4.26 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\u2019s frictionless
momentum equation (4.36) in streamline
coordinates becomes

1
s
pV VV g
t s s
\u2202\u2202 \u2202
\u2202 \u2202 \u3c1 \u2202+ = \u2212 + (1)

2 1
n
V pV g
t R n
\u2202\u3b8 \u2202
\u2202 \u3c1 \u2202\u2212 \u2212 = \u2212 + (2)

Fig. P4.26
270 Solutions Manual \u2022 Fluid Mechanics, Fifth Edition

Further show that the integral of Eq. (1) with respect to s is none other than our old friend
Bernoulli\u2019s equation (3.76).
Solution: This 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
\u2202 \u3b8\u3c1 \u3c1\u3c1\u2202 + = \u2212 + = \u2212 \u2212 = \u2212 \u2212
Integrate from 1 to 2: (Bernoulli) .Ans\ufffd \ufffd
\u2202
\u3c1\u2202
2 22 2
2 1
2 1
1 1
V V dpV ds g z z 0
t 2
\u2212
+ + + ( \u2212 ) =

4.27 A frictionless, incompressible steady-flow field is given by
V = 2xyi \u2013 y2j
in arbitrary units. Let the density be \u3c10 = constant and neglect gravity. Find an expression
for the pressure gradient in the x direction.
Solution: For this (gravity-free) velocity, the momentum equation is
2
ou v p, or: [(2xy)(2y ) ( y )(2x 2y )] p
x y
\u2202 \u2202\u3c1 \u3c1\u2202 \u2202
\ufffd \ufffd
+ = \u2212 + \u2212 \u2212 = \u2212\ufffd \ufffd\ufffd \ufffd
V V i i j\u2207 \u2207
2 3
oSolve for p (2xy 2y ), or: .Ans\u3c1= \u2212 + 2o
pi j 2xy
x
\u2207 = \u2212\u2202 \u3c1\u2202

4.28 If z is \u201cup,\u201d what are the conditions on constants a and b for which the velocity
field u = ay, \u3c5 = bx, w = 0 is an exact solution to the continuity and Navier-Stokes
equations for incompressible flow?
Solution: First examine the continuity equation:
?u v w 0 (ay) (bx) (0) 0 0 0 for all and
x y z x y z
a b\u2202 \u2202 \u2202 \u2202 \u2202 \u2202\u2202 \u2202 \u2202 \u2202 \u2202 \u2202+ + = = + + = + +
Given gx = gy = 0 and w = 0, we need only examine x- and y-momentum:
2 2
2 2
u u p u u p
u v [(ay)(0) (bx)(a)] (0 0)
x y x xx y
\u2202 \u2202 \u2202 \u2202 \u2202 \u2202\u3c1 µ \u3c1 µ\u2202 \u2202 \u2202 \u2202\u2202 \u2202
\ufffd \ufffd\ufffd \ufffd
+ = \u2212 + + = + = \u2212 + +\ufffd \ufffd\ufffd \ufffd
\ufffd \ufffd \ufffd \ufffd

2 2
2 2
v v p v v p
u v [(ay)(b) (bx)(0)] (0 0)
x y y yx y
\u2202 \u2202 \u2202 \u2202 \u2202 \u2202\u3c1 µ \u3c1 µ\u2202 \u2202 \u2202 \u2202\u2202 \u2202
\ufffd \ufffd\ufffd \ufffd
+ = \u2212 + + = + = \u2212 + +\ufffd \ufffd\ufffd \ufffd
\ufffd \ufffd \ufffd \ufffd

Chapter 4 \u2022 Differential Relations for a Fluid Particle 271
p pSolve for abx and aby, or:
x y
Ans.\u2202 \u2202\u3c1 \u3c1\u2202 \u2202= \u2212 = \u2212
2 2p ab(x y ) const
2
= \u2212 + +
\u3c1

The given velocity field, u = ay and v = bx, is an exact solution independent of a or b. It
is not, however, an \u201cirrotational\u201d flow.

4.29 Consider a steady, two-dimensional, incompressible flow of a newtonian fluid
with the velocity field u = \u20132xy, v = y2 \u2013 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.
Solution: Evaluate and check the incompressible continuity equation:
( )0 2 2 0 0 au v w y y
x y z
\u2202 \u2202 \u2202
\u2202 \u2202 \u2202+ + = = \u2212 + + \u2261 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 u
u v w or
x y z x x y z
\u2202 \u2202 \u2202 \u2202 \u2202 \u2202 \u2202\u3c1 µ\u2202 \u2202 \u2202 \u2202 \u2202 \u2202 \u2202
\ufffd \ufffd\ufffd \ufffd
+ + = \u2212 + + +\ufffd \ufffd \ufffd \ufffd\ufffd \ufffd \ufffd \ufffd

2 2 2 3[ 2 ( 2 ) ( )( 2 )] (0 0 0), : 2 ( )p pxy y y x x or xy x
x x
\u2202 \u2202\u3c1 µ \u3c1\u2202 \u2202\u2212 \u2212 + \u2212 \u2212 = \u2212 + + + = \u2212 +
and, similarly for the y-momentum relation,
2 2 2
2 2 2 , :
v v v p v v v
u v w or
x y z y x y z
\u2202 \u2202 \u2202 \u2202 \u2202 \u2202 \u2202\u3c1 µ\u2202 \u2202 \u2202