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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 
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 
p(t) t
o p o
CRT Ap dpIntroduce and separate variables: dt
RT p
= = \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
df df yThus, near the wall, v y dy . (b)
dx dx 2
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
\u3b8 \ufffd \ufffd= \u2212\ufffd \ufffd\ufffd \ufffd 
2v sin 1
\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 
pV VV g
t s s
\u2202\u2202 \u2202
\u2202 \u2202 \u3c1 \u2202+ = \u2212 + (1) 
2 1
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: 
V dp dp dpds V dV g ds gsin ds g dz
\u2202 \u3b8\u3c1 \u3c1\u3c1\u2202 + = \u2212 + = \u2212 \u2212 = \u2212 \u2212 
Integrate from 1 to 2: (Bernoulli) .Ans\ufffd \ufffd
2 22 2
2 1
2 1
1 1
V V dpV ds g z z 0
t 2
+ + + ( \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 
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
\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
= \u2212 + +
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