1 2(k 1) c 2 (5.33) Dividing the mass equation by the momentum
equation and combining it with the perfect gas model yields c1 2 kU1 +
U1 = c2 2 kU2 + U2 (5.34) Combining equation (5.33) and (5.34) results
in 1 kU1 k + 1 2 c 2 k 1 2 U1 + U1 = 1 kU2 k + 1 2
we allow RO to go to infinity in (6.45); O then tends to zero. In these
cases the potential vortex satisfies not only the Navier-Stokes equations
(this is so for all incompressible potential flows), but also the no slip
condition at the wall. Therefore we
the Navier-Stokes equations then reduces to p x = 2u y2 , (6.13)
and the y component reads 0 = 1 p y . (6.14) A consequence of the
last equation is that p can only be a function of x. However since by
assumption the right-hand side of (6.13) is not a func
coordinates are r : : z : 2(z) = 1 r p r + 1 r Prr r + 1 r2 (Prr
P) , (6.150) 0 = Pz z , and (6.151) 0 = p z + Pzz z . (6.152)
From (3.35) the friction stresses only depend on . But from (6.151) we
see that Pz = z is not a function of z and for symmetry
e /2 y ei(t /2 y) , (6.117) or u = U e /2 y cos(t
"/2 y) . (6.118) The separation h no longer appears in (6.118).
Measured in units = "2/ the upper wall is at infinity. Relative to the
variable y the solutions also have a wave form; we call these shearin
with a unidirectional flow. In a coordinate system whose z axis runs
parallel to the axis of the conduit, Poissons equation u = K , (6.72)
follows from (6.3) for the only nonvanishing velocity component (which
we shall denote by u) in steady flow. Since K
Unidirectional Flows 183 Two-dimensional channel flow corresponds to
c/b = 0, and we have f(c/b)=3/2. For c/b = 1 we obtain f(c/b)=0.89. For
an equilateral triangle of height h (Fig. 6.8), the velocity distribution is u
= K 1 4 h (y h) (3x2 y2) (6.93) and
given by an = 2 c c/2 c/2 K 2 1 4 c2 y2 cos(my) dy . (6.83)
Integrating leads to the Fourier expansion uP = 2 K c n=1 c m2
cos(m c/2) 2 m3 sin(m c/2) cos(m y) . (6.84) Because m c 2 = (2n 1)
2 , (6.85) the first term in brackets in (6.84) vanishes, and t
= 0) yields the Couette-Poiseuille flow (Fig. 6.1). As is directly obvious
from (6.19), the general case is a superposition of Couette flow and
Poiseuille flow. Since the unidirectional flows are described by linear
differential equations, the superpositi
(eit) has physical meaning. Instead of (6.12) we now have u = f(y, t) , v
= 0 (6.102) and instead of (6.13): u t = 1 p x + 2u y2 . (6.103)
We set p/x = 0 , i.e. the flow is only kept in motion by the wall
velocity through the no slip condition u(0, t) = u
pressure relative to the hydrostatic pressure distribution, we extract
from (2.38b) 0 = T . (6.134) In component representation (see
Appendix B) and noting Pij (r) we find for the r component p r = 1 r
r (r Prr) P , (6.135) for the component p = 0, (6.13
% eit "h2/2 (1 + i) (1 y/h) "h2/2 (1 + i) & , (6.115) and deduce
that u = U cos(t) (1 y/h) = U (1 y/h) . (6.116) Equation (6.116)
corresponds to (6.4) where the upper plate represents the moving wall.
We also obtain this limiting case if the kinematic vis
therefore C1 = 2/ ; (6.130) thus the solution reads u/U = 1 2/
0 e2 d for t 0 . (6.131) 188 6 Laminar Unidirectional Flows The
integral erf ()=2/ 0 e 2 d (6.132) is the error function. For t = 0
we have and u/U = 0; thus the initial condition is satisfie
Unidirectional Flows 187 Equation (6.119) is a linear equation and since
U enters the problem only linearly from the boundary condition (6.120),
the field u(y, t) must be proportional to U, so that the solution has to be
of the form u/U = f(y, t, ) . (6.1
easily confirmed if, starting with (6.22) we construct the formula (6.71).
As can be seen, the pressure drop for the circular tube is very different
from the pressure drop for the ring gap, even when the hydraulic
diameter is used as the reference length.
Since the pressure gradient is constant, we may write K = p l = p1 p2 l
(6.59) and mean by p the pressure drop in the pipe over the length l .
The pressure drop is positive if the pressure gradient p/z is negative.
It is appropriate to represent this pres
6.8. Channels with rectangular and triangular cross-section 6.1 Steady
Unidirectional Flow 181 with the boundary conditions u( b 2 , y)=0 ,
(6.74a) and u(x, c 2 )=0 . (6.74b) To solve the linear equation (6.73) we
set u = uP + uH , (6.75) where uH satisfi
Uy 0 c.v. Ux 0 (a) Stationary coordinates y Py Uy = Us + Uy 0 Ty
c.v. Ux = Us + Ux 0 Upstream (b) Moving coordinates Fig. 5.13: A
shock as a result of a sudden and partially a valve closing or a narrowing
the passage to the flow The totally closed valve i
temperature behind the shock. Assume the specific heat ratio is 1.3.
SOLUTION It can be observed that the gas behind the shock is moving
while the gas ahead of the shock is still. Thus, it is the case of a shock
moving into still medium (suddenly opened v
tube. Note however that exit Mach number, M2 < 1 and is not 1. A back
pressure that is at the critical point c results in a shock wave that is at the
exit. When the back pressure is below point c, the tube is clean of any
shock 13 . The back pressure belo
Approaching the shock location from the upstream (entrance) yields A =
A A A = 2.3396 3 = 7.0188[cm2 ] Note, as simple check this
value is larger than the value in the previous example. 6.1 Nozzle
efficiency Obviously nozzles are not perfectly efficient a
u x + v u y + w u z = 1 p x + 2u x2 + 2u y2 + 2u z2 .
(6.7) Because of (6.4), all the convective (nonlinear) terms on the lefthand side vanish. This is the case in all unidirectional flows. Of course
since we are dealing with a two-dimensional flow we cou
CouettePoiseuille flow, which as we shall see is here zero. The flow is
not driven by the pressure gradient but by the volume body force of
gravity, whose components are fx = kx = g sin , (6.25a) fy = ky = g
cos . (6.25b) Because of (6.6) and v = 0 the Na
Poiseuille flow if we subject the general solution (6.50) to the boundary
conditions (Fig. 6.7) u(RO)=0 , (6.64a) and u(RI ) = U . (6.64b) The
resulting flow is clearly the Couette-Poiseuille flow in a ring gap, and is
given by u(r) = K 4 % R2 O r2 R2 O R
then given by the coordinate surfaces r = RI and r = RO. In the axial
direction the flow extends to infinity. Changes in flow quantities in the
axial direction must therefore vanish or be periodic so that these
quantities do not take on infinite values at
From here on we shall write p in place of pdyn, and shall understand
that in all problems without free surfaces, p means the pressure
difference p pst . If the problem being dealt with does contain free
surfaces, we shall, without further explanation, mak
allow closed form solutions even for nonNewtonian fluids. As has
already been discussed in Sect. 4.4, this solvability rests on the
particularly simple kinematics of these flows. Here we shall restrict
ourselves to incompressible flows for which only pres
crosssection we have dh = 4 (R2 O R2 I ) 2 (RO + RI ) = dO dI .
(6.68) We first write the loss factor in the form = p (dO dI )2 2 U2
d2 h , (6.69) into which we replace one U by (6.66) (from (6.59) and
extract = 64 U dh l dh 1 dI dO 2 ln dI dO 1 dI dO 2 +
force is balanced. Equation (6.40) is coupled with (6.41): if the velocity
distribution is given by (6.41), then the pressure distribution
corresponding to it follows from (6.40). Equation (6.41) is a linear
ordinary differential equation with variable co
(1) = u y (2) . (6.32) If we ignore the effect of the friction in the air,
the left-hand side of (6.32) vanishes and this boundary condition reads
0 = u y y=h . (6.33) From integrating (6.27) we obtain p = gy cos
+ C(x) , (6.34) and with the boundary con
Einsteins theory of General Relativity. Fig. 1.7: The Photo of the bullet
in a supersonic flow that Mach made. Note it was not taken in a wind
tunnel Machs revolutionary experiment demonstrated the existence of
the shock wave as shown in Figure 1.7. It is
to the understanding of the unique phenomena of compressible flow.
These unique issues of compressible flow are to be emphasized and
shown. Their applicability to real world processes is to be 9Please read
the undersigneds book Fundamentals of Die Casting
in a very narrow width. The chemical reactions (even condensation) are
neglected, and the shock occurs at a very narrow section. Clearly, the
isentropic transition assumption is not appropriate in this case because
the shock wave is a discontinued area. T
pressure should be P = 21.76780 8.4826 = 2.566[bar] Note that tables
in this example are for k = 1.31 Example 8.2: A flow of gas was
considered for a distance of 0.5 [km] (500 [m]). A flow rate of 0.2
[kg/sec] is required. Due to safety concerns, the maxi
treated as verbatim copying in other respects. If the required texts for
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ones listed (as many as fit reasonably) on the actual cover, and continue
the rest onto adjacent pages. If yo