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PROBLEM 1.50 1.49 In the structure shown, an 3- mm diameter pin is used at A, and 12 mm
diameter pins are used at B and D. Knowing that the ultimate shearing stress is 100
MPa at al
CV_ENG 3700- Fluid Mechanics
University of Missouri
Example Exam
2017 SP
Note: This example exam is only used to show how the exam problems will be constructed and to allow
the students read the first page carefully before the exam. The same questions wil
(9.172) we extract the equation u t + (u a) u x 1 a p t + (u
a) p x = 0 , (9.178) from which follow the two ordinary differential
equations du 1 a dp = 0 along dx = (u a) dt . (9.179) Ds/Dt = 0 (cf.
(4.48) clearly implies that the change in the entropy o
characteristics; the path of a particle is therefore also a characteristic.
The differential equations which are valid along these characteristics are
called characteristic or sometimes compatibility relations. As an
example of their application we shall
this, solutions from electrostatics, in particular, can be carried over to
fluid mechanics. 10.3.1 Simple Examples of Potential Flows In these
indirect problems we shall first examine solutions in the form of
polynomials. In this manner we are led to thre
considerably larger than the maximum velocity in steady flow (cf.
(9.101). Of course the result (9.211) does not contradict the energy
equation. If the piston is moved even faster, a region in which vacuum
prevails is generated between the piston and the
fact that as the piston is set suddenly in motion at the position x = 0 at
time t = 0 (thus in an infinitesimally short time), the gas must pass
through the whole velocity interval, from the undisturbed velocity u = 0
up to the velocity given by the kinem
calculate the flow in an infinitely long duct. Since the duct has no ends
we are dealing with a pure initial value problem. At time t = 0 the initial
condition in the duct is given by u(x, 0) and a(x, 0) (Fig. 9.27). We are
looking for the state of the fl
Potential Flow 329 Fig. 10.7. Flow in a right-angled corner The choice b
= a, that is c = 2a, in (10.62) leads to the potential of rotationally
symmetric stagnation point flow (Fig. 10.8) = a 2 (x2 + y2 2z2) ,
(10.70) whose velocity components are u = ax
(1.13). The streamlines are fixed in space for unsteady stagnation point
flow too, as we see directly from the fact that a does not arise in the
equations for the streamlines. In order to determine the pressure field
we now have to apply Bernoullis equati
Plane stagnation point flow is met close to the stagnation point (or
more exactly, the stagnation line) of a plane body (Fig. 10.6), but it is
only realized locally there, as we see from the fact that the incident flow
velocity tends to infinity as y . Wi
exhaust jets of rocket engines, because the temperature of the fluid
particles is raised by passing through the shock and then lowered again
by passing through the expansion waves, where the intrinsic luminosity
of the exhaust is altered in a correspondin
critical or sonic and denote them with the superscript *. These values
differ from the total values only by 286 9 Stream Filament Theory
constant factors and are therefore often used as reference values. In
particular for diatomic gases ( = 1.4), we find
characteristics by their tangents at these points. At the point where the
tangents cross we can determine the values of u and a by the above
method; but by doing this we again know the directions of the
characteristics in these points and can approximate
(ds/dh)F = 0 and through which the isentrope goes. From Gibbs
relation T ds = dh dp , (9.129) it follows that for this point dp dh F =
= p h s . (9.130) Using (9.128) and (9.121) we further have u2 d = dh
(9.131) or dh d F = u2 = h s (9.132) and, because
cv = 2 1 122 p2 p1 p1 3 . (9.159) This shows directly that for
calorically perfect gases, p2 p1 must always be greater than zero so
that only compression shock waves can occur, since otherwise the
entropy would have to decrease across the shock. 9.3 Unste
= 0. In order to evaluate the behavior at the singular point we calculate
the volume flux (for simplicity) through the surface of a sphere with
radius r, which we call the strength m of the source: m = (Ssph) u n dS
= (Ssph) xi ni dS = (Ssph) r dS . (10.8
= 0.). Under this assumption, the continuity equation reads P t + a2 0
2 xixi = 0 , (10.5) while Bernoullis equation assumes the form
t + P = 0 , (10.6) where the constant has been absorbed into the
potential. (10.6) corresponds to the linearized form of
and U2 /2 the dynamic pressure. If we were to place a pressure tap
at the stagnation point of the body in Fig. 10.10, we would measure the
stagnation pressure, as in (10.116). At a pressure tap on the (almost)
cylindrical part of the body some distance be
Of course, starting for example from the subsonic branch, we can apply
heat until M = 1 is reached, and then remove heat to move back along
the supersonic branch. If the Mach number M = 1 is reached at the exit
(2) in the duct flow in Fig. 9.20, the great
and experimental results agree well for small supersonic Mach
numbers. However we shall not go into the calculation of the shock
structure here since in practice it is usually enough to know the change
in flow quantities across the shock. We assume that c