Geology Reference
In-Depth Information
and
v t = p t / c (5.15)
in Equation 5.12, we have
p r /p i = ( A 1 - A 2 )/(A 1 + A 2 ) (5.16)
Thus, in the limit of equal areas when A 1 = A 2 , the magnitude of the reflected
pressure wave vanishes. If the areas are unequal, p r is completely determined,
since the incident pressure p i and both cross-sectional areas are known. Then,
the magnitude of the transmitted pressure wave can be obtained from Equation
5.10. In fact, it is easily seen that p t = p i + p r = p i (1 + p r /p i ) or
p t /p i = 1 + ( A 1 - A 2 )/(A 1 + A 2 ) = 2A 1 /(A 1 + A 2 ) (5.17)
5.1.6 Example 5-2. Classical water hammer.
Equation 5.13 is the well known formula for one-dimensional “water
hammer,” and it can be interpreted as follows. If a piston mounted at the end of
a semi-infinite pipe containing a stationary fluid with density and speed of
sound c is struck with a velocity v, a propagating pressure pulse with magnitude
p = vc is created. The fluid, being compressible, does not move as a rigid
column, which would require infinite energy. Rather, it remains undisturbed
except during the passage of the wave, when local fluid particles are
momentarily displaced a distance u with the speed v = u/ t. Pressure, fluid
velocity, and particle displacement all propagate together at the speed of sound c
(note c >> v ! ).
5.1.7 Example 5-3. Acoustic pipe resonances.
Again, recall that the displacement u(x,t) satisfies 2 u/ t 2 - c 2
2 u/ x 2 = 0 ,
where p(x,t) = - B u/ x is the acoustic pressure, and c 2 = B /
. As in Chapter 1,
we take
u(x,t) = X(x)T(t)
(5.18)
This separation of variables leads to
T"(t)/c 2 T(t) = X"(x)/X(x) = - p 2 < 0 (5.19)
and the solution X(x) = A sin p x + B cos p x. The function X(x) is proportional
to the fluid displacement u(x,t), and it vanishes at closed ends. On the other
hand, note that
p(x,t) = - BX'(x)T(t) (5.20)
Thus, p(x,t) is proportional to the spatial derivative X'(x). Our finite system is
defined on 0 < x < L.
Search WWH ::




Custom Search