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SECTION LV. THE CALCULATIONS, 
Section | has described a step-by-step method by which the oropagation of the disturbance 
may be followed, provided that all parts of the fluid belong to the one adiabatic, This is not 
the case in the problem we wish to consider, because the shock wave raises the temperature more 
than does adiabatic compression to the same pressure. Thus, if thc pressure difference on the 
two sides of the shock wave were 30,000 kgn/ cm, the excess rise of temperature over that of the 
adiabatic compression, from Table 4, is 4c®, Tne shock wave, as it passes outwards, expands and 
rapidly becomes less intense; as a result, the excess rise In temperature becomes rapidly less. 
For example, when the difference In pressure has dropped to 20,000 kgm/om?, the excess rise in 
temperature is only 1°. 
We estimate that the shock wave has moved only about 12 cm. (and Is therefore at radius 
42 cm.) béfore the pressure difference on the two sides has dropped to 20,000 kg/cm’, 
The physica) properties of water at very high pressures in so far as they affect tne 
transmission of pressure waves Cannot vary very much in a few degrees. We therefore neglect tne 
heating caused by the shock wave in excess of the adiabatic heating, and this approximation 
enables us to make use of the simple mathematical process described in Section |. 
At time t = 0 seconds, therefore, we have a sphere of exploded gas at rest, of radius 
30 cm., at uniform pressure, temperature and density, and surrounded by water at 20°C and 
atmospheric oressure. A rarefaction wave sets off into the gas and a shock wave sets off into 
the water. we wish to find the pressure at the interface, Clearly, it is somewhere between 
92, the initial pressure In the gas, and 0.001, the initial pressure in the water, Since the 
Initial densities are about the same (1.565 and 1.0), we anticipate something about the average, 
Say p = 45. The proper value may be found as follows. 
Let us for the moment consider only the one-dimensional Riemann problem ano therefore 
ignore the changes in the functions P and Q caused by the "spherical" term 30u/r in the equation 
of continuity. Then in the narrow strip of jas in the neignbourhcod of the interface there wil) 
be very rapid changes of pressure and velocity, but the pressure and velocity are connected by 
the relation 
7 RON ae (a an 221 (15) 
where the suffix 9 stands for gas, since the value of P in this region is that which was originally 
in the undisturbed part of the gas. 
Now the pressure and velocity are continuous across the interface, and from our work on 
shock waves in water (Section II!) we know the values of u corresponding with values of p, We 
therefore try various values of p and get the corresponding value of us By trying these values 
in (15) we soon get tne proper solution. The solution ts 
p = 37.4, u = 960. 
The corresponding shock wave velocity is 
U = 4OuO, 
After time T seconds, where T is a very short interval (we took 107° seconds) the shock 
wave has moved to 30 + U7, and the interface has moved to 30 + u 7, and the pressure and mass 
velocity between these two radli are constant, and are given by (16). 
The pressure and velocity distribution in tnat part of tne jas which has been affected 
during the time T may ?e obtained by the following arguments. From (15), at any point In the 
gas where the pressure al a particular instant Is p, the mass velocity u is f,(92) - f,(p). 
Now the Ingoing Q wave at this point has velocity V — u, where V is the velocity of sound at 
this pressure. Hence the value of Q corresponding with a particular p is propagaged Inwaras 
with a velocity depending only on p, namely V— (92) + f {p) or 1950 + 35.20 + f (p) = f (92), 
making use of the expression for V obtained in Section 11. The form of the tunctYon Q after 
time T may therefore be found. The value of P everywhere ir the gas is f_(92). Hence p and 
u may everywhere be determined, 2 
AS @ sees 
