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Encyclopedia of Physical Science and Technology EN007E-968 June 30, 2001 17:35

High-Pressure Synthesis (Chemistry) 369

2
P 0 on the face area πR . Along the flank of the piston, larger than indicated by the equations. Also, it is neces-
0
the supporting pressure P(h) falls to zero at h = h 0 .We sary to develop the necessary gasket support at all cham-
require that at any h along the flank, the cross-sectional ber pressures higher than S as the pressure is raised and
2
2
area of the piston, πh sin α, bears a net stress less than lowered.
the simple compression strength of the material, S. The Instead of using two moving pistons to compress ma-
net stress is the total load on the cross section divided by terial inside a relatively stationary cylinder, one can re-
its area, minus the support pressure P(h). The total load place the cylinder with an array of pistons with gaskets
is the piston face load plus the piston flank load. between them so that the entire apparatus consists of pis-
In symbols this requirement is tons moving toward each other. The simplest example of
this device has four pistons arranged tetrahedrally. Six pis-
1
S + P(h) ≥ 2 tons can form a cube or a rectangular parallellepiped. The
2
πh sin pistons can be driven by separate rams or driven into ta-

h pered bearing rings, or the entire set, suitably sealed, can
2 2
· P 0 πR + 2π sin hP dh (2)
0 be immersed in a pressurized liquid. The rewards for this
complexity are higher chamber pressures (especially on
h 0
For the maximum allowable flank pressure gradient the very compressible materials), a greater number of inde-
equality holds in Eq. (2). pendent electrical connections (one per piston), and more
By differentiating with respect to h we obtain: nearly isotropic compression of the chamber contents.
If we assume that the pistons of a multiple piston ap-
dP/dh =−2S/h (3) paratus are relatively incompressible compared with the
and high-pressure-chamber contents or the gaskets, then the
chamber volume V is compressed because the gaskets are
P = P 0 − 2S ln(h/h 0 ) (4) compressed and/or extruded. If we exclude extrusion for
the moment and regard the chamber as a sphere of radius
Equation (4) is based on the boundary condition that
r, its surface area is made of two parts: the area taken by
P(h 0 ) = P 0 , the chamber pressure, since it is physically
2
compressing gaskets, a = 4πr f , where f changes with
difficult to avoid this situation even though the piston tip
r, and the area taken by incompressible piston faces,
is thereby given more support than necessary.
2
A = 4πr − a. For a change dr, dA/dr = 0, which yields:
At h = h 1 , P = 0 and we have
da/dr = 8πr (8)
P 0 = 2S ln(h 1 /h 0 ) (5)
From this, we find:
which tells us that in principle any chamber pressure can
be confined, but the logarithmic dependence makes it slow d ln a = 2dr/rf (9)
going. If we regard h 1 /h 0 as a measure of the size of the and the ratio of compressions of chamber and gaskets is
3
apparatus, its volume V goes as (h 1 /h 0 ) ,sowehave
d ln V/d ln a = 3rf dr/2rdr = 1.5f (10)
V = Bexp(3P 0 /2S). (6)
For the chamber contents, d lnV = Kd ln p and for the
2
The required piston force F will go as (h 1 /h 0 ) so that gasket material, d ln a = kd ln p, where K and k are com-
pressibilities and p is the same in both the chamber and
F = C exp(P 0 /S) (7)
the gasket next to the chamber (to avoid overstressing the
where B and C are geometrical constants. The advantages pistons), thus
of a high S are obvious, but practical piston materials are
d ln V/d lna = K/k = 1.5f (11)
limited mostly to cemented tungsten carbide (S = 5GPa)
or various forms of diamond (S =∼10 GPa). The latter which tells us what the ratio K/k must be to make the
are available only in small (1 cm) sizes. pressure in the chamber increase as fast as the pressure in
Equations (1) to (7) are rather general and apply to the gasket next to the chamber for any value of r, f ,or p.
conical or pyramidal pistons. For example, if α = 45 and For a cylinder with fixed ends and shrinking radius, the
◦
we let (h 1 − h 0 )be6R 0 , the maximum chamber pressure analog of Eq. (11) is
is about 4S and approximately 14% of the piston force is
d ln V/d ln a = K/k = 2f (12)
due to chamber pressure; 86% is ideally taken by gasket
load. One might expect f to be about 0.2 and this tells us that
In practice, the ideal support gradient is difficult to K/k should be 0.3 or 0.4 for the sphere or cylinder, re-
achieve and the apparatus and the required force will be spectively. So suitable gasket material should be about