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9 Feb 2004: 870.0 461.1A25 9 3.0 8.0 24.0 500.0 0.432 1.957 939.0 479.8. ... 675.6 0.350 1.776 761.2 428.A13 53 3.0 8.0 24.0 500.0 0.432 1.957 939.0 479.

9 Feb 2004: r = (F/πV )1/2. (2.24)In contrast to the instantaneous, fixed volume release of the last section, there is not a finite radius at whichthe mass of particulate matter ... Applying these concepts to the relationship for the density of deposit for a single

9 Feb 2004: H 0 1!5 2F P12 f P1 1! 4.24! FIG. 2. ... On applicationthe boundary conditions4.23! and 4.24!, we find that. a 5 4.523 102 3, F P15 2 8.42 for Fr5 11.9.

9 Feb 2004: The volume fration of particles in suspension evolves as23,24. df. dt5. ... X5 x/l , 24! T5 t/ v sl /q!1/ a 2 1!, 25!

9 Feb 2004: 5H1 2(H0U1)′=0, (24)3U1 H ′1 H ′0ϕ1. 1. 2H0ϕ. ′1=0, (25)ϕ1=2720. ... 4y2)], (38). and then subsequent substitution into (24) yields a second-order differential equation forH1(y), namely(1.

10 Feb 2004: 24 February. Steve Roper. (ITG). Magma fracture in tough materials. 2 March.

9 Mar 2004: H.) 3–24 (Springer, Berlin, 1990). 5. Nakada, S., Shimizu, H. & Ohta, K.

9 Mar 2004: 6.3 6.3 24.3 5.89 1.00 3.86 9.6 6.1 3.8 0.7417 6.3 6.3 24.3 10.71 1.00 ... 86 9.6 8.2 5.2 0.8118 6.3 6.3 24.3 20.37 1.00 3.86 9.6 11.3 7.1 0.8919 13.2

9 Mar 2004: rNt 0 rN0 24. M. Ungarish, H.E. Huppert / International Journal of Multiphase Flow 25 (1999) 715737 725. ... For this case. the box model yields [rN/rN(0)]max=[8Fr/(3 b)]1/4. (This is by about 24% smaller than for.

9 Mar 2004: These substitutions transform (2.24) into. uN = I(hN) = Fr(hN) h1/2N. ... Appendix. The functions I and χA.1. The function I(hN). From the definitions (2.19), (2.24) and (2.25) we obtain.