Search

27 Apr 2006: 2.18), which can be (2.23) and (2.24) more closely.used in (2.22) to show that Segregation is more sensitive to the value of. ... 4.24)mass reduce to and we use this condition together with (4.21)—.

30 May 2006: r 2 ) ( r < 1) 2.94 0.93 1.27 1.00 {(‘i ( r > 1 ). - I Cylindrical Gaussian e-T2 ( f < 00) 3.13 1-12 2.24 0.86. ... If h,R-l > Fmin (2.24) t It is interesting to compare these values with the relationship h0R-l = 2 deduced by.

26 Apr 2006: Leppäranta, pp. 24–104,Univ. of Helsinki Press, Helsinki, 1998. Wettlaufer, J. S., Heat flux at the ice-ocean interface, J. ... E. Huppert, The phase evolution ofyoung sea ice, Geophys. Res. Lett., 24, 1251–1254, 1997a.

22 Jan 2007: Inserting equations (2.23) and (2.24) into equations (2.12)–(2.16) andlinearizing in the perturbed quantities leads to the following nonlinear ordinarydifferential equation for the disturbance amplitude:.

11 Apr 2013: In a related context, the classical modelhas been modified to replace the sharp interface by a capillarytransition zone [24–26]. ... Porous Media 18, 107 (1995).[24] L. W. Lake, Enhanced Oil Recovery (Prentice-Hall, Englewood.

12 Jun 2006: g. [23,24]). The major difficulties then consist in the formulation of accurate internalboundary conditions and in following the evolution of the interfaces through time. ... 104 This study 19.6 16.2 3.53 0.585 2.24[2] 19.51 16.2 3.53 0.586 2.243[3]

30 May 2006: 24-2. 372 H. E. Huppert and 144. E. Stern. $1. , //,/,////,//////////////,,,/,////// / ///,/I/ // / / / , // , / , / // , / / / / // / /, /' I -. FIGURE 1. An elevation of the channel, bounded by vertical walls at. ... Moh(x/L) e--ikg. (3.24). While in

27 Apr 2006: Subsequently (ap- proximately 40-50 rain from the beginning),. 24 r. KOYa_GUCHI L, kt. ... grade particles with A T = 2 4 C and Co= 1.24 wt.%.

21 Nov 2014: 19:. 20:. 21:. 22:. 23:. 24:. 25:. 26:. 27:. 28:. 29:.

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.