## Example

a. Suppose that the magnetite of the previous example has been ground to give d50 = 0.008 in. (0.20 mm) and d85 = 0.012 in. (0.30 mm). As in the previous example, D = 1.64 ft (0.50 m) and Vm = 16.4 ft/s (5.0 m/s). In this case the volumetric concentration Cv = 0.20. Find the Vsm and the hydraulic gradient im.

Vsm is found as in the previous example. The pipe size is joined to the particle size of 0.008 in. (0.20 mm) and projected to the central axis to give the deposition velocity for sand-weight solids. Solids specific gravity is entered on the sloping line, and projected to the right-hand axis to give Vsm = 14.4 ft/s (4.4 m/s). As in the previous example this is less than the proposed operating velocity, which is satisfactory.

The flow of this slurry will be partially stratified, and from Eq. 24, M = [/n(0.30/0.20)]—1 = 2.5. However, the maximum limit of M is 1.70, which will be used here. From Eq. 25,

For this value of V50, with M = 1.7 and Vm = 16.4 ft/s (5.0 m/s), the right hand side of Eq. 22 becomes 0.098. (Sm — 1) equals (Ss — 1)(Cv) = (3.40)(0.2) = 0.68. Hence im = iw + (0.68)(0.098), and with iw of 0.033 (found in Example 2) im is found to be 0.100 (ft water per ft or m water/m). Note that this gradient is only about one-third that for the coarse particles of the previous example, despite the fact that the solids concentration (and the tonnage transported) is twice as much.

b. As in part (a) but with d85 = 0.016 in. (0.40 mm). The power M is re-evaluated as M = [/n(0.40/0.20)]—1 = 1.44. For the unchanged values of V50 and Vm, the right-hand side of Eq. 22 becomes 0.110 and im equals 0.033 + 0.68 (.110) or 0.108 (ft water/ft or m water/m).

c. As in part (a) but assume that there is a fine-particle fraction that increases the viscosity of the homogeneous component to 4 times that of water. From part (a), M = 1.70 and V50 = 10 ft/s (3.1 m/s) in water. Adjusting for the density difference from the sand-water case gives the abscissa of Figure 8 as 3.1/(3.40/1.65)1133 i.e. 7.88 ft/s (2.4 m/s). With this abscissa and a relative viscosity of 4, Figure 8 gives a multiple of 0.42, for a revised estimate of V50 as (0.42)(3.1) = 4.3 ft/s (1.3 m/s). Thus the right/hand side of Eq. 22 beocmes 0.022, and im = .033 + 0.015 or 0.048 (ft water/ft or m water/m). This revised estimate is only about one-half the value of 0.100 obtained in part (a) for pure water as the carrier fluid.

## Survival Treasure

This is a collection of 3 guides all about survival. Within this collection you find the following titles: Outdoor Survival Skills, Survival Basics and The Wilderness Survival Guide.

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