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E%HIHIT D - APPENDIB II <br />• <br />CALCIILATIOH OF CAVITY VOLIIME <br />AND ESTRACTION <br />BASIS: <br />1. Nahco:lite ore density = 135 pounds per cubic foot <br />2. Insolubles in ore = 15 percent <br />3. Cavity temperature upon completion = 165 degrees F. <br />4. Saturated brine at 165 degrees F, contains 15.6 ~ <br />nahco:Lite <br />5. The volume of the ends of the horizontal cavity <br />' would be that of a frustrum of a cone. The formula <br />being Vol. = 1/3(height)[A + sq root of[( A x B ) <br />+B] =Ve <br />Where A = Area of the .Base and <br />B = Area of the Plane parallel to the Base <br />6. The volume of the base of the horizontal cavity <br />would be that of a quadrantal prism. The formula <br />being Vol. = length [(height squared) + ( base x <br />• height)] = Vb <br />7. The total cavity volume = Ve + Vb = Vt <br />EXAMPLE: <br />Calculate the volume of a cavity 1000 feet long at the <br />base, with 45 degree sides, 26 feet in height, a base <br />width of 6 feet and top width of 56 feet. <br />For Ve: A = 3.14(29x29) = 2,640.7 Cubic feet <br />B = 3.14(1.91x1.91) = 11.4 cubic feet <br />Ve = 1/3(26)[2640.7 +sq. root of (2640.7x11.4) + <br />11.4] <br />= 24,489 cubic feet <br />For Vb = 1,000[(26x26) + (6x26)] <br />= 832,000 cubic feet <br />For Vt = 24,489 + 832,000 = 856,489 cubic feet <br />To calculate t:he tonnage to be removed, follow this <br />procedure: <br />1. Calculate the amount of nahcolite remaining in <br />solution upon cessation of cavity leaching. <br />• Given 'the cavity temperature determine the <br />specific gravity of the brine from Page 6.5 of the <br />