Question

7. Drilling sites for oil contain solids control equipment to remove the large, heavy drilled solids from the water and smaller, lighter solids in the drilling mud. The drilling mud enters the solids control equipment at a rate of 3000 lbm/hr and contains 80 wt% water, 15 wt% small solids, and 5 wt% heavy solids. 98% of the heavy solids fed to this equipment are removed and leave the equipment through a solid waste stream. 95% of the water fed to this equipment remains with the cleaned drilling mud and the concentration of small solids in the cleaned mud is 2 wt%. Draw and label a flowchart of this process. Perform a degree-of-freedom analysis and list the equations and their order of calculation to determine the small solids concentration (wt%) in the solid waste stream leaving the solids control equipment. Do not perform the calculations.

Answer 1

Xw, p= 9 Xits,p=& F = 3000lb/hr. Xss, p=0.07 XWE 0.8 SSF = 0,15 X45,F= 0.05 * HS, s = ? ss,s: Xw,s = ? Qimen, other conditions relating unknocons c) 98% of heavy solids fed to this equipment reagan . removed through solid waste stream (s). (4) 95% of Water fed to equipruut remains with cleansed drilling mud (P). oof = nou of unknowns - Independent Equations - non of other conditing and relating to len knowns = [ 5,8, *H5,5 - 455 sXw,p] - [3] 1 " three specis - [2] = 5-3-2 -O TIT Note: a Xas, st Yos, s* Xw, s = 1 Hare, we take only two of them as unknowns xss.pt xwptX Hg, pa

from condition () (3000 * 0.05 ) SXH5.5 ST HS, s = 147 days Heary sells Haury solids balance: 3000 (0.05) - s Xu5,5 + p (X145p) 3000 (0.05) = 147 + P MSP P Yasp = 150-147 = 3 (gle Hay solids from condition (2) [300 x 0.8] 1 o = P KWA poelp Twop = 2280 tals watir Water balance: 3000 (0.8) = S XwR + D xw, A S*w,5 = 3000 (0.8) - 1280 stus = 120 16 watter hr Total flow rate of p (cleaned drilling mud) P = p Xss,p + p Xw,p + P XHs,p P = P(0.0x) + 2?80 + 3 228.3 2329.6 16 P- 2283 0.98 p(0.95) 2283

Overall mas balance: Foś+P SE 3000 - 2329.6 = 670.4 lb/hr Small solids balances ! 3000x (0.15) = 670.4 (Xss,s) + 2399.6 (0.02) 450 - 670.4 (xsss) + 46.6 Xss, s - 403.4 - 670.4 = 0.6