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34 / March 2021 powderbulk.com Note the important conversion that's needed from pounds of mass, lb m , to pounds of force, lb f , which can be accomplished through the use of the gravitational constant, g c , which is shown as 32.2 lb m /ft ÷ lb f · s 2 below. See Equation 6. 2,404 lb m ΔP air = ft⋅s 2 = g c 6 2,404 lb m ft⋅s 2 = 74.7 lb f ft 2 = 0.52 lb f = 0.52 psi 32.2 lb m ⋅ft ft 2 12 in in 2 lb f ⋅s 2 Now we need to return to the solids-component pressure losses, as given in Equation 7. ΔP solids = ΔP accel. + ΔP friction + ΔP bends + ΔP lift 7 where ΔP accel. is the pressure loss from the solids acceleration, ΔP friction is the pressure loss from the sol- ids friction, ΔP bends is the pressure loss for the solids through the bends, and ΔP lift is the pressure loss for the solids lift or the vertical pipelines where solids need to be carried upward. The pressure loss from accelera- tion of the solids can be found using Equation 8, with an assumed slip ratio of solids-to-air of 0.5, shown as u air /2 below. The slip ratio refers to the ratio of the air- phase velocity to the liquid-phase velocity. Because of the drag forces between the flowing air and the solids, the particle velocity will always be lower than the air velocity. Commonly, the slip ratio ranges from 0.5 to 0.9. Heavier and more coarse particles (for example, gravel) will have a slip ratio of 0.5 or less, and lighter and finer powders (for example, talc) will have a value closer to 0.8. Again, use the gravitational constant, g c , to resolve the force and mass units. ΔP accel. = Øu air ρ air v solids 8 7 77 ft 2 0.075 lb m ΔP accel. =Øu air ρ air u air = Øu air 2 ρ air = s ft 3 =1,556 lb m 2 2 2 ft⋅s 2 1,556 lb m 1,556 lb m ΔP accel. = ft⋅s 2 = ft⋅s 2 = 48.3 lb f ft 2 =0.34 lb f = 0.34 psi g c 32.2 lb m ⋅ft ft 2 12 in in 2 lb f ⋅s 2 where u air is the new air velocity of 77 ft/s based on the pipeline's smaller internal diameter determined in Equation 2 and v solids is the solids-to-air slip ratio of u air /2. The pressure loss from the solids friction can be found using Equation 9. ΔP friction = λsØρ air u air 2 ΔL 2D 9 ΔP friction = (0.02)(7) 0.075 lb m 77 ft 2 130 ft = 13,686 lb m ft 3 s 2 3.55 in ft⋅s 2 12 in/ft 13,686 lb m ΔP friction = ft⋅s 2 = 425.0 lb f ft 2 = 2.95 lb f = 2.95 psi 32.2 lb m ⋅ft ft 2 12 in in 2 lb f ⋅s 2 where λs is the solids friction factor. Note the signifi- cant pressure loss from friction from moving the solids through the pipeline (2.95 psi) compared to only mov- ing the air (0.52 psi) — about one-fifth! Next, calculate the pressure loss for the solids through the bends, ΔP bends , (Equation 10) and for the sol- ids lift, ΔP lift , (Equation 11). Please note, now that we're calculating for solids and not air, we're estimating the ΔP bends as the product of friction through the pipeline divided into 130-foot increments, shown as 2.95 psi/130 ft below, then multiplied by the elbow equivalent length, shown as Leq below, of 9 feet. ΔP bends = ΔP friction Leq L 10 ΔP bends = 2.95 psi 9 ft = 0.20 psi 130 ft ΔP lift = Øρ air gΔz = Øρ air gΔz = Øρ air gΔz = 2Øρ air gΔz v solids /u air (u air /2)/u air 0.5 11 ΔP lift = 2(7) 0.075 lb m 32.2 ft (30 ft) = 1,014 lb m ft 3 s 2 ft⋅s 2 where g is universal gravity of 32.2 ft/s 2 and Δz is the vertical height needed to lift solids up 30 feet. Again, consider the units conversion needed for force and mass resolution using the gravitational constant, g c , to resolve.