MoldMaking Technology

APR 2017

Advertising in MoldMaking Technology offers

Issue link: https://mmt.epubxp.com/i/802173

Contents of this Issue

Navigation

Page 24 of 59

moldmakingtechnology.com 23 Because the 4s and πs in this last equation cancel out, we are left with: dh = D 2 / D dh = D. The results are as expected. In a round passage, the hydraulic diameter is the same as the actual diameter. Rectangle. If we then test the hydraulic diameter of a rect- angular duct or pipe (see Figure 3) in which "a" represents the width/height of the passage and "b" is the height/width of the passage, we get: dh = 4 A / p A = ab p = 2a + 2b = 2 (a + b) so dh = (4 (ab)) / (2 (a + b)) dh = 2 ab / (a + b). Notice the difference between calculating a feeder line for the rectangular passage using the equivalent area calculation and using the hydraulic diameter equation. Assume the rect- angle is 0.200 by 0.300 inch with an area of 0.060 square inch. The area calculation method would imply a diameter of 0.2764 inch, but the hydraulic diameter calculation indicates that the effective passage is smaller at 0.240 inch. The reason for this is that the circle has the most efficient perimeter-to-area ratio of all of the planar geometric shapes. The sphere is the most efficient volume. The fluid has a stationary boundary layer at the walls of the fluid passage, and the fluid shears as it reaches maximum velocity at the passage's center. This is similar to polymer flow, except the velocity profile is much less parabolic due to water's lower viscosity. This means that the fluid velocity goes from zero at the boundary to maximum velocity at a level not too far from the boundary in less-viscous materials. Viscous mate- rials, like plastic, reach their maximum velocity at the center of the flow passage. As the shape of passages deviate from round, the perimeter increases relative to the area of the opening. As the length of the perimeter increases, the pressure to force fluids through the passage increases, due to the increased shear forces. Annular. Using the industry standard method of equivalent area to calculate the size of bubbler tubes, and the holes that they reside in, results in undersized drilled holes or oversized bubblers with truly excessive pressure losses, clogged lines (especially with the smaller tubes) and longer-than-necessary cycle times. So, if we apply the hydraulic diameter to an annu- lar channel (see Figure 4, page 24): dh = 4 A / p A = π / 4 (D 2 - d 2 ) p = (πD - πd) so dh = (4 (π / 4 (D 2 - d 2 ))) / (πD + πd). Because the 4s and πs cancel out, we are left with: dh = (D 2 - d 2 ) / (D + d). You may recognize (D 2 - d 2 ) from high school algebra and the FOIL (first, outer, inner, last) method of multiplication: (D + d)(D - d) = (D 2 - d 2 ). If we use this substitution in the numerator, we arrive at: dh = (D + d)(D - d) / (D + d) (dividing numerator and denominator by (D + d)). Then we end up with a very simple calculation of: dh = (D - d). The hydraulic diameter of a round passage is equal to the hole's diameter. FIGURE 2 The hydraulic diameter of a rectangular passage equals four times the area of the rectangle divided by the perimeter. FIGURE 3

Articles in this issue

Links on this page

Archives of this issue

view archives of MoldMaking Technology - APR 2017