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Example 11.2. We now wish to apply a sufficient pressure difference to the water flowing through the packed bed in Fig. 11.3 for the water superficial velocity to be 2 ft/s. What pressure gradient is required? Applying B.E. as before, we find ΔΡ +g Az=-F Ρ (11.G) Here, however, the gravity term is negligible compared with the others, so, substituting from Eq. 11.16, we find -AP 1.75pV 1 - E Ax = Dp 1.75 62.3 lbm/ft³ (2 ft/s)² - 0.67 (0.03ft/12) 0.333 32.2 lbm ft/(lbf s²) 144 in²/ft² = 701 psi/ft = 15.9 MPa/m (11.H) = 690, so this is at the low end of the turbulent flow Here we check, finding RP.M. region (see Fig. 11.4, discussed below). 1/4 ft Large 2 in Water 1ft Ion-exchange resin Dp = 0.03 in = 0.76 mm Wire mesh support screen FIGURE 11.3 Gravity drainage of fluid through a porous medium.

Example 11.2. We now wish to apply a sufficient pressure difference to the water flowing through the packed bed in Fig. 11.3 for the water superficial velocity to be 2 ft/s. What pressure gradient is required? Applying B.E. as before, we find ΔΡ +g Az=-F Ρ (11.G) Here, however, the gravity term is negligible compared with the others, so, substituting from Eq. 11.16, we find -AP 1.75pV 1 - E Ax = Dp 1.75 62.3 lbm/ft³ (2 ft/s)² - 0.67 (0.03ft/12) 0.333 32.2 lbm ft/(lbf s²) 144 in²/ft² = 701 psi/ft = 15.9 MPa/m (11.H) = 690, so this is at the low end of the turbulent flow Here we check, finding RP.M. region (see Fig. 11.4, discussed below). 1/4 ft Large 2 in Water 1ft Ion-exchange resin Dp = 0.03 in = 0.76 mm Wire mesh support screen FIGURE 11.3 Gravity drainage of fluid through a porous medium.

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
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Calculate ℛP.M. in Example 11.2.

Example 11.2. We now wish to apply a sufficient pressure difference to the water flowing through the packed bed in Fig. 11.3 for the water superficial velocity to be 2 ft/s. What pressure gradient is required? Applying B.E. as before, we find ΔΡ +g Az=-F Ρ (11.G) Here, however, the gravity term is negligible compared with the others, so, substituting from Eq. 11.16, we find -AP 1.75pV 1 - E Ax = Dp 1.75 62.3 lbm/ft³ (2 ft/s)² - 0.67 (0.03ft/12) 0.333 32.2 lbm ft/(lbf s²) 144 in²/ft² = 701 psi/ft = 15.9 MPa/m (11.H) = 690, so this is at the low end of the turbulent flow Here we check, finding RP.M. region (see Fig. 11.4, discussed below).
Transcribed Image Text:Example 11.2. We now wish to apply a sufficient pressure difference to the water flowing through the packed bed in Fig. 11.3 for the water superficial velocity to be 2 ft/s. What pressure gradient is required? Applying B.E. as before, we find ΔΡ +g Az=-F Ρ (11.G) Here, however, the gravity term is negligible compared with the others, so, substituting from Eq. 11.16, we find -AP 1.75pV 1 - E Ax = Dp 1.75 62.3 lbm/ft³ (2 ft/s)² - 0.67 (0.03ft/12) 0.333 32.2 lbm ft/(lbf s²) 144 in²/ft² = 701 psi/ft = 15.9 MPa/m (11.H) = 690, so this is at the low end of the turbulent flow Here we check, finding RP.M. region (see Fig. 11.4, discussed below).
1/4 ft Large 2 in Water 1ft Ion-exchange resin Dp = 0.03 in = 0.76 mm Wire mesh support screen FIGURE 11.3 Gravity drainage of fluid through a porous medium.
Transcribed Image Text:1/4 ft Large 2 in Water 1ft Ion-exchange resin Dp = 0.03 in = 0.76 mm Wire mesh support screen FIGURE 11.3 Gravity drainage of fluid through a porous medium.
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