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ference to the water flowing through the packed bed in Fig 3 for the water superficial velocity to be 2 ft / s. Wha essure gradient is required? Applying B.E. as before, we find ΔΡ AP+8 Az=-F re, however, the gravity term is negligible compared with the ■ers, so, substituting from Eq. 11.16, we find -AP 1.75pV 1.75pV 1-€ Ax Ꭰ, 1.75-62.3 lbm/ft³ (2 ft/s)² -0.67

ference to the water flowing through the packed bed in Fig 3 for the water superficial velocity to be 2 ft / s. Wha essure gradient is required? Applying B.E. as before, we find ΔΡ AP+8 Az=-F re, however, the gravity term is negligible compared with the ■ers, so, substituting from Eq. 11.16, we find -AP 1.75pV 1.75pV 1-€ Ax Ꭰ, 1.75-62.3 lbm/ft³ (2 ft/s)² -0.67

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
Problem 1.1MA
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Answer 11.6 please! 

11.6. For the flows in Examples 11.1 and 11.2, calculate the magnitudes of the A 12 / 2 terms omitted in B.E., and compare these with the magnitude of the 7 terms.
Transcribed Image Text:11.6. For the flows in Examples 11.1 and 11.2, calculate the magnitudes of the A 12 / 2 terms omitted in B.E., and compare these with the magnitude of the 7 terms.
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 P Here, however, the gravity term is negligible compared with the others, so, substituting from Eq. 11.16, we find 1.75pV 1- 1/4ft -AP Ax Ꭰ, 1.75 - 62.3 lbm/ft³ (0.03ft/12) - 0.333 - 32.2 lbm (2 ft/s)² - 0.67 ft/(lbf · s²) · 144 in²/ft² = 701 psi/ft = 15.9 MPa/m Large <-2 in Water Ift Ion-exchange resin D,= 0.03 in=0.76 mm Wire mesh support screen FIGURE 11.3 Gravity drainage of fluid through a porous medium. Example 11.1. Figure 11.3 shows a water softener in which water trickles by gravity through a bed of spherical ion-exchange resin particles, each 0.03 in (0.76 mm) in diameter. The bed has a porosity of 0.33. Calculate the volumetric flow rate of water. Applying B.E. from the top surface of the fluid to the outlet of the packed bed and ignoring the kinetic-energy term and the pressure drop through the support screen, which are both small, we find 8(Az)=-F (11.C
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 P Here, however, the gravity term is negligible compared with the others, so, substituting from Eq. 11.16, we find 1.75pV 1- 1/4ft -AP Ax Ꭰ, 1.75 - 62.3 lbm/ft³ (0.03ft/12) - 0.333 - 32.2 lbm (2 ft/s)² - 0.67 ft/(lbf · s²) · 144 in²/ft² = 701 psi/ft = 15.9 MPa/m Large <-2 in Water Ift Ion-exchange resin D,= 0.03 in=0.76 mm Wire mesh support screen FIGURE 11.3 Gravity drainage of fluid through a porous medium. Example 11.1. Figure 11.3 shows a water softener in which water trickles by gravity through a bed of spherical ion-exchange resin particles, each 0.03 in (0.76 mm) in diameter. The bed has a porosity of 0.33. Calculate the volumetric flow rate of water. Applying B.E. from the top surface of the fluid to the outlet of the packed bed and ignoring the kinetic-energy term and the pressure drop through the support screen, which are both small, we find 8(Az)=-F (11.C
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