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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 P - (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 - € (11.H) 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 Here we check, finding RP.M. = 690, so this is at the low end of the turbulent flow region (see Fig. 11.4, discussed below). 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 g(Az) = -F Substituting from Eq. 11.14 and solving for Vs, we find Therefore, g(-Az) De³p Vs= 150μ (1-E)²Ax 32.2 ft/s 1.25 ft (0.03 ft / 12)² 0.333 . 62.3 lbm/ft³ 150 1.002 cP (10.33)2 1 ft 6.72 104 lbm/(ft s cP) = 0.0124 ft/s = 0.00379 m/s Q = AVS = (a) 0.01240.000277.6 cm³ (11.C). Page 414 Before accepting this as the correct solution, we check the Reynolds number, finding RP.M. = (0.03 ft/12) 0.0124 ft/s 62.3 lbm/ft³ 1.002 cP 0.67 6.72 10 = 4.29 lbm/(ft. s. cP) (11.D) (11.E) (11.F)

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 - (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 - € (11.H) 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 Here we check, finding RP.M. = 690, so this is at the low end of the turbulent flow region (see Fig. 11.4, discussed below). 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 g(Az) = -F Substituting from Eq. 11.14 and solving for Vs, we find Therefore, g(-Az) De³p Vs= 150μ (1-E)²Ax 32.2 ft/s 1.25 ft (0.03 ft / 12)² 0.333 . 62.3 lbm/ft³ 150 1.002 cP (10.33)2 1 ft 6.72 104 lbm/(ft s cP) = 0.0124 ft/s = 0.00379 m/s Q = AVS = (a) 0.01240.000277.6 cm³ (11.C). Page 414 Before accepting this as the correct solution, we check the Reynolds number, finding RP.M. = (0.03 ft/12) 0.0124 ft/s 62.3 lbm/ft³ 1.002 cP 0.67 6.72 10 = 4.29 lbm/(ft. s. cP) (11.D) (11.E) (11.F)

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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For the flows in Examples 11.1 and 11.2, calculate the magnitudes of the Δ V2 / 2 terms omitted in B.E., and compare these with the magnitude of the ℱ 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 - (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 - € (11.H) 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 Here we check, finding RP.M. = 690, so this is at the low end of the turbulent flow 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 P - (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 - € (11.H) 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 Here we check, finding RP.M. = 690, so this is at the low end of the turbulent flow region (see Fig. 11.4, discussed below).
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 g(Az) = -F Substituting from Eq. 11.14 and solving for Vs, we find Therefore, g(-Az) De³p Vs= 150μ (1-E)²Ax 32.2 ft/s 1.25 ft (0.03 ft / 12)² 0.333 . 62.3 lbm/ft³ 150 1.002 cP (10.33)2 1 ft 6.72 104 lbm/(ft s cP) = 0.0124 ft/s = 0.00379 m/s Q = AVS = (a) 0.01240.000277.6 cm³ (11.C). Page 414 Before accepting this as the correct solution, we check the Reynolds number, finding RP.M. = (0.03 ft/12) 0.0124 ft/s 62.3 lbm/ft³ 1.002 cP 0.67 6.72 10 = 4.29 lbm/(ft. s. cP) (11.D) (11.E) (11.F)
Transcribed Image Text: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 g(Az) = -F Substituting from Eq. 11.14 and solving for Vs, we find Therefore, g(-Az) De³p Vs= 150μ (1-E)²Ax 32.2 ft/s 1.25 ft (0.03 ft / 12)² 0.333 . 62.3 lbm/ft³ 150 1.002 cP (10.33)2 1 ft 6.72 104 lbm/(ft s cP) = 0.0124 ft/s = 0.00379 m/s Q = AVS = (a) 0.01240.000277.6 cm³ (11.C). Page 414 Before accepting this as the correct solution, we check the Reynolds number, finding RP.M. = (0.03 ft/12) 0.0124 ft/s 62.3 lbm/ft³ 1.002 cP 0.67 6.72 10 = 4.29 lbm/(ft. s. cP) (11.D) (11.E) (11.F)
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