11.4. Calculate RP.M. in Example 11.2.

Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
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Answer 11.4!!

11.4. Calculate RP.M. in Example 11.2.
Transcribed Image Text:11.4. Calculate RP.M. in Example 11.2.
F = 1.75
V Ax 1-e
Dp
Equation 11.16, the Burke-Plummer equation, is
satisfactory for RP.M. greater than about 1000.
1
1/4 ft
Ift
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
[turbulent flow, uniform size spheres]
Here, however, the gravity term is negligible compared with the
others, so, substituting from Eq. 11.16, we find
-AP 1.75pV3 1-E
Ax
Dp
Large
ΔΡ
P
2 in
1.75.62.3 lbm/ft³ · (2 ft/s)².0.67
(0.03ft/12) 0.333³ 32.2 lbm ft/ (lbf-s²) 144 in² / ft²
= 701 psi/ft = 15.9 MPa/m
+8 Az = -F
Wire mesh
support screen
Water
Ion-exchange
resin D₂ = 0.03 in = 0.76 mm
(11.16)
FIGURE 11.3
Gravity drainage of fluid through a porous medium.
.
Transcribed Image Text:F = 1.75 V Ax 1-e Dp Equation 11.16, the Burke-Plummer equation, is satisfactory for RP.M. greater than about 1000. 1 1/4 ft Ift 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 [turbulent flow, uniform size spheres] Here, however, the gravity term is negligible compared with the others, so, substituting from Eq. 11.16, we find -AP 1.75pV3 1-E Ax Dp Large ΔΡ P 2 in 1.75.62.3 lbm/ft³ · (2 ft/s)².0.67 (0.03ft/12) 0.333³ 32.2 lbm ft/ (lbf-s²) 144 in² / ft² = 701 psi/ft = 15.9 MPa/m +8 Az = -F Wire mesh support screen Water Ion-exchange resin D₂ = 0.03 in = 0.76 mm (11.16) FIGURE 11.3 Gravity drainage of fluid through a porous medium. .
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