F = 1.75V Ax 1-eDpEquation 11.16, the Burke-Plummer equation, issatisfactory for Rp.M. greater than about 1000.11/4 ftIftExample 11.2. We now wish to apply a sufficient pressuredifference to the water flowing through the packed bed in Fig.11.3 for the water superficial velocity to be 2 ft /s. Whatpressure gradient is required?Applying B.E. as before, we find[turbulent flow, uniform size spheres]Here, however, the gravity term is negligible compared with theothers, so, substituting from Eq. 11.16, we find-AP 1.75pV3 1-EAxDpLargeΔΡ2 inP1.75.62.3 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/mWire meshsupport screen+ g Az=-FWaterIon-exchangeresin D₂ = 0.03 in = 0.76 mm(11.16)FIGURE 11.3Gravity drainage of fluid through a porous medium.. 11.4. 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 A² +8 Az = -F P Here, however, the gravity term is negligible compared with the others, so, substituting from Eq. 11.16, we find

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 A² +8 Az = -F P Here, however, the gravity term is negligible compared with the others, so, substituting from Eq. 11.16, we find

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

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Question

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
2 in
P
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
Wire mesh
support screen
+ g Az=-F
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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