Chapter 7 Permeability

Chapter 7

7.1




head loss S tan α
From the figure: i = = = sin α
length ⎛ S ⎞
⎜ ⎟
⎝ cos α ⎠

q = k i A = (k)(sin α)(4.2 cos α)(1)

k = 4.8 × 10-3 cm/sec = 4.8 × 10-5 m/sec

q = (4.8 × 10−5 )(sin 6°)(4.2 cos 6°)(3600) = 0.0754 m3 /hr/m

to change
to m/hr

≈ 7.54 × 10- 2 m 3 /hr/m


h
7.2 i=
⎛ S ⎞
⎜ ⎟
⎝ cosα ⎠
⎛ ⎞
⎛ h cos α ⎞ ⎜ 0.075 ⎟⎛ 2.75 cos14 ⎞
q = kiA = k ⎜ ⎟( H 1 cos α × 1) = ⎜ 2 ⎟⎜ ⎟(2 cos14)
⎝ S ⎠ ⎜ 10
N ⎟⎝ 30 ⎠
⎝ m/sec ⎠
= 1.29 × 10 -4 m 3 /sec/m



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, QL (350 cm 2 )(30 cm)
7.3 a. k = = 2
= 3.96 × 10- 3 cm/sec
Aht (176.71 cm )(50 cm)(300 sec)

⎛1+ e ⎞
b. vs = v ⎜ ⎟; v = ki
⎝ e ⎠

⎛1+ e ⎞ ⎛ 50 cm ⎞⎛ 1 + 0.61 ⎞
vs = ki⎜ ⎟ = (0.00396)⎜ ⎟⎜ ⎟ = 0.0174 cm/sec
⎝ e ⎠ ⎝ 30 cm ⎠⎝ 0.61 ⎠


QL (160)(15)
7.4 k= ; 0.062 =
Aht (31.67)(h)(60)

h = 20.37 cm


⎛ aL ⎞ ⎛h ⎞ ⎡ (0.25)(15) ⎤ ⎛ 40 ⎞
7.5 a. k = 2.303⎜ ⎟ log10 ⎜⎜ 1 ⎟⎟ = ( 2.303) ⎢ ⎥ log10 ⎜ ⎟
⎝ At ⎠ ⎝ h2 ⎠ ⎣ (19.64)(8 × 60) ⎦ ⎝ 20 ⎠

= 2.75 × 10-4 cm/sec

⎛h ⎞
b. k = 2.303⎛⎜ aL ⎞⎟ log10 ⎜⎜ 1 ⎟⎟
⎝ At ⎠ ⎝ h2 ⎠

⎡ (0.25)(15) ⎤ ⎛ 40 ⎞ ⎛ 40 ⎞
0.000275 = ( 2.303) ⎢ ⎥ log10 ⎜⎜ ⎟⎟ = 0.00122 log⎜⎜ ⎟⎟
⎣ (19.64)(6 × 60) ⎦ ⎝ h2 ⎠ ⎝ h2 ⎠

h2 = 23.82 cm


⎛h ⎞
7.6 a. k = 2.303⎛⎜ aL ⎞⎟ log10 ⎜⎜ 1 ⎟⎟
⎝ At ⎠ ⎝ h2 ⎠

⎛ 0.97 × 50 ⎞ ⎛ 41 ⎞
= (2.303)⎜ ⎟ log10 ⎜ ⎟ = 0.241 cm/min = 0.00402 cm/sec
⎝ 16 × 10 ⎠ ⎝ 18.5 ⎠

kη ( 4.02 × 10−5 m/sec)(1.005 × 10−3 N ⋅ s/m2 )
K= = = 4.13 × 10-12 m2
γw 9.789 × 103 N/m3




42
© 2014 Cengage Learning. All rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.