Chapter 8
h1k1
8.1 Eq. (8.14): h2 =
⎛k k ⎞
H1 ⎜⎜ 1 + 2 ⎟⎟
⎝ H1 H 2 ⎠
(20 cm)(0.004 cm/sec)
8 cm =
(10 cm )⎛⎜ 0.004 cm/sec + k2 cm/sec ⎞⎟
⎝ 10 cm 15 cm ⎠
k2 = 0.009 cm/sec
8.2 The flow net is shown.
k = 4 × 10-4 cm/sec
H = H1 – H2
= 6.0 – 1.5 = 4.5 m.
So
⎛ 4 × 10−4 ⎞⎛ 4.5 × 4 ⎞
q=⎜ ⎟⎜ ⎟
⎝ 10 ⎠⎝ 8 ⎠
2
= 9 ×10−6 m3/m/sec
= 77.76× 10-6 m3 /m/day
8.3 The flow net is shown.
Nf = 3; Nd = 5
⎛Nf ⎞
q = kH ⎜⎜ ⎟⎟
⎝ Nd ⎠
51
© 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.
, ⎛ 4 × 10 −4 ⎞
m/sec ⎟(3 − 0.5)⎛⎜ ⎞⎟ = 6 × 10 −6 m 3 /m/sec = 0.518 m 3 /m/day
3
q=⎜
⎝ 10
2
⎠ ⎝ 5⎠
8.4 Based on the notations in Figure 8.10:
H = (4 – 1.5) m = 2.5 m; S = D = 3.6 m; T ' = D1 = 6 m; S/T ' = 3.6/6 = 0.6
q
From the figure, ≈ 0.44
kH
⎛ 4 × 10−4 ⎞
q = (0.44)( 2.5)⎜ 2
× 60 × 60 × 24 m/day ⎟ = 0.38 m3 /m/day
⎝ 10 ⎠
8.5 The flow net is shown.
⎛N ⎞
q = kH ⎜⎜ f ⎟⎟ = ⎛⎜ × 60 × 60 × 24 m/day ⎞⎟(10)⎛⎜ ⎞⎟ = 7.2 m 3 /m/day
0.002 5
2
⎝ N d ⎠ ⎝ 10 ⎠ ⎝ 12 ⎠
8.6 Refer to the flow net given in Problem 8.5 and the figure on the next page.
The flow net has 12 potential drops. Also, H = 10 m. So the head loss for each
drop = (10/12) m. Thus,
52
© 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.
h1k1
8.1 Eq. (8.14): h2 =
⎛k k ⎞
H1 ⎜⎜ 1 + 2 ⎟⎟
⎝ H1 H 2 ⎠
(20 cm)(0.004 cm/sec)
8 cm =
(10 cm )⎛⎜ 0.004 cm/sec + k2 cm/sec ⎞⎟
⎝ 10 cm 15 cm ⎠
k2 = 0.009 cm/sec
8.2 The flow net is shown.
k = 4 × 10-4 cm/sec
H = H1 – H2
= 6.0 – 1.5 = 4.5 m.
So
⎛ 4 × 10−4 ⎞⎛ 4.5 × 4 ⎞
q=⎜ ⎟⎜ ⎟
⎝ 10 ⎠⎝ 8 ⎠
2
= 9 ×10−6 m3/m/sec
= 77.76× 10-6 m3 /m/day
8.3 The flow net is shown.
Nf = 3; Nd = 5
⎛Nf ⎞
q = kH ⎜⎜ ⎟⎟
⎝ Nd ⎠
51
© 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.
, ⎛ 4 × 10 −4 ⎞
m/sec ⎟(3 − 0.5)⎛⎜ ⎞⎟ = 6 × 10 −6 m 3 /m/sec = 0.518 m 3 /m/day
3
q=⎜
⎝ 10
2
⎠ ⎝ 5⎠
8.4 Based on the notations in Figure 8.10:
H = (4 – 1.5) m = 2.5 m; S = D = 3.6 m; T ' = D1 = 6 m; S/T ' = 3.6/6 = 0.6
q
From the figure, ≈ 0.44
kH
⎛ 4 × 10−4 ⎞
q = (0.44)( 2.5)⎜ 2
× 60 × 60 × 24 m/day ⎟ = 0.38 m3 /m/day
⎝ 10 ⎠
8.5 The flow net is shown.
⎛N ⎞
q = kH ⎜⎜ f ⎟⎟ = ⎛⎜ × 60 × 60 × 24 m/day ⎞⎟(10)⎛⎜ ⎞⎟ = 7.2 m 3 /m/day
0.002 5
2
⎝ N d ⎠ ⎝ 10 ⎠ ⎝ 12 ⎠
8.6 Refer to the flow net given in Problem 8.5 and the figure on the next page.
The flow net has 12 potential drops. Also, H = 10 m. So the head loss for each
drop = (10/12) m. Thus,
52
© 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.
