53
Chapter 7
7.1
α
tan
loss head =
S
m/hr to
change to
7.2 For
°
=
5
α
:
54
7.3 a.
k
= 0.009 cm/sec
3.216
21
−
−
∆
HH
h
b. kiv
e
e
vv
s
=
+
=;
1
7.4 a.
k
= 0.0063 cm/sec
b. kiv
e
e
vv
s
=
+
=;
1
e
s
c. Quantity of seepage/day:
=
α
cos
5.7
S
h
i
/sec/mm 108.22
35−
×=
=×
==
)12cos75.1(
12cos5.2
055.0
)1cos(
cos
m/sec
2
1
α
α
H
h
kkiAq
55
7.6 a.
cm/sec 10 7.1
3−
×=
==
)(7)(60)55()7(
4
(450)(14)
2
π
Aht
QL
k
e
+
1
8.0
14
e
s
7.7
)(60)(78.5)(
(130)(20)
0.021 ; hAht
QL
k==
; h = 26.28 cm
7.9 a.
=
1
10
log303.2 h
h
aL
k
56
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From Eq. (7.13),
213
m 108.4
−
−−
×=
×
⋅××
==
33
236
N/m 10789.9
)s/mN 10005.1)(m/sec 1018.8(
w
γ
k
K
η
b.
=
2
1
10
log303.2 h
h
At
aL
k
××
×
=
2
10
45
log
60454.78
4095.0
)303.2( 000818.0 h
;
cm 30=
2
h
7.10
kiAq
=
158175 =
−
3
2
7.11 From Eq. (7.15) and Table 7.2 for T = 24° C,
°
°
°
CT
C20
C20
η
7.12 Eq. (7.31):
1871.0
3319.0
68.01
68.0
85.01
85.0
08.0
or ,
1
1
3
3
2
2
3
2
1
3
1
2
1
=
+
+
=
+
+
=k
e
e
e
e
k
k; k
2
=
0.045 cm/sec
+
462.01
7.14 e
1
= e
max
– (e
max
– e
min
)(D
r
) = 0.86 – (0.86 – 0.4)(0.7) = 0.538
63.01
63.0
1
3
2
2
3
2
2
+
+
e
e
k
2
7.15
388.0
28.01
28.0
1
;28.0
1
1
11 =
−
=
−
==
n
n
en
cm/sec 0.415=
=
3
388.0
818.0
818.1
388.1
)058.0(
7.16
)5.18)(88.0(
)max()field(
=
=
dd
Rγγ
= 16.28 kN/m3
1 0.602
+
7.17
Sieve
No.
Opening
(cm)
Percent
passing
Fraction between two
consecutive sieves (%)
200
0.0075
3
07.306
0425.006.0
15
40 and 30 Nos. sieve
between fractionFor
595.0404.0595.0404.0
=
×
=
×
sili
i
DD
f
+
5.6
7.18
n
n
n
e
e
e
e
e
e
e
e
k
k
+
+
=
+
+
=
2
1
1
2
2
2
1
1
2
1
1
11
1
97.1
3
7.19 Eq. (7.37): log k = A′ log e + B′
)1.1log()78.0log(
loglog
21
−
−
ee
59
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136.6)78.0log(95.1)1045.0log(loglog
6
11
−=−×=
′
−=
′
−
eAkB
log(k
3
) = (1.95)log(0.97) – 6.136 = -6.161
k
3
=
6.89
×
××
×
10
-7
cm/sec
7.20 PI (in fraction) = 0.27
7.21
….)(
1
2211)eq(
++=
HkHk
H
k
H
×
−
5
)eq(
104.9
V
k
7.22 From Eq. (7.49):
60
7.23 Referring to Figure 7.25a:
d = 4 – 1 = 3 m; h = 2 m; dh = 5 cm; dt = 8 min
CRITICAL THINKING PROBLEM
109.3
20
102.4
20
105
20
60
a.7.C.1
423
3
3
2
2
1
1
)eq(
−−−
×
+
×
+
×
=
++
=
k
H
k
H
k
H
H
k
v
γ
w
61
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x = 200 mm
Z =
−220 mm
11)eq(
ikik
v
=
=
20
∆
)005.0(
60
47
)001076.0( H
mm 33.71cm 3.371∆
=
=
H
Therefore,
mm 436.29
=
−
=
71.33470h
mm 656.29=−−= )220(29.436
w
γ
u
x = 400 mm
Z =
−220 mm
22)eq(
ikik
v
=
=
20
∆
)042.0(
60
47
)001076.0( H
mm 4cm 0.4∆
=
=
H
Therefore,
436.29 4h
= − =
432.29 mm
u
62
© 2018 Cengage Learning®. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
mm 220=−−= )220(0
w
γ
u
c. The variation of heads with distance is shown in the figure.
d.
000843.0
60
47
)001076.0(
)eq(
=
== ikv
v
cm/sec.
n
v
v
s
=
000843.0
v