Urban Drainage 2nd edition
SOLUTIONS MANUAL David Butler and John W Davies
This solutions manual is made available free of charge. Details of the accompanying textbook Urban Drainage 2nd edition (ISBN 0415-30606-X hbk; 0415-30607-8 pbk) are on the website of the publisher www.sponpress.com and can be ordered from
[email protected] or phone: +44 (0) 1264 343071 First published 2004 by Spon Press, an imprint of Taylor & Francis, 2 Park Square, Milton Park, Abingdon, Oxon OX14 4RN Simultaneously published in the USA and Canada by Taylor & Francis, 270 Madison Avenue, New York, NY 10016, USA @ 2004 David Butler and John W Davies
All rights reserved. No part of this book may be reprinted or reproduced or utilized in any form or by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying and recording, or in any information storage or retrieval system, except for the downloading and printing of a single copy from the website of the publisher, without permission in writing from the publishers. Publisher's note This book has been produced from camera ready copy provided by the authors Contents of this solutions manual Chapter
Page Number
Chapter 3--------------------------------------------- 2 Chapter 5--------------------------------------------- 3 Chapter 6--------------------------------------------- 3 Chapter 7--------------------------------------------- 5 Chapter 8--------------------------------------------- 5 Chapter 9--------------------------------------------- 6 Chapter 10 ------------------------------------------- 7 Chapter 11 ------------------------------------------- 8 Chapter 12 ------------------------------------------10 Chapter 13 ------------------------------------------11 Chapter 14 ------------------------------------------12 Chapter 15 ------------------------------------------12 Chapter 16 ------------------------------------------13 Chapter 17 ------------------------------------------15 Chapter 18 ------------------------------------------16 Chapter 21 ------------------------------------------16 Chapter 23 ------------------------------------------16
Note: For chapters 1, 2, 4, 19, 20, 22 and 24 the problems are simply issues to think about. Solutions are only provided to problems with numerical solutions for these chapters and all others.
1
CHAPTER 3 3.1
Volumetric concentration, Cv = Vs/V, where Vs is volume of solid and V is sample volume. So:
Cv =
Vs = 20 ppm = 20 × 10 −6 m 3 / m 3 V
Density of solid, ρ = M/V, where M is its mass and ρ = 2650 kg/m3. So:
C=
M = ρCv = 2650 × 20 × 10 −6 kg / m 3 = 53 mg / l V
3.2 Time (hrs) 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 Sum Av
Flow (l/s) 80 170 320 610 670 590 380 220 100 50 3190 319
COD (mg/l) 50 160 380 400 230 130 70 40 20 0
COD (mg/s) 4000 27200 121600 244000 154100 76700 26600 8800 2000 0 665000
148
From equation 3.1: C av =
ΣQC 665000 = = 208 mg / l Qav 3190
Flow (l/s) or COD (mg/l)
800 700 600 500 Flow (l/s)
400
COD (mg/l)
300 200 100 0 0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
Time (hrs)
2
3.3
Gram atomic weight of N is 14.0 g gram atomic weight of H is 1.0 g gram molecular weight of ammonium is 14 + (4 × 1) = 18 g From equation 3.2: 35 mgNH 4+ / l = 35 ×
14 = 27 mgNH 4+ − N / l 18
So, total N conc = 15 + 27 = 42 mgN / l 3.12
Taken over a 10-year period, there were 11 violations, whereas only up to 10 are permitted, so this water is not in compliance with the standard.
CHAPTER 5 Ministry of health (equation 5.2): i=
1000 a = = 20.0 mm / h D + b 30 + 20
Bilham (5.3): N = 1.25(I / 25.4 + 0.1) −3.55 = 10 = 1.25 × 0.5 × (I / 25.4 + 0.1) −3.55 I = 9.1mm 9 .1 I i= = = 18.2 mm / h D 0 .5
Holland (5.4) N = D(I / 25.4) −3.14 = 10 = 0.5 × (I / 25.4) −3.14 I = 9.8 mm i = 19.6 mm / h
CHAPTER 6 6.4
From equation 6.2: ft = fc + (fo − fc )e − k2t
ft = 1 + 9e −t Time (min) Time (hour) ft (mm/h) calc ft (mm/h) appl Net rain (mm/h) Eff rain (mm/h)
0-10 0.167 8.6 0 0
20 0.333 7.4 8.6 9 0.4
30 0.500 6.5 7.4 18 10.6
40 0.667 5.6 6.5 6 0 3
Assumes Horton infiltration starts when net rain begins. 6.5
Use the approach summarised in equation 6.8: Q(1) Q(2) Q(3) Q(4) Q(5) Q(6) Q(7) Q(8)
=1x0 = 1 x 250 = 1 x 500 = 1 x 375 = 1 x 250 = 1 x 125 =1x0 =1x0
+2x0 + 2 x 250 + 3 x 0 + 2 x 500 + 3 x 250 + 0 x 0 + 2 x 375 + 3 x 500 + 0 x 250 + 1 x 0 + 2 x 250 + 3 x 375 + 0 x 500 + 1 x 250 + 2 x 125 + 3 x 250 +………. + 2 x 0 + 3 x 125 + ………
=0 = 250 = 1000 = 2125 = 2500 = 2000
Peak = 2500 l/s after 40 mins. 6.7
From equation 6.13a: t p = (α − 1)K = (3 − 1) × 0.2 = 0.4 h = 24 min From equation 6.13b:
1 ⎛ α − 1⎞ Qp = ⎜ ⎟ K (α − 1)! ⎝ e ⎠
n −1
2
1 ⎛2⎞ −1 = ⎜ ⎟ = 1.35 h 0.2 × 2! ⎝ e ⎠
The units are h-1 because this is based on an instantaneous inflow of unit volume. So, actual volume in this case: V = 10 mm × 100,000 m2 [10 ha] = 1000 m3 Qp = 1000 m3/h = 278 l/s 6.10
Based on conditions in Example 6.3. From equation 6.20: M s (t ) = M s (0)e − k 4it M s (t ) = 18e −1.9t From equation 6.22: k 4 M s 0.19 × 10 −3 = c= Ms Ai 15000 Time (min) Time (hour) Ms (kg) C (mg/l)
0-10 0.167 13.1 166
20 0.333 9.6 122
30 0.500 7.0 88
4
CHAPTER 7 7.2
From co-ordinates, W-E and N-S distance between manholes are 68.114 m and 22.451 m respectively. From Pythagoras, length of sewer is 71.719 m. Invert level of MHA (ILA) = 16.438 – 0.375 = 16.063 m Change in height along sewer (∆H) = 71.719 / 258 = 0.278 m ILB1 (entry invert level) = 16.063 – 0.278 = 15.785 m ILB2 (exit) = 15.785 + 0.375 – 0.45 = 15.710 m Depth = 18.590 – 15.710 = 2.88 m
CHAPTER 8 8.1
ks Q = 0.002 v = = 1.41 m/s D A vD 1.41× 0.3 Re = = = 371000 ν 1.14 × 10 −6 from Moody diagram (Fig. 8.4), λ = 0.024 , transitional
λL v 2
0.024 × 100 1.412 ⋅ = 0 .8 m D 2g 0 .3 2g from Wallingford chart (Fig. 8.5), hydraulic gradient 0.8 in 100 hf =
8.2
⋅
Wallingford chart (Fig. 8.5): hydraulic gradient 0.18 in 100 needed for v of 1.0 m/s giving Q of 300 l/s Fig. 8.8: for
8.3
=
v = 0. 8 vf
d = 0.3 which gives D = 180 mm D
ks = 0.0033 D from Moody diagram (Fig. 8.4), for Re down to 500 000, λ = 0.026 LE k L 0 .5 × 0.45 = 8.7 m = so LE = 0.026 λ D
determine minimum v to ensure R e ≥ 500 000 v × 0.45 = 500 000 giving v = 1.27 m/s 1.14 × 10 −6
8.4
so assumption is valid
d 150 = = 0.25 D 600 from Fig. 8.8
and
Q = 0.13 Qf
v = 0.7 so v = 0.7 × 2.2 = 1.5 m/s vf so
Q = 80 l/s
5
Af = 0.283 m2 Pf = 1.885 m
R=
A = 0.195 Af P from Fig. 8.8 = 0.335 Pf from Fig. 8.8
A 0.055 = = 0.09 m P 0.63
a) b) c) d)
8.6
A = 0.055 m2
so
P = 0.63 m
(or determine from Rf and Fig. 8.8)
Τ0 = ρ gRS0 = 1000 × 9.81× 0.09 × 8.5
so
1 = 4.4 N/m2 200
from Fig. 8.9
1:190 d from Fig. 8.9 at = 0.83 giving depth of 250 mm D from Fig. 8.9 v would be 0.78 m/s from Fig. 8.9 for v of 1.0 m/s and Q of 10 l/s gradient must be 1:95
from Fig. 8.5 Qf is 400 l/s Q d 100 = = 0.22 from Fig. 8.8 = 0.12 giving Q 48 l/s Qf D 450 dc 0.048 0.506 = 0.567 = 0.33 D 0.451.264 therefore dc > d and so conditions are supercritical
equation (8.28)
8.7
d c = 0.33 × 0.45 = 0.15 m 1.93
⎡ 0.15 ⎤ = 2.19 equation (8.30) Fr 1 = ⎢ ⎥ ⎣ 0 .1 ⎦ since Fr1 > 1.7 use equation (8.32) 1. 8 d 0.151.8 d 2 = c0.73 = = 0.18 m 0.10.73 d1
CHAPTER 9 9.1
dia below 200 mm would not be suitable so try 200 mm and assume not drowned water level 1.5 m H = 1.5 − 0.2 = 1.3 m 0. 2 2 Q = 0.6π 2 g 1.3 = 95 l/s 4 water level 1.7 m gives Q of 102 l/s so 200 mm is the appropriate diameter check not drowned: Wallingford chart or table for ks 0.03 mm gives Qf of 140 l/s Q d = 0.73 for this, Fig. 8.8 gives = 0.62 water level 1.7 m: Qf D
6
Fig. 9.3: H > Hmin 9.2
D0 200 = = 0.44 D 450 so not drowned
giving
H min = 1.9 thus Hmin is 0.38 m Do
hydraulic gradient is 1.5 m in 25 m or 6 in 100 Wallingford chart or table for ks 0.03 mm gives Qf of 125 l/s inlet-controlled? Try orifice equation with H 1.5 m 0 .2 2 Q = 0.6π 2 g 1.5 = 102 l/s 4 yes – inlet-controlled (in fact H would be less than 1.5 m because the pipe itself would have a gradient, see Fig. 9.6, but that would make Q lower, therefore still inlet-controlled)
9.3
b 2.2 m
P 1.05 m
equation (9.5)
H 0.15 m
C d = 0.602 + 0.0832
0.15 = 0.614 1.05
3 2 Q = 0.614 × × 2.2 × 2g [0.15 + 0.0012]2 = 0.235 m3/s 3
9.4
From geometry in Figure 9.13: y=
B 2 .5 = = 0.0625 m z 40
From equation 9.9:
C=
1 2 o
1 2
zS 40 × 0.005 = = 217.57 n 0.013
From 9.8: 8
8
Q = 0.31Cy 3 = 0.31× 217.57 × 0.0625 3 = 0.042m 3 / s
CHAPTER 10 10.4
a)
b)
Population = 75 × 500 = 37, 500 Average domestic flow = 37500 × 160 Trade flow = 10000 × 500 × 0.1 Infiltration = 37500 × 20 Total From table 10.4:
= 6.0 × 106 l/d = 0.5 × 106 l/d = 0.75 × 106 l/d = 7.25 × 106 l/d = 84 l/s
7
PF =
5 P
1 5
5
=
37.5
1 5
= 2.42
Assuming the peak factor is applied to all flows, Qp = 203 l/s 10.5
a)
Using a Butler-Pinkerton chart gives d/D= 0.65 d = 0.65 × 500 = 325 mm
b)
Using the same chart gives a flow capacity of 240 l/s at d/D = 0.75. So solve for P (population) in: 5
(P 1000 )1
5
⎛ P × 160 + 0.5 × 10 6 + P × 20 ⎞ ⎟⎟ = 240 × ⎜⎜ 3600 × 24 ⎠ ⎝
P= 40422 ∆P = 2922 10.8
p = 10/300 = 0.0333
N=5
J=0.999
Using equation 10.5b e.g. 5! P (0,5) = 0.0333 0 0.967 5 0!5! And following equation 10.6: r 0 1 2 3
P(r,N) 0.845 1.44 0.010 ….
Σ P(r,N) 0.845 0.989 0.999 ….
So r = 2 (max discharging simultaneously) 10.10 150 mm dia pipe @ 1:300 flowing d/D = 0.75 Æ Q = 8 l/s (B-P charts) From equation 10.7: 8 = 0 .5 N × 3 .5 8 = 0.046 × N
N = 73 N = 174
CHAPTER 11 2
11.4
a)
Equation 11.2:
⎛ 1⎞ r = 1 − ⎜1 − ⎟ = 0.75 ⎝ 2⎠
50
b) c)
⎛ 1⎞ r = 1 − ⎜1 − ⎟ = 1.00 ⎝ 2⎠ p = 1/T = 1/11 = 0.10 8
1⎞ ⎛ r = 1 − ⎜1 − ⎟ ⎝ 10 ⎠
d) 11.8
10
= 0.65
Use equation 11.5b (C= 1.0): Equation 5.2 (MoH rainfall)
Sewer
L (m) 180 90 90 90
1.0 2.0 3.0 1.1
A (ha) 0.2 0.6 0.9 0.4
Q = 2.78 i Ai
ΣA (ha) 0.2 0.6 0.9 2.1
tc = tf + 4 (min) 6 5 5 7
and
i (mm/h) 46.19 50.0 50.0 44.1
Q (l/s) 26.1 83.4 125 257
11.11 2.5
2
1.1
1.5 A (ha)
1.0 2.0 3.0
1
SUM
0.5
0 1
2
3
4
5
6
7
8
9
Time (min)
The time area diagram shows a tangent (dashed line) can be drawn from t = 6mins, giving A = 20.5 ha. So using the Rational method and MoH rainfall: ⎛ A ⎞ ⎛ 2.05 ⎞ Q = 2.78 × 750⎜ ⎟ = 2.78 × 750⎜ ⎟ = 267 l / s ⎝ 6 + 10 ⎠ ⎝ t + 10 ⎠ 11.12 Using incremental areas read from time-area diagram: Q(1) Q(2) Q(3) Q(4) Q(5) Q(6)
= 0.05 x 20 = 0.40 x 20 = 0.40 x 20 = 0.40 x 20 = 0.45 x 20 = 0.35 x 20
+ 0.05 x 28 + 0.40 x 28 + 0.05 x 64 + 0.40 x 28 + 0.40 x 64 + 0.40 x 28 + 0.40 x 64 + 0.45 x 28 + 0.40 x 64
=1 = 9.4 = 22.4 = 44.8 = 45.8 = 45.2 9
Q(7) = 0.05 x 20 + 0.35 x 28 + 0.45 x 64 + ……….
= 39.6
Qp = 2.78 × 45.8 = 127 l/s
CHAPTER 12 12.1
Equation 8:10 ⎛ 1.5 × 10 −3 2.51× 1.14 × 10 −6 ⎞⎟ ⎜ + v f = −2 2g 0.002 × 0.75 log10 = 0.493 m 3 / s ⎜ 3.7 × 0.75 0.75 2g 0.002 × 0.75 ⎟ ⎝ ⎠ So, no, there is not sufficient capacity. Using B-P charts or other Colebrook-White part full pipe flow solution methods for Q = 15 l/s gives v = 0.44 m/s, so no this would not self cleanse.
12.5
E 2 l/s or 172 800 l/day DWF = PG + I + E = 5000 × 180 × 1.1 + 172800 = 1 162 800 l/day or 13.5 l/s setting = DWF + 1360P + 2E
= 1162 800 + 1360 × 5000 + 2 × 172 800 = 8 308 400 l/day or 96.2 l/s i.e. 7.1× DWF
12.6
Table 12.1 gives storage of 80P i.e. 400 000 l
or 400 m3
rate storage would fill = 500 − 96.2 = 403.8 l/s 400 000 = 16.5 minutes time to fill = 403.8
12.7
Table 12.2 gives multiplying factor of 2.0 for suspended solids so average storm concentration can be taken as 800 mg/l flow-rate to receiving water is 403.8 l/s mass in 10 minutes = 403.8 × 10 × 60 × 800 = 194 kg
12.9
D is 0.6 m, which fixes main dimensions: width 1.4 × 0.6 = 0.84 m length of weirs 8 × 0.6 = 4.8 m etc (as Fig. 12.7) use Fig.9.11 (all data confirms that this is appropriate) Bu = Bd = 0.84 m Qu = 350 l/s
Qu
2
gBu
5
= 0.03
L = 5.7 Bu 10
Yu − Pu = 0.06 Bu so upstream depth relative to weir crest (Yu − Pu ) ≈ 50 mm
Fig. 9.11 gives
Yd − Pd = 0 .1 Bu so downstream depth relative to weir crest (Yd − Pd ) ≈ 80 mm
Fig. 9.11 gives
57 = 15% 380 for total efficiency of 40% Table 12.6 gives K of 1.28 equation (12.1) Dmin = KQ 0.4 = 1.28 × 0.38 0.4 = 0.87 m so make diameter of inflow pipe 900 mm from Fig. 12.4: length inlet to scumboard 7D 6.3 m width of chamber 2.5D 2.25 m height of weir crest above inlet invert 1.2D 1.08 m
12.10 Flow ratio =
CHAPTER 13 13.1 Storm duration D (minutes) 8 12 16
13.2
VI (iAiD) (m3) 100.1 122.8 138.2
Intensity I (mm/h) 41.7 34.1 28.8
VO (QOD) (m3) 33.6 50.4 67.2
S (VI - VO) (m3) 66.5 72.4 71
2 O = 0.63 × × 1.5 2g × H 1.5 = 2.79 H 1.5 3 H
O
(m) 0 0.1 0.2 0.3 0.4 0.5
(m3/s) 0 0.088 0.25 0.458 0.706 0.986
S
(25 × 4 × H) (m3) 0 10 20 30 40 50
S O + ∆t 2 (m3/s) 0 0.127 0.292 0.479 0.686 0.910
S O against O + ∆t 2 follow procedure in Table 13.3
plot
11
CHAPTER 14 14.1
plot pump characteristic against system characteristic (static lift + losses) at operating point: flow-rate 0.105 m3/s head 19 m efficiency 55% ρ g × 0.105 × 19 = 36 kW power supplied = 0.55 in rising main, v = 2.1 m/s, which comfortably exceeds 0.75 m/s so OK
14.2
plot characteristic for pumps in parallel (as Figs 14.5 and 14.6) at operating point: flow-rate 0.14 m3/s head 23 m efficiency for each pump (flow-rate 0.07 m3/s) is 49% ρ g × 0.14 × 23 power supplied = = 64 kW 0.49 one pump is more efficient
14.4
equation (14.4) … outflow rate should be 60 l/s 900 × 0.06 i) equation (14.5) = 9 m3 V= 6 900 × 0.06 = 4 . 5 m3 ii) V = 12 4 .5 operating (emptying) = 2.5 minutes 0.06 − 0.03 4 .5 = 2.5 minutes idle (filling) 0.03
CHAPTER 15
15.2
DN 450 OD is 550 mm from Table 15.5 min trench width is OD + 0.7 giving 1.25 m depth of 2 m won’t affect this (Table 15.6) DN 200 OD is 244 mm Table 15.5 gives min trench width is OD + 0.4 giving 0.644 m however Table 15.6 (for depth 2 m) gives 0.9 m so this is minimum
15.3
from Table 15.4, the lower of the values for backfill and soil: Kµ ′ = 0.13 minimum trench width: Table 15.5 gives OD + 0.5 so 0.78 m; Table 15.6 gives 0.9 m so Bd = 0.9 m −2×0.13×3
equation (15.3) equation (15.2)
0. 9 1− e Cd = = 2.23 2 × 0.13 Wc = 2.23 × 19.6 × 0.9 2 = 35.4 kN/m
12
15.4
from Table 15.4: Kµ = 0.19 Bc is outside diameter of pipe, 0.28 m for complete projection, equation (15.5) Cc =
e
2×0.19×3
0.28
−1
= 152 2 × 0.19 for incomplete projection (Table 15.2) 3 − 0.09 = 16.9 Cc = 1.59 so use Cc = 16.9 (incomplete projection) 0.28 equation (15.4) Wc = 16.9 × 19.6 × 0.28 2 = 26 kN/m the lower value of Wc should be used – ‘wide trench’ 15.5
Fig. 15.6, light road, for H of 3 m, Ps is 10 kN/m2 equation (15.6) Wcsu = 10 × 0.28 = 2.8 kN/m equation (15.7) equation (15.1)
0.225 2 = 0.3 kN/m 4 We = 26 + 2.8 + 0.3 = 29.1 kN/m
Ww = 0.75 × 1000g × π
equation (15.8) Wt Fm ≥ We Fse so Wt Fm ≥ 29.1× 1.25 so … Wt (kN/m) Appropriate bedding Fm 28 1.3 Class F 36 1.0 Class D 48 0.76
CHAPTER 16
16.7
Using data from table 16.4:
Type
Depth (mm)
Vol (m3/m)
A C Total
0-300 300-320
0.252 0.024
Bulk density (kg/m3) 1720 1170
BOD (g/kg)
Unit BOD (g/m length)
BOD (kg)
3.1 5.4
1344 152
2016 228 2244
Volume of storm: V = 2.2 × 30 × 60 = 3960 m3 BOD (conc) = 2244/3960 = 567 mg/l
13
Type
Depth (mm)
Vol (m3/m)
A C Total
0-300 300-320
0.252 0.024
Bulk density (kg/m3) 1720 1170
COD (g/kg)
Unit COD (g/m length)
COD (kg)
16.9 20.5
7325 576
10988 863 11851
COD = 2993 mg/l 16.10 A bed roughness of 1.2 mm can be used to estimate bed friction factor from equation 16.10. Pipe runs half full so
λb =
1 ⎡ ⎛ k ⎞⎤ 4 ⎢log10 ⎜ b ⎟⎥ ⎝ 3.7D ⎠⎦ ⎣
2
=
1 ⎡ 1. 2 ⎞⎤ ⎛ 4 ⎢log10 ⎜ ⎟⎥ ⎝ 3.7 × 1000 ⎠⎦ ⎣
2
= 0.0205
Equation 16.9: vf =
8τ b
ρλb
=
8×2 = 0.88 m / s 1000 × 0.0205
For kb = 50 mm , λb = 0.0716 (from equation 16.10). Then rearranging 16.9 gives:
τb =
ρλbv 2 8
=
1000 × 0.0716 × 0.88 2 = 6 .9 N / m 2 8
16.12 Pot sediment accumulation rate can be calculated form equation 16.17: Note one sweep per month is 0.0125 sweeps per week e=
εκbr bs + br
=
0.65 × 14 × 0.05 = 7.28 g / m 2 .wk 0.0125 + 0.05
From equation 16.18: Tc =
(hmax Ap )Sd ε ' Ai e
=
0.090 × 1400 × 10 3 × 0.70 = 48 wk 250 × 7.28
16.13 Equation 16.19 can be rearranged to determine Q directly: 2 2 −3 2 2 ⎞ 0.60πg (0.5 × 10 ) 0.45 (2.65 − 1) ⎛ 1 ⎞ απgd ' D p (SG − 1) ⎛ 1 Q = ⎜ − 1⎟ =⎜ − 1⎟ 72υ 72 × 10 −6 ⎠ ⎝ 0.90 ⎠ ⎝ε = 2 .2 l / s
14
16.14 Using equations 16.14 and 16.15 applied to current conditions: f 0.25 b = br + s = 0.06 + = 0.085 wk −1 10 10 Equation 16.13: Xu =
κ b
=
5 = 58.8 g / m 2 0.085
Now for the new situation of quadrupled sediment supply: 20 κ b= = = 0.34 wk −1 X u 58.8 fs = 10(b − br ) = 10(0.34 − 0.06) = 2.8 wk −1
CHAPTER 17
17.7
Calculate hydraulic properties from equations 8.17 table 8.5 2d ⎞ 2 × 240 ⎞ −1 ⎛ ⎟ = 2 cos ⎜1 − ⎟ = 4.429 rad D ⎠ 300 ⎠ ⎝ 2 D2 (θ − sinθ ) = 0.30 (4.429 − sin 4.429 ) = 0.061m 2 A= 8 8 Dθ 0.30 × 4.429 P= = = 0.66 m 2 2 A R = = 0.091m P ⎛ θ⎞ B = D⎜ sin ⎟ = 0.24 m 2⎠ ⎝ ⎛ ⎝
θ = 2 cos −1 ⎜1 −
So from Manning’s equation (8.23): ⎛ ⎜ vn So = ⎜ 2 ⎜ 3 ⎝R
2
2
⎞ ⎛ ⎞ ⎜ 0.75 × 0.012 ⎟ ⎟ = 2 ⎟ = 0.0020 ⎜ ⎟ ⎟ ⎟ ⎜ ⎠ ⎠ ⎝ 0.0913
Q = vA = 0.75 × 0.061 = 46 l / s Given EBOD = 750 × 1.0710 = 1475 mg / l , from equation 17.4: Z=
3(EBOD ) P 3(1475 ) 0.66 = ≈ 7600 1 1 1 1 B 0 . 24 So2 Q 3 0.20 2 46 3
15
CHAPTER 18 500 000
? 10 r ⎞ ⎛ ⎜1 + ⎟ ⎝ 100 ⎠ 5% annual discount rate gives £307 000 – so rehab is appropriate 7% gives £254 000 – so rehabilitation is not appropriate in cost terms
18.1
Is cost of rehabilitation (£280 000) is less than
18.2
Renovation cost, plus replacement cost discounted over 25 years gives: 850 000 500 000 + = 751 000 1.05 25 This is less than the replacement cost (£850 000) so it is more cost-effective to renovate.
CHAPTER 21 21.4
Follow procedure in example 21.2, calculating L from equations (20.2) to (20.4), based on the worst case infiltration of 25 mm/h: D (h) 0.083 0.166 0.25 0.5 1 2 4 6 10 24 48
i (mm/h) 112.8 80.4 62 38.2 24.8 14.9 8.6 6.1 4 2 1.1
L (m) 3.73 5.27 6.06 7.30 9.09 10.09 10.10 9.46 8.32 5.80 3.50
The critical case is 10.1 m at 4h.
CHAPTER 23 23.8
For this cross section, A = 2.5 d2 and R = 0.5d. So from continuity: Q = vA = 0.3 = 1.0 × 2.5d2 d = 0.346 m From Manning’s equation (8.23):
16
⎛ ⎜ vn So = ⎜ 2 ⎜ 3 ⎝ 0.5d
2 ⎛ ⎞ ⎜ 1.0 × 0.025 ⎟ 2 ⎟ =⎜ ⎟ ⎜ 3 ⎠ ⎝ (0.5 × 0.346)
2
⎞ ⎟ ⎟ = 0.0065 ≈ 1:150 ⎟ ⎠
17
