SOLUTIONS TO SELECTED PROBLEMS

Transcription

1 URBAN DRAINAGE, rd edition, David Butler and John W Davies SOLUTIONS TO SELECTED PROBLEMS This solutions manual is made available free of charge. Details of the accompanying textbook Urban Drainage rd edition (ISBN hbk; pbk) are on the website of the publisher and can be ordered from Book.orders@tandf.co.uk or phone: +44 (0) First edition published 000 by E & FN Spon Second edition published 004 by Spon Press Third edition published 00 by Taylor & Francis Park Square, Milton Park, Abingdon, Oxon OX4 4RN Simultaneously published in the USA and Canada by Taylor & Francis, 70 Madison Avenue, New York, NY 006, USA Spon Press is an imprint of the Taylor & Francis Group, an informa business 000, 004, 00 David Butler and John W. Davies All rights reserved. No part of this book may be reprinted or reproduced or utilised 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, without permission in writing from the publishers. The publisher makes no representation, express or implied, with regard to the accuracy of the information contained in this book and cannot accept any legal responsibility or liability for any efforts or omissions that may be made. Publisher's note: This solutions manual has been produced from camera ready copy provided by the authors CONTENTS Chapter Page Number Note: Solutions not provided for all questions, and typically only for a selection of problems with numerical solutions. In other cases problems are simply issues to think about, and no solutions are provided.

2 CHAPTER. Volumetric concentration, C v V s /V, where V s is volume of solid and V is sample volume. So: C v Vs V 6 0 ppm 0 0 m / m Density of solid, ρ M/V, where M is its mass and ρ 650 kg/m. So: M 6 C ρ Cv kg / m 5 mg / l V. Time (hrs) Flow (l/s) COD (mg/l) COD (mg/s) Sum Av 9 48 From equation.: ΣQC C 08 mg l av Q 90 / 800 av Flow (l/s) or COD (mg/l) Time (hrs) Flow (l/s) COD (mg/l). Gram atomic weight of N is 4.0 g gram atomic weight of H is.0 g

3 gram molecular weight of ammonium is 4 + (4 ) 8 g From equation.: mgnh 4 / l 5 7 mgnh 4 N / l 8 So, total N conc mgn / l.4 Taken over a 0-year period, there were violations, whereas only up to 0 are permitted, so this water is not in compliance with the standard. CHAPTER 5 5. Ministry of health (equation 5.): a 000 i 0.0 mm / h D + b Bilham (5.): N.5( I / ) I 9.mm i I D mm / h ( I / ).55 Holland (5.4) N D( I / 5.4) I 9.8mm i 9.6mm / h ( I / 5.4).4 CHAPTER From equation 6.: kt f f + ( f f ) e f t t c + 9e o t c Time (min) Time (hour) f t (mm/h) calc f t (mm/h) appl Net rain (mm/h) Eff rain (mm/h) Assumes Horton infiltration starts when net rain begins.

4 6.5 Use the approach summarised in equation 6.8: Q() x 0 0 Q() x 50 + x 0 50 Q() x x 50 + x Q(4) x 75 + x x x 0 5 Q(5) x 50 + x 75 + x x 50 + x Q(6) x 5 + x 50 + x x x Q(7) x 0 + x 5 + x Q(8) x 0 + x 0 + x 5 + Peak 500 l/s after 40 mins. 6.7 From equation 6.a: t p ( α ) K ( ) h 4min From equation 6.b: Q p n α.5 K( α )! e 0.! e h The units are h- because this is based on an instantaneous inflow of unit volume. So, actual volume in this case: V 0 mm 00,000 m [0 ha] 000 m Q p 000 m /h 78 l/s 6.0 Based on conditions in Example 6.. From equation 6.0: k4it M ( t) M (0) e s s.9t Ms ( t) 8e From equation 6.: k 4Ms c A 5000 i M s Time (min) Time (hour) M s (kg) C (mg/l)

5 CHAPTER 7 7. From co-ordinates, W-E and N-S distance between manholes are 68.4 m and.45 m respectively. From Pythagoras, length of sewer is 7.79 m. Invert level of MHA (IL A ) m Change in height along sewer (ΔH) 7.79 / m IL B (entry invert level) m IL B (exit) m Depth m CHAPTER 8 8. k Q s v. 4 m/s D A vd.4 0. R e ν from Moody diagram (Fig. 8.4), λ 0. 04, transitional λl v h f 0.8 m D g 0. g from Wallingford chart (Fig. 8.5), hydraulic gradient 0.8 in Wallingford chart (Fig. 8.5): hydraulic gradient 0.8 in 00 needed for v of.0 m/s giving Q of 00 l/s v d Fig. 8.8: for v D f which gives D 80 mm k s D from Moody diagram (Fig. 8.4), for R e down to , λ L E k L 0.5 so L E m D λ 0.06 determine minimum v to ensure R e v giving v. 7 m/s so assumption is valid 5

6 d D 600 v from Fig Q and 0. Q f v f so so v m/s Q 80 l/s m A A 0.8 from Fig f P P.885 m from Fig f A f P f so A m so P 0. 6 m A R 0.09 m (or determine from R f and Fig. 8.8) P 0.6 Τ0 ρ grs N/m a) from Fig. 8.9 :90 b) d from Fig. 8.9 at 0. 8 D giving depth of 50 mm c) from Fig. 8.9 v would be 0.78 m/s d) from Fig. 8.9 for v of.0 m/s and Q of 0 l/s gradient must be : from Fig. 8.5 Q f is 400 l/s d 00 Q 0. from Fig D 450 d c Q f giving Q 48 l/s equation (8.8) D 0.45 therefore d c > d and so conditions are supercritical m d c equation (8.0) F r since F r >.7 use equation (8.).8.8 d c 0.5 d 0.8 m 0.7 d

7 CHAPTER 9 9. dia below 00 mm would not be suitable so try 00 mm and assume not drowned water level.5 m H m 0. Q 0.6π g. 95 l/s 4 water level.7 m gives Q of 0 l/s so 00 mm is the appropriate diameter check not drowned: Wallingford chart k s 0.6 mm (Fig 8.5) gives Q f of 40 l/s Q d water level.7 m: 0. 7 for this, Fig. 8.8 gives 0. 6 Q f D D0 00 Hmin Fig. 9.: giving. 9 thus H min is 0.8 m D 450 D o H > H min so not drowned 9. hydraulic gradient is.5 m in 5 m or 0 in 00 Wallingford chart for k s 0.6 mm (Fig 8.5) gives Q f of 0 l/s inlet-controlled? Try orifice equation with H.5 m 0. Q 0. 6π g. 5 l/s 4 so not inlet-controlled (in fact H would be less than.5 m because the pipe itself would have a gradient, see Fig. 9.6, but still unlikely to be inlet controlled) answer yes in the book is an error 9. b. m P.05 m H 0.5 m 0.5 equation (9.5) C d Q g [ ] 0. 5 m /s 9.4 From geometry in Figure 9.: B.5 y m z 40 From equation 9.9: zs C n o From 9.8: 8 Q 0.Cy m / s 8 7

8 CHAPTER The book refers here to a Problem 0.4 which has been omitted in error. This Problem 0.4 should have been: An urban catchment is drained by a separate foul sewer network and has an area of 500 ha and a population density of 75 hd/ha. At the outfall of this catchment, calculate: a) the average dry weather flow (in l/s) assuming water consumption is 60 l/hd.d, trade effluent is 0 m /ha.d over 0% of the catchment and infiltration is 0 l/hd.d b) the peak dry weather flow using Babbitt s formula. Solution to this is: a) Population , 500 Average domestic flow Trade flow Infiltration Total b) From table 0.4: l/d l/d l/d l/d 84 l/s P F P Assuming the peak factor is applied to all flows, Q p 0 l/s Solution to 0.5: a) Using a Butler-Pinkerton chart gives d/d 0.65 d mm b) Using the same chart gives a flow capacity of 40 l/s at d/d So solve for P (population) in: 5 ( P 000) P 404 ΔP 9 6 P P

9 0.8 p 0/ N 5 J0.999 Using equation 0.5b e.g. 5! 0 5 P ( 0,5) !5! And following equation 0.6: r P(r,N) Σ P(r,N) So r (max discharging simultaneously) mm dia :00 flowing d/d 0.75 Q 8 l/s (B-P charts) From equation 0.7: N.5 N N N 74 CHAPTER.4 a) Equation.: r b) r. 00 c) p /T / 0.0 d) r Use equation.5b (C.0): Q.78 i A i and Equation 5. (MoH rainfall) Sewer L (m) A (ha) ΣA (ha) t c t f + 4 (min) i (mm/h) Q (l/s)

10 . question should refer to pipe..5 A (ha) SUM Time (min) The time area diagram shows a tangent (dashed line) can be drawn from t 6mins, giving A 0.5 ha. So using the Rational method and MoH rainfall: A.05 Q l / s t should be pipe. Using incremental areas read from time-area diagram: Q() 0.05 x 0 Q() 0.40 x x Q() 0.40 x x x 64.4 Q(4) 0.40 x x x Q(5) 0.45 x x x Q(6) 0.5 x x x Q(7) 0.05 x x x Q p l/s 0

11 CHAPTER. Equation 8:0 v f g log g m / s So, no, there is not sufficient capacity. Using B-P charts or other Colebrook-White part full pipe flow solution methods for Q 5 l/s gives v 0.44 m/s, so no this would not self cleanse..5 E l/s or l/day DWF PG + I + E l/day or.5 l/s setting DWF +60 P + E l/day or 96. l/s i.e. 7. DWF.6 Table. gives storage of 80P i.e l or 400 m rate storage would fill l/s time to fill 6. 5 minutes Ignore this problem. The table referred to is not in edition. This problem should not have been included..9 D is 0.6 m, which fixes main dimensions: width m length of weirs m etc (as Fig..7) use Fig.9. (all data confirms that this is appropriate) B u B 0.84 m d Qu Qu 50 l/s gbu Yu Pu Fig. 9. gives B u L B u 5.7 so upstream depth relative to weir crest ( P ) 50 Yd Pd Fig. 9. gives 0. B u Y mm so downstream depth relative to weir crest ( P ) 80 u d u Y mm d

12 57. Flow ratio 5% 80 for total efficiency of 40% Table. gives K of equation (.) Dmin KQ m so make diameter of inflow pipe 900 mm from Fig..8: length inlet to scumboard 7D 6. m width of chamber.5d.5 m height of weir crest above inlet invert.d.08 m CHAPTER 4 4. Storm duration Intensity V I (ia i D) V O (Q O D) S (V I - V O ) D (minutes) I (mm/h) (m ) (m ) (m ) O g H.79 H H O S (5 4 H) S + Δt (m) (m /s) (m ) (m /s) S O plot + against O Δt follow procedure in Table 4. O CHAPTER 5 5. plot pump characteristic against system characteristic (static lift + losses) at operating point: flow-rate 0.05 m /s head 9 m efficiency 55% ρ g power supplied 6 kw 0.55 in rising main, v. m/s, which comfortably exceeds 0.75 m/s so OK

13 5. plot characteristic for pumps in parallel (as Figs 5.5 and 5.6) at operating point: flow-rate 0.4 m /s head m efficiency for each pump (flow-rate 0.07 m /s) is 49% ρ g 0.4 power supplied 64 kw 0.49 one pump is more efficient 5.4 equation (5.4) outflow rate should be 60 l/s i) equation (5.5) V 9 m ii) V 4. 5 m 4.5 operating (emptying). 5 minutes idle (filling). 5 minutes 0.0 CHAPTER 6 6. DN 450 OD is 550 mm from Table 6.5 min trench width is OD giving.5 m depth of m won t affect this (Table 6.6) DN 00 OD is 44 mm Table 6.5 gives min trench width is OD giving m however Table 6.6 (for depth m) gives 0.9 m so this is minimum 6. from Table 6.4, the lower of the values for backfill and soil: K μ 0. minimum trench width: Table 6.5 gives OD so 0.78 m; Table 6.6 gives 0.9 m so 0.9 m B d 0. equation (6.) 0. 9 e C d. 0. equation (6.) kn/m W c 6.4 from Table 6.4: K μ 0. 9 B c is outside diameter of pipe, 0.8 m for complete projection, equation (6.5) e C c for incomplete projection (Table 6.) C c so use C c 6. 9 (incomplete projection) 0.8

14 equation (6.4) W c 6 kn/m the lower value of W c should be used wide trench 6.5 Fig. 6.6, light road, for H of m, P s is 0 kn/m equation (6.6) W kn/m csu 0.5 equation (6.7) W w g π 0. kn/m 4 equation (6.) W kn/m e equation (6.8) so WtFm WeFse so W tf m W t (kn/m) F m Appropriate bedding 8. Class F 6.0 Class D Using data from table 7.4: CHAPTER 7 Type Depth (mm) Vol (m /m) Bulk density (kg/m ) BOD (g/kg) Unit BOD (g/m length) BOD (kg) A C Total 44 Volume of storm: V m BOD (conc) 44/ mg/l Type Depth (mm) Vol (m /m) Bulk density (kg/m ) COD (g/kg) Unit COD (g/m length) COD (kg) A C Total 85 COD 99 mg/l 4

15 7.0 A bed roughness of. mm can be used to estimate bed friction factor from equation 7.0. Pipe runs half full so λ b 4 log 0 k b.7d 4 log Equation 7.9: 8τ b 8 v f 0.88 m / s ρλ b For k b 50 mm, λ b (from equation 7.0). Then rearranging 7.9 gives: b τ ρλ bv 8 6.9N / m 7. Pot sediment accumulation rate can be calculated from equation 7.7: Note one sweep per month is 0.05 sweeps per week br e εκ b + b s r g / m. wk From equation 7.8: ( hmax Ap ) Sdε' Tc 48wk A e i 7. Equation 7.9 can be rearranged to determine Q directly: απgd' D ( S Q ε 7υ.l / s ) 0.60πg(0.5 0 ) p G 7.4 Using equations 7.4 and 7.5 applied to current conditions: fs 0.5 b br wk 0 0 Equation 7.: 5 X u κ 58.8g / m b (.65 ) Now for the new situation of quadrupled sediment supply: κ b 0 0.4wk X 58.8 f s u 0( b b ) 0( ).8wk r 5

16 CHAPTER Calculate hydraulic properties from equations 8.7 table 8.5 d 40 cos θ cos 4.49rad D 00 D 0.0 A θ sinθ 4.49 sin m 8 8 D P θ 0. 66m A R 0. 09m P θ B D sin 0. 4m ( ) ( ) So from Manning s equation (8.): S o vn R Q va l / s Given EBOD mg / l, from equation 8.4: Z ( EBOD) P ( 475) S B o Q

17 CHAPTER.4 Follow procedure in example., calculating L from equations (.) to (.4), based on the worst case infiltration of 5 mm/h: D (h) i (mm/h) L (m) The critical case is 0. m at 4h. CHAPTER.8 For this cross section, A.5 d and R 0.5d. So from continuity: Q va d d 0.46 m From Manning s equation (8.): vn S o : d ( ) 7