Solutions Manual Water-resources Engineering 3rd Edition David A. Chin

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Solutions Manual Water-Resources Engineering 3rd Edition David A. Chin Download full at: https://testbankdata.com/download/solut https://testbankda ta.com/download/solutions-manual-water-resources-en ions-manual-water-resources-en gineering-3rd-edition-david-chin/ Chapter 3 Design of Water-Distribution Systems 3.1. (a) For geometric geometric growth: growth: dP  =  k 1 P  dt dP  = P  �  �  Which gives k1 dt = ⇒ ln P  = k1t + C  P  =  P 0 ek1 t (b) For arithmetic growth: growth: dP  =  k 2 dt �  �  dP  = (c) For declining growth: growth: k2 dt = ⇒ dP  =  k 3 (P sat sat dt �  which gives dP  P sat sat − P  = �  k3 dt ′ 3.2. P ) − P ) ln(P sat P ) = −k3 t − C ′ ⇒ ln(P  sat − P ) = P  =  P sat sat where C  where  C  =  e −C  . P  =  k 2 t + P 0 − C e−k t 3 (a) By graphic graphical al extensi extension on P 2030 100, 000 2030  = 100, (b) For arithmetic growth: growth: P  =  kt  k t + P 0 where k  = P 1990 1990   61000 − 52000 − P 1980 1980 = = 900 10 10 Therefore P 2030 900t + P 1990 900(40) + 61000 61000 = 97, 97, 000 2030  = 900t 1990  = 900(40) 49 (c) For geometric growth: growth: P  =  P 0 ekt Therefore k(20) P 1990 1990  =  P 1970 1970 e 61000 = 40000e 40000e20k k  = 0.021 and hence k(40) P 2030  =  P 1990 = 61000e 61000e(0.021)(40) = 141, 141, 298 2030  = P  1990 e (d) For declining growth: growth: P  =  P sat sat where P  where  P sat sat  = 100,000 and − C e−kt (1) 1970 : t : t =  = 0, P  = P  = 40000 1990 : t : t =  = 20 20,, P  = P  = 61000 Substituting 1970 population into Equation 1 gives 40000 = 100000 − C e−k(0) = 100000 − C  which gives C  = 60 60,, 000 Substituting 1990 population into Equation 1 gives 61000 = 100000 60000ee−k(20) − 60000 which gives k  = 0.0215 Equation 1 can now be used to predict the 2030 population ( t  = 60) as P 2030 2030  = 100000 60000ee−0.0215(60) = − 60000 83 83,, 483 (e) Equations Equations 3.11 and 3.12 give the logistic-curv logistic-curvee parameters a parameters a  and  b  as P sat P 0 100 40 sat = = 1 .5 P 0 40 1 P 0 (P sat P 1 ) 1 40(100 sat b  =  ln  =  ln ∆t 10 52(100 P 1 (P sat P 0 ) sat a  = − − � − − � � − 52) − 40) �  = −0.0486 The logistic curve for predicting the population is given by Equation 3.9 as P  = P sat 100,, 000 100 sat = 1 + aebt 1 + 1. 1 .5e−0.0486 t In 2030, t 2030,  t  = 60 years and the population given by Equation 3.19 is P  = 92 92,, 487 people 50 (2) 3.3.  From the given data: P 0  = 13,000, P  13,000,  P 1  = 125,000, and P  and  P 2  = 300,000. The saturation population, P  tion,  P sat sat , is given by Equation 3.10 as 2P 0 P 1 P 2 P 12 (P 0 + P 2 )   2(13)(125)(300) (13)2 (13 + 300) P sat = = sat  = (13)(300) (125)2 P 0 P 2 P 12 − − − − −78 thousand people Since the calculated value of  P sat sat  is negative, estimation of the saturation population using Equati Equation on 3.10 is not possibl possiblee . 3.4. average daily demand = 580(100000) = 58, 58, 000 000,, 000 L/d = 0. 0.671 m3 /s × 108 L/d 2.18 m3 /s = 1. 1.89 × 108 L/d maximum daily demand = 1. 1.8(0. 8(0.671) 671) = 1.21 m3 /s = 1. 1.04 maximum hourly demand = 3. 3 .25(0. 25(0.671) 671) = 3.5.  The NFF can be estimated by Equation 3.20 as NFFi  = C   =  C i Oi (X  +  + P ) P )i where the construction factor, C  factor,  C i , is given by √  C i  = 220F  220F  Ai For the 5-story building, F  building,  F  =  = 1.0 (Table 3.3, Class 2 construction), and  A i  = 5000 m2 , hence √  C i  = 220(1. 220(1.0) 5000 = 16000 L/min where C i  has b een rounded rounded to the nearest nearest 1000 L/m L/min. in. The occupanc occupancy y factor, factor, Oi , is given by Table 3.4 as 0.85 (C-2 Limited Combustible, Office), ( X  + P   +  P ))i  can be estimated by the median value of 1.4, and hence the needed fire flow, NFF, is given by NFFi  = (16000)(0. (16000)(0.85)(1. 85)(1.4) = 1900 190000 L/min L/min This flow must be maintained for a duration of 5 hours (Table 3.6), hence the required volume, V , V , of water is given by V  = V  = 19000 3.6.   The × 5 × 60 = 5.5.7 × 106 L = 5700 5700 m3 needed needed fire flow flow for for any any buil buildi ding ng shoul should d not not exce exceed ed 45, 45,000 000 L/m L/min in for a dura durati tion on of  10 hours . 3.7. The (low-lift) (low-lift) supply pumps and the water-treatmen water-treatmentt plant plant should be designed designed for a capacity capacity equal to the maximum daily demand (Table (Table 3.7). With a demand factor of 1.8 (Table (Table 3.2), the per capita demand on the maximum day is equal to 1.8  600 = 1080 L/day/capita. Since the population served is 200 000 people, the design capacity, Qdesign  of the pumps and water-treatment plant is given by  × Qdesign  = 1080 L/d d= × 200000 = 2.2.16 × 108 L/ 51 2.50 m3 /s The required fire flow, Qfire, is 28000 L/min = 0.467 m 3 /s. Accordi According ng to Table able 3.6 the fire flow flow must must be able to be sustaine sustained d for 7 hours hours . The volum volume, e, V fire fire , required for the fire flow will be stored in the service reservoir, and is given by V fire fire  = 0.467 3600 = × 7 × 3600 11800 118 00 m3 The required flow rate in the distribution pipes is equal to the maximum daily plus fire demand or the maximum hourly demand, whichever is greater. maximum daily + fire demand = 2. 2 .50 + 0. 0.467 = 2. 2.98 m3 /s 3.25 1.80  × 2.50 = 4.4.51 m3/s maximum maximum hourly demand = where where a demand demand factor factor of 3.25 has been assumed assumed for the maximu maximum m hourly hourly demand. demand. The design capacity of the supply pipes in the distribution system should therefore be taken as 4.51 m3 /s . 3.8.  According to Table 3.8, the minimum acceptable pressure under average daily demand condition ditionss is 240 kPa kPa . 3.9.  From the given data: P  = P  = 200,000, per capita demand = 500 (L/d)/person = 5. 5 .79 3 3 (m /s)/person, and fire flow = 30,000 L/min = 0.5 m /s. × 10−6 (a) The required required flow rate in the main distribution distribution pipeline is the maximum daily demand demand plus fire demand or maximum maximum hourly demand, whicheve whicheverr is greater. greater. Taking a peaking factor of 1.8 for maximum daily demand and 3.25 for maximum hourly demand gives × 10−6) + 0.0.5 = 2.58 m3/s maximum hourly demand = 3. 3 .25(200000)(5. 25(200000)(5.79 × 10−6 ) = 3.76 m3 /s maximum daily demand + fire demand = 1. 1 .8(200000)(5. 8(200000)(5.79 Therefore, Therefore, the design capacit capacity y of the distribution pipeline pipeline should be 3.76 m3 /s . For a maximum allowable velocity of 1.5 m/s, the minimum required pipe diameter,  D,  D , requires that π 2 Q 3.76 D = = 4 V  1 .5 which yields D  = 1.78 m The The pipe mate materia riall to be used used in duct ductil ilee iron iron pipe pipe (DIP) (DIP) and and a cov cover of 1.5 1.5 m would ould be appropriate. appropriate. (b) For DIP with ks  = 0.26 mm, D  = 1.8 m, A  = 2.544 m2 , Q  = 3.80 m3 /s, V  =  Q/A = 1.4944 m/s, 1.49 m/s, and Re = V =  V D/ D/ν  ν = = (1. (1.494)(1. 494)(1.80)/ 80)/10−6 = 2.69 106 , × f  = �− � 2log ks 5.74 + 3.7D Re0.9 −2 �� �− � = 0.26 5.74 2log  + 3.7(1800) (2. (2.69 106 )0.9 52 × −2 �� = 0.0134 and, taking γ  taking  γ  =  = 9.79 kN/m3 , the Darcy-Weisbach equation gives  p1  p2  L V 2 =  f  γ  γ  D 2g 550 350 L (1. (1.494)2 = 0.0134 9.79 9.79 1.8 2(9. 2(9.81) −  − which yields L  = 24 24..1 km For old DIP, take C  take  C H  H  = 80 (Table 2.2) and the Hazen-Williams formula gives  p1 γ  550 9.79  − − � � � �  p2 L =  h f  = 6.82 1.17 γ  D 350 L = 6.82 1.17 9.79 1.8 V  C H  H  1.494 80 1.85 1.85 which yields L  = 9.4 km The Hazen-Williams equation yields a significantly shorter distance. The Darcy-Weisbach Darcy-Weisbach equation is preferable since it covers all flow regimes, while the Hazen-Williams equation is restricte restricted d to a narrow narrow range of flow flow conditi conditions ons.. Also, Also, the value value of  C H  H  might not be comparable to the value of  k  k s  used in the Darcy-Weisbach equation. 3.10. The require required d storage storage is the sum of three three component components: s: (1) volume volume to supply supply the demand demand in excess of the maximum daily demand; (2) fire storage; and (3) emergency storage. The volume to supply the peak demand can be taken as 25% of the maximum daily demand volum volume. e. Problem Problem 3.7 gives gives the maximu maximum m daily daily flow rate as 2.50 m3 /s, hence the storage volume to supply the peak demand fluctuations over a day,  V peak peak , is given by V peak (0.25)(2. 25)(2.50 peak  = (0. × 86400) = 54000 m3 The required fire flow,  Q f , is calculated in Problem 3.7 as 0.467 m 3 /s (= 28,029 L/min) and, according to Table 3.6, this fire flow must be maintained for at least 7 hours. The volume to supply the fire demand,  V fire fire , is therefore given by V fire (0.467 fire  = (0. × 3600)(7) = 11768 m3 The emergency storage, V  storage,  V emer emer , can be taken as the average daily demand, in which case V emer emer  = 200000 × 600 = 1.1.2 × 108 L = 120000 m3 The required volume, V  volume,  V ,, of the service reservoir is therefore given by 3 V  =  V peak peak  + V fire fire + V emer emer  = 54000 + 11768 + 120000 = 185768 m The service service reservoir reservoir should should be designed designed to store about 185 800 m 3 of water. 53 3.11. (a) For line line AB: population served = 50000 + 20000 = 70000 people average average demand = (70000)(0 (70000)(0.6) = 42000 m3 /d = 0. 0.486 m3 /s maximum daily demand = (0. (0 .486)(1. 486)(1.8) = 0. 0.875 m3 /s maximum hourly demand = (0. (0.486)(3. 486)(3.25) = 1. 1.58 m3 /s fire demand = 15000 + 10000 = 25000 L/min = 0. 0.417 m3 /s maximum daily + fire demand = 0. 0 .875 + 0. 0.417 = 1. 1.29 m3 /s design flow = max(1. max(1.29 29,, 1.58) 58) = 1.58 m3 /s For line BC: population served = 20000 people average demand = (20000)(0. (20000)(0.6) = 12000 m3 /d = 0. 0.139 m3 /s maximum daily demand = (0. (0 .139)(1. 139)(1.8) = 0. 0.250 m3 /s maximum hourly demand = (0. (0.139)(3. 139)(3.25) = 0. 0.452 m3 /s fire demand = 10000 L/min = 0. 0.167 m3 /s maximum daily + fire demand = 0. 0 .250 + 0. 0.167 = 0. 0.417 m3 /s design flow = max(0. max(0.417 417,, 0.452) 452) = 0.452 m3 /s (b) For the steel transmission transmission pipe (k (k  = 1 mm, D mm,  D  = 1200 mm): π π A  = D2 = 1.22 = 1.131 m2 4 4 1.58 V AB = 1.40 m/s AB  = 1.131 V AB (1. (1.40)(1. 40)(1.2) AB D ReAB  = = = 1.68 106 ν  10−6 1 k/D 5.74 1/1200 5.74 = 2log +  = 2log + 0.9 3.7 3.7 (1. (1.68 106 )0.9 f AB ReAB AB × √  − � � � − × � × � f AB AB  = 0.0191 0.452 V BC = 0.400 m/s BC  = 1.131 V BC (0. (0.400)(1. 400)(1.2) BC D ReBC  = = = 4.80 105 − 6 ν  10 1 k/D 5.74 1/1200 5.74 = 2log +  = 2log + 0.9 3.7 3 .7 (4. (4.80 105 )0.9 f BC ReBC BC f BC BC  = 0.0196 × √  − � � − � Applying the energy equation between A and B, neglecting minor losses, gives 2 0 0 − 2 LAB V AB pB  V AB + h p  = + + zB − f AB AB D 2g γ  2g 5000 1.402 550 1.402 (0. (0.0191)  + h p  =  +  + 20 1.2 2(9. 2(9.81) 9.79 2(9. 2(9.81) 54 which gives h gives  h p  = 84.2 m. The required specific speed,  n s , (where ω (where  ω  = 1200 rpm = 125.7 rad/s) is therefore given by 1 ns  = 1 ωQ 2 3 (gh p ) 4 =  (125.  (125.7)(1. 7)(1.58) 2 (9. (9.81 84..2) × 84 3 4 = 1.03 This This is a centri centrifuga fugall pump . Applying the energy equation between B and C, neglecting minor losses, gives 2 LBC V BC 0 f BC  + h p  = BC  D 2g 7000 0.4002 (0. (0.0196)  + h p  = 1.2 2(9. 2(9.81) − 0 − 2 pC   V BC + + zC  γ  2g 480 0.4002  +  + 20 9.79 2(9. 2(9.81) which gives h gives  h p  = 69.97 m. The required specific speed,  n s , is therefore given by 1 ns  = 1 ωQ 2 3 (gh p ) 4 =  (125.  (125.7)(0. 7)(0.452) 2 (9. (9.81 69..97) × 69 3 4 = 0.63 This This is a centri centrifuga fugall pump . (c) For the storage reservoir, reservoir, taking the daily service storage storage as 25% of the maximum daily demand volume and noting that 10000 L/min fire flow is to be maintained for 3 h, and 15000 L/min for 4 h, gives service storage = (0. (0.25)(1. 25)(1.8)(42000) = 18900 m3 fire storage = (10)(60)(3) + (15)(60)(4) = 5400 m3 emergency storage = 42000 m3 required required storage = 18900 + 5400 + 42000 = 66300 m3 3.12.  From the given data: Q  = 4.67 L/s, L  = 110 m, p1  = 380 kPa, and ∆z ∆z   = 3 m. The The pipe pipe velocity is given by Q 4.67 10−3 0.00595 V  = = = π 2 A D2 4D × Hence, for V for  V <  2.  2 .4 m/s, D> �  0.00595 = 0.0498 m = 49. 49.8 mm 2.4 Take D Take  D  = 50 mm and see if this is adequate for the pressure. For copper, ks  = 0.0023 mm, 55 × 10−6 m/s2, 4.67 × 10−3 V  = = 2.38 m/s and at 20◦ C, ν  C,  ν  = 1.00 π (0.050)2 4 (0. vD (2. (2.38)(0. 38)(0.050) = = 119 105 − 6 ν  1 10 0.25 0.25 f  = 2 = 2 = 0.0175 k 5.74 0.0023 5.74 log 3.7D + log 3.7(50)  + (1.19×105 )0 9 Re0 9  L V 2 110 2.382 hf  =  f  = 0.0175 = 11 11..12 m D 2g 0.050 2(9. 2(9.81)  p1  380  p2  = γ   =  γ  ∆z hf   = 9.79 3 11 11..12  = 241 kPa γ  9.79 Re = × × � � � �� � � s . − − � � . �� �  − − Since p Since  p 2  > 2  >  240 40 kPa, kPa, a 50 mm copper copper line line is (barely (barely)) adequat adequate. e. 3.13.  The design calculations for this problem are summarized in Table 3.1. the given data: Qref  = 200 L/min = 0.00333 m3 /s, L1  = 20 m, L2  = 5 m, ∆z ∆ z1 = 2 m, ∆z2   = 3 m, p0   = 380 kPa, and p2   = 240 kPa. For galvaniz galvanized ed iron, ks   = 0.15 mm = − 4 1.5 10 . From the supply supply pipe to the first floor: 3.14.  From ×  p0  V 02 p1  V 12 L1 V 12 + + z0  = + + z1  + f 1 γ  2g γ  2g D1 2g where π 2 D = 0.7854 7854D D2 4 2Qref  2(0. 2(0.00333) 0.008487 V 1  = = = A1 0.7854 7854D D2 D2 V 1 D 0.008487 D  8487 Re1  = = = ν  D2 10−6 D 0.25 f 1  = = 2 4 5.74 log log 1.53×.710 + 8487 09 D ( ) A1  = � � � � − . D �� � � 0.25 4.054×10 5 − D + 0. 0.001671 001671D D0.9 �� 2 Combining Combining the above above equations equations and substituting substituting known quantities quantities yields  p1 = 36 36..02 γ  − 0.25 � � log 4.054×10 5 − D From the first to the second floor, + 0. 0.001671 001671D D0.9 7.342 10−5 2 D5 ��  p1  V 12 p2  V 22 L2 V 22 + + z1  = + + z2  + f 2 γ  2g γ  2g D2 2g 56 × (1)      l    e    a    r      )    n    u    a     6     6     0       6     1       9      i    s      P     1     2     0       4     6       0    s    m    e     k     2     2     2       1     1       1    r    r      (    e      T     P    s      h     t    g    n    e      L    e    p      i      P     t    n    e      l    a    v      i    u    q      E      f    o    s    m    r    e      T    n      i    s    g    n      i     t     t      i      F      d    r    a      d    n    a     t      S    n      i    s    s    o      L      d    a    e      H    :      1  .      3    e      l      b    a      T      l    a    n     d     )      3     2     6     1     4     9      i      3    a  .      1  .      3  .      4  .      4  .      2  .    e    m     2    m      3      0      5      6      (    r      H      2     2     2     1     1     1      1    e      T      0    v     ff     )  .  .    e     i  .      0      0     1     5      0     0      l    m      2      4      (      D      E    −    r    s    e     )      2    e    s      h  .      0     0     0     0     0    s    m     1     t    o      (      5      O     L      1    n    o    s      )      7     1     6     5     2     5      i    s     t    o  .      2  .      1  .      1  .      0  .  .      1    m     3    c      i      (      1      0      0      5      0      5      L    r      F      l      h      2     6     5     2     3     2    a     t      5      5    g     )      1     t  .      3  .      1  .  .      5  .      7  .      7    m    n    o    e     (      2      7      5      6      4      4      T     L      2    g     h    n     t      6     5     2     3     2      i    g     )      2     t  .      6  .      0  .      5  .      1  .      5  .    m     1    n     (     t      i    e      5     4     3     1     2     1      F     L    y     )     t      i    s      2     7     7     2     7     2    c     /      1    o  .      1  .      1  .      6  .      3  .      5  .      l    m     2      1     1     2     0     1    e     (      V      )    m   m     4     1     1     5     1     5    a      i    m     6     5     5     2     5     2      D     (      h     t      )      0      0      0  .      7  .      0  .    g  .      5  .      6  .      6    m     7     2    n     (      2      4      1      4      4    e      L      )    n      i    w      9     4     4     5     5     5    o    m     0     4     4     4     4     4      l      /      F     L     4     1     1      (    g    n     d      2     3     2     6     6     4      i     t    a     )      8  .      4  .      4  .  .      3  .      1  .      4    r    e    m     8      2      3      0      0      6      (    a     t      H      3     2     2     2     2     1      S      ’      ’      ’      ’      ’    e      F     ’      E      C     ’      D     ’    p     B     B     ’      i      A      B      P      B     C     C     D 57 where π 2 D = 0.7854 7854D D2 4 Qref  0.00333 0.004244 V 2  = = = A2 0.7854 7854D D2 D2 V 2 D 0.004244 D  4244 Re2  = = = ν  D2 10−6 D 0.25 f 1  = 2 = 4 5.74 log log 1.53×.710 + 4244 09 D ( ) A2  = � � � � − . D �� � � 0.25 4.054×10 5 − D + 0. 0.003118 003118D D0.9 �� 2 Combining Combining the above above equations equations and substituting substituting known quantities quantities yields  p1 = 27 27..51 + γ  0.25 � � log 4.054×10 5 − D + 0. 0.003118 003118D D0.9 4.590 10−6 2 D5 �� × (2) Solving Equations 1 and 2 for D for  D  and taking the next larger available diameter yields D  = 0.0508 m = 2 in. as the required required pipe diameter. diameter. For this diameter, diameter, the actual actual pressure pressure on the second floor ( p ( p2 ) is 262 kPa and the pressure on the first floor ( p ( p1 ) is 295 295 kPa kPa . 58 Chapter 4 Fundamentals of Flow in Open Channels 4.1. V  = V  = 1 m/s A  = (b + my) my )y  = (5 + 2 × 2)2 = 18 m2 Therefore Q  = V  =  V A  = (1)( (1)(18 18)) = 18 m3 /s 4.2. Q  = 8 m3 /s, w /s,  w 1  = 4 m, V  m,  V 1  = 1 m/s, and Q = w  =  w1 y1 V 1 which leads to Q 8 = = 2m w1 V 1 (4)(1)  0.5 m = 1.5 m, and y1  = m,  y 2 =  y 1 w2  = 5 m, y  − Q = w  =  w2 y2 V 2 which leads to V 2  = 4.3.   The Q 8 = = 1.07 m/s w2 y2 (5)(1. (5)(1.5) hydraulic radius,  R,  R , is defined by R  = where, for circular pipes, Hence or πD 2 A  = 4 A P  and P  =  πD  π D πD 2 /4 D R  = = πD 4 D  = 4R 59 4.4.  The shear stress,  τ 0 , on the perimeter of the channel is given by τ 0  = γ  =  γRS  RS 0 (1) From the given data b data  b =  = 5 m, y m,  y =  = 1.8 m, m m,  m =  = 1.5, and the geometric geometric properties of the channel channel are A  = by  =  by +  + my 2 = 5(1. 5(1.8) + 1. 1.5(1. 5(1.8)2 = 13 13..86 m2 √  P  =  b + 2 1 + m2 y  = 5 + 2 A 13 13..86 R  = = = 1.21 m P  11 11..49 √  1 + 1. 1 .52 (1. (1.8) = 11. 11.49 m From the given data, τ  data,  τ 0  = 3.5 N/m2 , and since γ  since  γ  =  = 9790 N/m2 , Equation 1 gives the maximum allowable slope, S  slope,  S 0 , as τ 0 3. 5 S 0  = = = 0.00030 γR (9790)(1. (9790)(1.21) For the excav excavated channel, channel, ks   = 3 mm = 0.003 m, and ν  = 1.00 Substituting these data into Equation 4.38 gives the flow rate,  Q,  Q , as Q  = Q  = −2A √  8gRS 0 log10 ks 0.625 625ν  ν   + 12 12R R R 8gS 0 3 2 √  √  −2(13. 2(13.86) = 17 17..2 m3 /s � � � 8(9. 8(9.81)(1. 81)(1.21)(0. 21)(0.000 00030) 30) log10 × 10−6 m2 /s at 20◦ C. 0.003 0.625(1. 625(1.00 10−6 )  + 12(1. 12(1.21) (1. (1.21) 8(9. 8(9.81)(0. 81)(0.00030) 3 2 √  × � Therefore, for the given flow depth restrictions in the channel, the flow capacity of the channel is 17.2 17.2 m3 /s . the given data: b  = 8 m, S 0  = 0.0001, ks  = 2 mm = 0.002 m, and Q  = 15 m3 /s. /s. At ◦ − 6 2 20 C, µ C,  µ  = 1.00 10 m /s, and, for a rectangular channel, 4.5.  From × A  = by  =  by and R  = by 2y + b Substituting into Equation 4.38 gives Q = 15 = −2A √  �  8gRS 0 log10 −2(8y 2(8y) 8(9. 8(9.81)( � ks 0.625 625ν  ν  + 12 12R R R 8gS 0 3 2 √  � � −6 8y 0.002 0.625(1. 625(1.00 10 ) )(0. )(0.000 0001) 1) log10 + 8y 8y 2y + 8 12( 2y+8 ) ( 2y+8 ) 8(9. 8(9.81)(0. 81)(0.0001) which yields y  = 2.25 m Therefore, Therefore, the uniform-flow uniform-flow depth depth in the channel is 2.25 m . 60 3 2 √  × � 4.6.  Hydraulically rough flow conditions occur in open channels when u∗ ks ν  ≥ 70 (1) where √  u∗  = gRS f  f  (2) Equation 4.46 can be rearranged and put in the form ks  = n 0.039 6 � � = 2.84 × 108n6 (3) Substituting Equations 2 and 3 into Equation 1 and noting that ν  = 1.00 20◦C and g  = 9.81 m/s2 yields √ 9.81 √   × 2.84 × 108n6 ≥ 70 × 10−6 RS f  f  1.00 which simplifies to n6 × 10−6 m2/s at √   ≥ 7.9 × 10−14 RS f  f  (4) From the given data: b  = 5 m, S  m,  S 0  = 0.05% = 0.0005, and by definition: R  = A by 5y = = P  2y + b 2y + 5 (5) Equation 4, can be combined with Equation 5 to give the following condition for fully turbulent flow, 5y (0. (0.013)6 (0. (0.0005) 7.9 10−14 2y + 5 � � � This condition is satisfied when y 4.7.  The ≥ ×  ≥ 0.683 m . Darcy-Weisbach uniform-flow equation is given by Equation 4.38 as Q  = √  −2A 8gRS 0 log10 � ks 0.625 625ν  ν   + 3 12R 12 R R 2 8gS 0 √  � where the following variables are known: y  = 2.20 m S 0  = 0.0006 ks  = 2 mm = 0. 0.002 m ν  =  = 1.00 × 10−6 m2/s g  = 9.81 m/s2 A  = 3.6y + 2y 2 y 2 = 3.6(2. 6(2.20) + 2(2. 2(2.20)2 = 17 17..6 m2 R  = 3.6y + 2y 2y2 3.6(2. 6(2.20) + 2(2. 2(2.20)2 = = 1.31 m 3 .6 + 2 5 y 3.6 + 2 5(2. 5(2.20) √  √  61 (1) Substi Substitut tuting ing these these variable ariabless into into Equati Equation on 1 yields yields Q   = 34.0 m3 /s . Sinc Sincee y   = 2.20 m corresponds corresponds to A to  A  = 17.6 m2 , then V  then  V    = 34.0/17.6 34.0/17.6 = 1.93 m/s m/s . The Manning’s equation gives the average velocity,  V ,  V  , as 1 2 12 V  = R 3 S 0 n Table 4.2 indicates that a mid-range roughness coefficient for concrete is n   = 0.01 0.015. 5. The The average velocity given by the Manning equation is 2 1 1 (1. (1.31) 3 (0. (0.0006) 2 = 1.96 m/s 0.015 and the corresponding flow rate, Q rate,  Q,, is V  = Q  = AV   =  AV  =  = (17. (17.6)(1. 6)(1.96) 96) = 34. 34.5 m3 /s Hence, Hence, in this case, the Darcy-Weisba Darcy-Weisbach ch and Manning equations give give the similar similar results . 4.8.  The Darcy-Weisbach uniform-flow equation is given by Equation 4.38 as Q  = √  −2A 8gRS 0 log10 � ks 0.625 625ν  ν   + 3 12R 12 R R 2 8gS 0 √  � (1) where the following variables are either known or can be expressed in terms of the uniform-flow depth, y depth,  y : S 0  = 0.0001 ks  = 1 mm = 0. 0.001 m Q = 18 m3 /s ν  =  = 1.00 × 10−6 m2/s g  = 9.81 m/s2 A  = 5y + 2y 2 y2 R  = 5y + 2y 2y2 5 + 2 5y √  Substituting these variables into Equation 1 and solving for  y  yields y  yields  y  = 2.19 2.19 m . Check u∗ ks /ν  and R/ks  to determine the state of the flow and the validity of the Manning equation. Taking y Taking  y  = 2.19 m gives R gives  R  = 1.39 m and √  √  (9. (9.81)(1. 81)(1.39)(0. 39)(0.0001)(0. 0001)(0.001) u∗ ks gRS 0 ks = = = 37 ν  ν  1.00 10−6 1.39 R =  = 1390 ks 0.001 × Therefore, since 5 u∗ ks /ν  70 (i.e., 5 37 70) then according to Equation 5.19 the flow is in transit transition ion . Since Since u∗ ks /ν <  70 (i.e., 37 <  70) and R/ks >  500 (i.e., 1390 >  500), then the Manning Manning equation equation is not valid valid . ≤  ≤  ≤ ≤ ≤ 62 4.9.   Comparing the Manning and Darcy-Weisbach equations 1 �  8g R6 = f  n which gives √ f R n = √  8g 1 6 1 = 1 1 1 f 2 R 6 f 2 R 6 = 8.86 8(9. 8(9.81) √  If the friction factor, f , f , is taken as a constant, the above relation indicates that n  will be a 1 function of the depth (since  R is  R  is a function of the depth). If  f   f  R− 3 ,  n  would be a constant in the above above equation equation.. So the answer answer to the questio question n is no .  ∼  ∼ 4.10.   Given: Q  = 20 m3 /s, n /s,  n  = 0.015, S  0.015,  S 0  = 0.01 (a) Manning Manning equation is given given by 5 2 1 1 1 An3 12 Q  = An Rn3 S 02 = S  n n P  23 0 n where An  = [b + myn ]yn  = [2. [2.8 + 2y 2 y n ]y n P n  = b  =  b + 2 √  √  1 + m2 yn  = 2.8 + 2 5yn  = 2.8 + 4. 4 .472 472yyn Substituting into the Manning equation yields 5 1 1 [(2. [(2.8 + 2y 2 yn )yn ] 3 2 20 = (0. (0 . 01) 2 0.015 (2. (2.8 + 4. 4 .472 472yyn ) 3 or 5 [(2. [(2.8 + 2y 2 yn )yn ] 3 (2. (2.8 + 4. 4 .472 472yyn ) 2 3 = 3.0 Solving by trial and error yields yn  = 0.91 m (b) Comparing Comparing the Manning Manning and Darcy-Weisba Darcy-Weisbach ch equations gives 1 �  8g R6 = f  n which leads to f  = 63 8gn 2 1 R3 In this case A  = (2. (2.8 + 2y 2 y )y  = (2. (2.8 + 2 91)(0.91) = 4. 4.2 m2 × 0.91)(0. P  = 2.8 + 4. 4 .472(0. 472(0.91) = 6. 6.87 m A 4.20 R  = = = 0.611 m P  6.87 therefore f  = 8(9. 8(9.81)(0. 81)(0.015)2 1 (0. (0.611) 3 = 0.0208 For fully turbulent flow, where the Manning equation applies, 1 = f  √  ks 12R 12 R ks 2log 12(0. 12(0.611) 2log[0. 2log[0.136 136k ks ] −2log √ 0.10208 = − 6.93 = − � � � � which leads to 002499 m = 2.5 mm ks  = 0.0024 4.11.   From the given information, 1 n  = 0.039 039d d6 where d  is in m. In this this case, case, d  = 30 mm = 0.030 m, and a 70% error in d  is 0.7(0.030) = 0.021 0.02 1 m. Hence, Hence, d  = 0.030 m  0.021 m. Hence, Hence, the “best estimate” estimate” of  n, denoted by n ¯ , is given by 1 n ¯  = 0.039(0. 039(0.030) 6 = 0.022  ± The lower estimate of  n,  n ,  n L , is given by nL  = 0.039(0. 039(0.030 − 0.021) 1 6 = 0.018 and the upper estimate of  n,  n , nU , is given by 1 nU  = 0.039(0. 039(0.030 + 0. 0.021) 6 = 0.024 The maximum percentage error in estimating n  is therefore given by error = 4.12.  According 0.022 0.018 0.022 − 100 = × 100 18% 18% to Equation 4.45, n 1 6 ks = √ 18g 2.0log 64 1 6 �� � � R ks 12 kRs (1)