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2 New Website: M P E il Add Mr. Peterson s Address:

3

4 If 6 m 3 of oil weighs 47 kn calculate its If 6 m 3 of oil weighs 47 kn, calculate its specific weight Ƴ and specific gravity.

5 specific weight Ƴ = 47 kn = kn/m 3 6 m 3 specific gravity = Ƴoil = kn/m 3 = Ƴwater 9.79 kn/m 3

6 If 1 m 3 of concrete has a mass.4 Tons, calculate its specific weight Ƴ and specific gravity.

7 9.81 N / kg specific weight 300 kg.56 kn / m 3 1m 3 3 con.56 kn / m specific gravity.30 3 water 9.79 kn / m

8

9 Vacuum = space with less that atmospheric pressure Atmospheric pressure refers to prevailing pressure in the air around us At sea level, l standard d atmospheric pressure is: p h kpa, or 760 mm of mercury, or 1atmosphere

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11

12 h p Atmospheric pressure p kpa p kpa kn / m

13 Compute : Specific gravity of mercury = 13.6 Remember, weight of substance specific gravity = = weight of equal amount of water = specific gravity weight of equal amount of water = kn / m kn / m 3

14 p h p kn / m kn / m kn / / m h 0.76 m 760 mm kn / m

15 1.1m

16 Weight of glycerin 1.34kN m kn Pressure p at A h m m kN 13.57kPa m

17 Determine the gage pressure in kpa at a depth of 10.0 meters below the free surface of a body of water.

18

19 Weight of water 9.79 kn / m 3 p h 9.79kN 97.9kN p m kpa 3 m m

20 Find the pressure at the bottom of a tank containing glycerin under pressure as shown in following figure.

21

22 pressure at bottom p = 50kPa 50 kn m weight of glycerin = 50 h 1.34kN 3 m 50kN 1.34kN 74.68kN p m 74.68kPa 3 3 m m m

23 (a) Find the elevation of the liquid surface in Piezometer A (b) The elevation of mercury in Piezometer B (c) The pressure at the bottom, Elevation 0

24 (a) The liquid in Piezometer A will rise to the same elevation as the top of the tank.

25 979kN 9.79 ( b) pa h (0.7 )(1.7 m) 3 m 11.98kN 11.98kPa m kn / m ha p / ( kn / m ) h TOTAL m

26 () c p0 pa pb kpa ( kn / m )(0.3 m) kPa

27

28 F hcga hcg A specific weight of liquid depth of the center of gravity Area

29 4.0m A 0.1m Water 0.0m 7.0m Tank Width 3m Example

30 Total force (F BC ) on the bottom of the tank Total weight (W) of the water Explain the difference

31 4.0m A 0.1m Water 0.0m 7.0m Tank Width 3.0m Example

32 FBC pa ( h ) A 9.8kN (6 m ) (7m 3 m ) m 3 F BC 135 kn

33 4.0m A 0.1m Water 0.0m 7.0m Tank Width 3m Example

34 W ( Volume) 98kN 9.8 W [(7mm3 m(4m0.1 m 3 )] m 3 W 416kN

35 4.0m A 0.1m Water 0.0m 7.0m Tank Width 3m Example

36 The Force on the bottom of the tank is: FBC 135 kn But the total weight of the water is only: W 416 kn What is the source of the additional force?

37 4.0m A 0.1m Water 0.0m 7.0m Tank Width 3m Example

38 FAD pa ( h) A 9.8 kn 3 m (4 m ) (7m 3m 0.1 m ) F AD 819 kn

39 4.0m A 0.1m Water 0.0m 7.0m Tank Width 3m Example

40 135kN 416kN 819Kn and dtherefore: F BC W F AD

41 The large forces that can be developed with a small amount of liquid (the liquid in the tube) acting over a much larger surface. Hydraulic lift

42

43 A stone weighs 90N in air. When immersed in water it weighs 50N. Compute the volume of the stone and its specific gravity.

44 Y 0 W T FB 0 T 50N Wt Water FB W T W 90 N FB 90N 50N FB 90N 50N 40N FBF

45 buoyant force FB 40N 40 N 9.8 kn v 3 m 40 N 3 v m 4.1 liters N / m

46 weight of the stone specific gravity = weight of an equal volume of water 90N specific gravity =.5 40N

47 An object that is 0.m wide by 0.m thick by 0.4m wide is found to weigh 50N in water at a depth of 0.6m. What is its weight in air and what is its specific gravity.

48 T 50 N Y 0 W T FB W T F B 0 W 50 N F B W FB

49 buoyant force FB weight of displaced liquid 9.8kN FB (0.m 0.m 0.4 m) 0.157kN 157N 3 m W 50 N F B 50 N 157 N 07 N weight of the stone specific gravity = weight of an equal volume of water 07N specific gravity = NN

50 A hydrometer is an instrument used to measure the specific gravity of liquids. Remember, specific gravity is the ratio of the density of the liquid to the density of water.

51 A hydrometer weighs 0.016N 016N and has a stem at the upper end that is cylindrical and.8mm in diameter. How much deeper will it float in oil with sp gr than in alchohol of sp gr 0.81?

52 h alcohol oil sp gr 0.81 sp gr 0.780

53 For position 1, in alcohol, compute weight of hydrometer = weight of displaced liquid 9.8kN N 0.81 v 1 3 m N v m N / m 6 3 v m v 1 3

54 For position, in oil weight of hydrometer = weight of displaced liquid 9.8 kn 0.016N ( v1 Ah ) 3 m 0.016N ( v1 Ah) m m N / 6 3 ( v1 Ah ) m 3 3

55 ( v1 Ah ).8310 m m.6810 m h A 6 A (0.008 m) m m.6810 m h 0.04m 4mm m

56

57 A rectangular tank 6.4m long by.0m deep by.5m wide contains 1.0m of water. If horizontal acceleration is.45m/s, then: Compute the total force due to water acting on each end of the tank Show that the difference between these two forces is equal to the force needed to accelerate the mass.

58 .45m / s 1.0m 6.4m Tank is.0 0m deep

59 a acceleration of vessel m s tan g acceleration of gravity m s (, / ) (, / ).45.5 m / s tan m/ s tan 0.5

60 tan slope m / m.45m / s 1.0m 6.4m Tank is.0 0m deep

61 y 3.mtan 3.m m dcd 1.0m0.8m 0.m dab 1.0m0.8m 1.8m

62 .45m / s 1.0m 6.4m Tank is.0 0m deep

63 9.8kN 1.8m FAB hcga (1.8m.5 m) 39.7kN 3 m 9.8kN 0.m FCD hcga (0. m.5 m ) 0.5kN 3 m

64 .45m / s F AB 39.7 kn F CD kn 1.0m 6.4m Tank is.0 0m deep

65 Force needed to accelerate = mass of water X acceleration = 3 6.4m.5m1.0m9.8 kn / m.45m 39. kn 9.8 m/ s s check : FAB FCD 39.7kN 0.5kN 39.kN

66 Force needed to accelerate = mass of water X acceleration = 3 mass 6.4m.5m 1.0m 1000 kg / m 16, 000kg.45m 39, 00kgm force 16, 000kg s s

67 force newton force.45m 39,00kgm 16,000 kg s s kg m s 39.kN

68 .45m / s force 39. kn F AB 39.7 kn F CD kn 1.0m 6.4m Tank is.0 0m deep

69 check : FAB FCD 39.7kN 0.5kN 39.kN

70 A similar tank filled with water and accelerated at 1.5m/s. Compute how many liters of water are spilled.

71 .0m drop in water surface 7.0m

72 tan a acceleration of vessel m s (, / ) g acceleration of gravity m s (, / ) 1.5 m/ s tan m/ s tan slope of water surface

73 drop in surface = m

74 .0m 1.07m 7.0m

75 7.0m1.07m Volume.5 m( ) 9.36m Volume 9360liters 3

76 A 1.5m cubic tank is filled with oil with sp gr Find the force acting on the side of the tank with an acceleration of 4.9m/s up and with 4.9m/s down.

77

78 for acceleration up: p B a h(1 ) g m/ s p B ( kn / m )(1.5 m )(1 ) 9.8 m/ s pb kn / m 16.58kPa 1 FAB area of loading diagram ( kn / m 1.5 m)(1.5 m) FAB 18.65kN

79

80 for acceleration up: 4.9 m/ s FAB kn m m m m m/ s FAB 18.65kN ( / 3 )(0.75 )(1 )( )

81

82 for acceleration down: 4.9 m/ s FAB kn m m m m m/ s FAB 6.kN ( / 3 )(0.75 )(1 )( )

83

84 When 0.03m 03m 3 /s flows through a 300mm pipe that reduces to 150mm, calculate the average velocities in the two pipes. Q A V A V V V / Q m s 0.4 m/ s A m 4 3 Q 0.03 m / s 1.70 m/ s A m 4

85 If the velocity in a 300mm pipe is 0.50m/s, what is the velocity on a 75mm-dia jet from a nozzle attached to the pipe? Q A V A V m V 0.075m V m 0.50 m/ s 0.075m V V m 0.50 m / s m m/ s

86 Oil of sp gr 0.75 is flowing through a 150mm pipe under a pressure of 103kPa. If the total energy relative to a datum plane.4m below the center of the pipe is 17.9m, determine the flow of oil.

87 p 103kPa z.40 m pipe dia 150mm specific gravity 0.750

88 E PE KE FE V p E z g V 103 kpa 17.9m.40m 9.8 m/ s kn / m V 17.9m.40m14.0m 19.6 m/ s V m / s 3

89 p 103kPa V m/ s z.40 m pipe dia 150mm specific gravity 0.750

90 Q AV A V (0.150 m ) m/ s 0.018m Q m 5.4 m/ s 3 Q m / s

91 In the following figure, water flows from A to B at the rate of 0.40m 3 /s and the pressure head at A is 6.7m. Considering no loss in energy from A to B, find the pressure head at B. Draw the energy line.

92 V A gg???? p B V B???? g???? p A 6.70m Dia 600mm zb 8.00m za m Dia 300mm 3 Q 0.40 m / s

93 Use the Bernoulli theorem, from A to B: energy at + energy energy energy = energy at section 1 added lost extracted section pa V A pb V B za HA HL HE zb g g H 0 H H A L E 0 0

94 V A gg???? p B V B???? g???? p A 6.70m Dia 600mm zb 8.00m za m Dia 300mm 3 Q 0.40 m / s

95 pa VA pb VB z A zb g g V V A B 3 Q 0.40 m / s A (.300 m ) / 4 A 3 Q 0.40 m / s 5.66 m/ s m / s A (.600 m) / 4 B

96 V A gg???? p B V B???? g???? p A 6.70m Dia 600mm zb 8.00m za m Dia 300mm 3 Q 0.40 m / s

97 5.66 m p m B 6.7m 3.0m 8.0m g g 6.7m1.6m3.0m 0.1m8.0m p B p B 11.3m 8.1m p B 3. mwater

98 V A V B 1.6m g gg p B 3.m p A 6.7m Dia 600mm 0.1m zb 8.0m za m Dia 300mm 3 Q 0.40 m / s

99 The energy line The hydraulic grade line

100 KE V A 1.6m gg p B V B 3.m g 0.1m KE FE FE PE za p A 6.7m m Dia 300mm Dia 600mm zb 8.0m PE 3 Q 0.40 m / s

101 A pipe carrying oil of sp gr changes in size from 150mm at section E to 450mm at section R. Section E is 3.66m lower than R and the pressures are 91.0kPa and 60.3kPa, respectively. If the discharge is 0.146m 3 /s, determine the lost head and the direction of flow.

102 Draw a diagram to illustrate the problem

103 Calculate average velocity at each section: Q V V AV Q A m / 150 s (0.150 m ) / m/ s m / s V m/ s (0.450 m) / 4

104 Using lower section, E, as datum: E E E E EE p E V150 g z E 91.0 kn / m (8.6 m/ s) kn / m 9.8 m / s 14.1m 0

105 E R E R E R pr V150 g z R 60.3 kn / m (0.9 m / s) kn / m 9.8 m / s 10.7 m

106 E m E E R 10.7m Flow will occur from E to R because the energy head at E is greater Lost Head m m m

107 A 0.15m pipe 180m long carries water from A at elevation 4.0m to B at elevation 36.0m. The frictional stress between the liquid and the pipe walls is 0.6N/m. Determine the lost head and the pressure change.

108 First, draw a sketch of the problem

109 Use Bernoulli's Theorum: p A V A pb V B za HL zb g g First, calculate loss due to friction H L

110 An equation for loss due to friction H L H L L R shear stress L length area R Hydraulic Radius wetted perimeter

111 For a round pipe flowing full: A d /4 d R P d 4 So : H H L L L 4L R d / 180 N m m / 0.15 N m m 14.7 m

112 p V V A A pb B za 14.7 m zb g g Velocity V is the same at both ends of the pipe and z =4m, z =36m, so: p p A A A pb 14.7m 1m p B B 6.7m 9.8kN m pa pb 6.7m 6 kn / m 6kPa 3

113

114 A 1m diameter new cast iron pipe (C=130) is 845m long and has a head loss of 1.11m. Find the discharge capacity of the pipe according to the Hazen-Williams formula.

115 V 0.849CR S C 130 d 1m R hydraulic radius 4 4 head loss 1.11m S hydraulic grade line length 845

116 V 0.849CR S m m V 0.849(130) 1.81 m/ s m 3 Q AV 1.81 m / s 1.01 m / s 4

117 Solve Problem 8.8 using the Manning formula. V n /3 1/ R S n d 1m R hydraulic radius 4 4 head loss 1.11m S hydraulic grade line length 845

118 V V /3 1/ R S n /3 1/ 1 m m m/ s 1m 3 Q AV m/ s 0.94 m / s 4

119 m/s m 3 /s Hazen Williams Manning

120 A 0.9m diameter concrete pipe (C=10) is 10m long and has a head loss of 3.9m. Find the discharge capacity of the pipe using Hazen-Williams.

121 V 0.849CR S C 10 d 0.9m R hydraulic radius 4 4 head loss 3.9m S hydraulic grade line length 10

122 V 0.849CR S m m V 0.849(10) m/ s m 3 Q AV m / s m / s 4

123 Solve Problem 8.3 using the Manning formula. V n /3 1/ R S n d 0.9m R hydraulic radius 4 4 head loss 3.9m S hydraulic grade line length 10

124 V V R S n /3 1/ /3 1/ 0.9m 3.9m m/ s 0.9m 4 3 Q AV m/ s 1.0 m / s

125 m/s m 3 /s Hazen Williams Manning

126 What size square concrete conduit is needed to carry 4.0m 3 /s of water a distance of 45m with a head loss of 1.8m? Use Hazen- Williams.

127 find square dimension a V 0.849CR S Q m / s V A a C 10 (for concrete from Problem 8.3) a a R 4a 4 head loss m S hydraulic grade line length 45m

128 V 0.849CR S a a 4 45 a Specify 0.80m by 0.80m conduit 0.75m

129 Water is flowing in a 500mm diameter new cast iron pipe (C=130) at a velocity of.0m/s. Find the pipe friction loss per 100m of pipe. Use Hazen-Williams. V 0.849CR S V C R 0.0 m/ s m m

130 V CR S m/ s m S S S m/ s 063 m m/ m Total head loss for 100m pipe: pp m/ m100m 0.67m

131 To express the loss as pressure: 98kN 9.8 p h 0.67m 3 m 6.6kN p 6.6kPa m

132

133 For a lost head of 5.0m/1000m and C=100 for all pipes, how many 0cm pipes are equivalent to one 40cm pipe? How may are equivalent to a 60cm pipe?

134 Out of curiousity, let's compare the cross-sectional areas: 0 A 0 = A 40 = A 60 = If area was all that mattered: it would take 4 0cm pipes to equal 1-40cm pipe, and it would take 1 0cm pipes to equal 1-60cm pipe

135 But we must consider head loss. Let s use two equations: Q AV and V 0.849CR S

136 Q AV A d V 0.849CR S d R 4 5m S m / m 1000m C 100 d Q 0.63 d

137 Q m / s Q m / s Q m / s 4

138 3 Q m / s 3 Q m / s 6. 0cm pipes equivalent to a 40cm pipe Q m s cm pipes equivalent to 60cm pipe Q m s / /

139 A 300mm pipe that is 5m long and a 500m pipe that is 400m long are connected in series. Find the diameter of a 65m long equivalent pipe. Assume all pipes are concrete.

140 Q AV; A ; V 0.849CR S 4 d R ; C 10 for concrete 4 h1 3 S ; assume Q 0.1 m / s L d

141 0.1 d 0.63 d h L

142 for the 300mm dia, 5m long pipe: h h h m

143 for the 500mm dia, 400m long pipe: h h h 0.48m

144 total head loss m Therefore: for a 65m long equivalent pipe: d 1.96 d d 360mm

145 A 300mm pipe that is 5m long and a 500m pipe that is 400m long are connected in series. Find the diameter of a 65m long equivalent pipe. Assume all pipes are concrete. Use Diagram B-3 to solve.

146 3 Assume a flow rate, 0.1 / From Diagram B-3, 1 Q m s for 300mm pipe, h m/ m for 500mm pipe, pp h m/ m total head loss = m/ m 5m m/ m400m 1.91m for a 65m long gpp pipe, h 1.91 m/ 65m m from diagram B-3, d 360m 1

147 Water flows at a rate of m 3 /s from reservoir A to reservoir B through three concrete pipes connected in series, as shown on the following slide. Find the difference in water surface elevations in the reservoirs. Neglect all minor losses.

148

149 Use Diagram B-3 Q m / s from Diagram B-3: 3 for the 400mm pipe, h m/ m for the 300mm pipe, h m/ m for the 00mm pipe, h m/ m 1 1 1

150 total head loss m / m 600m m / m 1850 m m/ m970m 19.58m

151 Use Hazen-Williams Formula d Q AV; A ; V CR S 4 d R ; C 10 for concrete 4 h 3 S ; Q 0.05 m / s L

152 0.05 d d h L 4

153 for the 400mm dia, 600m long pipe: h h m

154 for the 300mm dia, 1850m long pipe: h h m

155 for the 00mm dia, 970m long pipe: h h 14.44m 300 total head 1.3m 3.8m 14.44m 19.58m 0.54

156 The flow in pipes AB and EF is 0.850m 3 /s. All pipes are concrete. Find the flow rate in pipes BCE and BDE.

157 Assume head loss from B to E 1.00m 1.00m for pipe BCE, h m/ m 1 340m from Diagram B-3, Q m / s BCE m for pipe pp BDE, h m/ m 1 300m from Diagram B-3 3, Q m / s BDE 3

158

159 if head loss from B to E = 1.00m is correct, then sum of the flow rates through BCE and BDE will 3 equal 0.850m /s but,

160 head loss of 1.00m is not correct, however, actual flow rates through BCE and BDE will be at the same proportion. So, Q BCE = m / Q 3 BDE = m / 171 s s

161 1500m m C 10 Compute the flow in each branch. 3 Q m / s W Z 900m C 10

162 Q AV; A ; V 0.849CR S 4 d h R ; C 10 for concrete; S 4 L d Q d d h L 4

163 1500m m C 10 Compute the flow in each branch. 3 Q m / s W Z 900m C 10

164 assume a head loss of 10m from W to Z then, for the 300mm pipe: Q m / s 4 and, for the 400mm pipe: Q m / s

165 3 Q 0.09 m / s 1500 m, 300 mm C 10 3 Q m / s W Z 900 m, 400mm C Q m / s m / s m / s m / s

166 / 0.6 / 0.35 / / m s m s m s m s Q m / s Q m / s

167 3 Q 0.13 m / s 1500 m, 300 mm C 10 3 Q m / s W Z 900 m, 400mm C Q m / s m / s m / s m / s

168 Next, let s solve using Hardy Cross Method

169 3 Q m / s W 1500 m, 300 mm C 10 Z Compute the flow in each branch using Hardy Cross Method. 900 m, 400mm C 10

170

171 Many pipes connected in a complex manner Many pipes connected in a complex manner with many entry and exit points.

172 Analyze using the Hardy Cross Method 1. assume flows for each individual pipe.. calculate the head loss thru each pipe using Hazen-Williams. 3. find the sum of head losses in each loop. 4. remember, head loss between two joints is the same for each branch. 5. sum of head losses in each loop must be zero. 6. compute a flow rate correction. 7. adjust the assumed flow rates for all pipes and repeat the process until all corrections are zero.

173 For this problem, we are looking at just one loop: Q Q

174 3 Q m / s W 1500 m, 300 mm C 10 Z Compute the flow in each branch using Hardy Cross Method. 900 m, 400mm C 10 3 Q m / s

175 Analyze using the Hardy Cross Method 1. assume flows for each individual pipe.. calculate the head loss thru each pipe using Hazen-Williams. 3. find the sum of head losses in each loop. 4. remember, head loss between two joints is the same for each branch. 5. sum of head losses in each loop must be zero. 6. compute a flow rate correction. 7. adjust the assumed flow rates for all pipes and repeat the process until all corrections are zero.

176 For correction, use: LH 1.85 ( LH / Q ) 0 3 flow adjustment, m / s LH lost head L S Q0 assumed initial flow

177 For Consistency: Clockwise, Q and LH are positive Counterclockwise, Q and LH are negative So, in: LH 1.85 ( LH / Q ) Sign (+ or -) is important in numerator But, denominator is always positive 0

178 In the table that follows: S is from the Hazen - Williams formula : V 0.849CR S also use : Q VA

179 3 Q m / s W 1500 m, 300 mm C 10 Z Compute the flow in each branch using the Hardy Cross Method. 900 m, 400mm C 10

181

182 Analyze using the Hardy Cross Method 1. assume flows for each individual pipe.. calculate the head loss thru each pipe using Hazen-Williams. 3. find the sum of head losses in each loop. 4. remember, head loss between two joints is the same for each branch. 5. sum of head losses in each loop must be zero. 6. compute a flow rate correction. 7. adjust the assumed flow rates for all pipes and repeat the process until all corrections are zero.

183 3 0.4 m / s A 600m B 600m 300 mm dia 300 mm dia C 400m 50 mm dia 400m 50 mm dia 400m 50 mm dia F 600m 600m 300 mm dia 300 mm dia E D m / s

184 Analyze using the Hardy Cross Method 1. assume flows for each individual pipe.. calculate the head loss thru each pipe using Hazen-Williams. 3. find the sum of head losses in each loop. 4. remember, head loss between two joints is the same for each branch. 5. sum of head losses in each loop must be zero. 6. compute a flow rate correction. 7. adjust the assumed flow rates for all pipes and h il ll i repeat the process until all corrections are zero.

185 For this problem, we are looking at only loops: Q Q

187 For S, use Hazen-Williams: V 0.849CR S LH total head loss in pipe L S LH 1.85 LH / Q

188 3 0.4 m / s A B 600m 600m 300 mm dia 300 mm dia 3 3 Q m / s Q m / s C 400m 50 mm dia 3 Q m / s 400m 50 mm dia 3 Q m / s 400m 50 mm dia Q m / s F 600m 300 mm dia 3 Q m / s E 600m 300 mm dia 3 Q m / s D m / s

189

190

191

192 3 0.4 m / s A B 600m 600m 300 mm dia 300 mm dia 3 3 Q m / s Q m / s 3 3 Q0 0.41m / s Q 0.159m / s C 400m 50 mm dia 3 Q m / s 3 Q m / s 400m 50 mm dia 3 Q m / s 3 Q m / s 400m 50 mm dia Q m / s 3 Q 0.159m / s 3 0 F 600m 300 mm dia 3 Q m / s 3 Q 0.160m / s E 600m 300 mm dia Q m / s 3 Q 0.41m / s 3 D m / s

193

194 Water flows in a rectangular concrete open channel that is 1.0m wide at a depth of.5m. The channel slope is Find the water velocity and the flow rate.

195 using the Manning equation: /3 1/ R S V= n area.51.0 R wetted perimeter /3 1/ V= m/ s Q AV m / s m

196 Water is to flow at a rate of 30 m 3 /s in the concrete channel shown on the following slide. What slope of channel will be required?

197 /3 1/ Q R S V A n 1.6 m 3.6 m A (3.6m.0 m).0m 1.40m R 1.40m 3.6m1.6m.0m.0m.0m 30 m / s 1.36m S 3 /3 1/ m or, m / kilometer m/ m 1.36m

198 On what slope should a 600mm concrete sewer pipe be laid in order that 0.17m 3 /s will flow when the sewer is one-half full? What slope if the sewer flows full? (use n=0.013) Vn use : S= /3 R

199 1 d 4 d 0.6m R m 1/ d R Full d m m

200 V Q A / 1.06 m/ s S Vn R / /3 / m/ m

201 V Full Q A m/ s S Full Vn /3 /3 R m/ m

202 A 600mm diameter concrete pipe on a 1/400 slope carries water at a depth of 40mm. Find the flow rate, Q.

203 depth 40mm diameter 600mm slope 1/ 400

204 depth 40mm diameter 600mm slope 1/ 400 r 300mm

205 depth 40mm 300mm 40mm 60mm cos 1.369rad sin(1.369 rad ) 0.94 m r 300mm 0.94m 60mm

206 Area of triangles (0.94 ) area of circle section r 300mm mm m 0.94m A A m

207 wetted perimeter R r 300mm 60mm A 0.105m R 0.18

208 /3 1/ R S V= n / V= V / r 300mm 60mm A 0.105m Q AV m / s R 0.18

209 After a flood had passed an observation station on a river, an engineer visited the site. By locating flood marks, performing appropriate survey and doing necessary computations, she determined that, at the time of peak flooding : The cross-sectional sectional area was 960m The wetted perimeter was 341m The water surface slope was m/m

210 The engineer also noted that the channel bottom was earth with grass and weeds, for which a handbook gave a Manning n value of Estimate the peak flood discharge. Q /3 960 (960) /3 1/ AR S 341 n Q 11,500 m / s 1/

211 A rectangular channel 6.1m wide carries 11.3m 3 /s and discharges onto a 6.1m wide apron with no slope at a mean velocity of 6.1m/s. What is the height of the hydraulic jump? What energy is absorbed (lost) in the jump?

212 q g y y 1 yy 1 3 Q 11.3 m / s v1 6.1 m/ s

213 q g y y 1 yy q flow / unit width 1.85 m / s y 1 q 1.85 V m y m y y

214 q g y y 1 yy 1 3 Q 11.3 m / s 3 q 1.85 m / s V1 6.1 m/ s y m y 1.37m jump 1.37m 0.303m 1.07m

215 critical depth: 3 3 y C q / g 1.85 / m Therefore: Flow depth before the jump (0.303m) is < 0.70 and is supercritical Flow depth after the jump (1.37m) is > 0.70 and is subcritical

216 q g y y 1 yy 1 3 Q 11.3 m / s 3 q 1.85 m / s V1 6.1 m/ s y m y c 0.70m y 1.37m jump 1.37m 0.303m 1.07m

217 Lost energy: m / s E1 y1 V1 / g 0.303m.0m / 1.37 / s E y V / g 1.37m 1.46m 9.8 lost head.0m 1.46m 0.74m

218 .0m 0.303m 1.90m 1.46m1.37m 0.09 V1 6.1 m/ s 1.90m 0.74m 0.09m y m y c 0.70m y 1.37m jump 1.37m 0.303m 1.07m

219

220 Measures stagnation pressure (at B), which exceeds the local static pressure (at A), to determine velocity head. ha hb

221 Velocity (V) at Point B is zero. Apply the Bernoulli equation, next slide h A h B

222 p A VA no loss pb VB za g assumed g V 0; z z B A B so, pa VA pb gg z B

223 A A B p V p A A B p p g B A p p V g g p p B A B A p p h h d With no friction: V gd

224 h h d B A h A h B

225 A small amount of friction normally occurs, so a coefficient of velocity c V (see discussion on following slides) is sometimes used: c V actual velocity theoretical velocity V c gd V to assume cv 1 provides sufficient i accuracy for most engineering problems involving Pitot tubes.

226 The ratio of the actual velocity in a stream to the theoretical velocity that would occur without friction. c V actual velocity theoretical velocity

227 A Pitot tube having a coefficient of 0.98 is used to measure the velocity of water at the center of a pipe. The stagnation pressure head is 5.67m and the static pressure head in thepipeis4.73m. is Whatisthevelocity? the

228 4.74m 5.67m

229 h B h A d 5.67m4.74m 0.94m 4.74m 5.67m

230 V c gd V d m m m c V g m/ s V m/ s 0.94m 4.1 m/ s

231 A 100mm diameter standard orifice discharges water under a 6.1m head. What is the flow?

232 Q ca gh A area; c 0.6 H total head causing flow 0.1m Q m/ s m 4 3 Q m / s

233 The tank in problem 1.9 is closed ant the air space above the water is under pressure, causing to flow to increase to 0.075m3/s. 075m3/s Find the pressure in the air space.

234 Q ca gh m / s m/ s H H 1.9m m h H h 1.9 m 6.1 m 6.8 m P Z 3 p hp 9.8 kn / m 6.8m 70 kn / m 70kPa

235 During a test on a.4m suppressed weir that was 0.9m high, the head was maintained constant at 0.3m. In 38 seconds, 9m 3 of water were collected. Find the weir factor m using equations A and B.

236 H m Z m

237 3 9m m 3 Q m / s 38s flow depth m03 0.3m 1 1.m V 3 Q m / s A.4m1.m 0.65 m/ s

238 using Eq. A: V V Q m b H g g 3/ 3/ 3/ m m m / 3/ 0.763m m /

239 using Eq. B: Q mbh 3/ m / m Equation B is OK for weirs placed dhighh

240 During a test on a.4m suppressed weir that was 0.0m high, the head was maintained constant at 0.3m. In 38 seconds, 9m 3 of water were collected. Find the weir factor m using equations A and B.

241 H m Z m

242 b H Z=0 From:

243 3 9m m 3 Q m / s 38s flow depth m m m V 3 Q m / s m / s A.4m 0.3m

244 using Eq. A: V V Q m b H g g 3/ 3/ 3/ m m m / 3/ 0.763m m /

245 using Eq. B: Q mbh 3/ m m / , Equation A must be used, q for shallow weirs

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