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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.
specific weight Ƴ = 47 kn = 7.833 kn/m 3 6 m 3 specific gravity = Ƴoil = 7.833 kn/m 3 = 0.800 Ƴwater 9.79 kn/m 3
If 1 m 3 of concrete has a mass.4 Tons, calculate its specific weight Ƴ and specific gravity.
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
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 101.3 kpa, or 760 mm of mercury, or 1atmosphere
h p Atmospheric pressure p 101.3 kpa p 101.3 kpa 103.3 3 kn / m
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 = 13.6 9.79 kn / m 3 133.1 kn / m 3
p h p 101.3 kn / m 133.1 kn / m 3 101.3 kn / / m h 0.76 m 760 mm 3 133.1 kn / m
1.1m
Weight of glycerin 1.34kN m 3 1.34 kn Pressure p at A h 11 1.1m m 3 13.57kN 13.57kPa m
Determine the gage pressure in kpa at a depth of 10.0 meters below the free surface of a body of water.
Weight of water 9.79 kn / m 3 p h 9.79kN 97.9kN p 10.00 m 97.99 kpa 3 m m
Find the pressure at the bottom of a tank containing glycerin under pressure as shown in following figure.
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
(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
(a) The liquid in Piezometer A will rise to the same elevation as the top of the tank.
979kN 9.79 ( b) pa h (0.7 )(1.7 m) 3 m 11.98kN 11.98kPa m 11.98 kn / m ha p / 0519 0.519 3 (.36 9.79 kn / m ) h TOTAL 03 0.30519 0.519 0819 0.819m
() c p0 pa pb 11.98 kpa (.36 9.79 kn / m )(0.3 m) 3 18.9kPa
F hcga hcg A specific weight of liquid depth of the center of gravity Area
4.0m A 0.1m Water 0.0m 7.0m Tank Width 3m Example
Total force (F BC ) on the bottom of the tank Total weight (W) of the water Explain the difference
4.0m A 0.1m Water 0.0m 7.0m Tank Width 3.0m Example
FBC pa ( h ) A 9.8kN (6 m ) (7m 3 m ) m 3 F BC 135 kn
4.0m A 0.1m Water 0.0m 7.0m Tank Width 3m Example
W ( Volume) 98kN 9.8 W [(7mm3 m(4m0.1 m 3 )] m 3 W 416kN
4.0m A 0.1m Water 0.0m 7.0m Tank Width 3m Example
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?
4.0m A 0.1m Water 0.0m 7.0m Tank Width 3m Example
FAD pa ( h) A 9.8 kn 3 m (4 m ) (7m 3m 0.1 m ) F AD 819 kn
4.0m A 0.1m Water 0.0m 7.0m Tank Width 3m Example
135kN 416kN 819Kn and dtherefore: F BC W F AD
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
A stone weighs 90N in air. When immersed in water it weighs 50N. Compute the volume of the stone and its specific gravity.
Y 0 W T FB 0 T 50N Wt Water FB W T W 90 N FB 90N 50N FB 90N 50N 40N FBF
buoyant force FB 40N 40 N 9.8 kn v 3 m 40 N 3 v 0.0041m 4.1 liters 3 9800 N / m
weight of the stone specific gravity = weight of an equal volume of water 90N specific gravity =.5 40N
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.
T 50 N Y 0 W T FB W T F B 0 W 50 N F B W FB
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 = 1.31 157NN
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.
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 0.780 than in alchohol of sp gr 0.81?
h alcohol oil sp gr 0.81 sp gr 0.780
For position 1, in alcohol, compute weight of hydrometer = weight of displaced liquid 9.8kN 0.016016 N 0.81 v 1 3 m 0.016016 N v1 0.0000068m 3 0.81 9800 N / m 6 3 v1.6810 m v 1 3
For position, in oil weight of hydrometer = weight of displaced liquid 9.8 kn 0.016N 0.780 ( v1 Ah ) 3 m 0.016N ( v1 Ah) m 0.0000083 0000083m 0.780 9800 N / 6 3 ( v1 Ah ).83 10 m 3 3
( v1 Ah ).8310 m 6 3 6 3 6 3.8310 m.6810 m h A 6 A (0.008 m) 6.16 10 m 4 6 3 6 3.8310 m.6810 m h 0.04m 4mm 6 6.16 10 m
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.
.45m / s 1.0m 6.4m Tank is.0 0m deep
a acceleration of vessel m s tan g acceleration of gravity m s (, / ) (, / ).45.5 m / s tan 05 0.5 9.8 m/ s tan 0.5
tan slope 0 0.5 m / m.45m / s 1.0m 6.4m Tank is.0 0m deep
y 3.mtan 3.m0.5 0.8m dcd 1.0m0.8m 0.m dab 1.0m0.8m 1.8m
.45m / s 1.0m 6.4m Tank is.0 0m deep
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
.45m / s F AB 39.7 kn F CD 05 0.5 kn 1.0m 6.4m Tank is.0 0m deep
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
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
force newton force.45m 39,00kgm 16,000 kg s s kg m s 39.kN
.45m / s force 39. kn F AB 39.7 kn F CD 05 0.5 kn 1.0m 6.4m Tank is.0 0m deep
check : FAB FCD 39.7kN 0.5kN 39.kN
A similar tank filled with water and accelerated at 1.5m/s. Compute how many liters of water are spilled.
.0m drop in water surface 7.0m
tan a acceleration of vessel m s (, / ) g acceleration of gravity m s (, / ) 1.5 m/ s tan 0.153 98 9.8 m/ s tan 0.153 slope of water surface
drop in surface = 7 0.153 1.07m
.0m 1.07m 7.0m
7.0m1.07m Volume.5 m( ) 9.36m Volume 9360liters 3
A 1.5m cubic tank is filled with oil with sp gr 0.75. Find the force acting on the side of the tank with an acceleration of 4.9m/s up and with 4.9m/s down.
for acceleration up: p B a h(1 ) g 3 4.9 m/ s p B (0.75 9.8 kn / m )(1.5 m )(1 ) 9.8 m/ s pb 16.58 kn / m 16.58kPa 1 FAB area of loading diagram ( 16.58 kn / m 1.5 m)(1.5 m) FAB 18.65kN
for acceleration up: 4.9 m/ s FAB kn m m m m 98 9.8 m/ s FAB 18.65kN (0.75 9.8 / 3 )(0.75 )(1 )(1.5 1.5 )
for acceleration down: 4.9 m/ s FAB kn m m m m 98 9.8 m/ s FAB 6.kN (0.75 9.8 / 3 )(0.75 )(1 )(1.5 1.5 )
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 300 150 300 300 150 150 3 0.03 / Q m s 0.4 m/ s A 300 0.300m 4 3 Q 0.03 m / s 1.70 m/ s A 150 0.150 m 4
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 300 300 75 75 0.300m V 0.075m V 4 4 300 75 0.300m 0.50 m/ s 0.075m V V 0.300 m 0.50 m / s 75 0.075m 75 8.00 m/ s
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.
p 103kPa z.40 m pipe dia 150mm specific gravity 0.750
E PE KE FE V p E z g V 103 kpa 17.9m.40m 9.8 m/ s 0.750 9.8 kn / m V 17.9m.40m14.0m 19.6 m/ s V 54 5.4 m / s 3
p 103kPa V 54 5.4 m/ s z.40 m pipe dia 150mm specific gravity 0.750
Q AV A V (0.150 m ) 4 5.4 m/ s 0.018m Q 0.018018 m 5.4 m/ s 3 Q 0.097 m / s
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.
V A gg???? p B V B???? g???? p A 6.70m Dia 600mm zb 8.00m za 300 3.00m Dia 300mm 3 Q 0.40 m / s
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
V A gg???? p B V B???? g???? p A 6.70m Dia 600mm zb 8.00m za 300 3.00m Dia 300mm 3 Q 0.40 m / s
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 141 1.41 m / s A (.600 m) / 4 B
V A gg???? p B V B???? g???? p A 6.70m Dia 600mm zb 8.00m za 300 3.00m Dia 300mm 3 Q 0.40 m / s
5.66 m p 141 1.41 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
V A V B 1.6m g gg p B 3.m p A 6.7m Dia 600mm 0.1m zb 8.0m za 30 3.0m Dia 300mm 3 Q 0.40 m / s
The energy line The hydraulic grade line
KE V A 1.6m gg p B V B 3.m g 0.1m KE FE FE PE za p A 6.7m 30 3.0m Dia 300mm Dia 600mm zb 8.0m PE 3 Q 0.40 m / s
A pipe carrying oil of sp gr 0.877 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.
Draw a diagram to illustrate the problem
Calculate average velocity at each section: Q V V AV Q A 3 0.0146 m / 150 s (0.150 m ) / 4 8.6 m/ s 3 0.0146 m / s V450 09 0.9 m/ s (0.450 m) / 4
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) 3 0.877 9.8 kn / m 9.8 m / s 14.1m 0
E R E R E R pr V150 g z R 60.3 kn / m (0.9 m / s) 366 3.66 3 0.877 9.8 kn / m 9.8 m / s 10.7 m
E 14.11 m E E R 10.7m Flow will occur from E to R because the energy head at E is greater Lost Head 141 14.1m107 10.7m 3.4 34m
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.
First, draw a sketch of the problem
Use Bernoulli's Theorum: p A V A pb V B za HL zb g g First, calculate loss due to friction H L
An equation for loss due to friction H L H L L R shear stress L length area R Hydraulic Radius wetted perimeter
For a round pipe flowing full: A d /4 d R P d 4 So : H H L L L 4L R d 4 0.6 / 180 N m m 3 9800 / 0.15 N m m 14.7 m
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
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.
V 0.849CR S 0.63 0.54 C 130 d 1m R hydraulic radius 4 4 head loss 1.11m S hydraulic grade line length 845
V 0.849CR S 0.63 0.54 0.63 0.54 1m 111 1.11m V 0.849(130) 1.81 m/ s 4 845 1m 3 Q AV 1.81 m / s 1.01 m / s 4
Solve Problem 8.8 using the Manning formula. V n /3 1/ R S n 0.0101 d 1m R hydraulic radius 4 4 head loss 1.11m S hydraulic grade line length 845
V V /3 1/ R S n /3 1/ 1 m 111 1.11 m 4 845 001 0.01 1.199 m/ s 1m 3 Q AV 1.199 m/ s 0.94 m / s 4
m/s m 3 /s Hazen Williams 1.81 1.01 Manning 1.199 0.94
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.
V 0.849CR S 0.63 0.54 C 10 d 0.9m R hydraulic radius 4 4 head loss 3.9m S hydraulic grade line length 10
V 0.849CR S 0.63 0.54 0.63 0.54 09 0.9m 3.9 39m V 0.849(10) 1.789 m/ s 4 10 0.9m 3 Q AV 1.789 m / s 1.1414 m / s 4
Solve Problem 8.3 using the Manning formula. V n /3 1/ R S n 0013 0.013 d 0.9m R hydraulic radius 4 4 head loss 3.9m S hydraulic grade line length 10
V V R S n /3 1/ /3 1/ 0.9m 3.9m 4 10 0.013 1.609 m/ s 0.9m 4 3 Q AV 1.609 m/ s 1.0 m / s
m/s m 3 /s Hazen Williams 1.789 1.14 Manning 1609 1.609 10 1.0
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.
find square dimension a V 0.849CR S 0.63 0.54 3 Q 40 4.0 m / s V A a C 10 (for concrete from Problem 8.3) a a R 4a 4 head loss 18 1.8 m S hydraulic grade line length 45m
V 0.849CR S 0.63 0.54 0.63 0.54 40 4.0 a 18 1.8 0.84910 a 4 45 a.63 0.63 0.54 4.0 4 45 0.849 10 1.8 0.54 Specify 0.80m by 0.80m conduit 0.75m
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 130 0.50m 4 063 0.63 0.54 054 0.15m
V 0.849 CR S 0.63 0.54.0 m/ s 0.849 130 0.15m S S S 0.54 0.63 0.54.0 m/ s 063 m 0.63 0.849 130 0.15 0.00670067 m/ m Total head loss for 100m pipe: pp 0.0067 m/ m100m 0.67m
To express the loss as pressure: 98kN 9.8 p h 0.67m 3 m 6.6kN p 6.6kPa m
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?
Out of curiousity, let's compare the cross-sectional areas: 0 A 0 = 100 4 40 A 40 = 400 4 60 A 60 = 100 4 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
But we must consider head loss. Let s use two equations: Q AV and V 0.849CR S 0.63 0.54
Q AV A d 4 0.63 0.54 V 0.849CR S d R 4 5m S 0005 0.005 m / m 1000m C 100 d Q 0.63 d 0.54 0.849 100 0.005 4 4
0.63 00 0.0 0.54 0.0 0.849 100 0.005 4 3 Q0 0.03 m / s 4 0.63 040 0.40 0.54 0.40 0.849 100 0.005 4 3 Q40 0.143 m / s 4 0.63 060 0.60 0.54 0.60 0.849 100 0.005 4 3 Q60 0.416 m / s 4
3 Q40 0.143 m / s 3 Q0 0.03 m / s 6. 0cm pipes equivalent to a 40cm pipe Q m s 18.11 0cm pipes equivalent to 60cm pipe Q m s 3 60 0.416 / 3 0 0.03 /
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.
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 0.63 0.54
0.1 d 0.63 d h1 0.849 10 4 L 4 0.54
for the 300mm dia, 5m long pipe: 0.1 0.63 0.30 h 1 0.30 0.849 10 4 5 4 h1 0.1 1.4087 5 h1 1.678m 0.54 0.54
for the 500mm dia, 400m long pipe: 0.1 0.63 0.50 h 0.50 0.849 10 4 400 4 h h 0.48m 0.1 5.3985 400 0.54 0.54
total head loss 1.678 0.48 1.96m Therefore: for a 65m long equivalent pipe: 0.63 0.54 d 1.96 d 0.849 10 0.1 d 360mm 4 65 4
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.
3 Assume a flow rate, 0.1 / From Diagram B-3, 1 Q m s for 300mm pipe, h 0.0074 m/ m for 500mm pipe, pp h 0.00064 m/ m total head loss = 0.0074 m/ m 5m0.00064 m/ m400m 1.91m for a 65m long gpp pipe, h 1.91 m/ 65m 0.00307m from diagram B-3, d 360m 1
Water flows at a rate of 0.05 05 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.
Use Diagram B-3 Q 005 0.05 m / s from Diagram B-3: 3 for the 400mm pipe, h 0.00051 m/ m for the 300mm pipe, h 0.000 m/ m for the 00mm pipe, h 0.015 m/ m 1 1 1
total head loss 0.0005100051 m / m 600m 0.000000 m / m 1850 m 0.015 m/ m970m 19.58m
Use Hazen-Williams Formula d 063 0.63 054 0.54 Q AV; A ; V 0849 0.849 CR S 4 d R ; C 10 for concrete 4 h 3 S ; Q 0.05 m / s L
0.05 d 0.63 0.54 d h 0.84910 4 L 4
for the 400mm dia, 600m long pipe: 063 0.63 0.40 h400 0.40 0.84910 4 600 0.05 4 h300 13 1.3 m 054 0.54
for the 300mm dia, 1850m long pipe: 063 0.63 0.30 h300 0.30 0.849 10 4 1850 0.05 4 h300 38 3.8 m 054 0.54
for the 00mm dia, 970m long pipe: 0.63 00 0.0 h00 0.0 0.84910 4 970 0.05 4 h 14.44m 300 total head 1.3m 3.8m 14.44m 19.58m 0.54
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.
Assume head loss from B to E 1.00m 1.00m for pipe BCE, h 0.00043 m/ m 1 340m from Diagram B-3, Q 0.133 m / s BCE 3 1.00m for pipe pp BDE, h 0.00031 m/ m 1 300m from Diagram B-3 3, Q 0.038 038 m / s BDE 3
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, 0.133 0.038038 0.171 0.850
head loss of 1.00m is not correct, however, actual flow rates through BCE and BDE will be at the same proportion. So, 0.133 3 Q BCE = 0.850 0.661 m / 171 0.038 Q 3 BDE = 0.850 0.189 m / 171 s s
1500m m C 10 Compute the flow in each branch. 3 Q 051 0.51 m / s W Z 900m C 10
Q AV; A ; V 0.849CR S 4 d h R ; C 10 for concrete; S 4 L d 0.63 0.54 Q d 0.63 0.54 d h 0.849 10 4 L 4
1500m m C 10 Compute the flow in each branch. 3 Q 051 0.51 m / s W Z 900m C 10
assume a head loss of 10m from W to Z then, for the 300mm pipe: 063 0.63 054 0.54 0.3 10 0.3 0.849 10 4 1500 3 Q300 0.09 m / s 4 and, for the 400mm pipe: Q 400 0.63 0.54 04 0.4 10 0.4 0.849 10 4 900 4 3 0.6 m / s
3 Q 0.09 m / s 1500 m, 300 mm C 10 3 Q 051 0.51 m / s W Z 900 m, 400mm C 10 3 3 3 3 Q 06 0.6 m / s 009 0.09 m / s 06 0.6 m / s 0.51 051 m / s
3 3 3 0.09 / 0.6 / 0.35 / 0.51 3 / m s m s m s m s 0.09 3 Q 300 0.51 0.13 m / s 035 0.35 0.6 3 Q400 0.51 0.38 m / s 035 0.35
3 Q 0.13 m / s 1500 m, 300 mm C 10 3 Q 051 0.51 m / s W Z 900 m, 400mm C 10 3 3 3 3 Q 038 0.38 m / s 013 0.13 m / s 038 0.38 m / s 0.51 051 m / s
Next, let s solve using Hardy Cross Method
3 Q 051 0.51 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
Many pipes connected in a complex manner Many pipes connected in a complex manner with many entry and exit points.
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.
For this problem, we are looking at just one loop: Q Q
3 Q 051 0.51 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 051 0.51 m / s
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.
For correction, use: LH 1.85 ( LH / Q ) 0 3 flow adjustment, m / s LH lost head L S Q0 assumed initial flow
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
In the table that follows: S is from the Hazen - Williams formula : V 0.849CR S also use : 0.63 0.54 Q VA
3 Q 051 0.51 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
See Excel Spreadsheet Or Pdf Version
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.
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 3 0.4 m / s
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.
For this problem, we are looking at only loops: Q Q
Solve using Excel Solve using Excel.pdf Version
For S, use Hazen-Williams: V 0.849CR S LH total head loss in pipe L S LH 1.85 LH / Q 0.63 0.54
3 0.4 m / s A B 600m 600m 300 mm dia 300 mm dia 3 3 Q0 0.00 m / s Q0 0.100 m / s C 400m 50 mm dia 3 Q0 0.00 m / s 400m 50 mm dia 3 Q0 0.100 m / s 400m 50 mm dia Q 3 0 0.100 m / s F 600m 300 mm dia 3 Q0 0.00 m / s E 600m 300 mm dia 3 Q0 0.300 m / s D 3 0.4 m / s
3 0.4 m / s A B 600m 600m 300 mm dia 300 mm dia 3 3 Q0 0.00 m / s Q0 0.100 m / s 3 3 Q0 0.41m / s Q 0.159m / s C 400m 50 mm dia 3 Q0 0.00 m / s 3 Q 016 0.16 0 m / s 400m 50 mm dia 3 Q0 0.100 m / s 3 Q0 0.0808m / s 400m 50 mm dia Q0 0.100 m / s 3 Q 0.159m / s 3 0 F 600m 300 mm dia 3 Q0 0.00 m / s 3 Q 0.160m / s E 600m 300 mm dia Q0 0.300 m / s 3 Q 0.41m / s 3 D 3 0.4 m / s
Water flows in a rectangular concrete open channel that is 1.0m wide at a depth of.5m. The channel slope is 0.08. 08 Find the water velocity and the flow rate.
using the Manning equation: /3 1/ R S V= n area.51.0 R wetted perimeter.5 1.0.5 /3 1/ 1.765 0.008008 V= 5.945 m/ s 0.013 3 Q AV.51.05.945 178 m / s 1765 1.765 m
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?
/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 0.013013 1.40 or, 0.746 m / kilometer 0.000746 m/ m 1.36m
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
1 d 4 d 0.6m R 015 0.15 m 1/ 1 4 4 d R Full d 06 0.6m 4 4 0.15m
V Q 017 0.17 A 1 0.6 4 1/ 1.06 m/ s S Vn 1.06 0.013 R 0.15 1/ /3 /3 0.0031 m/ m
V Full Q 017 0.17 A 0.6 4 0.601 m/ s S Full Vn 0.601 0.013 /3 /3 R 015 0.15 0.00077 m/ m
A 600mm diameter concrete pipe on a 1/400 slope carries water at a depth of 40mm. Find the flow rate, Q.
depth 40mm diameter 600mm slope 1/ 400
depth 40mm diameter 600mm slope 1/ 400 r 300mm
depth 40mm 300mm 40mm 60mm 1 0.060060 cos 1.369rad 0.30 0.30 sin(1.369 rad ) 0.94 m r 300mm 0.94m 60mm
Area of triangles 1 0.060 (0.94 ) 0.0176 area of circle section 1.369 0.60 r 300mm 0.13 60mm 4 0.94m 0.94m A 0.13 0.018 0.105 A 0.105 m
wetted perimeter 1.369 0.60 1.369 0.60 0.81 0.105 R 0.18 0.81 r 300mm 60mm A 0.105m R 0.18
/3 1/ R S V= n /3 1 0.18 400 V= 0.013 V 0.977 1/ r 300mm 60mm A 0.105m Q AV 3 0.1050.977 0.10 m / s R 0.18
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 0.00076m/m
The engineer also noted that the channel bottom was earth with grass and weeds, for which a handbook gave a Manning n value of 0.030. Estimate the peak flood discharge. Q /3 960 (960) 0.0007600076 /3 1/ AR S 341 n 0.030 3 Q 11,500 m / s 1/
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?
q g y y 1 yy 1 3 Q 11.3 m / s v1 6.1 m/ s
q g y y 1 yy 1 11.3 6.1 3 q flow / unit width 1.85 m / s y 1 q 1.85 V 61 6.1 1 0.303m 1.85 0.303 9.8 y 137 1.37 m 0.303 y y
q g y y 1 yy 1 3 Q 11.3 m / s 3 q 1.85 m / s V1 6.1 m/ s y1 0.303 m y 1.37m jump 1.37m 0.303m 1.07m
critical depth: 3 3 y C q / g 1.85 / 9.8 0.70m 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
q g y y 1 yy 1 3 Q 11.3 m / s 3 q 1.85 m / s V1 6.1 m/ s y1 0.303 m y c 0.70m y 1.37m jump 1.37m 0.303m 1.07m
Lost energy: 61 6.1 m / s E1 y1 V1 / g 0.303m.0m 9.8 1.85 / 1.37 / s E y V / g 1.37m 1.46m 9.8 lost head.0m 1.46m 0.74m
.0m 0.303m 1.90m 1.46m1.37m 0.09 V1 6.1 m/ s 1.90m 0.74m 0.09m y1 0.303 m y c 0.70m y 1.37m jump 1.37m 0.303m 1.07m
Measures stagnation pressure (at B), which exceeds the local static pressure (at A), to determine velocity head. ha hb
Velocity (V) at Point B is zero. Apply the Bernoulli equation, next slide h A h B
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
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
h h d B A h A h B
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.
The ratio of the actual velocity in a stream to the theoretical velocity that would occur without friction. c V actual velocity theoretical velocity
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
4.74m 5.67m
h B h A d 5.67m4.74m 0.94m 4.74m 5.67m
V c gd V d 567 5.67 m 4.73 473 m 0.94 094 m c V g 0.98 9.8 m/ s V 0.98 9.8 m/ s 0.94m 4.1 m/ s
A 100mm diameter standard orifice discharges water under a 6.1m head. What is the flow?
Q ca gh A area; c 0.6 H total head causing flow 0.1m Q 06 0.6 9.8 98 m/ s 6.1 61 m 4 3 Q 005 0.05 m / s
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.
Q ca gh 3 0.075 m / s 0.6 4 9.8 m/ s H H 1.9m 01 0.1 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
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.
H 03 0.3 m Z 09 0.9m
3 9m m 3 Q 0.763 m / s 38s flow depth 09 0.9m03 0.3m 1 1.m V 3 Q 0.763 m / s A.4m1.m 0.65 m/ s
using Eq. A: V V Q m b H g g 3/ 3/ 3/ 3 0.65 0.65 0.763m m.4 0.3 9.8 9.8 m 1.90 3 3/ 3/ 0.763m m.4 0.3 0.00358 0.00358 3/
using Eq. B: Q mbh 3/ 0.763 m.4 0.3 3/ m 1.93 1.90 1.93 Equation B is OK for weirs placed dhighh
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.
H 03 0.3 m Z 00 0.0m
b H Z=0 From: http://www.lmnoeng.com/weirs/cipoletti.htm
3 9m m 3 Q 0.763 m / s 38s flow depth 00 0.0m 03 0.3m 03 0.3m V 3 Q 0.763 m / s 106 1.06 m / s A.4m 0.3m
using Eq. A: V V Q m b H g g 3/ 3/ 3/ 3 1.06 1.06 0.763m m.4 0.3 9.8 9.8 m 1.53 3 3/ 3/ 0.763m m.4 0.3 0.0573 0.0573 3/
using Eq. B: Q mbh 3/ 0.763 m.4 0.3 m 193 1.93 3/ 1.53 1.93, Equation A must be used, q for shallow weirs