Brad Peterson, P.E.
New Website: New Website:http://njut2009fall.weebly.com
M P ’ E il Add Mr. Peterson’s Email Address: [email protected]
If 6 m3 of oil weighs 47 kN calculate itsIf 6 m3 of oil weighs 47 kN, calculate its specific weight Ƴ and specific gravity.
specific weight Ƴ = 47 kN = 7 833 kN/m3specific weight Ƴ = 47 kN = 7.833 kN/m3
6 m3
specific gravity = Ƴoil = 7.833 kN/m3 = 0.800Ƴwater 9 79 kN/m3Ƴwater 9.79 kN/m
If 1 m3 of concrete has a mass 2 4 Tons If 1 m3 of concrete has a mass 2.4 Tons, calculate its specific weight Ƴ and specific gravitygravity.
39.81 /ifi i ht 2300 22 56 /N kgk kN 33specific weight 2300 22.56 /
1gkg kN m
m
322.56 /specific gravity 2.30con kN m 3specific gravity 2.30
9.79 /water kn m
V i h l h h i Vacuum = space with less that atmospheric pressureA h i f ili Atmospheric pressure refers to prevailing pressure in the air around us
l l d d h At sea level, standard atmospheric pressure is:
101.3 , orp kPa760 of mercury, or
1 atmosphereh mm1 atmosphere
ph
101 3p kPa
Atmospheric pressure 101.3p kPaAtmospheric pressure
2101 3 103 3 /p kPa kN m 101.3 103.3 /p kPa kN m
:Compute :13.6
Compute Specific gravity of mercury = R b
Remember,
weight of substance specific gravity = =specific gravity = = weight of equal amount of water
= specific gravity weight of equal amount of water
313.6 9.79 /kN m = 3133.1 /kN m
ph
2101 3 /p kN m
3
101.3 /
133.1 /
p kN m
kN m 2101.3 / / 0 76 760kN mh m mm
3 0.76 760133.1 /
h m mmkN m
1.1m
12 34kN3
12.34Weight of glycerin kNm
12.34P A 1 1
mkNh 3
.3Pr essure at A 1.1kNp h mm
2
13.57 13.57kN kPa2 13.57kPam
Determine the gage pressure in kPa at a depthDetermine the gage pressure in kPa at a depth of 10.0 meters below the free surface of a body of waterbody of water.
39.79 /kN m Weight of water Ng
p h
9.79 97.910 0 97 9kN kNp m kPa 3 210.0 97.9p m kPam m
Find the pressure at the bottom of a tankFind the pressure at the bottom of a tank containing glycerin under pressure as shown in following figurein following figure.
5050
p hkN
pressure at bottom =
25050 kNkPam
312.34kN
mweight of glycerin =
m
2 3 3
50 12.34 74.682 74.68kN kN kNp m kPam m m
m m m
(a) Find the elevation of the liquid surface in Piezometer(a) Find the elevation of the liquid surface in Piezometer A(b) The elevation of mercury in Piezometer B(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 tankas the top of the tank.
9 79kN3
9.79( ) (0.72 )(1.7 )AkNb p h m
m
2
11.98 11.98kN kPa2
211.98 // 0 519
mkN mh p 3 / 0.519
(2.36 9.79 / )0 3 0 519 0 819
Ah pkN m
h
0.3 0.519 0.819TOTALh m
0( ) A Bc p p p 311.98 (2.36 9.79 / )(0.3 )kPa kN m m
18.9kPa
specific weight of liquidcgF h A
specific weight of liquiddepth of the center of gravitycgh
depth of the center of gravity
Areacgh
A
20.1A m4.0m
2 02.0mWater
7.0m
Tank Width 3m
Example
Total force (FBC) on the bottom of the tankof the tank
Total weight (W) of the waterTotal weight (W) of the waterExplain the difference
20.1A m4.0m
2 02.0mWater
7.0m
Tank Width 3.0m
Example
( )F A h A( )BCF pA h A
9.8 (6 ) (7 3 )kN3 (6 ) (7 3 )m m m
m
1235BCF kN 1235BCF kN
20.1A m4.0m
2 02.0mWater
7.0m
Tank Width 3m
Example
( )W Volume
9 8kN 33
9.8 [(7 2 3 (4 0.1 )]kNW m m m m mm
k416W kN
20.1A m4.0m
2 02.0mWater
7.0m
Tank Width 3m
Example
The Force on the bottom of the tank is:The Force on the bottom of the tank is:
1235 BCF kN
B h l i h f h i l
N
But the total weight of the water is only:
416W kN416W kN
What is the source of the additional force?
20.1A m4.0m
2 02.0mWater
7.0m
Tank Width 3m
Example
( )ADF pA h A
29.8 ( ) ( )kN 23
9.8 (4 ) (7 3 0.1 )kN m m m mm
819ADF kN 819ADF kN
20.1A m4.0m
2 02.0mWater
7.0m
Tank Width 3m
Example
1235 416 819d th f
kN kN Kn and therefore:
BC ADF W F BC ADF W F
The large forces that can be developed with a The large forces that can be developed with a small amount of liquid (the liquid in the tube) acting over a much larger surfaceacting over a much larger surface.
Hydraulic lift Hydraulic lift
A stone weighs 90N in air When immersed inA stone weighs 90N in air. When immersed in water it weighs 50N. Compute the volume of the stone and its specific gravitythe stone and its specific gravity.
0Y 50T N
W t0BW T F
F W T
90W N
Water
90 50B
B
F W TF N N
90W N
90 50 40BF N N N BFBF
buoyant force 40BF N
3
y
40 9.8 kNN v 3
3
40 9.8
40
N vmN 3
3
40 0.0041 4.1 9800 /
Nv m litersN m
weight of the stonespecific gravity = weight of an equal volume of waterweight of an equal volume of water
90specific gravity = 2.2540
NN
An object that is 0 2m wide by 0 2m thick by An object that is 0.2m wide by 0.2m thick by 0.4m wide is found to weigh 50N in water at a depth of 0 6m What is its weight in air anda depth of 0.6m. What is its weight in air and what is its specific gravity.
50T N50 T N
0Y 00B
YW T F
50B
B
W T FW N F
W50 BW N F
BF
buoyant force weight of displaced liquidBF
3
9.8 (0.2 0.2 0.4 ) 0.157 157BkNF m m m kN N
m
50 50 157 207B
m
W N F N N N 50 50 157 207BW N F N N N
weight of the stonespecific gravity = weight of an equal volume of water207specific gravity = 1.31157
NN
157N
A hydrometer is an A hydrometer is an instrument used to measure the specific gravity of liquids.
Remember, specific gravity is the ratio of the density of thethe density of the liquid to the density of water.water.
A hydrometer weighs 0 0216N and has a stemA hydrometer weighs 0.0216N and has a stem at the upper end that is cylindrical and 2 8mm in diameter How much deeper will it2.8mm in diameter. How much deeper will it float in oil with sp gr 0.780 than in alchohol of sp gr 0.821?of sp gr 0.821?
h
alcohol0 821
oil0 780sp gr 0.821 sp gr 0.780
1For position 1, in alcohol, compute weight of hydrometer = weight of displaced liquid
v
1
weight of hydrometer weight of displaced liquid9.80 0216 0 821 kNN v 13
3
0.0216 0.821
0 0216
N vm
N
31 3
0.0216 0.000002680.821 9800 /
Nv mN m
6 31 2.68 10 v m
For position 2 in oilFor position 2, in oilweight of hydrometer = weight of displaced liquid
13
9.80.0216 0.780 ( )kNN v Ah 3
3 3
( )
0.0216( ) 0 00000283
mNAh 3 3
1
6 3
( ) 0.000002830.780 9800 /
( ) 2 83 10
v Ah m mN
Ah
6 31( ) 2.83 10 v Ah m
6 3( )h 6 31
6 3 6 3
( ) 2.83 10 2.83 10 2.68 10
v Ah mm mh
hA
2 6 2
6 3 6 3
(0.0028 ) 6.16 10 4
A m m
6 3 6 3
6 2
2.83 10 2.68 10 0.024 246.16 10
m mh m mmm
A rectangular tank 6 4m long by 2 0m deep A rectangular tank 6.4m long by 2.0m deep by 2.5m wide contains 1.0m of water. If horizontal acceleration is 2 45m/s2 then:horizontal acceleration is 2.45m/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.
2
1.0m
22.45m / s
6.4m
T k i 2 0 dTank is 2.0 deepm
2
2
( , / )tan a acceleration of vessel m s 2
2
( , / )2.45 / 0 25
g acceleration of gravity m sm s 2
. 5 /tan 0.259.8 /
tan 0 25
m sm s
tan 0.25
tan
0 2 /slope
2
0.25 /m m
1.0m
22.45m / s
6.4m
T k i 2 0 dTank is 2.0 deepm
3.2 tan 3.2 0.25 0.8y m m m 1.0 0.8 0.2CD
yd m m m
1.0 0.8 1.8ABd m m m
2
1.0m
22.45m / s
6.4m
T k i 2 0 dTank is 2.0 deepm
3
9.8 1.8 (1.8 2.5 ) 39.72
AB cgkN mF h A m m kN
m
3
29.8 0.2 (0.2 2.5 ) 0.5
2CD cg
mkN mF h A m m kN 3 ( )
2m
2
1.0m
22.45m / s
39 7ABF kN 0 5CDF kN39.7ABF kN 0.5CDF kN
6.4m
T k i 2 0 dTank is 2.0 deepm
Force needed to accelerate =Force needed to accelerate = mass of water X acceleration =
3
2 2
6.4 2.5 1.0 9.8 / 2.45 39.2m m m kN m m kN 2 2 39.2
9.8 /:
kNm s s
check39.7 0.5 39.2AB CDF F kN kN kN
Force needed to accelerate =Force needed to accelerate = mass of water X acceleration =
36.4 2.5 1.0 1000 / 16,000mass m m m kg m kg
2 2
2.45 39,20016,000 m kg mforce kgs s
s s
2 2
2.45 39,20016,000 m kg mforce kg
2 2,f gs s
kg m2
kg mnewtons
39.2force kN
1.0m
22.45m / s
39 7ABF kN 0 5CDF kN39 2force kN 39.7ABF kN 0.5CDF kN39.2force kN
6.4m
T k i 2 0 dTank is 2.0 deepm
:check39.7 0.5 39.2AB CDF F kN kN kN
A similar tank filled with water and A similar tank filled with water and accelerated at 1.5m/s2.◦ Compute how many liters of water are spilled◦ Compute how many liters of water are spilled.
drop in watersurface
2.0m
7.0m
2( / )a acceleration of vessel m s2
( , / )tan( , / )
a acceleration of vessel m sg acceleration of gravity m s
2
2
1.5 /tan 0.1539 8 /
m sm s
9.8 /
tan 0.153 slope of water surfacem s
drop in surface = 7 0.153 1.07m
1.07m
2.0m
7.0m
37.0 1.072.5 ( ) 9.362
m mVolume m m
29360Volume liters
A 1 5m cubic tank is filled with oil with sp gr A 1.5m cubic tank is filled with oil with sp gr 0.752.◦ Find the force acting on the side of the tank with an◦ Find the force acting on the side of the tank with an
acceleration of 4.9m/s2 up and with 4.9m/s2 down.
for acceleration up:for acceleration up:
(1 )Bap h
23 4.9 /(0.752 9.8 / )(1.5 )(1 )B
gm sp kN m m 2
2
(0.752 9.8 / )(1.5 )(1 )9.8 /
16.58 / 16.58
B
B
p kN m mm s
p kN m kPa
21area of loading diagram ( 16.58 / 1.5 )(1.5 )2
ABF kN m m m
18.65ABF kN
for acceleration up:2
32
for acceleration up:4.9 /(0.752 9.8 / )(0.75 )(1 )(1.5 1.5 )9 8 /
ABm sF kN m m m m 29.8 /
18.65AB
m sF kN
for acceleration down:2
32
for acceleration down:4.9 /(0.752 9.8 / )(0.75 )(1 )(1.5 1.5 )9 8 /
ABm sF kN m m m m 29.8 /
6.22AB
m sF kN
When 0 03m3/s flows through a 300mm pipe When 0.03m3/s flows through a 300mm pipe that reduces to 150mm, calculate the average velocities in the two pipesvelocities in the two pipes.
300 300 150 150Q A V A V 3
3002
0.03 / 0.42 /Q m sV m sA
2300
3
0.3004
A m
3
1502150
0.03 / 1.70 /0 150
Q m sV m sA
150 0.150
4m
If the velocity in a 300mm pipe is 0.50m/s, h h l d fwhat is the velocity on a 75mm-dia jet from a
nozzle attached to the pipe?
300 300 75 75
2 2
Q A V A V
2 2300 75
2 2
0.300 0.0754 4
m V m V
2 275
2
0.300 0.50 / 0.075
0 300 0 50 /
m m s m V
75 2
0.300 0.50 /8.00 /
0.075
m m sV m s
m
Oil of sp gr 0 75 is flowing through a 150mm 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 2 4m belowenergy relative to a datum plane 2.4m below the center of the pipe is 17.9m, determine the flow of oil.the flow of oil.
103p kPap
2.40z m 150pipe dia mm
2.40z m0.750specific gravity
E PE KE FE 2
E PE KE FEV pE z
2
2103
E zg
V kP
2
2 3
10317.9 2.402 9.8 / 0.750 9.8 /
V kPam mm s kN m
2 217.9 2.40 14.0 19.6 /5 4 /
V m m m m sV
5.4 /V m s
103p kPap
5 4 /V m s5.4 /V m s
2.40z m 150pipe dia mm
2.40z m0.750specific gravity
2(0 150 )Q AV
m
2(0.150 ) 0.0184
mA m
2
5.4 /0 018 5 4 /
V m sQ m m s
3
0.018 5.4 /0.097 /
Q m m sQ m s
In the following figure water flows from A to In the following figure, water flows from A to B at the rate of 0.40m3/s and the pressure head at A is 6 7m Considering no loss inhead at A is 6.7m. Considering no loss in energy from A to B, find the pressure head at B. Draw the energy line.B. Draw the energy line.
2
????2
BVg2
????2
AVg
????Bp
2g
p
600Dia
mm
6.70Apm
Dia
600mm
8.00Bz m
3 00300Dia
mm3.00Az m
30.40 /Q m s
Use the Bernoulli theorem from A to B: Use the Bernoulli theorem, from A to B:
energy at + energy energy energy = energy atenergy at + energy – energy – energy = energy atsection 1 added lost extracted section 2
2 2 2 2
2 2A A B B
A A L E Bp V p V
z H H H zg g
2 2 0A
g gH
00
LH 0EH
2
????2
BVg2
????2
AVg
????Bp
2g
p
600Dia
mm
6.70Apm
Dia
600mm
8.00Bz m
3 00300Dia
mm3.00Az m
30.40 /Q m s
2 2A A B Bp V p V
z z
3
2 2A Bz zg g
3
2
0.40 / 5.66 /( 300 ) / 4A
Q m sV m sA m
3
(.300 ) / 40.40 / 1 41 /
AA mQ m sV m s
2 1.41 /(.600 ) / 4B
B
V m sA m
2
????2
BVg2
????2
AVg
????Bp
2g
p
600Dia
mm
6.70Apm
Dia
600mm
8.00Bz m
3 00300Dia
mm3.00Az m
30.40 /Q m s
2 25.66 1 41m mp
5.66 1.416.7 3.0 8.0
2 2B
m mpm m m
g g
6.7 1.6 3.0 0.1 8.0Bpm m m m m
Bp
11.3 8.1Bpm m
3.2 Bpm water
2
0.12
BVm
g2
1.62
AVm
g
3.2Bpm
2g
p
600Dia
mm
6.7Apm
Dia
600mm
8.0Bz m
3 0300Dia
mm3.0Az m
30.40 /Q m s
The energy line The energy line The hydraulic grade line
2
0.12
BVm
g2
1.62
AVm
gKE
KE3.2Bp
m
2g
p
FE
600Dia
mm
6.7Apm
FE
Dia
600mm
8.0Bz m
3 0PEPE
300Dia
mm3.0Az mPE
30.40 /Q m s
A pipe carrying oil of sp gr 0 877 changes in 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 Rsection R. Section E is 3.66m lower than R and the pressures are 91.0kPa and 60.3kPa, respectively. If the discharge is 0.146m3/s,respectively. If the discharge is 0.146m /s, determine the lost head and the direction of flow.flow.
Draw a diagram to illustrate the problem Draw a diagram to illustrate the problem
Calculate average velocity at each section:g yQ AV
Q
QVA
3
150 2
0.0146 / 8.26 /(0 150 ) / 4
m sV m s 2
3
(0.150 ) / 40.0146 / 0 92 /
mm sV
450 2 0.92 /(0.450 ) / 4
V m sm
Using lower section, E, as datum:2
150
Using lower section, E, as datum:
E VpE z
2 2
2
91 0 / (8 26 / )
E EE zg
kN
2 2
3 2
91.0 / (8.26 / ) 00.877 9.8 / 2 9.8 /E
kN m m sEkN m m s
14.1EE m
2 2150
2R
R RVp
E zg
2 260.3 / (0.92 / ) 3 66
g
kN m m sE
3 2 3.660.877 9.8 / 2 9.8 /
10 7
REkN m m s
E
10.7RE m
14 1E 14.110.7
E
R
E mE m
Flow will occur from E to R because
R
the energy head at E is greater
14 1 10 7 3 4L H d 14.1 10.7 3.4Lost Head m m m
A 0 15m pipe 180m long carries water from AA 0.15m pipe 180m long carries water from A at elevation 24.0m to B at elevation 36.0m. The frictional stress between the liquid andThe frictional stress between the liquid and the pipe walls is 0.26N/m2. Determine the lost head and the pressure change.lost head and the pressure change.
First draw a sketch of the problemFirst, draw a sketch of the problem
2 2
Use Bernoulli's Theorum:
p V p V 2 2
A A B BA L B
p V p Vz H z
g g
First, calculate loss due to friction LHL
An equation for loss due to friction HL
LH L
LHR
h
shear stressL length g
areaR Hydraulic Radiusd
ywetted perimeter
For a round pipe flowing full:2
For a round pipe flowing full:/ 4A d dR
4
:
RP d
So
4L
L LHR d
24 0.26 / 180 14 7
L R dN m mH
3
4 0.26 / 180 14.79800 / 0.15L
N m mH mN m m
2 2
14 7A A B Bp V p V 14.7
2 2Velocity V is the same at both ends of the pipe
A A B BA B
p pz m z
g g
A B
Velocity V is the same at both ends of the pipeand z =24m, z =36m, so:
14.7 12A Bp pm m
26.7A Bp pm
23
9.8 26.7 262 / 262A BkNp p m kN m kPa
m
A 1m diameter new cast iron pipe (C=130) is 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 pipeFind the discharge capacity of the pipe according to the Hazen-Williams formula.
0.63 0.540.8492V CR S130C
1 4 4d mR hydraulic radius 4 4
1.11head loss mS hydraulic grade line 845
S hydraulic grade linelength
0.63 0.54
0.63 0.54
0.8492
1 1 11
V CR S
0.63 0.541 1.110.8492(130) 1.281 /
4 845m mV m s
2
31
1 281 / 1 01 /m
Q AV m s m s
1.281 / 1.01 /4
Q AV m s m s
Solve Problem 8 28 using the Manning formulaSolve Problem 8.28 using the Manning formula.2/3 1/2R SV
0 012
Vn
n
0.0121
nd mR hydraulic radius
4 4
1.11
R hydraulic radius
head loss mS hydraulic grade line 845
S hydraulic grade linelength
2/3 1/2R SV
2/3 1/21 1 11
Vn
m m
1 1.114 845 1.199 /
0 012
m m
V m s
2
0.0121m
31.199 / 0.94 /
4Q AV m s m s
m/s m3/sHazen‐Williams 1.281 1.01Manning 1.199 0.94
A 0 9m diameter concrete pipe (C=120) is A 0.9m diameter concrete pipe (C=120) is 1220m long and has a head loss of 3.9m. Find the discharge capacity of the pipe usingFind the discharge capacity of the pipe using Hazen-Williams.
0.63 0.540.8492V CR S120C
0.9 4 4d mR hydraulic radius 4 4
3.9head loss mS hydraulic grade line 1220
S hydraulic grade linelength
0.63 0.54
0.63 0.54
0.8492
0 9 3 9
V CR S
0.63 0.540.9 3.90.8492(120) 1.789 /
4 1220m mV m s
2
30.9
1 789 / 1 14 /m
Q AV m s m s
1.789 / 1.14 /4
Q AV m s m s
Solve Problem 8 32 using the Manning formulaSolve Problem 8.32 using the Manning formula.2/3 1/2R SV
0 013
Vn
n
0.0130.9
nd mR hydraulic radius
4 4
3.9
R hydraulic radius
head loss mS hydraulic grade line 1220
S hydraulic grade linelength
2/3 1/2S2/3 1/2R SVn
2/3 1/20.9 3.94 1220m m
2
4 1220 1.609 /0.013
V m s
2
30.9
1.609 / 1.02 /4
mQ AV m s m s
4
m/s m3/sm/s m /sHazen‐Williams 1.789 1.14M i 1 609 1 02Manning 1.609 1.02
What size square concrete conduit is needed What size square concrete conduit is needed to carry 4.0m3/s of water a distance of 45m with a head loss of 1 8m? Use Hazen-with a head loss of 1.8m? Use HazenWilliams.
find square dimension a0.63 0.54
3
0.84924 0 /
V CR SQ
3
2
4.0 /Q m sVA a
2
120 (for concrete from Problem 8.32)Ca aR
4 41 8
Ra
head loss m
1.8 45
head loss mS hydraulic grade linelength m
0.63 0.54
0.63 0.54
0.8492
4 0 1 8
V CR S
a
2
4.0 1.80.8492 1204 45a
a
0.63 0.542.63
0.54
4.0 4 450.75
0 8492 120 1 8a m
0.8492 120 1.8Specify 0.80m by 0.80m conduit
Water is flowing in a 500mm diameter new Water is flowing in a 500mm diameter new cast iron pipe (C=130) at a velocity of 2 0m/s Find the pipe friction loss per 100m2.0m/s. Find the pipe friction loss per 100m of pipe. Use Hazen-Williams.
0 63 0 540.63 0.540.84922 0 /
V CR SV m s 2.0 /
130V m sC
0.50 0.1254
mR m 4
0.63 0.540.8492V CR S
0.63 0.54
0.8492
2.0 / 0.8492 130 0.125
V CR S
m s m S 0.54
0 63
2.0 /m sS 0.630.8492 130 0.125
0 0067 /m
S m m 0.0067 /Total head loss for 100m pipe: S m m
p p0.0067 / 100 0.67m m m m
To express the loss as pressure:9 8kN
3
9.8 0.67kNp h mm
2
6.6 6.6kNp kPam
m
For a lost head of 5 0m/1000m and C=100 For a lost head of 5.0m/1000m and C=100 for all pipes, how many 20cm pipes are equivalent to one 40cm pipe? How may areequivalent to one 40cm pipe? How may are equivalent to a 60cm pipe?
Out of curiousity, let's compare the cross-sectional areas:2
2020A = 1004
2
4040A = 4004
2
6060A = 12004
If area was all that mattered:it would take 20cm pipes to equal 1-40cm pipe, and4 it would take 20cm pipes to equal 1-60c12 m pipe
But we must consider head loss But we must consider head loss. Let’s use two equations:
Q AVQand
0.63 0.540.8492V CR S
Q AV2
4dA
0.63 0.540.8492V CR SdR
4
5 0 005 /
R
mS m m
0.005 /1000100
S m mm
C 0.63
2 0.540.8492 100 0.0054dd
Q
4
Q
0.632 0 540 20 2 0.54
320
0.200.20 0.8492 100 0.0054 0.023 /
4Q m s
0.63
4
0 40 2 0.54
340
0.400.40 0.8492 100 0.0054 0.143 /
4Q m s
0.63
4
0 60 2 0.54
360
0.600.60 0.8492 100 0.0054 0.416 /
4Q m s
4
340
320
0.143 / 6.2 20cm pipes equivalent to a 40cm pipe0.023 /
Q m sQ m s
360 0.416 / 18 1 20 i i l t t 60 i
Q m s603
20
18.1 20cm pipes equivalent to 60cm pipe0.023 /
QQ m s
A 300mm pipe that is 225m long and a 500m A 300mm pipe that is 225m long and a 500m pipe that is 400m long are connected in series Find the diameter of a 625m longseries. Find the diameter of a 625m long equivalent pipe. Assume all pipes are concrete.concrete.
2d 20.63 0.54; ; 0.8492
4dQ AV A V CR S
; 120 for concrete4dR C
31
4
; assume 0.1 /h
S Q m s ; QL
0.540.632 10.8492 120
4hddL
40.14
L
for the 300mm dia 225m long pipe:0.540.63
2 1
for the 300mm dia, 225m long pipe:
0.300 30 0 8492 120h
2 10.300.30 0.8492 120
4 2250.1
0.544
h 10.1 1.4087225h
1 1.678h m
for the 500mm dia, 400m long pipe:0.540.63
2 2
for the 500mm dia, 400m long pipe:
0.500 50 0 8492 120h
20.50 0.8492 120
4 4000.14
0.542
4h 20.1 5.3985
400h
2 0.248h m
total head loss 1.678 0.248 1.926m
Th fTherefore:for a 625m long equivalent pipe:
0.63 0.542
for a 625m long equivalent pipe:
1.9260 8492 120 dd 0.8492 1204 6250.1
4
d 4
360d mm
A 300mm pipe that is 225m long and a 500m A 300mm pipe that is 225m long and a 500m pipe that is 400m long are connected in series Find the diameter of a 625m longseries. Find the diameter of a 625m long equivalent pipe. Assume all pipes are concrete.concrete.
Use Diagram B-3 to solve Use Diagram B-3 to solve.
3Assume a flow rate, 0.1 /Q m sAssume a flow rate, 0.1 /From Diagram B-3,
Q m s
1
2
for 300mm pipe, 0.0074 /for 500mm pipe, 0.00064 /
h m mh m m2p p ,
total head loss =
1
0.0074 / 225 0.00064 / 400 1.921for a 625m long pipe, 1.921 / 625 0.00
m m m m m m mh m m
307m1g p p ,
from diagram B-3, 360d m
Water flows at a rate of 0 05 m3/s from Water flows at a rate of 0.05 m3/s from reservoir A to reservoir B through three concrete pipes connected in series as shownconcrete pipes connected in series, as shown on the following slide. Find the difference in water surface elevations in the reservoirs.water surface elevations in the reservoirs. Neglect all minor losses.
Use Diagram B-3 Use Diagram B-3
30 05 /Q m s 0.05 /from Diagram B-3:Q m s
1
gfor the 400mm pipe, 0.00051 /h m m
1for the 300mm pipe, 0.0020 /h m m
1for the 200mm pipe, 0.015 /h m m
total head loss0 00051 / 26000.00051 / 2600
0 0020 / 1850m m m
m m m 0.0020 / 1850
0.015 / 970m m m
m m m
19.58m
Use Hazen-Williams Formula Use Hazen-Williams Formula2
0 63 0 540 8492dQ AV A V CR S 0.63 0.54; ; 0.84924
Q AV A V CR S
d
; 120 for concrete4dR C
3; 0.05 /hS Q m sL
L
0.63 0.542 0.8492 120
4d hd
L
40.054
L
0 540 63
for the 400mm dia, 2600m long pipe:0.540.63
2 4000.400.40 0.8492 1204 2600
h 4 26000.05
41 32h m
300 1.32h m
0 540 63
for the 300mm dia, 1850m long pipe:0.540.63
2 3000.300.30 0.8492 1204 1850
h 4 18500.05
43 82h m
300 3.82h m
0.540.63
for the 200mm dia, 970m long pipe:
0 20 h 2 2000.200.20 0.8492 1204 9700.05
h
300
0.054
14.44h m300
1.32 3.82 14.44 19.58total head m m m m
The flow in pipes AB and EF is 0 850m3/s All The flow in pipes AB and EF is 0.850m3/s. All pipes are concrete. Find the flow rate in pipes BCE and BDEpipes BCE and BDE.
Assume head loss from B to E 1.00m
11.00for pipe BCE, 0.00043 /2340
mh m m
3
2340from Diagram B-3, 0.133 /BCE
mQ m so g 3, 0. 33 /BCEQ m s
11.00for pipe BDE, 0.00031 /3200
mh m m 1
3
p p ,3200
from Diagram B-3 0 038 /m
Q m sfrom Diagram B 3, 0.038 /BDEQ m s
if head loss from B to E = 1 00m is correct thenif head loss from B to E = 1.00m is correct, thensum of the flow rates through BCE and BDE will
3
gequal 0.850m /s
but 0 133 0 038 0 171 0 850 but, 0.133 0.038 0.171 0.850
head loss of 1.00m is not correct, however,actual flow rates through BCE and BDE willbe at the same proportion Sobe at the same proportion. So,
3BCE
0.133Q = 0.850 0.661 /m s BCE
3
Q171
0.038Q 0 850 0 189 /3BDEQ = 0.850 0.189 /
171m s
C t th
1500m
Compute the flow in each 1500
120m
C
30 51 /Q
branch.
W Z
30.51 /Q m s
900120m
C
2d 20.63 0.54; ; 0.8492
4dQ AV A V CR S
; 120 for concrete; 4d hR C S
L
4 L
0.63 0.542 0.8492 120
4d hd
L
44
LQ
C t th
1500m
Compute the flow in each 1500
120m
C
30 51 /Q
branch.
W Z
30.51 /Q m s
900120m
C
h d l f 10 f W t Z
0 63 0 54
assume a head loss of 10m from W to Zthen, for the 300mm pipe:
0.63 0.54
2
3300
0.3 100.3 0.8492 1204 1500 0.09 /
4Q m s
0.63 0.54
4and, for the 400mm pipe:
0 4 10 2
400
0.4 100.4 0.8492 1204 900
4Q
30.26 /m s4
1500 , 300m mm
30.09 /Q m s1500 , 300
120m mm
C
30 51 /QW Z
30.51 /Q m s
900 , 400120m mm
C 30 26 /Q m s 3 3 30 09 / 0 26 / 0 51 /m s m s m s 0.26 /Q m s 0.09 / 0.26 / 0.51 /m s m s m s
3 3 3 33 3 3 3
3300
0.09 / 0.26 / 0.35 / 0.51 /0.09 0.51 0.13 /0 35
m s m s m s m s
Q m s
300
3400
0.350.26 0.51 0.38 /0 35
Q
Q m s 0.35
1500 , 300m mm
30.13 /Q m s1500 , 300
120m mm
C
30 51 /QW Z
30.51 /Q m s
900 , 400120m mm
C 30 38 /Q m s 3 3 30 13 / 0 38 / 0 51 /m s m s m s 0.38 /Q m s 0.13 / 0.38 / 0.51 /m s m s m s
Next let’s solve using Hardy Cross Method Next, let s solve using Hardy Cross Method
C t th
1500 , 300m mm
Compute the flow in each 1500 , 300
120m mm
C
30 51 /Q
branch using Hardy‐Cross
W Z
30.51 /Q m s yMethod.
900 , 400120m mm
C
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• Analyze using the Hardy Cross Method– 1. assume flows for each individual pipe.
2 calculate the head loss thru each pipe using– 2. calculate the head loss thru each pipe using Hazen-Williams.
– 3. find the sum of head losses in each loop.p– 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 dj t th d fl t f ll i d– 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 For this problem, we are looking at just one loop:
Q
Q
C t th
1500 , 300m mm
Compute the flow in each 1500 , 300
120m mm
C
30 51 /Q
branch using Hardy‐Cross
W Z
30.51 /Q m s yMethod.
30 51 /Q900 , 400120m mm
C
30.51 /Q m s
Analyze using the Hardy Cross Method• Analyze using the Hardy Cross Method– 1. assume flows for each individual pipe.
2 calculate the head loss thru each pipe using– 2. calculate the head loss thru each pipe using Hazen-Williams.
– 3. find the sum of head losses in each loop.p– 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 dj t th d fl t f ll i d– 7. adjust the assumed flow rates for all pipes and repeat the process until all corrections are zero.
For correction use: For correction, use:
LH 01.85 ( / )
LHLH Q
3, /flow adjustment m s
LH lost head L S
0 Q assumed initial flow
For Consistency: For Consistency:◦ Clockwise, Q and LH are positive◦ Counterclockwise Q and LH are negativeCounterclockwise, Q and LH are negative◦ So, in:
LH
01.85 ( / )LH Q
◦ Sign (+ or -) is important in numerator◦ But, denominator is always positive
In the table that follows:
- :S is from the Hazen Williams formula
0.63 0.54 0.8492V CR S
l
:also use
Q VA Q VA
C t th
1500 , 300m mm
Compute the flow in each 1500 , 300
120m mm
C
30 51 /Q
branch using the Hardy Cross
W Z
30.51 /Q m s yMethod.
900 , 400120m mm
C
See Excel Spreadsheet See Excel Spreadsheet Or
Pdf Version Pdf Version
Analyze using the Hardy Cross Method• Analyze using the Hardy Cross Method– 1. assume flows for each individual pipe.
2 calculate the head loss thru each pipe using– 2. calculate the head loss thru each pipe using Hazen-Williams.
– 3. find the sum of head losses in each loop.p– 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 dj t th d fl t f ll i d– 7. adjust the assumed flow rates for all pipes and repeat the process until all corrections are zero.
30.4 /m s600m 600mBA C300 mm dia 300 mm dia
400250
mmm dia
400250
mmm dia
400250
mmm dia
600300
mmm dia
600300
mdiEF D
30.4 /m s300 mm dia 300 mm diaEF D
• Analyze using the Hardy Cross Method– 1. assume flows for each individual pipe.– 2. calculate the head loss thru each pipe using
Hazen WilliamsHazen-Williams.– 3. find the sum of head losses in each loop.– 4 remember head loss between two joints is the4. 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 irepeat the process until all corrections are zero.
For this problem we are looking at only For this problem, we are looking at only loops:
Q
Q
Solve using Excel Solve using Excel .pdf Version
0.63 0.54For , use Hazen-Williams: 0.8492S V CR S
LH total head loss in pipe L S
LH
1.85 /LH Q
30.4 /m s600m 600mBA C300 mm dia 300 mm dia
30 0.200 /Q m s 3
0 0.100 /Q m s
400250
mmm dia 400
250 mmm dia 400
250 mmm dia
30 0.100 /Q m s
30 0.100 /Q m s
30 0.200 /Q m s
600300
mmm dia 600
300mmm diaEF D
0Q
30.4 /m s300 mm diaEF D3
0 0.300 /Q m s3
0 0.200 /Q m s
30.4 /m s600m 600mBA C300 mm dia 300 mm dia
3
30
0 0.2410.200 /
/Q sQ s
mm
3
030.15 /
.100 /9
0Q m sQ m s
400250
mmm dia 400
250 mmm dia 400
250 mmm dia
3
30
0 0820.100 /
/QQ sm
30
30 16 /.200 /
00Q m s
Q m s 3
0 .100 /0Q m s
600300
mmm dia 600
300mmm diaEF D
30 0.082 /Q sm0.16 /0Q m s 0
30.15 /9QQ m s
30.4 /m s300 mm diaEF D3
030.16 /
.200 /0
0Q m sQ m s
30
30.24 /.300 /
10Q m s
Q m s
Water flows in a rectangular concrete open Water flows in a rectangular concrete open channel that is 12.0m wide at a depth of 2 5m The channel slope is 0 028 Find the2.5m. The channel slope is 0.028. Find the water velocity and the flow rate.
using the Manning equation:2/3 1/2
g g qRV= S
2.5 12.0 1 765
nareaR m
2/3 1/2
1.765 2.5 12.0 2.5
1 765 0 0028
R mwetted perimeter
3
1.765 0.0028V= 5.945 /0.013
m s
32.5 12.0 5.945 178 /Q AV m s
Water is to flow at a rate of 30
3m3/s in the concrete channel shown on the following slide. What slope of channel will be required?
2/3 1/2RQ SV
21.6 3.6
VA n
m m
2
2
1.6 3.6(3.6 2.0 ) 2.0 12.402
m mA m m m m
2
2 2
12.40 1.2363.6 1.6 2.0 2.0 2.0
mR mm m m m m
3 2/3 1/2
2
30 / 1.236 0.000746 /0 01312 40
m s m S m mm
0.01312.40
or, 0.746 /m
m kilometer
On what slope should a 600mm concrete On what slope should a 600mm concrete sewer pipe be laid in order that 0.17m3/s will flow when the sewer is one-half full? Whatflow when the sewer is one half full? What slope if the sewer flows full? (use n=0.013)
2V 2
2/3: =RVnuse S
R
21 d 2 4 0.6 0 15d mR m
1/2 0.15
1 4 4R m
d
2
0 6
d
d
0.6 0.154 4Fulld mR m 4 4
0 17Q1/2 2
0.17 1.206 /1 0.6
QV m sA
2 4
2 2
1/2 2/3 2/3
1.206 0.013 0.0031 /VnS m m
1/2 2/3 2/30.15R
0 17Q2
0.17 0.601 /0.6Full
QV m sA
42 2
2/3 2/3
0.601 0.013 0.00077 /0 15Full
VnS m mR
2/3 2/30.15Full R
A 600mm diameter concrete pipe on a 1/400 A 600mm diameter concrete pipe on a 1/400 slope carries water at a depth of 240mm. Find the flow rate QFind the flow rate, Q.
240depth mm600
1 / 400
pdiameter mmslope
p
240depth mm600
1 / 400
pdiameter mmslope
p
300r mm
240300 240 60depth mm
mm mm mm
300r mm 60mm0 060 1 0.060cos 1.3690.30
0 30 sin(1 369 ) 0 294
rad
rad m
0.294m
0.30 sin(1.369 ) 0.294rad m
Area of triangles g1 0.060 (0.294 2) 0.01762
f i l ti300r mm
60mm2
area of circle section1.369 2 0.60 0.123
2 4
2 4
0.123 0.018 0.105A
20 105A m
0.294m 0.294m
0.123 0.018 0.105A 0.105A m
tt d i twetted perimeter1.369 2 0.60
2
2
1.369 0.60 0.821
300r mm60mm
20.105A m0.105R 0.1280.821
0.128R
2/3 1/2RV= S
1/22/3 10.128
400
n 400V=
0.0130.977V
300r mm60mm
0.977V
Q AV 20.105A m
30.105 0.977 0.10 /Q
m s 0.128R
After a flood had passed an observation After a flood had passed an observation station on a river, an engineer visited the site.
By locating flood marks performing By locating flood marks, performing appropriate survey and doing necessary computations she determined that at thecomputations, she determined that, at the time of peak flooding :◦ The cross-sectional area was 2960m2The cross sectional area was 2960m◦ The wetted perimeter was 341m◦ The water surface slope was 0.00076m/m p /
The engineer also noted that the channel The engineer also noted that the channel bottom was “earth with grass and weeds,” for which a handbook gave a Manning n value ofwhich a handbook gave a Manning n value of 0.030.
Estimate the peak flood discharge Estimate the peak flood discharge.
2/3
1/22960(2960) 0 00076 2/3 1/2 (2960) 0.00076
3410.030
AR SQn
311,500 /Q m s
A rectangular channel 6 1m wide carries A rectangular channel 6.1m wide carries 11.3m3/s and discharges onto a 6.1m wide apron with no slope at a mean velocity ofapron 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 thejump? What energy is absorbed (lost) in the jump?
21 2
1 2y yq y y 1 2 2
y yg
311.3 /Q m s 11.3 /Q m s
1 6.1 /v m s
21 2
1 2y yq y y 1 2 2
y yg
311.3/ 1.85 /6.1
q flow unit width m s
11.85 0.3036 1
qy mV
1
22
6.10.3031.85 0 303
Vy
y 20.303
9.8 21 37
y
y m
2 1.37y m
21 2
1 2y yq y y 1 2 2
y yg
311.3 /Q m s
311.3 /1.85 /
Q m sq m s
1 6.1 /V m s
0 303y m 2 1.37y m1 0.303y m
1.37 0.303 1.07jump m m m
critical depth:32 23
critical depth:
/ 1.85 / 9.8 0.70Cy q g m Cy q g
Therefore:◦ Flow depth before the jump (0.303m) is < 0.70 and
i iti lis supercritical◦ Flow depth after the jump (1.37m) is > 0.70 and is
subcriticalsubcritical
21 2
1 2y yq y y 1 2 2
y yg
311.3 /Q m s
311.3 /1.85 /
Q m sq m s
1 6.1 /V m s
0 303y m 2 1.37y m0.70cy m1 0.303y m c
1.37 0.303 1.07jump m m m
Lost energy: Lost energy:
26 1 / 21 1 1
6.1 // 2 0.303 2.20
2 9.8m s
E y V g m m
2
22 2 2
1.85 / 1.37 // 2 1.37 1.46
sE y V g m m 2 2 2 2 9.8
2.20 1.46 0.74
y g
lost head m m m
2.20 0.303 1.90m m m 1.46 1.37 0.09m m
1 6.1 /V m s 1.90m0.09m0.74m
0 303y m 2 1.37y m0.70cy m1 0.303y m c
1.37 0.303 1.07jump m m m
Measures stagnation pressure (at B) which Measures stagnation pressure (at B), which exceeds the local static pressure (at A), to determine velocity headdetermine velocity head.
Ah Bh
Velocity (V) atVelocity (V) at Point B is zerozero.
A l th Ah
Bh
Apply the Bernoulli
tiequation, next slide
2 2V Vl 2 2
2 2A A B B
A Bp V p Vno lossz z
g assumed g
0; zB A B
g gV z
2
,so2
2A A Bp V p
g
2g
2A A Bp V p
2A A Bp p
g
2 B Ap pV g g
p p
B AB A
p ph h d
With no friction:2V gd
h h dB Ah h d
Ah Bh
A small amount of friction normally occurs, ff f l ( dso a coefficient of velocity cV (see discussion
on following slides) is sometimes used:
V
actual velocityc
2
V theoretical velocity
V d2VV c gd
1 id ffi ito assume 1 provides sufficient accuracy for most engineering problems involving Pitot tubes
vc
involving Pitot tubes.
The ratio of the actual velocity in a stream to h h l l h ldthe theoretical velocity that would occur
without friction.
actual velocity V
actual velocityctheoretical velocity
A Pitot tube having a coefficient of 0 98 is 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 pressurecenter of a pipe. The stagnation pressure head is 5.67m and the static pressure head in the pipe is 4.73m. What is the velocity?the pipe is 4.73m. What is the velocity?
4.74m5.67m
B Ah h d 5.67 4.74 0.94
B Ah h dm m m
4.74m 5.67m
2V d25 67 4 73 0 94
VV c gdd m m m
5.67 4.73 0.940.98V
d m m mc
29.8 /V
g m s20.98 2 9.8 / 0.94 4.21 /V m s m m s
A 100mm A 100mm diameter standard orificestandard orifice discharges water under a 6.1munder a 6.1m head. What is the flow?flow?
2area; 0 6
Q cA gHA c
area; 0.6total head causing flow
A cH
22
g0.10 6 2 9 8 / 6 1mQ m s m
3
0.6 2 9.8 / 6.14
0 05 /
Q m s m
Q
30.05 /Q m s
The tank in problem 12 9 is closed ant the air The tank in problem 12.9 is closed ant the air space above the water is under pressure, causing to flow to increase to 0 075m3/scausing to flow to increase to 0.075m3/s. Find the pressure in the air space.
2
2
0 1
Q cA gH
m
3 20.10.075 / 0.6 2 9.8 /
4m
m s m s H
12.912.9 6.1 6.8P Z
H mh H h m m m
3 2
12.9 6.1 6.8
9.8 / 6.8 70 / 70P Z
P
h H h m m m
p h kN m m kN m kPa
During a test on a 2 4m suppressed weir that During a test on a 2.4m suppressed weir that was 0.9m high, the head was maintained constant at 0 3m In 38 seconds 29m3 ofconstant at 0.3m. In 38 seconds, 29m of water were collected. Find the weir factor m using equations A and B.using equations A and B.
0 3H m0.3H m
0 9Z 0.9Z m
329m 329 0.763 /38
0 9 0 3 1 2
mQ m ss
fl d h
3
0.9 0.3 1.20.763 / 0 265 /
flow depth m m mQ m sV
0.265 /2.4 1.2
QV m sA m m
3/2 3/22 2
using Eq. A:
V VQ b H
3/2
2 2Q b H
gm
g
3/2 3/22 23
2 2
0.265 0.2650.763 2.4 0.32 9.8 2 9.8
mm
3/2 3/230.763 2.4 0.3 0.00358 0.00358mm
1.90m
i E B3/2
using Eq. B:Q bHm
3/20.763 2.4 0.3Q
mbHm
1.93m
1.90 1.93 Equation B is OK for weirsl d hi h
placed high
During a test on a 2 4m suppressed weir that During a test on a 2.4m suppressed weir that was 0.0m high, the head was maintained constant at 0 3m In 38 seconds 29m3 ofconstant at 0.3m. In 38 seconds, 29m of water were collected. Find the weir factor m using equations A and B.using equations A and B.
0 3H m0.3H m
0 0Z 0.0Z m
b Hb
Z=0
From: http://www.lmnoeng.com/Weirs/cipoletti.htm
329m 329 0.763 /38
0 3 0 30 0
mQ m ss
fl d h
3
0.3 0.30.763 / 1 06 /
0.0flow depth m mQ m sV
m
1.06 /2.4 0.3
QV m sA m m
3/2 3/22 2
using Eq. A:
V VQ b H
3/2
2 2Q b H
g gm
3/2 3/22 23
2 2
1.06 1.060.763 2.4 0.32 9.8 2 9.8
m m
3/2 3/230.763 2.4 0.3 0.0573 0.0573mm
1.53m
using Eq. B:3/2
3/2
Q bHm3/20.763 2.4 0.3
1 93m
m
1.93m
1.53 1.93, Equation A must be used , q for shallow weirs