1
Flow Measurement
Dr. James C.Y. Guo, P.E., Professor and Director Civil Engineering, U. of Colorado Denver
Pitot-Static Tube
2
Pitot-tube for Velocity Measurement
Dr. James C.Y. Guo, Professor and P.E., Civil Engineering, U. of Colorado Denver
Operation of Pitot-Tube
Dr. James C.Y. Guo, Professor and P.E., Civil Engineering, U. of Colorado Denver
3
Manometer and Pressure Differential
Example of Flow Velocity Measurement
The air flow is carried in a 12-inch circular pipe. The air flow is at 70oF and under an atmospheric pressure of 12.4 psi. The pitot-tube is used to
th i fl l it t th t li Th diff ti l imeasure the air flow velocity at the centerline. The pressure differential is measured to be 3 inches in a water manometer. Find the centerline velocity.
wa
wa hh )1( −=
γγ
4
Measurement of Flow Rate
0velocity in feet/sec
20
18
12
8
0 2 4 6 10 ftRadius
Dr. James C.Y. Guo, Professor and P.E., Civil Engineering, U. of Colorado Denver
Example of Air Flow Velocity ContoursLocation Measured Average Circle Ring Incremental Accum
from Air Flow Flow Area Area Flow FlowPipe Cntr Velocity Velocity
cm mps mps sq m sq m mps mps58 0.000 1.06
0.875 0.107 0.093 0.09355 1 750 0 9555 1.750 0.95
1.885 0.133 0.251 0.34451 2.020 0.82
2.145 0.236 0.507 0.85143 2.270 0.58
2.315 0.127 0.295 1.14638 2.360 0.45
2.485 0.352 0.874 2.02018 2.610 0.10
2.670 0.102 0.272 2.2920 2 730 0 00
Calculate the cross sectional average velocityCalculate the ratio of the center velocity to average velocity
0 2.730 0.00
5
Review of Turbulent Flow Velocity Profile
rRRuU
rRRuUu mm −
−=−
−= log76.5ln5.2 ** (22)
Eq 22 has two unknown: Um and u*.
VfVU m 2.1)326.11( ≈+= (23)
8*fVu ==
ρτ (24)
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Example of Flow Velocity Measurement4.1 A 10-ft pipe carries a water flow of 1000 cfs. The friction factor for this case is 0.03. Analyze this flow.
(1) Cross section area A = 78.5 sq ft (2) Flow velocity V = 12.73 fps( ) y p(3) Centerline velocity 66.1523.1)33.11( ==+= VVfUm fps (V/Um = 0.81)
(4) Shear velocity 78.08
* ===fVu
ρτ fps
(5) Shear stress 17.1=τ lb/ft2
(6) )5
5ln(95.166.15r
u−
−=
Flow Measurement: Centerline Velocity Approach
A pitot-tube is used to measure the centerline velocity in a 12-inch circular pipe that has a friction factor f of 0.025• Calculate the centerline velocity• Find the average velocity• Calculate the flow rate.
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Venturi Meter
iiii ghAVAQ == 2
i
a
ia
iiii
QQK
ghKAQ
ghVQ
=
= 2
Orifice Meter
ia
iiii
ghKAQ
ghAVAQ
=
==
2
2
Q
i
a
QQK =
Qi
Qa
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Orifice OutletOrifice = an opening on a wallNozzle = a short converting pipe Jet = a high-speed stream of flow issued from a nozzle
Dr. James C.Y. Guo, Professor and P.E., Civil Engineering, U. of Colorado Denver
9
Orifice Outlet Equation
c
i
cc
VC
AAC =
vc
iviccca
iiii
i
cv
CCKghKAghCACVAQ
ghAVAQ
VVC
=
===
==
=
22
2
)(2 oo YYgKAQ −=
Dr. James C.Y. Guo, Professor and P.E., Civil Engineering, U. of Colorado Denver
Examples of Orifice Outlet
Dr. James C.Y. Guo, Professor and P.E., Civil Engineering, U. of Colorado Denver
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Weir Hydraulics
Dr. James C.Y. Guo, Professor and P.E., Civil Engineering, U. of Colorado Denver
End Contraction
5.15.1)1.0(232 HLCHHnLgCQ ewd =−=
Cd=0.67
Dr. James C.Y. Guo, Professor and P.E., Civil Engineering, U. of Colorado Denver
11
Examples of Orifice Outlet
An opening area in the flow field can be vertical, horizontal, or inclined. This opening area may be operated like an orifice when the entire opening area is submerged or operated like a weir if the opening area is partially submerged.
Q=min (Qweir, Qorifice) for a given depth
Dr. James C.Y. Guo, Professor and P.E., Civil Engineering, U. of Colorado Denver
Example for Weir Flow
Find the release flow rate through the vertical orifice(1)when the water surface elevation at 5002 ft(2)when the water surface elevation at 5006 ft
Dr. James C.Y. Guo, Professor and P.E., Civil Engineering, U. of Colorado Denver
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Sluice Gate
gYACQ s 2= Cs=0.5 to 0.6
Dr. James C.Y. Guo, Professor and P.E., Civil Engineering, U. of Colorado Denver
OPEN CHANNEL FLOW 1. Bottom Width B 2. Side Slope Z3. Top Width T= B + 2 Z Y4. Area A = 0.5 (T+B)Y5. Wetted Perimeter P = B + 2 Y(1+Z2)0.5
6. Hydraulic Radius R = A/P7. Hydraulic Depth D = A/T8. Froude Number Fr = U/(gD)0.5
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Example: Channel Section ElementA symmetrical trapezoidal channel has a bottom width of 20 ft, side slope of1V:5H (Z= 5), and flow depth of 2 feet. Find the flow area, top width, wettedperimeter, hydraulic radius, and hydraulic depth. A = BY +ZY2 = 20.0*2.0+5.0*2.0*2.0=60.0 ft2 T= B+2ZY=20 + 2.0*5.0*2.0 = 40.0 ft 39.400.51*0.2*22012 22 =++=++= ZYBP ft
49.139.400.60===
PAR ft
5.10.400.60===
TAD ft
Dr. James C.Y. Guo, Professor and P.E., Civil Engineering, U. of Colorado Denver
Definition of Slope
1. Channel Bottom Slope So2. Water Surface Slope Sw3. Energy Slope Se4. Friction Slope Sf
Dr. James C.Y. Guo, Professor and P.E., Civil Engineering, U. of Colorado Denver
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Manning’s Eq is dimensionally inconsistent, but most widely accepted and used for channel design.K= 1.486 for feet-sec or K=1 for meter-sec.
Empirical Equations
SPAnKAS
PA
nKUAQ 3
235
21
32 −
=⎟⎠⎞
⎜⎝⎛==2
132
SRnKU =
Use the bottom slope to produce the normal depth (uniform flow)Use the energy slope to predict the flow depth (non-uniform flow)
Man-made Channels
Cross Waves
Super Elevation on Outer Bank
Uniform Flow
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Prismatic Channel
Uniformity inAlignment StraightConstant Cross SectionConstant Channel SlopeConstant Channel RoughnessIt takes a long distance to develop uniform flow
Normal Flow1. Let Se=So to find Y=Yn2. Froude Number
A t id l h l i di h f 300 f Th h l h l fA trapezoidal channel carries a discharge of 300 cfs. The channel has a slope of0.005, roughness of 0.03, bottom width of 5 ft, and side slope of 1V:ZH, Z=2.0.Determine the normal depth. )25( nnn YYA +=
nnn YYP 47.40.52125 2 +=++=
005.0)47.45()]25([03.049.1 3/23/5 −++= nnn YYYQ
So e ha e Y 3 91 ft A 50 08 ft2 T 20 63 ft d U 5 99 fSo we have Yn= 3.91 ft, An=50.08 ft2, Tn=20.63 ft, and Un=5.99 fps. 43.2
63.2008.50
===TAD feet
68.043.22.32
99.5=
×=rF < 1.0. Subcritical flow!
Dr. James C.Y. Guo, Professor and P.E., Civil Engineering, U. of Colorado Denver
17
Example of Weir and Channel Flow
Determine the flow from the weir measurementCalculate the normal flow condition downstream
Dr. James C.Y. Guo, Professor and P.E., Civil Engineering, U. of Colorado Denver
Specific Energy=Y+U2/2g1. Illustration-- Sluice gate (ΔE=0)2. For a given Q, the curve for Y vs Es3. Flow regimes4 C iti l Fl4. Critical Flow 5. Increase or decrease of Q
Dr. James C.Y. Guo, Professor and P.E., Civil Engineering, U. of Colorado Denver
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Example for Specific Energy Curve
SPECIFIC ENERGY CURVE FOR TRAPEZOIDAL CHANNEL
Bottom Width B= 10.00 feetLeft Side Slope Z1= 3.00 ft/ftRight Side Slope Z2= 3.00 ft/ft
Design Flow Q= 100 cfsDesign Flow Q= 100 cfsBeginning Flow Depth Y= 0.50 ftIncremental Depth Dy= 0.25 ft
Flow Flow Top Flow Kinetic Sp. FroudeDepth Area Width Velocity Energy Energy Numer
Y A T u u^2/2g Es Fr ft sq ft ft fps ft ft
0.50 5.75 13.00 17.39 4.70 5.20 4.610.75 9.19 14.50 10.88 1.84 2.59 2.411.00 13.00 16.00 7.69 0.92 1.92 1.501.25 17.19 17.50 5.82 0.53 1.78 1.031.50 21.75 19.00 4.60 0.33 1.83 0.761.75 26.69 20.50 3.75 0.22 1.97 0.582.00 32.00 22.00 3.13 0.15 2.15 0.462 25 37 69 23 50 2 65 0 11 2 36 0 37
Specific Energy
2 00
2.50
3.00
3.50
4.00
4.50
5.00
w D
epth
in ft
2.25 37.69 23.50 2.65 0.11 2.36 0.372.50 43.75 25.00 2.29 0.08 2.58 0.302.75 50.19 26.50 1.99 0.06 2.81 0.263.00 57.00 28.00 1.75 0.05 3.05 0.223.25 64.19 29.50 1.56 0.04 3.29 0.193.50 71.75 31.00 1.39 0.03 3.53 0.163.75 79.69 32.50 1.25 0.02 3.77 0.144.00 88.00 34.00 1.14 0.02 4.02 0.124.25 96.69 35.50 1.03 0.02 4.27 0.114.50 105.75 37.00 0.95 0.01 4.51 0.10
0.00
0.50
1.00
1.50
2.00
0.00 1.00 2.00 3.00 4.00 5.00 6.00
Es in feet
Flow
Application of Energy Eq---Sluice Gate
Flow Flow Wetted Hy- Flow Flow Froude Sp Depth to SpDepth Area P-meter Radius Velocity rate Numer Energy Centroid Force
Y A P R U Q Fr Es Yh Fsft sq ft ft ft fps cfs ft ft klb10.00 500.0 92.5 5.41 9.30 4650.7 0.70 11.34 3.64 197.49
Flow Flow Top Flow Kinetic Sp. FroudeDepth Area Width Velocity Energy Energy Numer
Y A T u u^2/2g Es Fr ft sq ft ft fps ft ft
6.00 204.00 58.00 22.80 8.07 14.07 2.147.35 289.59 68.80 16.06 4.00 11.35 1.388.00 336.00 74.00 13.84 2.97 10.97 1.149.00 414.00 82.00 11.23 1.96 10.96 0.88
10.00 500.00 90.00 9.30 1.34 11.34 0.7011.00 594.00 98.00 7.83 0.95 11.95 0.5612.00 696.00 106.00 6.68 0.69 12.69 0.46
Dr. James C.Y. Guo, Professor and P.E., Civil Engineering, U. of Colorado Denver
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Critical Flow Fr=1.01. Froude number = 12. Critical flow depth 3. Critical Slope is solved by Manning’s eq with Yn=Yc4. Mild Slope and Steep Slope
012
== cTQF
1. Bottom Width B 2. Side Slope Z3. Top Width Tc= B + 2 Z Yc4. Area Ac = 0.5 (Tc+B)Yc5. Hydraulic Depth Dc = Ac/Tc6. Froude Number Fr = Uc/(gDc)0.5 =1.0
0.13 ==c
r gAF
Dr. James C.Y. Guo, Professor and P.E., Civil Engineering, U. of Colorado Denver
cccccc
c SPAnKAS
PA
nKUAQ 3
2353
2−
=⎟⎟⎠
⎞⎜⎜⎝
⎛==
Example for Critical FlowGiven Design Information: Find Critical Slope
Bottom Width B= 5.00 feet Bottom Width B= 5.00 feetLeft Side Slope Z1= 3.00 ft/ft Left Side Slope Z1= 3.00 ft/ftRight Side Slope Z2= 3.00 ft/ft Right Side Slope Z2= 3.00 ft/ftManning's n N= 0.035 Manning's n N= 0.014Longitudinal Slope S= 0.0075 ft/ft Longitudinal Slope S= 0.00237 ft/ftDesign Flow Q= 500.00 cfs Design Flow Q= 500.00 cfs
Normal Flow ConditionNormal Flow Depth Yn= 4.27 feet Normal Flow Depth Yn= 3.68 feetWetted Periemeter P= 32.02 ft/ft Wetted Periemeter P= 28.27 ft/ftNormal Flow Area An= 76.12 sq ft Normal Flow Area An= 59.03 sq ftHydraulic Radius R= 2.38 feet Hydraulic Radius R= 2.09 feetFroude Number Fr= 0.73 Froude Number Fr= 1.01Difference In Q dQ= 0.00 ft/ft Difference In Q dQ= 0.00 ft/ft
Critical Flow Condition:Critical Depth Yc= 3.68 feetCritical Top Width Tc= 27.06 feetCritical Flow Area Ac= 58.93 sq ftFroude Number Fr= 1.01Froude Number Fr 1.01
0.13
2
==c
cr gA
TQF
cccccc
c SPAnKAS
PA
nKUAQ 3
2353
2−
=⎟⎟⎠
⎞⎜⎜⎝
⎛==
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Specific Force 1. Illustration -- Jump (ΔF=0)2. For a given Q, the curve for Y vs Fs3. Flow regimes4. Critical Flow
Yo
4. Critical Flow 5. Increase or decrease of Q
Dr. James C.Y. Guo, Professor and P.E., Civil Engineering, U. of Colorado Denver
Example for Specific Force Curve
SPECIFIC FORCE CURVE FOR TRAPEZOIDAL CHANNEL
Bottom Width B= 10.00 feetLeft Side Slope Z1= 3.00 ft/ftRight Side Slope Z2= 3.00 ft/ft
Design Flow Q= 100 cfs
Yo
Design Flow Q= 100 cfsBeginning Flow Depth Y 0.25 ftIncremental Depth Dy= 0.25 ft
Flow Flow Top Central Flow Sp. Froude SPDepth Area Width Depth Velocity Force Numer Energy
Y A T Yo u Fs Fr Esft sq ft ft ft fps Klbs ft
0.25 2.69 11.50 0.12 37.21 7.24 13.56 21.750.50 5.75 13.00 0.24 17.39 3.46 4.61 5.200.75 9.19 14.50 0.35 10.88 2.31 2.41 2.591.00 13.00 16.00 0.46 7.69 1.87 1.50 1.921.25 17.19 17.50 0.57 5.82 1.74 1.03 1.781.50 21.75 19.00 0.67 4.60 1.80 0.76 1.831.75 26.69 20.50 0.77 3.75 2.01 0.58 1.972 00 32 00 22 00 0 87 3 13 2 35 0 46 2 15
Specific Force
1.50
2.00
2.50
3.00
3.50
4.00
Flow
Dep
th in
ft
2.00 32.00 22.00 0.87 3.13 2.35 0.46 2.152.25 37.69 23.50 0.97 2.65 2.80 0.37 2.362.50 43.75 25.00 1.07 2.29 3.36 0.30 2.582.75 50.19 26.50 1.16 1.99 4.03 0.26 2.813.00 57.00 28.00 1.26 1.75 4.82 0.22 3.053.25 64.19 29.50 1.35 1.56 5.72 0.19 3.293.50 71.75 31.00 1.45 1.39 6.74 0.16 3.53
0.00
0.50
1.00
0.00 2.00 4.00 6.00 8.00
Fs in Klb
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Application of Force Eq---Jump
Flow Flow Wetted Hy- Flow Flow Froude Sp Depth to SpDepth Area P-meter Radius Velocity rate Numer Energy Centroid Force
Y A P R U Q Fr Es Yh Fsft sq ft ft ft fps cfs ft ft klb
2.50 31.3 17.1 1.83 35.59 1112.1 4.35 22.11 1.17 79.06
Flow Flow Top Central Flow Sp. Froude SPDepth Area Width Depth Velocity Force Numer Energy
Y A T Yo u Fs Fr EsY A T Yo u Fs Fr Esft sq ft ft ft fps Klbs ft
2.50 31.25 15.00 1.17 35.59 79.05 4.34 22.175.00 75.00 20.00 2.22 14.83 42.37 1.35 8.41
10.00 200.00 30.00 4.15 5.56 63.79 0.38 10.4811.27 239.71 32.54 4.62 4.64 79.11 0.30 11.6012.00 264.00 34.00 4.89 4.21 89.60 0.27 12.28
Dr. James C.Y. Guo, Professor and P.E., Civil Engineering, U. of Colorado Denver
Hydraulic Jump: from supercritical to subcritical flow
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PUMP DESIGN
Pump
Pump Design by Dr. James Guo, UC-Denver
Pump = mechanic energy converted to fluid energyOutput energy = efficiency x Input energy
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Pump Head and Units fp HHhH +=+ 21
A pump is identified by its rotating speed, power, and efficiency.
Horse Power η
γ550
pQhHp = for ft-second units where Hp in horse power = 550 lb-ft,
Kilowatts η
γ1000
pQhKw = for meter-second units. One Horse Power = 0.746 Kw
Pump Net Positive Suction Head
PumptoLossesHdSuctionHeaH i +=+Fi d S ti H d PumptoLossesHdSuctionHeaH pumpin ++
gVZ
PH pump
pumpvapor
pump 2
2++=
γ
inatm
in ZP
H +=γ
Find Suction Head.
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Pump Net Required Discharge Head
PumpFromLossesHeHeadDischH +=+ argFi d Di h H d PumpFromLossesHeHeadDischH outpump +=+ arg
gVZ
PH pump
pumpvapor
pump 2
2++=
γ
Outatm
Out ZP
H +=γ
Find Discharge Head.
Design Pump Head and Horsepower
Find Pump Head
sOutflowLosInflowLossHhH outpin ++=+
Find the pump head with Q=2 cfsIs the pump head equal to suction head + discharge head?Why do we have to calculate suction and discharge heads?
d u p ead
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Affinity Law
Pump Model and Prototype
1. Basic DataModel Pump Diameter D1 1.00 feet
Pump Rotational Speed n1 1800.00 rpmProto-type Pump Diameter D2 2.00 feet
P R t ti l S d 2 1200 00
Performance Curve (Model Pump)
4045
Pump Rotational Speed n2 1200.00 rpm
2. Pump Affiniity and Similarity Laws: Proto-type to Model RatiosRotational speed ratio n*=n2/n1 0.67 Diameter ratio D*=D2/D1 2.00Head ratio h*=h2/h1 1.78Flow ratio Q*=Q2/Q1 5.33 Power ratio P*=P2/P1 9.48
3. Pump Performance Curve Enter model pump curve:
Flowrate cfs 0.10 0.50 0.75 1.00 1.50 2.00 2.50
05
101520253035
0
0 1 2 3Flow rate in cfs
Pum
p he
ad in
feet
Pump Design by Dr. James Guo, UC-Denver
Pumphead feet 40.00 39.00 37.00 33.00 22.00 10.00 2.00
Calculate Proto-type pump curveFlowrate cfs 0.53 2.67 4.00 5.33 8.00 10.67 13.33Pumphead feet 71.11 69.33 65.78 58.67 39.11 17.78 3.56
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Pump in series or in parallel
Calculate Proto-type pump curveFlowrate cfs 0.53 2.67 4.00 5.33 8.00 10.67 13.33Pumphead feet 71.11 69.33 65.78 58.67 39.11 17.78 3.56
3. Pump Performance Curve for multiple pumps Flowrate Number of Pump in Series Pumphead Number of Pump in Parallel
Q 1 2 3 4 Hp 1 2 3 4cfs Pump head in feet, Hp feet Flow rate in cfs, Q
0.53 71.11 142.22 213.33 284.44 71.11 0.53 1.07 1.60 2.132.67 69.33 138.67 208.00 277.33 69.33 2.67 5.33 8.00 10.674.00 65.78 131.56 197.33 263.11 65.78 4.00 8.00 12.00 16.005.33 58.67 117.33 176.00 234.67 58.67 5.33 10.67 16.00 21.338.00 39.11 78.22 117.33 156.44 39.11 8.00 16.00 24.00 32.00
10.67 17.78 35.56 53.33 71.11 17.78 10.67 21.33 32.00 42.6713.33 3.56 7.11 10.67 14.22 3.56 13.33 26.67 40.00 53.33
A and Q
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