007c (PPT) Pitot tube, Notches & Weirs.pdf

007c (PPT) Pitot tube, Notches & Weirs.pdf

1. Flow measurement
Session 3
Pitot tube
&
Notches/Weirs
Dr. Vijay G. S.
Professor
Dept. of Mech. & Mfg. Engg.
MIT, Manipal
email: [email protected]
Mob.: [phone]
1

2. 2
Pitot tube
•ThePitottube(namedaftertheFrenchscientistPitot)isoneofthesimplestandmost
usefulinstrumentseverdevisedforflowmeasurement.
•Itisgenerallyusedtomeasurethelocalvelocityofflowinpipesoropenchannels.
•ThePitottubeisasmalltubebentatrightangles
andisplacedintheflowsuchthatlowerend
(calledtheprobe),isbentthrough90
o
andis
directedintheupstreamdirection.
•Theprobeismaintainedsuchthattheaxisofits
hole(probe)isparalleltothevelocityofthefluid
flowing.
•Theliquidrisesinthetubebyadistancehdueto
conversionofkineticenergyintopotentialenergy.
•Thevelocitycanbedeterminedbymeasuringthe
riseofliquidinthetube.
At point (1), Static pressure, p
1= 
ogz
At point (2), Stagnation pressure, p
2= 
og(z + h)
Free surface of liquid
Pitot
tube
Flow

o
Pitot tube used in an open channel
Tube diashould not
influence capillarity
Probe

3. 3
Pitot tube for a closed conduit (pipe)
Pitot tube used in a closed conduit (a pipe)
•Inaclosedconduit(likeapipe),thereissomeinternalpressure,whichresultsinthe
staticpressureheadzatpoint(1),asindicatedbythepiezometerat(1).
•Duetostagnationoffluidatpoint(2),thestagnationpressureheadindicatedbythepitot
tubewillbez+h.
•ThustheDynamicpressurehead,
h=(Stagnationpressurehead–
Staticpressurehead)

4. 4
V
2= 0
•Thestreamlinealongtheprobeaxiscomestoastopat(2).i.e.,theflowstagnatesat(2).
•Thispointisknownasthe“Stagnationpoint”wherethefluidvelocityiszero,i.e.,V
2=0.
•Sincetheflowissuddenlybroughttorest,thepressureatthatpointrisessharplydueto
theconversionofthekineticheadintopressurehead.
•Thepressurep
2at(2)isgreaterthanp
1at(1)andisknownasthe“Stagnationpressure”.
Pitot tube –contd…
V
1= ?

5. 5
Pitot tube –Stagnation pressure
•TheStagnationpressureatapointinthefluidflowisdefinedasthetotalpressurewhich
wouldresultifthefluidwassuddenlybroughttorestisentropically.
•TherelationshipbetweentheStagnationpressure(p
2)andtheStaticpressure(p
1)canbe
deducedbyapplyingtheBernoulli’sequationatpoints(1)and(2).
•Let p
1, V
1and p
2, V
2be the pressures and velocities at points (1) and (2) respectively.22
1 1 2 2
12
22
    
oo
p V p V
zz
g g g g
•Atthestagnationpoint(2),thevelocityV
2=0.
•Alsoz
1=z
22
1 1 2
2

oo
p V p
g g g 2
1
21
2


o
V
pp

•Stagnation pressure (p
2
) = Static pressure (p
1
) + Dynamic pressure2
1
2



o
V

o

6. 6
Pitot tube –Stagnation pressure –contd…
•Stagnation pressure (p
2
) = Static pressure (p
1
) + Dynamic pressure2
1
2



o
V
•The stagnation pressure (p
2) is more than the static pressure (p
1) by an amount equal to
the dynamic pressure2
1
2



o
V
The dynamic pressure head is2
1
2
1
2
2
(2)







o
o
V
hg
V
g
Kineticheadlostat
(head = p/g)

7. 7
Pitot tube –Finding flow velocity2
1
21
2


o
V
pp
If the stagnation pressure p
2and the static pressure
p
1are known, then the velocity of the flow V
1can
be determined.2
1
21
2


o
V
pp  
212
1
2



o
pp
V  
 
21 21
1 2 1
2
2 2 2
 
 
    
o o o
gpp pp
V g ghhgh
g gg 
1
2
th
V gh
V
1
Static
pressure
head (h
1)
Stagnation
pressure
head (h
2)

8. (V
1)
act<(V
1)
th(V
1)
act= C
v(V
1)
th
where, C
vis known as “Coefficient of velocity”
(V
1)
act= C
v(V
1)
th
This equation gives the actual local velocity of flow measured by a pitot tube.
C
v0.98
8
Pitot tube –Finding flow velocity –contd…
v
Actualvelocity
C
Theoreticalvelocity 
1
2
vact
VCgh 
1
2
th
V gh

9. Pitot-Prandtlprobe OR Pitot-Static probe
Probe opening for sensing
stagnation pressure p
2
Connections
to differential
manometer
9
Holes for
sensing static
pressure p
1


V
max
V
1

10. 10
Central
velocity V
1
Pitot-Prandtlprobe OR Pitot-Static probe –contd…
Importance of positioning the pitot-static probe (for a pipe flow)

11. 11
Pitot-Prandtlprobe OR Pitot-Static probe –contd…11
mm
oo
S
hx x
S




 11
mm
oo
S
hx x
S





If 
m> 
oOR S
m> S
othen
an upright differential
manometer is used.
If 
m< 
oOR S
m< S
othen
an inverted differential
manometer is used.

m

m

o

o
Computing the dynamic pressure head from manometer reading

12. 12

13. PROBLEMS ON PITOT TUBE
13

14. Problem1:Findthevelocityoftheflowofanoilthroughapipe,whenthe
differenceofmercurylevelinadifferentialU-tubemanometerconnectedtothe
twotappingsofthepitot-tubeis100mm.Takeco-efficientofpitot-tube=0.98
andspecificgravityofoil=0.8.

o= 800 kg/m
3
(Flowing oil)

m= 13600 kg/m
3
(Manometricfluid)
x = 100 mm = 0.1 m of mercury
C
v= 0.982
0.9829.811.6
5.49/



act v
VCgh
ms
141
13600
0.1 11.6
800






 


m
o
hx
h mofoil

15. Problem2:Apitot-statictubeisusedtomeasurethevelocityofwaterinapipe.The
stagnationpressureheadis6mofwaterandstaticpressureheadis5mofwater.
Calculatethevelocityofflowassumingthatthecoefficientofpitottubeisequalto0.98.
15
Stagnation pressure head, h
2= 6 m of water (Flowing water)
Static pressure head, h
1= 5 m of water (Flowing water)
C
v= 0.98
Dynamic pressure head, h= h
2–h
1= 6 –5 = 1 m of water2
0.9829.811
4.34/



act v
VCgh
ms

16. Problem3:Asub-marinemoveshorizontallyinseaandhasitsaxis15mbelow
thefreesurfaceofwater.Apitot-tubeproperlyplacedjustinfrontofthesub-
marineandalongitsaxisisconnectedtothetwolimbsofaU-tubecontaining
mercury.Thedifferenceofmercurylevelisfoundtobe170mm.Findthespeed
ofthesub-marineknowingthatthespecificgravityofmercuryis13.6andthatof
seawateris1.026withrespecttothatoffreshwater.TakeC
v=0.98.
16
h
1= 15 m (Static pressure head)

o= 1026 kg/m
3
(Flowing sea water)

m= 13600 kg/m
3
(Manometricfluid)
x = 170 mm = 0.17 m of mercury
C
v= 0.981
13600
0.17 12.083
1026






 


m
o
hx
h mofseawater 2
0.9829.812.083
6.265/



act v
VCgh
ms
V
act= 6.265 ×60 ×60/1000
= 22.55 km/hr

17. Problem4:Apitot-statictubeplacedinthecenterofa300mmpipelinehasone
orificepointingupstreamandotherperpendiculartoit.Themeanvelocityinthe
pipeis80%ofthecentralvelocity.Findthedischargethroughthepipeifthe
pressuredifferencebetweenthetwoorificesis60mmofwater.Takethe
coefficientofpitottubeas0.98.
17
d
1= 300 mm = 0.3 m (pipe dia) a
1= (d
1
2
/4) = 0.0707 m
2
V
avg= 0.8×V
1(Mean velocity, V
1= Central velocity)

o= 1000 kg/m
3
(Flowing water)
h = 60 mm = 0.06 m of water (Dynamic pressure head)
C
v= 0.98
Q= ?
1
2
0.9829.810.06
1.063/



vact
VCgh
ms

18. 18
Problem 4: contd…
V
avg= 0.8×V
1
= 0.8×1.063
= 0.851 m/s
Discharge through the pipe,
Q = a
1×V
avg= 0.0707×0.851 = 0.06 m
3
/s
Q =60 lps
Central
velocity V
1

19. Problem5:Apitotstaticprobeisinsertedinapipeof300mmdiameter.The
staticpressureinthepipeis100mmofmercury(vacuum).Thestagnation
pressureatthecenterofthepipe,recordedbypitot-tubeis0.981N/cm
2
.
Calculatetherateofflowofwaterthroughpipe,ifthemeanvelocityofflowis
0.85timesthecentralvelocity.TakeC
v=0.98.
19
d
1= 300 mm = 0.3 m (pipe dia) a
1= (d
1
2
/4) = 0.0707 m
2
V
avg= 0.85×V
1(Mean velocity, V
1= Central velocity)

o= 1000 kg/m
3
(Flowing water)
h
1= 100 mm (vacuum) = -100 mm = -0.1 m of Hg (Static pressure head)
p
2= 0.981 N/cm
2
= 0.981×10
4
N/m
2
(Stagnation pressure)
C
v= 0.98
Q= ?
p
1= 
Hg×g ×h
1= 13600×9.81×-0.1 = -13341.6 N/m
2

20. 20  
21
4
098110133416
1000981
236
..
.
.








o
pp
h
g
mofwater 
1
2
0.9829.812.36
6.668/



vact
VCgh
ms
V
avg= 0.85×V
1
= 0.85×6.668
= 5.668 m/s
Discharge through the pipe,
Q = a
1×V
avg= 0.0707×5.668 = 0.4 m
3
/s
Q =400 lps
Problem 5: contd…

21. Notches/Weirs
Measurement of discharge through open channels
21

22. 22
Notches/Weirs
Rectangular weir/notch
Triangular weir/notch
ANotchoraWeirisusedformeasuringtherateofflowofliquidthrough
openchannelsorrivers.

23. Notches/Weirs –contd…
23
Notch Weir
•Itisastructurewhichobstructstheflow
fromatankorreservoir
•Itisastructurewhichobstructstheflow
inanopenchannelorriver
•Itisanopeningprovidedononesideof
thetankorreservoir,withfreesurfaceof
liquidpresentbelowthetopedgeofthe
opening
•Itisanopeningprovidedonthe
obstructingstructure,suchthatflow
occursonthetopedgeoftheweir
•Notchisnormallymadeofametallic
plate
•Weirismadeofconcreteormasonry
•Comparativelysmallerinsize •Comparativelylargerinsize
•Itisusedtomeasuredischargeinsmall
channelsorinlaboratoryreservoirs.
•Itisusedtomeasuredischargeinlarger
bodieslikeriversandlargechannels.

24. 24
Rate of flow through a rectangular notch/weir
Crest or Sill
Top edge of notch

25. 25
Rate of flow through a rectangular notch/weir –contd…
H=Headofwateroverthecrest
L= Length of the notch or weir
•Consider a horizontal strip of water
of thickness dhand length Lat a
depth hbelow the free surface of
water
•The area of strip = Ldh
•Theoretical velocity of water flowing through strip = 2gh
•ThedischargedQ,throughstripis
dQ=C
dAreaofstripTheoreticalvelocity2
d
dQ C L dh gh   
where C
d= Co-efficient of discharge

26. 26
Rate of flow through a rectangular notch/weir –contd…
•The total discharge, Q, for the whole notch or weir is determined by integrating the above
equation between the limits 0 and H.00
2
HH
d
Q dQ C L dh gh     
0
3/2
0
3/2
2
2
3/2
2
2
3
 






H
d
H
d
d
QCLghdh
h
CLg
CLgH 
3/22
2
3
d
QCLgH

27. 27
Rate of flow through a triangular notch/weir
Apex
Top edge of notch
•LetH=headofwaterabovetheapexof
theV-notchortriangularweir.
•=angleofnotch
•Considerahorizontalstripofwaterof
thicknessdhatadepthofhfromthefree
surfaceofwater

28. 28
Rate of flow througha triangular notch/weir –contd…tan
2
AC AC
OC H h



  tan
2
AC H h

  


Width of the strip AB =  2 2 tan
2
AC H h




Area of the strip =  2 tan
2
H h dh




Theoretical velocity of water flowing through strip = 2gh
ThedischargedQ,throughstripis
dQ=C
dAreaofstripTheoreticalvelocity 2 tan 2
2
d
dQ C H h dh gh

    


where C
d= Co-efficient of discharge

29. 29
Rate of flow througha triangular notch/weir –contd…
The total discharge, Q, for the whole notch or weir is determined by integrating the above
equation between the limits 0 and H. 
00
2 tan 2
2
HH
d
Q dQ C H h dh gh

     


    
1/2 3/2
00
22tan 22tan
22
 
   
 
 

HH
dd
QCg HhhdhCg Hhhdh 3/2 5/2 5/2 5/2
0
22
2 2 tan 2 2 tan
2 3/ 2 5/ 2 2 3 5
H
dd
Hh h H H
Q C g C g
      
   
      
       5/28
2tan
15 2
d
QCg H

  

 5/ 28 90
0.6 2 9.81 tan
15 2
o
QH

      

For a right angled notch (weir) (= 90
o
) and C
d= 0.6,5/2
1.417QH

30. PROBLEMS ON NOTCHES/WEIRS
30

31. Problem 1: Find the discharge of water flowing over a rectangular notch of 2 m
length when the constant head over the notch is 300 mm. Take C
d= 0.60.
31
L= 2 m (length of notch)
H= 300 mm = 0.3 m (head over the notch)
C
d= 0.60

3/2
3/2
3
2
2
3
2
0.6229.810.3
3
0.582/



d
QCLgH
ms
Q = 582lps

32. Problem2:Theheadofwateroverarectangularnotchis900mm.The
dischargeis300liters/s.Findthelengthofthenotch,whenC
dis0.62.
32
H= 900 mm = 0.9 m (head over the notch)
Q = 300lps= 0.3 m
3
/s
C
d= 0.62
L= ? (length of notch)

3/2
3/2
2
2
3
2
0.30.6229.810.9
3
0.192



d
QCLgH
L
Lm

33. Problem3:Determinetheheightofarectangularweiroflength6mtobebuilt
acrossarectangularchannel.Themaximumdepthofwaterontheupstreamside
oftheweiris1.8manddischargeis2000litres/s.TakeC
d=0.6andneglectend
contractions.
33
L = 6 m
H
1= 1.8 m (depth of water on upstream)
Q = 2000lps= 2 m
3
/s
C
d= 0.6
H
2= ? (Height of weir)
Let H
2= height of weir
H
2= H
1–H
H
2= 1.8 –0.328
= 1.472 m

3/2
3/2
2
2
3
2
20.6629.81
3
0.328



d
QCLgH
H
Hm

34. Problem4:Findthedischargeoveratriangularnotchofangle60
o
whenthe
headovertheV-notchis0.3m.AssumeC
d=0.6.
34
= 60
o
(V notch angle)
H= 0.3 m (head of water above apex)
C
d= 0.6
Q= ? (Flow rate)
Q = 40 lps

35. Problem5:Waterflowsoverarectangularnotch1mwideatadepthof150
mmandafterwardspassesthroughatriangularright-anglednotch.TakingC
dfor
therectangularandtriangularweiras0.62and0.59respectively,findthedepth
overthetriangularweir.
35
H
R= 150 mm = 0.15 m (head over the rectangular notch)
L= 1 m (length of rectangular notch)
C
d= 0.62 (rectangular notch)
C
d= 0.59 (triangular notch)
= 90
o
(V notch angle)
H
T= ? (head over the triangular notch)
H
T
H
R
90
o
Q
Q
Rectangular= Q
Triangular
3/2
5/22 8 90
2 2tan
3 15 2
o
dR R dT T
CLgH Cg H

   


36. Problem 5: Contd…
362
3
2
dR
CLg 
3/28
R
H 
4
15
2
5
dT
Cg
5/290
tan
2
o
T
H


 

3/2
5/2
5
4tan45
dR R
T o
dT
CLH
H
C


 
 
3/2
5/2
2/5
50.6210.15
40.59tan45
0.0763
0.07630.3573
o
T
T
H
Hm




