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FLOW THROUGH PIPESDefinition of flow through pipesA pipe is a closed conduit carrying a fluid under pressure. Fluid motion in a pipe
is subjected to a certain resistance. Such a resistance is assumed to be due to
Friction. In reality this is mainly due to the viscous property of the fluid.
Reynolds Number (Re)
It is defined as the ratio of Inertia force of a flowing fluid and the Viscous force.
Re=(Inertia force/Viscous force) =( V D/ )Classification of pipe flow:
Based on the values of Reynolds number (Re), flow is classified as Follows:
Laminar flow or Viscous FlowIn such a flow the viscous forces are more predominent compared to inertia Forces.
Stream lines are practically parallel to each other or flow takes place In the form of
telescopic tubes. This type of flow occurs when Reynolds number Re< 2000. In laminarflow velocity increases gradually from zero at the boundary to Maximum at the center.
Laminar flow is regular and smooth and velocity at any point practically remains constant
in magnitude & direction. Therefore, the flow is also known as stream Line flow.There will be no exchange of fluid particles from one layer to another. Thus there
will be no momentum transmission from one layer to another. Ex: Flow of thick oil
in narrow tubes, flow of Ground Water, Flow of Blood in blood vessels.
Transition flow:
In such a type of flow the stream lines get disturbed a little. This type of flowoccurs when 2000< Re < 4000.
Laminar flow Transition flow Turbulent flow
Water
Dye
Glass tube
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Hydraulic Grade Line & Energy Grade Line
A Line joining the peizometric heads at various points in a flow is known as Hydraulic
Grade Line (HGL)Energy Grade Line (EGL)
It is a line joining the elevation of total energy of a flow measured above a datum, i.e.
EGL Line lies above HGL by an amount V2/2g.
Losses in Pipe Flow
Losses in pipe flow can be two types viz:-a)Major Loss
b)Minor Loss
a)Major Loss: As the name itself indicates, this is the largest of the losses in a pipe. Thisloss occurs due to friction only. Hence, it is known as head loss due to friction (hf)
b)Minor Loss: Minor losses in a pipe occurs due to change in magnitude or direction offlow.
Minor losses are classified as (i) Entry Loss, (ii) Exit loss, (iii) Sudden expansion
loss (iv) Sudden contraction loss (v) Losses due to bends & pipe fittings.
Head Loss due to Friction
Consider the flow through a straight horizontal pipe of diameter D, Length L, between two sections (1) & (2)as shown. Let P1 & P2 be the pressures at these sections. To is the shear stress acting along the pipe
boundary.
(1)
(1) (2)
(2)
L
Dp1p2
Flow (V)
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From II Law of NewtonForce = Mass x accn. But acceleration = 0, as there is no change in velocity, however the reasonthat pipe diameter is uniform or same throughout.
Applying Bernoullis equation between (1) & (2) with the centre line of the pipe asdatum & considering head loss due to friction hf,.
Substituting eq (2) in eq.(1)
From Experiments, Darcy Found that
( )
( ) )1(4
4
44..
0
021
0
2
21
0
2
2
2
1
=
=
+
=
D
LPPor
DLD
PP
DLxD
PD
Pei
forces
fhg
VpZ
g
VpZ +++=++
22
2
222
2
111
21 ZZ =
21 VV =
Pipe is horizontal
Pipe diameter is
same throughout
)2(21 =
fhPP
)4(8
2
0 = Vf
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f=Darcys friction factor (property of the pipe materials Mass density of the liquid.V = velocityEquations (3) & (4)
But,
from Continuity equation
& (5) & (6) are known as DARCY WEISBACH Equation
Pipes in Series or Compound PipeD1, D2, D3, D4 are diameters.
L1, L2,L3, L4 are lengths of a number of Pipes connected in series
(hf)1, (hf)2, (hf)3 & (hf)4 are the head loss due to friction for each pipe.
The total head loss due to friction hffor the entire pipe system is given by
orD
VLfhf
84 2=
)5(2
2
=
=
gD
fLVh
g
f
)6(8
52
2
=
Dgh
fLQhf
D2
D3
D4D1Q
L2 L4L1
4321 hfhfhfhfhf +++=
5
3
2
2
3
5
2
2
2
2
5
1
2
2
1 888
Dg
QfL
Dg
QfL
Dg
QfLhf
+++=
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D1, D2 and D3 are the pipe diameters. Length of each pipe is same, that is, L1=L2=L3For pipes in parallel hf1=hf2=hf3 i.e
Equivalent pipe
In practice adopting pipes in series may not be feasible due to the fact that they may be of
unistandard size (ie. May not be comemercially available) and they experience otherminor losses. Hence, the entire system will be replaced by a single pipe of uniform
diameter D, but of the same length L=L1+ L2+ L3 such that the head loss due to friction
for both the pipes, viz equivalent pipe & the compound pipe are the same.For a compound pipe or pipes in series.
for an equivalent pipe
D1
D2
D3
Q Q
Q1
Q2
Q3
L = L1 = L2 = L3
Pipes in Parallel
D2D1 D3 == Q D
L= L1+L2+L3L
1L
3
Q
321 hfhfhfhf ++=
)1(888
5
3
2
2
3
5
2
2
2
2
5
1
2
2
1 ++=Dg
QfL
Dg
QfL
Dg
QfLhf
)2(8
5
1
2
2
=Dg
fLQhf
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Equating (1) & (2) and simplifying
Or
5
3
35
2
25
1
15 D
LDL
DL
DL ++=
5
1
5
3
3
5
2
2
5
1
1
++=
D
L
D
L
D
L
LD
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Problems
1) Find the diameter of a Galvanized iron pipe required to carry a flow of 40lps of water, if the loss of head
is not to exceed 5m per 1km. Length of pipe, Assume f=0.02.
Solution:-
D=?, Q=40lps = 40x10-3 m3/shf=5m, L=1km = 1000m. f=0.02
Darcys equation is
2) Two tanks are connected by a 500mm diameter 2500mm long pipe. Find the rate of
flow if the difference in water levels between the tanks is 20m. Take f=0.016. Neglect
minor losses.Solution:-
Applying Bernoullis equation between (1) & (2) with (2) as datum & considering head
loss due to friction hfonly,
Z1 = 20m, Z2 = 0 (Datum); V1=V2 = 0 (tanks are very large)
p1=p2=0 (atmospheric pressure)
Therefore From (1)
20+0+0=0+0+0+hf
Or, hf= 20m. But
mmmD 22022.0 =
5
28
Dg
fLQhf
=
2
152
2500016.08
5.081.920
=xx
xxxQ
lpsmQ 8.434sec/4348.0 3 ==
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3) Water is supplied to a town of 0.5million inhabitants from a reservoir 25km away and
the loss of head due to friction in the pipe line is measured as 25m. Calculate the size of
the supply main, if each inhabitant uses 200 litres of water per day and 65% of the dailysupply is pumped in 8 hours. Take f=0.0195.
Solution:-
Number of inhabitants = 5million = 5,00,000Length of pipe = 25km = 25,000m.
Hf= 25m, D=?
Per capita daily demand = 200litres.
Total daily demand = 5,00,000x200= 100x106 litres.
Daily supply = 65/100 x 100x106 = 65,000m3.
Supply rate
4) An existing pipe line 800m long consists of four sizes namely, 30cm for 175m, 25cm
dia for the next 200m, 20cm dia for the next 250m and 15cm for the remaining length.
Neglecting minor losses, find the diameter of the uniform pipe of 800m. Length toreplace the compound pipe.
Solution:-
L=800mL1=175m D1=0.3m
L2=200m D2=0.25m
L3=250m D3=0.20m
L4=175m D4=0.15m
For an equivalent pipe
D = Diameter of equivalent pipe = 0.189m less than or equal to 19cm.
=52
28
Dg
fLQhf
5
1
2
2
2581.9
)1248.2(000,25195.08
=xx
xxxD
mD 487.1
=
+++=5
4
4
5
3
3
5
2
2
5
1
1
5 D
L
D
L
D
L
D
L
D
L
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5) Two reservoirs are connected by four pipes laid in parallel, their respective diametersbeing d, 1.5d, 2.5d and 3.4d respectively. They are all of same length L & have the same
friction factors f. Find the discharge through the larger pipes, if the smallest one carries
45lps.Solution:-
D1=d, D2 =1.5d, D3=2.5d, D4=3.4d
L1=L2=L3=L4= L.
f1=f2=f3=f4=f.
Q1=45x10-3m3/sec, Q2=? Q3=? Q4=?For pipes in parallel hf1=hf2=hf3=hf4 ,i.e.
5
1
5555 15.0
175
2.0
250
25.0
200
3.0
175
800
+++
=D
5
4
2
4
5
3
2
3
5
2
2
2
5
1
2
1
D
Q
D
Q
D
Q
D
Q===
( ) sec/124.010455.1 32
1
23
5
2 mxxd
dQ =
=
( ) sec/4446.010455.2 32
1
23
5
2 mxxd
dQ =
=
( ) sec/9592.010454.3 32
1
23
5
2 mxxd
dQ =
=
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6) Two pipe lines of same length but with different diameters 50cm and 75cm are made
to carry the same quantity of flow at the same Reynolds number. What is the ratio of
head loss due to friction in the two pipes?Solution:-
D1=0.5m, D2 =0.75m
L1=L2Q1=Q2
(Re)1 = (Re)2,
Reynolds number Re=
7) A 30cm diameter main is required for a town water supply. As pipes over 27.5cmdiameter are not readily available, it was decided to lay two parallel pipes of same
diameter. Find the diameter of the parallel pipes which will have the combined discharge
equal to the single pipe. Adopt same friction factor for all the pipes.Solution:-
222 VD
2
222
1
111
VDVD=
2211 DVDV =
21 75.05.0 VV =
( )21 = ( )21 =
21 5.1 VV =
gD
fLVhf
2
2
=
2
2
2
1
1
2
2
1
V
Vx
D
D
hf
hf=
375.35.1
5.0
75.02
2
2 =
=
V
Vx
From Darcys equation
)1(8
52
2
=Dg
fLQhf
)2(2
82
52
=Dg
QfL
hf
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Equating
8) Two reservoirs are connected by two parallel pipes. Their diameter are 300mm &350mm and lengths are 3.15km and 3.5km respectively of the respective values of
coefficient of friction are 0.0216 and 0.0325. What will be the discharge from the larger
pipe, if the smaller one carries 285lps?
Solution:-
D1=300mm=0.3m, D2=-.350m
L1=3150m L2=3500m
F1=0.0216 f 2=0.0325
Q1=0.285m3/sec Q2=?
For parallel pipes
5
15
4
275.0
=D
mmD 275.0205.0 =
or
=
=5
2
2
2
222
5
1
2
2
111 88
Dg
QLf
Dg
QLfhf
2
1
5
122
5
2
2
1112
=DLf
DQLfQ
21
5
52
23.035000325.0
35.0285.031500216.0
=xx
xxxQ
sec/324.0 32 mQ =
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9) Consider two pipes of same lengths and having same roughness coefficient, but with
the diameter of one pipe being twice the other. Determine (I) the ratio of discharges
through these pipes, if the head loss due to friction for both the pipes is the same. (ii) theratio of the head loss due to friction, when both the pipes carry the same discharge.
Solution:-
f1=f2 D1=2D2 L1=L2
(i)Given hf1=hf2 Q1/Q2=?
From Darcys equation
(ii) Given Q1/Q2, hf1/hf2=?
10) Two sharp ended pipes are 50mm & 105mm diameters and 200m length are
connected in parallel between two reservoirs which have a water level difference of 15m.If the coefficient of friction for each pipes of 0.0215. Calculate the rate of flow in each
pipe and also diameter of a single pipe 200m long which would give the same discharge,if it were substituted for the Original two pipes.Solution
D1=0.015m, D2=0.105m, L1=L2=200m
H=15m, f1=f2=0.0215,
a) Q1=?, Q2=?
(b) D=?, when Q=Q1+Q2
a) For parallel pipes
656.52 2
5
2
22
5
2
1
2
1 =
=
=
D
D
D
D
Q
Q
03125.02
85
2
2
5
1
2
5
1
2
2
211
2
1 =
=
==
D
D
D
D
Dg
QLf
hf
hf
=
=5
2
2
2
222
5
1
2
2
111 88
Dg
QLf
Dg
QLfhf
sec/1063.32000215.08
05.081.915 3321
52
1 mxxx
xxhxQ =
=
sec/023.02000215.08
105.081.915 3221
52
1 mxx
xxhxQ =
= ( ) sec/02684.00232.01063.3 2321 mxQQQ =+=+= b
52
28
Dg
fLQhf
=
( ) 51
2
2
1581.902684.02000215.08
=
xxxxxD
cmmD 12.111112.0 ==
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11) Two pipes with diameters 2D and D are first connected in parallel and when a
discharge Q passes the head loss is H1, when the same pipes are Connected in series for
the same discharge the loss of head is H2. Find the relationship between H1 and H2.
Neglect minor losses. Both the pipes are of same length and have the same frictionfactors.
SolutionH1 = head loss due to friction = hf= hf2
i.e.
Case(iii)
12) Two reservoirs are connected by a 3km long 250mm diameter. The difference in
water levels being 10m. Calculate the discharge in lpm, if f=0.03. Also find the
percentage increase in discharge if for the last 600m a second Pipe of the same diameteris laid parallel to the first.
Solution
Applying Bernoullis equation between (1) & (2) with (2) as datum and
considering head loss due to friction hf
52
2
52
2
)2(88
DgfLQ
DgfLQhf
+
=
+=
552
2
22
1
1
18
Dg
fLQH
)4(80312.1
52
2
2 =Dg
xflQxH
2
52
52
2
2
1
80312.1
802256.0
flQx
Dgx
Dg
xflQx
H
H
=
021876.00312.1
02256.0
2
1 ==H
H71.45
1
2 =H
H
Or
fhg
VpZ
g
VpZ +++=++
22
2
122
2
111
mhh ff 100000010 =+++=++
52
28
Dg
fLQhf
=
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Case (ii)
Change in discharge =
( ) 21
52
300003.08
1025.081.9
=xx
xxxQ
sec/03624.0 3mQ =
321 orhfhfhfhf +=
( )
+=5
2
1
5
2
1
2 25.0
2/600
25.0
2400
81.9
03.0810
x
x
2
166.647210 Q=
sec/0393.0 31 mQ =
( )QQQ = 1
( )03624.00393.0 =
sec/10066.3 33 mxQ =
1001
xQ
Q% increase in discharge =
%46.810003624.0
10066.3 3
==
x
x
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MINOR LOSSES IN PIPES
Minor losses in a pipe flow can be either due to change in magnitude or direction of flow.
They can be due to one or more of the following reasons.i)Entry loss
ii)Exit loss
iii)Sudden expansion lossiv)Sudden contraction loss
v)Losses due to pipe bends and fittings
vi)Losses due to obstruction in pipe.
Equation for head loss due to sudden enlargement or expansion of a pipe
Consider the sudden expansion of flow between the two section (1) (1)& (2) (2) as
shown.P1 & P2 are the pressure acting at (1) (1) and (2) (2), while V1 and V2 are the velocities.
From experiments, it is proved that pressure P1 acts on the area (a2 a1) i.e. at the point
of sudden expansion.
From II Law of Newton Force = Mass x Acceleration.
Consider LHS of eq(1)
Consider RHS of eq(1)
Mass x acceleration = x vol x change in velocity /time
=volume/time x change in velocity
Substitution (ii) & (iii) in eq(i)
Both sides by (sp.weight)
Applying Bernoullis equation between (1) and (2) with the centre line of the pipe as
datum and considering head loss due to sudden expansion hLonly.
( ) )(21 iiiVVxQx
( ) ( )21212 VVpQppa =
( ) ( )21221 VVVpp = or
zontalpipeishoriCZZ 21 =
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In Eq(V) hL is expressed in meters similarly, power (P) lost due to sudden expansion is
Equations for other minor losses
Loss due to entrance and exit
Loss due to bends & fittings
Problems
g
VpZ
g
VpZ
22
2
222
2
111 ++=++
( ) ( )g
VVVVVhL
2
2 222
1212 +=
g
VVVVVhL
2
22 222
1
2
221 +=
g
VVVVVhL
2
22 222
121
2
2 +=
g
VVVVhL
2
2 212
1
2
2 +=
( )
g
VVhL
2
2
21 =
g
VhL
25.0
2
2=Sudden contraction loss
gVh
exitL2
2
=
g
KVhL
2
2
=
K=coefficient
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1) A 25cm diameter, 2km long horizontal pipe is connected to a water tank. The pipe
discharges freely into atmosphere on the downstream side. The head over the centre line
of the pipe is 32.5m, f=0.0185. Considering the discharge through the pipe
Applying Bernoullis equation between (A) and (B) with (B) as datum & considering all
losses.
2) The discharge through a pipe is 225lps. Find the loss of head when the pipe issuddenly enlarged from 150mm to 250mm diameter.
Solution :
D1=0.15m, D2 = 0.25m Q=225lps = 225m3/secHead loss due to sudden expansion is
3) The rate of flow of water through a horizontal pipe is 350lps. The diameter of the pipe
is suddenly enlarge from 200mm to 500mm. The pressure intensity in the smaller pipe is
15N/cm2. Determine (i) loss of head due to sudden enlargement. (ii) pressure intensity inthe larger pipe (iii) power lost due to enlargement.
SolutionQ=350lps=0.35m3/s
D1=0.2m, D2=0.5m, P1=15N/cm2
hL=?, p2=?, P=?
From continuity equation
( )
g
VVhL
2
12
=
gX
D
Q
D
Q
2
1442
2
2
1
=
2
2
2
2
1
2
2 11
2
16
=
DDg
Q
mhL 385.3=
2
222
2
25.0
1
15.0
1
81.92
225.016
=
xx
x
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Applying Bernoullis equation between (1) (1) and (2) (2) with the central line ofthe pipe as datum and considering head loss due to sudden expansion h L only.
4) At a sudden enlargement of an horizontal pipe from 100 to 150mm, diameter, the
hydraulic grade line raises by 8mm. Calculate the discharge through the pipe system.
Solution
Applying Bernoullis equation between (1) & (2) with the central line of the pipe asdatum and neglecting minor losses (hL) due to sudden expansion.
From continuity equation
Lhg
VpZ
g
VpZ +++=++
22
2
222
2
111
( )ntalpipehorizoZZ 021 ==
463.462.19
78.1
81.90
62.19
14.11
81.9
1500
2
2
2
+++=++p
22
2 /67.16/68.166 cmNmkNp ==
LQhP =
463.435.081.9 xx=
kWP 32.15=
( ))1(
2
2
21
=g
VVhL
)2(108,31
12
2 =
+
+ mx
pZ
pZGiven
Lhg
VpZ
g
VpZ +++=++
22
222
2
211
1
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Discharge
5) Two reservoirs are connected by a pipe line which is 125mm diameter for the first 10m
and 200mm in diameter for the remaining 25m. The entrance and exit are sharp and thechange of section is sudden. The water surface in the upper reservoir is 7.5m above that
in the lower reservoir. Determine the rate of flow, assuming f=0.001 for each of the
types.
Solution
From continuity equation
Applying Bernoullis equation between (1) & (2) in both the reservoirs with the water in
the lower reservoir as datum and considering all losses
2
2
1
2
4
15.0
4
1.0xV
xV
x =
21 25.2 VV =
smx
V /25.01274.0
108 21
3
2 =
=
25.04
15.0
4
2
2
2
2 xx
VD
Q
==
2
2
1
2
4
2.0
4
125.0V
xV
x =
21 56.2 VV =
{ }1434.2243.52768.362.19
5.72
2 +++=V
( ) ( ) ( )
+++=++ g
Vg
VVg
Vxxg
V22
56.22
56.21001.02
5.25.0}005.7
2
2
2
2
2
2
2
1
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FLOW MEASUREMENTS
Flow Through Orifices
An orifice is an opening of any cross section, at the bottom or on the side walls of a
container or vessel, through which the fluid is discharged. If the geometric characteristicsof the orifice plus the properties of the fluid are known, then the orifice can be used to
measure the flow rates.
Classification of orifices
Flow through an orifice
As the fluid passes through the orifice under a head H, the stream lines converge andtherefore the jet contracts. The stream lines which converge are mostly those from near
the walls and they do so because stream lines cannot make right angled bend in motion.
This phenomenon occurs just down stream of the orifice, and such a section where thearea of cross section of the jet is minimum is know as VENA CONTRACTA.
The pressure at Vena Contracta is assumed to be atmospheric and the velocity is assumed
to be the same across the section since the stream lines will be parallel and equallyspaced. Downstream of Vena contracta the jet expands and bends down. Figure shows
the details of free flow through a vertical orifice.
Applying Bernoulli's equation between (B) & (C) with the horizontal through BC as
datum and neglecting losses (hL)
( ) smV /6.416.21 21
2 ==
sec/1445.06.4
4
2.0 22
mxx
Q =
=
Based on shape circulartriangular rectangular
Based on sizeSmall orifice
(when the headover the orifice is
more than fivetimes its size I.e.
H>5d, Largeorifice
Based on shape ofthe u/s edgeSharp edgeBell mouth
Based onflow FreeSubmerged
VVV == 21 ,0
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Velocity V in Eq(1) is known as TORRICELLIS VELOCITY.
Hydraulic Coefficients of an orifice
i)Coefficient of discharge (Cd): It is defined as the ratio of actual discharge (Qact) to thetheoretical discharge (Qth)
Value of Cd varies in the range of 0.61 to 0.65
ii) Coefficient of Velocity (Cv): It is defined as the ratio of actual velocity (Vact) to the
theoretical velocity (Vth).
Value of Cv varies in the range of 0.95 to 0.99
Coefficient of Contraction (Cc): It is defined as the ratio of the area of cross section of
the jet at Vena of cross section of the jet at Vena Contracta (a c) to the area of the orifice
(a).
Value of Cc will be generally more than 0.62.
Relationship between the Hydraulic Coefficients of an orifice
From continuity equation
Actual discharge Qact = ac x Vact
Theoretical discharge Qth = a x Vth
Equation for energy loss through an orifice
Applying Bernoullis equation between the liquid surface (A) and the centre of jet andVena Contracta (C) and considering losses (hL).
=
th
actV
V
VC
=
a
aC cC
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Torricellis equation
Equation for Coefficient of Velocity (CV) (Trajectory method)
Consider a point P on the centre line of the jet, such that its horizontal and vertical
coordinates are x and y respectively.By definition, velocity
Since, the jet falls through a vertical distance y under the action of gravity during this
time (t)
Equating equations (1) & (2)
,HZA =
,0=AV)(0 cityactualvelopp BA ==
Lhg
VaH +++=++
20000
2
)2
(2
g
VaHhL =
gHCButV Va 2=
)(2
VL HxCHh =
)1( 2VL CHh =
t
xVa =
aV
xt=
Or
Or
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But
2
1
2
=
g
y
V
x
a
gHCV Va 2=
2
1
2
2
=
g
y
gHC
x
V
2
1
2
1
2
1
2
1
2
1
2
1
22 y
gx
Hg
xCV =
Hy
xCV
2=
=
yH
xCV
4
2
Or
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Problems
1. The head of water over the centre of an orifice 30mm diameter is 1.5m. If thecoefficient of discharge for the orifice is 0.613, Calculate the actual discharge.
Solution:
d=30mm = 3x10-3H=1.5m
Cd=0.613
2. Compensation water is to be discharge by two circular orifices under a constant head
of 1.0m, measured from the centre of the orifices. What diameter will be required to give
a discharge of 20x103 m3 per day? Assume Cd for each notch as 0.615.
Solution: d=? H=1m. Qtotal = 20x103 m3/day Cd=0.615.
we know
3. A jet of water issuing from an orifice 25mm diameter under a constant head of 1.5m
falls 0.915m vertically before it strikes the ground at a distance of 2.288m measured
;th
actd
Q
QC =
gHxaCd 2=
smxQact /1035.233=
lpsQact 35.2=
sm /1157.03
=
gHaCQ dact 2=
181.924
615.01157.02
xxxxdx =
mmmd 5.2322325.0 ==
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horizontally from the Vena Contracta. The discharge was found to be 102lpm. Determine
the hydraulics coefficients of the orifice and the head due to resistance.
Solution: d=25mm=25x10-3
H=1.5m, y=0.915m, x=2.288m
Qact=102lpm = 102/60 = 1.7lps = 1.7x10-3m3/sec, Cd=?, Cc=?, hL=?
4. The head of water over a 100mm diameter orifice is 5m. The water coming out of the
orifice is collected in a circular tank 2m diameter. The time taken to collect 45cm ofwater is measured as 30secs. Also the coordinates of the jet at a point from Vena Contract
are 100cm horizontal and 5.2cm vertical. Calculate the hydraulic coefficients of theorifice.Solution:
D=100mm=0.1m, H=5m
Qact = Area of collecting tankxheight of water collected / time
X=100cm = 1m, y=5.2cm = 0.052m
Cd=?, Cv=?, Cc=?
5. The coordinates of a point on the jet issuing from a vertical orifice are 0.4m & 0.003m.
Neglecting air resistance, determine the velocity of the jet and the height of water above
the orifice in the tank.Solution.
smxx
/0471.030
45.0
4
2 32
==
98.0
5052.04
1
4
22
=
=
=
xxyH
xCv
605.0581.921.0
40471.02
=
==xxx
x
Q
QC
th
actd
618.098.0
605.0===
V
dC
C
CC
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X=0.4m, y=0.3m, V=? H=?
Assume
We know
6. A vertical orifice is fitted 0.2m above the bottom of a tank containing water to a depthof 2m. If G=0.98. What is the vertical distance from the orifice of a point on the jet 0.6m
away from the Vena Contracta?
SolutionHead over the orifice H=(2-0.2)=1.8m
CV=0.98, y=?, x=0.6m
7. A closed tank contains water to a height of 2m above a sharp edged orifice 1.5cm
diameter, made in the bottom of the tank. If the discharge through the orifice is to be 4lps.
1=VC
yH
xCV
4
2
=
==
=
2
2
2
2
22
103.04
4.0
4
4
xxyxG
xH
xyHxG
H=1.33m
mmmxx
y
xyxor
yH
xCV
52052.098.08.14
6.0
8.14
6.0)98.0(,
4
2
2
22
2
==
=
=
=
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Workout the pressure at which air should be pumped into the tank above water. Take
Cd=0.6.
Solution
Q=4lps = 4x10-3m3/s
D=1.5x10-2m, Cd=0.6
PA=?
Total head over the orifice
8. A closed tank contains 3m depth of water and an air space at 15kpa pressure. A 5cm
diameter orifice at the bottom of the tank discharge water to the tank B containingpressurized air at 25kpa. If Cd = 0.61 for the orifice. Calculate the discharge of water
from tank A.Solution
d=5cm = 5x10-2m Cd=0.61.
Total head over the orifice
H=1.9806m
9. A tank has two identical orifices in one of its vertical sides. The upper orifice is 4m
below the water surface and the lower one 6m below the water surface. If the value of Cvfor each orifice is 0.98, find the point of intersection of the two jets.Solution.
333 /10772.11/772.11 mkNxmNair==
+=
AphH
gHaCQ dact 2=
( )
+=
3
223
10772.11281.92
4
105.16.0104
x
Pxxx
xxxx A
)(/83.02
GaugemkNPA =
( )
+=
+=81.9
25153
BA pphH
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Given Cv is same for both the orifices
from figure
Substituting eq(1) in eq(2) and simplifying
Again
10. Two orifices have been provided in the side of the tank, one near the bottom and the
other near the top. Show that the jets from these two orifices will intersect a planethrough the base at the same distance from the tank if the head on the upper orifice is
equal to the height of the lower orifice above the base. Assume Cv to be the same for
both the orifices.
22
2
2
11
2
1
44 Hy
x
Hy
x=
)(44
21
22
2
11
2
1 xxHy
x
Hy
x=
)1(5.164 2121 == yoryyy
( )
)2(2
46
21
21
+=
+=
yy
yy
my
y
yy
4
25.0
25.1
2
2
22
=
=
+=
givesHy
xCV
22
2
2
4=
mx
xx
x
6.9
64498.0
2
2
2
=
=
(points of intersection of the jets from theVena contracts)
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Solution.
To show that x1=x2 when H1=y2
from figure y1=[y2+(H2-H1)---(1)
Problems on Orifices
A 4cm dia orifice in the vertical side of a tank discharges water. The water surface in the
tank is at a constant level of 2m above the centre of the orifice. If the head loss in the
orifice is 0.2m and coefficient of contraction can be assumed to be 0.63. Calculate (I) thevalues of coefficient of velocity & coefficient of discharge, (ii) Discharge through the
orifice and (iii) Location of the point of impact of the jet on the horizontal plane located
0.5m below the centre of the orifice.
Solution
Head loss
22
2
2
11
2
1
1 44, 2 Hy
x
Hy
x
CCGiven VV ==
Or
Or
22
2
12112 HyHHHHy =+
0)( 122212
1 =+ HHyHHH
00
0)(; 222222
221
=
=+= yHyyHyyH
substituting
=
g
VaHhL
2
2
=
81.9222.0
2
x
VaOr
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Coefficient of Velocity
Coefficient of discharge
(ii) Discharge through the orifice
(iii) Coefficient of velocity
An orifice has to be placed in the side of a tank so that the jet will be at a maximumhorizontal distance at the level of its base. If the depth of the liquid int the tank is D, what
is the position of the orifice? Show that the jets from the two orifices in the side of the
tank will intersect at the level of the base if the head on the on the upper orifice is equal
to the height of the orifice above the base.Solution:
943.0246.6
943.5===
V
VC av
Cvd xCCC = 63.0949.0 x=
598.0=dC
gHaCQ dact 2=
281.9204.04
598.0 2 xxxxx
=
yH
xCv
4
2
=
24 xyHCV =
x
Jet
Orifice
D
h
y
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By definition, Velocity V=x/t
But
and
For x to be maximum
Vt =
gHV 2=
2
2
1gty =
2
22
1
)(
= gHx
gHD
( )HDHx = 42
)(4 HDHx =
Or
0=dH
dx
0)2(4 = HD
2/DH =
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We know, x=Vt,
12gHV =
2
22
1gtHy =+
2
122
1
=
gH
xg
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FLOW THROUGH PIPESDefinition of flow through pipes
A pipe is a closed conduit carrying a fluid under pressure. Fluid motion in a pipeis subjected to a certain resistance. Such a resistance is assumed to be due to
Friction. In reality this is mainly due to the viscous property of the fluid.
Reynolds Number (Re)It is defined as the ratio of Inertia force of a flowing fluid and the Viscous force.
Re=(Inertia force/Viscous force) =( V D/ )Classification of pipe flow:
Based on the values of Reynolds number (Re), flow is classified asFollows:Laminar flow or Viscous Flow
In such a flow the viscous forces are more predominent compared to inertiaForces. Stream lines are practically parallel to each other or flow takes placeIn the form of telescopic tubes.This type of flow occurs when Reynolds number Re< 2000.In laminar flow velocity increases gradually from zero at the boundary toMaximum at the center.Laminar flow is regular and smooth and velocity at any point practically remainsconstant in magnitude & direction. Therefore, the flow is also known as stream
Line flow.There will be no exchange of fluid particles from one layer to another.
Thus there will be no momentum transmission from one layer to another.Ex: Flow of thick oil in narrow tubes, flow of Ground Water, Flow of
Blood in blood vessels.
Transition flow:In such a type of flow the stream lines get disturbed a little.
This type of flow occurs when 2000< Re < 4000.
Turbulent Flow: This is the most common type of flow that occurs in nature( flow in rivers, pipes).This flow will be random,erratic,unpredictable. Thus motion of fluid particles result in eddy currents
& they mix up. Streamlines are totally disturbed or cross each other.
The velocity changes in direction and magnitude from point to point.There will be transfer of momentum between the particles as they are continuously colliding witheach other.
There will be considerable loss of energy in this type of flow.This type of flow cannot be truly mathematically analysed and any analysis is possible by stastical
evaluation.For this type of flow in a pipe Re> 4000.
(REYNOLDS EXPERIMENT:Refer Fig.(1)
Hydraulic Grade Line & Energy Grade LineA Line joining the peizometric heads at various points in a flow is known as HydraulicGrade Line (HGL)Energy Grade Line (EGL)
It is a line joining the elevation of total energy of a flow measured above a datum, i.e.
.2
2
g
VpZ ++
EGL Line lies above HGL by an amount V2/2g.(Refer Fig.(2))
Losses in Pipe FlowLosses in pipe flow can be two types viz:-a)Major Loss
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b)Minor Lossa)Major Loss: As the name itself indicates, this is the largest of the losses in a pipe. This lossoccurs due to friction only. Hence, it is known as head loss due to friction (hf)
b)Minor Loss: Minor losses in a pipe occurs due to change in magnitude or direction of flow.Minor losses are classified as (i) Entry Loss, (ii) Exit loss, (iii) Sudden expansion loss (iv)
Sudden contraction loss (v) Losses due to bends & pipe fittings.
Head Loss due to Friction (DARCY-WEISBACH Equation)Consider the flow through a straight horizontal pipe of diameter D, Length L, between two sections (1) & (2)
as shown in fig.(3). Let P1 & P2 be the pressures at these sections. o is the shear stress acting along thepipe boundary.
From II Law of NewtonForce = Mass x accn. But acceleration = 0, as there is no change in velocity,however the reason that pipe diameter is uniform or same throughout.
( )
( ) )1(4
4
44..
0
021
0
2
21
0
2
2
2
1
==
+
=
D
LPPor
DLD
PP
DLxD
PD
Pei
forces
Applying Bernoullis equation between (1) & (2) with the centre line of the pipeas datum & considering head loss due to friction hf,.
fhg
VpZ
g
VpZ +++=++
22
2
222
2
111
21 ZZ =Pipe is horizontal
21 VV =Pipe diameter is same throughout
)2(21 =
fhPP
Substituting eq (2) in eq.(1)
)3(4
40
0 ==l
Dhor
D
Lxh
f
f
From Experiments, Darcy Found that
)4(8
2
0 = Vf
f=Darcys friction factor (property of the pipe materialsMass density of the liquid.
V = velocityEquations (3) & (4)
L
DhV
f f
48
2
=
or ,D
VLfhf
8
4 2=
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But,
)5(2
2
=
=
gD
fLVh
g
f
from Continuity equation2
4D
QV
=
)6(8
52
2
=
Dgh
fLQhf
& (5) & (6) are known as DARCY WEISBACH Equation
Pipes in Series or Compound PipeD1, D2, D3, D4 are diameters.(fig.4)
L1, L2,L3, L4 are lengths of a number of Pipes connected in series
(hf)1, (hf)2, (hf)3 & (hf)4 are the head loss due to friction for each pipe.
The total head loss due to friction hf for the entire pipe system is given by
4321 hfhfhfhfhf +++=
5
4
2
2
4
5
3
2
2
3
5
2
2
2
2
5
1
2
2
1 8888
Dg
QfL
Dg
QfL
Dg
QfL
Dg
QfLhf
+++=
Pipes in parallel
D1, D2 and D3 are the pipe diameters.(Fig.5)
Length of each pipe is same, that is, L1=L2=L3For pipes in parallel hf1=hf2=hf3
i.e
)1(
888
5
3
2
5
2
2
5
1
2
5
3
2
2
3
5
2
2
2
2
5
1
2
2
1
321
321
==
==
D
Q
D
Q
D
Q
orDg
QfL
Dg
QfL
Dg
QfL
From continuity equation Q= Q1+Q2+Q3--------(2)
Equivalent pipeIn practice adopting pipes in series may not be feasible due to the fact thatthey may be of unistandard size (ie. May not be comemercially available)and they experience other minor losses. Hence, the entire system will be
replaced by a single pipe of uniform diameter D, but of the same lengthL=L1+ L2+ L3 such that the head loss due to friction for both the pipes, viz
equivalent pipe & the compound pipe are the same (Fig.6).For a compound pipe or pipes in series
321 hfhfhfhf ++=
)1(888
5
3
2
2
3
5
2
2
2
2
5
1
2
2
1 ++=Dg
QfL
Dg
QfL
Dg
QfLhf
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for an equivalent pipe )2(8
5
1
2
2
=Dg
fLQhf
Equating (1) & (2) and simplifying 53
3
5
2
2
5
1
1
5 D
L
D
L
D
L
D
L++=
or
5
1
5
3
3
5
2
2
5
1
1
++=
D
L
D
L
D
L
LD
PROBLEMS
1) Find the diameter of a Galvanized iron pipe required to carry a flow of 40lps of water, if the lossof head is not to exceed 5m per 1km. Length of pipe, Assume f=0.02.
Solution:-D=?, Q=40lps = 40x10-3 m3/shf=5m, L=1km = 1000m. f=0.02
Darcys equation is5
28
Dg
fLQhf
=
=fhg
fLQD
2
28
5
1
2
23
581.9
)1040(100002.08
=
xx
xxxxD
mmmD 22022.0 ==
2) Two tanks are connected by a 500mm diameter 2500mm long pipe. Find the rate of flow if thedifference in water levels between the tanks is 20m. Take f=0.016. Neglect minor losses.Solution:-
Applying Bernoullis equation between (1) & (2) with (2) as datum & considering head loss due tofriction hfonly, (Fig.7).
)1(22
2
222
2
111 +++=++ fh
g
VpZ
g
VpZ
Z1 = 20m, Z2 = 0 (Datum); V1=V2 = 0 (tanks are very large)
p1=p2=0 (atmospheric pressure)
Therefore From (1)20+0+0=0+0+0+hfOr, hf= 20m.
But5
28
Dg
fLQhf
=
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2
152
2500016.08
5.081.920
=xx
xxxQ
lpsmQ 8.434sec/4348.0 3 ==
3) Water is supplied to a town of 0.5million inhabitants from a reservoir 25km away and the lossof head due to friction in the pipe line is measured as 25m. Calculate the size of the supply main,if each inhabitant uses 200 litres of water per day and 65% of the daily supply is pumped in 8 hours. Take f=0.0195.Solution:-Number of inhabitants = 5million = 5,00,000Length of pipe = 25km = 25,000m.Hf= 25m, D=?
Per capita daily demand = 200litres.Total daily demand = 5,00,000x200= 100x106 litres.Daily supply = 65/100 x 100x106 = 65,000m3.
Supply rate sec/1248.260605.8
000,65 3
mxxQ =
=
=52
28
Dg
fLQhf
5
1
2
2
2581.9
)1248.2(000,25195.08
=xx
xxxD
mD 487.1=
4) An existing pipe line 800m long consists of four sizes namely, 30cm for 175m, 25cm dia for thenext 200m, 20cm dia for the next 250m and 15cm for the remaining length. Neglecting minor
losses, find the diameter of the uniform pipe of 800m. Length to replace the compound pipe.Solution:-L=800mL1=175m D1=0.3m
L2=200m D2=0.25m
L3=250m D3=0.20m
L4=175m D4=0.15m
For an equivalent pipe
+++=5
4
4
5
3
3
5
2
2
5
1
1
5D
L
D
L
D
L
D
L
D
L
5
1
555515.0
175
2.0
250
25.0
200
3.0
175
800
+++
=D
D = Diameter of equivalent pipe = 0.189m less than or equal to 19cm.
5) Two reservoirs are connected by four pipes laid in parallel, their respective diameters being d,1.5d, 2.5d and 3.4d respectively. They are all of same length L & have the same friction factors f.Find the discharge through the larger pipes, if the smallest one carries 45lps.
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Solution:-D1=d, D2 =1.5d, D3=2.5d, D4=3.4d
L1=L2=L3=L4= L.
f1=f2=f3=f4=f.
Q1=45x10-3m3/sec, Q2=? Q3=? Q4=?
For pipes in parallel hf1=hf2=hf3=hf4
i.e.
5
4
2
4
5
3
2
3
5
2
2
2
5
1
2
1
D
Q
D
Q
D
Q
D
Q===
( ) sec/124.010455.1 32
1
23
5
2 mxxd
dQ =
=
( ) sec/4446.010455.2 32
1
23
5
2 mxxd
dQ =
=
( ) sec/9592.010454.3 321
23
5
3 mxxd
dQ =
=
6) Two pipe lines of same length but with different diameters 50cm and 75cm are made to carrythe same quantity of flow at the same Reynolds number. What is the ratio of head loss due tofriction in the two pipes?Solution:-D1=0.5m, D2 =0.75m
L1=L2Q1=Q2
(Re)1 = (Re)2, ?2
1 =hf
hf
Reynolds number Re=
VD
2
222
1
111
VDVD=
2211 DVDV = ( )21 = ( )21 =
21 75.05.0 VV = 21 5.1 VV =
From Darcys equationgD
fLVhf
2
2
=
2
2
21
1
2
2
1
V
Vx
D
D
hf
hf =
375.35.1
5.0
75.02
2
2 =
=
V
Vx
7) A 30cm diameter main is required for a town water supply. As pipes over 27.5cm diameter arenot readily available, it was decided to lay two parallel pipes of same diameter. Find the diameter
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of the parallel pipes which will have the combined discharge equal to the single pipe. Adoptsame friction factor for all the pipes.
Fig.(8)
Solution:- )1(8 52
2
= DgfLQhf
)2(2
8
2
52
= Dg
QfL
hf
Equating
=
52
2
52
22
88
Dg
QfL
Dg
fLQ
55
1 4
11
DD=
or51
5
4
275.0
=D
mmD 275.0205.0 =
8) Two reservoirs are connected by two parallel pipes. Their diameter are 300mm & 350mm andlengths are 3.15km and 3.5km respectively of the respective values of coefficient of friction are0.0216 and 0.0325. What will be the discharge from the larger pipe, if the smaller one carries285lps?Solution:-D1=300mm=0.3m, D2=-.350m
L1=3150m L2=3500m
F1=0.0216 f 2=0.0325
Q1=0.285m3/sec Q2=?
For parallel pipes
=
=5
2
2
2
222
5
1
2
2
111 88
Dg
QLf
Dg
QLfhf
2
1
5
122
5
2
2
1112
=DLf
DQLfQ
2
1
5
52
2 3.035000325.0
35.0285.031500216.0
= xx
xxxQ
sec/324.03
2 mQ =
9) Consider two pipes of same lengths and having same roughness coefficient, but with thediameter of one pipe being twice the other. Determine (I) the ratio of discharges through thesepipes, if the head loss due to friction for both the pipes is the same. (ii) the ratio of the head lossdue to friction, when both the pipes carry the same discharge.
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Solution:-f1=f2 D1=2D2 L1=L2(i)Given hf1=hf2 Q1/Q2=?
From Darcys equation
=52
28
Dg
fLQhf
5
2
2
2222
5
1
2
2111 88
Dg
QLf
Dg
QLf
=
656.52 2
5
2
22
5
2
1
2
1 =
=
=
D
D
D
D
Q
Q
(ii) Given Q1/Q2, hf1/hf2=?
03125.02
85
2
2
5
1
2
5
1
2
2
211
2
1 =
=
==
D
D
D
D
Dg
QLf
hf
hf
10) Two sharp ended pipes are 50mm & 105mm diameters and 200m lengthare connected in parallel between two reservoirs which have a water leveldifference of 15m. If the coefficient of friction for each pipes of 0.0215.Calculate the rate of flow in each pipe and also diameter of a single pipe200m long which would give the same discharge, if it were substituted for theOriginal two pipes.SolutionD1=0.015m, D2=0.105m, L1=L2=200m
H=15m, f1=f2=0.0215, a) Q1=?, Q2=? (b) D=?, when Q=Q1+Q2a) For parallel pipes
=
=5
2
2
2
222
5
1
2
2
111 88
Dg
QLf
Dg
QLfhf
sec/1063.32000215.08
05.081.915 332
152
1 mxxx
xxhxQ =
=
sec/023.02000215.08
105.081.915 3221
52
2 mxx
xxhxQ =
=
b)
( ) sec/02684.00232.01063.3 2321 mxQQQ =+=+=
52
2
8Dg
fLQhf
= ( ) 51
2
2
1581.902684.02000215.08
=
xxxxxD
cmmD 12.111112.0 ==
11) Two pipes with diameters 2D and D are first connected in parallel and
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when a discharge Q passes the head loss is H1, when the same pipes are
Connected in series for the same discharge the loss of head is H2. Find the
relationship between H1 and H2. Neglect minor losses. Both the pipes are of
same length and have the same friction factors.
Solution (Fig.9)
H1 = head loss due to friction = hf= hf2
i.e.
=52
2
1
)2(
8
Dg
fLQhf
)1(
)(
852
2
2
2
=Dg
fLQhf
)2(21 =+ QQQ
QQQ =+ 2266.5 QQ66.6
12 =
)3(02256.0866.6
18
52
2
52
2
1 =
=Dg
QfLx
Dg
QfL
H
Case(iii) 212 hfhfH +=
52
2
52
2
)2(
88
Dg
fLQ
Dg
fLQhf
+
=
+=
552
2
22
1
1
18
Dg
fLQH
)4(80312.1
52
2
2 =Dg
xflQxH
2
52
52
2
2
1
80312.1
802256.0
flQx
Dgx
Dg
xflQx
H
H
=
021876.00312.1
02256.0
2
1 ==H
H
or 71.451
2 =H
H
12) Two reservoirs are connected by a 3km long 250mm diameter. Thedifference in water levels being 10m. Calculate the discharge in lpm, if f=0.03.
Also find the percentage increase in discharge if for the last 600m a secondPipe of the same diameter is laid parallel to the first.
SolutionApplying Bernoullis equation between (1) & (2) with (2) as datum andconsidering head loss due to friction hf (Fig.11)
fhg
VpZ
g
VpZ +++=++
22
2
122
2
111
mhh ff 100000010 =+++=++
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52
28
Dg
fLQhf
= ( )
2
152
300003.08
1025.081.9
=xx
xxxQ
sec/03624.0 3mQ =
Case (ii) 321 orhfhfhfhf += 321 orhfhfhfhf += (Fig.12)
( )
+=5
2
1
5
2
1
2 25.0
2/600
25.0
2400
81.9
03.0810
x
x
2
166.647210 Q= sec/0393.03
1 mQ =Change in discharge = ( )QQQ = 1
( )03624.00393.0 = sec/10066.3 33 mxQ =
% increase in discharge = 1001
xQ
Q
%46.810003624.0
10066.3 3==
xx
MINOR LOSSES IN PIPESMinor losses in a pipe flow can be either due to change in magnitude or directionof flow. They can be due to one or more of the following reasons.i)Entry lossii)Exit lossiii)Sudden expansion lossiv)Sudden contraction lossv)Losses due to pipe bends and fittingsvi)Losses due to obstruction in pipe.
Equation for head loss due to sudden enlargement or expansion of a pipe
Consider the sudden expansion of flow between the two section (1) (1)& (2) (2)as shown in Fig.13P1 & P2 are the pressure acting at (1) (1) and (2) (2), while V1 and V2 are the
velocities.From experiments, it is proved that pressure P1 acts on the area (a2 a1) i.e. at
the point of sudden expansion.From II Law of Newton Force = Mass x Acceleration. ---------------(1)
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Consider LHS of eq(1)
( ) )(1212211 iaapapapforces ++=( ) )(, 212 iippaforcesor =
Consider RHS of eq(1)
Mass x acceleration =
x vol x change in velocity /time =volume/time x change in velocitySubstitution (ii) & (iii)
( ) ( )21212 VVQppa = or, ( ) ( )21221 VVVpp = Both sides by (sp.weight)
( ))(21221 iv
g
VVVpp
=
Applying Bernoullis equation between (1) and (2) with the centre line of the pipeas datum and considering head loss due to sudden expansion hLonly
gVpZ
gVpZ
22
2
222
2
111 ++=++
zontalpipeishoriCZZ 21 =
Lhg
VVpp=
=
2
2
2
2
121
( ) ( )g
VVVVVhL
2
2 222
1212 +=
g
VVVVVhL
2
22 222
1
2
221 +=
g
VVVVVhL
2
22 22212122 +=
g
VVVVhL
2
2 212
1
2
2 += OR( )
g
VVhL
2
2
21 =
In Eq(V) hL is expressed in meters similarly, power (P) lost due to sudden
expansion is )(viQhP f =
Equations for other minor losses (Fig.14 a,b,c)
Sudden contraction lossg
VhL
2
5.02
2=
Loss due to entrance and exitg
Vh
entryL2
5.0 2=
g
Vh
exitL2
2
=
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Loss due to bends & fittingsg
KVhL
2
2
=
K=coefficient
Problems1) A 25cm diameter, 2km long horizontal pipe is connected to a water tank. Thepipe discharges freely into atmosphere on the downstream side. The head overthe centre line of the pipe is 32.5m, f=0.0185. Considering the discharge throughthe pipe
Applying Bernoullis equation between (A) and (B) with (B) as datum &considering all losses.(Fig.15)
exitlossssfrictionloentryloss
g
vpZ
g
VPZ BBB
AAA +++++=++
22
22
g
V
gD
fLV
g
V
g
V
222
5.0
200005.32
2222
+++++=++
+++= 1
25.0
20000185.05.01
25.32
2 X
g
V
267.75.32 V=
smV /06.2= 4
2DQ
= sec/101.006.24
25.0 32 mxx =
lpsQ 101=
2) The discharge through a pipe is 225lps. Find the loss of head when the pipe issuddenly enlarged from 150mm to 250mm diameter.Solution: D1=0.15m, D2 = 0.25m Q=225lps = 225m3/secHead loss due to sudden expansion is
gX
D
Q
D
Q
2
1442
2
2
1
=
2
2
2
2
1
2
2 11
2
16
=
DDg
Q
2
222
2
25.0
1
15.0
1
81.92
225.016
=
xx
x
mhL 385.3=
3) The rate of flow of water through a horizontal pipe is 350lps. The diameter ofthe pipe is suddenly enlarge from 200mm to 500mm. The pressure intensity inthe smaller pipe is 15N/cm2. Determine (i) loss of head due to suddenenlargement. (ii) pressure intensity in the larger pipe (iii) power lost due toenlargement.
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Solution : (Fig.16)Q=350lps=0.35m3/sD1=0.2m, D2=0.5m, P1=15N/cm2
hL=?, p2=?, P=?
From continuity equationsm
x
x
D
QV /14.11
2.0
35.044
221
1
===
smx
x
D
QV /78.1
5.0
35.04422
2
2 ===
( ) ( )mofwater
xg
VVhL 463.4
81.92
78.114.11
2
2
21 =
=
=
Applying Bernoullis equation between (1) (1) and (2) (2) with the central line ofthe pipe as datum and considering head loss due to sudden expansion h L only.
Power lost
Lhg
VpZg
VpZ +++=++ 22
2
222
2
111
( )ntalpipehorizoZZ 021 ==
463.462.19
78.1
81.90
62.19
14.11
81.9
1500
2
2
2
+++=++p
22
2 /67.16/68.166 cmNmkNp ==Power lost LQhP = 463.435.081.9 xx=
kWP 32.15=
4) At a sudden enlargement of an horizontal pipe from 100 to 150mm, diameter,
the hydraulic grade line raises by 8mm. Calculate the discharge through the pipesystem.
Solution( )
)1(2
2
21
=g
VVhL (Fig.17)
)2(108, 3112
2 =
+
+ mx
pZ
pZGiven
Applying Bernoullis equation between (1) & (2) with the central line of the pipe asdatum and neglecting minor losses (hL) due to sudden expansion.
Lhg
VpZg
VpZ +++=++ 22
2
222
2
111
02
2
1
2
211
22 =
+
+
+
+ Lh
g
VVpZ
pZ
From continuity equation
2
2
1
2
4
15.0
4
1.0xV
xV
x =
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21 25.2 VV =
( ) ( )0
81.92
25.2
81.92
25.2108
2
22
2
2
2
23 =
+
+ x
VV
x
VVx
01274.0108 223 = Vx
smx
V /25.01274.0
108 21
3
2 =
=
Discharge 25.04
15.0
4
2
2
2
2 xx
VD
Q
==
smx /10428.4 33= or lpsQ 425.4=
5) Two reservoirs are connected by a pipe line which is 125mm diameter for thefirst 10m and 200mm in diameter for the remaining 25m. The entrance and exitare sharp and the change of section is sudden. The water surface in the upperreservoir is 7.5m above that in the lower reservoir. Determine the rate of flow,assuming f=0.001 for each of the types.
Solution
From continuity equation 2
2
1
2
4
2.0
4
125.0V
xV
x =
21 56.2 VV =
Applying Bernoullis equation between (1) & (2) in both the reservoirs with thewater in the lower reservoir as datum and considering all losses
ansionlosssuddenssfrictionloentrylossg
VpZg
VpZ BBBAA
A exp22
22
+++++=++
( )
+++++=++g
VV
g
VfL
g
V
222
5.0000}005.7
2
21
2
11
2
1
( ) ( ) ( )
+
++=++g
V
g
VV
g
Vxx
g
V
22
56.2
2
56.21001.0
2
5.25.0}005.7
2
2
2
2
2
2
2
1
{ }1434.2243.52768.362.19
5.72
2 +++=V
( ) smV /6.416.21 21
2 ==sec/1445.06.4
4
2.0 22
mxx
Q =
=
Additional ProblemsFlow through Pipes
1) Water flows upwards through a vertical pipeline. A mercury manometerconnected between two points 10m apart shows a reading of 40cm of mercury
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when discharge is 450lpm. If the friction factor is 0.02. Determine the size of thepipe.
2) A town having a population of 1.2lakhs is to be supplied with water from areservoir 4km away, and it is stipulated that half the daily supply at the rate of
140lpcd should be delivered in 8 hours. Determine the size of the concrete pipesto be laid, if the available head is 12m K for concrete pipes = 0.3mm.
3) Two reservoirs are connected by three pipes of same length laid in parallel,and the diameters are D, 2D & 3D respectively. If the coefficients of friction of allthe three pipes is same, and the discharge in the smallest pipe is 30lps,determine the flow rates in the other two pipes.
4) Two reservoirs are connected by a long pipes 300mm, diameter carrying150lps. If another pipe of the same material is to be laid in parallel to carry twicethis discharge, what should be its diameter? Neglect minor losses.
5) Three pipes are connected in parallel between two points and the totaldischarge is 3 cumecs. If the pipes are of length 1200m, 1400m & 1600mdiameter 1m, 0.8m & 1.2m respectively, and friction factor is the same for all thepipes, determine the discharge in each pipe and the pressure difference requiredto maintain the flow, assuming f=0.02.
6) A 450mm concrete pipe 1800m long connects two reservoirs whose differencein water level is 15m. What is the discharge? If another concrete pipe line300mm diameter is introduced in parallel what would be the percentage increasein discharge and the discharge in each pipe. If the parallel pipe is introduced.
a)In the first half of the length. b)In the second half of the lengthc)In the middle one third of the length. Assume f=0.03 for all pipes and samedifference in the reservoir levels.
7) A 450mm, concrete pipeline 200m long connects two reservoirs whosedifference in water levels is 15m. What is the discharge?a)What is the percentage increase in discharge if another pipe line of the samediameter is introduced is parallel for the second half of the length?b)If a 30% increase in discharge is desired, what diameter pipe should beintroduced in parallel for the second half of the length? Assume f=0.03 for all thepipes and the difference in reservoir levels same in the both the cases. Neglectminor losses.
8) Two pipes of 5cm diameter and 10cm diameter are connected in series. Theyhave the same length and friction factor. If the head loss in the 10cm pipes is 1m,what is the head loss in the 5cm pipe? If the discharge through the 10cm pipe is10lps, what is the discharge through the 5cm pipe?
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P1V1P2
V2
flow
Area = a1Area = a2
9) A pipe has D=40cm, L=10m, f=0.02. What is the length of an equivalent pipewhich has D=20cm and f=0.02.
10) An 8cm diameter pipe carrying water has an abrupt expansion 12cmdiameter at a section. If a differential mercury manometer connected to upstream
and downstream sections of the expansion indicate a gauge reading of 2cm.Estimate the discharge in the pipe.
11) When a sudden contraction form 50cm diameter to 25cm is introduced in ahorizontal pipe line the pressure changes from 105kps to 69kps. Assuming acoefficient of contraction of 0.65, calculate the flow rate. Following thiscontraction if there is a sudden enlargement to 50cm and if the pressure in the25cm diameter section is 69kps, what is the pressure in the 50cm section?
12) Three pipes A, B & C with details as given in the following are connected inseries.
Calculate a) the size of a pipe of length 125m and f=0.020, equivalent to the pipeline ABC b) the length of an 8cm diameter (f=0.015) pipe equivalent to the pipeline ABC.
13) A horizontal pipe line carrying water at 0.03 m3/s reduces abruptly from 15cmto 10cm diameter. Taking contraction coefficient CC=0.60. Determine the
pressure loss across the contraction. How this pressure loss compares with theloss that would result if the flow direction is reversed?
14) Two pipes of diameter 40cm and 20cm are 300m each in length. When thepipes are connected in series and the discharge through the pipe line is 0.1m3/s ,
find the loss of head incurred. What would be the loss of head incurred. Whatwould be the loss of head in the system to pass the same total discharge whenthe pipes are connected in parallel, take f=0.03 for each pipe.
(P.S: FOR ANSWERS TO THE ABOVE PROBLEMS, DOWNLOAD THECONTENTS OF SESSION-9, VTU.AC.IN (E-LEARNING).
1 2
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V1
V
entry
V
Fitting collar
V
entry
Fig.13
Fig 14.(a)
Fig 14.(b)
Fig 14. (c)
Fig 15
V2
exit
exit
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V1
V2
flow
V1 V2
flow
Z1+p1/ Z2+p2/
Fig 17
Fig 16
1
1
2
2
1
1
2
2
Q
P2
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FLOW MEASUREMENTSFlow Through Orifices
An orifice is an opening of any cross section, at the bottom or on the side walls ofa container or vessel, through which the fluid is discharged.If the geometric characteristics of the orifice plus the properties of the fluid areknown, then the orifice can be used to measure the flow rates.
Classification of orificesBased on shape: circular, triangular, rectangular
Based on size :Small orifice (when the head over the orifice is more than five times its size I.e. H>5d,Large orificeBased on shape of the u/s edge :Sharp edge, Bell mouth
Based on flow: Free, Submerged
Flow through an orificeAs the fluid passes through the orifice under a head H, the stream lines convergeand therefore the jet contracts. The stream lines which converge are mostlythose from near the walls and they do so because stream lines cannot make rightangled bend in motion. This phenomenon occurs just down stream of the orifice,and such a section where the area of cross section of the jet is minimum is knowas VENA CONTRACTA.
The pressure at Vena Contracta is assumed to be atmospheric and the velocityis assumed to be the same across the section since the stream lines will beparallel and equally spaced.
Downstream of Vena contracta the jet expands and bends down.Figure(18) shows the details of free flow through a vertical orifice.Applying Bernoulli's equation between (B) & (C) with the horizontal through BCas datum and neglecting losses (hL)
Lhg
VpZ
g
VpZ +++=++
22
2
222
2
111
;21 ZZ = ,1 H
p=
VVV == 21 ,0
02
00002
+++=++g
VH
)1(2 = gHorV Theoretical velocityVelocity V in Eq(1) is known as TORRICELLIS VELOCITY.
Hydraulic Coefficients of an orifice
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i)Coefficient of discharge (Cd): It is defined as the ratio of actual discharge (Q act)
to the theoretical discharge (Qth).
=
th
actd
Q
QC .
Value of Cd varies in the range of 0.61 to 0.65
ii) Coefficient of Velocity (Cv): It is defined as the ratio of actual velocity (Vact) to
the theoretical velocity (Vth).
=
th
actV
V
VC Value of Cv varies in the range of
0.95 to 0.99Coefficient of Contraction (Cc): It is defined as the ratio of the area of cross
section of the jet at Vena of cross section of the jet at Vena Contracta (a c) to the
area of the orifice (a).
=
a
aC cC
Value of Cc will be generally more than 0.62.Relationship between the Hydraulic Coefficients of an orificeFrom continuity equation
Actual discharge Qact = ac x VactTheoretical discharge Qth = a x Vth
th
actc
th
act
V
Vx
a
a
Q
Q=
or
ccd xCCC =Equation for energy loss through an orifice
Applying Bernoullis equation between the liquid surface (A) and the centre of jetand Vena Contracta (C) and considering losses (hL).
LCC
CAA
A hg
VpZ
g
VpZ +++=++
22
22
,HZA = )(0 atmospherepp BA ==,0=AV )(0 cityactualvelopp BA ==
Lhg
VaH +++=++
20000
2
)2
(2
g
VaHhL =
gHCButV Va 2=Torricellis equation
)(2
VL HxCHh =
)1(2
VL CHh =Equation for Coefficient of Velocity (CV) (Trajectory method)
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Consider a point P on the centre line of the jet, such that its horizontal andvertical coordinates are x and y respectively.By definition, velocity
t
xVa = or,
aV
xt=
Since, the jet falls through a vertical distance y under the action of gravity duringthis time (t)
2
2gty = or
)2(2 2
1
=
g
yt
Equating equations (1) & (2)
2
1
2
=
g
y
V
x
a
But ,
gHCV Va 2=
2
1
2
2
=
g
y
gHC
x
V
2
1
2
1
2
1
2
1
2
1
2
1
22 y
gx
Hg
xCV =
Hy
xCV
2=
or
=
yH
xCV
4
2
Problems1. The head of water over the centre of an orifice 30mm diameter is 1.5m. If thecoefficient of discharge for the orifice is 0.613, Calculate the actual discharge.Solution: d=30mm = 3x10-3 H=1.5m
Cd=0.613
;th
actd
QQC = thdact xQCQ = gHxaCd 2=
( ) 213
5.181.924
)1030(613.0 xxx
xxx
=
smxQact /1035.233= or lpsQact 35.2=
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2. Compensation water is to be discharge by two circular orifices under aconstant head of 1.0m, measured from the centre of the orifices. What diameterwill be required to give a discharge of 20x103 m3 per day? Assume Cd for each
notch as 0.615.Solution: d=?H=1m. Qtotal = 20x103 m3/day Cd=0.615.
60602411020
21 3
xxxxxQact = sm /1157.0 3=
we know gHaCQ dact 2=
181.924
615.01157.02
xxxxd
x
=
mmmd 5.2322325.0 ==
3. A jet of water issuing from an orifice 25mm diameter under a constant head of1.5m falls 0.915m vertically before it strikes the ground at a distance of 2.288m
measured horizontally from the Vena Contracta. The discharge was found to be102lpm. Determine the hydraulics coefficients of the orifice and the head due toresistance.Solution: d=25mm=25x10-3
H=1.5m, y=0.915m, x=2.288mQact=102lpm = 102/60 = 1.7lps = 1.7x10-3m3/sec, Cd=?, Cc=?, hL=?
976.05.1915.04
288.2
4
22
===xxyH
xCV
( )638.0
5.181.921025
4107.123
3
=
==
xxxxx
xx
Q
QC
th
actd
999.0976.0
638.0=
===
v
dCVCd
C
CCxCCC
( 21 vL CHheadlossh =( ) 2976.015.1 = mmmhL 2.710712.0 ==
4. The head of water over a 100mm diameter orifice is 5m. The water coming outof the orifice is collected in a circular tank 2m diameter. The time taken to collect45cm of water is measured as 30secs. Also the coordinates of the jet at a pointfrom Vena Contract are 100cm horizontal and 5.2cm vertical. Calculate the
hydraulic coefficients of the orifice.Solution:D=100mm=0.1m, H=5mQact = Area of collecting tankxheight of water collected / time
smxx
/0471.030
45.0
4
2 32
==
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X=100cm = 1m, y=5.2cm = 0.052mCd=?, Cv=?, Cc=?
98.05052.04
1
4
22
=
=
=
xxyH
xCv
605.0581.921.0
40471.02
=
==
xxxx
QQC
th
actd
618.098.0
605.0===
V
dC
C
CC
5. The coordinates of a point on the jet issuing from a vertical orifice are 0.4m &0.003m. Neglecting air resistance, determine the velocity of the jet and the heightof water above the orifice in the tank.Solution.X=0.4m, y=0.3m, V=? H=?
Assume
We knowyH
xCV
4
2
=
==
=
2
2
2
2
22
103.04
4.0
4
4
xxyxG
xH
xyHxG
H=1.33m
smxxxgHGV /115.533.181.9212 ===
6. A vertical orifice is fitted 0.2m above the bottom of a tank containing water to a
depth of 2m. If G=0.98. What is the vertical distance from the orifice of a point onthe jet 0.6m away from the Vena Contracta?SolutionHead over the orifice H=(2-0.2)=1.8mCV=0.98, y=?, x=0.6m
mmmxx
y
xyx
or
yH
xCV
52052.098.08.14
6.08.14
6.0)98.0(,
4
2
2
22
2
==
=
=
=
7. A closed tank contains water to a height of 2m above a sharp edged orifice1.5cm diameter, made in the bottom of the tank. If the discharge through the
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orifice is to be 4lps. Workout the pressure at which air should be pumped into thetank above water. Take Cd=0.6.
Solution: (fig.19)Q=4lps = 4x10-3m3/sD=1.5x10-2m, Cd=0.6
PA=? 333 /10772.11/772.11 mkNxmNair ==
Total head over the orifice
+=
AphH
gHaCQ dact 2=
( )
+=
3
223
10772.11281.92
4
105.16.0104
x
Pxxx
xxxx A
)(/83.0 2 GaugemkNPA =
8. A closed tank contains 3m depth of water and an air space at 15kpa pressure.
A 5cm diameter orifice at the bottom of the tank discharge water to the tank Bcontaining pressurized air at 25kpa. If Cd = 0.61 for the orifice. Calculate the
discharge of water from tank A.Solution: fig(20)d=5cm = 5x10-2m Cd=0.61.
Total head over the orifice
( )
+=
+=81.9
25153
BA pphH
H=1.9806m
9806.181.924
05.0
61.02
2
xxx
x
xgHaCQ dact
==lpssmxQact 47.7/1047.7
33 ==
9. A tank has two identical orifices in one of its vertical sides. The upper orifice is4m below the water surface and the lower one 6m below the water surface. If thevalue of Cv for each orifice is 0.98, find the point of intersection of the two jets.
Solution.
yH
xCV
4
2
=
Given Cv is same for both the orifices
22
2
2
11
2
1
44 Hy
x
Hy
x=
)(44
21
22
2
11
2
1 xxHy
x
Hy
x=
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)1(5.164 2121 == yoryyy
from figure(21)( )
)2(2
46
21
21
+=+=
yy
yy
Substituting eq(1) in eq(2) and simplifying
my
yyy
4
25.025.1
2
2
22
== +=
Again givesHy
xCV
22
2
2
4=
mx
xx
x
6.9
64498.0
2
2
2
=
=
(points of intersection of the jets from the Vena contracts)
10. Two orifices have been provided in the side of the tank, one near the bottomand the other near the top. Show that the jets from these two orifices willintersect a plane through the base at the same distance from the tank if the headon the upper orifice is equal to the height of the lower orifice above the base.
Assume Cv to be the same for both the orifices.
Solution.To show that x1=x2 when H1=y2from figure(22)y1=[y2+(H2-H1)---(1)
22
2
2
11
2
11 44
,2 Hy
xHy
xCCGiven VV ==
or 2211 44 HyHy =])([ 221122 HyHHHy =+
22
2
12112 HyHHHHy =+0)( 12221
2
1 =+ HHyHHH substituting
00
0)(; 222222
221
==+= yHyyHyyH
11. A 4cm dia orifice in the vertical side of a tank discharges water. The watersurface in the tank is at a constant level of 2m above the centre of the orifice. Ifthe head loss in the orifice is 0.2m and coefficient of contraction can be assumedto be 0.63. Calculate (I) the values of coefficient of velocity & coefficient ofdischarge, (ii) Discharge through the orifice and (iii) Location of the point ofimpact of the jet on the horizontal plane located 0.5m below the centre of theorifice.
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Solution
2 2 9.81 2V gH x x= =6.264 /V m s= Head loss
2
2
L
Vah H
g
=
20.2 2
2 9.81
Va
x
=
5.943 /aV m s=
Coefficient of Velocity5.943
0.9436.246
av
VC
V= = =
Coefficient of discharge
d v CC C xC = 0.949 0.63x= 0.598dC =(ii) Discharge through the orifice
2act d Q C a gH =
20.598 0.04 2 9.81 24
x x x x x
= 34.707 10 / 4.71x m s lps= =
(iii) Coefficient of velocity2
4v
xC
yH= or
24 VyHC x= 24 0.5 2(0.949)x x x = 1.898x m=
12. An orifice has to be placed in the side of a tank so that the jet will be at amaximum horizontal distance at the level of its base. If the depth of the liquid intthe tank is D, what is the position of the orifice? Show that the jets from the twoorifices in the side of the tank will intersect at the level of the base if the head onthe on the upper orifice is equal to the height of the orifice above the base.Solution: fig(23)By definition, Velocity V=x/t
Vt=But
2V gH= and 21
2y gt=
2
1( )2 2
xD H ggH
=
or
( )2 4x H D H= or
4 ( )x H D H= For x to be maximum
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0dx
dH=
4( 2 ) 0D H =/ 2H D =
We know, x=Vt, 12V gH=
2
2
1
2y H gt+ =
2
1
1
2 2
xg
gH
=
( )2 1 24 (1)x H y H= +
( )2 14 / 2 (2)x H y L= + Equating (1) & (2)
1 1 2 1 2 24 4 4 4H y H H H H yH+ = +
1 2 1 24 4H y yH H H= =
13. Two tanks with orifices in the same vertical plane are shown in figure.What should be the spacing x for the jets to intersect in the plane of the base?
Assume CV=0.98 for each orifices. x=?
Solution: fig(25)2 2
1 2
1 1 2 2
(1)4 4
x x
y H y H=
Assuming the coefficient of velocity CV to be the same for both the orifices, we
have (CV1) = (CV2)
where
1 12 2 0.4 1.6y H m= = =
2 22 2 1.6 0.4y H m= = =
1 20.4 , 1.6H m H m= =2 2
1 2
4 1.6 0.4 4 0.4 1.6
x x
x x x x =
2
1
11 14
V
xC
y H=
2
10.98
4 1.6 0.4
x
x x
= 1 21.568x m x= =
[ ]1 2 2 1.568 3.136x x x x m = + = =
14. A large tank has a circular sharp edged orifice 25mm diameter in thevertical side. The water level in the tank is 0.6m above the centre of the orifice.The diameter of the jet at Vena contracta is measured as 20mm. The water of
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the jet is collected in a tank 1.2m long x 0.6m wide and the water level rised from0.15 to 0.75 in 7 minutes. Calculate the orifice coefficients.Solution: Dia of jet at vena contracta dc=20mm
Dia of orifice = d = 25mmHead over the orifice H = 0.6m
Depth of water collected in the measuring tank h=(0.75-0.15)=0.6mDepth of water collected in the measuring tank A = 1.2x0.6 = 0.72m 2
Time taken for collecting 0.6m of water t=7min=7x60=420secTherefore actual discharge Qact = Area of measuring tank x depth of water
collected / time taken
0.72 0.6. .
420act
Ah xi e Q
t
= =
3 31.0286 10 /x m s=
2thQ a gH =( )
23
25 102 9.81 0.6
4
xx x x x
=
3 21.684 10 /thQ x m s=
Coefficient of discharge
actd
th
QC
Q=
3
3
1.0286 10
1.684 10
x
x
=
0.61dC =
Coefficient of contractionCc = area of jet at vena contracta / area of orifice
2
2
4
4C
c
d
C x d
=
2 220
0.6425c
d
d
= = =
We know, Cd=CcxCV
Therefore, Coefficient of velocity0.61
0.64
dV
c
CC
C= =
0.953VC = Coefficient of resistance 21
C
V
CC
=
2
11
0.953
=
0.1008rC =
15. A jet of water issuing from a vertical orifice in a tank under a constanthead of 4m. If the depth of water in the tank is 12m, at what depth another orificeto be mounted vertically below the former one, so that both the jets meet at acommon point on the horizontal at the bottom of the tank? Assume C vto be the
same for both the orifices = 0.98.
Solution: fig(26)
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2 2
1 2
1 21 1 2 2
,4 4
V V
x xC C
y H y H= =
From figure, (y1-y2)=(H2-H1)
( ) ( )1 1 12 4 8y y H m= = =
( ) ( )2 2 212y y H H m= = Equating the values of CV
2 2
1 2
1 1 2 24 4
x x
y H y H=
1 1 2 24 4y H y H=
1 2x x=Q
1 1 2 2y H y H=
( )2 2
8 4 12x H H= 2
2 212 32 0H H + =2
2
12 12 4 1 32
2 1
x xH
x
+ =
8 4mor m=H2 = 8m is the correct answer.
Hence, the second orifice should be 4m below the first orifice.16.
Water is to be discharged by two circular orifices under a constant head of 1mabove their centres. What should be the diameter of the orifices to give a
discharge of 20Mlpd? Assume a coefficient of discharge of 0.62.Solution.Total discharge=20Mlpd(million litres per day)
620 10231.48
24 60 60
xlps
x x= =
Therefore Discharge per orifices231.48
115.742
Q lps= =
20.11574 /orQ m s=
But,2dQ C a gH =
0.11574
0.62 2 9.81 1a
x x x
=
20.04214m=
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12
24 0.04214
4
d xa d
= =
=0.2316mTherefore, Diameter of each orifice d = 231.6mm
17. What is the discharge through the 60mm diameter orifice shown in figure,assuming the oil level remains constantSolution. Fig(27)
Head of the orifice H100
20.9 9.81x
= +
13.326H mofoil = 2dQ C a gH =
( )30.65 60 10 2 9.81 13.3264
x x x x x x =
30.02972 /m s= 29.72lps=
18. What is the discharge through a sharp edged slot 0.2 long x 10mm wideat the bottom of a tank 0.5m diameter with 3m depth of water constant?Solution.Head over the orifice H=(2-0.2)=1.8m
0.98, ?, 0.6VC y x m= = =2
4V
xC
yH=
22 0.6
(0.98) 4 1.8xyx
= 0.052 52y m mm = =
19. A vertical orifice is fitted 0.2m above the bottom of a tank containingwater to a depth of 2m. If CV=0.98. What is the vertical distance from the orifice
of a point on the jet 0.6m away from the Vena contracta?Solution.
100.61 0.2 2 9.81 3
100x x x x x
=
3 29.36 10 /x m s=9.36Q lps =
20. The coordinates of a point on the jet issuing from a vertical orifice are0.4m & 0.3m. Neglecting air resistances, determine the velocity of the jet and theheight of water above the orifice in the tank. Assume CV=0.98.
Solution.X=0.4m, y=0.3m, V=?, H=?
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2
4V
xC
yH=
20.40.98
4 0.3x xH=
0.1388H m =
2 2 9.81 0.1388 1.65 /V gH x x m s= = =
Mouth PiecesA mouth piece is a short tube or pipe connected in extension with an orifice
Classification of Mouth PiecesDepending on the position with respect to the tank: External, Internal
Depending on shape :Cylindrical,Convergent, Divergent
Nature of flow: Running Full,Running Free
External Cylindrical Mouthpiece fig(28)
It is a short pipe whose length is two or three times the diameter.H=Head over the centre of the mouth pieceVO=Velocity of the liquid at Vena Contracta
ac=Area of flow at Vena Contracta
V1=Velocity of liquid at outlet
a1=Area of mouth piece at outlet.
Cc=coefficient of contraction
Applying continuity equation between & (1) &(1)accc=a1v1
1
1c
c
a
V Va =
1
0.62cac
c coefficientofcontractiona
= = =
1
1(1)
0.62cV V =
As the jet flows from to (1) (1) there will be loss of head due to suddenenlargement of flow, and this value can be calculated from the relation.
( )
2
12 1
1 0.62
2 2
C
L
VV
V Vh
g g
= =
2
1
0.375(2)
2Lh V
g=
Applying Bernoullis equation between (A) and (1) (1) with the centre line of themouth piece as datum and considering head loss hL due to sudden expansion.
2 2
1 11
2 2
A AA L
p V p VZ Z h
g g + + = + + +
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11 0, , 0( )A
ppAZ Z H atmosphericpressure
= = = =
0( )AV Negligible=2 2
1 10.3750 0 0 0
2 2
V VH
g g
+ + = + + +
2
1
1.375
2H V
g=
or
1
2
1.375
gHV =
1 0.853 2 (3)V gH = By definition, Coefficient of velocityCV=Actual velocity/Theoretical velocity
0.853 2. ,2
V gHi e CgH=
0.853VC =At the exit of the mouth piece CC=1
1 0.853 0.853d c vC C xC x = = =Hence, for an external cylindrical mouth piece Cd=(=0.853) is more than that of
an orifice.Pressure head at Vena contracta
Applying Bernoullis equation between (A) & with the centre line of the mouthpiece as datum & neglecting losses.
22
2 2
c cA AA c L
p Vp VZ Z hg g
+ + = + + +
, 0, 0, 0, ?cA A C LPpA
H V Z Z h
= = = = = =
2
0 0 0 02
c cp VHg
+ + = + + +
2
2
c cp VHg
=
But,2
1 11.375 ,2 0.62
CV VH V
g= =
1
20.853 2
1.375
gHV gH = =
0.853 2&
0.62C
gHV =
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2
0.853 2 1
0.62 2
CgHp
H xg
=
1.893Cp
H H
=
0.893Cp
H
=
Negative sign indicates that the pressure at the Vena contracta is less th