6-7-8.Fluid Flow in pipes couse of fluid mechanics 2
abdulmoeedmalik043
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Feb 27, 2025
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About This Presentation
Fluid flow in mechanics course of civil engineering
Slide Content
Nikuradse
’
sexperiment
Nikuradses
experiment
’
sexperiment
Nikuradses
experiment
No scientific way o
f
specifying pipe
roughness
f
Experiments have shown that
f
riction not
only depends on size and shape of the
roughness
projections
but
also
on
their
roughness
projections
but
also
on
their
spacing
No scientific way o
f
specifying pipe
roughness
f
Experiments have shown that
f
riction not
only depends on size and shape of the
roughness
projections
but
also
on
their
roughness
projections
but
also
on
their
spacing
An expression of frictional factor based on fluid properties is required F L i fl i i i b ????
F
or
L
am
i
nar
fl
ow,
i
t
i
s g
i
ven
b
y
????
(equating Darcy-Wesbach and Hagen-Poiseuille eqs)
f=64/R
N
f=64/R
N
An expression of frictional factor based on fluid properties is required F L i fl i i i b ????
F
or
L
am
i
nar
fl
ow,
i
t
i
s g
i
ven
b
y
????
(equating Darcy-Wesbach and Hagen-Poiseuille eqs)
f=64/R
N
f=64/R
N
Calculationofheadlossduetofrictionin
Calculation
of
head
loss
due
to
friction
in
pipes
To analyse turbulent velocity profile
To compute shear stress on pipe wall
Calculationofheadlossduetofrictionin
Calculation
of
head
loss
due
to
friction
in
pipes
To analyse turbulent velocity profile
To compute shear stress on pipe wall
J
.Nikuradse
(
1933
)
conducted an ex
p
eriment to
J
(
)
p
develop the relationship for frictional factor
Coated several different sizes
(
2.5
,
5and10cm
(
,
dia) of pipes with sand grains by sieving them into
different grain sizes of reasonable uniform
diameters (0.1 to 1.6 mm).
Denoted the sand grain diameter by
e
(the arbitrary
absolute roughness)
Also defined arbitrary relative roughness over pipe
diameter by
e/
D
and produced graphs of‘f’against
R
for
relative
roughness
ranging
from
1
/
1014
to
R
e
for
relative
roughness
ranging
from
1
/
1014
to
1/30.
J
.Nikuradse
(
1933
)
conducted an ex
p
eriment to
J
(
)
p
develop the relationship for frictional factor
Coated several different sizes
(
2.5
,
5and10cm
(
,
dia) of pipes with sand grains by sieving them into
different grain sizes of reasonable uniform
diameters (0.1 to 1.6 mm).
Denoted the sand grain diameter by
e
(the arbitrary
absolute roughness)
Also defined arbitrary relative roughness over pipe
diameter by
e/
D
and produced graphs of‘f’against
R
for
relative
roughness
ranging
from
1
/
1014
to
R
e
for
relative
roughness
ranging
from
1
/
1014
to
1/30.
The identified regions will comprise of
fl
Laminar
fl
ow(f=64
/
R
e
),
Smooth turbulent region where,
) (R
e
f
) (R
e
f
316
0
Transitional turbulent (region where
f
varies both
25.0
e
R
316
.
0
f
f
with R
e
and relative roughness , most pipes lie in
this region)
)
/
(R
D
f
Rou
g
h turbulent
(
f
remains constant for a
g
iven
)
/
,
(R
e
D
e
f
g
(
f
g
relative roughness, it is independent of R
e
)
)
/
(
D
e
f
)
/
(
D
e
f
fl
Laminar
fl
ow(f=64
/
R
e
),
Smooth turbulent region where,
) (R
e
f
) (R
e
f
316
0
Transitional turbulent (region where
f
varies both
25.0
e
R
316
.
0
f
f
with R
e
and relative roughness , most pipes lie in
this region)
)
/
(R
D
f
Rou
g
h turbulent
(
f
remains constant for a
g
iven
)
/
,
(R
e
D
e
f
g
(
f
g
relative roughness, it is independent of R
e
)
)
/
(
D
e
f
)
/
(
D
e
f
If the thickness of the viscous sublayer
(i
e
)
where
e
is
the
height
of
e
5
/
v
eu
(i
.
e
.
)
,
where
e
is
the
height
of
the roughness projection, the viscous sublayer
completely submerges the effect ofe.
e
L
5
/
*
v
eu
Prandtl gave the following equation for the calculation
of
frictional
factor
for
such
a
case
calculation
of
frictional
factor
for
such
a
case
51
.
2
Re
log 2
1f
f
The above equation is implicit infand is difficult to
solve
and
can
only
be
solved
by
iteration
or
a
51
.
2
f
to
solve
,
and
can
only
be
solved
by
iteration
or
a
graph offversus R
e
.
If the thickness of the viscous sublayer
(i
e
)
where
e
is
the
height
of
e
5
/
v
eu
(i
.
e
.
)
,
where
e
is
the
height
of
the roughness projection, the viscous sublayer
completely submerges the effect ofe.
e
L
5
/
*
v
eu
Prandtl gave the following equation for the calculation
of
frictional
factor
for
such
a
case
calculation
of
frictional
factor
for
such
a
case
51
.
2
Re
log 2
1f
f
The above equation is implicit infand is difficult to
solve
and
can
only
be
solved
by
iteration
or
a
51
.
2
f
to
solve
,
and
can
only
be
solved
by
iteration
or
a
graph offversus R
e
.
C.F. Colebrook (1939) suggested the followingexplicitrelationforcalculating
f
Re
1
9.6
Re
log 8.1
1
f
C.F. Colebrook (1939) suggested the followingexplicitrelationforcalculating
f
Re
1
9.6
Re
log 8.1
1
f
For fully rough pipe flow, where R
e
is quite
high and correspondingly is quite small,
ih d
L
or
i
n ot
h
er wor
d
s,
70 /
*
v eu
7
3
1
(von Karman eq)
De f/
7.
3
log 2
1
For fully rough pipe flow, where R
e
is quite
high and correspondingly is quite small,
ih d
L
or
i
n ot
h
er wor
d
s,
70 /
*
v eu
7
3
1
(von Karman eq)
De f/
7.
3
log 2
1
For transitional case For
transitional
case
70 / 5
*
v eu
neither smooth not fully rough pipe equation
applies. Colebrook (1939) gave the following
relation
for
such
cases
relation
for
such
cases
f
De
f
Re
51.2
7
3
/
log 2
1
fore=0, it reduces to smooth pipe flow equation
and
for
large
R
it
reduces
to
rough
pipe
flow
f
f
Re
7
.
3
and
for
large
R
e
it
reduces
to
rough
pipe
flow
equation. But it is implicit inf.
transitional
case
70 / 5
*
v eu
neither smooth not fully rough pipe equation
applies. Colebrook (1939) gave the following
relation
for
such
cases
relation
for
such
cases
f
De
f
Re
51.2
7
3
/
log 2
1
fore=0, it reduces to smooth pipe flow equation
and
for
large
R
it
reduces
to
rough
pipe
flow
f
f
Re
7
.
3
and
for
large
R
e
it
reduces
to
rough
pipe
flow
equation. But it is implicit inf.
ld
hfll l
Haa
l
an
d
S. E. gave t
h
e
f
o
ll
owing exp
l
icit
relation for f
e
11.1
R
9.6
7.3
/
8.1
1De
f
f
hfll l
Haa
l
an
d
S. E. gave t
h
e
f
o
ll
owing exp
l
icit
relation for f
e
11.1
R
9.6
7.3
/
8.1
1De
f
f
So for commercially available pipes, frictional
factor
f
can be calculated using
Ui ii l i (l d did)
U
s
i
ng emp
i
r
i
ca
l
equat
i
ons
(
a
l
rea
d
y stu
di
e
d)
Using Moody’s diagrams ◦
Simple approach
◦
Simple
approach
◦Hit and Trial
factor
f
can be calculated using
Ui ii l i (l d did)
U
s
i
ng emp
i
r
i
ca
l
equat
i
ons
(
a
l
rea
d
y stu
di
e
d)
Using Moody’s diagrams ◦
Simple approach
◦
Simple
approach
◦Hit and Trial
EXERCISE
EXERCISE 8.2.1 8.5.3 8.9.2 8.11.1 8.13.2 8.16.1 8.18.3 822
854
8101
8112
8133
8162
8
.
2
.
2
8
.
5
.
4
8
.
10
.
1
8
.
11
.
2
8
.
13
.
3
8
.
16
.
2
8.2.3 8.6.1 8.10.2 8.11.3 8.13.4 8.17.1 831
862
8103
8121
8151
8172
8
.
3
.
1
8
.
6
.
2
8
.
10
.
3
8
.
12
.
1
8
.
15
.
1
8
.
17
.
2
8.3.2 8.8.1 8.10.4 8.12.2 8.15.2 8.17.3 8.5.1
8.8.2
8.10.5
8.12.3
8.15.3
8.18.1
8
.
5
.
1
8
.
8
.
2
8
.
10
.
5
8
.
12
.
3
8
.
15
.
3
8
.
18
.
1
8.5.2 8.9.1 8.10.6 8.13.1 8.15.4 8.18.2
EXERCISE 8.2.1 8.5.3 8.9.2 8.11.1 8.13.2 8.16.1 8.18.3 822
854
8101
8112
8133
8162
8
.
2
.
2
8
.
5
.
4
8
.
10
.
1
8
.
11
.
2
8
.
13
.
3
8
.
16
.
2
8.2.3 8.6.1 8.10.2 8.11.3 8.13.4 8.17.1 831
862
8103
8121
8151
8172
8
.
3
.
1
8
.
6
.
2
8
.
10
.
3
8
.
12
.
1
8
.
15
.
1
8
.
17
.
2
8.3.2 8.8.1 8.10.4 8.12.2 8.15.2 8.17.3 8.5.1
8.8.2
8.10.5
8.12.3
8.15.3
8.18.1
8
.
5
.
1
8
.
8
.
2
8
.
10
.
5
8
.
12
.
3
8
.
15
.
3
8
.
18
.
1
8.5.2 8.9.1 8.10.6 8.13.1 8.15.4 8.18.2
PROBLEMS
PROBLEMS All the problems from8.1 to847849till853 to
8
.
47
,
8
.
49
till
8
.
53
PROBLEMS All the problems from8.1 to847849till853 to
8
.
47
,
8
.
49
till
8
.
53
Assumptions
All
pipes
are
sufficiently
long
All
pipes
are
sufficiently
long
Minor head losses and velocity head can be
neglected
Cii
d
i
i
h
fl
f L
h h
C
ont
i
nu
i
ty an
d
energy equat
i
ons requ
i
re t
h
at
fl
ow
entering the junction equals the flow leaving it
Pressure
head
at
J
is
common
to
all
pipes
(P
is
Pressure
head
at
J
is
common
to
all
pipes
(P
is
common to all)
0
Q
Q
Q
0
Q
out in
Q
Q
3
2
1
Q
Q
Q
3
2
1
Q
Q
Q
All
pipes
are
sufficiently
long
All
pipes
are
sufficiently
long
Minor head losses and velocity head can be
neglected
Cii
d
i
i
h
fl
f L
h h
C
ont
i
nu
i
ty an
d
energy equat
i
ons requ
i
re t
h
at
fl
ow
entering the junction equals the flow leaving it
Pressure
head
at
J
is
common
to
all
pipes
(P
is
Pressure
head
at
J
is
common
to
all
pipes
(P
is
common to all)
0
Q
Q
Q
0
Q
out in
Q
Q
3
2
1
Q
Q
Q
3
2
1
Q
Q
Q
Elevation of P must lie between the surfaces of
Elevation
of
P
must
lie
between
the
surfaces
of
reservoirs A and C.
h
2
and Q
2
will be zero if P is at the same elevation
iB
as reservo
i
r
B
.
If P is above level of B then water must flow into B and
3 2 1
Q Q Q
If P is below the level of B then flow must be out of
B and
3 2 1
Q Q Q
Elevation of P must lie between the surfaces of
Elevation
of
P
must
lie
between
the
surfaces
of
reservoirs A and C.
h
2
and Q
2
will be zero if P is at the same elevation
iB
as reservo
i
r
B
.
If P is above level of B then water must flow into B and
3 2 1
Q Q Q
If P is below the level of B then flow must be out of
B and
3 2 1
Q Q Q
Given:
all pipe lengths
all pipe diameters
surface elevation of two reservoirs (A and B) surface
elevation
of
two
reservoirs
(A
and
B)
flow to or from one of these (Q
1
)
Re
q
uired:
q
flow from two reservoirs (Q
2
and Q
3
)
surface elevation of third reservoir (C)
all pipe lengths
all pipe diameters
surface elevation of two reservoirs (A and B) surface
elevation
of
two
reservoirs
(A
and
B)
flow to or from one of these (Q
1
)
Re
q
uired:
q
flow from two reservoirs (Q
2
and Q
3
)
surface elevation of third reservoir (C)
Solution:
determine
h
using
pipe
friction
equation
and
determine
h
1
using
pipe
friction
equation
and
Moody’s diagram This
fixes
the
elevation
of
P
so
h
can
be
This
fixes
the
elevation
of
P
so
h
2
can
be
determined. Knowing
h
one
can
determine
flow
Q
using
Knowing
h
2
one
can
determine
flow
Q
2
using
Colebrook/Haaland and Darcy Weisbach equation
Q
3
can be determined from continuity
3 2 1
Q Q Q
This enables the determination of h
3
and surface
elevation of C.
determine
h
using
pipe
friction
equation
and
determine
h
1
using
pipe
friction
equation
and
Moody’s diagram This
fixes
the
elevation
of
P
so
h
can
be
This
fixes
the
elevation
of
P
so
h
2
can
be
determined. Knowing
h
one
can
determine
flow
Q
using
Knowing
h
2
one
can
determine
flow
Q
2
using
Colebrook/Haaland and Darcy Weisbach equation
Q
3
can be determined from continuity
3 2 1
Q Q Q
This enables the determination of h
3
and surface
elevation of C.
Solve Sample Problem 8.14
Page No 329
Fluid Mechanics with Engineering Alii (10
th
Edi i )
A
pp
li
cat
i
ons
(10
th
Edi
t
i
on
)
E. John Finnemoreand Joseph B. Franzini
Solve Sample Problem 8.14
Page No 329
Fluid Mechanics with Engineering Alii (10
th
Edi i )
A
pp
li
cat
i
ons
(10
th
Edi
t
i
on
)
E. John Finnemoreand Joseph B. Franzini
Given:
all pipe lengths
all pipe diameters
surface elevation of two reservoirs
(e g A and C)
surface
elevation
of
two
reservoirs
(e
.
g
.
A
and
C)
flow to or from third reservoir (say Q
2
).
Re
q
uired:
q
surface elevation of third reservoir (B)
flow from two reservoirs (A and C)
all pipe lengths
all pipe diameters
surface elevation of two reservoirs
(e g A and C)
surface
elevation
of
two
reservoirs
(e
.
g
.
A
and
C)
flow to or from third reservoir (say Q
2
).
Re
q
uired:
q
surface elevation of third reservoir (B)
flow from two reservoirs (A and C)
Solution We know that
Assume an elevation of P, which yields values of h
1
and
h
13 3 1
h h h
and
h
3
Q
1
and Q
3
can be determined as in Case 1.
If the relation at J is not satisfied, adjust the value of
P
Thi
b
d
b
li
h
l
f
h
P
.
Thi
s can
b
e
d
one
b
yp
l
ott
i
ng t
h
eresu
l
to
f
eac
h
assumption on a graph (ΣQvsP)
Inflows to J will be taken as positive and outflows as
negative
Select the value of P forΣQ=0.
Determine
h
2
from
Q
2
with
Haaland
and
Darcy
Determine
h
2
from
Q
2
with
Haaland
and
Darcy
Weisbach equation
Assume an elevation of P, which yields values of h
1
and
h
13 3 1
h h h
and
h
3
Q
1
and Q
3
can be determined as in Case 1.
If the relation at J is not satisfied, adjust the value of
P
Thi
b
d
b
li
h
l
f
h
P
.
Thi
s can
b
e
d
one
b
yp
l
ott
i
ng t
h
eresu
l
to
f
eac
h
assumption on a graph (ΣQvsP)
Inflows to J will be taken as positive and outflows as
negative
Select the value of P forΣQ=0.
Determine
h
2
from
Q
2
with
Haaland
and
Darcy
Determine
h
2
from
Q
2
with
Haaland
and
Darcy
Weisbach equation
Sample Problem 8.15
Page No 330
Fluid Mechanics with Engineering Alii (10
th
Edi i )
A
pp
li
cat
i
ons
(10
th
Edi
t
i
on
)
E. John Finnemoreand Joseph B. Franzini
Sample Problem 8.15
Page No 330
Fluid Mechanics with Engineering Alii (10
th
Edi i )
A
pp
li
cat
i
ons
(10
th
Edi
t
i
on
)
E. John Finnemoreand Joseph B. Franzini
Three reservoir problem Given
all pipes lengths
all pipe diameters all
pipe
diameters
elevations of all three reservoirs
Required
Flows in all three pipes
all pipes lengths
all pipe diameters all
pipe
diameters
elevations of all three reservoirs
Required
Flows in all three pipes
Solution It is not clear whether the flow is into or out of reservoir
B
At first it is assumed that there is no flow in pipe 2 and
the peizometeric level in P is same as B
Q 1
and Q
3
can be determined using head losses h
1
and h
3
in the Haaland and Darcy Weisbach equations
If elevation at P must be raised to satisfy
continuity at
J
and
3 1
Q Q
3
2
1
Q
Q
Q
If then P must be lowered to satisfy
continuit
y
at
J
causin
g
water to flow out from B.
3
2
1
Q
Q
Q
3 1
Q Q
y
J
g
B
At first it is assumed that there is no flow in pipe 2 and
the peizometeric level in P is same as B
Q 1
and Q
3
can be determined using head losses h
1
and h
3
in the Haaland and Darcy Weisbach equations
If elevation at P must be raised to satisfy
continuity at
J
and
3 1
Q Q
3
2
1
Q
Q
Q
If then P must be lowered to satisfy
continuit
y
at
J
causin
g
water to flow out from B.
3
2
1
Q
Q
Q
3 1
Q Q
y
J
g
Then
3 2 1
Q Q Q
From here on, the solution proceeds along the same
lines as case 2.
3 2 1
Q Q Q
From here on, the solution proceeds along the same
lines as case 2.
Sample Problem 8.16
Page No 332
Fluid Mechanics with Engineering Alii (10
th
Edi i )
A
pp
li
cat
i
ons
(10
th
Edi
t
i
on
)
E. John Finnemoreand Joseph B. Franzini
Sample Problem 8.16
Page No 332
Fluid Mechanics with Engineering Alii (10
th
Edi i )
A
pp
li
cat
i
ons
(10
th
Edi
t
i
on
)
E. John Finnemoreand Joseph B. Franzini
For municipal distribution systems, pipes are
frequently
interconnected
so
that
the
flow
to
any
frequently
interconnected
so
that
the
flow
to
any
given outlet may come from several different paths
It is impossible to tell the direction of flow in such
i
p
i
pes
Flow in a network of pipes, however complicated it
must satisfy continuity and energy relations as
1.
Flow through any
j
unction must equal the flow out
of it (Q
in
=Q
out
)
2
Flow
in
each
pipe
must
satisfy
pipe
friction
laws
2
.
Flow
in
each
pipe
must
satisfy
pipe
friction
laws
3.
Algebraic sum of head losses around any closed loop must be zero (Σh
L
=0)
frequently
interconnected
so
that
the
flow
to
any
frequently
interconnected
so
that
the
flow
to
any
given outlet may come from several different paths
It is impossible to tell the direction of flow in such
i
p
i
pes
Flow in a network of pipes, however complicated it
must satisfy continuity and energy relations as
1.
Flow through any
j
unction must equal the flow out
of it (Q
in
=Q
out
)
2
Flow
in
each
pipe
must
satisfy
pipe
friction
laws
2
.
Flow
in
each
pipe
must
satisfy
pipe
friction
laws
3.
Algebraic sum of head losses around any closed loop must be zero (Σh
L
=0)
h
L
is positive clockwise and negative anticlockwise
Pipe networks are too complicated to solve
analytically
Method
of
successive
approximations
introduced
by
Method
of
successive
approximations
introduced
by
Hardy Cross (1936) is used.
The method consists of following steps i
By
careful
inspection
assume
the
most
reasonable
i
.
By
careful
inspection
,
assume
the
most
reasonable
distribution of flows that satisfies condition (1)
ii.
Write condition (2) for each pipe
n
L
KQ h
L
is positive clockwise and negative anticlockwise
Pipe networks are too complicated to solve
analytically
Method
of
successive
approximations
introduced
by
Method
of
successive
approximations
introduced
by
Hardy Cross (1936) is used.
The method consists of following steps i
By
careful
inspection
assume
the
most
reasonable
i
.
By
careful
inspection
,
assume
the
most
reasonable
distribution of flows that satisfies condition (1)
ii.
Write condition (2) for each pipe
n
L
KQ h
K and n are constant for each pipe. If minor losses
id d
h
are cons
id
ere
d
,t
h
en
2
1
C
K
And n=2 for constantf(minor losses are included
only within pipe or loop but neglected at junctions)
2
C
iii.
Compute sum of head losses around each
elementary loop to satisfy (3).
n
h
0
L
h
n
L
KQ
h
id d
h
are cons
id
ere
d
,t
h
en
2
1
C
K
And n=2 for constantf(minor losses are included
only within pipe or loop but neglected at junctions)
2
C
iii.
Compute sum of head losses around each
elementary loop to satisfy (3).
n
h
0
L
h
n
L
KQ
h
Clockwise-positive, anticlockwise-negative Th
diti
iht
t
b
ti fi d
i
fi t
til
Th
econ
diti
on m
i
g
ht
no
t
b
esa
ti
s
fi
e
d
i
n
fi
rs
t
t
r
i
a
l
iv.
Ad
j
ust flow in each loo
p
b
y
a correctionΔ
Q
to
j
p
y
Q
balance the head in that loop
So
h
f
h
0
n
KQ
Q
Q
Q
t
h
en
f
or eac
h
case
Q
Q
Q
0
diti
iht
t
b
ti fi d
i
fi t
til
Th
econ
diti
on m
i
g
ht
no
t
b
esa
ti
s
fi
e
d
i
n
fi
rs
t
t
r
i
a
l
iv.
Ad
j
ust flow in each loo
p
b
y
a correctionΔ
Q
to
j
p
y
Q
balance the head in that loop
So
h
f
h
0
n
KQ
Q
Q
Q
t
h
en
f
or eac
h
case
Q
Q
Q
0
......) ) ( ) (
1
0
Q nQ QK Q QK KQ
n
o
n
o
n n
IfΔQ is small compared with Q
0
, the terms after the
second
one
can
be
ignored
so
second
one
can
be
ignored
so 1
0
0
n n
L
Q
Kn
Q
KQ
h
For a loop,
0
0
L
QQ
Q
and becauseΔQissamefor
all
p
i
p
es in loo
p
0
n
L
KQ h
pp
p
1
0
Q nQ QK Q QK KQ
n
o
n
o
n n
IfΔQ is small compared with Q
0
, the terms after the
second
one
can
be
ignored
so
second
one
can
be
ignored
so 1
0
0
n n
L
Q
Kn
Q
KQ
h
For a loop,
0
0
L
Q
and becauseΔQissamefor
all
p
i
p
es in loo
p
0
n
L
KQ h
pp
p
0
1
0 0
n n
KnQ Q KQ
n
h
KQ
L
L
n
n
h
n
h
KnQ
KQ
Q
1
0
0
o
Q
n
0
1
0 0
n n
KnQ Q KQ
n
h
KQ
L
L
n
n
h
n
h
KnQ
KQ
Q
1
0
0
o
Q
n
-ve sign shows that whenever there is an excess of
hd
l
d
l
i
lki
di i
Δ
Q
h
ea
d
l
oss aroun
d
a
l
oop
i
nc
l
oc
k
w
i
se
di
rect
i
on,
Δ
Q
must be subtracted from clockwise Q
0
and added
to counter clockwise ones and vice versa.
v.
The losses are still not balanced even after the application of first correction to each loop due to the
interaction
of
one
loop
with
another
So
the
the
interaction
of
one
loop
with
another
.
So
the
procedure is repeated until the corrections become negligible.
hd
l
d
l
i
lki
di i
Δ
Q
h
ea
d
l
oss aroun
d
a
l
oop
i
nc
l
oc
k
w
i
se
di
rect
i
on,
Δ
Q
must be subtracted from clockwise Q
0
and added
to counter clockwise ones and vice versa.
v.
The losses are still not balanced even after the application of first correction to each loop due to the
interaction
of
one
loop
with
another
So
the
the
interaction
of
one
loop
with
another
.
So
the
procedure is repeated until the corrections become negligible.
Sample problem 8.19, page 342.
If the flow into and out of a two loop pipe system is as
shown
in
the
figure
determine
the
flow
in
each
as
shown
in
the
figure
,
determine
the
flow
in
each
pipe. n=2
If the flow into and out of a two loop pipe system is as
shown
in
the
figure
determine
the
flow
in
each
as
shown
in
the
figure
,
determine
the
flow
in
each
pipe. n=2
Fi
g
ure shows a
p
i
p
e made
gpp up of sections of different
diameters
This pipe must satisfy the This
pipe
must
satisfy
the
equations of continuity
and energy given by
3 2 1
Q Q Q Q
3 2 1
h h h h
g
ure shows a
p
i
p
e made
gpp up of sections of different
diameters
This pipe must satisfy the This
pipe
must
satisfy
the
equations of continuity
and energy given by
3 2 1
Q Q Q Q
3 2 1
h h h h
For i
f
the Q is given the problem is straight
forward. Head loss may be directly found by
adding contributions from various sections
i
e
by
using
the
Equation
of
the
head
loss
i
.
e
.
by
using
the
Equation
of
the
head
loss
For
if
Pipe
Material
or
‘
e
’
is
given
use
Darcy
’
s
For
if
Pipe
Material
or
e
is
given
use
Darcys
Weisbach Approach to calculate individual headlosscontributionsafterfindinge/D,V,R and
f
foreachpipe.
For i
f
the Q is given the problem is straight
forward. Head loss may be directly found by
adding contributions from various sections
i
e
by
using
the
Equation
of
the
head
loss
i
.
e
.
by
using
the
Equation
of
the
head
loss
For
if
Pipe
Material
or
‘
e
’
is
given
use
Darcy
’
s
For
if
Pipe
Material
or
e
is
given
use
Darcys
Weisbach Approach to calculate individual headlosscontributionsafterfindinge/D,V,R and
f
foreachpipe.
Two approaches are used for the solution
equivalent velocity head method (only this will be equivalent
velocity
head
method
(only
this
will
be
studied)
equivalent length method (Le)
According to the first approach
2
2
2
3
2
333
2
2
22 2
1
2
111
2 2 2gD
V
L
f
gD
V
L
f
gD
V
L
f
h
2
1
Q
Q
2
2
1
1
V
A
V
A
3 12
1
Q QQ
Q
33 112
2
1
1
VA VAV
V
equivalent velocity head method (only this will be equivalent
velocity
head
method
(only
this
will
be
studied)
equivalent length method (Le)
According to the first approach
2
2
2
3
2
333
2
2
22 2
1
2
111
2 2 2gD
V
L
f
gD
V
L
f
gD
V
L
f
h
2
1
Q
Q
2
2
1
1
V
A
V
A
3 12
1
Q QQ
Q
33 112
2
1
1
VA VAV
V
A
1
h
2
B 3
1
h
2
B 3
Used to increase the
discharge capacity of a
system
Governing equations Governing
equations
Q
Q
Q
Q
3 2 1
Q
Q
Q
Q
3 2 1
h h h h
discharge capacity of a
system
Governing equations Governing
equations
Q
Q
Q
Q
3 2 1
Q
Q
Q
Q
3 2 1
h h h h
A
B
B
Two approaches are used for solution
Approximate approach Exactly or relatively exact approach (will be
studied)
Approximate approach Exactly or relatively exact approach (will be
studied)
Forthecaseswhereheadlossh
L
isgiven
Total discharge can be calculated by
calculating the individual discharge from all
the
pipes
using
the
pipes
using
n
L
K
Q
h
If pipe material or in other words if e is given then
the
solution
becomes
more
accurate
L
Q
then
the
solution
becomes
more
accurate
Forthecaseswhereheadlossh
L
isgiven
Total discharge can be calculated by
calculating the individual discharge from all
the
pipes
using
the
pipes
using
n
L
K
Q
h
If pipe material or in other words if e is given then
the
solution
becomes
more
accurate
L
Q
then
the
solution
becomes
more
accurate
Neglecting minor losses and inserting values o
f
allindividualdischarges
n
f
n
f
n
f
h
h
h
/1 /1 /1
f
f
f
Kh
Kh
Kh
Q
3 2 13 2 1
Sinceallheadlossesareequal
n
n
f
n n n
n
f
K
h
K
K
K
h Q
/1
/1
/1
3
/1
2
/1
1
/1
1 1 1 1
3
2
1
Neglecting minor losses and inserting values o
f
allindividualdischarges
n
f
n
f
n
f
h
h
h
/1 /1 /1
f
f
f
Kh
Kh
Kh
Q
3 2 13 2 1
Sinceallheadlossesareequal
n
n
f
n n n
n
f
K
h
K
K
K
h Q
/1
/1
/1
3
/1
2
/1
1
/1
1 1 1 1
3
2
1
Darcy Weisbachequation (considering minor losses) will be written as
V
k
L
f
h
2
SolvingforVandQ
g
k
D
f
h
L
2
Solving
for
V
and
Q
L
h
C
g
h
A
V
A
Q
2
L
L
h
C
k
D
L
f
g
A
V
A
Q
1
1
1
1
1
1 11 1
1
Darcy Weisbachequation (considering minor losses) will be written as
V
k
L
f
h
2
SolvingforVandQ
g
k
D
f
h
L
2
Solving
for
V
and
Q
L
h
C
g
h
A
V
A
Q
2
L
L
h
C
k
D
L
f
g
A
V
A
Q
1
1
1
1
1
1 11 1
1
h
C
C
C
h
C
h
C
h
C
Q
L L L L
h
C
C
C
h
C
h
C
h
C
Q
3 2 1 3 2 1
h
C
C
C
h
C
h
C
h
C
Q
L L L L
h
C
C
C
h
C
h
C
h
C
Q
3 2 1 3 2 1
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