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Minor Losses (Local)
Pump
Tee Valve
Outlet
Elbow
Inlet
Pipe
(b)
Vena contracta
Flow separationat corner
Separated flow
Separatedflow
Q
Pipe entrance or exit
Sudden expansion or contraction
Bends, elbows, tees, and other fittings
Valves, open or partially closed
Gradual expansions or contractions
h h h mf +=λtotal loss = H1 – H2
friction loss: hf = f * (L/D) * (V2/2g)
minor loss: hm = KL (V2/2g)
KL is the loss coefficient
For each pipe segment (i.e. reaches along which pipe diameter remains constant) there may be several minor losses.
Total Head Loss
Flow in an elbow.
Expanding Flows
2211 AVAVQ ==
Experiments show that the pressureAt 3-3 is equal to P1.
Ignoring the friction force
Expanding flow
Continuity eq.:
Momentum eq. In x-direction:
2211 AVAVQ ==
)VVV(g
1)pp(
)VV(QA)pp(
2122
21
12221
−=γ
−
−ρ=−
Energy eq.:
m
222
2
211
1 hg2
Vpz
g2
Vpz ++
γ+=+
γ+
g2
VVpph
22
2121
m
−+
γ
−=
g2
VKh,
g2
V1
D
Dh
D
DV
A
AVV ,
g2
)VV(h
21
mm
21
22
2
1m
22
211
2
112
212
m
=
−
=
==−
=
g2
VKh
2
mm =
Deterination of Local
Loss (hm):
Friction loss for non-circular conduits
P
ARh =
DDh =
g2
V
D
Lfh
2
f =
ν=
VDRe
ε=
D,Rff e
Circular
P
ARh =
hh R4D =
g2
V
D
Lfh
2
hf =
ν=
VDR h
e
ε=
he
D,Rff
Non-circular
Friction loss for non-circular conduits
1 212 PPfor =
12 AA <
DDh <
12 VV >
121e2eh
ffRRandDD
>⇒≈ε
>ε
1f2fhh >
Hydraulic
Diameter
Example
Water flows from the basement to the second floor through the 2 cm diameter copper pipe (a drawn tubing) at a rate of Q = 0.8 lt/s and exits through a faucet of diameter 1.3 cm as shown in figure. Determine the pressure at point 1 if :
a) viscous effects are neglected,
b) the only losses included are major losses
c) all losses are included
s/m546.210*1416.3
0008.0
A
QV
4===
−
4
610*5.445062
10*13.1
02.0*546.2VDRe ===
ν=
− flow is turbulent
a) Energy equation between (1) and (9)
91 HH =
9
9
2
9
11
2
1 zP
g2
Vz
P
g2
V+
γ+=+
γ+ s/m546.2VV1 == , s/m027.6
A
QV
9
9 ==
+
−γ= 9
2
1
2
9
1 zg2
VVP
kPa6.94Pa5.9457812.881.9*2
546.2027.69810P
22
1 ==
+
−=
m12.8zwithComparem64.9P1 ==γ
Q = 0.8 lt/s
Pipe diameter = 2cm
Viscosity = 1.13 * 10-6 m2/sn
Pressure at point 1?????
γ
b) 9f1 HhH =−
++
−γ=
g2
V
D
Lfz
g2
VVP
2
9
2
1
2
9
1
drawn tubing, ε = 0.00016 cm ε /D =8.10
-5 , Re =45062
Ltotal = 4.6 + 3*6 = 22.6 m
f = 0.0215
m17.67γ
PkPa173.3173320PaP
9.81*2
2.546*
0.02
22.60.02158.12
9.81*2
2.5466.0279810P
11
222
1
===
++
−=
Moody diagram. (From L.F. Moody, Trans. ASME, Vol.66,1944.) (Note: If e/D = 0.01
and Re = 104, the dot locates ƒ = 0.043.)
........................
m23.38γ
PkPa229.4229390.3PaP
9.81*2
2.54617.3
9.81*2
2.546
0.02
22.60.02158.12
9.81*2
2.5466.0279810P
17.32101.50.40.41.51.5K
2g
VK
2g
V
D
Lfz
2g
VVγP
H2g
VKhHc)
11
2222
1
i
2
i
2
9
2
1
2
9
1
9
8
2
2
iL1
===
+⋅++
−=
=++++++=
+++
−=
=−−
∑
∑
∑
This pressure drop (229.4 kPa) calculated by
including all losses should be the most realistic
answer of the three cases considered.
Comparison:
hl=0 hl=hf hl=hf+hm
P1 (kPa) 94.6 173.3 229.4
P1/γγγγ (m) 9.64 17.67 23.38
Piping Networks
• Two general types of networks
– Pipes in series
• Volume flow rate is constant
• Head loss is the summation of components
– Pipes in parallel
• Volume flow rate is the sum of the components
• Pressure loss across all branches is the same
Pipes in Series
BA QQ =
Pipes in Series
• For pipes connected in series:
• Q=constant QA=Q1=Q2= ----- =Qn=QB
• But the head loss is additive:
• hl=∑hfi+ ∑hmi, i=1,2,....,n
Where hfi is the frictional loss in i-th pipe, and hmi is the local loss in i-th pipe.
Example • Given a three-pipe series system as shown below. The total
pressure drop is pA-pB = 150 kPa, and the elevation drop is zA-zB = 5 m. The pipe data are
• The fluid is water (ρ= 1000 kg/m3 , ν= 1.02x10-6 m2/s). Calculate the flow rate Q (m3/hr). Neglect minor losses.
pipe L (m) D (cm) ε (mm) 1 100 8 0.24 2 150 6 0.12 3 80 4 0.20
∑ ∑++= mfBA hhHH
g2
V
D
Lf
g2
V
D
Lf
g2
V
D
Lfz
P
g2
Vz
P
g2
V2
3
3
33
2
2
2
22
2
1
1
11B
B
2
BA
A
2
A ++++γ
+=+γ
+
( )g2
V1
D
Lf
g2
V
D
Lf
g2
V1
D
Lfzz
pp2
3
3
33
2
2
2
22
2
1
1
11BA
BA
+++
−=−+
γ−
γ
g2
V2
1
g2
V2
3
321
( )g2
V1
D
Lf
g2
V
D
Lf
g2
V1
D
Lfzz
pp2
3
3
33
2
2
2
22
2
1
1
11BA
BA
+++
−=−+
γ−
γ
from continuity: 332211 AVAVAV ==
3
2
2
2
1 A4
8A
6
8A
=
= and 321 A4A78.1A ==
V4V and V78.1V 1312 ==
( ) ( )
( )g2
V1
D
Lf4
g2
V
D
Lf78.1
g2
V1
D
Lfzz
pp
2
1
3
33
2
2
1
2
22
22
1
1
11BA
BA
++
+
−=−+
γ−
γ
( )g2
V15f32000f7920f12503.20
2
1321 +++=
20.3 = (1250 f1
+ 7920 f2
+ 32000 f3
+ 15) V12 / 2g
V2
= 1.78 V1
V3
= 4.0 V1
Velocities and f not known. Thus, assume rough flow
Pipe ε D ε/D fi V Re f1
(mm) (cm) m/sec
1 0.24 8 0.003 0.026 0.579 45381 0.029
2 0.12 6 0.002 0.023 1.030 60584 0.027
3 0.20 4 0.005 0.030 2.314 90762 0.031
Pipe ε D ε/D fi V
(mm) (cm) m/sec
1 0.24 8 0.003 0.026 0.579
2 0.12 6 0.002 0.023 1.030
3 0.20 4 0.005 0.030 2.314
20.3 = (1250 f1
+ 7920 f2
+ 32000 f3
+ 15) V12 / 2g
V2
= 1.78 V1
V3
= 4.0 V1
Pipe ε D ε/D fi V Re f1
(mm) (cm) m/sec
1 0.24 8 0.003 0.029 0.563 44147 0.029
2 0.12 6 0.002 0.027 1.002 58937 0.027
3 0.20 4 0.005 0.031 2.252 88295 0.031
Close enough. Thus,
V1 = 0.563 m/sec or Q= 0.0028 m3 /sec
Equivalent Pipe Concept
For pipes in series:
• Consider two pipes connected in series:
A, LA,DA,εA
B, LA,DB,εB
1 2
• We want to replace these two pipes with an equivalent pipe C:
• The head loss between sections (1) and (2) is:
• hf12= hfA + hfB = hfC (1)
• and
• QA = QB =QC (2)
C, Leq,Deq, εeq1 2
• The Darcy-Weissbach Equation:
• Therefore Eq.(1) can be written as:
2
2
5f
2
2
f
πg
Q
D
Lf8=h
:Dπ
4Q=V inserting and
g2
V
D
Lf=h
5
eq
eq
eq5
C
C
C5
B
B
B5
A
A
A
2
2
C
5
C
C
C2
2
B
5
B
B
B2
2
A
5
A
A
A
D
Lf=
D
Lf=
D
Lf+
D
Lf
or πg
Q
D
Lf8=
πg
Q
D
Lf8+
πg
Q
D
Lf8
• If fA=fB=fC, then
• Generalizing for n pipes connected in series
• We choose either Deq, or Leq, then compute the
other from the equation.
D
L+
D
L=
D
L
5
B
B
5
A
A
5
eq
eq
‡”n
1
5
i
i
5
eq
eq
D
L=
D
L
Example • Given a three-pipe series system as shown below. The total
pressure drop is pA-pB = 150 kPa, and the elevation drop is zA-zB = 5 m. The pipe data are
• The fluid is water (ρ= 1000 kg/m3 , ν= 1.02x10-6 m2/s). Calculate the flow rate Q (m3/hr). Neglect minor losses.
pipe L (m) D (cm) ε (mm) 1 100 8 0.24 2 150 6 0.12 3 80 4 0.20
‡”n
1
5
i
i
5
eq
eq
D
L=
D
L choose either Deq, or Leq, then compute the other from the equation
Pipe ε D ε/D L
(mm) (cm) (m)
1 0.24 8 0.003 100
2 0.12 6 0.002 150
3 0.20 4 0.005 80
Let’s use Leq = 330m
330 / D5 = (100/0.085) + (150/0.065) + (80/0.045) = 1*109
Solving for Deq= 0.0505 m
Assume ε = 0.2mm and hydraulically rough flow, ε/D = 0.004
initial f = 0.028 using 20.3 = =====� V = 1.475m/s Re = 75997
Recalculating f = 0.03 V = 1.425 m/sec…………………
Discharge Q = 0.002855 m3 / sec
g2
V
D
Lfh
2
f =
‡”n
1
5
i
i
5
eq
eq
D
L=
D
L choose either Deq, or Leq, then compute the other from the equation
Pipe ε D ε/D L
(mm) (cm) (m)
1 0.24 8 0.003 100
2 0.12 6 0.002 150
3 0.20 4 0.005 80
Let’s use Deq = 0.06m
L / (0.06)5 = (100/0.085) + (150/0.065) + (80/0.045) = 1* 109
Solving for Leq= 778 m
Assume ε = 0.2mm and hydraulically rough flow, ε/D = 0.0033
initial f = 0.027 using 20.3 = ====� V = 1.067 m/s Re = 65275
Recalculating f = 0.03 V = 1.012 m/sec…………………
Discharge Q = 0.00286 m3 / sec
g2
V
D
Lfh
2
f =
• If desired minor losses may be expressed in terms of equivalent lengths and added to the actual length of pipe as:
Where
• Km=local loss coefficient, and
• f=fricton factor of the pipe
• Then the pipe length should be taken as:
• L=Lac+Leq
f
DK=L Hence
g2
V
D
Lf=
g2
VK=h meq
2eq
2
mm
Pipes in Paralel
• In order to increase the capacity of a pipeline
system, pipes might be connected in parallel.
For pipes connected in parallel:
• The discharges are additive:
• QA = QB = ∑Qi, i=1,2,....,n
• The Total head at junctions must have single
value. Therefore the head loss in each
branch must be the same:
• hf1=hf2=hf3 = ----- =hfn
Example
• Assume that the same three pipes of previous example are now in parallel with the same total loss of 20.3 m. Compute the total rate Q(m3/hr), neglecting the minor losses.
pipe L (m) D (cm) ε (mm) 1 100 8 0.24 2 150 6 0.12 3 80 4 0.20
• Solution-I:Energy equation b/w A and B:
HA = HB + hL = HB + hf + hm
no matter which route is followed b/w A and B
HA – HB = 20.3 = g2
V
D
Lf
21
1
11 =
g2
V
D
Lf
22
2
22 =
g2
V
D
Lf
23
3
33
• Substituting the known L’s and D’s ,
20.3 = 1250g2
Vf
21
1 =2500g2
Vf
22
2 =2000g2
Vf
23
3
Since Vi’s and fi’s are not known, assume hydraulically rough regime
Pipe ε/D F0 V Re f1
1 0.003 0.0262 3.49 273726 0.268
2 0.002 0.0234 2.61 153529 0.247
3 0.005 0.0304 2.56 100392 0.315
ν=
VDRe
Tabular presentation
f’s have converged
Pipe ε/D f1 V Re f2
1 0.003 0.268 3.46 271373 0.268
2 0.002 0.247 2.55 150000 0.247
3 0.005 0.315 2.52 98823 0.315
Q= 100 m3/hr
Pipe V(m/s) Q(m3/s) Q(m3/hr)
1 3.46 0.0174 62.6
2 2.55 0.0072 26.0
3 2.52 0.0032 11.4
TOTAL 100
Equivalent pipe concept for parallel pipes
• Consider two pipes, A and B, connected in parallel:
• For the equivalent pipe with length, Leq, and diameter, Deq:
A, LA,DA,εA
B, LB,DB,εB
1 2
Q1 Q2
QA
QB 12
Leq, Deq
• hf1-2= hfA = hfB and Q1=QA+QB=Q2
• From the Darcy-Weissbach equation:
• Therefore:
fL8
Dπgh=Q
52
f
hence and Lf8
Dπgh=Q and
Lf8
Dπgh=Q
BB
5
B
2
fB
B
AA
5
A
2
fA
A
BB
5
B
2
fB
AA
5
A
2
fA
CC
5
C
2
fC
Lf8
Dπgh+
Lf8
Dπgh=
Lf8
Dπgh
• Simplifying:
Furthermore if fC=fA=fB, then
Generalizing for n pipes connected in parallel:
BB
5
B
AA
5
A
CC
5
C
Lf
D+
Lf
D=
Lf
D
B
5
B
A
5
A
C
5
C
L
D+
L
D=
L
D
‡”n
1 i
5
i
eq
5
eq
L
D=
L
D(1)
(n)
(i)Q Q
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