Minor Losses (Local)
Separ ated flow
Vena contracta
( b)
Outlet
Flow separation
at cor ner
Q
Pipe
Separated
flow
Pump
Elbow
Tee
Valve
Pipe entrance or exit
Sudden expansion or contraction
Inlet
Bends, elbows, tees, and other fittings
Valves, open or partially closed
Gradual expansions or contractions
Total Head Loss
total loss = H1 – H2
hλ = hf + hm
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.
Flow in an elbow.
Expanding Flows
Experiments show that the pressure
At 3-3 is equal to P1.
Ignoring the friction force
Q = V1A1 = V2 A 2
Expanding flow
Continuity eq.:
Q = V1A1 = V2 A 2
(p1 − p2 )A 2 = ρQ( V2 − V1)
Momentum eq. In x-direction:
Energy eq.:
(p1 − p2 ) 1 2
= ( V2 − V1V2 )
γ
g
2
2
p1 V1
p2 V2
z1 + +
= z2 +
+
+ hm
γ 2g
γ
2g
2
p −p
V − V2
hm = 1 2 + 1
2g
γ
2
2
( V2 − V1)2
hm =
2g
,
2
VA
VD
V2 = 1 1 = 1 21
A2
D2
2
 D 2  V 2
V1
1
1
hm =   − 1
, hm = K m
2g
 D2 
 2g
Deterination of Local
Loss (hm):
2
V
hm = K m
2g
Friction loss for non-circular conduits
Circular
Non-circular
A
Rh =
P
A
Rh =
P
Dh = D
Dh = 4Rh
L V2
hf = f
Dh 2g
2
LV
hf = f
D 2g
Re =
Dh V
Re =
ν
DV
ν

ε

f = f Re ,

D






ε 
f = f  Re , 
Dh 

Friction loss for non-circular conduits
1
for P2 = P1
2
A 2 < A1
Dh < D
V2 > V1
ε
ε
> and R e 2 ≈ R e1 ⇒ f2 > f1
Dh D
hf2 > hf1
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
Q = 0.8 lt/s
Pipe diameter = 2cm
Viscosity = 1.13 * 10-6 m2/sn
Pressure at point 1?????
V =
Re =
a)
Q
0 .0008
=
= 2 .546 m / s
A 3 .1416 *10 − 4
VD 2 .546 * 0 .02
=
= 45062 = 4 .5 * 10 4
−6
ν
1 .13 *10
flow is turbulent
Energy equation between (1) and (9)
H1 = H 9
V 92 P9
V12 P1
+
+ z1 =
+
+ z9
2g
γ
2g
γ
V1 = V = 2 .546 m / s , V9 =
 V 92 − V12

P1 = γ 
+ z9 
 2g

 6 .027 2 − 2 .546 2

P1 = 9810 
+ 8 .12  = 94578 .5 Pa = 94 .6 kPa
2 * 9 .81


P1
= 9 .64 m Compare with z = 8 .12 m
γ
Q
= 6 .027 m / s
A9
γ
b) H 1 − h f = H 9
 V92 − V12
L V2 
+ z9 + f
P1 = γ 

2
g
D
2
g


drawn tubing, ε = 0.00016 cm
ε /D =8.10-5 , Re =45062
f = 0.0215
Ltotal = 4.6 + 3*6 = 22.6 m
 6.027 2 − 2.546 2
22.6 2.546 2 
+ 8.12 + 0.0215
P1 = 9810
*

2
*
9.81
0.02
2
*
9.81


P1
P1 = 173320Pa = 173.3 kPa
= 17.67 m
γ
........................
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.)
V2
H1 − h L − ∑ K i
= H9
2g
2
8
c)
2
 V92 − V12
LV
V2 
P1 = γ 
+ z9 + f
+ ∑ Ki

2g
D
2g
2g


∑ K i = 1.5 + 1.5 + 0.4 + 0.4 + 1.5 + 10 + 2 = 17.3
 6.027 2 − 2.546 2
22.6 2.546 2
2.546 2 
P1 = 9810 
+ 8.12 + 0.0215
⋅
+ 17.3

2 * 9.81
0.02 2 * 9.81
2 * 9.81

P1
P1 = 229390.3Pa = 229.4 kPa
= 23.38 m
γ
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/γγ (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
QA = QB
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
pipe
1
2
3
L (m)
100
150
80
D (cm)
8
6
4
ε (mm)
0.24
0.12
0.20
• The fluid is water (ρ= 1000 kg/m3 , ν= 1.02x10-6 m2/s). Calculate the
flow rate Q (m3/hr). Neglect minor losses.
H A = HB + ∑ h f + ∑ h m
2
2
V3
V1
2g
2g
2
2
2
2
2
L 3 V3
VA
PA
VB
PB
L 1 V1
L 2 V2
+
+ zA =
+
+ z B + f1
+ f2
+ f3
2g
γ
2g
γ
D1 2g
D 2 2g
D 3 2g
2
 L3
 V3 2
 L1
 V12
 p A pB 
L 2 V2

−
 + (z A − z B ) =  f1
− 1
+ f2
+  f3
+ 1
γ 
D 2 2g  D 3
 γ
 D1
 2g
 2g
1
2
3
2
2
 L3
 V3 2
 L1
 V1
 p A pB 
L 2 V2

−
 + (z A − z B ) =  f1
− 1
+ f2
+  f3
+ 1
γ
γ
D
D
2
g
D
2
g



1

2
3

 2g
from continuity: V1 A 1 = V2 A 2 = V3 A 3
2
2
8
8
A 1 =   A 2 =   A 3 and
6
 4
V2 = 1.78 V1 and V3 = 4 V1
2
 L1
 V12
 p A pB 
L2 V1
2
−  + (z A − zB ) =  f1

− 1
+ (1.78 ) f2
γ 
D 2 2g
 γ
 D1  2g
 L3
 V12
+ (4 )  f3
+ 1
D

3
 2g
2
2
V
20.3 = (1250 f1 + 7920f 2 + 32000f 3 + 15 ) 1
2g
A 1 = 1.78 A 2 = 4 A 3
V2 = 1.78 V1
V3 = 4.0 V1
20.3 = (1250 f1 + 7920 f2 + 32000 f3 + 15) V12 / 2g
Velocities and f not known. Thus, assume rough flow
Pipe
1
2
3
ε
(mm)
0.24
0.12
0.20
Pipe
1
2
3
D
(cm)
8
6
4
ε/D
fi
0.003
0.002
0.005
0.026
0.023
0.030
D
fi
ε
ε/D
(mm) (cm)
0.24 8 0.003 0.026
0.12 6 0.002 0.023
0.20 4 0.005 0.030
V
m/sec
0.579
1.030
2.314
V
m/sec
0.579
1.030
2.314
Re
f1
45381 0.029
60584 0.027
90762 0.031
V2 = 1.78 V1
V3 = 4.0 V1
20.3 = (1250 f1 + 7920 f2 + 32000 f3 + 15) V12 / 2g
Pipe
1
2
3
D
ε
(mm) (cm)
0.24
8
0.12
6
0.20
4
ε/D
fi
0.003 0.029
0.002 0.027
0.005 0.031
V
m/sec
0.563
1.002
2.252
Re
f1
44147
58937
88295
0.029
0.027
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:
B, LA,DB,εB
1
A, LA,DA,εA
2
• We want to replace these two pipes with an
equivalent pipe C:
1
C, Leq,Deq, εeq
2
• The head loss between sections (1) and (2) is:
•
hf12= hfA + hfB = hfC
(1)
• and
•
QA = QB =QC
(2)
• The Darcy-Weissbach Equation:
L V2
4Q
hf = f
and inserting V =
2 :
D 2g
πD
L Q2
h f = 8f 5
D gπ 2
• Therefore Eq.(1) can be written as:
L C Q C2
L A Q 2A
L B Q 2B
8f A 5
2 + 8f B
5
2 = 8f C
D A gπ
D B gπ
D 5C gπ 2
L eq
LC
LA
LB
f A 5 + f B 5 = f C 5 = f eq 5
DA
DB
DC
D eq
or
• If fA=fB=fC, then
L eq
D 5eq
LA LB
= 5 + 5
DA DB
• Generalizing for n pipes connected in series
L eq
D 5eq
n
Li
= ‡” 5
1 Di
• We choose either Deq, or Leq, then compute the
other from the equation.
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
pipe
1
2
3
L (m)
100
150
80
D (cm)
8
6
4
ε (mm)
0.24
0.12
0.20
• The fluid is water (ρ= 1000 kg/m3 , ν= 1.02x10-6 m2/s). Calculate the
flow rate Q (m3/hr). Neglect minor losses.
L eq
D 5eq
n
Li
= ‡” 5
1 Di
choose either Deq, or Leq, then compute
the other from the equation
ε/D
1
D
ε
(mm) (cm)
0.24
8
0.003
L
(m)
100
2
3
0.12
0.20
0.002
0.005
150
80
Pipe
6
4
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
L V2
initial f = 0.028 using 20.3 = hf = f
===== V = 1.475m/s Re = 75997
D 2g
Recalculating f = 0.03 V = 1.425 m/sec…………………
Discharge Q = 0.002855 m3 / sec
L eq
D 5eq
choose either Deq, or Leq, then compute
the other from the equation
n
Li
= ‡” 5
1 Di
ε/D
1
D
ε
(mm) (cm)
0.24
8
0.003
L
(m)
100
2
3
0.12
0.20
0.002
0.005
150
80
Pipe
6
4
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
L V2
initial f = 0.027 using 20.3 = hf = f
==== V = 1.067 m/s Re = 65275
D 2g
Recalculating f = 0.03 V = 1.012 m/sec…………………
Discharge Q = 0.00286 m3 / sec
• If desired minor losses may be expressed
in terms of equivalent lengths and added to
the actual length of pipe as:
L eq V 2
V2
D
hm = Km
=f
Hence L eq = K m
2g
D 2g
f
Where
• Km=local loss coefficient, and
• f=fricton factor of the pipe
• Then the pipe length should be taken as:
• L=Lac+Leq
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
1
2
3
L (m)
100
150
80
D (cm)
8
6
4
ε (mm)
0.24
0.12
0.20
• Solution-I:
Energy equation b/w A and B:
H A = H B + hL = H B + hf + hm
no matter which route is followed b/w A and B
L 1 V12
L 2 V22
L 3 V32
HA – HB = 20.3 = f1
= f2
= f3
D1 2g
D 2 2g
D 3 2g
• Substituting the known L’s and D’s ,
V32
V12
V22
20.3 = 1250 f1
=2500 f2
=2000 f 3
2g
2g
2g
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
VD
Re =
ν
Tabular presentation
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
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
f’s have converged
Q= 100 m3/hr
Equivalent pipe concept for parallel pipes
• Consider two pipes, A and B, connected in
parallel:
A, LA,DA,εA
QA
Q1
Leq, Deq
Q2
1
2
QB
1
B, LB,DB,εB
• For the equivalent pipe with length, Leq, and
diameter, Deq:
2
• hf1-2= hfA = hfB and Q1=QA+QB=Q2
• From the Darcy-Weissbach equation:
Q=
h f gπ 2 D 5
8fL
• Therefore:
QA =
h fA gπ 2 D 5A
and
8f A L A
h fC gπ 2 D 5C
=
8f C L C
QB =
h fB gπ 2 D 5B
and hence
8f B L B
h fA gπ 2 D 5A
h fB gπ 2 D 5B
+
8f A L A
8f B L B
• Simplifying:
D 5C
=
fCLC
D 5B
D 5A
+
fALA
f BLB
Furthermore if fC=fA=fB, then
D 5C
=
LC
D 5A
D 5B
+
LA
LB
Generalizing for n pipes connected in
(1)
parallel:
5
D eq
L eq
n
= ‡”
1
5
i
D
Li
Q
Q
(i)
(n)