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